![\"MATLAB](\"https:\/\/blogs.mathworks.com\/videos\/..\/images\/videos\/2008\/09\/target.jpg\")
<\/p>\nI have been working on a side project and I found I needed the following algorithm:<\/p>\n
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Given a binary five by five matrix, I need to find the ‘1’ value that is in the position that has the lowest value in the spiral map above.\u00a0 I call that the target square. Once that target is found, then I need to return a delta x<\/em> and delta y<\/em> that will get me closer to that\u00a0 target.\u00a0 Delta x<\/em> and delta y<\/em> must be taken relative to the center square, and can only be from the set [-1, 0, 1].<\/p>\n As an example, given the matrix:<\/p>\n The ‘1’ in the (2,2) position in the matrix is the target because it maps to 9, and the other two ‘1’ values map to larger values, 11 and 18.\u00a0 From here, the delta x<\/em> is -1 and delta y<\/em> is 1.<\/p>\n If the target was 18, then the values would have been delta x<\/em> of 0 and delta y<\/em> of -1.<\/p>\n A final edge case, if the input matrix were all zeros, then delta x<\/em> and delta y<\/em> should be chosen randomly, but still pulled from the set [-1, 0, 1].<\/p>\n This is not a particularly difficult algorithm to program if you use 25 if, elseif<\/strong> statements. However, I am looking to be a bit more clever with this.\u00a0 Please post your solution in the comments, make it a function of this form:<\/p>\n \n[deltaX, deltaY] = puzzler(inMatrix)<\/strong><\/p>\n As always with these puzzlers, be sure to use the proper tags in the comments.<\/p>\n I have been working on a side project and I found I needed the following algorithm:<\/p>\n![\"MATLAB](\"https:\/\/blogs.mathworks.com\/images\/videos\/2008\/09\/targetbinary.jpg\")
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\n\r\n<pre> <code>\r\n[deltaX, deltaY] = puzzler(inMatrix)<\/strong>\r\n%% All the code so someone can just copy and paste it from the comments.\r\n<\/code> <\/pre>\r\n<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"