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Write a comment outlining a design for a rewritten faster solver.
[originally from svn r9544]
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63
bridges.c
63
bridges.c
@ -3,7 +3,68 @@
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*
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* Things still to do:
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*
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* * write a recursive solver?
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* - The solver's algorithmic design is not really ideal. It makes
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* use of the same data representation as gameplay uses, which
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* often looks like a tempting reuse of code but isn't always a
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* good idea. In this case, it's unpleasant that each edge of the
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* graph ends up represented as multiple squares on a grid, with
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* flags indicating when edges and non-edges cross; that's useful
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* when the result can be directly translated into positions of
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* graphics on the display, but in purely internal work it makes
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* even simple manipulations during solving more painful than they
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* should be, and complex ones have no choice but to modify the
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* data structures temporarily, test things, and put them back. I
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* envisage a complete solver rewrite along the following lines:
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* + We have a collection of vertices (islands) and edges
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* (potential bridge locations, i.e. pairs of horizontal or
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* vertical islands with no other island in between).
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* + Each edge has an associated list of edges that cross it, and
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* hence with which it is mutually exclusive.
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* + For each edge, we track the min and max number of bridges we
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* currently think possible.
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* + For each vertex, we track the number of _liberties_ it has,
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* i.e. its clue number minus the min bridge count for each edge
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* out of it.
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* + We also maintain a dsf that identifies sets of vertices which
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* are connected components of the puzzle so far, and for each
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* equivalence class we track the total number of liberties for
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* that component. (The dsf mechanism will also already track
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* the size of each component, i.e. number of islands.)
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* + So incrementing the min for an edge requires processing along
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* the lines of:
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* - set the max for all edges crossing that one to zero
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* - decrement the liberty count for the vertex at each end,
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* and also for each vertex's equivalence class (NB they may
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* be the same class)
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* - unify the two equivalence classes if they're not already,
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* and if so, set the liberty count for the new class to be
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* the sum of the previous two.
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* + Decrementing the max is much easier, however.
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* + With this data structure the really fiddly stuff in stage3()
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* becomes more or less trivial, because it's now a quick job to
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* find out whether an island would form an isolated subgraph if
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* connected to a given subset of its neighbours:
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* - identify the connected components containing the test
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* vertex and its putative new neighbours (but be careful not
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* to count a component more than once if two or more of the
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* vertices involved are already in the same one)
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* - find the sum of those components' liberty counts, and also
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* the total number of islands involved
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* - if the total liberty count of the connected components is
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* exactly equal to twice the number of edges we'd be adding
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* (of course each edge destroys two liberties, one at each
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* end) then these components would become a subgraph with
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* zero liberties if connected together.
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* - therefore, if that subgraph also contains fewer than the
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* total number of islands, it's disallowed.
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* - As mentioned in stage3(), once we've identified such a
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* disallowed pattern, we have two choices for what to do
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* with it: if the candidate set of neighbours has size 1 we
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* can reduce the max for the edge to that one neighbour,
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* whereas if its complement has size 1 we can increase the
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* min for the edge to the _omitted_ neighbour.
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*
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* - write a recursive solver?
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*/
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#include <stdio.h>
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