Unshred image using PIL. ======================== Courtesy of Instagram : http://instagram-engineering.tumblr.com/post/12651721845/instagram-engineering-challenge-the-unshredder This algorithm is very simple, it splits an image using shred objects. It starts from the left shred, searching the fitting shred on its right. Algorithm --------- I adopt a naive algorithm (with no signal theory). I compare each time one shred with every other shreds not ordered yet. If two shreds are very similar on their border, they must be consequently neighbor. So I just compare pixels of their border. The current implementation can use a mean approach with the border pixels to not just compare one pixel at a time but more. Some tests show that 3 pixels by each side is a good value (it also depends of the shred's width). E[P(x-2)P(x-1)P(x)] | E[P(x+1)P(x+2)P(x+3)] E1 E2 We subtract E1 and E2 (the closer they are, the more the result is close to 0). We iterate through the whole height (an optimization can be done here to only test some points). It gives us a score that we can compare. The final smaller score is the neighbor. With this algorithm, we can find neighbor. Strategy -------- For example (the good order is the crescent order): 4 -> 5 -> 1 -> 2 -> 3 So in this example 5 is the right border and 1 is the left border. The algorithm starts with : who is the right side of '4'. We found 5. We perform an inverse search to found the left side of 5. It is 4. We can continue. 4 -> 5 who is the right side of '5' We found '2' (can be anything, it will find something but this is meaningless, it is an edge case not solved here). We perform an inverse search which is not 2 (if we are unlucky, it is 2 again, another edge case here, not solved). So we can nearly say (see edge case) that '5' is the right border. We reverse the current order to perform an inverse search. We do not need to validate left side now. 5 -> 4 Who is the left side of '4' We found '3'. 5 -> 4 -> 3 And so on ... 5 -> 4 -> 3 -> 2 -> 1 Finally we reverse again. 1 -> 2 -> 3 -> 4 -> 5 It works. It can fail if it is unable to determine a neighbor (it will wrongly determine a border). Plus ---- Performance tip: all scoring is done with color (+alpha) tuple. It can be avoided to gain speed and memory to just sum them into a single int.