An exercise on polyglossy: the same problem solved on multiple languages
Further optimized the Prolog implementation.
Thanks to the awesome help of @flunschlik @wellmeaningtroll on redit the Prolog implementation is now much cleaner, more optimized and ideomatic. Check out the discussion in https://www.reddit.com/r/prolog/comments/ct266d/error_stack_limit_10gb_exceeded_help/.
1 file changed, 15 insertions, 22 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | script.pl | 37 | 15 | 22 |
diff --git a/script.pl b/script.pl @@ -3,34 +3,27 @@ % Let S: N -> N be the sum of the digits of a positive integer. % For all A and B in N, S(A + B) = S(A) + S(B) - 9k, where k is an integer. -sum_digits_acc(X, X, A) :- X < 10, A is 0, !. -sum_digits_acc(X, Y, A) :- - X >= 10, - X1 is div(X, 10), - X2 is mod(X, 10), - sum_digits_acc(X1, A, _), - Y is A + X2. +sum_digits(X, Y) :- sum_digits_acc(X, 0, Y). -sum_digits(X, Y) :- sum_digits_acc(X, Y, _). +sum_digits_acc(X, A, Y) :- X < 10, !, Y is A + X. +sum_digits_acc(X, A, Y) :- + Q = X div 10, + R = X mod 10, + A2 = A + R, + sum_digits_acc(Q, A2, Y). test_pair(A, B) :- - R is 0, - AB is A + B, + R = 0, + AB = A + B, sum_digits(A, SA), sum_digits(B, SB), sum_digits(AB, SC), - SAB is SA + SB, - D is SC - SAB, - R is mod(D, 9). + SAB = SA + SB, + D = SC - SAB, + R =:= D mod 9. -iter(A, 0) :- test_pair(A, 0), !. iter(A, B) :- - test_pair(A, B), - C is B - 1, - iter(A, C). + forall(between(0, B, X), test_pair(A, X)). -conjecture(0) :- test_pair(0, 0), !. conjecture(M) :- - iter(M, M), - N is M - 1, - conjecture(N).- \ No newline at end of file + forall(between(0, M, X), iter(X, X)).+ \ No newline at end of file