Re: [algogeeks] Re: Directi question

[algo bard]

int no_of_steps[arr_length] = {MAX};

if ( (arr_length==0) || (arr[0] == 0) )   //if there are no elements or
the very first element is 0 -> we can't jump anywhere
return MAX;

no_of_steps[0] = 0; //no jumps required to jump from element 0 to itself.

for (int i=1; i= (i - j) )//if it is possible for us to jump from
jth element to ith element.
  {
 if( no_of_steps[i] > no_of_steps[j] + 1)// if no. of steps to
reach the ith element recorded till now is greater than the no of
 {   //
steps reqd to reach jth element + 1, then replace.

 no_of_steps[i] > no_of_steps[j] + 1;
 }
  }
}

}

cout< wrote:

> There is a greedy solution discussion about this approach. I don't have a
> formal proof for this.
> Any counter example will be helpful.
>
> at every place 'k' .. do the following.
>
> --> find max ( a[k+i]+i )  where 1 <= i <= a[i]
>
> for the given example:
> A = {4 0 0 3 6 5 4 7 1 0 1 2}
>
> initially a 4, the loop will be.
>   0+1,0+2,3+3,6+4 -- 10 is max. jump to 6.
> now at 6. (since, you can't reach end.)
> 5+1, 4+2, 7+3, 1+4, 0+5, 1 + 6, ==> 10 is max. jump to 7.
> make another step. (but you can reach the end.) so jump to last.
>
> this is greedy approach.. and a linear time soultion.
>
>
> On Monday, 9 July 2012 09:32:08 UTC-4, Akshat wrote:
>>
>> Given an array A and the elements stored in an array denotes how much
>> jump an element can make from that array position. For example there is an
>> array A = {4 0 0 3 6 5 4 7 1 0 1 2}
>>
>> Now Ist element which is 4 can make a jump to element 0, 0, 3 and 6. You
>> are stuck if you end up at 0.
>> You have to output the minimum number of jumps that can be made from
>> starting position to end position of an array.
>>
>> --
>>
>>
>> Akshat Sapra
>> Under Graduation(B.Tech)
>> IIIT-Allahabad(Amethi Campus)
>> *--*
>> sapraaks...@gmail.com
>> akshatsapr...@gmail.com
>> rit20009008@ iiita.ac.in
>>
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Re: [algogeeks] Re: Directi question

[algo bard]

^  cout<< no_of_steps[arr_length -1];

On Mon, Jul 9, 2012 at 8:44 PM, algo bard  wrote:

> int no_of_steps[arr_length] = {MAX};
>
> if ( (arr_length==0) || (arr[0] == 0) )   //if there are no elements
> or the very first element is 0 -> we can't jump anywhere
> return MAX;
>
> no_of_steps[0] = 0; //no jumps required to jump from element 0 to itself.
>
> for (int i=1; i {
>no_of_steps [i] = MAX;
>
>for(int j=0; j to jump to.
>{
>   if( arr[j] >= (i - j) )//if it is possible for us to jump from
> jth element to ith element.
>   {
>  if( no_of_steps[i] > no_of_steps[j] + 1)// if no. of steps to
> reach the ith element recorded till now is greater than the no of
>  {   //
> steps reqd to reach jth element + 1, then replace.
>
>  no_of_steps[i] > no_of_steps[j] + 1;
>  }
>   }
> }
>
> }
>
> cout<
>
> On Mon, Jul 9, 2012 at 8:32 PM, Mr.B  wrote:
>
>> There is a greedy solution discussion about this approach. I don't have a
>> formal proof for this.
>> Any counter example will be helpful.
>>
>> at every place 'k' .. do the following.
>>
>> --> find max ( a[k+i]+i )  where 1 <= i <= a[i]
>>
>> for the given example:
>> A = {4 0 0 3 6 5 4 7 1 0 1 2}
>>
>> initially a 4, the loop will be.
>>   0+1,0+2,3+3,6+4 -- 10 is max. jump to 6.
>> now at 6. (since, you can't reach end.)
>> 5+1, 4+2, 7+3, 1+4, 0+5, 1 + 6, ==> 10 is max. jump to 7.
>> make another step. (but you can reach the end.) so jump to last.
>>
>> this is greedy approach.. and a linear time soultion.
>>
>>
>> On Monday, 9 July 2012 09:32:08 UTC-4, Akshat wrote:
>>>
>>> Given an array A and the elements stored in an array denotes how much
>>> jump an element can make from that array position. For example there is an
>>> array A = {4 0 0 3 6 5 4 7 1 0 1 2}
>>>
>>> Now Ist element which is 4 can make a jump to element 0, 0, 3 and 6. You
>>> are stuck if you end up at 0.
>>> You have to output the minimum number of jumps that can be made from
>>> starting position to end position of an array.
>>>
>>> --
>>>
>>>
>>> Akshat Sapra
>>> Under Graduation(B.Tech)
>>> IIIT-Allahabad(Amethi Campus)
>>> *--*
>>> sapraaks...@gmail.com
>>> akshatsapr...@gmail.com
>>> rit20009008@ iiita.ac.in
>>>
>>  --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
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>> https://groups.google.com/d/msg/algogeeks/-/QRWxd8B2DzcJ.
>>
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>>
>
>

