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(require both main diagonals to have one of every digit in addition to all the usual constraints) and Jigsaw Sudoku (replace the array of rectangular sub-blocks with the sub-blocks being random polyominoes). To implement the latter, I've moved my `divvy.c' library routine out of the `unfinished' subdirectory. Jigsaw mode is currently an undocumented feature: you enable it by setting the rows parameter to 1 (and the columns parameter to your desired grid size, which unlike normal Sudoku can be anything you like including a prime number). The reason it's undocumented is because generation times are not yet reliably short: sometimes generating a jigsaw-type puzzle can hang for hours and still get nowhere. (The algorithm should terminate in principle, but not in any time you're prepared to wait.) I _think_ I know how to solve this, but have yet to try it. Until then, jigsaw mode will remain a hidden feature. Printing of X-type puzzles is also substandard at present, because the current print-colour API replaces the desired light shading of the X-cells with heavy diagonal hatching. I plan to adjust the API imminently to address this. [originally from svn r7974]
782 lines
22 KiB
C
782 lines
22 KiB
C
/*
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* Library code to divide up a rectangle into a number of equally
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* sized ominoes, in a random fashion.
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*
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* Could use this for generating solved grids of
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* http://www.nikoli.co.jp/ja/puzzles/block_puzzle/
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* or for generating the playfield for Jigsaw Sudoku.
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*/
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/*
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* This code is restricted to simply connected solutions: that is,
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* no single polyomino may completely surround another (not even
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* with a corner visible to the outside world, in the sense that a
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* 7-omino can `surround' a single square).
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*
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* It's tempting to think that this is a natural consequence of
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* all the ominoes being the same size - after all, a division of
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* anything into 7-ominoes must necessarily have all of them
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* simply connected, because if one was not then the 1-square
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* space in the middle could not be part of any 7-omino - but in
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* fact, for sufficiently large k, it is perfectly possible for a
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* k-omino to completely surround another k-omino. A simple
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* example is this one with two 25-ominoes:
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*
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* +--+--+--+--+--+--+--+
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* | |
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* + +--+--+--+--+--+ +
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* | | | |
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* + + + +
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* | | | |
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* + + + +--+
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* | | | |
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* + + + +--+
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* | | | |
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* + + + +
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* | | | |
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* + +--+--+--+--+--+ +
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* | |
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* +--+--+--+--+--+--+--+
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*
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* I claim the smallest k which can manage this is 23. More
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* formally:
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*
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* If a k-omino P is completely surrounded by another k-omino Q,
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* such that every edge of P borders on Q, then k >= 23.
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*
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* Proof:
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*
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* It's relatively simple to find the largest _rectangle_ a
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* k-omino can enclose. So I'll construct my proof in two parts:
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* firstly, show that no 22-omino or smaller can enclose a
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* rectangle as large as itself, and secondly, show that no
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* polyomino can enclose a larger non-rectangle than a rectangle.
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*
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* The first of those claims:
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*
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* To surround an m x n rectangle, a polyomino must have 2m
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* squares along the two m-sides of the rectangle, 2n squares
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* along the two n-sides, and must fill in at least three of the
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* corners in order to be connected. Thus, 2(m+n)+3 <= k. We wish
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* to find the largest value of mn subject to that constraint, and
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* it's clear that this is achieved when m and n are as close to
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* equal as possible. (If they aren't, WLOG suppose m < n; then
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* (m+1)(n-1) = mn + n - m - 1 >= mn, with equality only when
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* m=n-1.)
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*
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* So the area of the largest rectangle which can be enclosed by a
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* k-omino is given by floor(k'/2) * ceil(k'/2), where k' =
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* (k-3)/2. This is a monotonic function in k, so there will be a
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* unique point at which it goes from being smaller than k to
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* being larger than k. That point is between 22 (maximum area 20)
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* and 23 (maximum area 25).
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*
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* The second claim:
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*
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* Suppose we have an inner polyomino P surrounded by an outer
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* polyomino Q. I seek to show that if P is non-rectangular, then
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* P is also non-maximal, in the sense that we can transform P and
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* Q into a new pair of polyominoes in which P is larger and Q is
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* at most the same size.
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*
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* Consider walking along the boundary of P in a clockwise
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* direction. (We may assume, of course, that there is only _one_
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* boundary of P, i.e. P has no hole in the middle. If it does
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* have a hole in the middle, it's _trivially_ non-maximal because
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* we can just fill the hole in!) Our walk will take us along many
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* edges between squares; sometimes we might turn left, and
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* certainly sometimes we will turn right. Always there will be a
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* square of P on our right, and a square of Q on our left.
