[algogeeks] Re: Google-Puzzle
buddy i said that kadane's algo(max subsum) wouldn't work.. On Feb 25, 1:31 pm, Ashish Goel ashg...@gmail.com wrote: max subsum problem Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 On Sat, Feb 25, 2012 at 1:03 PM, karthikeya s karthikeya.a...@gmail.comwrote: You have a circular track containing fuel pits at irregular intervals. The total amount of fuel available from all the pits together is just sufficient to travel round the track and finish where you started. Given the the circuit perimeter, list of each fuel pit location and the amount of fuel they contain, find the optimal start point on the track such that you never run out of fuel and complete circuit. my logic: we can use an array having element as fuel(in km)-dist to next pit so now aim is to traverse the array as always having some +ve resultant sum.nd plz we cant use here kadane's algo.there are cases in which it will not hold here -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Google-Puzzle
it will the diff is of fuel and dist forms the content of array which moves from 1 to 2n-1 elements(break the circle and instead of elem like 1,2,n have 1,2,n,1,2,...n-1 i.e. total 2n-1 so that mod stuff is not required. now find maxsubSum such that sum=0 and count of nodes is n not clear ehy it wont work. Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 On Sat, Feb 25, 2012 at 2:54 PM, karthikeya s karthikeya.a...@gmail.comwrote: buddy i said that kadane's algo(max subsum) wouldn't work.. On Feb 25, 1:31 pm, Ashish Goel ashg...@gmail.com wrote: max subsum problem Best Regards Ashish Goel Think positive and find fuel in failure +919985813081 +919966006652 On Sat, Feb 25, 2012 at 1:03 PM, karthikeya s karthikeya.a...@gmail.com wrote: You have a circular track containing fuel pits at irregular intervals. The total amount of fuel available from all the pits together is just sufficient to travel round the track and finish where you started. Given the the circuit perimeter, list of each fuel pit location and the amount of fuel they contain, find the optimal start point on the track such that you never run out of fuel and complete circuit. my logic: we can use an array having element as fuel(in km)-dist to next pit so now aim is to traverse the array as always having some +ve resultant sum.nd plz we cant use here kadane's algo.there are cases in which it will not hold here -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: C Puzzle
This is not a C puzzle. It depends on operating system. On Sep 17, 3:46 pm, teja bala pawanjalsa.t...@gmail.com wrote: you have to print the list of all the files in a directory and all its sub directories? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: C Puzzle
i tried it for Windows.
#includeiostream
using namespace std;
int main()
{
char ch;
int i=system(dir /s |more);
ch=getchar();
}
On Sep 17, 6:46 pm, teja bala pawanjalsa.t...@gmail.com wrote:
you have to print the list of all the files in a directory and all
its sub directories?
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[algogeeks] Re: C Puzzle
it's running..
http://ideone.com/XjGIo
wat is does is that through the system function u can run commands as
u do in command prompt.. nd dis is d command to show the files n all
as ur question asked.
try it on dev C++.
On Sep 17, 9:45 pm, teja bala pawanjalsa.t...@gmail.com wrote:
@ pooja
in borland c++ 4.5 compiler version its giving linker error,
wat actually ur code does?
On Sat, Sep 17, 2011 at 9:27 PM, pooja pooja27tan...@gmail.com wrote:
i tried it for Windows.
#includeiostream
using namespace std;
int main()
{
char ch;
int i=system(dir /s |more);
ch=getchar();
}
On Sep 17, 6:46 pm, teja bala pawanjalsa.t...@gmail.com wrote:
you have to print the list of all the files in a directory and all
its sub directories?
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[algogeeks] Re: Math Puzzle
It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
Don is right if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!! On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta guptaabhinav...@gmail.comwrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Math Puzzle
No, not at all. Here is a trivial counterexample: P = Q = R = 0 Don On Sep 15, 11:46 am, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
u cnt divide a number by 0..that thing is self undrstod On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover piyush4u.iit...@gmail.comwrote: Don is right if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!! On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
dude dats outside the domain of the qs...dont be oversmart. On Thu, Sep 15, 2011 at 9:49 AM, Don dondod...@gmail.com wrote: No, not at all. Here is a trivial counterexample: P = Q = R = 0 Don On Sep 15, 11:46 am, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Math Puzzle
Right, and in every proof above, at some point there is a possible division by zero. Therefore the proof is not valid in cases where R or P or Q are zero, and there are infinitely many such cases. The problem states P+Q+R=0 as the only constraint. There are infinitely many cases which fit that constraint where the expression is not equal to 3. Don On Sep 15, 11:57 am, abhinav gupta guptaabhinav...@gmail.com wrote: u cnt divide a number by 0..that thing is self undrstod On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover piyush4u.iit...@gmail.comwrote: Don is right if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!! On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
