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PO
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13304237
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2012-11-09T07:57:10.417
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2012-11-09T08:04:40.413
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You could increment the threshold until you find a square, and use a 2D FSM to detect the square. This will produce a match in <code>O(width * height * bpp)</code> (binary search on the lowest possible threshold, assuming a power-of-two range): <pre><code>- set threshold to its maximum value - for every bit of the threshold - clear the bit in the threshold - if there is a match - record the set of matches as a result - else - set the bit - if there is no record, then the threshold is its maximum. to detect a square: - for every pixel: - if the pixel is too bright, set its line-len to 0 - else if it's the first column, set its line-len to 1 - else set its line-len to the line-len of the pixel to the left, plus one - if the pixel line-len is less than N, set its rect-len to 0 - else if it's the first row, set its rect-len to 1 - else set its rect-len to the rect-len of the pixel above, plus one - if the rect-len is at least N, record a match. </code></pre> <code>line-len</code> represents the number of consecutive pixels that are dark enough. <code>rect-len</code> represents the number of consecutive rows of dark pixels that are long enough and aligned. For video-capture, replace the binary search by a linear search from the threshold for the previous frame. Obviously, you can't get better than <code>theta(width/N * height/N)</code> best case (as you'll have to rule out every possible position for a darker square) and the bit depth can be assumed constant, so this algorithm is asymptotically optimal for a fixed N. It's probably asymptotically optimal for N as a part of the input as well, as (intuitively) you have to consider almost every pixel in the average case.
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1. This table or related slice is empty.
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1. POdetection of the darkest fixed-size square from a picture
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1. PTAnswer
UserLastEditorUserId
1. USJohn Dvorak
UserOwnerUserId
1. USJohn Dvorak
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