Is it possible to declare an ascending list? - list

I can make lists of ascending integer like so:
?- findall(L,between(1,5,L),List).
I know I can also generate values using:
?- length(_,X).
But I don't think I can use this in a findall, as things like the following loop:
?- findall(X,(length(_,X),X<6),Xs).
I can also generate a list using clpfd.
:- use_module(library(clpfd)).
list_to_n(N,List) :-
length(List,N),
List ins 1..N,
all_different(List),
once(label(List)).
or
list_to_n2(N,List) :-
length(List,N),
List ins 1..N,
chain(List,#<),
label(List).
The last method seems best to me as it is the most declarative, and does not use once/1 or between/3 or findall/3 etc.
Are there other ways to do this? Is there a declarative way to do this in 'pure' Prolog? Is there a 'best' way?

The "best" way depends on your concrete use cases! Here's another way to do it using clpfd:
:- use_module(library(clpfd)).
We define predicate equidistant_stride/2 as suggested by #mat in a comment to a previous answer of a related question:
equidistant_stride([],_).
equidistant_stride([Z|Zs],D) :-
foldl(equidistant_stride_(D),Zs,Z,_).
equidistant_stride_(D,Z1,Z0,Z1) :-
Z1 #= Z0+D.
Based on equidistant_stride/2, we define:
consecutive_ascending_integers(Zs) :-
equidistant_stride(Zs,1).
consecutive_ascending_integers_from(Zs,Z0) :-
Zs = [Z0|_],
consecutive_ascending_integers(Zs).
consecutive_ascending_integers_from_1(Zs) :-
consecutive_ascending_integers_from(Zs,1).
Let's run some queries! First, your original use case:
?- length(Zs,N), consecutive_ascending_integers_from_1(Zs).
N = 1, Zs = [1]
; N = 2, Zs = [1,2]
; N = 3, Zs = [1,2,3]
; N = 4, Zs = [1,2,3,4]
; N = 5, Zs = [1,2,3,4,5]
...
With clpfd, we can ask quite general queries—and get logically sound answers, too!
?- consecutive_ascending_integers([A,B,0,D,E]).
A = -2, B = -1, D = 1, E = 2.
?- consecutive_ascending_integers([A,B,C,D,E]).
A+1#=B, B+1#=C, C+1#=D, D+1#=E.
An alternative implementation of equidistant_stride/2:
I hope the new code makes better use of constraint propagation.
Thanks to #WillNess for suggesting the test-cases that motivated this rewrite!
equidistant_from_nth_stride([],_,_,_).
equidistant_from_nth_stride([Z|Zs],Z0,N,D) :-
Z #= Z0 + N*D,
N1 #= N+1,
equidistant_from_nth_stride(Zs,Z0,N1,D).
equidistant_stride([],_).
equidistant_stride([Z0|Zs],D) :-
equidistant_from_nth_stride(Zs,Z0,1,D).
Comparison of old vs new version with #mat's clpfd:
First up, the old version:
?- equidistant_stride([1,_,_,_,14],D).
_G1133+D#=14,
_G1145+D#=_G1133,
_G1157+D#=_G1145,
1+D#=_G1157. % succeeds with Scheinlösung
?- equidistant_stride([1,_,_,_,14|_],D).
_G1136+D#=14, _G1148+D#=_G1136, _G1160+D#=_G1148, 1+D#=_G1160
; 14+D#=_G1340, _G1354+D#=14, _G1366+D#=_G1354, _G1378+D#=_G1366, 1+D#=_G1378
... % does not terminate universally
Now let's switch to the new version and ask the same queries!
?- equidistant_stride([1,_,_,_,14],D).
false. % fails, as it should
?- equidistant_stride([1,_,_,_,14|_],D).
false. % fails, as it should
More, now, again! Can we fail earlier by tentatively employing redundant constraints?
Previously, we proposed using constraints Z1 #= Z0+D*1, Z2 #= Z0+D*2, Z3 #= Z0+D*3 instead of Z1 #= Z0+D, Z2 #= Z1+D, Z3 #= Z2+D
(which the 1st version of code in this answer did).
Again, thanks to #WillNess for motivating this little experiment by
noting that the goal equidistant_stride([_,4,_,_,14],D) does not fail but instead succeeds with pending goals:
?- Zs = [_,4,_,_,14], equidistant_stride(Zs,D).
Zs = [_G2650, 4, _G2656, _G2659, 14],
14#=_G2650+4*D,
_G2659#=_G2650+3*D,
_G2656#=_G2650+2*D,
_G2650+D#=4.
Let's add some redundant constraints with equidistantRED_stride/2:
equidistantRED_stride([],_).
equidistantRED_stride([Z|Zs],D) :-
equidistant_from_nth_stride(Zs,Z,1,D),
equidistantRED_stride(Zs,D).
Sample query:
?- Zs = [_,4,_,_,14], equidistant_stride(Zs,D), equidistantRED_stride(Zs,D).
false.
Done? Not yet! In general we don't want a quadratic number of redundant constraints. Here's why:
?- Zs = [_,_,_,_,14], equidistant_stride(Zs,D).
Zs = [_G2683, _G2686, _G2689, _G2692, 14],
14#=_G2683+4*D,
_G2692#=_G2683+3*D,
_G2689#=_G2683+2*D,
_G2686#=_G2683+D.
?- Zs = [_,_,_,_,14], equidistant_stride(Zs,D), equidistantRED_stride(Zs,D).
Zs = [_G831, _G834, _G837, _G840, 14],
14#=_G831+4*D,
_G840#=_G831+3*D,
_G837#=_G831+2*D,
_G834#=_G831+D,
14#=_G831+4*D,
_G840#=_G831+3*D,
_G837#=_G831+2*D,
_G834#=_G831+D,
D+_G840#=14,
14#=2*D+_G837,
_G840#=D+_G837,
14#=_G834+3*D,
_G840#=_G834+2*D,
_G837#=_G834+D.
But if we use the double-negation trick, the residuum remains in cases that succeed ...
?- Zs = [_,_,_,_,14], equidistant_stride(Zs,D), \+ \+ equidistantRED_stride(Zs,D).
Zs = [_G454, _G457, _G460, _G463, 14],
14#=_G454+4*D,
_G463#=_G454+3*D,
_G460#=_G454+2*D,
_G457#=_G454+D.
... and ...
?- Zs = [_,4,_,_,14], equidistant_stride(Zs,D), \+ \+ equidistantRED_stride(Zs,D).
false.
... we detect failure in more cases than we did before!
Let's dig a little deeper! Can we detect failure early in even more generalized uses?
With code presented so far, these two logically false queries do not terminate:
?- Zs = [_,4,_,_,14|_], \+ \+ equidistantRED_stride(Zs,D), equidistant_stride(Zs,D).
... % Execution Aborted
?- Zs = [_,4,_,_,14|_], equidistant_stride(Zs,D), \+ \+ equidistantRED_stride(Zs,D).
... % Execution Aborted
Got fix? Got hack!
?- use_module(library(lambda)).
true.
?- Zs = [_,4,_,_,14|_],
\+ ( term_variables(Zs,Vs),
maplist(\X^when(nonvar(X),integer(X)),Vs),
\+ equidistantRED_stride(Zs,D)),
equidistant_stride(Zs,D).
false.
The hack doesn't guarantee termination of the redundant constraint "part", but IMO it's not too bad for a quick first shot. The test integer/1 upon instantiation of any variable in Zs is meant to allow the clpfd solver to constrain variable domains to singletons, while the instantiation with cons-pairs (which directly leads to non-termination of list-based predicates) is suppressed.
I do realize that the hack can be broken quite easily in more than one way (e.g., using cyclic terms). Any suggestions and comments are welcome!

