intended behaviour of classifying multiplication clues as low-quality
if they only left one possible pair of multiplicands has never
actually worked, because I should have compared the possible clue
count against 2 rather than 1 since the multiplicands can occur either
way round.
[originally from svn r9165]
for (k = 1; k <= w; k++)
if (v % k == 0 && v / k <= w && v / k != k)
n++;
- if (n > 1)
+ if (n > 2)
singletons[j] |= F_MUL;
else
singletons[j] |= F_MUL_BAD;