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Re: [algogeeks] Re: Directi question-centre of the tree

[sanjay pandey]

@ashish it wont be...first we r finding one end from any node dat is "r" n
den frm dat end we r traversing to other deepest end...
it is possible dat r b d intermediate node...n distance from r to v2 is
smaller than from r to v1

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Re: [algogeeks] Re: Directi question-centre of the tree

[Ashish Goel]

farthest from

2: Find a vertex v1 | the farthest form r.
3: Find a vertex v2 | the farthest form v1.



won't v2 be farthest from  r? or we are talking about alternate pats also



Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652


On Wed, Jun 20, 2012 at 6:25 PM, adarsh kumar  wrote:

> As you traverse along and find the diameter of the tree, keep track of the
> number of nodes thereby traversed. Let that be equal to n.
> Now, centre is the node corresponding to floor((n+1)/2).
>
>
> On Wed, Jun 20, 2012 at 5:19 PM, Nishant Pandey <
> nishant.bits.me...@gmail.com> wrote:
>
>> I am asking again .. can any one please suggest better method for getting
>> the median on the longest path of the tree ..
>> efficient method .
>>
>>
>> On Tue, Jun 19, 2012 at 5:08 PM, Nishant Pandey <
>> nishant.bits.me...@gmail.com> wrote:
>>
>>>
>>> for getting diameter we can simply add the max height of left subtree
>>> and max height of the right sub tree .
>>>
>>> the main issue is how efficiently we find median on that longest path to
>>> get the center of the tree .
>>> can any bdy sugest optimized algo for that ?
>>>
>>>
>>> On Mon, Jun 18, 2012 at 6:08 PM, rajesh pandey <
>>> rajesh.pandey.i...@gmail.com> wrote:
>>>
 I found it in some paper ;)

  Diameter and center
 De nition 4. The diameter of tree is the length of the longest path.
 De nition 5. A center is a vertex v such that the longest path from v
 to a leaf is minimal
 over all vertices in the tree.Tree center(s) can be found using simple
 algorithm.
 Algorithm 1. (Centers of tree)
 1: Choose a random root r.
 2: Find a vertex v1 | the farthest form r.
 3: Find a vertex v2 | the farthest form v1.
 4: Diameter is a length of path from v1 to v2.
 5: Center is a median element(s) of path from v1 to v2.

 This is O(n) algorithm. It is clear that we can't determine tree
 isomorphism faster
 than O(n). So, if we nd a O(f(n)) algorithm for rooted trees
 isomorphism we can also
 obtain O(f(n)) algorithm for ordinary trees.

