Related
This is what I have:
values(Count, A, B) :-
A is Count,
B is 0.
values(Count, A, B) :-
values(Count, Aa, Bb),
A is Aa-2,
B is Bb+2,
\+ A < 0;
B < 0.
So I want by output to be:
A = 6,
B = 0
A = 4,
B = 2
A = 2,
B = 4
A = 0,
B = 6
I'm getting that, but then the interpreter just keeps going and runs out of Stack space, because the recursive values(Count, Aa, Bb) is at the start. I don't know how to rewrite this so that the recursion isn't endless, I want it to end after I get the above output. Would anyone be able to help?
I would do it like this:
val(A,A,0):-
A>=0.
val(C,A,B):-
CC is C-2,
CC >=0,
val(CC,A,BB),
B is BB+2.
The output is
?- val(6, A, B).
A = 6,
B = 0 ;
A = 4,
B = 2 ;
A = 2,
B = 4 ;
A = 0,
B = 6 ;
false.
So what is the difference? I use the Count variable as counter. It has to decrease in every step to force the recursion to end. The downside is I have to make sure the counter is never less than zero.
It might be easier if you use between/3. Do you know that the argument is a positive even number?
p(To0, A, B) :-
To is To0 div 2,
between(0, To, X),
A is (To - X) * 2,
B is X * 2.
?- p(6, A, B).
A = 6,
B = 0 ;
A = 4,
B = 2 ;
A = 2,
B = 4 ;
A = 0,
B = 6.
I trying to create compact code in prolog that sovles systems of equations.
For example, in this case, the assumptions must be
A+B-C-D=4, A+B+C+D=14, A-B+C-D=2.
I'm trying to have it where it solves all combinations possible for A, B, C, and D but satisfies all 3 equations. They can only be #'s 0-9 though but somehow show all possible solutions/combinations.
So after running the query, it would output something like
Crypto(A,B,C,D)
A = 8, B = 1, C = 0, D = 5.
^That would be one solution. But I need to show all possible.
I'm kind of lost as to how to satisfy all 3 in Prolog. Thank you.
You can solve it by taking out one element from the domain of the variables and assign it to them such that every variable has a different number assigned to it. It's a brute force method.
takeout(X, [X|R], R).
takeout(X, [Y|Xs], [Y|Ys]):- takeout(X, Xs, Ys).
aas(X,L,L1):- takeout(X,L,L1).
crypto(A,B,C,D):-
L=[0,1,2,3,4,5,6,7,8,9],
aas(A,L,L1),aas(B,L1,L2),aas(C,L2,L3),aas(D,L3,_),
A+B-C-D=:=4,
A+B+C+D=:=14,
A-B+C-D=:=2,
nl.
aas(X,L,L1). used for assigning values to the variables.
takeout function is used for taking out one element and return a list excluding the element taken out.
OUTPUT
?- crypto(A,B,C,D).
A = 3,
B = 6,
C = 5,
D = 0
A = 5,
B = 4,
C = 3,
D = 2
A = 7,
B = 2,
C = 1,
D = 4
A = 8,
B = 1,
C = 0,
D = 5
This program prints all the possible solutions to that equation A+B-C-D=4, A+B+C+D=14, A-B+C-D=2. Hope this answers your question.
Let's say that I want to find two numbers where the sum of these are 8, are from 1-9 and must be different(it is obvious that these numbers are (7,1),(6,2),etc).So I wrote.
dif_list([H|T]):- \+ member(H,T),dif_list(T).
dif_list([]).
check1_9([H|T]):-H>=1,H=<9,check1_9(T).
check1_9([]).
find_number([A,B],N) :- N =:= A+B,dif_list([A,B]),check1_9([A,B]).
Afterwards I will ask prolog
?-find_number([A,B],8).
ERROR: =:=/2: Arguments are not sufficiently instantiated
My goal is that prolog will print for me the results.For example:
?-find_number([A,B],8).
A = 7,
B = 1 ;
A = 6,
B = 2 ;
...
The best way to handle this kind of problem in Prolog is to use the CLP(FD) library:
:- [library(clpfd)].
sum_of(A, B, Sum) :-
A #> 0,
B #> 0,
A + B #= Sum.
