Obliged to use Constraints in Prolog - prolog

Question

I'm currently trying to solve this puzzle houses using only constraints provided by the clpfd prolog library, which means I cannot use backtracking!
Basically I want to find out which pairs of houses should be made in order to only have 2 distances between all the connections.
My input is a list of coordinates like this [[0,0],[0,3],[2,0],[3,0],[2,1],[3,1],[2,2],[3,3]] And a solution for it would be:
[
[[0,0],[0,3]],
[[2,0],[3,1]],
[[2,1],[3,0]],
[[2,2],[3,3]]
]
My current progress is this one:
connect(Houses):-
%There are only 2 distances and they're different
length(Distances, 2),
all_distinct(Distances),
%One connection per 2 houses (pairs) means half the number of houses as connections
length(Houses, NHouses),
NConnections #= NHouses // 2,
length(Connections, NConnections),
restrictDistances(Connections, Distances), %restrict every connection to have one of the two distances
%All the houses must be connected
append(Connections, ConnectedHouses),
ensureAllConnected(Houses, ConnectedHouses), %table
removeSymmetries(Connections), %avoid symmetries
%flatten list and labeling
append(ConnectedHouses, HousesCoordinates),
labeling([], HousesCoordinates),
write(Connections).
/*
All distances of all connections are one of the two distances
Distance is kept squared to keep it an integer i.e. dist(connection) = dist([[x1, y1], [x2, y2]]) = (x2-x1)^2 + (y2-y1)^2
*/
restrictDistances([], _).
restrictDistances([[[X1, Y1], [X2, Y2]]|Connections], Distances):-
DiffX #= X2 - X1,
DiffY #= Y2 - Y1,
Dis #= DiffX * DiffX + DiffY * DiffY,
% element(Idx, Distances, Dis), %element
member(Dis, Distances), %element
restrictDistances(Connections, Distances).
/*
Ensures all houses are connected
*/
ensureAllConnected([], _).
ensureAllConnected([H|Houses], ConnectedHouses):-
member(H, ConnectedHouses),
% element(_, ConnectedHouses, H),
ensureAllConnected(Houses, ConnectedHouses).
/*
Remove symmetries and connection permutations in final result
*/
removeSymmetries([_]).
removeSymmetries([[[X1, _], [X2, _]], [[X3, Y3], [X4, Y4]]|Connections]):-
X1 #=< X2,
X1 #=< X3,
X3 #=< X4,
removeSymmetries([[[X3, Y3], [X4, Y4]]|Connections]).
The worst part is that this code works, however the predicate member cannot be used because it uses backtracking... And yes, the predicate element exists, but I am unable to replace with it because if I replace first one the output is different, and if I replace the second one I get an instantiation error.

Answer 1

Strictly speaking, the problem is underspecified, because there are more than one kind of distance, e.g. Euclidean distance and Hamiltonian distance. Apparently, Euclidean distances are intended, otherwise you get multiple solutions for this instance.
For this puzzle, it is useful to think about what subtasks might be encoded with global constraints. Here are some hints:
You need to find a matching - that can be encoded with
assignment(Xs,Xs).
You can use table/2 to encode the (house,house,distance) relation.
You can use nvalue/2 to constrain
the number of distinct distances.
These are global constraints in SICStus Prolog.