[algogeeks]
how are you -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@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]
Is it any puzzle On 11 Dec 2010 16:07, parth panchal parthpancha...@gmail.com wrote: how are you -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@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 algoge...@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] Longest interval with maximum sum
@amir can u explain a bit more... On Tue, Dec 7, 2010 at 10:09 PM, Amir hossein Shahriari amir.hossein.shahri...@gmail.com wrote: @jai : since sum of all values in C is between -n and n the last step can be done in O(n) time and O(n) space On Sun, Dec 5, 2010 at 12:44 PM, jai gupta sayhelloto...@gmail.comwrote: @fenghuang: the last step will take O(n logn ) . Or there is some better way? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@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 algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.comalgogeeks%2bunsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Aditya Kumar B-tech 3rd year Computer Science Engg. MNNIT, Allahabad. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algoge...@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] find the number.
Given range of numbers between A and B (A= B)
Find the number within given range which has more number of iterations as
per the following
n { stop ; return iteration number } if n=1;
n = 3n+1 if n is odd
n = n/2 if n is even
for eg :
n=3 odd
n=10;
n=5;
n=16;
n=8;
n=4;
n=2;
n=1;
iterations : 7
--
*
Time complexity= (n^2)
*
*NARESH ,A*
**
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[algogeeks] Re: find the number.
@Naresh: The sequence of numbers generated by this rule for any given
starting number is called a Collatz Sequence. Try googling it.
Here is a list of the number of iterations required for n between 1
and 10,000: http://oeis.org/A006577/b006577.txt. Maybe that will help.
Dave
On Dec 11, 7:20 am, Naresh A suryanar...@gmail.com wrote:
Given range of numbers between A and B (A= B)
Find the number within given range which has more number of iterations as
per the following
n { stop ; return iteration number } if n=1;
n = 3n+1 if n is odd
n = n/2 if n is even
for eg :
n=3 odd
n=10;
n=5;
n=16;
n=8;
n=4;
n=2;
n=1;
iterations : 7
--
*
Time complexity= (n^2)
*
*NARESH ,A*
**
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algoge...@googlegroups.com.
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algogeeks+unsubscr...@googlegroups.com.
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RE: [algogeeks] Re: program for evaluation of remainders
Hi, I agree with ankit sablok. And if we get the factorial of n in 1!, 2!, 3! Etc. Then we can find the number easily. In its complexity will be O(N) -Original Message- From: algogeeks@googlegroups.com [mailto:algoge...@googlegroups.com] On Behalf Of Dave Sent: Friday, December 10, 2010 8:10 PM To: Algorithm Geeks Subject: [algogeeks] Re: program for evaluation of remainders @Ankit: Why not just use the algorithm I proposed in http://groups.google.com/group/algogeeks/msg/2941ab071a39517c: x = 0; for( i = (n N ? n : N) ; i 0 ; --i ) x = (i * x + i) % n; Dave On Dec 10, 4:23 am, ankit sablok ankit4...@gmail.com wrote: @Dave we will use residues then i think the property of modulus 1!mod997 + 2!mod997 + 3!mod997 .. + 997!mod997 i just proposed the solution using congruences for the case nN can u generalize the problem using congruences if so then please post it thnanx in advance On Dec 9, 2:13 am, Dave dave_and_da...@juno.com wrote: @Ankit: So how does that work with, e.g., N = n = 997? I.e., what is the calculation? Dave On Dec 8, 11:33 am, ankit sablok ankit4...@gmail.com wrote: @ all the authors thanx for the suggestions actually wt i know about the problem is i think we can solve the problem mathematically if we know about congruences for instance if N=100 1! + 2! + . + 100! and n=12 we find that 4!mod24=0 hence the above equation reduces to the (1!+2!+3!)mod 12 =9 hence the answer is 9 so can anyone write a program for this logic On Dec 8, 6:19 pm, ankit sablok ankit4...@gmail.com wrote: Q) can anyboy find me the solution to this problem Given an integer N and an another integer n we have to write a program to find the remainder of the following problems (1! + 2! + 3! + 4! + . + N!)mod(n) N=100 n=1000; please help me write a program for this problem thanx in advance- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- 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 algoge...@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 algoge...@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 for evaluation of remainders
Are all of you talking about getting the result in closed form, so that no loop is involved? Other than mine, I haven't seen an implementation of a working algorithm. Let's see your code! My algorithm avoids calculating the factorials, which overflow 32-bit integers for N 12, and is O(min(N,n)). When you beat that, let me know. Dave On Dec 11, 10:15 am, Shiv Shankar mca.shivshan...@gmail.com wrote: Hi, I agree with ankit sablok. And if we get the factorial of n in 1!, 2!, 3! Etc. Then we can find the number easily. In its complexity will be O(N) -Original Message- From: algogeeks@googlegroups.com [mailto:algoge...@googlegroups.com] On Behalf Of Dave Sent: Friday, December 10, 2010 8:10 PM To: Algorithm Geeks Subject: [algogeeks] Re: program for evaluation of remainders @Ankit: Why not just use the algorithm I proposed inhttp://groups.google.com/group/algogeeks/msg/2941ab071a39517c: x = 0; for( i = (n N ? n : N) ; i 0 ; --i ) x = (i * x + i) % n; Dave On Dec 10, 4:23 am, ankit sablok ankit4...@gmail.com wrote: @Dave we will use residues then i think the property of modulus 1!mod997 + 2!mod997 + 3!mod997 .. + 997!mod997 i just proposed the solution using congruences for the case nN can u generalize the problem using congruences if so then please post it thnanx in advance On Dec 9, 2:13 am, Dave dave_and_da...@juno.com wrote: @Ankit: So how does that work with, e.g., N = n = 997? I.e., what is the calculation? Dave On Dec 8, 11:33 am, ankit sablok ankit4...@gmail.com wrote: @ all the authors thanx for the suggestions actually wt i know about the problem is i think we can solve the problem mathematically if we know about congruences for instance if N=100 1! + 2! + . + 100! and n=12 we find that 4!mod24=0 hence the above equation reduces to the (1!+2!+3!)mod 12 =9 hence the answer is 9 so can anyone write a program for this logic On Dec 8, 6:19 pm, ankit sablok ankit4...@gmail.com wrote: Q) can anyboy find me the solution to this problem Given an integer N and an another integer n we have to write a program to find the remainder of the following problems (1! + 2! + 3! + 4! + . + N!)mod(n) N=100 n=1000; please help me write a program for this problem thanx in advance- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- 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 algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group athttp://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 algoge...@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.
