Finding the first node of the cycle using the Floyd algorithm with moving "fast pointer" faster than two nodes per operation?

Question

In Floyd's algorithm, everyone uses a "slow pointer" that moves to the next node per operation, and a "fast pointer" that moves to the next.next node. When they meet, we move the "slow pointer" to the beginning of the List and then move both (by one node per time) until we reach the common node again. This node will be the start of the cycle.

That works with speeds like "slow=1" and "fast=2", but can we do the same but with "fast>2"?

I believe it's fine to have ```fast > 2```. Just need to adjust the speed of the slow pointer accordingly. — Daniel Hao, Mar 10 '23 at 22:13

score 0 · Answer 1 · answered Mar 11 '23 at 08:43

This is the same question as this one:

Why increase pointer by two while finding loop in linked list, why not 3,4,5?

Short answer: you can do with fast>2 but it actually makes total runtime slower so it's not really beneficial.

You can check out the link for a detailed proof.

Finding the first node of the cycle using the Floyd algorithm with moving "fast pointer" faster than two nodes per operation?

1 Answers1