Towers of Hanoi Problem in C++: Solving the Classic Puzzle with Recursion

[NickPatey]

I have to write a C++ program for the problem listed below, but i can't get it working properly, can anyone find the problem:

Question:
5.36 (Towers of Hanoi) Every budding computer scientist must grapple with certain classic problems,
and the Towers of Hanoi (see Fig. 5.19) is one of the most famous of these. Legend has it that
in a temple in the Far East, priests are attempting to move a stack of disks from one peg to another.
The initial stack had 64 disks threaded onto one peg and arranged from bottom to top by decreasing
size. The priests are attempting to move the stack from this peg to a second peg under the constraints
that exactly one disk is moved at a time, and at no time may a larger disk be placed above a smaller
disk. A third peg is available for temporarily holding the disks. Supposedly the world will end when
the priests complete their task, so there is little incentive for us to facilitate their efforts.

Let’s assume that the priests are attempting to move the disks from peg 1 to peg 3. We wish to
develop an algorithm that will print the precise sequence of disk-to-disk peg transfers.
If we were to approach this problem with conventional methods, we’d rapidly find ourselves
hopelessly knotted up in managing the disks. Instead, if we attack the problem with recursion in
mind, it immediately becomes tractable. Moving n disks can be viewed in terms of moving only
n – 1 disks (and hence the recursion) as follows:
a) Move n – 1 disks from peg 1 to peg 2, using peg 3 as a temporary holding area.
b) Move the last disk (the largest) from peg 1 to peg 3.
c) Move the n – 1 disks from peg 2 to peg 3, using peg 1 as a temporary holding area.
The process ends when the last task involves moving n = 1 disk, i.e., the base case. This is
accomplished by trivially moving the disk without the need for a temporary holding area.
Write a program to solve the Towers of Hanoi problem. Use a recursive function with four
parameters:
a) The number of disks to be moved
b) The peg on which these disks are initially threaded
c) The peg to which this stack of disks is to be moved
d) The peg to be used as a temporary holding area
Your program should print the precise instructions it will take to move the disks from the
starting peg to the destination peg. For example, to move a stack of three disks from peg 1 to peg 3,
your program should print the following series of moves:
1 →3 (This means move one disk from peg 1 to peg 3.)
1 → 2
3 → 2
1 →3
2 →1
2 →3
1 →3

What i have so far:

// towers.cpp : Defines the entry point for the console application.//

#include "stdafx.h"


int main(void)
{
	int n = 0;
	int sp = 0;
	int ep = 0;
	int hp = 0;
	int move (int n, int sp, int ep, int hp);
	int wait;
	printf("Enter number of disks");
	scanf("%d", &n);
	printf("Enter start peg (1, 2, or 3): ");
	scanf("%d", &sp);
	printf("Enter end peg number (1, 2, or 3): ");
	scanf("%d", &ep);

	if ( sp == ep ){
		printf("Nice try");
		return 0;
	}
	if (( sp == 1 && ep ==2) || (sp ==2 && ep == 1)) hp = 3;
	if (( sp == 1 && ep ==3) || (sp ==3 && ep == 1)) hp = 2;
	if (( sp == 2 && ep ==3) || (sp ==3 && ep == 2)) hp = 1;

	printf("Now to blow your mind as we move %d disks from peg %d to peg %d using %d as a holding peg never allowing a larger disk to be on top\n\n", n, sp , ep , hp);

	move (n , sp , ep , hp);
	scanf_s("%d", &wait);
	return 0;
}
int move ( int n, int sp, int ep, int hp )
{
	if (n==1)
	{
		printf("Move disk from %d--->%d", sp , ep);
	}
	else
	{
		move ( (n-1), sp, hp, ep );
		printf("Move disk from %d--->%d", sp , ep);
		move ( (n-1), hp, ep, sp );
	}
}

 

[NickPatey]

Nvm figured it out haha:

// towers.cpp : Defines the entry point for the console application.//

#include <stdafx.h>

int move (int n, int sp, int ep, int hp);

int main(void)
{
	int move (int n, int sp, int ep, int hp);
	int n = 0;
	int sp = 0;
	int ep = 0;
	int hp = 0;
	int wait;

	printf("Enter number of disks: ");
	scanf_s("%d", &n);
	printf("Enter start peg (1-3): ");
	scanf_s("%d", &sp);
	printf("Enter end peg (1-3):   ");
	scanf_s("%d", &ep);

	if ( sp == ep ){
		printf("Nice try");
		return 0;
	}

	if (( sp == 1 && ep ==2) || (sp ==2 && ep == 1)) hp = 3;
	if (( sp == 1 && ep ==3) || (sp ==3 && ep == 1)) hp = 2;
	if (( sp == 2 && ep ==3) || (sp ==3 && ep == 2)) hp = 1;

	printf("\n\nMove %d disks from peg %d to peg %d using %d as a holding peg\n\n", n, sp , ep , hp);

	move (n , sp , ep , hp);
	scanf_s("%d", &wait);

	return 0;
}
int move ( int n, int sp, int ep, int hp )
{

	if (n==1)
	{
		printf("Move disk from %d--->%d\n", sp , ep);
	}
	else
	{
		move ( (n-1), sp, hp, ep );
		printf("Move disk from %d--->%d\n", sp , ep);
		move ( (n-1), hp, ep, sp );
	}

	return n;

}

 

[TaxOnFear]

Note, this is C.