 On Saturday, June 16, 2012 12:04:32 PM UTC+5:30, achala sharma wrote:
>
> I think this algorithm is used for calculating poset in graph.
>
> On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh 
> wrote:
>
>> + 1 for DK's solution. Is that a standard algorithm coz I feel like I
>> have heard it somewhere ??
>>
>>
>> On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:
>>
>>> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
>>> Could you please state how you can use the traversals directly to
>>> get the center? (And prove your correctness too?)
>>>
>>> The solution given by Wladimir (& expanded upon by me) is O(N) and
>>> uses (somewhat) the inverse of a BFS as a traversal.
>>>
>>> --
>>> DK
>>>
>>> http://twitter.com/divyekapoor
>>> http://gplus.to/divyekapoor
>>> http://www.divye.in
>>>
>>> --
>>> You received this message because you are subscribed to the Google
>>> Groups "Algorithm Geeks" group.
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>>> https://groups.google.com/d/**msg/algogeeks/-/HnMOZtOrkqwJ.
>>>
>>>
>>> To post to this group, send email to algogeeks@googlegroups.com.
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>>> **googlegroups.com .
>>> For more options, visit this group at http://groups.google.com/**
>>> group/algogeeks?hl=en
>>> .
>>>
>>
>>
>>
>> --
>> Hemesh singh
>>
>> --
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>>>
>>>
>>>
>>> --
>>> Cheers,
>>>
>>> Nishant Pandey |Specialist Tools Development  |  
>>> npan...@google.com |
>>> +91-9911258345
>>>
>>>
>>>
>>
>>
>> --
>> Cheers,
>>
>> Nishant Pandey |Specialist Tools Development  |  
>> npan...@google.com |
>> +91-9911258345
>>
>>
>>  --

Re: [algogeeks] Re: Directi question-centre of the tree

[adarsh kumar]

As you traverse along and find the diameter of the tree, keep track of the
number of nodes thereby traversed. Let that be equal to n.
Now, centre is the node corresponding to floor((n+1)/2).

On Wed, Jun 20, 2012 at 5:19 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:

> I am asking again .. can any one please suggest better method for getting
> the median on the longest path of the tree ..
> efficient method .
>
>
> On Tue, Jun 19, 2012 at 5:08 PM, Nishant Pandey <
> nishant.bits.me...@gmail.com> wrote:
>
>>
>> for getting diameter we can simply add the max height of left subtree and
>> max height of the right sub tree .
>>
>> the main issue is how efficiently we find median on that longest path to
>> get the center of the tree .
>> can any bdy sugest optimized algo for that ?
>>
>>
>> On Mon, Jun 18, 2012 at 6:08 PM, rajesh pandey <
>> rajesh.pandey.i...@gmail.com> wrote:
>>
>>> I found it in some paper ;)
>>>
>>>  Diameter and center
>>> De nition 4. The diameter of tree is the length of the longest path.
>>> De nition 5. A center is a vertex v such that the longest path from v to
>>> a leaf is minimal
>>> over all vertices in the tree.Tree center(s) can be found using simple
>>> algorithm.
>>> Algorithm 1. (Centers of tree)
>>> 1: Choose a random root r.
>>> 2: Find a vertex v1 | the farthest form r.
>>> 3: Find a vertex v2 | the farthest form v1.
>>> 4: Diameter is a length of path from v1 to v2.
>>> 5: Center is a median element(s) of path from v1 to v2.
>>>
>>> This is O(n) algorithm. It is clear that we can't determine tree
>>> isomorphism faster
>>> than O(n). So, if we nd a O(f(n)) algorithm for rooted trees isomorphism
>>> we can also
>>> obtain O(f(n)) algorithm for ordinary trees.
>>>
>>> On Saturday, June 16, 2012 12:04:32 PM UTC+5:30, achala sharma wrote:

 I think this algorithm is used for calculating poset in graph.

 On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh 
 wrote:

> + 1 for DK's solution. Is that a standard algorithm coz I feel like I
> have heard it somewhere ??
>
>
> On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:
>
>> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
>> Could you please state how you can use the traversals directly to get
>> the center? (And prove your correctness too?)
>>
>> The solution given by Wladimir (& expanded upon by me) is O(N) and
>> uses (somewhat) the inverse of a BFS as a traversal.
>>
>> --
>> DK
>>
>> http://twitter.com/divyekapoor
>> http://gplus.to/divyekapoor
>> http://www.divye.in
>>
>> --
>> 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/-/HnMOZtOrkqwJ.
>>
>>
>> To post to this group, send email to algogeeks@googlegroups.com.
>> To unsubscribe from this group, send email to algogeeks+unsubscribe@*
>> *googlegroups.com .
>> For more options, visit this group at http://groups.google.com/**
>> group/algogeeks?hl=en
>> .
>>
>
>
>
> --
> Hemesh singh
>
> --
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> .
>

