Reversible numerical calculations in Prolog

Question

While reading SICP I came across logic programming chapter 4.4. Then I started looking into the Prolog programming language and tried to understand some simple assignments i

Answer 1

In this answer we use clpfd, just like this previous answer did.

:- use_module(library(clpfd)).

For easy head-to-head comparison (later on), we call the predicate presented here n_fac/2:

n_fac(N_expr,F_expr) :-
   N #= N_expr,                 % eval arith expr
   F #= F_expr,                 % eval arith expr
   n_facAux(N,F).

Like in this previous answer, n_fac/2 admits the use of arithmetic expressions.

n_facAux(0,1).                  % 0! = 1
n_facAux(1,1).                  % 1! = 1
n_facAux(2,2).                  % 2! = 2
n_facAux(N,F) :- 
   N #> 2,
   F #> N,                      % redundant constraint
                                %   to help `n_fac(N,N)` terminate
   n0_n_fac0_fac(3,N,6,F).      % general case starts with "3! = 6"

The helper predicate n_facAux/2 delegates any "real" work to n0_n_fac0_fac/4:

n0_n_fac0_fac(N ,N,F ,F).
n0_n_fac0_fac(N0,N,F0,F) :-
   N0 #< N,
   N1 #= N0+1,                  % count "up", not "down"
   F1 #= F0*N1,                 % calc `1*2*...*N`, not `N*(N-1)*...*2*1`
   F1 #=< F,                    % enforce redundant constraint
   n0_n_fac0_fac(N1,N,F1,F).

---

Let's compare n_fac/2 and n_factorial/2!

?- n_factorial(47,F).
  F = 258623241511168180642964355153611979969197632389120000000000
; false.
?- n_fac(47,F).
  F = 258623241511168180642964355153611979969197632389120000000000
; false.

?- n_factorial(N,1).
  N = 0
; N = 1
; false.
?- n_fac(N,1).
  N = 0
; N = 1
; false.

?- member(F,[3,1_000_000]), ( n_factorial(N,F) ; n_fac(N,F) ).
false.                          % both predicates agree

OK! Identical, so far... Why not do a little brute-force testing?

?- time((F1 #\= F2,n_factorial(N,F1),n_fac(N,F2))).
% 57,739,784 inferences, 6.415 CPU in 7.112 seconds (90% CPU, 9001245 Lips)
% Execution Aborted
?- time((F1 #\= F2,n_fac(N,F2),n_factorial(N,F1))).
% 52,815,182 inferences, 5.942 CPU in 6.631 seconds (90% CPU, 8888423 Lips)
% Execution Aborted

?- time((N1 #> 1,N2 #> 1,N1 #\= N2,n_fac(N1,F),n_factorial(N2,F))).
% 99,463,654 inferences, 15.767 CPU in 16.575 seconds (95% CPU, 6308401 Lips)
% Execution Aborted
?- time((N1 #> 1,N2 #> 1,N1 #\= N2,n_factorial(N2,F),n_fac(N1,F))).
% 187,621,733 inferences, 17.192 CPU in 18.232 seconds (94% CPU, 10913552 Lips)
% Execution Aborted

No differences for the first few hundred values of N in 2..sup... Good!

Moving on: How about the following (suggested in a comment to this answer)?

?- n_factorial(N,N), false.
false.
?- n_fac(N,N), false.
false.

Doing fine! Identical termination behaviour... More?

?- N #< 5, n_factorial(N,_), false.
false.
?- N #< 5, n_fac(N,_), false.
false.

?- F in 10..100, n_factorial(_,F), false.
false.
?- F in 10..100, n_fac(_,F), false.
false.

Alright! Still identical termination properties! Let's dig a little deeper! How about the following?

?- F in inf..10, n_factorial(_,F), false.
... % Execution Aborted                % does not terminate universally
?- F in inf..10, n_fac(_,F), false.
false.                                 % terminates universally

D'oh! The first query does not terminate, the second does. What a speedup! :)

---

Let's do some empirical runtime measurements!

?- member(Exp,[6,7,8,9]), F #= 10^Exp, time(n_factorial(N,F)) ; true.
%     328,700 inferences,  0.043 CPU in  0.043 seconds (100% CPU, 7660054 Lips)
%   1,027,296 inferences,  0.153 CPU in  0.153 seconds (100% CPU, 6735634 Lips)
%   5,759,864 inferences,  1.967 CPU in  1.967 seconds (100% CPU, 2927658 Lips)
%  22,795,694 inferences, 23.911 CPU in 23.908 seconds (100% CPU,  953351 Lips)
true.

?- member(Exp,[6,7,8,9]), F #= 10^Exp, time(n_fac(N,F)) ; true.
%       1,340 inferences,  0.000 CPU in  0.000 seconds ( 99% CPU, 3793262 Lips)
%       1,479 inferences,  0.000 CPU in  0.000 seconds (100% CPU, 6253673 Lips)
%       1,618 inferences,  0.000 CPU in  0.000 seconds (100% CPU, 5129994 Lips)
%       1,757 inferences,  0.000 CPU in  0.000 seconds (100% CPU, 5044792 Lips)
true.

Wow! Some more?

?- member(U,[10,100,1000]), time((N in 1..U,n_factorial(N,_),false)) ; true.
%      34,511 inferences,  0.004 CPU in  0.004 seconds (100% CPU, 9591041 Lips)
%   3,091,271 inferences,  0.322 CPU in  0.322 seconds (100% CPU, 9589264 Lips)
% 305,413,871 inferences, 90.732 CPU in 90.721 seconds (100% CPU, 3366116 Lips)
true.

?- member(U,[10,100,1000]), time((N in 1..U,n_fac(N,_),false)) ; true.
%       3,729 inferences,  0.001 CPU in  0.001 seconds (100% CPU,  2973653 Lips)
%      36,369 inferences,  0.004 CPU in  0.004 seconds (100% CPU, 10309784 Lips)
%     362,471 inferences,  0.036 CPU in  0.036 seconds (100% CPU,  9979610 Lips)
true.

---

The bottom line?

• The code presented in this answer is as low-level as you should go: Forget is/2!
• Redundant constraints can and do pay off.
• The order of arithmetic operations (counting "up" vs "down") can make quite a difference, too.
• If you want to calculate the factorial of some "large" N, consider using a different approach.
• Use clpfd!