Bipartite Testing When No Odd-Length Cycles Exist: Best Known Time Complexity

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Sep 13, 2026

To decide whether a graph $G=(V,E)$ is bipartite you can attempt a 2-coloring consistent with the edges. For a general graph, the standard approach is BFS/DFS over each connected component, assigning colors; any conflict implies the graph is not bipartite.

In particular, a graph is bipartite iff it contains no odd cycle. Therefore, if you are already guaranteed the graph “doesn’t contain cycle of odd length,” the correct conclusion is that the graph is bipartite; however, the question asks for the time complexity of the best known algorithm to find out whether it is bipartite or not.

That best known decision algorithm is the linear-time bipartite test via two-coloring:

Running time: $\mathbf{O(n+m)}$ , where $n=|V|$ and $m=|E|$ .

This matches the cost of visiting each vertex and scanning each edge a constant number of times in BFS/DFS-based two-coloring.

Below is a structured view of the algorithm and why it is linear.

Bipartite Graph Check (BFS/DFS Two-Coloring) — Time Complexity

Key concepts

Bipartiteness is equivalent to absence of odd cycle.
A BFS/DFS-based test is essentially constraint propagation via coloring.
The work is linear because each edge is examined during traversal.

Best-known bipartite decision algorithm (two-coloring with BFS/DFS)

1
Step 1
Create an array color[ ] initialized to "uncolored" for all $n$ vertices.
2
Step 2
For every vertex $s$ with color[s] = uncolored, run BFS/DFS starting from $s$ and set color[s] = 0.
3
Step 3
When exploring an edge $(u,v)$ , if color[v] is uncolored, set color[v] = 1 − color[u].
4
Step 4
If color[v] is already assigned and color[v] = color[u], you found an odd cycle witness via the conflict; return "not bipartite".
5
Step 5
If all components are processed without conflict, return "bipartite".

Time complexity reasoning

The BFS/DFS two-coloring algorithm performs:

$O(n)$ work to iterate over vertices/components.
$O(m)$ edge examinations across all traversals (each edge is considered a constant number of times, up to representation details like adjacency lists).

Hence total runtime is:

O(n+m).

Callout: Even if the graph is guaranteed to have no odd cycle (and thus is bipartite), the best-known decision algorithm for bipartiteness is still the same linear-time two-coloring method. The guarantee can let you skip verification only if you accept the premise as absolute input information rather than a property to test.

Why $O(n+m)$ is optimal (decision setting)

Any correct algorithm must at least read enough of the input to observe edges that could create a coloring conflict, which naturally leads to linear traversal in an adjacency-list model; BFS/DFS two-coloring achieves this bound.

Be careful about the wording 'given that...'

If you are guaranteed there is no odd cycle, the answer is logically immediate: the graph is bipartite. But the question asks for the time complexity of the best-known algorithm to determine bipartiteness in general—whose runtime is $O(n+m)$ .

Work Breakdown for BFS/DFS Two-Coloring Bipartite Test

Typical adjacency-list implementation costs

How the test finds an odd cycle (conceptual)

Pick a component root

Start

Assign an arbitrary color to the root vertex."

Color neighbors via edges

Propagation

Enforce opposite colors along every edge."

Edge constraint conflict

Contradiction

A conflict implies the existence of an odd cycle."

No conflict found

Completion

All components can be consistently 2-colored => bipartite."

Common interpretations and edge cases

Knowledge Check

Question 1 of 4

Q1Single choice

What is the best-known time complexity to decide whether a graph is bipartite (given $n$ vertices and $m$ edges) using BFS/DFS two-coloring with an adjacency-list representation?

O(n+m)

O(n^2)

O(m log n)

O(2^n)

Final answer

The best-known algorithm to decide bipartiteness runs in $O(n+m)$ time using BFS/DFS-based two-coloring on an adjacency-list representation.

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