In the following we discuss the code presented in this previous answer.
The goal consecutive_ascending_integers_from_1([2,3,5,8|non_list]) fails, but why?
Let's proceed step-by-step:
Here's the code we start with:
:- use_module(library(clpfd)).
equidistant_from_nth_stride([],_,_,_).
equidistant_from_nth_stride([Z|Zs],Z0,I0,D) :-
Z #= Z0 + I0*D,
I1 #= I0 + 1,
equidistant_from_nth_stride(Zs,Z0,I1,D).
equidistant_stride([],_).
equidistant_stride([Z0|Zs],D) :-
equidistant_from_nth_stride(Zs,Z0,1,D).
consecutive_ascending_integers(Zs) :-
equidistant_stride(Zs,1).
consecutive_ascending_integers_from(Zs,Z0) :-
Zs = [Z0|_],
consecutive_ascending_integers(Zs).
consecutive_ascending_integers_from_1(Zs) :-
consecutive_ascending_integers_from(Zs,1).
First, we make (some) unifications more explicit:
equidistant_from_nth_stride([],_,_,_).
equidistant_from_nth_stride([Z|Zs],Z0,I0,D) :-
Z #= Z0 + I0*D,
I1 #= I0 + 1,
equidistant_from_nth_stride(Zs,Z0,I1,D).
equidistant_stride([],_).
equidistant_stride([Z0|Zs],D) :-
I = 1,
equidistant_from_nth_stride(Zs,Z0,I,D).
consecutive_ascending_integers(Zs) :-
D = 1,
equidistant_stride(Zs,D).
consecutive_ascending_integers_from(Zs,Z0) :-
Zs = [Z0|_],
consecutive_ascending_integers(Zs).
consecutive_ascending_integers_from_1(Zs) :-
Z0 = 1,
consecutive_ascending_integers_from(Zs,Z0).
We follow the method and conventions introduced in this fine answer:
By removing goals, we can generalize a program. Here is my favorite way to do it. By adding a predicate (*)/1 like so:
:- op(950,fy, *).
*_.
#WillNess rightly noted that:
consecutive_ascending_integers_from_1([2|_]) fails, so its specialization consecutive_ascending_integers_from_1([2,3,5,8|non_list]) must fail too.
If maximally generalize the code so that consecutive_ascending_integers_from_1([2|_]) fails, we "know for sure: something in the visible remaining part of the program has to be fixed."
consecutive_ascending_integers_from(Zs,Z0) :-
Zs = [Z0|_],
* consecutive_ascending_integers(Zs).
consecutive_ascending_integers_from_1(Zs) :-
Start = 1,
consecutive_ascending_integers_from(Zs,Start).
Let's have another explanation!
With version #2 (see above), we observe the following generalized goal fails, too:
?- consecutive_ascending_integers_from_1([_,_,_,_|non_list]).
false.
Why does this fail? Let's maximally generalize the code suchthat the goal fails:
equidistant_from_nth_stride([],_,_,_).
equidistant_from_nth_stride([Z|Zs],Z0,I0,D) :-
* Z #= Z0 + I0*D,
* I1 #= I0 + 1,
equidistant_from_nth_stride(Zs,Z0,I1,D).
equidistant_stride([],_).
equidistant_stride([Z0|Zs],D) :-
* I = 1,
equidistant_from_nth_stride(Zs,Z0,I,D).
consecutive_ascending_integers(Zs) :-
* D = 1,
equidistant_stride(Zs,D).
consecutive_ascending_integers_from(Zs,Z0) :-
* Zs = [Z0|_],
consecutive_ascending_integers(Zs).
consecutive_ascending_integers_from_1(Zs) :-
* Start = 1,
consecutive_ascending_integers_from(Zs,Start).
Why does the goal consecutive_ascending_integers_from_1([2,3,5,8|non_list]) fail?
Up to now, we have seen two explanations, but there may be more...
The truth is out there: Join the hunt!