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>>
>>
>>
>> --
>> Cheers,
>>
>> Nishant Pandey |Specialist Tools Development  |  
>> npan...@google.com |
>> +91-9911258345
>>
>>
>>
>
>
> --
> Cheers,
>
> Nishant Pandey |Specialist Tools Development  |  
> npan...@google.com |
> +91-9911258345
>
>
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Re: [algogeeks] Re: Directi question-centre of the tree

[Nishant Pandey]

I am asking again .. can any one please suggest better method for getting
the median on the longest path of the tree ..
efficient method .

On Tue, Jun 19, 2012 at 5:08 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:

>
> for getting diameter we can simply add the max height of left subtree and
> max height of the right sub tree .
>
> the main issue is how efficiently we find median on that longest path to
> get the center of the tree .
> can any bdy sugest optimized algo for that ?
>
>
> On Mon, Jun 18, 2012 at 6:08 PM, rajesh pandey <
> rajesh.pandey.i...@gmail.com> wrote:
>
>> I found it in some paper ;)
>>
>>  Diameter and center
>> De nition 4. The diameter of tree is the length of the longest path.
>> De nition 5. A center is a vertex v such that the longest path from v to
>> a leaf is minimal
>> over all vertices in the tree.Tree center(s) can be found using simple
>> algorithm.
>> Algorithm 1. (Centers of tree)
>> 1: Choose a random root r.
>> 2: Find a vertex v1 | the farthest form r.
>> 3: Find a vertex v2 | the farthest form v1.
>> 4: Diameter is a length of path from v1 to v2.
>> 5: Center is a median element(s) of path from v1 to v2.
>>
>> This is O(n) algorithm. It is clear that we can't determine tree
>> isomorphism faster
>> than O(n). So, if we nd a O(f(n)) algorithm for rooted trees isomorphism
>> we can also
>> obtain O(f(n)) algorithm for ordinary trees.
>>
>> On Saturday, June 16, 2012 12:04:32 PM UTC+5:30, achala sharma wrote:
>>>
>>> I think this algorithm is used for calculating poset in graph.
>>>
>>> On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:
>>>
 + 1 for DK's solution. Is that a standard algorithm coz I feel like I
 have heard it somewhere ??


 On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:

> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
> Could you please state how you can use the traversals directly to get
> the center? (And prove your correctness too?)
>
> The solution given by Wladimir (& expanded upon by me) is O(N) and
> uses (somewhat) the inverse of a BFS as a traversal.
>
> --
> DK
>
> http://twitter.com/divyekapoor
> http://gplus.to/divyekapoor
> http://www.divye.in
>
> --
> You received this message because you are subscribed to the Google
> Groups "Algorithm Geeks" group.
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> *msg/algogeeks/-/HnMOZtOrkqwJ.
>
>
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> group/algogeeks?hl=en 
> .
>



 --
 Hemesh singh

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>>>
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>
>
>
> --
> Cheers,
>
> Nishant Pandey |Specialist Tools Development  |  
> npan...@google.com |
> +91-9911258345
>
>
>


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npan...@google.com |
+91-9911258345

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Re: [algogeeks] Re: Directi question-centre of the tree

[Nishant Pandey]

for getting diameter we can simply add the max height of left subtree and
max height of the right sub tree .

the main issue is how efficiently we find median on that longest path to
get the center of the tree .
can any bdy sugest optimized algo for that ?

On Mon, Jun 18, 2012 at 6:08 PM, rajesh pandey  wrote:

