The Tower of Hanoi is a mathematical puzzle. It consists of three poles and a number of disks of different sizes which can slide onto any pole. The puzzle starts with the disk in a neat stack in ascending order of size in one pole, the smallest at the top thus making a conical shape. The objective of the puzzle is to move all the disks from one pole (say ‘source pole’) to another pole (say ‘destination pole’) with the help of the third pole (say auxiliary pole).
The puzzle has the following two rules:
1. You can’t place a larger disk onto a smaller disk
2. Only one disk can be moved at a time
We’ve already discussed a recursive solution for the Tower of Hanoi. We have also seen that for n disks, a total of 2n – 1 moves are required.
Iterative Algorithm:
1. Calculate the total number of moves required i.e. "pow(2, n)
- 1" here n is number of disks.
2. If number of disks (i.e. n) is even then interchange destination
pole and auxiliary pole.
3. for i = 1 to total number of moves:
if i%3 == 1:
legal movement of top disk between source pole and
destination pole
if i%3 == 2:
legal movement top disk between source pole and
auxiliary pole
if i%3 == 0:
legal movement top disk between auxiliary pole
and destination pole
Example:
Let us understand with a simple example with 3 disks: So, total number of moves required = 7
S A D When i= 1, (i % 3 == 1) legal movement between‘S’ and ‘D’
When i = 2, (i % 3 == 2) legal movement between ‘S’ and ‘A’
When i = 3, (i % 3 == 0) legal movement between ‘A’ and ‘D’ ’
When i = 4, (i % 3 == 1) legal movement between ‘S’ and ‘D’
When i = 5, (i % 3 == 2) legal movement between ‘S’ and ‘A’
When i = 6, (i % 3 == 0) legal movement between ‘A’ and ‘D’
When i = 7, (i % 3 == 1) legal movement between ‘S’ and ‘D’
So, after all these destination poles contains all the in order of size.
After observing above iterations, we can think that after a disk other than the smallest disk is moved, the next disk to be moved must be the smallest disk because it is the top disk resting on the spare pole and there are no other choices to move a disk.
C++
// C++ Program for Iterative Tower of Hanoi using STL#include <iostream>#include <vector>#include <stack>using namespace std;char rod[]={'S', 'A', 'D'};vector<stack<int>> stacks(3); // 3 stacks for 3 rodsvoid moveDisk(int a, int b){ if (stacks[b].empty() || (!stacks[a].empty() && stacks[a].top() < stacks[b].top())) { cout << "Move disk " << stacks[a].top() << " from rod " << rod[a] << " to rod " << rod[b] << "\n"; stacks[b].push(stacks[a].top()); stacks[a].pop(); } else moveDisk(b, a);}void towerOfHanoi(int n){ cout << "Tower of Hanoi for " << n << " disks:\n"; int src = 0, aux = 1, dest = 2; for (int i = n; i > 0; i--) stacks[src].push(i); int totalMoves = (1 << n) - 1; if (n % 2 == 0) swap(aux, dest); for (int i = 1; i <= totalMoves; i++) { if (i % 3 == 0) moveDisk(aux, dest); else if (i % 3 == 1) moveDisk(src, dest); else moveDisk(src, aux); }}int main(){ int n = 3; // number of disks towerOfHanoi(n); return 0;} |
C
// C Program for Iterative Tower of Hanoi #include <stdio.h> #include <math.h> #include <stdlib.h> #include <limits.h> // A structure to represent a stack struct Stack { unsigned capacity; int top; int *array; }; // function to create a stack of given capacity. struct Stack* createStack(unsigned capacity) { struct Stack* stack = (struct Stack*) malloc(sizeof(struct Stack)); stack -> capacity = capacity; stack -> top = -1; stack -> array = (int*) malloc(stack -> capacity * sizeof(int)); return stack; } // Stack is full when top is equal to the last index int isFull(struct Stack* stack) { return (stack->top == stack->capacity - 1); } // Stack is empty when top is equal to -1 int isEmpty(struct Stack* stack) { return (stack->top == -1); } // Function to add an item to stack. It increases // top by 1 void push(struct Stack *stack, int item) { if (isFull(stack)) return; stack -> array[++stack -> top] = item; } // Function to remove an item from stack. It // decreases top by 1 int pop(struct Stack* stack) { if (isEmpty(stack)) return INT_MIN; return stack -> array[stack -> top--]; } //Function to show the movement of disks void moveDisk(char fromPeg, char toPeg, int disk) { printf("Move the disk %d from \'%c\' to \'%c\'\n", disk, fromPeg, toPeg); }// Function to implement legal movement between // two poles void moveDisksBetweenTwoPoles(struct Stack *src, struct Stack *dest, char s, char d) { int pole1TopDisk = pop(src); int pole2TopDisk = pop(dest); // When pole 1 is empty