If we want to detect and **odd cycle** if an undirected graph $G=<V,E>$. Suppose we run BFS algorithm from CLRS book as follows,

**Q**: Now my question is suppose we have the following graphs:

The figure to the left above:

- $d[v] bmod{2} == d[u] bmod{2}$ equal and gives odd cycle.
- $d[v] bmod{2} == d[a] bmod{2}$ not equal remainder, so won’t give odd cycle.

The figure to the right above:

- $d[x] bmod{2} == d[u] bmod{2}$ not equal remainder, so won’t give odd cycle.
- $d[x] bmod{2} == d[v] bmod{2}$ not equal remainder, so won’t give odd cycle.

**Problem**: so the only case to discover odd cycle is when we have the left right graph and when exactly we compare vertex $v$ distance with $u$ distance. What do you think please? If the mod operator works, can you give your interpretation as why this should work in general please if possible?