> I found it in some paper ;)
>
>  Diameter and center
> De nition 4. The diameter of tree is the length of the longest path.
> De nition 5. A center is a vertex v such that the longest path from v to a
> leaf is minimal
> over all vertices in the tree.Tree center(s) can be found using simple
> algorithm.
> Algorithm 1. (Centers of tree)
> 1: Choose a random root r.
> 2: Find a vertex v1 | the farthest form r.
> 3: Find a vertex v2 | the farthest form v1.
> 4: Diameter is a length of path from v1 to v2.
> 5: Center is a median element(s) of path from v1 to v2.
>
> This is O(n) algorithm. It is clear that we can't determine tree
> isomorphism faster
> than O(n). So, if we nd a O(f(n)) algorithm for rooted trees isomorphism
> we can also
> obtain O(f(n)) algorithm for ordinary trees.
>
> On Saturday, June 16, 2012 12:04:32 PM UTC+5:30, achala sharma wrote:
>>
>> I think this algorithm is used for calculating poset in graph.
>>
>> On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:
>>
>>> + 1 for DK's solution. Is that a standard algorithm coz I feel like I
>>> have heard it somewhere ??
>>>
>>>
>>> On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:
>>>
 @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
 Could you please state how you can use the traversals directly to get
 the center? (And prove your correctness too?)

 The solution given by Wladimir (& expanded upon by me) is O(N) and uses
 (somewhat) the inverse of a BFS as a traversal.

 --
 DK

 http://twitter.com/divyekapoor
 http://gplus.to/divyekapoor
 http://www.divye.in

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>>>
>>>
>>>
>>> --
>>> Hemesh singh
>>>
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Re: [algogeeks] Re: Directi question-centre of the tree

[rajesh pandey]

I found it in some paper ;)

 Diameter and center
De nition 4. The diameter of tree is the length of the longest path.
De nition 5. A center is a vertex v such that the longest path from v to a 
leaf is minimal
over all vertices in the tree.Tree center(s) can be found using simple 
algorithm.
Algorithm 1. (Centers of tree)
1: Choose a random root r.
2: Find a vertex v1 | the farthest form r.
3: Find a vertex v2 | the farthest form v1.
4: Diameter is a length of path from v1 to v2.
5: Center is a median element(s) of path from v1 to v2.

This is O(n) algorithm. It is clear that we can't determine tree 
isomorphism faster
than O(n). So, if we nd a O(f(n)) algorithm for rooted trees isomorphism we 
can also
obtain O(f(n)) algorithm for ordinary trees.

On Saturday, June 16, 2012 12:04:32 PM UTC+5:30, achala sharma wrote:
>
> I think this algorithm is used for calculating poset in graph.
>
> On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:
>
>> + 1 for DK's solution. Is that a standard algorithm coz I feel like I 
>> have heard it somewhere ?? 
>>
>>
>> On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:
>>
>>> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).  
>>> Could you please state how you can use the traversals directly to get 
>>> the center? (And prove your correctness too?)
>>>
>>> The solution given by Wladimir (& expanded upon by me) is O(N) and uses 
>>> (somewhat) the inverse of a BFS as a traversal.
>>>
>>> --
>>> DK
>>>
>>> http://twitter.com/divyekapoor
>>> http://gplus.to/divyekapoor
>>> http://www.divye.in
>>>  
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>>
>>
>>
>> -- 
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>>
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Re: [algogeeks] Re: Directi question-centre of the tree

[achala sharma]

I think this algorithm is used for calculating poset in graph.

On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:

> + 1 for DK's solution. Is that a standard algorithm coz I feel like I have
> heard it somewhere ??
>
>
> On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:
>
>> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
>> Could you please state how you can use the traversals directly to get the
>> center? (And prove your correctness too?)
>>
>> The solution given by Wladimir (& expanded upon by me) is O(N) and uses
>> (somewhat) the inverse of a BFS as a traversal.
>>
>> --
>> DK
>>
>> http://twitter.com/divyekapoor
>> http://gplus.to/divyekapoor
>> http://www.divye.in
>>
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>
>
>
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Re: [algogeeks] Re: Directi question-centre of the tree

[Hemesh Singh]

+ 1 for DK's solution. Is that a standard algorithm coz I feel like I have
heard it somewhere ??

On Mon, Aug 8, 2011 at 1:37 AM, DK  wrote:

> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
> Could you please state how you can use the traversals directly to get the
> center? (And prove your correctness too?)
>
> The solution given by Wladimir (& expanded upon by me) is O(N) and uses
> (somewhat) the inverse of a BFS as a traversal.
>
> --
> DK
>
> http://twitter.com/divyekapoor
> http://gplus.to/divyekapoor
> http://www.divye.in
>
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Re: [algogeeks] Re: Directi Question

[Vipul Sharma]

@kunal patil:
U were proceeding in correct way...in next series u cud hav seen a
formation of arithmatico geometric series...