@abhinav... it's not about being over smart or to show someone or to prove someone anything. It's just that you should not take any assumptions by yourself or if you do you should specify clearly. If u r asked this question in an interview and you give the answer 3 without telling your assumption, u r done!! And if you are living in the programming world, you need to take care of all the possible scenarios otherwise u will end up throwing exceptions and segmentation faults. On Thu, Sep 15, 2011 at 10:32 PM, Don dondod...@gmail.com wrote: Right, and in every proof above, at some point there is a possible division by zero. Therefore the proof is not valid in cases where R or P or Q are zero, and there are infinitely many such cases. The problem states P+Q+R=0 as the only constraint. There are infinitely many cases which fit that constraint where the expression is not equal to 3. Don On Sep 15, 11:57 am, abhinav gupta guptaabhinav...@gmail.com wrote: u cnt divide a number by 0..that thing is self undrstod On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover piyush4u.iit...@gmail.comwrote: Don is right if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!! On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Math Puzzle
solution is =3 with the condition p!=0 and q!=0 and r!=0 Ashima M.Sc.(Tech)Information Systems 4th year BITS Pilani Rajasthan On Thu, Sep 15, 2011 at 10:38 PM, Piyush Grover piyush4u.iit...@gmail.comwrote: @abhinav... it's not about being over smart or to show someone or to prove someone anything. It's just that you should not take any assumptions by yourself or if you do you should specify clearly. If u r asked this question in an interview and you give the answer 3 without telling your assumption, u r done!! And if you are living in the programming world, you need to take care of all the possible scenarios otherwise u will end up throwing exceptions and segmentation faults. On Thu, Sep 15, 2011 at 10:32 PM, Don dondod...@gmail.com wrote: Right, and in every proof above, at some point there is a possible division by zero. Therefore the proof is not valid in cases where R or P or Q are zero, and there are infinitely many such cases. The problem states P+Q+R=0 as the only constraint. There are infinitely many cases which fit that constraint where the expression is not equal to 3. Don On Sep 15, 11:57 am, abhinav gupta guptaabhinav...@gmail.com wrote: u cnt divide a number by 0..that thing is self undrstod On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover piyush4u.iit...@gmail.comwrote: Don is right if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!! On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta guptaabhinav...@gmail.com wrote: Shut up...its 3,, On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote: It might be 3, but it doesn't have to be 3. Don On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN naga4...@gmail.com wrote: if P+Q+R= 0 then P2 /QR + Q2/PR + R2/PQ = ?? how to solve this?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- @ |3 # ! /\/ @ \./ -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C puzzle
but o/p is -15 -13 -11 -9
On Sun, Sep 4, 2011 at 3:38 AM, vikas vikas.rastogi2...@gmail.com wrote:
f(19222)
|
|
f(1922) -11 -2 =-13
|
|
f(192) = -9 -2 = -11
|
|
f(19) = f(1) - 9 = -9
|
|
f(1) = 0
output -9 -11 -13
On Sep 3, 10:29 pm, teja bala pawanjalsa.t...@gmail.com wrote:
Find the output of the following code - plzzz xplain the o/p
int find(int j)
{
if(j1)
{
j=find(j/10)-(j%10);
printf(%d,j);}
else
{
j=0;}
return j;
}
int main()
{
int i=19222;
int k;
k=find(i);
}
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Re: [algogeeks] Re: C puzzle
o/p : -9-11-13-15 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C puzzle
@neha could u xplain it? On Sun, Sep 4, 2011 at 12:05 PM, Neha Gupta nehagup...@gmail.com wrote: o/p : -9-11-13-15 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C puzzle
output is : -9-11-13-15 first j=19222 then find(19222)=find(1922)-2 find(1922)=find(192)-2 find(192)=find(19)-2 find(19)=find(1)-9 find(1) returns 0 then find(19)=0-9=-9 find(192)=-9-2=-11 find(1922)=-11-2=-13 find(19222)=-13-2=-15 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C puzzle
@teja ...nitesh hv correctly explained it -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: C puzzle
@nitesh good explanation brother -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: C puzzle
f(19222)
|
|
f(1922) -11 -2 =-13
|
|
f(192) = -9 -2 = -11
|
|
f(19) = f(1) - 9 = -9
|
|
f(1) = 0
output -9 -11 -13
On Sep 3, 10:29 pm, teja bala pawanjalsa.t...@gmail.com wrote:
Find the output of the following code - plzzz xplain the o/p
int find(int j)
{
if(j1)
{
j=find(j/10)-(j%10);
printf(%d,j);}
else
{
j=0;}
return j;
}
int main()
{
int i=19222;
int k;
k=find(i);
}
--
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[algogeeks] Re: math puzzle
@Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: math puzzle
3x+4y = 60 it's a straight line equation whose x intercept is 20 and y intercept is 15. Draw it in first quadrant (as x, y are positive integers) now x = (60 - 4y)/3 = 4(15-y)/3 now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or not. It's simple y = 3, 6, 9, 12 -Piyush On Sun, Aug 28, 2011 at 6:38 PM, Dave dave_and_da...@juno.com wrote: @Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: math puzzle
maximum value of y satisfying this is y=15 and for that x=0; now decrease y by 3 and increase x by 4 ,you will have x and y satisfying the equation. keep on doing this till you reach minimum value of y i.e 0 this you can do 5 times decreasing y=15 by 3 every time so there will be 5 solutions . On 8/28/11, Piyush Grover piyush4u.iit...@gmail.com wrote: 3x+4y = 60 it's a straight line equation whose x intercept is 20 and y intercept is 15. Draw it in first quadrant (as x, y are positive integers) now x = (60 - 4y)/3 = 4(15-y)/3 now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or not. It's simple y = 3, 6, 9, 12 -Piyush On Sun, Aug 28, 2011 at 6:38 PM, Dave dave_and_da...@juno.com wrote: @Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: math puzzle
