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, return the length of the longest diagonal sequence 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 1
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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.
Rescoring based on corrected test suite. Sorry about that.
"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)"
Like Jonathan Campelli, I still feel the problem statement could be clearer. From the example I eventually figure that reference is being made to the diagonally oriented elements running between elements (1,5) and (4,2) of isprime(spiral(n)). But arguably these are not on a/the "diagonal" of that matrix, per http://mathworld.wolfram.com/Diagonal.html (cf. https://en.wikipedia.org/wiki/Main_diagonal ).
Furthermore, there is no mention in the current online MATLAB documentation that "spiral" is actually a function that is already provided with MATLAB — i.e. that the user does not have to write their own function to make the spiral matrix. Apparently, "spiral can be found in the demo folder" [ref: https://stackoverflow.com/questions/29466058/spiral-loop-on-a-matrix-from-a-point].
hard
eh, I think the problem statement made sense. especially if you just run 'isprime(spiral(n))' in your command window. I liked this problem. Ended up using the technique I almost used on another problem awhile back but it made more sense to use here, so I had the basic framework already started :D
for i=1:50
spy(isprime(spiral(i)))
pause(0.1)
end
Nice Problem