prologfailure-slice

Even and odd cases for the pivot/2 predicate aren't running properly when placed together


My goal is to implement the pivot(Before,After) predicate that returns true when Before's first half is equal to After's second half and Before's second half is equal to After's first half.

  1. With an even number of elements in input list, the lengths of the two halves are same.
  2. With an odd number of elements, the lengths of the two halves are same with the center element doesn't change.

I tried to implement the predicate and it is working but not perfectly. Below is my predicate.

pivot(Before,After) :-
    append(A,B,Before),
    length(A,N),
    length(B,N),
    append(B,A,After).
pivot(Before,After) :-
    append(A,B,Before),
    length(A,N),
    N1 is N + 1,
    length(B,N1),
    append(C,Tail,B),
    length(C,1),
    append(Tail,C,F),
    append(F,A,After).

The pivot(A,[1,2,3,4,5]) is running, but I do not get output.
I am expecting A = [4,5,3,1,2].


Solution

  • A possible solution is:

    pivot(In, Out) :-
        pivot(In, In, Left, Middle, Right),
        prepend(Middle, Left, New),
        append(Right, New, Out).
    
    pivot([], Ys, [], [], Ys).
    pivot([_], [Y|Ys], [], [Y], Ys).
    pivot([_,_|Xs], [Y|Ys], [Y|Left], Middle, Right) :-
        pivot(Xs, Ys, Left, Middle, Right).
    
    prepend([], Xs, Xs).
    prepend([X], Xs, [X|Xs]).
    

    How does the predicate work?

    ?- trace(pivot,+all), pivot([1,2,3,4], L), trace(pivot,-all).
    %         pivot/2: [all]
    %         pivot/5: [all]
     T [11] Call: pivot([1, 2, 3, 4], _15324)
     T [20] Call: pivot([1, 2, 3, 4], [1, 2, 3, 4], _19198, _19200, _19202)
     T [29] Call: pivot([3, 4], [2, 3, 4], _20104, _19200, _19202)
     T [38] Call: pivot([], [3, 4], _21006, _19200, _19202)
     T [38] Exit: pivot([], [3, 4], [], [], [3, 4])
     T [29] Exit: pivot([3, 4], [2, 3, 4], [2], [], [3, 4])
     T [20] Exit: pivot([1, 2, 3, 4], [1, 2, 3, 4], [1, 2], [], [3, 4])
     T [11] Exit: pivot([1, 2, 3, 4], [3, 4, 1, 2])
    %         pivot/2: Not tracing
    %         pivot/5: Not tracing
    L = [3, 4, 1, 2].
    
    ?- trace(pivot,+all), pivot([1,2,3,4,5], L), trace(pivot,-all).
    %         pivot/2: [all]
    %         pivot/5: [all]
     T [11] Call: pivot([1, 2, 3, 4, 5], _27180)
     T [20] Call: pivot([1, 2, 3, 4, 5], [1, 2, 3, 4, 5], _222, _224, _226)
     T [29] Call: pivot([3, 4, 5], [2, 3, 4, 5], _778, _224, _226)
     T [38] Call: pivot([5], [3, 4, 5], _1680, _224, _226)
     T [38] Exit: pivot([5], [3, 4, 5], [], [3], [4, 5])
     T [29] Exit: pivot([3, 4, 5], [2, 3, 4, 5], [2], [3], [4, 5])
     T [20] Exit: pivot([1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2], [3], [4, 5])
     T [11] Exit: pivot([1, 2, 3, 4, 5], [4, 5, 3, 1, 2])
    %         pivot/2: Not tracing
    %         pivot/5: Not tracing
    L = [4, 5, 3, 1, 2].
    

    Another examples:

    ?- length(In, _), pivot(In, Out).
    In = Out, Out = [] ;
    In = Out, Out = [_] ;
    In = [_A, _B],
    Out = [_B, _A] ;
    In = [_A, _B, _C],
    Out = [_C, _B, _A] ;
    In = [_A, _B, _C, _D],
    Out = [_C, _D, _A, _B] ;
    In = [_A, _B, _C, _D, _E],
    Out = [_D, _E, _C, _A, _B] ;
    In = [_A, _B, _C, _D, _E, _F],
    Out = [_D, _E, _F, _A, _B, _C] ;
    In = [_A, _B, _C, _D, _E, _F, _G],
    Out = [_E, _F, _G, _D, _A, _B, _C] 
    