?- sum_of(A, B, 8), label([A, B]).
A = 1,
B = 7 ;
A = 2,
B = 6 ;
A = 3,
B = 5 ;
A = B, B = 4 ;
A = 5,
B = 3 ;
A = 6,
B = 2 ;
A = 7,
B = 1.
?-
If you want the addends to be unique, you can further constrain it:
sum_of(A, B, Sum) :-
A #> 0,
B #>= A,
A + B #= Sum.
There's really no need to use a list to manage the variables A and B, but you can if you wish: sum_of([A,B], Sum).
Prolog is not that declarative: there are indeed answer set programming (ASP) or constraint logic programming (clp) languages where you can simply define a set of constraints and a finite domain solver aims to solve it (but these will take considerable time).
I would suggest that you define your program as follows:
find_number(A,B,N) :-
member(A,[1,2,3,4,5,6,7,8,9]),
member(B,[1,2,3,4,5,6,7,8,9]),
N is A+B,
A \= B.
Here member/2 will instantiate A and B to values that are provided by the list, so 1..9, next you use is/2 to calculate the sum and verify that the sum is equal to N. You can only call N is A+B if A and B are given a proper value. Finally we say A \= B (A is not equal to B).
When you run this predicate, it produces:
?- find_number(A,B,8).
A = 1,
B = 7 ;
A = 2,
B = 6 ;
A = 3,
B = 5 ;
A = 5,
B = 3 ;
A = 6,
B = 2 ;
A = 7,
B = 1 ;
false.
You can however also query with A and B already filled in, or one of them filled in, or where the sum is left open. So:
?- find_number(A,2,8).
A = 6 ;
false.
or:
?- find_number(A,2,N).
A = 1,
N = 3 ;
A = 3,
N = 5 ;
A = 4,
N = 6 ;
A = 5,
N = 7 ;
A = 6,
N = 8 ;
A = 7,
N = 9 ;
A = 8,
N = 10 ;
A = 9,
N = 11 ;
false.
I'm working with SWI-Prolog to get clpfd generate the list of all distinct integers between 5 and 10:
q1(Answer) :-
length(Xs, Answer),
Xs ins 0..20,
chain(Xs, #<),
maplist(q1constraints, Xs).
q1constraints(X) :-
X #>= 5,
X #=< 10.
Kind of works, but generates a solution for each of the lengths 0, 1, ... 6 and then hangs seeking a solution of length 7:
?- q1(Answer).
Answer = 0 ;
Answer = 1 ;
Answer = 2 ;
Answer = 3 ;
Answer = 4 ;
Answer = 5 ;
Answer = 6 ;
<hangs>
Is there a good way to generate the list of all integers that satisfy the desired constraints?
Your question is not that clear to me. But I will try:
?- length(Xs,6), Xs ins 5..10, chain(Xs,#<), Xs = [5|_], last(Xs, 10).
Xs = [5,6,7,8,9,10].
Note that with these elements, it is necessary to fix the length of the list first. Otherwise:
?- length(Xs,N), Xs ins 5..10, chain(Xs,#<), Xs = [5|_], last(Xs, 10).
Xs = [5,10], N = 2
; Xs = [5,_A,10], N = 3, _A in 6..9
; Xs = [5,_A,_B,10], N = 4, _A#=<_B+ -1, _A in 6..8, _B in 7..9
; Xs = [5,_C,_A,_B,10], N = 5, _A#=<_B+ -1, _C#=<_A+ -1, _A in 7..8, _C in 6..7, _B in 8..9
; Xs = [5,6,7,8,9,10], N = 6
; loops.
In fact, even the (ins)/2 is not needed:
?- length(Xs,6), chain(Xs,#<), Xs = [5|_], last(Xs, 10).
Xs = [5,6,7,8,9,10].
(In newer versions of clpfd called clpz the argument order of chain/2` is reversed adhering to the common argument ordering.)
