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
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.
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?
this is really awesome and impressive.I do learn something
thank you very much
a good precise algorithm
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
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.
197 Solvers
Project Euler: Problem 5, Smallest multiple
605 Solvers
390 Solvers
263 Solvers
Calculate Amount of Cake Frosting
11291 Solvers
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!