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Jun 13, 2017 at 9:42 history tweeted twitter.com/StackMath/status/874562677649137664
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S May 20, 2017 at 0:07 history suggested PJTraill CC BY-SA 3.0
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May 19, 2017 at 23:56 review Suggested edits
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May 19, 2017 at 16:58 answer added David K timeline score: 43
May 19, 2017 at 16:54 history edited Gallifreyan CC BY-SA 3.0
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May 19, 2017 at 14:08 comment added Servaes @pjs Yes I do, thanks for the correction. Unfortunately I can not edit the comment anymore.
May 19, 2017 at 14:06 comment added pjs36 @Servaes Do you mean 1x1 and 2x1, instead of 2x2 for the latter?
May 19, 2017 at 14:03 history edited Gallifreyan CC BY-SA 3.0
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May 19, 2017 at 13:57 comment added Servaes Not all 4096 combinations of border segments make sense; how do would a panel with only 1 out of 12 border segments make sense? Also; should each of the groups be a rectangle, or are non-rectangular panels also allowed? As a start, there are 127 different panels using only 1x1 and 2x2 blocks, so your picture is missing a lot of panels.
May 19, 2017 at 13:52 history edited Gallifreyan CC BY-SA 3.0
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May 19, 2017 at 13:44 comment added B. Mehta I don't think all 4096 possibilities work, for instance especially because we are restricting each panel to be a rectangle.
May 19, 2017 at 13:44 comment added mlk The 4096 ways also includes all kinds of non-rectangular panels. (Think of Tetris blocks, if you want an image) Naturally there are more of them, then if you only allow rectangles. However your 96 combinations are probably also off, since you overcount some symmetric and leave out others like a single big panel.
May 19, 2017 at 13:43 comment added badjohn Do you just want to enumerate all 4096 possibilities? If so, label the interior borders as 1, ..., 12. Write the numbers from 0 to 4095 in binary with leading zeros to give a consistent length. Include border 1 if bit 1 is 1, include border 2 if bit 2 is 1 etc. 0 will give you no internal borders hence one 9x9 region. 4095 will give you all internals borders hence 9 1x1 regions.
May 19, 2017 at 13:38 history asked Gallifreyan CC BY-SA 3.0