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Accepted
Ruslan López
Asked: 2022-07-18 12:57:43 +0800 CST 2022-07-18 12:57:43 +0800 CST

detect consecutive duplicates in a list

  • 772

I want to detect all consecutive duplicate numbers in a list. I am trying this with:

sublist( [], _ ):-!.
sublist( [X|XS], [X|XSS] ) :- sublist( XS, XSS ),!.
sublist( [X|XS], [_|XSS] ) :- sublist( [X,XS], XSS ).

consecutive_duplicates(Y,LS):-
    member(Y,LS)
    , sublist([Y,Y],LS)
    .

Departure:

consecutive_duplicates(X,[1,1,4,3,2,2,4]).

X = 1 ;

X = 1 ;

X = 2 ;

X = 2 ;

false.

However I expected the following output:

X = 1;

X = 2;

What do I need to change?

prolog

1 Answers

  1. Best Answer

    You can use append/3:

    consecutive_duplicates(X,L):-
 append(_, [X, X|_], L).

    However this solution can still include duplicates if the same duplicate number appears more than once:

    ?- consecutive_duplicates(X,[1,1,4,3,2,2,4]).
X = 1 ;
X = 2 ;
false.

?- consecutive_duplicates(X,[1,1,1,4,4,1,1]).
X = 1 ;
X = 1 ;
X = 4 ;
X = 1 ;
false.

    To avoid it you can use setof/3:

    consecutive_duplicates(X,L):-
 setof(X, H^T^append(H, [X, X|T], L), L1),
 member(X, L1).

    And now the tests get the following output:

    ?- consecutive_duplicates(X,[1,1,4,3,2,2,4]).
X = 1 ;
X = 2.

?- consecutive_duplicates(X,[1,1,1,4,4,1,1]).
X = 1 ;
X = 4.
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