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Dominosa: another forcing-chain based deduction.

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We've already spotted when a domino occurs twice in the _same_ forcing
chain. But now we also spot when it occurs twice in the same _pair_ of
complementary forcing chains, one in each of the two options. Then it
must appear in one of those two places, and hence, can't appear
anywhere else.
This commit is contained in:
Simon Tatham
2019-04-13 11:26:54 +01:00
parent d96298ed01
commit 1114a2af33
1 changed files with 62 additions and 0 deletions
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62
dominosa.c
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@ -1347,6 +1347,10 @@ static bool deduce_forcing_chain(struct solver_scratch *sc)
* Now read out the whole dsf into pc_scratch, flattening its * Now read out the whole dsf into pc_scratch, flattening its
* structured data into a simple integer id per chain of dominoes * structured data into a simple integer id per chain of dominoes
* that must occur together. * that must occur together.
*
* The integer ids have the property that any two that differ only
* in the lowest bit (i.e. of the form {2n,2n+1}) represent
* complementary chains, each of which rules out the other.
*/ */
for (pi = 0; pi < sc->pc; pi++) { for (pi = 0; pi < sc->pc; pi++) {
bool inv; bool inv;
@ -1518,6 +1522,64 @@ static bool deduce_forcing_chain(struct solver_scratch *sc)
} }
} }
/*
* Another thing you can do with forcing chains, besides ruling
* out a whole one at a time, is to look at each pair of chains
* that overlap each other. Each such pair gives you two sets of
* domino placements, such that if either set is not placed, then
* the other one must be.
*
* This means that any domino which has a placement in _both_
* chains of a pair must occupy one of those two placements, i.e.
* we can rule that domino out anywhere else it might appear.
*/
for (di = 0; di < sc->dc; di++) {
struct solver_domino *d = &sc->dominoes[di];
if (d->nplacements <= 2)
continue; /* not enough placements to rule one out */
for (j = 0; j+1 < d->nplacements; j++) {
int ij = d->placements[j]->index;
int cj = sc->pc_scratch[ij];
for (k = j+1; k < d->nplacements; k++) {
int ik = d->placements[k]->index;
int ck = sc->pc_scratch[ik];
if ((cj ^ ck) == 1) {
/*
* Placements j,k of domino d are in complementary
* chains, so we can rule out all the others.
*/
int i;
#ifdef SOLVER_DIAGNOSTICS
if (solver_diagnostics) {
printf("domino %s occurs in both complementary "
"forced chains:", d->name);
for (i = 0; i < sc->pc; i++)
if (sc->pc_scratch[i] == cj)
printf(" %s", sc->placements[i].name);
printf(" and");
for (i = 0; i < sc->pc; i++)
if (sc->pc_scratch[i] == ck)
printf(" %s", sc->placements[i].name);
printf("\n");
}
#endif
for (i = d->nplacements; i-- > 0 ;)
if (i != j && i != k)
rule_out_placement(sc, d->placements[i]);
done_something = true;
goto done_this_domino;
}
}
}
done_this_domino:;
}
return done_something; return done_something;
} }