Russ Abbott wrote:
In answer to the question in the previous message, it appears that
FD.distinct doesn't figure out that the cell between 7 and 6 must be 2.
declare A B C V = [A B C]
V ::: 1#3
{FD.distinct V}
A \=: 2
B \=: 2
{Show V}
FD.distinct is well known for not being clever in that case. Use
FD.distinctD instead. This one will detect that the set {A,B}={1,3},
and propagate that C does not belong to that set, giving C=2.
{FD.distinctD V}
Cheers,
raph
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