We'll define ascending lists as such that contain at least two elements which are increasing integer numbers (non-decreasing lists could be empty, or singleton, but "ascending" is a more definite property). It's a somewhat arbitrary determination.
In SWI Prolog:
ascending( [A,B|R] ):-
freeze(A,
freeze(B, (A < B, freeze(R, (R=[] -> true ; ascending([B|R])))) )).
To easily fill'em up, we could use
mselect([A|B],S,S2):- select(A,S,S1), mselect(B,S1,S2).
mselect([], S2, S2).
Testing:
15 ?- ascending(LS),mselect(LS,[10,2,8,5],[]).
LS = [2, 5, 8, 10] ;
false.
16 ?- mselect(LS,[10,2,8,5],[]), ascending(LS).
LS = [2, 5, 8, 10] ;
false.
As to the bounty question, according to https://stackoverflow.com/tags/logical-purity/info,
Only monotonic (also called: "monotone") predicates are pure: If the predicate succeeds for any arguments, then it does not fail for any generalization of these arguments, and if it fails for any combination of arguments, then it does not succeed for any specialization of these arguments.
consecutive_ascending_integers_from_1([2|B]) fails, so its specialization consecutive_ascending_integers_from_1([2,3,5,8|non_list]) must fail too.
For the extended bounty " consecutive_ascending_integers_from_1([2,3,5,8|non_list]) fails, but why?", additional failing goals are: ( 1 )
consecutive_ascending_integers_from_1([_,3|_])
for the code
equidistant_from_nth_stride([],_,_,_).
equidistant_from_nth_stride([Z|Zs],Z0,I0,D) :-
Z #= Z0 + I0*D, % C1
*( I1 #= I0 + 1 ),
equidistant_from_nth_stride(Zs,Z0,I1,D).
and the rest unchanged, because C1 becomes 3 #= 1 + 1*1. Also, ( 2 and 3 )
consecutive_ascending_integers_from_1([A,B,5|_])
consecutive_ascending_integers_from_1([A,B,C,8|_])
both fail with the unchanged code, because the 1st defines
A = 1, B #= 1 + 1*1, 5 #= 1 + 2*1
and the 2nd defines
A = 1, B #= 1 + 1*1, C #= 1 + 2*1, 8 #= 1 + 3*1
Yet another possibility ( 4 ) is
consecutive_ascending_integers_from_1([_,3,5|_])
with the generalized
consecutive_ascending_integers_from_1(Zs) :-
*( Z0 = 1 ),
consecutive_ascending_integers_from(Zs,Z0).
consecutive_ascending_integers_from(Zs,Z0) :-
*( Zs = [Z0|_] ),
consecutive_ascending_integers(Zs).
because
26 ?- 3 #= Z + 1*1, 5 #= Z + 2*1.
false.
Similarly, with the like modified code, the goal ( 5 )
consecutive_ascending_integers_from_1([_,3,_,8|_])
because
27 ?- 3 #= Z + 1*1, 8 #= Z + 3*1.
false.
and also the ( 6 ... 9 )
consecutive_ascending_integers_from_1([2,3,_,8|_])
consecutive_ascending_integers_from_1([2,_,_,8|_])
consecutive_ascending_integers_from_1([2,_,5,8|_])
consecutive_ascending_integers_from_1([2,_,5|_])
for the same reason. Yet another possible code generalization is to leave D uninitialized (with the rest of the original code unchanged):
consecutive_ascending_integers(Zs) :-
*( D = 1 ),
equidistant_stride(Zs,D).
so that the goal ( 5 ) ...[_,3,_,8|_]... again fails, because of
49 ?- 3 #= 1 + 1*D, 8 #= 1 + 3*D.
false.
but,
50 ?- 3 #= 1 + 1*D, 5 #= 1 + 2*D.
D = 2.
so ...[_,3,5|_]... would succeed (indeed it does). ( 10 )
consecutive_ascending_integers_from_1([_,_,5,8|_])
fails, as well, for the same reason.
There might be some more combinations, but the general gist of it becomes clearer: it all depends on how the constraints created by this predicate operate.

Related

DCG and inversion of a list in Prolog

I'm trying to count the numer of inversions in a list. A predicate inversion(+L,-N) unifies N to the number of inversions in that list. A inversion is defined as X > Y and X appears before Y in the list (unless X or Y is 0). For example:
?- inversions([1,2,3,4,0,5,6,7,8],N).
N = 0.
?- inversions([1,2,3,0,4,6,8,5,7],N).
N = 3.
For what I'm using this for, the list will always have exacly 9 elements, and always containing the numbers 0-8 uniquely.
I'm quite new to Prolog and I'm trying to do this as concise and as elegant as possible; It seems like DCG will probably help a lot. I read into the official definition and some tutorial sites, but still don't quit understand what it is. Any help would be greatly appreciated.
Here is another solution that doesn't leave choice points using if_/3:
inversions([],0).
inversions([H|T], N):-
if_( H = 0,
inversions(T,N),
( find_inv(T,H,N1),inversions(T, N2), N #= N1+N2 )
).
find_inv([],_,0).
find_inv([H1|T],H,N1):-
if_( H1=0,
find_inv(T,H,N1),
if_( H#>H1,
(find_inv(T,H,N2),N1 #= N2+1),
find_inv(T,H,N1)
)
).
#>(X, Y, T) :-
( integer(X),
integer(Y)
-> ( X > Y
-> T = true
; T = false
)
; X #> Y,
T = true
; X #=< Y,
T = false
).
I'm not so sure a DCG would be helpful here. Although we're processing a sequence, there's a lot of examination of the entire list at each point when looking at each element.
Here's a CLPFD approach which implements the "naive" algorithm for inversions, so it's transparent and simple, but not as efficient as it could be (it's O(n^2)). There's a more efficient algorithm (O(n log n)) involving a divide and conquer approach, which I show further below.
:- use_module(library(clpfd)).
inversions(L, C) :-
L ins 0..9,