sorry 6 solutions y=15,12,9,6,3,0 and x=0,4,8,12,16,20 respectively On 8/28/11, harshit sethi hshoneyma...@gmail.com wrote: maximum value of y satisfying this is y=15 and for that x=0; now decrease y by 3 and increase x by 4 ,you will have x and y satisfying the equation. keep on doing this till you reach minimum value of y i.e 0 this you can do 5 times decreasing y=15 by 3 every time so there will be 5 solutions . On 8/28/11, Piyush Grover piyush4u.iit...@gmail.com wrote: 3x+4y = 60 it's a straight line equation whose x intercept is 20 and y intercept is 15. Draw it in first quadrant (as x, y are positive integers) now x = (60 - 4y)/3 = 4(15-y)/3 now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or not. It's simple y = 3, 6, 9, 12 -Piyush On Sun, Aug 28, 2011 at 6:38 PM, Dave dave_and_da...@juno.com wrote: @Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: math puzzle
@Harhsit: Normally, 0 is not considered positive. Dave On Aug 28, 10:45 am, harshit sethi hshoneyma...@gmail.com wrote: sorry 6 solutions y=15,12,9,6,3,0 and x=0,4,8,12,16,20 respectively On 8/28/11, harshit sethi hshoneyma...@gmail.com wrote: maximum value of y satisfying this is y=15 and for that x=0; now decrease y by 3 and increase x by 4 ,you will have x and y satisfying the equation. keep on doing this till you reach minimum value of y i.e 0 this you can do 5 times decreasing y=15 by 3 every time so there will be 5 solutions . On 8/28/11, Piyush Grover piyush4u.iit...@gmail.com wrote: 3x+4y = 60 it's a straight line equation whose x intercept is 20 and y intercept is 15. Draw it in first quadrant (as x, y are positive integers) now x = (60 - 4y)/3 = 4(15-y)/3 now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or not. It's simple y = 3, 6, 9, 12 -Piyush On Sun, Aug 28, 2011 at 6:38 PM, Dave dave_and_da...@juno.com wrote: @Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: math puzzle
3x+4y = 60 can be expressed as 15 -y = 3x+3y -45 i.e, 15-y = 3(x+y-15) which implies tht for every value of x,y in the above eq 15-y is divisible by 3 On Sun, Aug 28, 2011 at 10:03 PM, Dave dave_and_da...@juno.com wrote: @Harhsit: Normally, 0 is not considered positive. Dave On Aug 28, 10:45 am, harshit sethi hshoneyma...@gmail.com wrote: sorry 6 solutions y=15,12,9,6,3,0 and x=0,4,8,12,16,20 respectively On 8/28/11, harshit sethi hshoneyma...@gmail.com wrote: maximum value of y satisfying this is y=15 and for that x=0; now decrease y by 3 and increase x by 4 ,you will have x and y satisfying the equation. keep on doing this till you reach minimum value of y i.e 0 this you can do 5 times decreasing y=15 by 3 every time so there will be 5 solutions . On 8/28/11, Piyush Grover piyush4u.iit...@gmail.com wrote: 3x+4y = 60 it's a straight line equation whose x intercept is 20 and y intercept is 15. Draw it in first quadrant (as x, y are positive integers) now x = (60 - 4y)/3 = 4(15-y)/3 now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or not. It's simple y = 3, 6, 9, 12 -Piyush On Sun, Aug 28, 2011 at 6:38 PM, Dave dave_and_da...@juno.com wrote: @Sivaviknesh: The smallest values of x and y are 1. The largest value of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3. Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values of x. Dave On Aug 28, 7:46 am, sivaviknesh s sivavikne...@gmail.com wrote: *Find the number of solutions for 3x+4y=60, if x and y are positive integers.* Is there any standard method for solving these type of ques ..or only trial and error ??? -- Regards, $iva -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Rishabbh A Dua -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
I think there is some ambiguity in the question. (All this time you don't know you were tossing a fair coin or not). 1) Does the above statement mean that the thower don't know whether he or she threw a fair coin even after throwing? Or is the thrower not informed beforehand that one of them is not a fair coin? 2) Does the coin count reduce after every throw or should it be put back? 3) Depending on 1) and 2), there will be different answers. On Aug 9, 12:13 am, Maddy madhu.mitha...@gmail.com wrote: I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
I'm little late but I too got 17/18. On Tue, Aug 16, 2011 at 10:47 PM, Jacob Ridley jridley2...@gmail.comwrote: I think there is some ambiguity in the question. (All this time you don't know you were tossing a fair coin or not). 1) Does the above statement mean that the thower don't know whether he or she threw a fair coin even after throwing? Or is the thrower not informed beforehand that one of them is not a fair coin? 2) Does the coin count reduce after every throw or should it be put back? 3) Depending on 1) and 2), there will be different answers. On Aug 9, 12:13 am, Maddy madhu.mitha...@gmail.com wrote: I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- regards, chinna. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: program puzzle
#include ctype.h
#include string.h
int main(int argc, char* argv[])
{
char line[500];
char tmp[500];
char *words[100];
int wordCount = 0;
char *p, *wordStart=0;
printf(Enter string:);
fgets(line,500,stdin);
for(p = line; *p; ++p)
{
if (!wordStart isalpha(*p)) wordStart = p;
else if (wordStart !isalpha(*p))
{
words[wordCount++] = wordStart;
*p = 0;
wordStart = 0;
}
}
p = tmp;
for(int i = wordCount-1; i = 0; --i) p += sprintf(p, %s ,
words[i]);
strcpy(line,tmp);
printf(%s\n, line);
return 0;
}
On Aug 15, 6:18 am, programming love love.for.programm...@gmail.com
wrote:
write a program to reverse the words in a give string.
also state the time complexity of the algo.
