Why Tower of Hanoi Problem in DSA C?

The Big Idea

What if a simple rule could solve a puzzle that seems impossible to plan by hand?

The Scenario

Imagine you have three pegs and a stack of different sized disks on one peg. You want to move all disks to another peg, but you can only move one disk at a time and never place a larger disk on top of a smaller one.

Trying to figure out the steps manually for many disks is confusing and easy to mess up.

The Problem

Manually planning each move for many disks is slow and error-prone. You might forget a step or break the rules by placing a bigger disk on a smaller one.

It becomes very hard to keep track of all moves as the number of disks grows.

The Solution

The Tower of Hanoi problem uses a simple set of rules and recursion to break the big problem into smaller ones. This method gives a clear, step-by-step solution that always works, no matter how many disks.

It saves time and avoids mistakes by following a repeatable pattern.

Before vs After

✗ Before

/* Manually list moves for 3 disks */
move disk 1 from A to C
move disk 2 from A to B
move disk 1 from C to B
move disk 3 from A to C
move disk 1 from B to A
move disk 2 from B to C
move disk 1 from A to C

✓ After

void towerOfHanoi(int n, char from, char to, char aux) {
  if (n == 1) {
    printf("Move disk 1 from %c to %c\n", from, to);
    return;
  }
  towerOfHanoi(n-1, from, aux, to);
  printf("Move disk %d from %c to %c\n", n, from, to);
  towerOfHanoi(n-1, aux, to, from);
}

What It Enables

This problem teaches how to solve complex tasks by breaking them into smaller, repeatable steps using recursion.

Real Life Example

Moving boxes between shelves with size limits or organizing tasks where some must be done before others can be solved using the same step-by-step approach.

Key Takeaways

Manual planning is hard and error-prone for many disks.

Recursion breaks the problem into smaller, manageable parts.

The Tower of Hanoi solution is a clear, repeatable method for moving disks correctly.

Practice

1. What is the minimum number of moves required to solve the Tower of Hanoi problem with 3 disks?

easy

A. 6

B. 5

C. 9

D. 7

Solution

1. Step 1: Understand the formula for minimum moves

   The minimum moves required for n disks is 2^n - 1.

2. Step 2: Calculate for 3 disks

   For 3 disks, minimum moves = 2^3 - 1 = 8 - 1 = 7.

3. Final Answer:

   7 -> Option D

4. Quick Check:

   2^3 - 1 = 7 [OK]

Hint: Use formula 2^n - 1 for minimum moves [OK]

Common Mistakes:

• Using n instead of 2^n - 1
• Calculating 2*n - 1 instead of 2^n - 1
• Confusing number of disks with moves

2. Which of the following is the correct recursive function signature in C for solving Tower of Hanoi?

easy

A. void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod);

B. int towerOfHanoi(int n, char from_rod, char to_rod);

C. void towerOfHanoi(int n, int from_rod, int to_rod);

D. void towerOfHanoi(int n, char from_rod, char to_rod);

Solution

1. Step 1: Identify parameters needed for Tower of Hanoi

   We need number of disks and names of three rods: source, destination, auxiliary.

2. Step 2: Check function signature matching these parameters

   void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod); has int n and three char rods, matching the problem requirements.

3. Final Answer:

   void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod); -> Option A

4. Quick Check:

   Function needs 3 rods and disk count [OK]

Hint: Tower of Hanoi needs 3 rods as parameters [OK]

Common Mistakes:

• Missing auxiliary rod parameter
• Using int instead of char for rods
• Returning int instead of void

3. What will be the output of the following C code snippet for Tower of Hanoi with 2 disks?

void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 1) {
    printf("Move disk 1 from rod %c to rod %c\n", from_rod, to_rod);
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}

int main() {
  towerOfHanoi(2, 'A', 'C', 'B');
  return 0;
}

medium

A. Move disk 2 from rod A to rod C
Move disk 1 from rod A to rod B
Move disk 2 from rod B to rod C

B. Move disk 1 from rod A to rod C
Move disk 2 from rod A to rod B
Move disk 1 from rod C to rod B

C. Move disk 1 from rod A to rod B
Move disk 2 from rod A to rod C
Move disk 1 from rod B to rod C

D. Move disk 1 from rod B to rod C
Move disk 2 from rod A to rod C
Move disk 1 from rod A to rod B

Solution

1. Step 1: Trace recursive calls for n=2

   First move disk 1 from A to B, then disk 2 from A to C, then disk 1 from B to C.

2. Step 2: Match output with options

   Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C matches the exact sequence of moves printed by the code.

3. Final Answer:

   Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C -> Option C

4. Quick Check:

   Recursive moves for 2 disks = Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C [OK]

Hint: For 2 disks, move smaller disk to aux first [OK]

Common Mistakes:

• Swapping auxiliary and destination rods
• Printing moves in wrong order
• Confusing disk numbers

4. Identify the error in the following Tower of Hanoi C function and select the correct fix:

void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 0) {
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}

medium

A. Change base case condition to if (n == 1) instead of if (n == 0)

B. Add else before recursive calls

C. Change printf to print n-1 instead of n

D. Swap aux_rod and to_rod in recursive calls

Solution

1. Step 1: Analyze base case condition

   Base case should stop recursion when n == 1, not 0, to print the last move.

2. Step 2: Understand effect of current base case

   With n == 0, no moves are printed for the smallest disk, causing no output.

3. Final Answer:

   Change base case condition to if (n == 1) instead of if (n == 0) -> Option A

4. Quick Check:

   Base case must be n == 1 to print moves [OK]

Hint: Base case must be n == 1 to print last move [OK]

Common Mistakes:

• Using n == 0 as base case
• Changing print statement unnecessarily
• Swapping rods incorrectly

5. You want to solve Tower of Hanoi for 4 disks but print only the moves where disk 1 is moved. Which modification to the recursive function is correct?

void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 1) {
    printf("Move disk 1 from rod %c to rod %c\n", from_rod, to_rod);
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}

hard

A. Remove all recursive calls except for n == 1 case

B. Add condition to print only if n == 1 inside the function

C. Print all moves but filter disk 1 moves outside the function

D. Change base case to n == 0 and print only disk 1 moves

Solution

1. Step 1: Understand requirement to print only disk 1 moves

   We must print moves only when disk number is 1 inside the recursive function.

2. Step 2: Modify print statements accordingly

   Keep recursive calls intact but print only when n == 1, skipping other prints.

3. Final Answer:

   Add condition to print only if n == 1 inside the function -> Option B

4. Quick Check:

   Print condition inside recursion for disk 1 only [OK]

Hint: Print moves only when disk number is 1 inside recursion [OK]

Common Mistakes:

• Removing recursive calls breaks solution
• Filtering outside loses recursion context
• Changing base case incorrectly