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Dominosa: allow set analysis even with adjacency.

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I've always had the vague idea that the usual set analysis deduction
goes wrong when there are two adjacent squares, because they might be
opposite ends of the same domino and mess up the count. But I just
realised that actually you can correct for that by adjusting the
required count by one: if you have four 0 squares which between them
can only be parts of 0-0, 0-1 and 0-2, then the only way this can work
is if two of them are able to be the 0-0 - but in that case, you can
still eliminate those dominoes from all placements elsewhere. So set
analysis _can_ work in that situation; you just have to compensate for
the possible double.

(This enhanced form _might_ turn out to be something that needs
promoting into a separate difficulty level, but for the moment, I'll
try leaving it in Hard and seeing if that's OK.)
This commit is contained in:
Simon Tatham
2019-04-03 18:26:42 +01:00
parent e2f52df5ec
commit 453a2c1ca8
1 changed files with 35 additions and 6 deletions
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41
dominosa.c
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@ -909,6 +909,7 @@ static bool deduce_set_simple(struct solver_scratch *sc)
for (squares = 0; squares < (1UL << nsq); squares++) {
unsigned long dominoes = 0;
int bitpos, nsquares, ndominoes;
bool got_adj_squares = false;
bool reported = false;
/*
@ -936,7 +937,7 @@ static bool deduce_set_simple(struct solver_scratch *sc)
continue; /* this bit isn't set in the mask */
if (adjacent[bitpos] & squares)
goto skip_subset; /* contains two adjacent squares */
got_adj_squares = true;
dominoes |= domino_sets[bitpos];
nsquares++;
@ -947,8 +948,33 @@ static bool deduce_set_simple(struct solver_scratch *sc)
for (bitpos = 0; bitpos < nsq; bitpos++)
ndominoes += 1 & (dominoes >> bitpos);
/* Are the two sets equal in size? */
if (nsquares != ndominoes)
/*
* Do the two sets have the right relative size?
*
* In the normal case, analogous to set analysis in many
* other puzzles, you want N squares which between them
* have to account for N distinct dominoes, with a 1-1
* correspondence between them.
*
* But if any two squares in this set can possibly both be
* covered by the same double domino (i.e. if they are
* adjacent, and moreover, the placement containing both
* is not yet ruled out), then that argument doesn't hold
* up, because the N squares might be covered by N-1
* dominoes - or, put another way, if you list the
* containing domino for each of the squares they aren't
* all distinct.
*
* In that situation, we can only do the set analysis if
* there is one _more_ square than there are dominoes. For
* example, suppose we had four squares which between them
* could contain only the 0-0, 0-1 and 0-2 dominoes. Then
* that can only work at all if the 0-0 covers two of
* those squares - and in that situation that _must_ be
* what's happened, so we can rule out those three
* dominoes anywhere else they might look possible.
*/
if (ndominoes != (got_adj_squares ? nsquares - 1 : nsquares))
continue;
/* Skip sets of size 1, or whose complement has size 1.
@ -994,6 +1020,7 @@ static bool deduce_set_simple(struct solver_scratch *sc)
if (!reported) {
reported = true;
done_something = true;
#ifdef SOLVER_DIAGNOSTICS
if (solver_diagnostics) {
int b;
@ -1010,7 +1037,11 @@ static bool deduce_set_simple(struct solver_scratch *sc)
printf("%s%s", sep, ds[b]->name);
sep = ",";
}
printf("}\n");
printf("}");
if (got_adj_squares)
printf(" (including both ends of the "
"double)");
printf("\n");
}
#endif
}
@ -1020,8 +1051,6 @@ static bool deduce_set_simple(struct solver_scratch *sc)
skip_placement:;
}
}
skip_subset:;
}
}