all_distinct(L),
count_inv(L, C).
% Count inversions
count_inv([], 0).
count_inv([X|T], C) :-
count_inv(X, T, C1), % Count inversions for current element
C #= C1 + C2, % Add inversion count for the rest of the list
count_inv(T, C2). % Count inversions for the rest of the list
count_inv(_, [], 0).
count_inv(X, [Y|T], C) :-
( X #> Y, X #> 0, Y #> 0
-> C #= C1 + 1, % Valid inversion, count it
count_inv(X, T, C1)
; count_inv(X, T, C)
).
?- inversions([1,2,3,4,0,5,6,7,8],N).
N = 0 ;
false.
?- inversions([1,2,3,0,4,6,8,5,7],N).
N = 3 ;
false.
?- inversions([0,2,X],1).
X = 1 ;
false.
It does leave a choice point, as you can see, which I haven't sorted out yet.
Here's the O(n log n) solution, which is using the sort/merge algorithm.
inversion([], [], 0).
inversion([X], [X], 0).
inversion([HU1, HU2|U], [HS1, HS2|S], C) :- % Ensure list args have at least 2 elements
split([HU1, HU2|U], L, R),
inversion(L, SL, C1),
inversion(R, SR, C2),
merge(SL, SR, [HS1, HS2|S], C3),
C #= C1 + C2 + C3.
% Split list into left and right halves
split(List, Left, Right) :-
split(List, List, Left, Right).
split(Es, [], [], Es).
split(Es, [_], [], Es).
split([E|Es], [_,_|T], [E|Ls], Right) :-
split(Es, T, Ls, Right).
% merge( LS, RS, M )
merge([], RS, RS, 0).
merge(LS, [], LS, 0).
merge([L|LS], [R|RS], [L|T], C) :-
L #=< R,
merge(LS, [R|RS], T, C).
merge([L|LS], [R|RS], [R|T], C) :-
L #> R, R #> 0 #<==> D, C #= C1+D,
merge([L|LS], RS, T, C1).
You can ignore the second argument, which is the sorted list (just a side effect if all you want is the count of inversions).
Here is another possibility to define the relation. First, #</3 and #\=/3 can be defined like so:
:- use_module(library(clpfd)).
bool_t(1,true).
bool_t(0,false).
#<(X,Y,Truth) :- X #< Y #<==> B, bool_t(B,Truth).
#\=(X,Y,Truth) :- X #\= Y #<==> B, bool_t(B,Truth).
Based on that, if_/3 and (',')/3 a predicate inv_t/3 can be defined, that yields true in the case of an inversion and false otherwise, according to the definition given by the OP:
inv_t(X,Y,T) :-
if_(((Y#<X,Y#\=0),X#\=0),T=true,T=false).
And subsequently the actual relation can be described like so:
list_inversions(L,I) :-
list_inversions_(L,I,0).
list_inversions_([],I,I).
list_inversions_([X|Xs],I,Acc0) :-
list_x_invs_(Xs,X,I0,0),
Acc1 #= Acc0+I0,
list_inversions_(Xs,I,Acc1).
list_x_invs_([],_X,I,I).
list_x_invs_([Y|Ys],X,I,Acc0) :-
if_(inv_t(X,Y),Acc1#=Acc0+1,Acc1#=Acc0),
list_x_invs_(Ys,X,I,Acc1).
Thus the example queries given by the OP succeed deterministically:
?- list_inversions([1,2,3,4,0,5,6,7,8],N).
N = 0.
?- list_inversions([1,2,3,0,4,6,8,5,7],N).
N = 3.
Such application-specific constraints can often be built using reified constraints (constraints whose truth value is reflected into a 0/1 variable). This leads to a relatively natural formulation, where B is 1 iff the condition you want to count is satisfied:
:- lib(ic).
inversions(Xs, N) :-
( fromto(Xs, [X|Ys], Ys, [_]), foreach(NX,NXs) do
( foreach(Y,Ys), param(X), foreach(B,Bs) do
B #= (X#\=0 and Y#\=0 and X#>Y)
),
NX #= sum(Bs) % number of Ys that are smaller than X
),
N #= sum(NXs).
This code is for ECLiPSe.
Using clpfd et automaton/8 we can write
:- use_module(library(clpfd)).
inversions(Vs, N) :-
Vs ins 0..sup,
variables_signature(Vs, Sigs),
automaton(Sigs, _, Sigs,
[source(s),sink(i),sink(s)],
[arc(s,0,s), arc(s,1,s,[C+1]), arc(s,1,i,[C+1]),
arc(i,0,i)],
[C], [0], [N]),
labeling([ff],Vs).
variables_signature([], []).
variables_signature([V|Vs], Sigs) :-
variables_signature_(Vs, V, Sigs1),
variables_signature(Vs, Sigs2),
append(Sigs1, Sigs2, Sigs).
variables_signature_([], _, []).
variables_signature_([0|Vs], Prev, Sigs) :-
variables_signature_(Vs,Prev,Sigs).
variables_signature_([V|Vs], Prev, [S|Sigs]) :-
V #\= 0,
% Prev #=< V #<==> S #= 0,
% modified after **false** remark
Prev #> V #<==> S,
variables_signature_(Vs,Prev,Sigs).
examples :
?- inversions([1,2,3,0,4,6,8,5,7],N).
N = 3 ;
false.
?- inversions([1,2,3,0,4,5,6,7,8],N).
N = 0 ;
false.
?- inversions([0,2,X],1).
X = 1.
in SWI-Prolog, with libraries aggregate and lists:
inversions(L,N) :-
aggregate_all(count, (nth1(P,L,X),nth1(Q,L,Y),X\=0,Y\=0,X>Y,P<Q), N).
both libraries are autoloaded, no need to explicitly include them.
If you want something more general, you can see the example in library(clpfd), under the automaton section, for some useful ideas. But I would try to rewrite your specification in simpler terms, using element/3 instead of nth1/3.
edit
after #false comment, I tried some variation on disequality operators, but none I've tried have been able to solve the problematic query. Then I tried again with the original idea, to put to good use element/3. Here is the result:
:- use_module(library(clpfd)).
inversions(L) :-
L ins 0..8,
element(P,L,X),
element(Q,L,Y),
X #\= 0, Y #\= 0, X #> Y, P #< Q,
label([P,Q]).
inversions(L,N) :-
aggregate(count, inversions(L), N) ; N = 0.
The last line label([P,Q]) it's key to proper reification: now we can determine the X value.
?- inversions([0,2,X],1).
X = 1.

