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Solution to the problem WWWDOT - GOOGLE = DOTCOM (GLAT question 1) in prolog
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% solution to the problem WWWDOT - GOOGLE = DOTCOM in prolog | |
carry(0). | |
carry(1). | |
digit(A) :- carry(A). | |
digit(2). | |
digit(3). | |
digit(4). | |
digit(5). | |
digit(6). | |
digit(7). | |
digit(8). | |
digit(9). | |
solve(W,D,O,T,G,L,E,C,M):- | |
carry(ALPHA), | |
carry(BETA), | |
carry(GAMMA), | |
carry(DELTA), | |
carry(EPSILON), | |
digit(T), | |
digit(E), | |
T \== E, | |
digit(M), | |
T \== M, | |
E \== M, | |
(T+10*ALPHA) - E =:= M, | |
digit(O), | |
T \== O, | |
E \== O, | |
M \== O, | |
digit(L), | |
T \== L, | |
E \== L, | |
M \== L, | |
O \== L, | |
(O-ALPHA+10*BETA) - L =:= O, | |
digit(W), | |
T \== W, | |
E \== W, | |
M \== W, | |
O \== W, | |
L \== W, | |
(W-GAMMA+10*DELTA) - O =:= T, | |
(W-DELTA+10*EPSILON) - O =:= O, | |
digit(D), | |
T \== D, | |
E \== D, | |
M \== D, | |
O \== D, | |
L \== D, | |
W \== D, | |
digit(G), | |
T \== G, | |
E \== G, | |
M \== G, | |
O \== G, | |
L \== G, | |
W \== G, | |
D \== G, | |
digit(C), | |
T \== C, | |
E \== C, | |
M \== C, | |
O \== C, | |
L \== C, | |
W \== C, | |
D \== C, | |
G \== C, | |
(D-BETA+10*GAMMA) - G =:= C, | |
(W-EPSILON) - G =:= D. |
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