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Chaotic Puzzles
@chaotic_puzzles
Ivan Koswara, puzzle author on several places. The puzzle-only Twitter account of .
Joined February 2018

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This entry won the Haisu installment Logic Showcase in Puzzlers Club, where people submit a Haisu puzzle and then vote for their favorite entries. This entry also happens to be the easiest among all 13 entries; there was a 29x29 entry.
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If this looks impenetrable, try looking at the penalty method that I just posted a moment ago. (The second half of the thread walks through this puzzle, so spoiler warning.)
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This thread is about a global Heyawake/Heyawacky deduction, when you have a near-maximum number of black squares on the grid. This is not mine; I'll quote the the original at the end of this thread (to avoid spamming mentions). This applies to Heyawake on rectangular grids.
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This thread is taken from 's observation about this exact same penalty method, although there it's only used for even x even grid. After Puzzlers Club shared about this, I translated, generalized, and reorganized it. (The 7x8 puzzle is my creation.)
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もしかしたらへやわけで使えるかも知れない考え方について、既出かも知れませんが、自分用に書いておこうかなと思います ひとまず10x10の盤面を扱うことにします 与えられたへやわけの盤面に対して、白マスを頂点、隣り合う白マスを辺に持つようなグラフを考えてみます。こんな感じに。
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Now, this means the left side is missing at least one black square from the maximum possible. This is our penalty. Now we have matched all our penalties, so we can't make any more: - Corners are all black - Top and right side are determined (to fit maximum blacks)
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Recap: R,C = dimensions X = # corner blacks Y = # edge blacks Z = # internal blacks Threshold = (R-1)(C-1)+2ceil(R/2)+2ceil(C/2) Penalties: K = # corner whites M = # wasted edges N = # non-redundant loops # penalties = threshold - 3 * # blacks
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(Formally, N is the minimum number of edges you need to remove to make it a tree. Equivalently, record loops you find in the graph, but each loop must use a new edge that hasn't been used by previous loops, otherwise it doesn't count.)
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Recall that N = # of edges - # of vertices + 1. This counts the number of "non-redundant" loops of white cells. Here (lower-left), there are 3 loops, but only 2 are non-redundant. Pick any 2, and the third is deducible from the two. So here N=2.
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K is simply corners that fail to be black, i.e. white corners. M is # of blacks less from the maximum. This means: - each L-trimino on corner that is all white - each domino on even-length side is all white - each missing/misplaced black on odd-length side
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