How to debug non-termination - prolog

I'm trying to implement n_factors/2 predicate that works in all directions.
:- use_module(library(clpz)).
n_factors(N, Fs) :-
integer(N),
N > 1,
primes(Ps),
n_factors0(N, Fs, Ps),
!.
n_factors(N, Fs) :-
var(N),
primes(Ps),
N #> 1,
above(2, N),
n_factors0(N, Fs, Ps).
above(I, I).
above(I, N) :- I1 is I + 1, above(I1, N).
n_factors0(N, [F|Fs], [P|Ps]) :-
N #> 1,
F #=< N,
P #=< N,
( P * P #> N ->
F = N, Fs = []
; ( N #= N1 * P ->
F #= P, n_factors0(N1, Fs, [P|Ps])
; F #> P, n_factors0(N, [F|Fs], Ps)
)
).
When I am issuing the following query I get:
?- C #> 6, C #< 12, n_factors(A, [B,C]).
C = 7, A = 14, B = 2
; C = 7, A = 21, B = 3
; C = 11, A = 22, B = 2
; C = 11, A = 33, B = 3
; C = 7, A = 35, B = 5
; C = 7, A = 49, B = 7
; C = 11, A = 55, B = 5
; C = 11, A = 77, B = 7
; C = 11, A = 121, B = 11
;
before the program moves on to exploring the realm of rather large numbers. So the question I've go is the following: knowing for certain that the mathematical problem is constraint enough to terminate, how do I find the missing constraint in my program? What I am doing right now is staring at the screen before trying to add "invariant" conditions here and there and see if they help.
primes(Ps) is a "frozen" infinite list with all prime numbers. I don't think the implementation thereof is important for this question but just in case
primes(Ps) :-
Ps = [2,3|T],
primes0(5, Ps, Ps, T),
!.
primes0(C, [D|Ds], Ps, T) :-
( D * D > C ->
T = [C|T1], C1 is C + 2, freeze(T1, primes0(C1, Ps, Ps, T1))
; ( C mod D =:= 0 ->
C1 is C + 2, primes0(C1, Ps, Ps, T)
; primes0(C, Ds, Ps, T)
)
).

Related

Finding the minimum value out of all the outcomes of a predicate

I have the following predicates:
slow(a , 10 , b).
slow(b , 17 , c).
slow(c , 12 , d).
slow(d , 19 , e).
slow(e , 13 , f).
fast(a , 20 , c).
fast(b , 20 , d).
fast(d , 20 , f).
line(X,T,Y) :- slow(X,T,Y).
line(X,T,Y) :- fast(X,T,Y).
journey(X,N,Z) :-
line(X,L,Y),
journey(Y,M,Z),
N is L+M.
journey(X,0,X) :-
!.
I now need to create a new predicate of 2arity to find the minimum value of N defined in the journey predicate.
I've so far defined the new predicate as:
minimum(X,Y) :-
journey(X,N,Y),
write(N).
I've tried to use findall/3 and bagof/3 to make a list and find the minimum atom of that list, but I am not sure how to use them.
I'm not sure how else to go about it, so any advice would be greatly appreciated.
Thanks
(First, remove the !)
?- journey(A,N,B), false.
false. % your definition always terminates, good
?- journey(A,N,B), \+ ( journey(A,M,B), M < N ).
A = a, N = 49, B = e
; A = a, N = 50, B = f
; A = a, N = 30, B = d
; A = a, N = 10, B = b
; A = b, N = 17, B = c
; A = c, N = 31, B = e
; ... .
?- setof(N,journey(A,N,B),[N|_]).
N = 0, A = B
; N = 10, A = a, B = b
; N = 20, A = a, B = c
; N = 30, A = a, B = d
; N = 49, A = a, B = e
; N = 50, A = a, B = f
; N = 17, A = b, B = c
; ... .
?- setof(N,journey(A,N,B),[N|_]), journey(A,N,B).
... . % same as above but in general safer

