Related
I am a new learner for prolog. Here is the question of our workshop, I have no idea where start it.
Would really appreciate any help with this.
sublist(Xs, Ys)
This holds when Xs is a list containing some of the elements of Ys, in the same order they appear in the list Ys. This should work whenever Ys is a proper list. For example:
sublist([a,c,e],[a,b,c,d,e]) should succeed.
sublist([a,e,c],[a,b,c,d,e]) should fail.
sublist([a,X,d],[a,b,c,d,e]) should have the two solutions X=b and X=c.
sublist(X,[a,b,c]) should have the eight solutions X=[]; X=[c]; X=[b]; X=[b,c]; X=[a]; X=[a,c]; X=[a,b]; and X=[a,b,c].
My implementation:
sublist([], []).
sublist([H| Rest1], [H| Rest2]) :-sublist(Rest1, Rest2).
sublist(H, [_ | Rest2]) :-sublist(H, Rest2).
Examples:
?- sublist(X,[a,b,c]).
X = [a, b, c] ;
X = [a, b] ;
X = [a, c] ;
X = [a] ;
X = [b, c] ;
X = [b] ;
X = [c] ;
X = [].
?- sublist([a,c,e],[a,b,c,d,e]) .
true ;
false.
?- sublist([a,e,c],[a,b,c,d,e]) .
false.
?- sublist([a,X,d],[a,b,c,d,e]).
X = b ;
X = c ;
false.
Please note that a sublist needs its elements to be consecutive in the original list.
With that definition it would be easier to define a sublist through defining auxiliary predicates prefix and suffix, examples are taken from Shapiro's book:
prefix([], _).
prefix([X|Xs], [X,Ys]) :-
prefix(Xs, Ys).
suffix(Xs, Xs).
suffix(Xs, [_|Ys]) :-
suffix(Xs, Ys).
– and then it is just a matter of defining a sublist as either a suffix of a prefix:
sublist(Xs, Ys) :-
prefix(Ps, Ys),
suffix(Xs, Ps).
Result:
?- sublist(X, [1,2,3]).
X = [] ;
X = [1] ;
X = [] ;
X = [1, 2] ;
X = [2] ;
X = [] ;
X = [1, 2, 3] ;
X = [2, 3] ;
X = [3] ;
X = [] ;
false.
– or as a prefix of a suffix:
sublist(Xs, Ys) :-
prefix(Xs, Ss),
suffix(Ss, Ys).
Result:
?- sublist(X, [1,2,3]).
X = [] ;
X = [] ;
X = [] ;
X = [] ;
X = [1] ;
X = [2] ;
X = [3] ;
X = [1, 2] ;
X = [2, 3] ;
X = [1, 2, 3] ;
But one could also do a recursive definition:
sublist(Xs, Ys) :-
prefix(Xs, Ys).
sublist(Xs, [_|Ys]) :-
sublist(Xs, Ys).
Result:
?- sublist(X, [1,2,3]).
X = [] ;
X = [1] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [] ;
X = [2] ;
X = [2, 3] ;
X = [] ;
X = [3] ;
X = [] ;
false.
I have to write a predicate: double(X,Y) to be true when Y is the list consisting of each element of X
repeated twice (e.g. double([a,b],[a,a,b,b]) is true).
I ended with sth like this:
double([],[]).
double([T],List) :- double([H|T],List).
double([H|T],List) :- count(H, List, 2).
Its working fine for lists like [a,a,b] but it shouldnt... please help.
And i need help with another predicate: repeat(X,Y,N) to be true when Y is the list consisting of each element of X
repeated N times (e.g. repeat([a,b], [a,a,a,b,b,b],3) is true).
double([],[]).
double([I|R],[I,I|RD]) :-
double(R,RD).
Here's how you could realize that "repeat" predicate you suggested in the question:
:- use_module(library(clpfd)).
Based on if_/3 and (=)/3 we define:
each_n_reps([E|Es], N) :-
aux_n_reps(Es, E, 1, N).
aux_n_reps([], _, N, N). % internal auxiliary predicate
aux_n_reps([E|Es], E0, N0, N) :-
if_(E0 = E,
( N0 #< N, N1 #= N0+1 ), % continue current run
( N0 #= N, N1 #= 1 )), % start new run
aux_n_reps(Es, E, N1, N).
Sample queries1 using SICStus Prolog 4.3.2:
?- each_n_reps(Xs, 3).
Xs = [_A,_A,_A]
; Xs = [_A,_A,_A,_B,_B,_B] , dif(_A,_B)
; Xs = [_A,_A,_A,_B,_B,_B,_C,_C,_C], dif(_A,_B), dif(_B,_C)
...
How about fair enumeration?
?- length(Xs, _), each_n_reps(Xs, N).
