CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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A connected finite undirected multigraph has an Euler trail if and only if it has zero or two odd-degree vertices; an open Euler trail occurs exactly in the two-vertex case

Statement

A connected finite undirected multigraph has an Euler trail if and only if the number of odd-degree vertices is zero or two. It has an open Euler trail, one whose endpoints are distinct, if and only if it has exactly two odd-degree vertices; those vertices are the endpoints.

Facts & Assumptions

Given: A connected finite undirected multigraph $M$ .

[F1]

An Euler trail uses every edge exactly once and is an Euler circuit when closed (Euler trails and Euler circuits in multigraphs and digraphs).

[F2]

Loops count twice toward undirected degree (Degree in a multigraph, indegree and outdegree in a digraph, and their underlying connectivity).

[L1]

A connected finite undirected multigraph has an Euler circuit exactly when all degrees are even (Euler's theorem and Hierholzer's construction: a connected finite undirected multigraph has an Euler circuit if and only if every degree is even).

Proof

technique · direct

1.1

Along any trail, every vertex other than two distinct endpoints uses incident edge ends in pairs. If the trail is open, its two endpoints each use one unpaired end; if it is closed, every used incidence is paired. Loops contribute pairs by [F2].

F1F2algebra

1.2

If every degree is even, [L1] supplies an Euler circuit, which is an Euler trail.

L1F1

1.3

If exactly $u$ and $v$ have odd degree, add one distinguished edge between $u$ and $v$ , allowing it to be parallel to an existing edge. The augmented connected multigraph has all degrees even, so [L1] supplies an Euler circuit. Rotate that circuit to begin with the distinguished edge and delete it; what remains is an Euler trail in $M$ from $v$ to $u$ .

L1F1F2algebra

2.1

Therefore an Euler trail forces exactly two odd-degree vertices when open and zero odd-degree vertices when closed.

step 1.1F1

3.1

Steps 2.1, 1.2, and 1.3 prove the criterion and show that openness occurs exactly in the two-odd-vertex case.

step 2.1step 1.2step 1.3∎

Depends on

Used by

The Konigsberg bridges multigraph has four odd-degree vertices and no Euler trail Example

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Applied Combinatorics, Euler Tours and Trails (standard reference, not scraped)