It doesnt matter what the value of no of faces in a dice is..ans will
be always 2...:)

My simplified soln: o+e+1

o=probability of odd number coming in 1
 throw of dice
 In case o & e are mutually exclusive the
ans is always 2...(and in even odd case like
 here e+o=1...so ans is always 2...let n be
 anything)

On 8/8/11, Kunal Patil  wrote:
> @Shady :
> No...we can say this only at the time when following constraints are
> satisfied:
> 1) *Outcome* of event *should be* *binary*. (In above example Sum can have
> binary outcomes only i.e. EVEN or ODD)
>
> 2) Random variable x in P(x) should be supported on set {1,2,3,4,} i.e.
> It *should start from 1* and then take on *contiguous values*.
>
> 3) *Sequence* *of probabilities* for x,x+1,x+2,... *must form a GP*.
>(In above sum P(1)=1/2; P(2)=1/4; P(3)=1/8 forms GP)
>
> Taking another simple example having biased coin:
> P(H) --> 1/4
> P(T) --> 1 - 1/4 = 3/4.
> Say we want to find out expected number of coin tosses required to get first
> head.
> (Outcome is binary and random variable starts from 1 & can take contiguous
> values.Also, sequence of probabilities form a GP.)
>
> You may verify that answer comes out to be 1/P(H) --> 1/(1/4) --> 4
>
> @ never_smile: If I understand your question correctly then Probability
> of getting even sum in one roll is P(2) + P(4)
> For e.g. say
> P(1) --> 1/5
> P(2) --> 2/5
> P(3) --> 1/5
> P(4) --> 0
> P(5) --> 1/5
>
> P(even in one roll) --> 2/5 + 0 --> 2/5.
>
> P(even in one roll) = 2/5;
> P(even in 2 rolls) = (3/5)*(3/5);
> P(even in 3 rolls)=(3/5)*(2/5)*(3/5)
> This probability sequence doesn't form a GP.
> Thus, above formula should not be applied & you should calculate E[x] by
> trivial method.
>
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nEvEr sMiLe bUt kEEp lAuGhIn'

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Re: [algogeeks] Re: Directi question-centre of the tree

[DK]

@Everyone: Wladimir has posted the correct solution to the problem and it is 
an O(N) bottom up solution.

*The original solution:*
On Wednesday, 6 October 2010 21:10:40 UTC+5:30, wladimirufc wrote:
>
> Find the leaf of tree then put all leaf in a queue.
>
> While queue not empty:
> remove u from queue
> remove u of tree
> if some v adjacent a u become leaf 
>  insert v in queue
>
> the last vertice is the center of the tree
>   



Let me expand on the solution for a bit.

*Definition:*
1. For the purposes of this solution, a *leaf node* is defined as a node 
having only one edge connected to it.

*Visualization:*
Think about the tree being flattened onto a table and laid out in a circular 
fashion.
eg.
1 
 /  /   \  \ 
4  2   3  7
/\ 
   5 6 
   \
   8

can be laid out as:


   4   3
\  /
 1 -- 7
  |
 2
   /\ 
  5 6
  \
   8
Now, wrap a rubber band around the layout and start tightening it. 
At every stage, you'll be getting rid of nodes with only a single edge (so 
called leaf nodes)
Stop when you end up with either a single node or a pair of nodes connected 
by an edge.

*For example, *
remove leaf nodes as:
Step 1: 8,5,4,3,7
Step 2: 6,1 (Note: 2 is not removed as it has 2 connecting edges)

The center of the tree is thus 2.

*Proof of correctness:*

Let T be a tree for which D(u, v) = D(v, u) = C is the maximum possible.
It is obvious that both u and v must be leaf nodes since u..v is a diameter 
of the graph.
Also, any non-leaf node w on the path u...v has M(w) < M(u) or M(v) since 
D(u, v) = D(u, w) + D(w, v)

Consider all such diameters of the tree graph T.
Let the pairs (u1, v1), (u2, v2)... (uk, vk) represent the nodes of a 
diameter of the graph.
Consider the subgraph G' = G - {u1, u2, ..., uk, v1, v2, .. vk}

The diameter of G' is C-2. The center(s) of the graph is/are unchanged.*
Applying this reduction multiple times, any tree's undirected graph 
representation
can be reduced to a graph of diameter 0 or 1.