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*
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* The net angle through which we turn during the entire walk must
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* add up to 360 degrees rightwards. So if there are no left
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* turns, then we must turn right exactly four times, meaning we
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* have described a rectangle. Hence, if P is _not_ rectangular,
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* then there must have been a left turn at some point. A left
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* turn must mean we walk along two edges of the same square of Q.
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*
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* Thus, there is some square X in Q which is adjacent to two
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* diagonally separated squares in P. Let us call those two
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* squares N and E; let us refer to the other two neighbours of X
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* as S and W; let us refer to the other mutual neighbour of S and
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* W as D; and let us refer to the other mutual neighbour of S and
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* E as Y. In other words, we have named seven squares, arranged
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* thus:
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*
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* N
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* W X E
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* D S Y
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*
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* where N and E are in P, and X is in Q.
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*
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* Clearly at least one of W and S must be in Q (because otherwise
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* X would not be connected to any other square in Q, and would
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* hence have to be the whole of Q; and evidently if Q were a
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* 1-omino it could not enclose _anything_). So we divide into
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* cases:
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*
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* If both W and S are in Q, then we take X out of Q and put it in
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* P, which does not expose any edge of P. If this disconnects Q,
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* then we can reconnect it by adding D to Q.
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*
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* If only one of W and S is in Q, then wlog let it be W. If S is
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* in _P_, then we have a particularly easy case: we can simply
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* take X out of Q and add it to P, and this cannot disconnect X
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* since X was a leaf square of Q.
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*
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* Our remaining case is that W is in Q and S is in neither P nor
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* Q. Again we take X out of Q and put it in P; we also add S to
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* Q. This ensures we do not expose an edge of P, but we must now
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* prove that S is adjacent to some other existing square of Q so
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* that we haven't disconnected Q by adding it.
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*
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* To do this, we recall that we walked along the edge XE, and
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* then turned left to walk along XN. So just before doing all
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* that, we must have reached the corner XSE, and we must have
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* done it by walking along one of the three edges meeting at that
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* corner which are _not_ XE. It can't have been SY, since S would
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* then have been on our left and it isn't in Q; and it can't have
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* been XS, since S would then have been on our right and it isn't
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* in P. So it must have been YE, in which case Y was on our left,
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* and hence is in Q.
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*
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* So in all cases we have shown that we can take X out of Q and
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* add it to P, and add at most one square to Q to restore the
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* containment and connectedness properties. Hence, we can keep
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* doing this until we run out of left turns and P becomes
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* rectangular. []
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*
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* ------------
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*
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* Anyway, that entire proof was a bit of a sidetrack. The point
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* is, although constructions of this type are possible for
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* sufficiently large k, divvy_rectangle() will never generate
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* them. This could be considered a weakness for some purposes, in
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* the sense that we can't generate all possible divisions.
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* However, there are many divisions which we are highly unlikely
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* to generate anyway, so in practice it probably isn't _too_ bad.
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*
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* If I wanted to fix this issue, I would have to make the rules
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* more complicated for determining when a square can safely be
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* _removed_ from a polyomino. Adding one becomes easier (a square
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* may be added to a polyomino iff it is 4-adjacent to any square
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* currently part of the polyomino, and the current test for loop
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* formation may be dispensed with), but to determine which
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* squares may be removed we must now resort to analysis of the
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* overall structure of the polyomino rather than the simple local
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* properties we can currently get away with measuring.
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*/
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/*
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* Possible improvements which might cut the fail rate:
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*
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* - instead of picking one omino to extend in an iteration, try
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* them all in succession (in a randomised order)
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*
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* - (for real rigour) instead of bfsing over ominoes, bfs over
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* the space of possible _removed squares_. That way we aren't
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* limited to randomly choosing a single square to remove from
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* an omino and failing if that particular square doesn't
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* happen to work.
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*
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* However, I don't currently think it's necessary to do either of
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* these, because the failure rate is already low enough to be
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* easily tolerable, under all circumstances I've been able to
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* think of.
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*/
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#include <assert.h>
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#include <stdio.h>
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#include <stdlib.h>
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#include <stddef.h>
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#include "puzzles.h"
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/*
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* Subroutine which implements a function used in computing both
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* whether a square can safely be added to an omino, and whether
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* it can safely be removed.