if the string is i am a programmer
the output should be programmer a am i
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Re: [algogeeks] Re: program puzzle
First reverse the whole sentence and then reverse every word of the
sentence
Example : I am a programmer
Step 1 Reverse entire sentence
remmargorp a ma I
Step 2 Now reverse every word in a sentence
programmer a am I
Complexity O(n)
On Mon, Aug 15, 2011 at 10:22 PM, Don dondod...@gmail.com wrote:
#include ctype.h
#include string.h
int main(int argc, char* argv[])
{
char line[500];
char tmp[500];
char *words[100];
int wordCount = 0;
char *p, *wordStart=0;
printf(Enter string:);
fgets(line,500,stdin);
for(p = line; *p; ++p)
{
if (!wordStart isalpha(*p)) wordStart = p;
else if (wordStart !isalpha(*p))
{
words[wordCount++] = wordStart;
*p = 0;
wordStart = 0;
}
}
p = tmp;
for(int i = wordCount-1; i = 0; --i) p += sprintf(p, %s ,
words[i]);
strcpy(line,tmp);
printf(%s\n, line);
return 0;
}
On Aug 15, 6:18 am, programming love love.for.programm...@gmail.com
wrote:
write a program to reverse the words in a give string.
also state the time complexity of the algo.
if the string is i am a programmer
the output should be programmer a am i
--
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Re: [algogeeks] Re: Jumping Puzzle
HI Rohit, Although i haven't checked test many cases also I am not saying algo is wrong but DP will always Gives us Optimal Solution, even for large data-set,but Greedy Will Fail in That Case.I was aware of the same Greedy Algo/Code That you posted but found DP Will Excite Interviewer :) Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra On Sat, Aug 13, 2011 at 11:08 PM, rohit jangid rohit.nsi...@gmail.comwrote: ok check this, https://ideone.com/hZboG there may be bugs in coding, but I'm quite sure that algo is correct need to check more cases though but working on all the cases discussed here is there any proof that greedy won't work in this case? On Sat, Aug 13, 2011 at 11:03 PM, rohit jangid rohit.nsi...@gmail.com wrote: found some bugs , will repost it On Sat, Aug 13, 2011 at 10:55 PM, rohit jangid rohit.nsi...@gmail.com wrote: I can only say that above code is wrong, check this code of mine, I have tested more cases and all are working, https://ideone.com/pEBs8 see if you can find any bug in this one . thanks. On Sat, Aug 13, 2011 at 8:03 PM, WgpShashank shashank7andr...@gmail.com wrote: @rohit , I think we will get some cases where greedy won't work check out its giving 3 jumps still https://ideone.com/6UWW1 checked in hurray , if anything wrong do send me over gmail ? Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/Rerf5mzR7XcJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
A=p(biased coin/5 heads)=8/9 probability that the coin is biased given 5 heads (bayes theorem) B=p(unbiased coin/5 heads)=1/9 P(6th head)=A*1+B*1/2=17/18 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Jumping Puzzle
@shashank acc to me the problem can be done easily using greedy approach https://ideone.com/dYbUd ( it is your non optimal greedy solution with just one line added ) couldn't find any case where the above code is giving wrong answer . if you know any case, please let me know. I wrote a dp O(n^2) solution before that but found that same can be done using greedy as well On Fri, Aug 12, 2011 at 7:57 PM, WgpShashank shashank7andr...@gmail.com wrote: A Top Down Memoization DP Algorithm Will be As Follow Step 1. Declare an array of size N say jump[N]; where ith position indicates the number of jumps requires to reach from start to this (ith) position in array . Step 2. Initialize jump[0]=0; indicates to reach oth location form starting location (3which is obviuosly 0) we don't requires any jump. Step 3. Check size of array and value at 0th location in array For first index, optimum number of jumps will be zero. Please note that if value at first index is zero, we can’t jump to any element and return infinite.so these tow cases are A.If size of array==0 means array is of zero size; B.If value at zero then we can't jump or we can't proceed to next location Step 4. Run Through Loop for remaining elements in array and Initialize jump[i] as infinite. where 1=i=N. Sub-step 4A. (Please Note j runs in inner loop until i=j+a[j] ) if jump[i]jump[j]+1 update jump[i] repeat until ji (this whole processing will happen in inner loop). e.g. for each ith element this loop will run tries to figure out optimal/minimum number of jumps required to reach this ith position from starting of array . Step 5. After running above algorithm we will return array[n-1] that will show number of jumps required to reach last elemnt in array from start. Time Complexity O(N^2) Space Complexity O(N) Hope You Can Convert it into Production Code Do Notify me via mail if i missed anything ?? Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/37L_lAEKGVkJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Jumping Puzzle
I can only say that above code is wrong, check this code of mine, I have tested more cases and all are working, https://ideone.com/pEBs8 see if you can find any bug in this one . thanks. On Sat, Aug 13, 2011 at 8:03 PM, WgpShashank shashank7andr...@gmail.com wrote: @rohit , I think we will get some cases where greedy won't work check out its giving 3 jumps still https://ideone.com/6UWW1 checked in hurray , if anything wrong do send me over gmail ? Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/Rerf5mzR7XcJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Jumping Puzzle
