[Paper Review] Note on Counting Eulerian Circuits

Graham Brightwell, Peter Winkler|ArXiv.org|May 18, 2004

Markov Chains and Monte Carlo Methods13 references21 citations

TL;DR

This paper proves that counting Eulerian circuits in undirected graphs is #P-complete by reducing the #P-hard problem of counting Eulerian orientations to counting orbs (Eulerian orientations with arborescences) in multigraphs. Using modular counting via prime moduli and the Chinese Remainder Theorem, the authors show that an oracle for counting orbs enables efficient computation of the number of Eulerian orientations, thereby establishing #P-completeness for Eulerian circuit counting.

ABSTRACT

We show that the problem of counting the number of Eulerian circuits in an undirected graph is complete for the class #P.

Motivation & Objective

• To resolve the long-open complexity question of whether counting Eulerian circuits in undirected graphs is #P-complete.
• To establish a connection between counting Eulerian circuits and the #P-hard problem of counting Eulerian orientations.
• To demonstrate that an oracle for counting orbs (Eulerian orientations with arborescences) enables efficient computation of the number of Eulerian orientations modulo primes.
• To use modular counting and the Chinese Remainder Theorem to reconstruct the exact number of Eulerian orientations from modular counts.
• To show that the reduction technique applies not only to Eulerian orientations but also to other #P-complete problems like not-all-equal 3-SAT counting.

Proposed method

• Construct a multigraph $G_p$ from a given undirected graph $G$ by replacing each edge with $p$ parallel edges and adding a new root vertex connected to all original vertices via two parallel edges.
• Define an 'orb' as a pair of an Eulerian orientation and an arborescence rooted at the new vertex, which corresponds to a unique Eulerian circuit up to a factor of $\prod (d_v - 1)!$.
• Classify orbs by type $\tau$, where $\tau(e) = (k, X)$ indicates how many of the $p$ parallel edges are oriented from lower to higher vertex and whether any is a tree edge.
• Show that non-special types (where $k \notin \{0,p\}$ or $X = T$) contribute counts divisible by $p$, so only special types (with $k \in \{0,p\}$ and $X = F$) contribute modulo $p$.
• Prove that the number of special-type orbs is exactly $2^n \times N$, where $N$ is the number of Eulerian orientations of the original graph $G$, so $N \mod p$ can be recovered.
• Repeat the process for all primes $p$ in the interval $[m, 2m]$ where $m = |E|$, and apply the Chinese Remainder Theorem to reconstruct $N$ exactly.

Experimental results

Research questions

• RQ1 Is counting Eulerian circuits in undirected graphs #P-complete?
• RQ2 Can the number of Eulerian orientations of a graph be computed efficiently using an oracle for counting orbs in multigraphs?
• RQ3 Does modular counting via prime moduli and the Chinese Remainder Theorem allow exact reconstruction of the number of Eulerian orientations?
• RQ4 Can the reduction technique used for Eulerian orientations be extended to other #P-complete problems such as not-all-equal 3-SAT counting?
• RQ5 Is there a fully polynomial randomized approximation scheme (fpras) for counting Eulerian orientations, and does a Markov chain on orbs mix rapidly?

Key findings

• Counting Eulerian circuits in undirected graphs is #P-complete, resolving a 25-year-old open problem.
• The number of orbs of special type in the constructed multigraph $G_p$ is exactly $2^n \times N$, where $N$ is the number of Eulerian orientations of the original graph $G$.
• For any prime $p$ in $[m, 2m]$ with $m = |E|$, the number of orbs in $G_p$ is congruent to $2^n \times N \mod p$, enabling modular computation of $N$.
• The number of orbs of non-special type is divisible by $p$, so their contribution vanishes modulo $p$, ensuring that only special types affect the modular count.
• By repeating the modular computation for all primes in $[m, 2m]$ and applying the Chinese Remainder Theorem, the exact value of $N$ can be reconstructed.
• The same reduction technique applies to not-all-equal 3-SAT counting, showing that it too is #P-complete via reduction to Eulerian orientation counting.