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*
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* We enumerate the eight squares 8-adjacent to this one, in
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* cyclic order. We go round that loop and count the number of
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* times we find a square owned by the target omino next to one
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* not owned by it. We then return success iff that count is 2.
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*
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* When adding a square to an omino, this is precisely the
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* criterion which tells us that adding the square won't leave a
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* hole in the middle of the omino. (If it did, then things get
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* more complicated; see above.)
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*
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* When removing a square from an omino, the _same_ criterion
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* tells us that removing the square won't disconnect the omino.
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* (This only works _because_ we've ensured the omino is simply
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* connected.)
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*/
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static int addremcommon(int w, int h, int x, int y, int *own, int val)
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{
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int neighbours[8];
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int dir, count;
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for (dir = 0; dir < 8; dir++) {
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int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1);
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int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1);
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int sx = x+dx, sy = y+dy;
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if (sx < 0 || sx >= w || sy < 0 || sy >= h)
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neighbours[dir] = -1; /* outside the grid */
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else
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neighbours[dir] = own[sy*w+sx];
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}
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/*
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* To begin with, check 4-adjacency.
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*/
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if (neighbours[0] != val && neighbours[2] != val &&
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neighbours[4] != val && neighbours[6] != val)
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return FALSE;
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count = 0;
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for (dir = 0; dir < 8; dir++) {
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int next = (dir + 1) & 7;
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int gotthis = (neighbours[dir] == val);
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int gotnext = (neighbours[next] == val);
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if (gotthis != gotnext)
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count++;
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}
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return (count == 2);
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}
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/*
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* w and h are the dimensions of the rectangle.
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*
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* k is the size of the required ominoes. (So k must divide w*h,
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* of course.)
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*
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* The returned result is a w*h-sized dsf.
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*
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* In both of the above suggested use cases, the user would
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* probably want w==h==k, but that isn't a requirement.
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*/
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static int *divvy_internal(int w, int h, int k, random_state *rs)
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{
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int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf;
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int wh = w*h;
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int i, j, n, x, y, qhead, qtail;
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n = wh / k;
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assert(wh == k*n);
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order = snewn(wh, int);
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tmp = snewn(wh, int);
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own = snewn(wh, int);
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sizes = snewn(n, int);
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queue = snewn(n, int);
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addable = snewn(wh*4, int);
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removable = snewn(wh, int);
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/*
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* Permute the grid squares into a random order, which will be
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* used for iterating over the grid whenever we need to search
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* for something. This prevents directional bias and arranges
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* for the answer to be non-deterministic.
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*/
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for (i = 0; i < wh; i++)
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order[i] = i;
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shuffle(order, wh, sizeof(*order), rs);
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/*
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* Begin by choosing a starting square at random for each
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* omino.
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*/
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for (i = 0; i < wh; i++) {
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own[i] = -1;
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}
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for (i = 0; i < n; i++) {
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own[order[i]] = i;
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sizes[i] = 1;
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}
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/*
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* Now repeatedly pick a random omino which isn't already at
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* the target size, and find a way to expand it by one. This
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* may involve stealing a square from another omino, in which
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* case we then re-expand that omino, forming a chain of
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* square-stealing which terminates in an as yet unclaimed
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* square. Hence every successful iteration around this loop
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* causes the number of unclaimed squares to drop by one, and
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* so the process is bounded in duration.
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*/
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while (1) {
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#ifdef DIVVY_DIAGNOSTICS
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{
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int x, y;
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printf("Top of loop. Current grid:\n");
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for (y = 0; y < h; y++) {
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for (x = 0; x < w; x++)
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printf("%3d", own[y*w+x]);
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printf("\n");
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}
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}
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#endif
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/*
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* Go over the grid and figure out which squares can
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* safely be added to, or removed from, each omino. We
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* don't take account of other ominoes in this process, so
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* we will often end up knowing that a square can be
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* poached from one omino by another.
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*
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* For each square, there may be up to four ominoes to
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* which it can be added (those to which it is
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* 4-adjacent).
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*/
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for (y = 0; y < h; y++) {
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for (x = 0; x < w; x++) {
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int yx = y*w+x;
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int curr = own[yx];
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int dir;
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if (curr < 0) {
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removable[yx] = FALSE; /* can't remove if not owned! */
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} else if (sizes[curr] == 1) {
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removable[yx] = TRUE; /* can always remove a singleton */
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} else {
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/*
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* See if this square can be removed from its
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* omino without disconnecting it.