ok check this, https://ideone.com/hZboG there may be bugs in coding, but I'm quite sure that algo is correct need to check more cases though but working on all the cases discussed here is there any proof that greedy won't work in this case? On Sat, Aug 13, 2011 at 11:03 PM, rohit jangid rohit.nsi...@gmail.com wrote: found some bugs , will repost it On Sat, Aug 13, 2011 at 10:55 PM, rohit jangid rohit.nsi...@gmail.com wrote: I can only say that above code is wrong, check this code of mine, I have tested more cases and all are working, https://ideone.com/pEBs8 see if you can find any bug in this one . thanks. On Sat, Aug 13, 2011 at 8:03 PM, WgpShashank shashank7andr...@gmail.com wrote: @rohit , I think we will get some cases where greedy won't work check out its giving 3 jumps still https://ideone.com/6UWW1 checked in hurray , if anything wrong do send me over gmail ? Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/Rerf5mzR7XcJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- Rohit Jangid Under Graduate Student, Deptt. of Computer Engineering NSIT, Delhi University, India -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Jumping Puzzle
A Top Down Memoization DP Algorithm Will be As Follow Step 1. Declare an array of size N say jump[N]; where ith position indicates the number of jumps requires to reach from start to this (ith) position in array . Step 2. Initialize jump[0]=0; indicates to reach oth location form starting location (3which is obviuosly 0) we don't requires any jump. Step 3. Check size of array and value at 0th location in array For first index, optimum number of jumps will be zero. Please note that if value at first index is zero, we can’t jump to any element and return infinite.so these tow cases are A.If size of array==0 means array is of zero size; B.If value at zero then we can't jump or we can't proceed to next location Step 4. Run Through Loop for remaining elements in array and Initialize jump[i] as infinite. where 1=i=N. Sub-step 4A. (Please Note j runs in inner loop until i*=j+a[j] ) if jump[i]jump[j]+1 update jump[i] repeat until ji (this whole processing will happen in inner loop). e.g. for each ith element this loop will run tries to figure out optimal/minimum number of jumps required to reach this ith position from starting of array . Step 5. After running above algorithm we will return array[n-1] that will show number of jumps required to reach last elemnt in array from start. Time Complexity O(N^2) Space Complexity O(N) Hope You Can Convert it into Production Code Do Notify me via mail if i missed anything ?? **Regards Shashank Mani Computer Science Is Awesome So Why I Write Code Computer Science Birla institute of Technology Mesra * * * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/37L_lAEKGVkJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Jumping Puzzle
Hi, probably you did'nt got my approach :
A={2,2,3,4,1,1,1,1,1}
first i=0 A[i] = 2
so candidates are A[1] ,A[2]
we choose max(A[1] +1 , 2 + A[2]) = A[2]
now we jump to A[2].
now candidates are A[3],A[4],A[5]
we choose max(A[3] + 3 , A[4] + 4, A[5] + 5)
= A[3]
so now we jump to A[3]
now candidates are A[4],A[5],A[6],A[7],
Clearly A[7] + 7 is maximum so we jump to A[7] and then we jump to
A[8]
so we went from 0-2-3-7-8.
Note i am not choosing max A[i] I am choosing max(i + A[i]).
Can anyone find a flaw in this approach ?
On Aug 11, 9:42 pm, Tharun Damera tharun1...@gmail.com wrote:
Yes, greedy algo doesn't work.. U shud hv DP.. plz post the soln which u
read..
A={2,2,3,4,1,1,1,1,1}
Ur algo gives 0,2,5,6,7,8(these are indices... index starts frm 0)
best soln is 0,1,3,7,8..
On Thu, Aug 11, 2011 at 2:37 PM, Arun Vishwanathan
aaron.nar...@gmail.comwrote:
I did not get the optimal solution part..how is that u jump 1 to index 1?
On Thu, Aug 11, 2011 at 10:07 AM, Algo Lover algolear...@gmail.comwrote:
Given an array, start from the first element and reach the last by
jumping. The jump length can be at most the value at the current
position in the array. Optimum result is when you reach the goal in
minimum number of jumps.
For ex:
Given array A = {2,3,1,1,4}
possible ways to reach the end (index list)
i) 0,2,3,4 (jump 2 to index 2, then jump 1 to index 3 then 1 to index
4)
ii) 0,1,4 (jump 1 to index 1, then jump 3 to index 4)
Since second solution has only 2 jumps it is the optimum result.
My solution is for any index i loop from i to i + A[i]
find an index j where
(j + A[j]) is maximum for all j.
make i=j;
This solution in O(n) i suppose coz we are picking each element twice
in the worst case.
I have read a O(n^2) DP solution for this problem.Is there any
case where my approach will fail ?
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[algogeeks] Re: Jumping Puzzle
int A[100];
int dist[100];
int N;
void findDist(int p, int d)
{
if (d dist[p])
{
dist[p] = d;
for(int i = 0; i p; ++i)
if ((i+A[i]) = p)
findDist(i,d+1);
}
}
int main(int argc, char* argv[])
{
int i;
int location = 0;
printf(Number of elements:);
scanf(%d, N);
for(i = 0; i N; ++i)
{
printf(Element %d:, i);
scanf(%d,A[i]);
}
for(i = 0; i N; ++i)
dist[i] = N;
findDist(N-1, 0);
for(i = 1; i N; ++i)
if (dist[i] == (dist[location]-1))
{
printf(Move %d to location %d\n, i-location, i);
location = i;
}
return 0;
}
On Aug 11, 3:07 am, Algo Lover algolear...@gmail.com wrote:
Given an array, start from the first element and reach the last by
jumping. The jump length can be at most the value at the current
position in the array. Optimum result is when you reach the goal in
minimum number of jumps.
For ex:
Given array A = {2,3,1,1,4}
possible ways to reach the end (index list)
i) 0,2,3,4 (jump 2 to index 2, then jump 1 to index 3 then 1 to index
4)
ii) 0,1,4 (jump 1 to index 1, then jump 3 to index 4)
Since second solution has only 2 jumps it is the optimum result.