N = 1, Xs = [_A]
; N = 2, Xs = [_A,_A]
; N = 1, Xs = [_A,_B] , dif(_A,_B)
; N = 3, Xs = [_A,_A,_A]
; N = 1, Xs = [_A,_B,_C] , dif(_A,_B), dif(_B,_C)
; N = 4, Xs = [_A,_A,_A,_A]
; N = 2, Xs = [_A,_A,_B,_B], dif(_A,_B)
; N = 1, Xs = [_A,_B,_C,_D], dif(_A,_B), dif(_B,_C), dif(_C,_D)
...
How can [A,B,C,D,E,F] be split into runs of equal length?
?- each_n_reps([A,B,C,D,E,F], N).
N = 6, A=B , B=C , C=D , D=E , E=F
; N = 3, A=B , B=C , dif(C,D), D=E , E=F
; N = 2, A=B , dif(B,C), C=D , dif(D,E), E=F
; N = 1, dif(A,B), dif(B,C), dif(C,D), dif(D,E), dif(E,F).
Footnote 1: Answers were reformatted to improve readability.
Ok for repeat/3 i have sth like this:
repeat1([],[],0).
repeat1([A|B],[X|T],Y):- repeat1(B,T,Z), Y is 1+Z.
repeat1([A1|B],[X1|T], Z) :- A1\=A, X1\=X, repeat1(B,T,Z).
I am trying to compute arrangements of K elements in Prolog, where the sum of their elements is equal to a given S. So, I know that arrangements can be computed by finding the combinations and then permute them. I know how to compute combinations of K elements, something like:
comb([E|_], 1, [E]).
comb([_|T], K, R) :-
comb(T, K, R).
comb([H|T], K, [H|R]) :-
K > 1,
K1 is K-1,
comb(T, K1, R).
The permutations of a list, having the property that the sum of their elements is equal to a given S, I know to compute like this:
insert(E, L, [E|L]).
insert(E, [H|T], [H|R]) :-
insert(E, T, R).
perm([], []).
perm([H|T], P) :-
perm(T, R),
insert(H, R, P).
sumList([], 0).
sumList([H], H) :-
number(H).
sumList([H|Tail], R1) :-
sumList(Tail, R),
R1 is R+H.
perms(L, S, R) :-
perm(L, R),
sumList(R, S1),
S = S1.
allPerms(L, LP) :-
findall(R, perms(L,R), LP).
The problem is that I do not know how to combine them, in order to get the arrangements of K elements, having the sum of elements equal to a given S. Any help would be appreciated.
Use clpfd!
:- use_module(library(clpfd)).
Using SWI-Prolog 7.3.16 we query:
?- length(Zs,4), Zs ins 1..4, sum(Zs,#=,7), labeling([],Zs).
Zs = [1,1,1,4]
; Zs = [1,1,2,3]
; Zs = [1,1,3,2]
; Zs = [1,1,4,1]
; Zs = [1,2,1,3]
; Zs = [1,2,2,2]
; Zs = [1,2,3,1]
; Zs = [1,3,1,2]
; Zs = [1,3,2,1]
; Zs = [1,4,1,1]
; Zs = [2,1,1,3]
; Zs = [2,1,2,2]
; Zs = [2,1,3,1]
; Zs = [2,2,1,2]
; Zs = [2,2,2,1]
; Zs = [2,3,1,1]
; Zs = [3,1,1,2]
; Zs = [3,1,2,1]
; Zs = [3,2,1,1]
; Zs = [4,1,1,1].
To eliminate "redundant modulo permutation" solutions use chain/2:
?- length(Zs,4), Zs ins 1..4, chain(Zs,#=<), sum(Zs,#=,7), labeling([],Zs).
Zs = [1,1,1,4]
; Zs = [1,1,2,3]
; Zs = [1,2,2,2]
; false.
I use SWI-Prolog.
You can write that
:- use_module(library(lambda)).
arrangement(K, S, L) :-
% we have a list of K numbers
length(L, K),
% these numbers are between 1 (or 0) and S
maplist(between(1, S), L),
% the sum of these numbers is S
foldl(\X^Y^Z^(Z is X+Y), L, 0, S).
The result
?- arrangement(5, 10, L).
L = [1, 1, 1, 1, 6] ;
L = [1, 1, 1, 2, 5] ;
L = [1, 1, 1, 3, 4] ;
L = [1, 1, 1, 4, 3] .
You can use also a CLP(FD) library.
Edited after the remark of #repeat.
This response is similar to response of #repeat
predicates that below are implemented using the SICStus 4.3.2 tool
after simple modification of gen_list(+,+,?)
edit Code
gen_list(Length,Sum,List) :- length(List,Length),
domain(List,0,Sum),
sum(List,#=,Sum),
labeling([],List),
% to avoid duplicate results
ordered(List).