A length/2 predicate that works with constraints [duplicate]

Here is the problem:
$ swipl
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?- use_module(library(clpfd)).
true.
?- N in 1..3, length(L, N).
N = 1,
L = [_G1580] ;
N = 2,
L = [_G1580, _G1583] ;
N = 3,
L = [_G1580, _G1583, _G1586] ;
ERROR: Out of global stack % after a while
(I can switch the order of the subqueries, the result is the same).
I guess I need to label N before I can use it, but I wonder what the problem is? I have not managed to choke up length/2 before.
What's probably more useful than a slightly less nondeterministic length/2 is a proper list-length constraint. You can find an ECLiPSe implementation of it here, called len/2. With this you get the following behaviour:
?- N :: 1..3, len(Xs, N).
N = N{1 .. 3}
Xs = [_431|_482] % note it must contain at least one element!
There is 1 delayed goal.
Yes (0.00s cpu)
You can then enumerate the valid lists either by enumerating N:
?- N :: 1..3, len(Xs, N), indomain(N).
N = 1
Xs = [_478]
Yes (0.00s cpu, solution 1, maybe more)
N = 2
Xs = [_478, _557]
Yes (0.02s cpu, solution 2, maybe more)
N = 3
Xs = [_478, _557, _561]
Yes (0.02s cpu, solution 3)
or by generating lists with good old standard length/2:
?- N :: 1..3, len(Xs, N), length(Xs, _).
N = 1
Xs = [_488]
Yes (0.00s cpu, solution 1, maybe more)
N = 2
Xs = [_488, _555]
Yes (0.02s cpu, solution 2, maybe more)
N = 3
Xs = [_488, _555, _636]
Yes (0.02s cpu, solution 3)
Let's start with the most obvious one. If you switch the goals, you have:
?- length(L, N), N in 1..3.
which has the same termination properties as:
?- length(L, N), false, N in 1..3.
So obviously, this must not terminate with Prolog's execution mechanism.
However, if you put N in 1..3 in front, this might affect termination. To do so, it must be possible with finite means to prove that there is no N from 4 on. How can you prove this in a system without constraints - that is, only with syntactic unification present? Well, you can't. And length/2 is commonly defined just without constraints present.
With library(clpfd) things are trivial, for N #>= 4, N in 1..3 simply fails1. Note also that library(clpfd) does not collaborate much with library(clpq) which might be an interesting candidate, too.
As a consequence you would need to define your own length — for each constraint package you are interested in. That's a bit of a pity, but currently there is no generic way to do so in sight. ((That is, if you are interested and think a bit about it, you might come up with a nice API that every constraint system should adhere to. Alas, this will take some more decades, I suspect. Currently, there is much too much divergence.))
So here is a first naive way for fd_length/2:
fd_length([], N) :-
N #= 0.
fd_length([_|L], N0) :-
N0 #>= 1,
N1 #= N0-1,
fd_length(L, N1).
OK, this could be optimized to avoid the superfluous choicepoint. But there is a more fundamental problem: If you are determining the length of a list of length N, this will create N constraint variables! But we do need only one.
fd_length(L, N) :-
N #>= 0,
fd_length(L, N, 0).
fd_length([], N, N0) :-
N #= N0.
fd_length([_|L], N, N0) :-
N1 is N0+1,
N #>= N1,
fd_length(L, N, N1).
Again, this is not perfect for so many reasons: It could use Brent's algorithm like current systems do ; and combine it with all the fd properties. Also, arithmetic expressions are probably not a good idea to permit ; but I would have to wait for (#)/1 in SWI...
1: Strictly speaking, this "simply fails" only for SICStus, SWI, and YAP. For in those systems, there is no accidental failure due to exhaustion of the current representation. That is, their failure can always be taken as an honest no.
How about the following baroque work-around based on clpfd and meta-predicate tcount/3?
:- use_module([library(clpfd), library(lambda)]).
list_FDlen(Xs, N) :-
tcount(\_^ =(true), Xs, N).
Let's query!
?- N in 1..3, list_FDlen(Xs, N).
N = 1, Xs = [_A]
; N = 2, Xs = [_A,_B]
; N = 3, Xs = [_A,_B,_C]
; false. % terminates universally
?- N in inf..2, list_FDlen(Xs, N).
N = 0, Xs = []
; N = 1, Xs = [_A]
; N = 2, Xs = [_A,_B]
; false. % terminates universally, too
What about this particular query?
?- N in 2..sup, list_FDlen(Xs, N).
N = 2, Xs = [_A,_B]
; N = 3, Xs = [_A,_B,_C]
; N = 4, Xs = [_A,_B,_C,_D]
... % does not terminate (as expected)
We present a clpfd-ish variant of
length/2 that's tailored to #mat's clpfd implementation.
:- use_module(library(clpfd)).
:- use_module(library(dialect/sicstus)).
:- multifile clpfd:run_propagator/2.
The "exported" predicate lazy_len/2 is defined like this:
lazy_len(Es, N) :-
N in 0..sup, % lengths are always non-negative integers
lazylist_acc_len(Es, 0, N),
create_mutable(Es+0, State),
clpfd:make_propagator(list_FD_size(State,N), Propagator),
clpfd:init_propagator(N, Propagator),
clpfd:trigger_once(Propagator).
The global constraint handler list_FD_size/3 incrementally modifies its internal state as constraint propagation occurs. All modifications are trailed and are un-done upon backtracking.
clpfd:run_propagator(list_FD_size(State,N), _MState) :-
get_mutable(Es0+Min0, State),
fd_inf(N, Min),
Diff is Min - Min0,
length(Delta, Diff),
append(Delta, Es, Es0),
( integer(N)
-> Es = []
; Delta = []
-> true % unchanged
; update_mutable(Es+Min, State)
).
lazy_len/2 tackles the problem from two sides; the clpfd constraint part of it was shown above. The tree side uses prolog-coroutining to walk down the list as far as the partial instantiation allows1:
lazylist_acc_len(_, _, N) :-
integer(N),
!.
lazylist_acc_len(Es, N0, N) :-
var(Es),
!,
when((nonvar(N);nonvar(Es)), lazylist_acc_len(Es,N0,N)).
lazylist_acc_len([], N, N).
lazylist_acc_len([_|Es], N0, N) :-
N1 is N0+1,
N in N1..sup,
lazylist_acc_len(Es, N1, N).
Sample queries:
?- lazy_len(Xs, N).
when((nonvar(N);nonvar(Xs)), lazylist_acc_len(Xs,0,N)),
N in 0..sup,
list_FD_size(Xs+0, N).
?- lazy_len(Xs, 3).
Xs = [_A,_B,_C].
?- lazy_len([_,_], L).
L = 2.
?- lazy_len(Xs, L), L #> 0.
Xs = [_A|_B],
when((nonvar(L);nonvar(_B)), lazylist_acc_len(_B,1,L)),
L in 1..sup,
list_FD_size(_B+1, L).
?- lazy_len(Xs, L), L #> 2.
Xs = [_A,_B,_C|_D],
when((nonvar(L);nonvar(_D)), lazylist_acc_len(_D,3,L)),
L in 3..sup,
list_FD_size(_D+3, L).
?- lazy_len(Xs, L), L #> 0, L #> 2.
Xs = [_A,_B,_C|_D],
when((nonvar(L);nonvar(_D)), lazylist_acc_len(_D,3,L)),
L in 3..sup,
list_FD_size(_D+3, L).
And, at long last, one more query... well, actually two more: one going up—the other going down.
?- L in 1..4, lazy_len(Xs, L), labeling([up], [L]).
L = 1, Xs = [_A]
; L = 2, Xs = [_A,_B]
; L = 3, Xs = [_A,_B,_C]
; L = 4, Xs = [_A,_B,_C,_D].
?- L in 1..4, lazy_len(Xs, L), labeling([down], [L]).
L = 4, Xs = [_A,_B,_C,_D]
; L = 3, Xs = [_A,_B,_C]
; L = 2, Xs = [_A,_B]
; L = 1, Xs = [_A].
Footnote 1:
Here, we focus on preserving determinism (avoid the creation of choice-points) by using delayed goals.