My solution is for any index i loop from i to i + A[i]
find an index j where
(j + A[j]) is maximum for all j.
make i=j;
This solution in O(n) i suppose coz we are picking each element twice
in the worst case.
I have read a O(n^2) DP solution for this problem.Is there any
case where my approach will fail ?
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Re: [algogeeks] Re: Probability Puzzle
@dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
go through the posts before posting anything :) On Tue, Aug 9, 2011 at 6:29 PM, arpit.gupta arpitg1...@gmail.com wrote: it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
ans is 16/17 + 1/2*1/17 = 33/34 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Arpit: No. The probability of getting 6 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 17/80, while the probability of getting 5 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 9/40. Thus, the probability of getting a head on the sixth roll given that you have gotten heads on all five previous rolls is (17/80) / (9/40), which is 17/18. Dave On Aug 9, 7:59 am, arpit.gupta arpitg1...@gmail.com wrote: it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@dave- calculation mistake on my part - method is right. getting 17/18 only thanks anyways. On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote: @Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The probability of getting n consecutive heads is P(n heads) = 1/5 * 1 + 4/5 * (1/2)^n, Thus, the probability of getting a head on the n+1st roll given that you have gotten heads on all n previous rolls is P(n+1 heads | n heads) = P(n+1) / P(n) = ( 1/5 * 1 + 4/5 * (1/2)^(n+1) ) / ( 1/5 * 1 + 4/5 * (1/2)^n ). Multiplying numerator and denominator by 5* 2^(n-1) and recognizing 4 as 2^2 gives P(n+1 heads | n heads) = (2^(n-1) + 1) / (2^(n-1) + 2). Dave On Aug 9, 12:30 am, programming love love.for.programm...@gmail.com wrote: @Dave: I guess 17/18 is correct. Since we have to *calculate the probability of getting a head in the 6th flip given that first 5 flips are a head*. Can you please explain how you got the values of consequent flips when you said this? *In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18.* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@dave - method is right, calculation mistake on my part, getting 17/18 only. thanks anyways. On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote: @Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@Dave: Thanks for the explanation :) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@dave: thank you.. nice explanation :) On Wed, Aug 10, 2011 at 3:24 AM, Dave dave_and_da...@juno.com wrote: @Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@dave : nice explanationsthank you for pointing out :) On Wed, Aug 10, 2011 at 3:39 AM, Prakash D cegprak...@gmail.com wrote: @dave: thank you.. nice explanation :) On Wed, Aug 10, 2011 at 3:24 AM, Dave dave_and_da...@juno.com wrote: @Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Amazon Puzzle
I ve a small assumption, if we consider flight go in straight direction. Then in one paper, they can write Walk towards flight direction and jump first. In another paper Walk towards opposite to flight direction. and jump jump at any random location in that stright line.. then they can meet at middle point. Its may be a soultion, but not sure. Thanks Venkat http://cloud-computation.blogspot.com/ On Aug 7, 10:59 pm, Algo Lover algolear...@gmail.com wrote: Two people are travelling through flight. Both have parachute and jump anywhere randomly i.e none of them knows who has jumped where.(Assume there's a big desert and they jump at any random location). Now, both of them have a single piece of paper on which they can write instructions before jumping and that's the only way they can meet each other. What would they write on paper before jumping ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Amazon Puzzle
i feel, 1st person who jumps he just writes the time at he jumped. second person(my assumption) may be having the compass and watch to calculate the direction(of the 1st person) on his page/paper. Thank you, Siddharam On Mon, Aug 8, 2011 at 11:30 AM, Venkat venkataharishan...@gmail.comwrote: I ve a small assumption, if we consider flight go in straight direction. Then in one paper, they can write Walk towards flight direction and jump first. In another paper Walk towards opposite to flight direction. and jump jump at any random location in that stright line.. then they can meet at middle point. Its may be a soultion, but not sure. Thanks Venkat http://cloud-computation.blogspot.com/ On Aug 7, 10:59 pm, Algo Lover algolear...@gmail.com wrote: Two people are travelling through flight. Both have parachute and jump anywhere randomly i.e none of them knows who has jumped where.(Assume there's a big desert and they jump at any random location). Now, both of them have a single piece of paper on which they can write instructions before jumping and that's the only way they can meet each other. What would they write on paper before jumping ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
(3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Amazon Puzzle
Person jumping first faces in the direction of flight, person jumping second faces in the direction opp. to the flight, and will just drop down and they'll walk in the facing direction after jump. On Aug 7, 10:59 pm, Algo Lover algolear...@gmail.com wrote: Two people are travelling through flight. Both have parachute and jump anywhere randomly i.e none of them knows who has jumped where.(Assume there's a big desert and they jump at any random location). Now, both of them have a single piece of paper on which they can write instructions before jumping and that's the only way they can meet each other. What would they write on paper before jumping ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Amazon Puzzle
I think if we use the relevance of the flight we would not be able to do it, because we dont when they jumped or where they jumped, as far as i think, any one can just write an angle and the app. length of the shadow( they may take the same as their heights) and i think that will take them to the same height, i know it sounds a bit ridiculous :P and the sunlight must be there On Mon, Aug 8, 2011 at 1:34 PM, sumit gaur sumitgau...@gmail.com wrote: Person jumping first faces in the direction of flight, person jumping second faces in the direction opp. to the flight, and will just drop down and they'll walk in the facing direction after jump. On Aug 7, 10:59 pm, Algo Lover algolear...@gmail.com wrote: Two people are travelling through flight. Both have parachute and jump anywhere randomly i.e none of them knows who has jumped where.(Assume there's a big desert and they jump at any random location). Now, both of them have a single piece of paper on which they can write instructions before jumping and that's the only way they can meet each other. What would they write on paper before jumping ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- --- Puneet Goyal Student of B. Tech. III Year (Software Engineering) Delhi Technological University, Delhi --- -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Amazon Puzzle
