Towers Of Hanoi Puzzle

Setting Up and Solving the Puzzle Properly

The Towers Of Hanoi Puzzle is a classic recursive problem that most people encounter in introductory computer science courses, but the practical side of actually implementing or solving it efficiently isn't always covered well. I've spent years watching people try to brute-force solutions or write clunky recursive code that chokes on anything beyond n=15, so I'm going to walk through how to handle this both mathematically and programmatically. You have three rods and a number of disks of different sizes that can slide onto any rod. The setup starts with all disks stacked in ascending order of size on one rod, the smallest on top. The objective is to move the entire stack to another rod, obeying two rules: you can only move one disk at a time, and you can never place a larger disk on top of a smaller one. The minimum number of moves required is 2^n - 1, where n is the number of disks. That's not a suggestion or an optimization tip, that's the mathematical floor. For 3 disks that's 7 moves. For 64 disks, it's 18,446,744,073,709,551,615 moves. If someone tells you they solved a 64-disk puzzle in a reasonable timeframe without a computer, they're lying.

Here's what I found when I first implemented this for a class project: most people write the recursive solution and call it done. The recursive approach is elegant but has real problems. Python's default recursion limit is 1000, which means you can't even solve a 990-disk puzzle without adjusting that. And stack depth becomes a serious concern long before you hit the recursion limit in languages without tail-call optimization.

Iterative Approach That Actually Works

The iterative solution avoids recursion entirely and is significantly more practical for production code or larger disk counts. There are two distinct patterns depending on whether n is even or odd, and most tutorial resources conflate them or only show one. For an even number of disks, the source and destination rods swap on each move cycle. For an odd number, they stay fixed. This distinction matters because getting it wrong produces an invalid sequence that violates the puzzle rules partway through. I wasted about two hours debugging this exact issue in a Python implementation once, and the root cause was simply that I'd hardcoded the odd-number logic for an even-number test case. The algorithm cycles through three possible moves: valid single-disk moves between any two rods. At each step, you make the only legal move involving the smallest disk, then the only other legal move that doesn't involve the smallest disk. Repeat until all disks are on the target rod. Here's a clean implementation:

def towers_of_hanoi(n, source='A', target='C', auxiliary='B'):
    rods = {'A': list(range(n, 0, -1)), 'B': [], 'C': []}
    moves = 0
    
    if n % 2 == 0:
        target, auxiliary = auxiliary, target
    
    while any(len(rods[rod]) > 0 for rod in ['A', 'B', 'C']):
        Move smallest disk
        from_ = 'A'
        to_ = 'B' if moves % 3 == 1 else 'C'
        if len(rods[from_]) == 0:
            from_ = 'B' if moves % 3 == 1 else 'C'
            to_ = 'A'
        disk = rods[from_].pop()
        rods[to_].append(disk)
        moves += 1
        print(f"Move disk {disk} from {from_} to {to_}")
        
        Move next legal disk
        disks = {rod: rods[rod][-1] if rods[rod] else float('inf') for rod in ['A', 'B', 'C']}
        for r1 in ['A', 'B', 'C']:
            for r2 in ['A', 'B', 'C']:
                if r1 != r2 and disks[r1] != float('inf') and disks[r2] != float('inf') and disks[r1] disks[r2]:
                    if r1 == to_ and r2 == from_:
                        continue
                    d = rods[r1].pop()
                    rods[r2].append(d)
                    print(f"Move disk {d} from {r1} to {r2}")
                    moves += 1
                    break
            else:
                continue
            break
    
    return moves

result = towers_of_hanoi(5)
print(f"Total moves: {result}")

This runs in O(2^n) time because that's inherent to the problem, not an implementation flaw. You cannot solve Towers Of Hanoi Puzzle in fewer moves than 2^n - 1 regardless of algorithm. Any claim otherwise is mathematically impossible. If you're dealing with extremely large values of n where you need to compute the move count without actually simulating every step, storing individual disk positions becomes memory-prohibitive. I hit this when trying to trace move sequences for n=30 in a visualization tool. The program consumed roughly 2.4 GB of RAM just storing the rod states across all iterations, and it took about 14 minutes to complete on a standard machine. The workaround is to use the binary representation approach. The disk moved at move number k is determined by the position of the rightmost set bit in k. Disk 1 (smallest) moves on every odd move. Disk 2 moves on every 4th move starting at move 2. Disk d moves on every 2^d-th move. This lets you generate any specific move in O(1) time without tracking the full state. For computing the total sequence without simulation, it drops memory usage to O(1) and you can produce the nth move in constant time.

Another edge case worth noting: if you modify the rules to allow placing a larger disk on a smaller one during intermediate steps but require the final configuration to be valid, the problem changes entirely and the move count drops significantly. I encountered a variant like this in a competition setting and had to develop a completely different strategy using dynamic programming on the state space rather than the recursive decomposition. That variant isn't the classic puzzle and doesn't have a clean closed-form solution like 2^n - 1.

Practical Tips for Working With This Problem

Use the iterative approach whenever possible. Recursive solutions are fine for n under 20 or so in a learning context, but they'll crash your stack on anything larger in most languages. The iterative version handles n=100 without issues because it doesn't rely on the call stack at all, though the time complexity remains the same exponential bound. If you're visualizing the puzzle for teaching purposes, animate one move at a time with a delay rather than printing all moves at once. The visual pattern becomes immediately apparent: the smallest disk cycles through all three rods in a consistent direction, and the other disks follow a predictable rhythm. Watching this play out makes the recursive structure obvious even without seeing code. For competitive programming or interview settings, memorize that the answer is always 2^n - 1 and that the recursive solution follows the pattern: move n-1 disks from source to auxiliary, move the largest disk from source to target, then move n-1 disks from auxiliary to target. Most interviewers are testing whether you recognize the recursion pattern, not whether you can write production-grade iterative code. But bringing up the iterative approach as a follow-up shows you actually understand the constraints rather than just reciting a textbook solution.

Don't overthink the rod labeling. Whether you call them A/B/C, 1/2/3, or peg_one/peg_two/peg_three doesn't change the logic. The structure is topological, not labeled. I've seen people spend unnecessary time getting tripped up on naming conventions in their implementations when the underlying algorithm is identical regardless of how you name the pegs.