如何用迭代法求解 Hanoi 问题

Question

我正在尝试使用迭代方法解决河内问题。我尝试通过使用两个 for 嵌套循环来做到这一点，以便在每个循环中重复 n - 1 个步骤，其中 n 是移动次数。我想我已经通过使用两个 for 很好地提出了这个问题，但我不明白如何在每次通过时更改塔的顺序。谁能帮我完成这个任务？

inizio 正在启动，罚款是结束和 supp是支持

#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

步数为 2−1，因此您的嵌套循环无法实现，因为它只会产生 (+1)/2 步。

移动哪个圆盘的逻辑也不像 6 个不同移动的循环那么简单。

有多种方法可以实现迭代解决方案，Wikipedia 上对其中一些进行了说明。但是，此处列出的那些要求您保持每个桩（桩）的状态，因此您可以检查什么构成有效移动。

您还可以查看着法序列号的位模式，并从中得出正确的着法。

下面是后一个想法的实现：

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;
}

如何用迭代法求解 Hanoi 问题

How Solve Iterative Method For Hanoi Problem

c

algorithm

towers-of-hanoi