Problem 659. How long is the longest prime diagonal?
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9 Comments
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
Solution Comments
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1 Comment
Dear Bora,
You asked for help resolving this but I was "far away from Cody" for a long time... ;-)
I celebrate you were at least successful with it.
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1 Comment
Interesting! This error stems from using "Ix" in both function and sub-function. As I test my code in scripts not contained within a function I was missing it.
Why are parameters made available to sub-functions by default?
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2 Comments
this is really awesome and impressive.I do learn something
What a wonderfull solution!!!
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1 Comment
thank you very much
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1 Comment
a good precise algorithm
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2 Comments
Best solution that doesn't cheat with a lookup or regexp
Think I might have found a problem with merging the original and rotated matrices. In the case of n=7, if you change (1,6) and (2,7) to ones, the output becomes 6 when it should still be 4. If you evaluate the matrices separately then it should work fine, but apart from that this is a really nice solution
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1 Comment
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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