Why Tower of Hanoi Problem in DSA C++?

The Big Idea

Can a simple set of rules 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. Your goal is to move all disks to another peg, but you can only move one disk at a time and never place a bigger disk on top of a smaller one. Trying to figure out the steps manually for many disks becomes confusing very fast.

The Problem

Manually planning each move is slow and easy to mess up. As the number of disks grows, the number of moves grows exponentially, making it almost impossible to keep track without errors or forgetting rules.

The Solution

The Tower of Hanoi problem uses a simple, clear set of rules and recursion to break down the big problem into smaller, manageable steps. This method ensures every move is valid and the puzzle is solved efficiently without guesswork.

Before vs After

✗ Before

void moveDisk(int fromPeg, int toPeg) {
  // Manually move disks one by one
  // Hard to track and error-prone
}

✓ After

void towerOfHanoi(int n, int fromPeg, int toPeg, int auxPeg) {
  if (n == 1) {
    std::cout << "Move disk 1 from peg " << fromPeg << " to peg " << toPeg << std::endl;
    return;
  }
  towerOfHanoi(n-1, fromPeg, auxPeg, toPeg);
  std::cout << "Move disk " << n << " from peg " << fromPeg << " to peg " << toPeg << std::endl;
  towerOfHanoi(n-1, auxPeg, toPeg, fromPeg);
}

What It Enables

This problem teaches how to solve complex tasks by breaking them into smaller, repeatable steps, enabling efficient problem solving in many areas.

Real Life Example

Moving a stack of plates from one shelf to another without breaking or mixing their order, or organizing tasks that depend on each other in the right sequence.

Key Takeaways

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

Recursion breaks the problem into smaller, manageable steps.

Ensures all moves follow the rules and solve the puzzle efficiently.

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 base case condition in the Tower of Hanoi recursive function in C++?

easy

A. if (n == 1) move disk from source to destination;

B. if (n == 0) return;

C. if (n > 1) call recursive function;

D. if (n < 0) return;

Solution

1. Step 1: Identify the base case for recursion

   The base case is when there is only 1 disk to move, so move it directly.

2. Step 2: Check the options

   if (n == 1) move disk from source to destination; correctly states the base case: if (n == 1) move disk from source to destination.

3. Final Answer:

   if (n == 1) move disk from source to destination; -> Option A

4. Quick Check:

   Base case is n == 1 [OK]

Hint: Base case moves single disk directly [OK]

Common Mistakes:

• Using n == 0 as base case
• Not defining a base case
• Using incorrect condition like n < 0

3. What will be the output sequence of moves for Tower of Hanoi with 2 disks from peg A to peg C using peg B as auxiliary?

medium

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

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

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

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

Solution

1. Step 1: Understand the recursive move sequence for 2 disks

   Move top disk (1) from A to B, move bottom disk (2) from A to C, then move disk (1) from B to C.

2. Step 2: Match the sequence with options

   Move disk 1 from A to B Move disk 2 from A to C Move disk 1 from B to C matches the correct sequence exactly.

3. Final Answer:

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

4. Quick Check:

   Correct move order for 2 disks [OK]

Hint: For 2 disks: move top to aux, bottom to dest, top to dest [OK]

Common Mistakes:

• Swapping auxiliary and destination pegs
• Moving larger disk before smaller disk
• Incorrect move order

4. Identify the error in the following Tower of Hanoi recursive function snippet in C++:

void towerOfHanoi(int n, char source, char aux, char dest) {
  if (n == 1) {
    cout << "Move disk 1 from " << source << " to " << dest << endl;
    return;
  }
  towerOfHanoi(n - 1, source, dest, aux);
  cout << "Move disk " << n << " from " << source << " to " << dest << endl;
  towerOfHanoi(n - 1, aux, source, dest);
}

medium

A. The order of auxiliary and destination pegs is swapped in recursive calls.

B. Base case condition should be n == 0 instead of n == 1.

C. The function does not print the moves correctly.

D. The function lacks a return statement after recursive calls.

Solution

1. Step 1: Analyze recursive calls parameters

   The first recursive call should move n-1 disks from source to auxiliary using destination as auxiliary, so parameters should be (n-1, source, dest, aux).

2. Step 2: Identify incorrect parameter order

   In the code, the first call uses (n-1, source, aux, dest) which swaps aux and dest incorrectly.

3. Final Answer:

   The order of auxiliary and destination pegs is swapped in recursive calls. -> Option A

4. Quick Check:

   Correct peg order is crucial in recursion [OK]

Hint: Check peg order in recursive calls carefully [OK]

Common Mistakes:

• Swapping auxiliary and destination pegs
• Wrong base case condition
• Missing return statements

5. If the Tower of Hanoi problem is modified to have 4 pegs instead of 3, which of the following statements is true about the minimum number of moves required to move n disks?

hard

A. It remains the same as 3 pegs: 2^n - 1 moves.

B. It is always more than 2^n - 1 moves due to extra peg.

C. It is always less than 2^n - 1 moves but no exact formula is known.

D. It is exactly n^2 moves for any n.

Solution

1. Step 1: Understand the classic formula for 3 pegs

   For 3 pegs, minimum moves = 2^n - 1.

2. Step 2: Consider 4 pegs case

   With 4 pegs, the problem is more complex; fewer moves than 2^n - 1 are possible but no simple closed formula exists.

3. Final Answer:

   It is always less than 2^n - 1 moves but no exact formula is known. -> Option C

4. Quick Check:

   4 pegs reduce moves but formula unknown [OK]

Hint: More pegs reduce moves but no simple formula [OK]

Common Mistakes:

• Assuming formula 2^n - 1 applies to 4 pegs
• Thinking more pegs increase moves
• Assuming quadratic moves for 4 pegs