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

Question

I had a look at question already which talk about algorithm to find loop in a linked list. I have read Floyd\'s cycle-finding algorithm solution, mentioned at lot of places

Answer 1

Consider a cycle of size L, meaning at the kth element is where the loop is: x_k -> x_k+1 -> ... -> x_k+L-1 -> x_k. Suppose one pointer is run at rate r₁=1 and the other at r₂. When the first pointer reaches x_k the second pointer will already be in the loop at some element x_k+s where 0 <= s < L. After m further pointer increments the first pointer is at x_{k+(m mod L)} and the second pointer is at x_{k+((m*r2+s) mod L)}. Therefore the condition that the two pointers collide can be phrased as the existence of an m satisfying the congruence

m = m*r₂ + s (mod L)

This can be simplified with the following steps

m(1-r₂) = s (mod L)

m(L+1-r₂) = s (mod L)

This is of the form of a linear congruence. It has a solution m if s is divisible by gcd(L+1-r₂,L). This will certainly be the case if gcd(L+1-r₂,L)=1. If r₂=2 then gcd(L+1-r₂,L)=gcd(L-1,L)=1 and a solution m always exists.

Thus r₂=2 has the good property that for any cycle size L, it satisfies gcd(L+1-r₂,L)=1 and thus guarantees that the pointers will eventually collide even if the two pointers start at different locations. Other values of r₂ do not have this property.