Here is what your program does:
Generate a list of increasing lengths, starting with an empty list
For each element X in the list, pose a constraint that X is in [0, 20]
For the whole list, pose a constraint that values are strictly increasing in magnitude
For each element in the list, pose an additional constraint that X is in [5, 10].
You then ask for the length of the generated list.
There are 6 values that are in [0, 20] and in [5,10]: 5, 6, 7, 8, 9, 10. For the empty list you generate first, there are no constrained variables; for the list with 1 variable, there would be 6 possible values of the variable, but you don't ask for these values, only for the length of the list; for the list with 2 variables, you will have 5 possible combinations: {5,6}, {6,7}, ..., {9,10}, but again, you don't ask for them, just for the length of the list.
Eventually, you get to list with 7 values. Since there are only 6 values that each element could have, there are no solutions.
So what is your goal here? Maybe you should try and explain better. To get all values between 5 and 10 by backtracking, you could say: between(5, 10, X), or, with CLPFD, X in 5..10, label([X]). If it is neither of these, you need to re-write your question.
If you want the total number of even integers in 5..10 (SPOILER: there are 3 of them!):
?- aggregate(count, X^(X in 5..10, X mod 2 #= 0, indomain(X)), Answer).
Answer = 3.
Breaking it down:
X in 5..10, X mod 2 #= 0 just constrains X to be an even integer between 5 and 10:
?- X in 5..10, X mod 2 #= 0.
X in 6..10,
X mod 2#=0.
indomain(X) does the actual search, succeeding once for each feasible value of X:
?- X in 5..10, X mod 2 #= 0, indomain(X).
X = 6 ;
X = 8 ;
X = 10.
X^(...) existentially quantifies X within the parentheses, limiting its scope. If we instead leave it as a free variable, aggregation will respect it:
?- aggregate(count, (X in 5..10, X mod 2 #= 0, indomain(X)), Answer).
X = 6,
Answer = 1 ;
X = 8,
Answer = 1 ;
X = 10,
Answer = 1.
How to define a as a integer/float number ?
I want to find the results of a+b+c+d=10 where a,b,c,d is integer and >=0.
Here is a simple, modern, pure Prolog, non-CLP-library solution:
range(X):-
member(X,[0,1,2,3,4,5,6,7,8,9,10]).
ten(A,B,C,D):-
range(A),
range(B),
range(C),
range(D),
10 =:= A + B + C + D.
with SWI-Prolog you can use CLP(FD) library
1 ?- use_module(library(clpfd)).
% library(error) compiled into error 0.00 sec, 9,764 bytes
% library(clpfd) compiled into clpfd 0.05 sec, 227,496 bytes
true.
2 ?- Vars=[A,B,C,D],Vars ins 0..10,sum(Vars,#=,10),label(Vars).
Vars = [0, 0, 0, 10],
A = 0,
B = 0,
C = 0,
D = 10 ;
Vars = [0, 0, 1, 9],
A = 0,
B = 0,
C = 1,
D = 9 ;
Vars = [0, 0, 2, 8],
A = 0,
B = 0,
C = 2,
D = 8 ;
Vars = [0, 0, 3, 7],
A = 0,
B = 0,
C = 3,
D = 7 ;
...
Here is GNU-Prolog piece of code with constraint solving over finite domains :
$ gprolog
| ?- [user].
compiling user for byte code...
ten(A,B,C,D) :- fd_domain([A,B,C,D],0,9999999), 10 #= A + B + C + D.
Ctrl + D
| ?- ten(A,B,C,D), fd_labeling([A,B,C,D]).
As you can see, it solves problem of big ranges like 0-9999999
A = 0
B = 0
C = 0
D = 10 ? ;
A = 0
B = 0
C = 1
D = 9 ? ;
A = 0
B = 0
C = 2
D = 8 ? ;
...
P.S. Thanks for Przemysław Kobylański for his blog with clear, very nice Prolog examples, where I've found inspiring examples.
P.P.S. When playing with finite domains, you might like to use fd_set_vector_max/1 . In above case it's not needed, but depending on constraint might be usefull - more details when Gnu-Prolog operates on ranges, when on vectors of possible values, can be found at manual "Finite domain solver and built-in predicates - Introduction"