如何解决河内问题的迭代方法

Question

I'm trying to solve the Hanoi problem using an iterative method. I try to do this by using two for nested loops, so as to repeat n - 1 steps in each loop, where n is the number of moves. I think I have well posed the problem by using two for, but I don't understand how to change the order of the towers at each pass. Can anyone help me with this task?

inizio is start, fine is end and supp is support

#include <stdlib.h>
#include <stdio.h>

void tower_Organizer(int *inizio, int *fine, int *supp);
void hanoi_Iter(int n, int inizio, int fine, int supp);

int main(void){
    
    int n;
    
    printf("%s\n" ,"inserisci un nuero positivo");
    scanf("%d", &n);
    
    hanoi_Iter(n, 1, 2, 3);
    
    return 0;
    
}


void hanoi_Iter(int n, int inizio, int fine, int supp){
    
    for(int i = 1 ; i <= n ; i++){
        
        for(int j = 1 ; j <= i - 1 ; j++){
            
            printf("%d -> %d\n", inizio, fine);
            tower_Organizer(&inizio, &fine, &supp);

                
        }
        
        printf("%d -> %d\n", inizio, fine);
        tower_Organizer(&inizio, &fine, &supp);

    }       
        
}


void tower_Organizer(int *inizio, int *fine, int *supp){
    
    static int count = 1;
    
    int c = count % 6;
    
    switch( c ){
        
        case 0:
            *inizio = 1;
            *fine = 2;
            *supp = 3;
            break;
        case 1:
            *inizio = 1;
            *fine = 3;
            *supp = 2;
            break;
        case 2:
            *inizio = 2;
            *fine = 3;
            *supp = 1;
            break;
        case 3:
            *inizio = 1;
            *fine = 2;
            *supp = 3;
            break;
        case 4:
            *inizio = 3;
            *fine = 1;
            *supp = 2;
            break;
        case 5:
            *inizio = 3;
            *fine = 2;
            *supp = 1;
            break;
        
    }
    
    count++;
    
    
}  

  

Answer 1

The number of moves is 2 −1, so the nested loop you have cannot achieve that, as it will only produce (+1)/2 moves.

The logic for which disc to move is also not as simple as a cycle of 6 distinct moves.

There are several ways to implement an iterative solution, some of which are explained onWikipedia . However, the ones that are listed there, require that you maintain the state of each peg (pile), so you can perform checks on what constitutes a valid move.

You can also look at the bit-pattern of the move's sequence number and derive from that which is the correct move.

Here is an implementation of that latter idea:

void hanoi(int n, int start, int end, int helper) {
    // Create a converter for a pile index (0, 1 or 2) to the identifiers 
    //    that are given as arguments:
    int map[] = {start, end, helper};
    // Perform as many iterations as there are moves to perform:
    for (int last = 1<<n, i = 1; i < last; i++) {
        // Use bit pattern of i to determine the source peg.
        // First remove and count the trailing 0 bits in i:
        int j = i;
        int zeroCount = 0;
        while (j % 2 == 0) {
            j >>= 1;
            zeroCount++;
        }
        // Derive the source pile from that zero-stripped number
        int source = (j >> 1) % 3;
        // Get next pile as destination
        int target = (source + 1) % 3;
        // Depending on parity, the source/target pile should be mirrored
        if ((n + zeroCount) % 2 == 0) {
            source = (3 - source) % 3;
            target = (3 - target) % 3;
        }
        printf("Move from %d to %d\n", map[source], map[target]);
    }
}

int main(void) {
    int n;
    printf("%s\n" ,"Enter a positive number (of discs)");
    scanf("%d", &n);
    hanoi(n, 1, 2, 3);
    return 0;
}