• akash_rawal@lemmy.world
    link
    fedilink
    arrow-up
    7
    ·
    7 months ago

    I didn’t know the answer either, but usually you can compose solution from solutions of smaller problems.

    solution(0): There are no disks. Nothing to do. solution(n): Let’s see if I can use solution(n-1) here. I’ll use solution(n-1) to move all but last disk A->B, just need to rename the pins. Then move the largest disk A->C. Then use solution(n-1) to move disks B->C by renaming the pins. There we go, we have a stack based solution running in exponential time.

    It’s one of the easiest problem in algorithm design, but running the solution by hand would give you a PTSD.

    • CanadaPlus@lemmy.sdf.org
      link
      fedilink
      arrow-up
      2
      ·
      edit-2
      7 months ago

      Good for you. I think I’d figure it out eventually, but it would certainly take me a while.

      I’d probably be trying a number of approaches, including the recursive one. Renaming pegs is a critical piece that you’d have to realise you can do, and you can’t be sure you have a correct inductive solution unless you actually walk through the first few solutions from the base instance.