Generalizing Fibonacci sequence with SICStus Prolog

问题

I'm trying to find a solution for a query on a generalized Fibonacci sequence (GFS). The query is: are there any GFS that have 885 as their 12th number? The initial 2 numbers may be restricted between 1 and 10.

I already found the solution to find the Nth number in a sequence that starts at (1, 1) in which I explicitly define the initial numbers. Here is what I have for this:

fib(1, 1).
fib(2, 1).

fib(N, X) :-
    N #> 1,
    Nmin1 #= N - 1,
    Nmin2 #= N - 2,
    fib(Nmin1, Xmin1),
    fib(Nmin2, Xmin2),
    X #= Xmin1 + Xmin2.

For the query mentioned I thought the following would do the trick, in which I reuse the fib method without defining the initial numbers explicitly since this now needs to be done dynamically:

fib(N, X) :-
    N #> 1,
    Nmin1 #= N - 1,
    Nmin2 #= N - 2,
    fib(Nmin1, Xmin1),
    fib(Nmin2, Xmin2),
    X #= Xmin1 + Xmin2.

fib2 :-
    X1 in 1..10,
    X2 in 1..10,
    fib(1, X1),
    fib(2, X2),
    fib(12, 885).

... but this does not seem to work.

Is it not possible this way to define the initial numbers, or am I doing something terribly wrong? I'm not asking for the solution, but any advice that could help me solve this would be greatly appreciated.

回答1:

Under SWI-Prolog:

:- use_module(library(clpfd)).

fib(A,B,N,X):-
    N #> 0,
    N0 #= N-1,
    C #= A+B,
    fib(B,C,N0,X).
fib(A,B,0,A).

task(A,B):-
    A in 1..10,
    B in 1..10,
    fib(A,B,11,885).

回答2:

Define a predicate gfs(X0, X1, N, F) where X0 and X1 are the values for the base cases 0 and 1.

回答3:

Maybe not a solution in the strict sense but I will share it never the less. Probably the only gain is to show, that this does neither need a computer nor a calculator to be solved. If you know the trick it can be done on a bearmat.

If F_n ist the n-th Term of the ordinary Fibo-sequence, starting with F_1=F_2=1, then the n-th Term of the generalized sequence will be G_n = F_{n-2}*a+F_{n-1}*b. Define F_{-1}=1, F_0 = 0

(Indeed, by induction

G_1 = F_{-1}*a+F_0*b = 1*a+0*b=a
G_2 = F_0 * a + F_1 * b = 0*a + 1*b = b
G_{n+1} = F_{n-1}a + F_nb = (F_{n-3} + F_{n-2} )a + (F_{n-2} + F_{n-1})*b = G_n + G_{n-1}

)

Thus G_12 = F_10 * a + F_11 * b = 55a + 89b.

Now you can either search for solutions to the equation 55a + 89b = 885 with your computer

do the math:

Residues mod 11 (explanation):

55a + 89b = 0 + 88b + b = b; 885 = 880 + 5 = 80*11 + 5 = 5

So b = 5 mod 11, but since 1 <= b <= 10, b really is 5. 89 * 5 = 445 and 885-445 = 440. Now, divide by 55 and get a=8.

回答4:

I'd say you're doing something terribly wrong... When you call fib(1, X1), the variable X1 is the number that the function fib will return, in this case, it will be 1, because of the base case fib(1, 1)..