Numbers in a list smaller than a given number

xMenores(_,[],[]).
xMenores(X,[H|T],[R|Z]) :-
xMenores(X,T,Z),
X > H,
R is H.
xMenores takes three parameters:
The first one is a number.
The second is a list of numbers.
The third is a list and is the variable that will contain the result.
The objective of the rule xMenores is obtain a list with the numbers of the list (Second parameter) that are smaller than the value on the first parameter. For example:
?- xMenores(3,[1,2,3],X).
X = [1,2]. % expected result
The problem is that xMenores returns false when X > H is false and my programming skills are almost null at prolog. So:
?- xMenores(4,[1,2,3],X).
X = [1,2,3]. % Perfect.
?- xMenores(2,[1,2,3],X).
false. % Wrong! "X = [1]" would be perfect.
I consider X > H, R is H. because I need that whenever X is bigger than H, R takes the value of H. But I don't know a control structure like an if or something in Prolog to handle this.
Please, any solution? Thanks.
Using ( if -> then ; else )
The control structure you might be looking for is ( if -> then ; else ).
Warning: you should probably swap the order of the first two arguments:
lessthan_if([], _, []).
lessthan_if([X|Xs], Y, Zs) :-
( X < Y
-> Zs = [X|Zs1]
; Zs = Zs1
),
lessthan_if(Xs, Y, Zs1).
However, if you are writing real code, you should almost certainly go with one of the predicates in library(apply), for example include/3, as suggested by #CapelliC:
?- include(>(3), [1,2,3], R).
R = [1, 2].
?- include(>(4), [1,2,3], R).
R = [1, 2, 3].
?- include(<(2), [1,2,3], R).
R = [3].
See the implementation of include/3 if you want to know how this kind of problems are solved. You will notice that lessthan/3 above is nothing but a specialization of the more general include/3 in library(apply): include/3 will reorder the arguments and use the ( if -> then ; else ).
"Declarative" solution
Alternatively, a less "procedural" and more "declarative" predicate:
lessthan_decl([], _, []).
lessthan_decl([X|Xs], Y, [X|Zs]) :- X < Y,
lessthan_decl(Xs, Y, Zs).
lessthan_decl([X|Xs], Y, Zs) :- X >= Y,
lessthan_decl(Xs, Y, Zs).
(lessthan_if/3 and lessthan_decl/3 are nearly identical to the solutions by Nicholas Carey, except for the order of arguments.)
On the downside, lessthan_decl/3 leaves behind choice points. However, it is a good starting point for a general, readable solution. We need two code transformations:
Replace the arithmetic comparisons < and >= with CLP(FD) constraints: #< and #>=;
Use a DCG rule to get rid of arguments in the definition.
You will arrive at the solution by lurker.
A different approach
The most general comparison predicate in Prolog is compare/3. A common pattern using it is to explicitly enumerate the three possible values for Order:
lessthan_compare([], _, []).
lessthan_compare([H|T], X, R) :-
compare(Order, H, X),
lessthan_compare_1(Order, H, T, X, R).
lessthan_compare_1(<, H, T, X, [H|R]) :-
lessthan_compare(T, X, R).
lessthan_compare_1(=, _, T, X, R) :-
lessthan_compare(T, X, R).
lessthan_compare_1(>, _, T, X, R) :-
lessthan_compare(T, X, R).
(Compared to any of the other solutions, this one would work with any terms, not just integers or arithmetic expressions.)
Replacing compare/3 with zcompare/3:
:- use_module(library(clpfd)).
lessthan_clpfd([], _, []).
lessthan_clpfd([H|T], X, R) :-
zcompare(ZOrder, H, X),
lessthan_clpfd_1(ZOrder, H, T, X, R).
lessthan_clpfd_1(<, H, T, X, [H|R]) :-
lessthan_clpfd(T, X, R).
lessthan_clpfd_1(=, _, T, X, R) :-
lessthan_clpfd(T, X, R).
lessthan_clpfd_1(>, _, T, X, R) :-
lessthan_clpfd(T, X, R).
This is definitely more code than any of the other solutions, but it does not leave behind unnecessary choice points:
?- lessthan_clpfd(3, [1,3,2], Xs).
Xs = [1, 2]. % no dangling choice points!
In the other cases, it behaves just as the DCG solution by lurker:
?- lessthan_clpfd(X, [1,3,2], Xs).
Xs = [1, 3, 2],
X in 4..sup ;
X = 3,
Xs = [1, 2] ;
X = 2,
Xs = [1] ;
X = 1,
Xs = [] .
?- lessthan_clpfd(X, [1,3,2], Xs), X = 3. %
X = 3,
Xs = [1, 2] ; % no error!
false.
?- lessthan_clpfd([1,3,2], X, R), R = [1, 2].
X = 3,
R = [1, 2] ;
false.
Unless you need such a general approach, include(>(X), List, Result) is good enough.
This can also be done using a DCG:
less_than([], _) --> [].
less_than([H|T], N) --> [H], { H #< N }, less_than(T, N).
less_than(L, N) --> [H], { H #>= N }, less_than(L, N).
| ?- phrase(less_than(R, 4), [1,2,3,4,5,6]).
R = [1,2,3] ? ;
You can write your predicate as:
xMenores(N, NumberList, Result) :- phrase(less_than(Result, N), NumberList).
You could write it as a one-liner using findall\3:
filter( N , Xs , Zs ) :- findall( X, ( member(X,Xs), X < N ) , Zs ) .
However, I suspect that the point of the exercise is to learn about recursion, so something like this would work:
filter( _ , [] , [] ) .
filter( N , [X|Xs] , [X|Zs] ) :- X < N , filter(N,Xs,Zs) .
filter( N , [X|Xs] , Zs ) :- X >= N , filter(N,Xs,Zs) .
It does, however, unpack the list twice on backtracking. An optimization here would be to combine the 2nd and 3rd clauses by introducing a soft cut like so:
filter( _ , [] , [] ) .
filter( N , [X|Xs] , [X|Zs] ) :-
( X < N -> Zs = [X|Z1] ; Zs = Z1 ) ,
filter(N,Xs,Zs)
.
(This is more like a comment than an answer, but too long for a comment.)
Some previous answers and comments have suggested using "if-then-else" (->)/2 or using library(apply) meta-predicate include/3. Both methods work alright, as long as only plain-old Prolog arithmetics—is/2, (>)/2, and the like—are used ...
?- X = 3, include(>(X),[1,3,2,5,4],Xs).
X = 3, Xs = [1,2].
?- include(>(X),[1,3,2,5,4],Xs), X = 3.
ERROR: >/2: Arguments are not sufficiently instantiated
% This is OK. When instantiation is insufficient, an exception is raised.