On Mon, Aug 8, 2011 at 1:58 PM, Puneet Goyal puneetgoya...@gmail.comwrote: I think if we use the relevance of the flight we would not be able to do it, because we dont know when they jumped or where they jumped, as far as i think, any one can just write an angle and the app. length of the shadow( they may take the same as their heights) and i think that will take them to the same point, i know it sounds a bit ridiculous :P and the sunlight must be there -- --- Puneet Goyal Student of B. Tech. III Year (Software Engineering) Delhi Technological University, Delhi --- -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
I think 17/80 is wrong because if you say that while calculating the answer 3/5, you havent included the first 5 cases, then even after including it will only increase the probability of getting the biased coin in hand and thus increasing the overall probability of getting the heads and 17/80 is a way lesser than 3/5 although i am not sure about 3/5 even coz of the reasoning i just gave also, when you are calculating 4/5*1/2^6 you are not getting any benefit out of the first five tosses, like, they must have gave you some positive response towards that yes, you will get the head even next time, but doing this you are actually decreasing the probability as compared to the one you could have get without those 5 cases On Mon, Aug 8, 2011 at 1:06 PM, Shachindra A C sachindr...@gmail.comwrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- --- Puneet Goyal Student of B. Tech. III Year (Software Engineering) Delhi Technological University, Delhi --- -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Amazon Puzzle
Maybe those guys were having some gadgets like mobile phones or satellite phones.. So they would write their number on the other's paper... Or maybe they were carrying a GPS system.. So they would write a common (latitude,longitude) to meet at the same point... -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The answer is 17 in 18, because flipping 5 heads in a row is evidence that the probability is high that we have the coin with two heads. Don On Aug 7, 12:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@don: i too get yr answer 17/18 using conditional probability...does that make sense??i guess this is first new answer lol On Mon, Aug 8, 2011 at 9:29 PM, Don dondod...@gmail.com wrote: The answer is 17 in 18, because flipping 5 heads in a row is evidence that the probability is high that we have the coin with two heads. Don On Aug 7, 12:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@shady: 3/5 can be the answer to such a question: what is prob of getting head on nth toss if we have 4 coins fair and one biased...then at nth toss u choose 4/5 1/5 prob and then u get 3/5 @shady , don: i did this: P( 6th head | 5 heads occured)= P( 6 heads )/ P( 5 heads) answr u get is 17/18..i cud be wrong please correct if so On Mon, Aug 8, 2011 at 10:45 PM, shady sinv...@gmail.com wrote: answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Man, I feel so stupid. Yes, it is a case of conditional probability. We have to calculate the probability of six heads, given that 5 heads have occured. So answer is 17/18. On Tue, Aug 9, 2011 at 1:47 AM, Arun Vishwanathan aaron.nar...@gmail.comwrote: @shady: 3/5 can be the answer to such a question: what is prob of getting head on nth toss if we have 4 coins fair and one biased...then at nth toss u choose 4/5 1/5 prob and then u get 3/5 @shady , don: i did this: P( 6th head | 5 heads occured)= P( 6 heads )/ P( 5 heads) answr u get is 17/18..i cud be wrong please correct if so On Mon, Aug 8, 2011 at 10:45 PM, shady sinv...@gmail.com wrote: answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com http://twitter.com/ShuaibKhan -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
it's 3/5 On Tue, Aug 9, 2011 at 8:29 AM, Dave dave_and_da...@juno.com wrote: @Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Pls check the ques 8th This may remove misunderstanding... http://www.folj.com/puzzles/difficult-logic-problems.htm On Tue, Aug 9, 2011 at 10:21 AM, raj kumar megamonste...@gmail.com wrote: @all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Coder: You (and others) are saying that the probability of a head is 3/5 on the first flip, and that it doesn't change after any number of heads are flipped. Notice, however, that if the first flip were tails, you wouldn't say that the probability of getting heads on the next flip is 3/5. You would have learned that one of the four fair coins was chosen. So even though the probability of a head was 3/5 on the first flip, it changes to 1/2 on all flips subsequent to a tail. Since the probabililty changes if a tail is flipped, what makes you think it doesn't change if a head is flipped. In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18. Dave On Aug 8, 11:23 pm, coder dumca coder.du...@gmail.com wrote: it's 3/5 On Tue, Aug 9, 2011 at 8:29 AM, Dave dave_and_da...@juno.com wrote: @Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj. Granted that the first flip has a 3/5 probability of getting a head. But if it produces a tail, would you say that the second flip also has a 3/5 probability of getting a head? Or have you learned something from the tail? If you learn something from a tail, why don't you learn something from a head? Dave On Aug 8, 11:51 pm, raj kumar megamonste...@gmail.com wrote: @all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Just to resolve the issue what will be the probability of getting 6 consecutive heads -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj: After getting 5 consecutive heads, the probability of getting a 6th head is 17/18. Dave On Aug 9, 12:17 am, raj kumar megamonste...@gmail.com wrote: Just to resolve the issue what will be the probability of getting 6 consecutive heads -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj. Good. So now answer my last question? Dave On Aug 9, 12:21 am, raj kumar megamonste...@gmail.com wrote: no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@Dave: I guess 17/18 is correct. Since we have to *calculate the probability of getting a head in the 6th flip given that first 5 flips are a head*. Can you please explain how you got the values of consequent flips when you said this? *In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18.* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
