在序言中解决一个简单的难题

Question

I solved a puzzle in C and tried to do the same in Prolog but i'm having some trouble expressing the facts and goals in this language.

The very simplified version of the problem is this: there's two levers in a room. Each lever control a mechanism that can move either forward or backward in four different positions (which i noted 0, 1, 2 or 3). If you move a mechanism four times in the same direction, it'll be in the same position as before.

The lever n°1 move the mechanism n°1 two positions forward. The lever n°2 move the mechanism n°2 one position forward.

Initially, the mechanism n°1 is in position 2 and the mechanism n°2 is in position 1. The problem is to find the quickest way to move both mechanisms in position 0 and get the sequence of lever that lead to each solution.

Of course here the problem is trivial and you only need to pull the lever n°1 one time and the lever n°2 three times to have a solution.

Here's a simple code in C which gives the sequence of lever to pull to solve this problem by pulling less than 5 levers:

int pos1 = 2, pos2 = 1;

int main()
{
    resolve(0,5);
    return 0;
}

void lever1(){
    pos1 = (pos1 + 2) % 4;
}

void undolever1(){
    pos1 = (pos1 - 2) % 4;
}

void lever2(){
    pos2 = (pos2 + 1) % 4;
}

void undolever2(){
    pos2 = (pos2 - 1) % 4;
}

void resolve(l, k){
    if(k == 0){
        return;
    }
    if(pos1 == 0 && pos2 == 0){
        printf("Solution: %d\n", l);
        return;
    }
    if(k>0){
        k--;
        lever1();
        resolve(l*10+1,k);
        undolever1();
        lever2();
        resolve(l*10+2,k);
        undolever2();
    }
}

My code in Prolog looks like this so far:

lever(l1).
lever(l2).

mechanism(m1).
mechanism(m2).

position(m1,2).
position(m2,1).

pullL1() :- position(m1, mod(position(m1,X)+2,4)).
pullL2() :- position(m2, mod(position(m2,X)+1,4)).

solve(k) :- solve_(k, []).
solve_(0, r) :- !, postion(m1, p1), postion(m2, p2), p1 == 0, p2 == 0.
solve_(k, r) :- k > 0, pullL1(), k1 is k - 1, append(r, [1], r1), solve_(k1, r1).
solve_(k, r) :- k > 0, pullL2(), k1 is k - 1, append(r, [2], r2), solve_(k1, r2).

I'm pretty sure there's multiple problems in this code but I'm not sure how to fix it. Any help would be really appreciated.

Answer 1

I think this is a very interesting Problem. I suppose you want a general solution -> one lever can move multiple mechanisms. In the case the problem is like yours, where one lever only controls one mechanism the solution is trivial. You just move every lever for the amount of time until the mechanism is at state zero.

But I want to provide a more general solution, so where one lever can move multiple mechanisms. But first a little bit math. Don't worry i'll end up doing an example too.

Lets define

as being n levers and

being m mechanisms. Then lets define every lever by a vector:

whereis the amount of stepsmovesforward.

For the mechanisms we define:

beeing the bias of the mechanisms -> so is in initial stateand

being the amount of states for every mechanism. So now we can describe our whole system like this:

wheris the amount of times we have to activate. If we want to set all mechanisms to zero. If you are not familiar with the notation this just means that a%m = b%m.

can we rewritten as:

where k can be any natural number. So we can rewrite our system to an equation system:

prolog can solve for us such a equation system. (there are different solutions to solve diophantine equation systems look at https://en.wikipedia.org/wiki/Diophantine_equation )

Ok now lets make an example: let say we have two levers and three mechanisms with 4 states. The fist lever moves M1 one forward and M3 two forward. The second lever moves M2 one forward and M3 one forward. M1 is in State 2. M2 is in State 3. M3 is in State 3. So our equation system looks like this:

in prolog we can solve this with the clpfd libary.

?- [library(clpfd)].

and then solve like this:

?- X1+(-4)*K1+2 #= 0, 1*X2+(-4)*K2+3 #= 0, 2*X1+X2+(-4)*K3+3 #= 0,Vs = [X1,X2], Vs ins 0..100,label(Vs).

which gives us the solution

Vs = [2, 1]

-> so X1 = 2 and X2 = 1 which is correct. Prolog can give you more solutions.