append/3append(?List1, ?List2, ?List3)
A list consisting of List1 followed by List2.
Appends lists List1 and List2 to form List3:
| ?- append([a,b], [a,d], X).
X = [a,b,a,d]
| ?- append([a], [a], [a]).
no
| ?- append(2, [a], X).
no
Takes List3 apart:
| ?- append(X, [e], [b,e,e]).
X = [b,e]
| ?- append([b|X], [e,r], [b,o,r,e,r]).
X = [o,r]
| ?- append(X, Y, [h,i]).
X = [],
Y = [h,i] ;
X = [h],
Y = [i] ;
X = [h,i],
Y = [] ;
no
Suppose L is bound to a proper list. That is, it has the form [T1,...,Tn] for some n. In that instance, the following things apply:
append(L, X, Y) has at most one solution, whatever X and Y are, and
cannot backtrack at all.
append(X, Y, L) has at most n+1 solutions, whatever X and Y are, and
though it can backtrack over these it cannot run away without finding
a solution.
append(X, L, Y), however, can backtrack indefinitely if X and Y are
variables.
The following examples are
perfectly ordinary uses of append/3:
To enumerate adjacent pairs of elements from a list:
next_to(X, Y, (*in*) List3) :-
append(_, [X,Y|_], List3).
To check whether Word1 and Word2 are the same except for
a single transposition. (append/5 in library(lists) would be
better for this task.)
one_transposition(Word1, Word2) :-
append(Prefix, [X,Y|Suffix], Word1),
append(Prefix, [Y,X|Suffix], Word2).
| ?- one_transposition("fred", X).
X = "rfed" ;
X = "ferd" ;
X = "frde" ;
no
Given a list of words and commas, to backtrack through the phrases delimited by commas:
comma_phrase(List3, Phrase) :-
append(F, [','|Rest], List3),
!,
( Phrase = F
; comma_phrase(Rest, Phrase)
).
comma_phrase(List3, List3).
| ?- comma_phrase([this,is,',',um,',',an, example], X).
X = [this,is] ;
X = [um] ;
X = [an,example] ;
no
ref-lte-acl,
library(lists).