prolog improvement of an algorithm

Question

% SEND+MORE=MONEY
solve(VarList):-
    VarList=[D,E,M,N,O,R,S,Y],     % Οι μεταβλητές του προβλήματος
    Digits=[0,1,2,3,4,5,6,7,8,9],  % Οι τιμές των μεταβλητών (τα ψηφία)
    member(D,Digits),
    member(E,Digits),
    member(M,Digits),
    member(N,Digits),     % Ανάθεση τιμών στις μεταβλητές
    member(O,Digits),
    member(R,Digits),
    member(S,Digits),
    member(Y,Digits),
    M=0, S=0,           % Περιορισμοί
    E=D,
    M=D, M=E,
    N=D, N=E, N=M,
    O=D, O=E, O=M, O=N,
    R=D, R=E, R=M, R=N, R=O,
    S=D, S=E, S=M, S=N, S=O, S=R,
    Y=D, Y=E, Y=M, Y=N, Y=O, Y=R, Y=S,
    S*1000+E*100+N*10+D + M*1000+O*100+R*10+E =:= M*10000+O*1000+N*100+E*10+Y.

if i decrease the number of varriables VarList. does it improves its speed?

if i S*1000+E*100+N*10+D + M*1000+O*100+R*10+E =:= M*10000+O*1000+N*100+E*10+Y

before the checks does it improve its speed?

Answer 1

A clpfd approach, I am putting my solution in case someone is looking into this problem.

:- use_module( library( clpfd)).

puzzle(X):-
    X=([S,E,N,D]+[M,O,R,E]=[M,O,N,E,Y]), 
    
    Vars=[S,E,N,D,M,O,R,Y],Vars ins 0..9,
    
    S*1000 + E*100 + N*10 + D +
    M*1000 + O*100 + R*10 + E #=
    M*1000 + O*1000 + N*100 + E*10 + Y,
    
    S#\=0, M#\=0,
    all_different(Vars),
    labeling([],Vars).

?- puzzle(X).
X = ([1, 8, 0, 5]+[4, 2, 7, 8]=[4, 2, 0, 8, 3])
X = ([1, 8, 0, 5]+[6, 2, 7, 8]=[6, 2, 0, 8, 3])
X = ([1, 8, 0, 5]+[9, 2, 7, 8]=[9, 2, 0, 8, 3])
X = ([1, 8, 0, 6]+[3, 2, 7, 8]=[3, 2, 0, 8, 4])
X = ([1, 8, 0, 6]+[5, 2, 7, 8]=[5, 2, 0, 8, 4])
X = ([1, 8, 0, 6]+[9, 2, 7, 8]=[9, 2, 0, 8, 4])
X = ([2, 7, 0, 4]+[5, 3, 6, 7]=[5, 3, 0, 7, 1])....

Answer 2

No, if you move the line

S*1000+E*100+N*10+D + M*1000+O*100+R*10+E =:= M*10000+O*1000+N*100+E*10+Y

above what you call "Περιορισμοί" ("restrictions", according to Google Translate), it will only become slower because it will needlessly perform the arithmetic calculations which would have been avoided with the restrictions weeding out the illegal digits assignments first.

You also have erroneous equations S = 0, M = 0, E = D, ... when it should have been S =\= 0, M =\= 0, E =\= D, ..., since all the digits in these numbers are required to be unique and the first digits in the numbers can't be zeroes.

Overall your code's speed can be improved, by reducing the domain of available values with each choice of a digit value, using select/3, instead of making all the choices from the same unaffected domain Digits with member/2. This will much reduce the combinatorial choices space, and all the digits picked will be different by construction obviating the inequality checks. The tag cryptarithmetic-puzzle's info page and Q&A entries should have more discussion and / or examples of this technique (also, the tag zebra-puzzle).