if (pole1TopDisk == INT_MIN) { push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When pole2 pole is empty else if (pole2TopDisk == INT_MIN) { push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } // When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk) { push(src, pole1TopDisk); push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When top disk of pole1 < top disk of pole2 else { push(dest, pole2TopDisk); push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } } //Function to implement TOH puzzle void tohIterative(int num_of_disks, struct Stack *src, struct Stack *aux, struct Stack *dest) { int i, total_num_of_moves; char s = 'S', d = 'D', a = 'A'; //If number of disks is even, then interchange //destination pole and auxiliary pole if (num_of_disks % 2 == 0) { char temp = d; d = a; a = temp; } total_num_of_moves = pow(2, num_of_disks) - 1; //Larger disks will be pushed first for (i = num_of_disks; i >= 1; i--) push(src, i); for (i = 1; i <= total_num_of_moves; i++) { if (i % 3 == 1) moveDisksBetweenTwoPoles(src, dest, s, d); else if (i % 3 == 2) moveDisksBetweenTwoPoles(src, aux, s, a); else if (i % 3 == 0) moveDisksBetweenTwoPoles(aux, dest, a, d); } } // Driver Program int main() { // Input: number of disks unsigned num_of_disks = 3; struct Stack *src, *dest, *aux; // Create three stacks of size 'num_of_disks' // to hold the disks src = createStack(num_of_disks); aux = createStack(num_of_disks); dest = createStack(num_of_disks); tohIterative(num_of_disks, src, aux, dest); return 0; } |
Java
// Java program for iterative // Tower of Hanoiclass TOH{ // A structure to represent a stackclass Stack{ int capacity; int top; int array[];}// Function to create a stack of given capacity.Stack createStack(int capacity){ Stack stack = new Stack(); stack.capacity = capacity; stack.top = -1; stack.array = new int[capacity]; return stack;}// Stack is full when the top is equal// to the last indexboolean isFull(Stack stack){ return (stack.top == stack.capacity - 1);}// Stack is empty when top is equal to -1boolean isEmpty(Stack stack){ return (stack.top == -1);}// Function to add an item to stack.It // increases top by 1void push(Stack stack, int item){ if (isFull(stack)) return; stack.array[++stack.top] = item;}// Function to remove an item from stack.It// decreases top by 1int pop(Stack stack){ if (isEmpty(stack)) return Integer.MIN_VALUE; return stack.array[stack.top--];}// Function to implement legal movement between// two polesvoid moveDisksBetweenTwoPoles(Stack src, Stack dest, char s, char d) { int pole1TopDisk = pop(src); int pole2TopDisk = pop(dest); // When pole 1 is empty if (pole1TopDisk == Integer.MIN_VALUE) { push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When pole2 pole is empty else if (pole2TopDisk == Integer.MIN_VALUE) { push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } // When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk) { push(src, pole1TopDisk); push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When top disk of pole1 < top disk of pole2 else { push(dest, pole2TopDisk); push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); }}// Function to show the movement of disksvoid moveDisk(char fromPeg, char toPeg, int disk){ System.out.println("Move the disk " + disk + " from " + fromPeg + " to " + toPeg);}// Function to implement TOH puzzlevoid tohIterative(int num_of_disks, Stack src, Stack aux, Stack dest){ int i, total_num_of_moves; char s = 'S', d = 'D', a = 'A'; // If number of disks is even, then // interchange destination pole and // auxiliary pole if (num_of_disks % 2 == 0) { char temp = d; d = a; a = temp; } total_num_of_moves = (int)(Math.pow( 2, num_of_disks) - 1); // Larger disks will be pushed first for(i = num_of_disks; i >= 1; i--) push(src, i); for(i = 1; i <= total_num_of_moves; i++) { if (i % 3 == 1) moveDisksBetweenTwoPoles(src, dest, s, d); else if (i % 3 == 2) moveDisksBetweenTwoPoles(src, aux, s, a); else if (i % 3 == 0) moveDisksBetweenTwoPoles(aux, dest, a, d); }}// Driver codepublic static void main(String[] args){ // Input: number of disks int num_of_disks = 3; TOH ob = new TOH(); Stack src, dest, aux; // Create three stacks of size 'num_of_disks' // to hold the disks src = ob.createStack(num_of_disks); dest = ob.createStack(num_of_disks); aux = ob.createStack(num_of_disks); ob.tohIterative(num_of_disks, src, aux, dest);}}// This code is contributed by Sumit Ghosh |
Python3