Efficient Fibonacci in Prolog

I am trying to implement a Fibonacci predicate that can be efficiently used with CLP.
:- module(fibonacci, [fibonacci/2]).
fibonacci(N, F) :-
( var(N) ; integer(N) ),
( var(F) ; integer(F) ),
( var(F) ->
( integer(N) ->
fib_1(N, F), !
; fib_3(0, N, F)
)
; ( integer(N) ->
fib_1(N, F0), F0 = F, !
; fib_2(0, F, N0), N0 = N, !
)
).
fib_3(I, J, F) :-
( I = J, fib_1(I, F) ) ;
( I1 is I + 1, fib_3(I1, J, F) ).
fib_2(I, F, J) :-
fib_1(I, F0),
( F = F0 ->
J = I, !
; ( F0 > F -> !, fail
; I1 is I + 1,
fib_2(I1, F, J)
)
).
fib_1(0, 0).
fib_1(1, 1).
fib_1(2, 1).
fib_1(N, F) :-
var(F),
N > 2,
( N mod 2 =:= 0 ->
N0 is div(N, 2),
N1 is N0 + 1,
fib_1(N0, F0),
fib_1(N1, F1),
F is F0 * (2 * F1 - F0)
; N0 is div(N + 1, 2),
N1 is N0 - 1,
fib_1(N0, F0),
fib_1(N1, F1),
F is F0 * F0 + F1 * F1
).
This is not the prettiest code, but it does what I want it to do.
?- fibonacci(A, 10).
false.
?- fibonacci(A, 13).
A = 7.
?- fibonacci(12, A).
A = 144.
?- fibonacci(12, 144).
true.
?- fibonacci(12, 145).
false.
?- fibonacci(A, B).
A = B, B = 0 ;
A = B, B = 1 ;
A = 2,
B = 1 ;
A = 3,
B = 2 ;
A = 4,
B = 3 ;
A = B, B = 5 .
What's the magic potion that is missing for this query to work:
fibonacci(_, B), B #< 1000
Is it rectifiable at all, or is CLP a completely different beast altogether, and every predicate that is CLP-compatible needs to understand more than just integers and vars?
You should avoid using ! within an algorithm that uses clp(FD) as they don't mix well. Also if-then-else may backfire too. I'd also keep an eye on using var/1 within an algorithm that uses clp.
Here goes a solution that uses clp(FD) and accumulators to avoid double recursion:
fibonacci(0, 0).
fibonacci(1, 1).
fibonacci(N, F):-
N #> 1,
zcompare(C, 2, N),
fibonacci(C, 2, N, 0, 1, F).
fibonacci(=, N, N, F1, F2, F):-
F #= F1+F2.
fibonacci(<, N0, N, F1, F2, F):-
N1 #= N0+1,
F3 #= F1+F2,
F #> F3,
zcompare(C, N1, N),
fibonacci(C, N1, N, F2, F3, F).
Also for the test you should issue the constraint over the expected number before calling fibonacci/2. So instead of fibonacci(_, B), B #< 1000. use B #< 1000, fibonacci(_, B).
sample runs:
?- fibonacci(10, F).
F = 55.
?- B #< 1000, fibonacci(_, B).
B = 0 ;
B = 1 ;
B = 1 ;
B = 2 ;
B = 3 ;
B = 5 ;
B = 8 ;
B = 13 ;
B = 21 ;
B = 34 ;
B = 55 ;
B = 89 ;
B = 144 ;
B = 233 ;
B = 377 ;
B = 610 ;
B = 987 ;
false.

How can labeling/2 generate solutions starting from the midpoint of a domain?

