Inverse factorial in Prolog

Question

Can someone helping me to find a way to get the inverse factorial in Prolog...

For example inverse_factorial(6,X) ===> X = 3.

I have been working on it a lot of time.

I currently have the factorial, but i have to make it reversible. Please help me.

Answer 1

Prolog's predicates are relations, so once you have defined factorial, you have implicitly defined the inverse too. However, regular arithmetics is moded in Prolog, that is, the entire expression in (is)/2 or (>)/2 has to be known at runtime, and if it is not, an error occurs. Constraints overcome this shortcoming:

:- use_module(library(clpfd)).

n_factorial(0, 1).
n_factorial(N, F) :-
   N #> 0, N1 #= N - 1, F #= N * F1,
   n_factorial(N1, F1).

This definition now works in both directions.

?- n_factorial(N,6).
N = 3 ;
false.

?- n_factorial(3,F).
F = 6 ;
false.

Since SICStus 4.3.4 and SWI 7.1.25 also the following terminates:

?- n_factorial(N,N).
   N = 1
;  N = 2
;  false.

See the manual for more.

Answer 2

For reference, here is the best implementation of a declarative factorial predicate I could come up with.

Two main points are different from @false's answer:

• It uses an accumulator argument, and recursive calls increment the factor we multiply the factorial with, instead of a standard recursive implementation where the base case is 0. This makes the predicate much faster when the factorial is known and the initial number is not.

• It uses if_/3 and (=)/3 extensively, from module reif, to get rid of unnecessary choice points when possible. It also uses (#>)/3 and the reified (===)/6 which is a variation of (=)/3 for cases where we have two couples that can be used for the if -> then part of if_.

factorial/2

factorial(N, F) :-
    factorial(N, 0, 1, F).

factorial(N, I, N0, F) :-
    F #> 0,
    N #>= 0,
    I #>= 0,
    I #=< N,
    N0 #> 0,
    N0 #=< F,
    if_(I #> 2,
        (   F #> N,
            if_(===(N, I, N0, F, T1),
                if_(T1 = true,
                    N0 = F,
                    N = I
                ),
                (   J #= I + 1,
                    N1 #= N0*J,
                    factorial(N, J, N1, F)
                )
            )
        ),
        if_(N = I,
            N0 = F,
            (   J #= I + 1,
                N1 #= N0*J,
                factorial(N, J, N1, F)
            )
        )
    ).

(#>)/3

#>(X, Y, T) :-
    zcompare(C, X, Y),
    greater_true(C, T).

greater_true(>, true).
greater_true(<, false).
greater_true(=, false).

(===)/6

===(X1, Y1, X2, Y2, T1, T) :-
    (   T1 == true -> =(X1, Y1, T)
    ;   T1 == false -> =(X2, Y2, T)    
    ;   X1 == Y1 -> T1 = true, T = true
    ;   X1 \= Y1 -> T1 = true, T = false
    ;   X2 == Y2 -> T1 = false, T = true
    ;   X2 \= Y2 -> T1 = false, T = false
    ;   T1 = true, T = true, X1 = Y1
    ;   T1 = true, T = false, dif(X1, Y1)
    ).

Some queries

?- factorial(N, N).
N = 1 ;
N = 2 ;
false.          % One could probably get rid of the choice point at the cost of readability


?- factorial(N, 1).
N = 0 ;
N = 1 ;
false.          % Same


?- factorial(10, N).
N = 3628800.    % No choice point


?- time(factorial(N, 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000)).
% 79,283 inferences, 0.031 CPU in 0.027 seconds (116% CPU, 2541106 Lips)
N = 100.        % No choice point


?- time(factorial(N, 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518284253697920827223758251185210916864000000000000000000000000)).
% 78,907 inferences, 0.031 CPU in 0.025 seconds (125% CPU, 2529054 Lips)
false.


?- F #> 10^100, factorial(N, F).
F = 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000,
N = 70 ;
F = 850478588567862317521167644239926010288584608120796235886430763388588680378079017697280000000000000000,
N = 71 ;
F = 61234458376886086861524070385274672740778091784697328983823014963978384987221689274204160000000000000000,
N = 72 ;
...

Answer 3

a simple 'low tech' way: enumerate integers until

• you find the sought factorial, then 'get back' the number
• the factorial being built is greater than the target. Then you can fail...

Practically, you can just add 2 arguments to your existing factorial implementation, the target and the found inverse.

Answer 4

Just implement factorial(X, XFact) and then swap arguments

factorial(X, XFact) :- f(X, 1, 1, XFact).

f(N, N, F, F) :- !.
f(N, N0, F0, F) :- succ(N0, N1), F1 is F0 * N1, f(N, N1, F1, F).