Why Tower of Hanoi Problem in DSA Javascript?

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. Your goal is 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 moves manually gets confusing very fast as the number of disks grows.

The Problem

Manually planning each move is slow and easy to mess up. You might forget rules or repeat moves, causing frustration and mistakes.

It's like trying to solve a puzzle without a clear plan, which wastes time and effort.

The Solution

The Tower of Hanoi algorithm breaks the problem into smaller steps using a simple rule: move smaller stacks first, then the largest disk, then the smaller stacks again.

This step-by-step method ensures you never break the rules and solves the puzzle efficiently.

Before vs After

✗ Before

function moveDisk(from, to) {
  console.log(`Move disk from ${from} to ${to}`);
}
// Manually write each move for 3 disks
moveDisk('A', 'C');
moveDisk('A', 'B');
moveDisk('C', 'B');
// ... many more moves needed

✓ After

function towerOfHanoi(disks, from, to, aux) {
  if (disks === 1) {
    console.log(`Move disk from ${from} to ${to}`);
    return;
  }
  towerOfHanoi(disks - 1, from, aux, to);
  console.log(`Move disk from ${from} to ${to}`);
  towerOfHanoi(disks - 1, aux, to, from);
}
towerOfHanoi(3, 'A', 'C', 'B');

What It Enables

This algorithm lets you solve complex stacking puzzles quickly and correctly, no matter how many disks you have.

Real Life Example

It's like organizing boxes from one shelf to another using a helper shelf, making sure you never put a heavier box on a lighter one.

Key Takeaways

Manual moves get confusing as disks increase.

The algorithm uses a simple, repeatable rule to solve the puzzle.

It guarantees the fewest moves without breaking rules.

Practice

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

easy

A. 7

B. 6

C. 8

D. 9

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 A

4. Quick Check:

   2^3 - 1 = 7 [OK]

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

Common Mistakes:

• Forgetting to subtract 1
• Using n instead of 2^n - 1
• Confusing number of disks with moves

2. Which of the following is the correct base case condition in a recursive Tower of Hanoi function in JavaScript?

easy

A. if (n < 0) return;

B. if (n === 1) moveDisk();

C. if (n > 1) return;

D. if (n === 0) return;

Solution

1. Step 1: Identify base case for recursion

   The recursion stops when there are no disks to move, i.e., n === 0.

2. Step 2: Check the options

   if (n === 0) return; correctly returns when n is 0, stopping recursion.

3. Final Answer:

   if (n === 0) return; -> Option D

4. Quick Check:

   Base case is n === 0 [OK]

Hint: Base case stops recursion at zero disks [OK]

Common Mistakes:

• Using n === 1 as base case incorrectly
• Not stopping recursion causing infinite calls
• Using invalid conditions like n < 0

3. What will be the output sequence of moves for 2 disks in the Tower of Hanoi problem using this code snippet?

function towerOfHanoi(n, from, to, aux) {
  if (n === 0) return;
  towerOfHanoi(n - 1, from, aux, to);
  console.log(`Move disk ${n} from ${from} to ${to}`);
  towerOfHanoi(n - 1, aux, to, from);
}
towerOfHanoi(2, 'A', 'C', 'B');

medium

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

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

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

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

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 C

4. Quick Check:

   Correct move sequence for 2 disks = "Move disk 1 from A to B" "Move disk 2 from A to C" "Move disk 1 from B to C" [OK]

Hint: Follow recursion calls step-by-step [OK]

Common Mistakes:

• Mixing source and auxiliary rods
• Confusing disk numbers in moves
• Ignoring order of recursive calls

4. Identify the error in this Tower of Hanoi recursive function and select the fix:

function towerOfHanoi(n, from, to, aux) {
  if (n === 1) {
    console.log(`Move disk 1 from ${from} to ${to}`);
    return;
  }
  towerOfHanoi(n - 1, from, to, aux);
  console.log(`Move disk ${n} from ${from} to ${to}`);
  towerOfHanoi(n - 1, aux, to, from);
}

medium

A. Swap 'to' and 'aux' in the first recursive call

B. Change base case to n === 0

C. Remove the second recursive call

D. Add an extra console.log before recursion

Solution

1. Step 1: Analyze recursive calls

   The first recursive call should move n-1 disks from 'from' to 'aux' using 'to' as auxiliary, but here 'to' and 'aux' are swapped.

2. Step 2: Correct the argument order

   Swapping 'to' and 'aux' in the first recursive call fixes the logic to move disks correctly.

3. Final Answer:

   Swap 'to' and 'aux' in the first recursive call -> Option A

4. Quick Check:

   Correct recursive call order fixes error [OK]

Hint: Check recursive call parameters carefully [OK]

Common Mistakes:

• Mixing auxiliary and target rods
• Incorrect base case causing infinite recursion
• Removing necessary recursive calls

5. You want to solve the Tower of Hanoi problem for 4 disks but only want to print moves where disk 1 is moved. How would you modify this function?

function towerOfHanoi(n, from, to, aux) {
  if (n === 0) return;
  towerOfHanoi(n - 1, from, aux, to);
  console.log(`Move disk ${n} from ${from} to ${to}`);
  towerOfHanoi(n - 1, aux, to, from);
}

hard

A. Change base case to n === 1 and print only then

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

C. Remove recursive calls and print only disk 1 move

D. Print all moves but filter output outside the function

Solution

1. Step 1: Understand the requirement

   We want to print moves only when disk 1 moves, but still perform all recursive calls to solve the problem.

2. Step 2: Modify print statement with condition

   Add an if condition to print only when n === 1, so other moves are not printed but recursion continues.

3. Final Answer:

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

4. Quick Check:

   Conditional print inside recursion solves problem [OK]

Hint: Use if (n === 1) to print selectively [OK]

Common Mistakes:

• Changing base case breaks recursion
• Removing recursive calls stops solution
• Filtering output outside misses recursion logic