tower of hanoi - How to not skip over a peg every recursion

Question

My assignment is to solve the Towers of Hanoi for any number using recursion. I wrote my code in C++.

Rules:

1. Cannot stack a larger disk on

Answer 1

Basically on each call you need to do the following steps:

1. Move the n-1 stack of discs to the 4th peg (or 2nd when going backwards)

2. Move the nth disc to the middle (3rd peg)

3. Move the n-1 stack back to the 2nd peg (i.e. 4th when going backwards)

4. Move the nth disc to its destination

5. Move the n-1 stack to the destination

So you need 3 recursive calls without memoization.

function hanoi5(n) {
  let out = []
  move(n, 0, 4)
  console.log(out.length + " steps")
  console.log(out.join("\n"))
  function move(n, from, to) {
    if (n == 1) {
      let dir = from < to ? 1 : -1
      for (let i = from; i != to; i += dir)
        out.push("move disc 1 from peg " + (i+1) + " to peg " + (i+1+dir))
    } else if (from < to) {
      move(n - 1, from, 3)
      for (let i = from; i < 2; i++)
        out.push("move disc " + n + " from peg " + (i+1) + " to peg " + (i+2))
      move(n - 1, 3, 1)
      for (let i = 2; i < to; i++)
        out.push("move disc " + n + " from peg " + (i+1) + " to peg " + (i+2))
      move(n - 1, 1, to)
    } else {
      move(n - 1, 3, 1)
      out.push("move disc " + n + " from peg 3 to peg 2")
      move(n - 1, 1, 3)
      out.push("move disc " + n + " from peg 2 to peg 1")
      move(n - 1, 3, 1)
    }
  }
}

hanoi5(3)