..., but when doing the seemingly benign switch from (>)/2 to (#>)/2, we lose soundness!
?- X = 3, include(#>(X),[1,3,2,5,4],Xs).
X = 3, Xs = [1,2].
?- include(#>(X),[1,3,2,5,4],Xs), X = 3.
false.
% This is BAD! Expected success with answer substitutions `X = 3, Xs = [1,2]`.
No new code is presented in this answer.
In the following we take a detailed look at different revisions of this answer by #lurker.
Revision #1, renamed to less_than_ver1//2. By using dcg and clpfd, the code is both very readable and versatile:
less_than_ver1(_, []) --> [].
less_than_ver1(N, [H|T]) --> [H], { H #< N }, less_than_ver1(N, T).
less_than_ver1(N, L) --> [H], { H #>= N }, less_than_ver1(N, L).
Let's query!
?- phrase(less_than_ver1(N,Zs),[1,2,3,4,5]).
N in 6..sup, Zs = [1,2,3,4,5]
; N = 5 , Zs = [1,2,3,4]
; N = 4 , Zs = [1,2,3]
; N = 3 , Zs = [1,2]
; N = 2 , Zs = [1]
; N in inf..1, Zs = []
; false.
?- N = 3, phrase(less_than_ver1(N,Zs),[1,2,3,4,5]).
N = 3, Zs = [1,2] % succeeds, but leaves useless choicepoint
; false.
?- phrase(less_than_ver1(N,Zs),[1,2,3,4,5]), N = 3.
N = 3, Zs = [1,2]
; false.
As a small imperfection, less_than_ver1//2 leaves some useless choicepoints.
Let's see how things went with the newer revision...
Revision #3, renamed to less_than_ver3//2:
less_than_ver3([],_) --> [].
less_than_ver3(L,N) --> [X], { X #< N -> L=[X|T] ; L=T }, less_than_ver3(L,N).
This code uses the if-then-else ((->)/2 + (;)/2) in order to improve determinism.
Let's simply re-run the above queries!
?- phrase(less_than_ver3(Zs,N),[1,2,3,4,5]).
N in 6..sup, Zs = [1,2,3,4,5]
; false. % all other solutions are missing!
?- N = 3, phrase(less_than_ver3(Zs,N),[1,2,3,4,5]).
N = 3, Zs = [1,2] % works as before, but no better.
; false. % we still got the useless choicepoint
?- phrase(less_than_ver3(Zs,N),[1,2,3,4,5]), N = 3.
false. % no solution!
% we got one with revision #1!
Surprise! Two cases that worked before are now (somewhat) broken, and the determinism in the ground case is no better... Why?
The vanilla if-then-else often cuts too much too soon, which is particularly problematic with code which uses coroutining and/or constraints.
Note that (*->)/2 (a.k.a. "soft-cut" or if/3), fares only a bit better, not a lot!
As if_/3 never ever cuts more (often than) the vanilla if-then-else (->)/2, it cannot be used in above code to improve determinism.
If you want to use if_/3 in combination with constraints, take a step back and write code that is non-dcg as the first shot.
If you're lazy like me, consider using a meta-predicate like tfilter/3 and (#>)/3.
This answer by #Boris presented a logically pure solution which utilizes clpfd:zcompare/3 to help improve determinism in certain (ground) cases.
In this answer we will explore different ways of coding logically pure Prolog while trying to avoid the creation of useless choicepoints.
Let's get started with zcompare/3 and (#<)/3!
zcompare/3 implements three-way comparison of finite domain variables and reifies the trichotomy into one of <, =, or >.
As the inclusion criterion used by the OP was a arithmetic less-than test, we propose using
(#<)/3 for reifying the dichotomy into one of true or false.
Consider the answers of the following queries:
?- zcompare(Ord,1,5), #<(1,5,B).
Ord = (<), B = true.
?- zcompare(Ord,5,5), #<(5,5,B).
Ord = (=), B = false.
?- zcompare(Ord,9,5), #<(9,5,B).
Ord = (>), B = false.
Note that for all items to be selected both Ord = (<) and B = true holds.
Here's a side-by-side comparison of three non-dcg solutions based on clpfd:
The left one uses zcompare/3 and first-argument indexing on the three cases <, =, and >.
The middle one uses (#<)/3 and first-argument indexing on the two cases true and false.
The right one uses (#<)/3 in combination with if_/3.
Note that we do not need to define auxiliary predicates in the right column!
less_than([],[],_). % less_than([],[],_). % less_than([],[],_).
less_than([Z|Zs],Ls,X) :- % less_than([Z|Zs],Ls,X) :- % less_than([Z|Zs],Ls,X) :-
zcompare(Ord,Z,X), % #<(Z,X,B), % if_(Z #< X,
ord_lt_(Ord,Z,Ls,Rs), % incl_lt_(B,Z,Ls,Rs), % Ls = [Z|Rs],
less_than(Zs,Rs,X). % less_than(Zs,Rs,X). % Ls = Rs),
% % less_than(Zs,Rs,X).
ord_lt_(<,Z,[Z|Ls],Ls). % incl_lt_(true ,Z,[Z|Ls],Ls). %
ord_lt_(=,_, Ls ,Ls). % incl_lt_(false,_, Ls ,Ls). %
ord_lt_(>,_, Ls ,Ls). % %
Next, let's use dcg!
In the right column we use if_//3 instead of if_/3.
Note the different argument orders of dcg and non-dcg solutions: less_than([1,2,3],Zs,3) vs phrase(less_than([1,2,3],3),Zs).
The following dcg implementations correspond to above non-dcg codes:
less_than([],_) --> []. % less_than([],_) --> []. % less_than([],_) --> [].
less_than([Z|Zs],X) --> % less_than([Z|Zs],X) --> % less_than([Z|Zs],X) -->
{ zcompare(Ord,Z,X) }, % { #<(Z,X,B) }, % if_(Z #< X,[Z],[]),
ord_lt_(Ord,Z), % incl_lt_(B,Z), % less_than(Zs,X).
less_than(Zs,X). % less_than(Zs,X). %
% %
ord_lt_(<,Z) --> [Z]. % incl_lt_(true ,Z) --> [Z]. %
ord_lt_(=,_) --> []. % incl_lt_(false,_) --> []. %
ord_lt_(>,_) --> []. % %
OK! Saving the best for last... Simply use meta-predicate tfilter/3 together with (#>)/3!
less_than(Xs,Zs,P) :-
tfilter(#>(P),Xs,Zs).
The dcg variant in this previous answer is our starting point.
Consider the auxiliary non-terminal ord_lt_//2:
ord_lt_(<,Z) --> [Z].
ord_lt_(=,_) --> [].
ord_lt_(>,_) --> [].
These three clauses can be covered using two conditions:
Ord = (<): the item should be included.
dif(Ord, (<)): it should not be included.
We can express this "either-or choice" using if_//3:
less_than([],_) --> [].
less_than([Z|Zs],X) -->
{ zcompare(Ord,Z,X) },
if_(Ord = (<), [Z], []),
less_than(Zs,X).
Thus ord_lt_//2 becomes redundant.
Net gain? 3 lines-of-code !-)

