CLPZ having one variable's domain be the union of two others' domains

[Garklein]

If I have variables X, Y, Z, is there a nice way to make Z's domain be the union of X and Y's? Right now, I'm just doing

fd_dom(X, Xd),
fd_dom(Y, Yd),
Z in Xd \/ Yd.

but it feels clunky.

[UWN]

What is your intended meaning of this?

[Garklein]

Solving Advent of Code day 5 part 2 😁
I'm trying to map some ranges to other ranges, which I'm using constraints to represent.
I can take them away from the original range by constraining it, but I can't find a nice way to add the mapped range to a constrained variable.

[jasagredo]

I solved it using something similar to what you are trying to do I think, but I am not very sure it is the most idiomatic way. I create variables in domains, then extract domains from them and combine them again. What I meant is that I do exactly as you, manually create and carry D1\/D2 domains.

FTR, it is quite fast:

Part 1:
   % CPU time: 0.046s, 582_419 inferences
Part 2:
   % CPU time: 0.156s, 1_648_840 inferences

[Mousaka]

Please share code! I struggled on day 5 :p

[triska]

An interesting example, thank you for sharing this! Personally, I find the solution quite acceptable. One additional observation:

Reified disjunction (#\/)/2 could be made to do this, though currently it does not. For example:

?- A in 1..3, B in 2..5, X #= A #\/ X #= B.
   clpz:(A in 1..3),
   clpz:(X#=A#<==>_A),
   clpz:(X#=B#<==>_B),
   clpz:(B in 2..5),
   clpz:(_B in 0..1),
   clpz:(_A#\/_B#<==>1),
   clpz:(_A in 0..1).

Note that the domain of X is not constrained in this example, even though it is clear that it could be constrained to the union of that of A and B.

This is an interesting trade-off: Reified disjunction could be extended with reasoning that takes such situations into account, at the cost of slowing down cases that do not need this. In addition, it is not clear what guarantees then still hold. For example, what if I post, instead of X #= A #\/ X #= B, a variant such as A #= X #\/ X #= B? Should this still be detected then? At what cost?

Note that it already works in very specific situations:

?- A #= 3 #\/ 5 #= A.
   clpz:(A in 3\/5).