Stanislaw Ulam once observed that if the counting numbers are arranged in a spiral, the prime numbers contained in it form a surprising pattern. They appear to cluster along diagonals of the spiral matrix.
Given n, find the longest diagonal arrangement of primes in spiral(n).
Example:
Input n = 7 Output d = 4
Since isprime(spiral(n)) is
1 0 0 0 1 0 0
0 0 0 1 0 0 0
1 0 1 0 0 0 0
0 1 0 0 1 1 0
0 0 1 0 1 0 1
0 1 0 0 0 1 0
1 0 0 0 0 0 1Solution Stats
Problem Comments
Jacob
on 11 Jun 2012
I agree that the solution is mis-coded. isprime(spiral(4)) gives:
1 0 0 0
0 0 1 1
1 0 1 0
0 0 0 1
which means that d=2 for this case.
Ned Gulley
on 11 Jun 2012
Rescoring based on corrected test suite. Sorry about that.
Jonathan Campelli
on 11 May 2015
"find the longest diagonal arrangement of primes in spiral(n)" leaves room for misinterpretation.
I would have quickly understood:
"return the length of the longest diagonal sequence of primes in spiral(n)"
Solution Comments
3 players like this solution
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1 Comment
Ramoflaple
on 11 Aug 2015
a good precise algorithm
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1 Comment
Jon
on 21 Mar 2014
Best solution that doesn't cheat with a lookup or regexp
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1 Comment
Dirk Engel
on 26 Jun 2012
max(diff(find(~zeroOneVector)))-1 returns the longest run of ones in a vector of zeros and ones, but it FAILS IN SEVERAL CASES, e.g.:
- all ones: [1 1 1 1 1] returns empty
- just a single zero: [1 1 0 1 1] returns empty
- all zeros at one end: [1 1 1 0 0] returns 0
- longest run at one end: [1 1 0 1 0] returns the longest run between zeros, here 1
As a workaround for all cases, add zeros to both ends of the vector:
max(diff(find(~[0 zeroOneVector 0])))-1
However, this is not necessary for this solution because the matrix is very sparse.
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6 players like this problem
6 players like this problem