Create list of pair of values 0 to n-1 in lexicographical order

I'm doing a program with Result is a pair of values [X,Y] between 0 and N-1 in lexicographic order
I have this right now:
pairs(N,R) :-
pairsHelp(N,R,0,0).
pairsHelp(N,[],N,N) :- !.
pairsHelp(N,[],N,0) :- !.
pairsHelp(N,[[X,Y]|List],X,Y) :-
Y is N-1,
X < N,
X1 is X + 1,
pairsHelp(N,List,X1,0).
pairsHelp(N,[[X,Y]|List],X,Y) :-
Y < N,
Y1 is Y + 1,
pairsHelp(N,List,X,Y1).
I'm getting what I want the first iteration but Prolog keeps going and then gives me a second answer.
?-pairs(2,R).
R = [[0,0],[0,1],[1,0],[1,1]] ;
false.
I don't want the second answer (false), just the first. I want it to stop after it finds the answer. Any ideas?
Keep in mind that there is a much easier way to get what you are after. If indeed both X and Y are supposed to be integers, use between/3 to enumerate integers ("lexicographical" here is the same as the order of natural numbers: 0, 1, 2, .... This is the order in which between/3 will enumerate possible solutions if the third argument is a variable):
pairs(N, R) :-
succ(N0, N),
bagof(P, pair(N0, P), R).
pair(N0, X-Y) :-
between(0, N0, X),
between(0, N0, Y).
And then:
?- pairs(2, R).
R = [0-0, 0-1, 1-0, 1-1].
?- pairs(3, R).
R = [0-0, 0-1, 0-2, 1-0, 1-1, 1-2, 2-0, 2-1, ... - ...].
I am using the conventional Prolog way of representing a pair, X-Y (in canonical form: -(X, Y)) instead of [X,Y] (canonical form: .(X, .(Y, []))).
The good thing about this program is that you can easily re-write it to work with another "alphabet" of your choosing.
?- between(0, Upper, X).
is semantically equivalent to:
x(0).
x(1).
% ...
x(Upper).
?- x(X).
For example, if we had an alphabet that consists of b, a, and c (in that order!):
foo(b).
foo(a).
foo(c).
foo_pairs(Ps) :-
bagof(X-Y, ( foo(X), foo(Y) ), Ps).
and then:
?- foo_pairs(R).
R = [b-b, b-a, b-c, a-b, a-a, a-c, c-b, c-a, ... - ...].
The order of the clauses of foo/1 defines the order of your alphabet. The conjunction foo(X), foo(Y) together with the order of X-Y in the pair defines the order of pairs in the list. Try writing for example bagof(X-Y, ( foo(Y), foo(X) ), Ps) to see what will be the order of pairs in Ps.
Use dcg in combination with lambda!
?- use_module(library(lambda)).
In combination with meta-predicate init0/3 and
xproduct//2 ("cross product") simply write:
?- init0(=,3,Xs), phrase(xproduct(\X^Y^phrase([X-Y]),Xs),Pss).
Xs = [0,1,2], Pss = [0-0,0-1,0-2,1-0,1-1,1-2,2-0,2-1,2-2].
How about something a little more general? What about other values of N?
?- init0(=,N,Xs), phrase(xproduct(\X^Y^phrase([X-Y]),Xs),Pss).
N = 0, Xs = [], Pss = []
; N = 1, Xs = [0], Pss = [0-0]
; N = 2, Xs = [0,1], Pss = [0-0,0-1,
1-0,1-1]
; N = 3, Xs = [0,1,2], Pss = [0-0,0-1,0-2,
1-0,1-1,1-2,
2-0,2-1,2-2]
; N = 4, Xs = [0,1,2,3], Pss = [0-0,0-1,0-2,0-3,
1-0,1-1,1-2,1-3,
2-0,2-1,2-2,2-3,
3-0,3-1,3-2,3-3]
; N = 5, Xs = [0,1,2,3,4], Pss = [0-0,0-1,0-2,0-3,0-4,
1-0,1-1,1-2,1-3,1-4,
2-0,2-1,2-2,2-3,2-4,
3-0,3-1,3-2,3-3,3-4,
4-0,4-1,4-2,4-3,4-4]
...
Does it work for other terms, too? What about order? Consider a case #Boris used in his answer:
?- phrase(xproduct(\X^Y^phrase([X-Y]),[b,a,c]),Pss).
Pss = [b-b,b-a,b-c,a-b,a-a,a-c,c-b,c-a,c-c]. % succeeds deterministically

Segregating Lists in Prolog

I am having a really hard time understanding how to get my code to show my segregated lists consisting of even and odd numbers. I am not even sure what my understanding is lacking. I am new to this language obviously and must use it for school. My imperative and functional mind won't let me know what the hell is going on with this lol.
Now, no I am not asking you to do my homework! I am simply asking you to help me see what my lack of understanding is. I have also looked up similar answers but I cannot convert them to the way I am supposed to write this function.
Please, once more, do not bash me for this like I have previously usually been bashed. Please just help me see what my understanding is lacking. Do not just give me answers and code snippets without explaining it please.
Here it is:
is_even(H) :-
0 is mod(H, 2).
segregate(List, Even, Odd) :- segregator(List, Even, Odd).
segregator([], [], []).
segregator([H|T], E, O) :-
is_even(H),
% I feel here is where I am supposed to build the list,
% but I have no clue how since Even or Odd has not been unified.
segregator(T, E, O),
write('Even is '), write(E), nl.
segregator([H|T], E, O) :-
% Same here as above.
segregator(T, E, O),
write('Odd is '), write(O), nl.
A logically pure implementation is very straight-forward, thanks to clpfd:
:- use_module(library(clpfd)).
list_evens_odds([],[],[]).
list_evens_odds([X|Xs],[X|Es],Os) :-
X mod 2 #= 0,
list_evens_odds(Xs,Es,Os).
list_evens_odds([X|Xs],Es,[X|Os]) :-
X mod 2 #= 1,
list_evens_odds(Xs,Es,Os).
Some sample queries we expect to succeed (with a finite sequence of answers):
?- Xs = [1,2,3,4,5,6,7], list_evens_odds(Xs,Es,Os).
Xs = [1,2,3,4,5,6,7],
Es = [ 2, 4, 6 ],
Os = [1, 3, 5, 7] ;
false.
?- list_evens_odds(Ls,[2,4],[1,3]).
Ls = [2,4,1,3] ? ;
Ls = [2,1,4,3] ? ;
Ls = [2,1,3,4] ? ;
Ls = [1,2,4,3] ? ;
Ls = [1,2,3,4] ? ;
Ls = [1,3,2,4] ? ;
no
What about queries we expect to fail?
?- list_evens_odds(Ls,[2,4,5],[1,3]).
no
?- list_evens_odds(Ls,[2,4],[1,3,6]).
no
?- list_evens_odds([_,_,_],[2,4],[1,3]).
no
At last, the most general query:
?- assert(clpfd:full_answer).
yes
?- list_evens_odds(Ls,Es,Os).
Ls = [], Es = [], Os = [] ? ;
Ls = [_A], Es = [_A], Os = [], _A mod 2#=0, _A in inf..sup ? ...
Edit 2015-05-06
Here's another way to do it with logical-purity!
Use the meta-predicate tpartition/4 together with zeven_t/2 or zodd_t/2.
bool01_t(1,true).
bool01_t(0,false).
zeven_t(Z,Truth) :- Z mod 2 #= 0 #<==> B, bool01_t(B,Truth).
%zodd_t(Z,Truth) :- Z mod 2 #= 1 #<==> B, bool01_t(B,Truth).
zodd_t(Z,Truth) :- Z mod 2 #= B, bool01_t(B,Truth). % tweaked
zeven_t/2 reifies the evenness of an integer, zodd_t/2 the oddness.
With everything in place, let's run some queries!
?- tpartition(zeven_t,[1,2,3,4,5,6,7],Es,Os).
Es = [2,4,6], Os = [1,3,5,7].
?- tpartition(zodd_t ,[1,2,3,4,5,6,7],Os,Es). % argument order differs
Es = [2,4,6], Os = [1,3,5,7].
Both succeed deterministically. The equivalent query using list_evens_odds/3 does not.