# Python3 program for iterative Tower of Hanoiimport sys# A structure to represent a stackclass Stack: # Constructor to set the data of # the newly created tree node def __init__(self, capacity): self.capacity = capacity self.top = -1 self.array = [0]*capacity# function to create a stack of given capacity.def createStack(capacity): stack = Stack(capacity) return stack # Stack is full when top is equal to the last indexdef isFull(stack): return (stack.top == (stack.capacity - 1)) # Stack is empty when top is equal to -1def isEmpty(stack): return (stack.top == -1) # Function to add an item to stack.# It increases top by 1def push(stack, item): if(isFull(stack)): return stack.top+=1 stack.array[stack.top] = item # Function to remove an item from stack.# It decreases top by 1def Pop(stack): if(isEmpty(stack)): return -sys.maxsize Top = stack.top stack.top-=1 return stack.array[Top] # Function to implement legal# movement between two polesdef moveDisksBetweenTwoPoles(src, dest, s, d): pole1TopDisk = Pop(src) pole2TopDisk = Pop(dest) # When pole 1 is empty if (pole1TopDisk == -sys.maxsize): push(src, pole2TopDisk) moveDisk(d, s, pole2TopDisk) # When pole2 pole is empty else if (pole2TopDisk == -sys.maxsize): push(dest, pole1TopDisk) moveDisk(s, d, pole1TopDisk) # When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk): push(src, pole1TopDisk) push(src, pole2TopDisk) moveDisk(d, s, pole2TopDisk) # When top disk of pole1 < top disk of pole2 else: push(dest, pole2TopDisk) push(dest, pole1TopDisk) moveDisk(s, d, pole1TopDisk) # Function to show the movement of disksdef moveDisk(fromPeg, toPeg, disk): print("Move the disk", disk, "from '", fromPeg, "' to '", toPeg, "'") # Function to implement TOH puzzledef tohIterative(num_of_disks, src, aux, dest): s, d, a = 'S', 'D', 'A' # If number of disks is even, then interchange # destination pole and auxiliary pole if (num_of_disks % 2 == 0): temp = d d = a a = temp total_num_of_moves = int(pow(2, num_of_disks) - 1) # Larger disks will be pushed first for i in range(num_of_disks, 0, -1): push(src, i) for i in range(1, total_num_of_moves + 1): if (i % 3 == 1): moveDisksBetweenTwoPoles(src, dest, s, d) else if (i % 3 == 2): moveDisksBetweenTwoPoles(src, aux, s, a) else if (i % 3 == 0): moveDisksBetweenTwoPoles(aux, dest, a, d)# Input: number of disksnum_of_disks = 3# Create three stacks of size 'num_of_disks'# to hold the diskssrc = createStack(num_of_disks)dest = createStack(num_of_disks)aux = createStack(num_of_disks)tohIterative(num_of_disks, src, aux, dest)# This code is contributed by divyeshrabadiya07. |
C#
// C# program for iterative // Tower of Hanoi using System;class GFG { // A structure to represent a stack public class Stack { public int capacity; public int top; public int []array; } // function to create a stack of given capacity. Stack createStack(int capacity) { Stack stack = new Stack(); stack.capacity = capacity; stack.top = -1; stack.array = new int[capacity]; return stack; } // Stack is full when top is equal to the last index Boolean isFull(Stack stack) { return (stack.top == stack.capacity - 1); } // Stack is empty when top is equal to -1 Boolean isEmpty(Stack stack) { return (stack.top == -1); } // Function to add an item to stack. // It increases top by 1 void push(Stack stack,int item) { if(isFull(stack)) return; stack.array[++stack.top] = item; } // Function to remove an item from stack. // It decreases top by 1 int pop(Stack stack) { if(isEmpty(stack)) return int.MinValue; return stack.array[stack.top--]; } // Function to implement legal // movement between two poles void moveDisksBetweenTwoPoles(Stack src, Stack dest, char s, char d) { int pole1TopDisk = pop(src); int pole2TopDisk = pop(dest); // When pole 1 is empty if (pole1TopDisk == int.MinValue) { push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When pole2 pole is empty else if (pole2TopDisk == int.MinValue) { push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } // When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk) { push(src, pole1TopDisk); push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When top disk of pole1 < top disk of pole2 else { push(dest, pole2TopDisk); push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } } // Function to show the movement of disks void moveDisk(char fromPeg, char toPeg, int disk) { Console.WriteLine("Move the disk "+disk + " from "+fromPeg+" to "+toPeg); } // Function to implement TOH puzzle void tohIterative(int num_of_disks, Stack src, Stack aux, Stack dest) { int i, total_num_of_moves; char s = 'S', d = 'D', a = 