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*/
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removable[yx] = addremcommon(w, h, x, y, own, curr);
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}
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for (dir = 0; dir < 4; dir++) {
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int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0);
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int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0);
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int sx = x + dx, sy = y + dy;
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int syx = sy*w+sx;
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addable[yx*4+dir] = -1;
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if (sx < 0 || sx >= w || sy < 0 || sy >= h)
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continue; /* no omino here! */
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if (own[syx] < 0)
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continue; /* also no omino here */
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if (own[syx] == own[yx])
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continue; /* we already got one */
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if (!addremcommon(w, h, x, y, own, own[syx]))
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continue; /* would non-simply connect the omino */
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addable[yx*4+dir] = own[syx];
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}
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}
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}
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for (i = j = 0; i < n; i++)
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if (sizes[i] < k)
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tmp[j++] = i;
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if (j == 0)
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break; /* all ominoes are complete! */
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j = tmp[random_upto(rs, j)];
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#ifdef DIVVY_DIAGNOSTICS
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printf("Trying to extend %d\n", j);
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#endif
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/*
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* So we're trying to expand omino j. We breadth-first
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* search out from j across the space of ominoes.
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*
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* For bfs purposes, we use two elements of tmp per omino:
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* tmp[2*i+0] tells us which omino we got to i from, and
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* tmp[2*i+1] numbers the grid square that omino stole
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* from us.
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*
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* This requires that wh (the size of tmp) is at least 2n,
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* i.e. k is at least 2. There would have been nothing to
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* stop a user calling this function with k=1, but if they
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* did then we wouldn't have got to _here_ in the code -
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* we would have noticed above that all ominoes were
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* already at their target sizes, and terminated :-)
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*/
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assert(wh >= 2*n);
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for (i = 0; i < n; i++)
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tmp[2*i] = tmp[2*i+1] = -1;
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qhead = qtail = 0;
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queue[qtail++] = j;
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tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */
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while (qhead < qtail) {
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int tmpsq;
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j = queue[qhead];
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/*
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* We wish to expand omino j. However, we might have
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* got here by omino j having a square stolen from it,
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* so first of all we must temporarily mark that
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* square as not belonging to j, so that our adjacency
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* calculations don't assume j _does_ belong to us.
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*/
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tmpsq = tmp[2*j+1];
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if (tmpsq >= 0) {
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assert(own[tmpsq] == j);
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own[tmpsq] = -3;
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}
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/*
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* OK. Now begin by seeing if we can find any
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* unclaimed square into which we can expand omino j.
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* If we find one, the entire bfs terminates.
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*/
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for (i = 0; i < wh; i++) {
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int dir;
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if (own[order[i]] != -1)
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continue; /* this square is claimed */
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/*
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* Special case: if our current omino was size 1
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* and then had a square stolen from it, it's now
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* size zero, which means it's valid to `expand'
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* it into _any_ unclaimed square.
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*/
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if (sizes[j] == 1 && tmpsq >= 0)
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break; /* got one */
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/*
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* Failing that, we must do the full test for
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* addability.
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*/
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for (dir = 0; dir < 4; dir++)
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if (addable[order[i]*4+dir] == j) {
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/*
|
|
* We know this square is addable to this
|
|
* omino with the grid in the state it had
|
|
* at the top of the loop. However, we
|
|
* must now check that it's _still_
|
|
* addable to this omino when the omino is
|
|
* missing a square. To do this it's only
|
|
* necessary to re-check addremcommon.
|
|
*/
|
|
if (!addremcommon(w, h, order[i]%w, order[i]/w,
|
|
own, j))
|
|
continue;
|
|
break;
|
|
}
|
|
if (dir == 4)
|
|
continue; /* we can't add this square to j */
|
|
|
|
break; /* got one! */
|
|
}
|
|
if (i < wh) {
|
|
i = order[i];
|
|
|
|
/*
|
|
* Restore the temporarily removed square _before_
|
|
* we start shifting ownerships about.
|
|
*/
|
|
if (tmpsq >= 0)
|
|
own[tmpsq] = j;
|
|
|
|
/*
|
|
* We are done. We can add square i to omino j,
|
|
* and then backtrack along the trail in tmp
|
|
* moving squares between ominoes, ending up
|
|
* expanding our starting omino by one.