*Base cases: *
Graph of diameter 0: The node itself is the center.
Graph of diameter 1: The pair of connected nodes are the center of the 
graph.

*Note:* The connectedness of the graph is maintained at every stage of the 
reduction process as no bridge nodes are ever removed.

** Proof of reduction of diameter by 2:*
Let us assume that the diameter of G' is C-1 with the nodes making up the 
diameter being (p, q).
Therefore atleast one of p and q is a leaf node in G.
But since we removed all the leaf nodes in the reduction process, both p and 
q must be internal nodes.
Hence the diameter of G' cannot be C-1. 
By removing the 2 end nodes on a diameter of G, we are assured that there 
exists a path of length C-2 in G'.
Since the diameter is an integral value and G' cannot have a diameter >= 
C-1, therefore the diameter 
of G' is C-2.

** Proof of non-changing nature of the center:* *(slightly hand-wavy, I 
would appreciate slightly more formality)*
Since the reduced graph G' has all the nodes of the original diameter 
(except the leaf nodes) present and the
diameter of the graph G is also present as one of the diameters of graph G', 
the center of the graph is 
still present in G'.

Taking all these parts together proves the result and also provides the 
algorithm for obtaining the result 
in O(N).

Hope you enjoyed it!

--
DK

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Re: [algogeeks] Re: Directi Question

[shady]

@kunal patil
expected number of rolls required to get even sum is inverse of that
i.e. 2 """
can we say like this all the time ?

i have understood the alternative method, thanks :D

On Sun, Aug 7, 2011 at 1:33 PM, Kunal Patil  wrote:

> Probability of getting an even sum in one roll is 1/2..
> Thus, expected number of rolls required to get even sum is inverse of that
> i.e. 2.
>
> Alternatively, Going by basics...
>
> Let P(x) be probability of getting Even sum in x rolls.
> P(1) = 1/2(Even)
> P(2) = (1/2) * (1/2)  (Odd + Odd)
> P(3) = (1/2) * (1/2) * (1/2) (Odd + Even + Odd)
> P(4) = (1/2) * (1/2) * (1/2) * (1/2) (Odd + Even + Even + Odd)
> & So on
>
> Expected number is given by Summation( X*P(x) )...i.e.
> 1*(1/2) + 2*(1/4) + 3*(1/8) + 4*(1/16) + (theoretically infinite terms)
>
> Let this sum be S.
>
> Thus, S = (1/2) + (2/4) + (3/8) + (4/16) + (5/32) +
>
> S / 2 = (1/4) + (2/8) + (3/16) + (4/32) +
>
> S - (S/2) = (1/2) + (2-1)/4 + (3-2)/8 + (4-3)/16 + (5-4)/32 + 
>
> Thus, S/2 = (1/2) + 1/4 + 1/8 + 1/16 + 1/32 + (theoretically infinite
> terms)
> R.H.S. is GP which evaluates to 1.
> Thus, S = 2.
> Thus, expected number of rolls required to get even sum is 2.
>
>
>
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Re: [algogeeks] Re: Directi Question

[nEvEr sMiLe bUt kEEp lAuGhIn']

@kunal patil:
how can u say "Probability of getting an even sum in one roll is 1/2."...
if tht is the case..can u find the ans in one go if the dice is having *5 
faces*??
i mean the numbers on dice are 1 2 3 4 5 ...and it is unbiased...
what will be the *Probability of getting an even sum in one roll*??

PS:ur explanation for basic method(alternative one) was awesome..:)

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Re: [algogeeks] Re: Directi Question

[Kunal Patil]

Probability of getting an even sum in one roll is 1/2..
Thus, expected number of rolls required to get even sum is inverse of that
i.e. 2.

Alternatively, Going by basics...

