advanced28 min studyLesson 638 in Mathematics

Random regular graphs and configuration models

Distinguish uniform half-edge matchings, their multigraph projections, and conditioning on simplicity.

On this page

What you’ll understand

  • Count configurations using labeled half-edges and perfect matchings.
  • Compute loop expectations and exact small simplicity probabilities.
  • Explain when conditioning transfers high probability statements to uniform simple regular graphs.

Independent edges are convenient, but they do not fix the degrees. The configuration model fixes a degree sequence by pairing labeled half-edges. The resulting object is naturally a multigraph: loops and parallel edges are allowed. Confusing that distribution with the uniform simple regular graph can invalidate an otherwise correct calculation. Conditioning on simplicity repairs the distinction, provided the conditioning event is possible and not too rare for the intended transfer.

Uniform pairings live above graphs

For degrees d_v whose sum L is even, give vertex v exactly d_v labeled half-edges and choose a uniformly random perfect matching of all L half-edges. Each matched pair becomes an edge. Two half-edges at the same vertex produce a loop, which contributes two to that vertex’s degree. Different matched pairs between the same two vertices produce parallel edges. The sample space consists of matchings, not multigraphs; different multigraphs can correspond to different numbers of matchings, so their projected probabilities need not be equal.

For L positive and even, the number of perfect matchings is the product of the odd numbers through L−1. A specified half-edge’s partner is uniform among the other L−1 half-edges. Thus a specified pair is matched with probability 1/(L−1). Summing over pairs within each vertex gives the expected number of loops. This expectation is exact, but does not equal the probability of at least one loop unless one has ruled out multiple loops. Markov supplies only an upper bound on that probability.

#pairings=(L−1)!!,E[loops]=∑v(dv2)L−1,L≥2\#\text{pairings}=(L-1)!!,\qquad \mathbb E[\text{loops}]=\frac{\sum_v\binom{d_v}{2}}{L-1},\quad L\ge2
Half-edges are labeled, L is even, and each loop corresponds to exactly one matched pair within a vertex.↔ Scroll wide equations horizontally.

Why conditioning produces a uniform simple graph

Assume a simple graph with the prescribed labeled degrees exists. For each such graph, assign its incident edges bijectively to the labeled half-edges at each vertex. This gives exactly the product of d_v factorials different pairings projecting to that graph, independent of which simple graph was chosen. Consequently the configuration model conditioned on producing a simple graph is uniform over all simple labeled graphs with that degree sequence. This constant-multiplicity argument applies on the simple event, not indiscriminately across multigraphs.

For a fixed nonnegative integer d, along n with nd even and n large enough, the simplicity probability of the d-regular configuration model tends to exp⁡(−(d2−1)/4)\exp(-(d^2-1)/4) when d is positive; d=0 is trivially simple. In particular it stays bounded away from zero for fixed d. Therefore an event whose failure probability tends to zero in the configuration model also holds with high probability after conditioning on simplicity. If d grows with n, that fixed-d formula and the same transfer conclusion must not be asserted without additional hypotheses.

P(Bc∣S)≤P(Bc)P(S),S={configuration is simple}\mathbb P(B^{\mathrm c}\mid S)\le\frac{\mathbb P(B^{\mathrm c})}{\mathbb P(S)},\qquad S=\{\text{configuration is simple}\}
A positive lower bound on P(S) is sufficient to transfer an unconditional high probability statement B to the simple model.↔ Scroll wide equations horizontally.

Expose pairings rather than pretending edges are independent

A matching can be sampled by repeatedly choosing a partner uniformly for one unmatched half-edge. The available partner set shrinks after each step, so different pairing events are dependent. Conditional probabilities are still simple when expressed in terms of unmatched half-edges. This makes the model useful for neighborhood exploration and counting small configurations, but it does not justify independent-edge Chernoff calculations for arbitrary adjacency indicators. Fixed vertex degree itself demonstrates the dependence: one additional neighbor can constrain the remaining choices.

Exact small examples are useful checks because the limiting simplicity formula need not be accurate at small n. They also expose the distinction between a number of graphs and a number of pairings. For degree three on four labeled vertices there is only one simple graph, the complete graph. It still receives many pairing realizations because its incident edges can be assigned to labeled half-edges independently at each vertex. Counting only that one graph in the pairing sample space would severely understate its probability.

The unique simple cubic graph on four vertices

For n=4 and d=3, calculate the number of configurations, expected loops, and exact probability that the configuration is simple.

  1. There are twelve labeled half-edges. The number of perfect matchings is 11!!=11⋅9⋅7⋅5⋅3⋅1=1039511!!=11\cdot9\cdot7\cdot5\cdot3\cdot1=10395.

  2. Each vertex supplies (32)=3\binom32=3 potential loop pairs. Across four vertices, loop expectation is 12/1112/11, which can exceed one because it is a count, not a probability.

  3. The only simple cubic graph on four labeled vertices is K₄. It has (3!)4=1296(3!)^4=1296 pairing realizations. Hence simplicity probability is 1296/10395=48/385≈0.1246751296/10395=48/385\approx0.124675.

The result

There are 10395 configurations, expected loops 12/11, and exact simplicity probability 48/385. Uniformity applies after conditioning, rather than to projected multigraphs in general.

Explore the model

Interactive lab

Interactive lab · probability

Fixed degree versus a binomial marginal

Pause and Reset, then set p = 0.50 and n = 8. Predict mean four and variance two for the displayed degree-like count, as for one vertex in G(9,1/2). Compare it with a degree-four configuration model on nine vertices, where every degree is exactly four and has variance zero, even before simplicity conditioning. Reset and set both p = 0.25 and n = 8 for an independent-edge comparison. This instrument has no pairing or simplicity controls and does not sample regular graphs.

Fixed degree versus a binomial marginalThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMatching an expected degree is weaker than fixing every degree in the sample space.

Check your understanding

Why is the configuration model conditioned on simplicity uniform over simple graphs with the prescribed labeled degrees?

Try it yourself

For three labeled vertices each of degree two, count pairings, expected loops, and exact simplicity probability. Identify the simple graph after conditioning.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

There are six labeled half-edges. A simple degree-two graph on three vertices must be a triangle, and each vertex has two assignments of its incident edges to half-edges.

Compare with the solution

There are 5!!=155!!=15 pairings. The expected loop count is 3(22)/5=3/53\binom22/5=3/5. The unique simple graph is the triangle, whose pairing multiplicity is (2!)3=8(2!)^3=8. Thus the exact simplicity probability is 8/158/15. After conditioning the graph is certainly that labeled triangle. The expectation 3/5 is not the probability of nonsimplicity: nonsimplicity has probability 7/15 and includes configurations whose multiple defects make naive counting overlap. The calculation uses matching multiplicity directly, so it needs no assumption that loop events or possible adjacency events are independent.

Take it with you

  • Uniform labeled half-edge matchings produce multigraphs with fixed degrees.
  • Equal simple-graph multiplicity proves uniformity only after simplicity conditioning.
  • High probability transfer needs control of the conditioning probability, especially when degrees grow.

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