Re: Solving a traditional binary CSP in Eclipse

On Thu, May 01, 2003 at 11:58:37AM +0100, Tom Kelsey wrote:
> Another approach is one constraint for each variable pair:

Tom's idea is fine, but unfortunately his implementation is wrong in a
number of ways.

> 	(
> 	    Y #< 4 -> Z #= 3
> 	;
> 	    true
> 	),

Unless you really know what you're doing, don't put a constraint in the
condition of an if-then-else; it almost certainly does not do what you
want.  Probably what Tom meant was:

	Y #< 4 #=> Z #= 3

The difference is that the latter sets up the implication as a constraint,
whereas the former does not.  The former immediately imposes the constraint
Y #< 4, and then decides whether or not to impose Z #= 3 right away; it also
can never impose Y #>= 4 (if Z #\= 3, for instance).  The latter will only
impose Z #= 3 if at some point Y's domain is sufficiently reduced that Y is
guaranteed to be less than 4; and if Z is assigned a value other than 3 it
will impose the constraint Y #>= 4.  The point is that the former commits to
the choice immediately (which may be premature) while the latter does not.

In any event, what was really wanted was

	Y #=< 3, Z #= 3

since Y cannot be more than 3, regardless of the value of Z, and Z is 3
regardless of the value of Y.

> 	(
> 	    Z ## 3 -> X #= Z
> 	;
> 	    X #= 1
> 	),

This is less convenient to write correctly, and there are a number of ways
of doing it.  One way might be:

	#=(Z, 3, B),	% B = 1 iff Z = 3
	B #=> X #= 1,	% if B = 1 (i.e. Z = 3) then X = 1
	B #\/ X #= Z	% if B = 0 (i.e. Z =\= 3) then X = Z

But since the relation is actually a function from Z to X, an easier (and
more effective in terms of propagation strength) way of implementing it
would be to use the element/3 constraint:

	element(Z, [1, 2, 1, 4, 5], X)

Cheers,
Warwick