# Solving $UCYCLE$ in logspace - two possible approaches ? Why can't we one of them use to solve connectivity?

$$UCYLE = \mathcal \{ <G> ~:~ G \text{ is an undirected graph that contains a simple cycle}\}.$$
There are two possible approaches to this exercise: Solving cycle in undirected graph in log space?

These two solutions are different. The instresting solution is second one (on stackoverflow). In this case we generally implement DFS in logspace. I can see that it does work (using DFS). However the question is:
Why we can't use this DFS algorithm to solve USTCON problem ?

However, I can see that we can't easily use this DFS to solve bipartity problem. It is not easy to count number of nodes in cycle.

Edit
Modified algorithm from stackoverflow (mentioned link) which should solve ustcon problem:

    for v_j in neighbours(s)
current, prev = v_j, s
repeat
idx = neighbours(current).index(v_j)
idx = (idx + 1) % len(neighbours(current))
current, prev = neighbours(current)[idx], current
if current = t then return connection between s and t

• Why do you think we can't use it? What we can't is to show that USTCON is in $\mathsf{L}$ using it. Because you will need to keep in memory every adjacent cycle. Aug 9 '17 at 18:12