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Editorial Team
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Asked: 2026-06-14T07:33:12+00:00

Let’s say a variable number of lists has been given, 5 for example, but

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Let’s say a variable number of lists has been given, 5 for example, but one can enter any number of lists: [1,2] + [3,4,5] + [6,7,8,9] + [10,11] + [12,13,14] equals L. should give the following list inside the variable L: [1,2,3,4,5,6,7,8,9,10,11,12,13,14].

Here’s my code for the concatenation of two lists:

joinLists([FLH|FLT], SL, [FLH|RLT]):-
  joinLists(FLT, SL, RLT).
  joinLists([], H, H).

:-op(500, xfx, +).
:-op(600, yfx, equals).

X + Y equals Z:-
  joinLists(X, Y, Z).

[1,2,3] + [4,5,6,7] equals L. gives L = [1,2,3,4,5,6,7], but how to solve the problem for an indefinite number of lists?

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1 Answer

  1. Editorial Team

    I think your operators declaration is incorrect, the sum should be associative (i also change the symbols, just to fit my aesthetic preference, and to avoid changing the behaviour of standard +/2):

    :-op(500, xfy, ++).
:-op(600, xfx, cat).

    then you can write 

    L1 ++ R cat C :- !, R cat R1, append(L1, R1, C).
L cat L .

    test

    ?- [1,2,3]++[4,5,6]++[7,8,9] cat X.
X = [1, 2, 3, 4, 5, 6, 7, 8, 9].

    I don’t see any convenience in such definition, consider that SWI-Prolog append/2 source is available…

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