Finding all cube roots with a number as a limit

问题

I want to find all of the cube roots that their cubes + their remainder add up to a number to user inputs. So for example, the query:

?- smallerCube(X,20).

Would give the result:

1 remainder 19
2 remainder 12

Since 1^3 = 1 (remainder 19), 2^3 = 8(remainder 12) and 3^3 = 27 which is bigger than the initial input of 20, and hence it's not being calculated here.

So far this is my code:

cubeLess(X,B,R) :- 
                   X =< B,
                   X1 is X*X*X,
                   R is B-X1.  

smallerCube(X,B) :- int(X),
                    X2 is X*X*X,
                    X2 =< B,
                    cubeLess(X2,B,R),
                    write(X), write(' rest '), writeln(R).

int(1).
int(N) :- int(N1), N is N1+1.

I use cubeLess to get the remainder, int to generate numbers from 1 onward. However, when I run the following query:

?- smallerCube(X,130)

I get the following weird result:

1 rest 129
X = 1 
2 rest -382
X = 2 
3 rest -19553
X = 3 ;

Why did it work for X=1, but gave negative results for X=2,3?

回答1:

Use clpfd!

:- use_module(library(clpfd)).

No need to worry about using clpfd for the 1^st time—you'll get the meaning in a moment for sure!

smallerCube_(X, Remainder, Maximum) :-
   X               #>= 0,
   Remainder       #>= 0,
   Remainder + X^3  #= Maximum.

First, the most general query of smallerCube_/3:

?- smallerCube_(X, Remainder, 20).
X in 1..2, _A in 1..8, Remainder in 12..19, X^3 #= _A, Remainder+_A #= 20.

One answer—two solutions: let's see them separated!

?- smallerCube_(X, Remainder, 20), indomain(X).
   X = 1, Remainder = 19                           % 20 #= 1^3 + 19
;  X = 2, Remainder = 12.                          % 20 #= 2^3 + 12

Here's the second query the OP wanted to run:

?- smallerCube_(X, Remainder, 130), indomain(X).
   X = 1, Remainder = 129                          % 130 #= 1^3 + 129
;  X = 2, Remainder = 122                          % 130 #= 2^3 + 122
;  X = 3, Remainder = 103                          % ...
;  X = 4, Remainder = 66                           % ...
;  X = 5, Remainder = 5.                           % 130 #= 5^3 + 5

Done! So what's next? Of course, that is up to you, so:

Why not re-invest the time clpfd saved you?
Why not read this very compact CLP(FD) primer as a jumpstart?

来源：https://stackoverflow.com/questions/35087702/finding-all-cube-roots-with-a-number-as-a-limit

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