http://math.arizona.edu/~jwatkins/f-condition.pdf see this link now ithink the answer should be 65/66 bcoz the probability of selectting double headed coin after n heads =2^n/2^n+1 and fair coin is =1/2^n+1 so for 6th head it should be :2^n/2^n+1*1+((1/2^n+1)*1/2) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
0.6? On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com http://twitter.com/ShuaibKhan -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.com wrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com http://twitter.com/ShuaibKhan -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To
Re: [algogeeks] Re: Probability Puzzle
On Mon, Aug 8, 2011 at 12:47 AM, aseem garg ase.as...@gmail.com wrote: Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem Well you are thinking about it the wrong way. Question asks that what is the probability that heads will show up the first five times, plus a sixth time. Not just the sixth time. The first five times head showing up is part of the question. On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com http://twitter.com/ShuaibKhan -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group,
Re: [algogeeks] Re: Probability Puzzle
@Shuaib: **What is the probability that you toss *next time, heads turns up ***. Aseem On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan aries.shu...@gmail.com wrote: On Mon, Aug 8, 2011 at 12:47 AM, aseem garg ase.as...@gmail.com wrote: Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem Well you are thinking about it the wrong way. Question asks that what is the probability that heads will show up the first five times, plus a sixth time. Not just the sixth time. The first five times head showing up is part of the question. On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com . To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib
Re: [algogeeks] Re: Probability Puzzle
no fight.. lets mention both the answers :D On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan aries.shu...@gmail.com wrote: On Mon, Aug 8, 2011 at 12:47 AM, aseem garg ase.as...@gmail.com wrote: Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem Well you are thinking about it the wrong way. Question asks that what is the probability that heads will show up the first five times, plus a sixth time. Not just the sixth time. The first five times head showing up is part of the question. On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.comwrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com . To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com
Re: [algogeeks] Re: Probability Puzzle
On Mon, Aug 8, 2011 at 12:51 AM, aseem garg ase.as...@gmail.com wrote: @Shuaib: **What is the probability that you toss *next time, heads turns up***. Well if you interpret it your way, then you are right. Otherwise, not. Aseem On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:47 AM, aseem garg ase.as...@gmail.com wrote: Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem Well you are thinking about it the wrong way. Question asks that what is the probability that heads will show up the first five times, plus a sixth time. Not just the sixth time. The first five times head showing up is part of the question. On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.com wrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to
Re: [algogeeks] Re: Probability Puzzle
abe yaar kya farak padta hai... 3/5=0.6 , other one may be 0.4 ya 0.3, 0.3 ke difference ke liye lad rahe ho... Chill guys... On 8/8/11, Shuaib Khan aries.shu...@gmail.com wrote: On Mon, Aug 8, 2011 at 12:51 AM, aseem garg ase.as...@gmail.com wrote: @Shuaib: **What is the probability that you toss *next time, heads turns up***. Well if you interpret it your way, then you are right. Otherwise, not. Aseem On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:47 AM, aseem garg ase.as...@gmail.com wrote: Think it like this. I have tossed a coin 5 times and it showed heads all the times. What is the probabilty of it shoing a HEADS now? Aseem Well you are thinking about it the wrong way. Question asks that what is the probability that heads will show up the first five times, plus a sixth time. Not just the sixth time. The first five times head showing up is part of the question. On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan aries.shu...@gmail.comwrote: On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam puneet.nsi...@gmail.com wrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. It is not independent. Re read the question. The first five times, it HAS to be heads. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message
[algogeeks] Re: Probability Puzzle
@Puneet: So are you saying that 100 heads in a row wouldn't convince you that you had the unfair coin? How many heads in a row would it take? Dave On Aug 7, 2:40 pm, Puneet Gautam puneet.nsi...@gmail.com wrote: Sixth toss is independent of previous tosses and dependent only on coin selection...! 1/5 + 4/5(1/2)= 3/5 is the correct answer we want to calc. probability of getting heads the sixth time only even if it would have been 100 th time...3/5 would be the answer only.. On 8/8/11, Prakash D cegprak...@gmail.com wrote: 1.) coin is fair 2.) coin is unfair P(head) for unfair coin= 1/5 * 1= 1/5 P(head) for fair coin= 4/5* 1/2 = 2/5 the probability at any instant that the tossed coin is a head is 3/5 17/80 is the probability to get head at all the six times. the soln. for this problem will be 3/5 On Mon, Aug 8, 2011 at 12:45 AM, aseem garg ase.as...@gmail.com wrote: If the coin is unbiased then probability of heads: 1/2 irrespective of whether it is first time or nth time. So answer should be 3/5. Aseem On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra saurabh131...@gmail.comwrote: Even u dont get why u people are gettin 17/80...the probability that it will be a head 6th time will be same as the frst time...so it shud be 3/5... On Aug 7, 11:05 pm, Kunal Yadav kunalyada...@gmail.com wrote: @algo: We can get head in two cases:- 1.) coin is biases 2.) coin is not biased P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5 P(head) for unbiased= 4/5*(1/2)^6 hence combined probability is what nitish has already mentioned. Hope you get the point. On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover algolear...@gmail.com wrote: Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Kunal Yadav (http://algoritmus.in/) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