Test
| ?- gen_list(4,7,L).
L = [0,0,0,7] ? ;
L = [0,0,1,6] ? ;
L = [0,0,2,5] ? ;
L = [0,0,3,4] ? ;
L = [0,1,1,5] ? ;
L = [0,1,2,4] ? ;
L = [0,1,3,3] ? ;
L = [0,2,2,3] ? ;
L = [1,1,1,4] ? ;
L = [1,1,2,3] ? ;
L = [1,2,2,2] ? ;
no
I don't think that permutations could be relevant for your problem. Since the sum operation is commutative, the order of elements should be actually irrelevant. So, after this correction
sumList([], 0).
%sumList([H], H) :-
% number(H).
sumList([H|Tail], R1) :-
sumList(Tail, R),
R1 is R+H.
you can just use your predicates
'arrangements of K elements'(Elements, K, Sum, Arrangement) :-
comb(Elements, K, Arrangement),
sumList(Arrangement, Sum).
test:
'arrangements of K elements'([1,2,3,4,5,6],3,11,A).
A = [2, 4, 5] ;
A = [2, 3, 6] ;
A = [1, 4, 6] ;
false.
You already know how to use findall/3 to get all lists at once, if you need them.
With the following piece of code in Prolog I managed to extract every third item from a list:
third([_,_,Z|L], Z).
third([_,_,C|L], Y) :-
third(L, Y).
However, I still need to append every item that I am extracting to a new list, and finally create a list with only "every third" items.
I would appreciate any hint or help!
Thank you.
How about doing it like this?
list_thirds([] , []).
list_thirds([_] , []).
list_thirds([_,_] , []).
list_thirds([_,_,E|Es], [E|Xs]) :-
list_thirds(Es, Xs).
Sample queries using SICStus Prolog 4.3.2:
| ?- list_thirds([], Xs).
Xs = [] ? ;
no
| ?- list_thirds([a,b,c], Xs).
Xs = [c] ? ;
no
| ?- list_thirds([a,b,c,d,e,f], Xs).
Xs = [c,f] ? ;
no
| ?- list_thirds([a,b,c,d,e,f,g], Xs).
Xs = [c,f] ? ;
no
How about going the "other direction"?
| ?- list_thirds(List, [x,y]).
List = [_A,_B,x,_C,_D,y] ? ;
List = [_A,_B,x,_C,_D,y,_E] ? ;
List = [_A,_B,x,_C,_D,y,_E,_F] ? ;
no % terminates universally
Last, we look at the answer sequence we get from the most general query:
| ?- list_thirds(Es, Xs).
Es = [] , Xs = [] ? ;
Es = [_A] , Xs = [] ? ;
Es = [_A,_B] , Xs = [] ? ;
Es = [_A,_B,_C] , Xs = [_C] ? ;
Es = [_A,_B,_C,_D] , Xs = [_C] ? ;
Es = [_A,_B,_C,_D,_E] , Xs = [_C] ? ;
Es = [_A,_B,_C,_D,_E,_F], Xs = [_C,_F] ? ;
...
With SWI-Prolog, and module lambda.pl, you can write
:- use_module(library(lambda)).
third(In, Out) :-
foldl(\X^Y^Z^(Y=[N,L],
( N = 2
-> append(L, [X], NL),
Z = [0,NL]
; N1 is N+1,
Z = [N1, L])),
In, [0, []], [_, Out]).
With same queries :
?- third([1,2,3,4,5,6,7,8,9,10], Out).
Out = [3, 6, 9].
?- third(X, [3,6,9]).
X = [_G103, _G180, 3, _G352, _G438, 6, _G613, _G699, 9] ;
X = [_G103, _G180, 3, _G352, _G438, 6, _G613, _G699, 9|...] .
?- third(X, Y).
X = Y, Y = [] ;
X = [_G87231],
Y = [] ;
X = [_G87231, _G87317],
Y = [] ;
X = [_G87231, _G87317, _G87403],
Y = [_G87403] ;
X = [_G87231, _G87317, _G87403, _G87489],
Y = [_G87403] ;
X = [_G87231, _G87317, _G87403, _G87489, _G87575],
Y = [_G87403] ;
X = [_G87231, _G87317, _G87403, _G87489, _G87575, _G87661],
Y = [_G87403, _G87661] .
There's a DCG approach that would also work, analogous to #repeat's answer:
thirds([]) --> [].
thirds([]) --> [_].
thirds([]) --> [_,_].
thirds([X|T]) --> [_,_,X], thirds(T).
Or more concisely (thanks #repeat):
thirds([]) --> [] | [_] | [_,_].
thirds([X|T]) --> [_,_,X], thirds(T).