Having a list with independent variables, whose domain is 1..N, how can we use labeling/2 so it starts producing solutions starting from the middle?
The flags i tried are [bisect], [enum], [max], [min], [ff], but no matter which i picked, i can't make it work.
My code is:
:-use_module(library(clpfd)).
combos(EMPLOYEES,POSTS,LIST):-
LIMIT is POSTS-EMPLOYEES+1,
length(LIST,EMPLOYEES),
LIST ins 1..LIMIT,
sum(LIST,#=,POSTS),
labeling([bisect],LIST).
after setting a query, for example:
?-combos(2,10,LIST).
i want it to return:
L = [5,5];
L = [4,6];
L = [6,4] ...
instead of:
L = [1,9];
L = [2,8];
L = [3,7] ...
As a rule of thumb, whenever you try to extend the functionality of clpfd, try to reuse as much as possible. It seems that you want solutions first whose sum of distances to the center is as small as possible.
combos2(EMPLOYEES,POSTS,LIST):-
LIMIT is POSTS-EMPLOYEES+1,
length(LIST,EMPLOYEES),
LIST ins 1..LIMIT,
sum(LIST,#=,POSTS),
Mid is (LIMIT+1) div 2, %%
maplist(dist(Mid), LIST, DISTS), %%
sum(DISTS,#=,Totaldist), %%
labeling([],[Totaldist|LIST]).
dist(Mid, E, D) :-
D #= abs(Mid-E).
?- combos2(2,10,L).
L = [5,5]
; L = [4,6]
; L = [6,4]
; L = [3,7]
; L = [7,3]
; ... .
Here you go!
combos(2,S,L) :- b2(S,L).
combos(C,S,[A|L]) :-
C > 2,
b2(S,[A,B]),
D is C-1,
combos(D,B,L).
b2(S,L) :- B is S-1, bisector(B,L).
bisector(Y,[A,B]) :-
odd(Y),
M is div(1+Y,2),
Z is M-1,
range(D,0,Z),
bisec1(D,M,A,B).
bisector(Y,[A,B]) :-
even(Y),
M is 1+Y,
Z is Y/2-1,
range(D,0,Z),
bisec2(D,M,A,B).
bisec1(0,M,M,M).
bisec1(D,M,A,B) :- D > 0, A is M + D, A > 0, B is M - D, B > 0.
bisec1(D,M,A,B) :- D > 0, A is M - D, A > 0, B is M + D, B > 0.
bisec2(D,M,A,B) :- A is (M+2*D+1)/2, A > 0, B is (M-2*D-1)/2, B > 0.
bisec2(D,M,A,B) :- A is (M-2*D-1)/2, A > 0, B is (M+2*D+1)/2, B > 0.
even(X) :- 0 is mod(X, 2).
odd(X) :- 1 is mod(X, 2).
range(M,M,_).
range(X,M,N) :- P is M + 1, P =< N, range(X,P,N).

