this post was submitted on 05 May 2024
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[โ€“] akash_rawal@lemmy.world 7 points 6 months ago (1 children)

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.

[โ€“] CanadaPlus@lemmy.sdf.org 1 points 6 months ago* (last edited 6 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.