Called as follows:
| ?- phrase(thirds(T), [a,b,c,d,e,f,g,h]).
T = [c,f] ? a
no
| ?- phrase(thirds(T), L).
L = []
T = [] ? ;
L = [_]
T = [] ? ;
L = [_,_]
T = [] ? ;
L = [_,_,A]
T = [A] ? ;
L = [_,_,A,_]
T = [A] ? ;
...
This will work:
bagof(Third,third(List,Third),Result).
This collects all items Third such that Third is third in List; and binds the list of them to Result.
For more on bag, see http://www.swi-prolog.org/pldoc/doc_for?object=bagof/3 .
I'm looking for a predicate that works as this:
?- subset([1,2,3], X).
X = [] ;
X = [1] ;
X = [2] ;
X = [3] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [2, 3] ;
...
I've seen some subset implementations, but they all work when you want to check if one list is a subset of the another, not when you want to generate the subsets. Any ideas?
Here goes an implementation:
subset([], []).
subset([E|Tail], [E|NTail]):-
subset(Tail, NTail).
subset([_|Tail], NTail):-
subset(Tail, NTail).
It will generate all the subsets, though not in the order shown on your example.
As per commenter request here goes an explanation:
The first clause is the base case. It states that the empty list is a subset of the empty list.
The second and third clauses deal with recursion. The second clause states that if two lists have the same Head and the tail of the right list is a subset of the tail of the left list, then the right list is a subset of the left list.
The third clause states that if we skip the head of the left list, and the right list is a subset of the tail of the left list, then the right list is a subset of the left list.
The procedure shown above generates ordered sets. For unordered sets you might use permutation/3:
unordered_subset(Set, SubSet):-
length(Set, LSet),
between(0,LSet, LSubSet),
length(NSubSet, LSubSet),
permutation(SubSet, NSubSet),
subset(Set, NSubSet).
On http://www.probp.com/publib/listut.html you will find an implementation of a predicate called subseq0 that does what you want to:
subseq0(List, List).
subseq0(List, Rest) :-
subseq1(List, Rest).
subseq1([_|Tail], Rest) :-
subseq0(Tail, Rest).
subseq1([Head|Tail], [Head|Rest]) :-
subseq1(Tail, Rest).
A short explanation: subseq0(X, Y) checks whether Y is a subset subsequence of X, while subseq1(X, Y) checks whether Y is a proper subset subsequence of X.
Since the default representation of a set is a list with unique elements, you can use it to get all subsets as in the following example:
?- subseq0([1,2,3], X).
X = [1, 2, 3] ;
X = [2, 3] ;
X = [3] ;
X = [] ;
X = [2] ;
X = [1, 3] ;
X = [1] ;
X = [1, 2] ;
false.
Set is a collection of distinct objects by definition. A subset procedure shouldn't care about the order of elements in the set(in the arguments). A proper solution(swi prolog) may look like:
subset(_, []).
subset([X|L], [A|NTail]):-
member(A,[X|L]),
subset(L, NTail),
not(member(A, NTail)).
For the question ?- subset([1,2,3], E) it will generate:
E = [] ;
E = [1] ;
E = [1, 2] ;
E = [1, 2, 3] ;
E = [1, 3] ;
E = [2] ;
E = [2, 3] ;
E = [3] ;
E = [3, 2] ;
false.
Hope it will help!
we can also test by deleting subset's item from the super set.
% to delete : an item can be deleted it its in the head or in the tail of a list
delete(I,[I|L],L).
delete(I,[H|L],[H|NL]) :- delete(I,L,NL).
% an [] is an item of an set.A set is a subset of we can recursively delete its head item from the super set.
subset(_,[]).
subset(S,[I|SS]) :- delete(I,S,S1), subset(S1,SS).
example:
subset([a,b,c],S).
S = []
S = [a]
S = [a, b]
S = [a, b, c]
S = [a, c]
S = [a, c, b]
S = [b]
S = [b, a]
S = [b, a, c]
S = [b, c]
S = [b, c, a]
S = [c]
S = [c, a]
S = [c, a, b]
S = [c, b]
S = [c, b, a]
subset([a,b,a,d,e],[a,e]).
1true
append([],L,L).
append([H|T],L,[H|L1]):-append(T,L,L1).
subset([X|T],[X|L]) :-subset(T,L).
subset([X|T],[G|L]) :-subset([X],L),append(L2,[X|L3],[G|L]),append(L2,L3,L4),subset(T,L4).
subset([],_).
----------------------------------------------
?- subset([1,2],[1,2]).
yes
?- subset([1,2],[2,1]).
yes
?- subset([1,1],[1,2]).
no
?- subset(D,[1,2]).
D = [1,2] ;
D = [1] ;
D = [2,1] ;
D = [2] ;
D = '[]' ;
no