Tower of Hanoi

Problem Link

Problem

Simulate the Tower of Hanoi puzzle. Given N disks, return the sequence of moves as a tuple of [from, to] pairs.

type Hanoi1 = TowerOfHanoi<1>
// [['A', 'C']]

type Hanoi2 = TowerOfHanoi<2>
// [['A', 'B'], ['A', 'C'], ['B', 'C']]

type Hanoi3 = TowerOfHanoi<3>
// [['A', 'C'], ['A', 'B'], ['C', 'B'], ['A', 'C'], ['B', 'A'], ['B', 'C'], ['A', 'C']]

Solution

type Peg = 'A' | 'B' | 'C'

type TowerOfHanoi<
  N extends number,
  From extends Peg = 'A',
  To extends Peg = 'C',
  Via extends Peg = 'B',
  Count extends unknown[] = []
> =
  Count['length'] extends N
    ? []
    : [
        ...TowerOfHanoi<N, From, Via, To, [...Count, unknown]>,
        [From, To],
        ...TowerOfHanoi<N, Via, To, From, [...Count, unknown]>
      ]

How it works:

1. Classic Tower of Hanoi recursion:
   • Move N-1 disks from From to Via using To as buffer.
   • Move the largest disk directly from From to To.
   • Move N-1 disks from Via to To using From as buffer.
2. Count tracks recursion depth — when Count['length'] === N we've reached the base case (0 disks = no moves).
3. Results are concatenated with spread into a flat tuple of [from, to] pairs.

Key Takeaways

• Using Count as a depth counter is the standard technique for implementing "decrement N" recursion in TypeScript types.
• The algorithm structure is identical to the runtime Tower of Hanoi: the type system just evaluates it statically.
• Result tuples are assembled by spreading three pieces: left recursive result + current move + right recursive result.