Double factorial swi-prolog

help please I need to find ((2n-1)!!) \ (2n)!! but I only know how to write factorial not double.
factorial(0,1).
factorial(N,F) :-
N>0,
N1 is N-1,
factorial(N1,F1),
F is N * F1.
If you look up the Wikipedia entry on double factorial you can easily spot the two candidates for base cases: 0!! = 1 and 1!! = 1. Depending on the number n being odd or even you'll arrive at one of those base cases when calculating n!! recursively because, in the recursive rule, n is decreased by 2 instead of 1 (compared to n!). You can express that in Prolog like so:
doublefactorial(0,1).
doublefactorial(1,1).
doublefactorial(N,F) :-
N > 1,
N2 is N-2,
doublefactorial(N2,F1),
F is N * F1.
If you query this predicate you'll get the desired result:
?- doublefactorial(0,F).
F = 1 ;
false.
?- doublefactorial(1,F).
F = 1 ;
false.
?- doublefactorial(2,F).
F = 2 ;
false.
?- doublefactorial(3,F).
F = 3 ;
false.
?- doublefactorial(4,F).
F = 8 ;
false.
?- doublefactorial(5,F).
F = 15 ;
false.
.
.
.
However, due to the use of >/2 and is/2, this predicate can only be used in one direction:
?- doublefactorial(N,10395).
ERROR: >/2: Arguments are not sufficiently instantiated
?- doublefactorial(N,F).
N = 0,
F = 1 ;
N = F, F = 1 ;
ERROR: >/2: Arguments are not sufficiently instantiated
If you'd like to use the predicate in all directions you could opt to use CLP(FD):
:- use_module(library(clpfd)).
doublefactorial2(0,1).
doublefactorial2(1,1).
doublefactorial2(N,F) :-
N #> 1,
N2 #= N-2,
F #= N * F1,
doublefactorial2(N2,F1).
?- doublefactorial2(N,10395).
N = 11 ;
false.
?- doublefactorial2(N,46080).
N = 12 ;
false.
?- doublefactorial2(N,F).
N = 0,
F = 1 ;
N = F, F = 1 ;
N = F, F = 2 ;
N = F, F = 3 ;
N = 4,
F = 8 ;
N = 5,
F = 15 ;
N = 6,
F = 48 ;
N = 7,
F = 105 ;
N = 8,
F = 384 ;
N = 9,
F = 945 ;
.
.
.
Note how the goal F #= N * F1 can now be put in front of the recursive goal, thus making the predicate tail-recursive, because that equation is now propagated as a constraint. For more information see the SWI-Prolog documentation on CLP(FD).
To calculate (2n)!!/(2n-1)!! or (2n-1)!!/(2n)!! for a particular n you can query the the predicate like this:
?- N=5, X #= 2*N, doublefactorial2(X,F1), Y #= 2*N-1, doublefactorial2(Y,F2), RESULT is F1/F2.
N = 5,
X = 10,
F1 = 3840,
Y = 9,
F2 = 945,
RESULT = 4.063492063492063 ;
false.
?- N=5, X #= 2*N, doublefactorial2(X,F1), Y #= 2*N-1, doublefactorial2(Y,F2), RESULT is F2/F1.
N = 5,
X = 10,
F1 = 3840,
Y = 9,
F2 = 945,
RESULT = 0.24609375 ;
false.
Note that the result is actually calculated using is/2. This is because, in general, the result is not an integer but a rational number, hence you can use CLP(Q) to get the result as a fraction:
?- use_module(library(clpq)).
% library(clpq) compiled into clpq 0.08 sec, 1,189 clauses
true.
?- N=5, X #= 2*N, doublefactorial2(X,F1), Y #= 2*N-1, doublefactorial2(Y,F2), {RESULT = F1/F2}.
N = 5,
X = 10,
F1 = 3840,
Y = 9,
F2 = 945,
RESULT = 256 rdiv 63 ;
false.
?- N=5, X #= 2*N, doublefactorial2(X,F1), Y #= 2*N-1, doublefactorial2(Y,F2), {RESULT = F2/F1}.
N = 5,
X = 10,
F1 = 3840,
Y = 9,
F2 = 945,
RESULT = 63 rdiv 256 ;
false.
For more details see the SWI-Prolog documentation on CLP(Q). Due to the use of CLP(FD) you can also query for a range of n, say 1 to 5:
?- N in 1..5, X #= 2*N, doublefactorial2(X,F1), Y #= 2*N-1, doublefactorial2(Y,F2), {RESULT = F1/F2}.
N = Y, Y = F2, F2 = 1,
X = F1, F1 = RESULT, RESULT = 2 ;
N = 2,
X = 4,
F1 = 8,
Y = F2, F2 = 3,
RESULT = 8 rdiv 3 ;
N = 3,
X = 6,
F1 = 48,
Y = 5,
F2 = 15,
RESULT = 16 rdiv 5 ;
N = 4,
X = 8,
F1 = 384,
Y = 7,
F2 = 105,
RESULT = 128 rdiv 35 ;
N = 5,
X = 10,
F1 = 3840,
Y = 9,
F2 = 945,
RESULT = 256 rdiv 63 ;
false.

How to find two numbers where a restriction is applied

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.

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