|
|
*/
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
printf("(%d,%d)", i%w, i/w);
|
|
#endif
|
|
while (1) {
|
|
own[i] = j;
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
printf(" -> %d", j);
|
|
#endif
|
|
if (tmp[2*j] == -2)
|
|
break;
|
|
i = tmp[2*j+1];
|
|
j = tmp[2*j];
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
printf("; (%d,%d)", i%w, i/w);
|
|
#endif
|
|
}
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
printf("\n");
|
|
#endif
|
|
|
|
/*
|
|
* Increment the size of the starting omino.
|
|
*/
|
|
sizes[j]++;
|
|
|
|
/*
|
|
* Terminate the bfs loop.
|
|
*/
|
|
break;
|
|
}
|
|
|
|
/*
|
|
* If we get here, we haven't been able to expand
|
|
* omino j into an unclaimed square. So now we begin
|
|
* to investigate expanding it into squares which are
|
|
* claimed by ominoes the bfs has not yet visited.
|
|
*/
|
|
for (i = 0; i < wh; i++) {
|
|
int dir, nj;
|
|
|
|
nj = own[order[i]];
|
|
if (nj < 0 || tmp[2*nj] != -1)
|
|
continue; /* unclaimed, or owned by wrong omino */
|
|
if (!removable[order[i]])
|
|
continue; /* its omino won't let it go */
|
|
|
|
for (dir = 0; dir < 4; dir++)
|
|
if (addable[order[i]*4+dir] == j) {
|
|
/*
|
|
* As above, re-check addremcommon.
|
|
*/
|
|
if (!addremcommon(w, h, order[i]%w, order[i]/w,
|
|
own, j))
|
|
continue;
|
|
|
|
/*
|
|
* We have found a square we can use to
|
|
* expand omino j, at the expense of the
|
|
* as-yet unvisited omino nj. So add this
|
|
* to the bfs queue.
|
|
*/
|
|
assert(qtail < n);
|
|
queue[qtail++] = nj;
|
|
tmp[2*nj] = j;
|
|
tmp[2*nj+1] = order[i];
|
|
|
|
/*
|
|
* Now terminate the loop over dir, to
|
|
* ensure we don't accidentally add the
|
|
* same omino twice to the queue.
|
|
*/
|
|
break;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Restore the temporarily removed square.
|
|
*/
|
|
if (tmpsq >= 0)
|
|
own[tmpsq] = j;
|
|
|
|
/*
|
|
* Advance the queue head.
|
|
*/
|
|
qhead++;
|
|
}
|
|
|
|
if (qhead == qtail) {
|
|
/*
|
|
* We have finished the bfs and not found any way to
|
|
* expand omino j. Panic, and return failure.
|
|
*
|
|
* FIXME: or should we loop over all ominoes before we
|
|
* give up?
|
|
*/
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
printf("FAIL!\n");
|
|
#endif
|
|
retdsf = NULL;
|
|
goto cleanup;
|
|
}
|
|
}
|
|
|
|
#ifdef DIVVY_DIAGNOSTICS
|
|
{
|
|
int x, y;
|
|
printf("SUCCESS! Final grid:\n");
|
|
for (y = 0; y < h; y++) {
|
|
for (x = 0; x < w; x++)
|
|
printf("%3d", own[y*w+x]);
|
|
printf("\n");
|
|
}
|
|
}
|
|
#endif
|
|
|
|
/*
|
|
* Construct the output dsf.
|
|
*/
|
|
for (i = 0; i < wh; i++) {
|
|
assert(own[i] >= 0 && own[i] < n);
|
|
tmp[own[i]] = i;
|
|
}
|
|
retdsf = snew_dsf(wh);
|
|
for (i = 0; i < wh; i++) {
|
|
dsf_merge(retdsf, i, tmp[own[i]]);
|
|
}
|
|
|
|
/*
|
|
* Construct the output dsf a different way, to verify that
|
|
* the ominoes really are k-ominoes and we haven't
|
|
* accidentally split one into two disconnected pieces.
|
|
*/
|
|
dsf_init(tmp, wh);
|
|
for (y = 0; y < h; y++)
|
|
for (x = 0; x+1 < w; x++)
|
|
if (own[y*w+x] == own[y*w+(x+1)])
|
|
dsf_merge(tmp, y*w+x, y*w+(x+1));
|
|
for (x = 0; x < w; x++)
|
|
for (y = 0; y+1 < h; y++)
|
|
if (own[y*w+x] == own[(y+1)*w+x])
|
|
dsf_merge(tmp, y*w+x, (y+1)*w+x);
|
|
for (i = 0; i < wh; i++) {
|
|
j = dsf_canonify(retdsf, i);
|
|
assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i));
|
|
}
|
|
|
|
cleanup:
|
|
|
|
/*
|
|
* Free our temporary working space.