    A comparison of the various suggested solutions for this problem, using SWI-Prolog 8.4.3.

    slago(In, Out) :-
        pivot(In, In, Left, Middle, Right),
        prepend(Middle, Left, New),
        append(Right, New, Out).
    
    pivot([], Ys, [], [], Ys).
    pivot([_], [Y|Ys], [], [Y], Ys).
    pivot([_,_|Xs], [Y|Ys], [Y|Left], Middle, Right) :-
        pivot(Xs, Ys, Left, Middle, Right).
    
    prepend([], Xs, Xs).
    prepend([X], Xs, [X|Xs]).
    
    
    brebs([A,B|T], LstPivot) :-
        same_length([A,B|T], LstPivot),
        pivot_list_half_([A,B|T], [A,B|T], Half1, Half2),
        append(Half2, Half1, LstPivot).
    
    pivot_list_half_([], H2, [], H2).
    pivot_list_half_([_|T], [H|Sgl], H1, H2) :-
        pivot_list_half_dbl_(T, H, Sgl, H1, H2).
    
    pivot_list_half_dbl_([], H, H2, [], H2Full) :-
        append(H2, [H], H2Full).
    pivot_list_half_dbl_([_|T], H, Sgl, [H|H1], H2) :-
        pivot_list_half_(T, Sgl, H1, H2).
    
    
    gusbro(Before, After):-
     pivot_df(Before, Before, Before, LAcc-LAcc, LAcc2-LAcc2, After, After, After, RAcc-RAcc, RAcc2-RAcc2).
    
    pivot_df([], _, Before, Before-RightFirstHalf, LeftFirstHalf-[], [], _, After, After-LeftFirstHalf, RightFirstHalf-[]).
    pivot_df([_], _, Before, Before-[Middle|RightFirstHalf], LeftFirstHalf-[], [_], _, After, After-[Middle|LeftFirstHalf], RightFirstHalf-[]).
    pivot_df([_,_|L1], [A|L2], Before, LAcc-LTail, LAcc2-LTail2, [_,_|L3], [B|L4], After, RAcc-RTail, RAcc2-RTail2):-
      LTail=[A|LTail1], RTail=[B|RTail1],
      LTail2=[A|LTail21], RTail2=[B|RTail21],
      pivot_df(L1, L2, Before, LAcc-LTail1, LAcc2-LTail21, L3, L4, After, RAcc-RTail1, RAcc2-RTail21).
    
    
    damiano(L,LP):-
        length(L,N),
        N2 is div(N,2),
        findall(E, (nth1(I, L, E), I >= 1, I =< N2), S1),
        (   0 is N mod 2 ->
                findall(E, (nth1(I, L, E), I > N2, I =< N), S2),
                append(S2,S1,LP) ;
                N21 is N2 + 2,
                findall(E, (nth1(I, L, E), I >= N21, I =< N), S2),
                N11 is N2 + 1,
                nth1(N11,L,El),
                append([El],S1,S11),
                append(S2,S11,LP)
        ).
    
    
    comparison :-
        format('  Length  Slago    Brebs    Gusbro   Damiano\n', []),
        forall(between(1, 8, I),
               (   N is 10^6*I,
                   length(L, N),
                   findall(T,
                           ( member(P, [slago, brebs, gusbro, damiano]),
                             runtime(P, L, T)),
                           Ts),
                   format('~|~t~w~8+ ~|~t~5f~8+ ~|~t~5f~8+ ~|~t~5f~8+ ~|~t~5f~8+\n', [N|Ts]) )).
    
    
    runtime(P, L, T) :-
        garbage_collect,
        T0 is cputime,
        call(P, L, _),
        T is cputime - T0.
    

    Results:

    ?- comparison.
      Length  Slago    Brebs    Gusbro   Damiano
     1000000  0.04688  0.07812  0.68750  0.60938
     2000000  0.09375  0.17188  0.92188  1.23438
     3000000  0.15625  0.26562  0.67188  1.92188
     4000000  0.21875  0.35938  1.84375  2.60938
     5000000  0.28125  0.46875  1.10938  3.35938
     6000000  0.32812  0.56250  1.35938  3.87500
     7000000  0.39062  0.64062  3.21875  4.67188
     8000000  0.45312  0.70312  3.51562  5.20312
    true.
    

    Gusbro's solution caused stack overflow for list with length 9000000.