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Two-issue prologue

I have a prologue.

I need to look at the first item in the list, see if its next items match until they are gone, and separate the lists by the first item and its duplicates. for example, if my list was a, a, a, b, c, he would first split it: a, a, a. second: b, c.

My current solution works, except that the last matching element is on the second list, not the first. I don't seem to think about how to make him appear on the first list.

grab([],[],[]).
grab([A,A|L],[A|L2],Rest) :- grab([A|L],L2,Rest).
grab([A|L],Init,[A|L2]) :- grab(L,Init,L2).
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4 answers

When the first two elements are different from each other, you do not need a recursive target.

grab([], [], []).
grab([A,A|Rest], [A|As], L2):- !, grab([A|Rest], As, L2).
grab([A|Tail], [A], Tail).
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grab(Xs, Ys, Zs) :-
   eqprefix(Xs, Ys, Zs).

eqprefix([],[],[]).
eqprefix([X],[],[X]).
eqprefix([X,Y|Xs], [], [X,Y|Xs]) :-
    dif(X,Y).
eqprefix([X,X|Xs], [X|Ys], Zs) :-
   eqprefix2([X|Xs], Ys, Zs).

eqprefix2([X], [X], []).
eqprefix2([X,Y|Xs], [X], [Y|Xs]) :-
    dif(X,Y).
eqprefix2([X,X|Xs], [X|Ys], Zs) :-
    eqprefix2([X|Xs], Ys, Zs).

?- eqprefix([a,a,a,b,c],Ys,Zs).
Ys = [a, a, a],
Zs = [b, c] ;
false.

?- eqprefix(Xs,[a,a,a],[b,c]).
Xs = [a, a, a, b, c] ;
false.

?- eqprefix([A,B,C,D,E],[a|Ys],[b,c]).
A = B, B = C, C = a,
D = b,
E = c,
Ys = [a, a] ;
false.

In the definition you gave, you get many different answers:

?- grab([a,a,a,b,c],Ys,Zs).
  Ys = [a, a], Zs = [a, b, c]
; Ys = [a],    Zs = [a, a, b, c]
; Ys = [a],    Zs = [a, a, b, c]
; Ys = [],     Zs = [a, a, a, b, c].
+3

, , @Sarah . , grab/3 .

grab([A], [A], []).
grab([A,B|Bs], [A], [B|Bs]) :-
   dif(A,B).
grab([A,A|Xs], [A|As], Bs) :-
   grab([A|Xs], As, Bs).

?- Xs = [A,B], grab(Xs,Ys,Zs).
  Xs = [A, B], Ys = [A],    Zs = [B], dif(A, B)
; Xs = Ys,     Ys = [B, B], Zs = [],  A = B
; false.
+2

, dcg s. - , . , :

grab_tentative(Xs, Ys, Zs) :-
   phrase((seq(Ys),seq(Zs)), Xs).

seq([]) --> [].
seq([E|Es]) --> [E], seq(Es).

grab_tentative/3 , Xs Ys, Zs. , .

?- Xs = [A,B,C], grab_tentative(Xs,Ys,Zs).
   Xs = Zs, Zs = [A, B, C], Ys = []
;  Xs = [A, B, C], Ys = [A], Zs = [B, C]
;  Xs = [A, B, C], Ys = [A, B], Zs = [C]
;  Xs = Ys, Ys = [A, B, C], Zs = []
;  false.

, Ys = [], ( @Sarah), Ys , :

grab_tentative(Xs, Ys, Zs) :-
   Ys = [_|_],
   phrase((seq(Ys),seq(Zs)), Xs).

Xs = [A, B, C], Ys = [A, B], Zs = [C] Xs = Ys, Ys = [A, B, C], Zs = [] , A B . , :

grab_tentative(Xs, Ys, Zs) :-
   Ys = [A|_],
   phrase((all_seq(=(A),Ys),seq(Zs)), Xs).

all_seq(_, []) --> [].
all_seq(C_1, [C|Cs]) -->
   [C],
   {call(C_1,C)},
   all_seq(C_1, Cs).

:

?- Xs = [A,B,C], grab_tentative(Xs,Ys,Zs).
   Xs = [A, B, C], Ys = [A], Zs = [B, C]
;  Xs = [B, B, C], A = B, Ys = [B, B], Zs = [C]
;  Xs = Ys, Ys = [C, C, C], A = B, B = C, Zs = []
;  false.

, A = B. dif(A,B). . . , or_end//1 []//0, , .

grab_final(Xs, Ys, Zs) :-
   Ys = [A|_],
   phrase ((all_seq (= (A), Ys), or_end (dif (A)) , seq (Zs)), Xs).

or_end (C_1) ->
  call (cond_or_end (C_1)). % interface to predicates

cond_or_end (_C_1, [], []).
cond_or_end (C_1, [E | Es], [E | Es]): -

Now the answers respond as follows:

?- Xs = [A,B,C], grab_final(Xs, Ys, Zs).
   Xs = [A, B, C], Ys = [A], Zs = [B, C], dif(A, B)
;  Xs = [B, B, C], A = B, Ys = [B, B], Zs = [C], dif(B, C)
;  Xs = Ys, Ys = [C, C, C], A = B, B = C, Zs = []
;  false.
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