|
|
*/
|
|
sfree(order);
|
|
sfree(tmp);
|
|
sfree(own);
|
|
sfree(sizes);
|
|
sfree(queue);
|
|
sfree(addable);
|
|
sfree(removable);
|
|
|
|
/*
|
|
* And we're done.
|
|
*/
|
|
return retdsf;
|
|
}
|
|
|
|
#ifdef TESTMODE
|
|
static int fail_counter = 0;
|
|
#endif
|
|
|
|
int *divvy_rectangle(int w, int h, int k, random_state *rs)
|
|
{
|
|
int *ret;
|
|
|
|
do {
|
|
ret = divvy_internal(w, h, k, rs);
|
|
|
|
#ifdef TESTMODE
|
|
if (!ret)
|
|
fail_counter++;
|
|
#endif
|
|
|
|
} while (!ret);
|
|
|
|
return ret;
|
|
}
|
|
|
|
#ifdef TESTMODE
|
|
|
|
/*
|
|
* gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
|
|
*
|
|
* or to debug
|
|
*
|
|
* gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
|
|
*/
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int *dsf;
|
|
int i;
|
|
int w = 9, h = 4, k = 6, tries = 100;
|
|
random_state *rs;
|
|
|
|
rs = random_new("123456", 6);
|
|
|
|
if (argc > 1)
|
|
w = atoi(argv[1]);
|
|
if (argc > 2)
|
|
h = atoi(argv[2]);
|
|
if (argc > 3)
|
|
k = atoi(argv[3]);
|
|
if (argc > 4)
|
|
tries = atoi(argv[4]);
|
|
|
|
for (i = 0; i < tries; i++) {
|
|
int x, y;
|
|
|
|
dsf = divvy_rectangle(w, h, k, rs);
|
|
assert(dsf);
|
|
|
|
for (y = 0; y <= 2*h; y++) {
|
|
for (x = 0; x <= 2*w; x++) {
|
|
int miny = y/2 - 1, maxy = y/2;
|
|
int minx = x/2 - 1, maxx = x/2;
|
|
int classes[4], tx, ty;
|
|
for (ty = 0; ty < 2; ty++)
|
|
for (tx = 0; tx < 2; tx++) {
|
|
int cx = minx+tx, cy = miny+ty;
|
|
if (cx < 0 || cx >= w || cy < 0 || cy >= h)
|
|
classes[ty*2+tx] = -1;
|
|
else
|
|
classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx);
|
|
}
|
|
switch (y%2 * 2 + x%2) {
|
|
case 0: /* corner */
|
|
/*
|
|
* Cases for the corner:
|
|
*
|
|
* - if all four surrounding squares belong
|
|
* to the same omino, we print a space.
|
|
*
|
|
* - if the top two are the same and the
|
|
* bottom two are the same, we print a
|
|
* horizontal line.
|
|
*
|
|
* - if the left two are the same and the
|
|
* right two are the same, we print a
|
|
* vertical line.
|
|
*
|
|
* - otherwise, we print a cross.
|
|
*/
|
|
if (classes[0] == classes[1] &&
|
|
classes[1] == classes[2] &&
|
|
classes[2] == classes[3])
|
|
printf(" ");
|
|
else if (classes[0] == classes[1] &&
|
|
classes[2] == classes[3])
|
|
printf("-");
|
|
else if (classes[0] == classes[2] &&
|
|
classes[1] == classes[3])
|
|
printf("|");
|
|
else
|
|
printf("+");
|
|
break;
|
|
case 1: /* horiz edge */
|
|
if (classes[1] == classes[3])
|
|
printf(" ");
|
|
else
|
|
printf("--");
|
|
break;
|
|
case 2: /* vert edge */
|
|
if (classes[2] == classes[3])
|
|
printf(" ");
|
|
else
|
|
printf("|");
|
|
break;
|
|
case 3: /* square centre */
|
|
printf(" ");
|
|
break;
|
|
}
|
|
}
|
|
printf("\n");
|
|
}
|
|
printf("\n");
|
|
sfree(dsf);
|
|
}
|
|
|
|
printf("%d retries needed for %d successes\n", fail_counter, tries);
|
|
|
|
return 0;
|
|
}
|
|
|
|
#endif
|