Let P(x) be probability of getting Even sum in x rolls.
P(1) = 1/2(Even)
P(2) = (1/2) * (1/2)  (Odd + Odd)
P(3) = (1/2) * (1/2) * (1/2) (Odd + Even + Odd)
P(4) = (1/2) * (1/2) * (1/2) * (1/2) (Odd + Even + Even + Odd)
& So on

Expected number is given by Summation( X*P(x) )...i.e.
1*(1/2) + 2*(1/4) + 3*(1/8) + 4*(1/16) + (theoretically infinite terms)

Let this sum be S.

Thus, S = (1/2) + (2/4) + (3/8) + (4/16) + (5/32) +

S / 2 = (1/4) + (2/8) + (3/16) + (4/32) +

S - (S/2) = (1/2) + (2-1)/4 + (3-2)/8 + (4-3)/16 + (5-4)/32 + 

Thus, S/2 = (1/2) + 1/4 + 1/8 + 1/16 + 1/32 + (theoretically infinite
terms)
R.H.S. is GP which evaluates to 1.
Thus, S = 2.
Thus, expected number of rolls required to get even sum is 2.

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Re: [algogeeks] Re: Directi question-centre of the tree

[programming love]

Traverse the tree inorder. Store the values in an array. Find the element in
the middle of the array.

On Sun, Aug 7, 2011 at 8:57 AM, subramania jeeva
wrote:

>  5
>  /\
> 6 7
>/
>   8
>
> Here the centre is both 5 and 6 . we need to find both of them..:)
>
>
>
>
>
>
>
>
>
> Cheers
>   ~ Jeeva ~
>
>
>
>
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Re: [algogeeks] Re: Directi Question

[shady]

Dave shouldn't it be :: E(x) = Summation( xp(x) ) thus it should
be 1*1/2 + 2*(1/2 * 1/2) + 3*(1/2 * 1/2 * 1/2) .

On Sun, Aug 7, 2011 at 11:55 AM, Nitish Garg wrote:

> @Dave - Can you please explain how did you generate the series? Shouldn't
> it be : 1/2 + 2(1/4) + 3(1/8) and so on?
>
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Re: [algogeeks] Re: Directi question-centre of the tree

[subramania jeeva]

 5
 /\
6 7
   /
  8

Here the centre is both 5 and 6 . we need to find both of them..:)









Cheers
  ~ Jeeva ~

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Re: [algogeeks] Re: Directi question-centre of the tree

[Wladimir Tavares]

Find the leaf of tree then put all leaf in a queue.

While queue not empty:
remove u from queue
remove u of tree
if some v adjacent a u become leaf
 insert v in queue

the last vertice is the center of the tree



On Wed, Oct 6, 2010 at 8:08 AM, Ruturaj  wrote:

> I am thinking on these lines.
>
> Start from any node. DFS to the fartheset node from there. let the
> farthest node be B. Start a new DFS from B to reach the fartheset node
> from B , let that be C. It can be proved that BC is the longest path
> in the tree. Then, the node in the center will be the answer to the
> question. Incase the path length is even we will have two nodes.
>
> I havent proved it yet, the second part of my solution. this was a
> quick thought.
>
> On Sep 29, 9:41 pm, jayapriya surendran  wrote:
> > In graph theory, a tree is defined as a graph on N nodes,and (N-1)
> > un-directed edges such that there are no cycles in the graph.Each node
> > has a single unique path to every other node.
> > Let D(u,v) be the number of edges in the unique path from node 'u' to
> > node 'v' (or from node 'v' to 'u' since the edges are
> > un-directed).D(u,u) is 0 for all nodes 'u'.
> > M(u)=MAX(D(u,i):for all nodes i)
> > The center of a tree is the node (or nodes) 'u',for which M(u) is
> > minimum among all the nodes in the graph.
> > You'll be given a graph which has N nodes (1<=N<=20).The nodes are
> > labeled 1,2,3,..N.You will be provided with N-1 edges in the form of
> > "a b" pairs where 1<=a,b<=N.No edge will be repeated.You can assume
> > that the edges are specified such that the graph is a valid tree as
> > defined above.
> > Output the node labels of the center(or centers) of the tree.
> > Sample Input:
> > 6(value of N)
> > 1 3 (edges)
> > 1 4
> > 1 2
> > 2 5
> > 2 6
> >
> > Sample Output
> > 1
> > 2
> > Expected:O(N) complexity algo
> > can anyone plz help me out with O(N) algo?
>
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