'A'; // If number of disks is even, then interchange // destination pole and auxiliary pole if (num_of_disks % 2 == 0) { char temp = d; d = a; a = temp; } total_num_of_moves = (int) (Math.Pow(2, num_of_disks) - 1); // Larger disks will be pushed first for (i = num_of_disks; i >= 1; i--) push(src, i); for (i = 1; i <= total_num_of_moves; i++) { if (i % 3 == 1) moveDisksBetweenTwoPoles(src, dest, s, d); else if (i % 3 == 2) moveDisksBetweenTwoPoles(src, aux, s, a); else if (i % 3 == 0) moveDisksBetweenTwoPoles(aux, dest, a, d); } } // Driver code public static void Main(String []args) { // Input: number of disks int num_of_disks = 3; GFG ob = new GFG(); Stack src, dest, aux; // Create three stacks of size 'num_of_disks' // to hold the disks src = ob.createStack(num_of_disks); dest = ob.createStack(num_of_disks); aux = ob.createStack(num_of_disks); ob.tohIterative(num_of_disks, src, aux, dest); } } // This code is Contributed by Arnab Kundu |
Javascript
<script> // Javascript program for iterative Tower of Hanoi // A structure to represent a stack class Stack { constructor(capacity) { this.capacity = capacity; this.top = -1; this.array = new Array(capacity); } } // function to create a stack of given capacity. function createStack(capacity) { let stack = new Stack(capacity); return stack; } // Stack is full when top is equal to the last index function isFull(stack) { return (stack.top == (stack.capacity - 1)); } // Stack is empty when top is equal to -1 function isEmpty(stack) { return (stack.top == -1); } // Function to add an item to stack. // It increases top by 1 function push(stack, item) { if(isFull(stack)) return; stack.array[++stack.top] = item; } // Function to remove an item from stack. // It decreases top by 1 function pop(stack) { if(isEmpty(stack)) return Number.MIN_VALUE; return stack.array[stack.top--]; } // Function to implement legal // movement between two poles function moveDisksBetweenTwoPoles(src, dest, s, d) { let pole1TopDisk = pop(src); let pole2TopDisk = pop(dest); // When pole 1 is empty if (pole1TopDisk == Number.MIN_VALUE) { push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When pole2 pole is empty else if (pole2TopDisk == Number.MIN_VALUE) { push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } // When top disk of pole1 > top disk of pole2 else if (pole1TopDisk > pole2TopDisk) { push(src, pole1TopDisk); push(src, pole2TopDisk); moveDisk(d, s, pole2TopDisk); } // When top disk of pole1 < top disk of pole2 else { push(dest, pole2TopDisk); push(dest, pole1TopDisk); moveDisk(s, d, pole1TopDisk); } } // Function to show the movement of disks function moveDisk(fromPeg, toPeg, disk) { document.write("Move the disk "+disk + " from '"+fromPeg+"' to '"+toPeg + "'</br>"); } // Function to implement TOH puzzle function tohIterative(num_of_disks, src, aux, dest) { let i, total_num_of_moves; let s = 'S', d = 'D', a = 'A'; // If number of disks is even, then interchange // destination pole and auxiliary pole if (num_of_disks % 2 == 0) { let temp = d; d = a; a = temp; } total_num_of_moves = parseInt(Math.pow(2, num_of_disks) - 1, 10); // Larger disks will be pushed first for (i = num_of_disks; i >= 1; i--) push(src, i); for (i = 1; i <= total_num_of_moves; i++) { if (i % 3 == 1) moveDisksBetweenTwoPoles(src, dest, s, d); else if (i % 3 == 2) moveDisksBetweenTwoPoles(src, aux, s, a); else if (i % 3 == 0) moveDisksBetweenTwoPoles(aux, dest, a, d); } } // Input: number of disks let num_of_disks = 3; let src, dest, aux; // Create three stacks of size 'num_of_disks' // to hold the disks src = createStack(num_of_disks); dest = createStack(num_of_disks); aux = createStack(num_of_disks); tohIterative(num_of_disks, src, aux, dest); // This code is contributed by decode2207</script> |
Output
Tower of Hanoi for 3 disks: Move disk 1 from rod S to rod D Move disk 2 from rod S to rod A Move disk 1 from rod D to rod A Move disk 3 from rod S to rod D Move disk 1 from rod A to rod S Move disk 2 from rod A to rod D Move disk 1 from rod S to rod D
Time Complexity: O(2ᴺ) Since the iterative solution needs to generate and process all possible combinations of moves for n disks, the number of iterations grows exponentially with the number of disks.
Auxiliary Space: O(n)
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References:
http://en.wikipedia.org/wiki/Tower_of_Hanoi#Iterative_solution
This article is contributed by Anand Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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