advanced28 min studyLesson 635 in Mathematics

The Lovász local lemma and dependency graphs

Replace a global small-error requirement with local control of rare interacting events.

On this page

What you’ll understand

  • Construct a valid dependency graph using sets of underlying independent variables.
  • Check the symmetric local-lemma condition with the correct maximum degree.
  • Explain why pairwise independence is insufficient and why an inconclusive bound does not show nonexistence.

The union bound charges for every bad event, even when most bad events live in unrelated parts of a large object. The Lovász local lemma can certify simultaneous avoidance using a local interaction bound instead of the total number of constraints. Its power comes with a stronger structural obligation: the dependency graph must encode independence from whole collections of nonneighbors, not merely pairwise independence.

What a dependency graph actually certifies

Let the vertices of a graph index bad events. It is a dependency graph when each event is independent of the sigma-algebra generated by all its nonneighbor events other than itself. Independence from each nonneighbor separately is weaker and does not suffice. An event can be pairwise independent from two others while determined by their joint information. For example, two independent fair bits and their parity are pairwise independent random bits, but the parity is fixed once both original bits are known.

The common safe construction begins with mutually independent underlying variables. Record the variables read by each bad event, and join two events whenever their recorded variable sets intersect. Disjoint supports ensure independence from the entire collection of nonneighbor events, since that collection is measurable using variables disjoint from the event’s own support. Extra edges are harmless for validity but worsen the degree bound. Missing a genuine collective dependence can invalidate the whole theorem, even when every marginal probability was computed correctly.

P(Ai)≤xi∏j∈N(i)(1−xj),0<xi<1⟹P ⁣(⋂iAic)≥∏i(1−xi)>0\mathbb P(A_i)\le x_i\prod_{j\in N(i)}(1-x_j),\quad0<x_i<1\quad\Longrightarrow\quad\mathbb P\!\left(\bigcap_i A_i^{\mathrm c}\right)\ge\prod_i(1-x_i)>0
This finite asymmetric local lemma assumes a valid dependency graph; the displayed condition must hold separately for every event.↔ Scroll wide equations horizontally.

The symmetric criterion is a sufficient test

Suppose each event has probability at most p and the dependency graph has maximum degree at most d. The convenient symmetric criterion is ep(d+1)≤1ep(d+1)\le1. It follows by choosing comparable x-values in the asymmetric statement and bounding the resulting product. The number d counts other events adjacent to a given event, not the size of that event’s variable support and not the total number of events. A huge instance can therefore satisfy the same local criterion as a small one.

For monochromatic edges in a k-uniform hypergraph with fair independent vertex colors, a bad event has probability 21−k2^{1-k}. One may join two hyperedges if they intersect. A bound on how many other edges intersect a given edge supplies d directly. A vertex-degree bound can also supply an upper bound on d, but should account for all k vertices and possible overcounting. Overcounting increases d safely; undercounting can falsely certify a coloring. No algorithmic running-time guarantee is asserted by this existence version.

max⁡iP(Ai)≤p,max⁡i∣N(i)∣≤d,ep(d+1)≤1⟹P(no bad event)>0\max_i\mathbb P(A_i)\le p,\quad\max_i|N(i)|\le d,\quad ep(d+1)\le1\Longrightarrow\mathbb P(\text{no bad event})>0
The maximum degree excludes the event itself; the extra one belongs to the sufficient criterion.↔ Scroll wide equations horizontally.

Conditioning is the heart of the improvement

The proof controls the probability of one bad event after conditioning on avoidance of many others. Nonneighbors can be removed from the difficult conditioning through the dependency property, while neighbors contribute factors bounded using the chosen x-values. This is why a list of pairwise correlations is insufficient: the argument repeatedly conditions on intersections of complements. The asymmetric formulation allows very rare events to compensate for neighbors with different probabilities, rather than forcing every event into one worst-case p.

If the symmetric criterion fails, try the asymmetric version, a tighter support graph, a different random experiment, or another proof. The local lemma is sufficient rather than necessary. It can also give a very small positive lower bound, so simple repeated sampling might still be impractical. Constructive resampling theorems provide additional conclusions in appropriate variable models, but they need their own algorithmic hypotheses and analysis. Here the goal is a valid finite existence certificate with all dependency assumptions exposed.

Many constraints, only twelve local interactions

A seven-uniform hypergraph has one thousand edges, and each edge intersects at most twelve other edges. Certify a proper two-coloring with the symmetric local lemma.

  1. Use independent fair vertex colors. A specified edge is monochromatic with probability 21−7=1/642^{1-7}=1/64.

  2. Join two bad events when their hyperedges intersect. Their supports are sets of independent vertex colors, so this is a valid dependency graph with maximum degree at most twelve.

  3. Compute ep(d+1)=13e/64≈0.552151<1ep(d+1)=13e/64\approx0.552151<1. The local lemma guarantees positive probability that none of the one thousand edges is monochromatic. The union-bound sum 1000/64=15.625 would have been inconclusive.

The result

A proper two-coloring exists. The certificate relies on at most twelve intersecting neighbors per edge, not on independence of all one thousand bad events.

Explore the model

Interactive lab

Interactive lab · probability

A zero-dependency baseline

Pause and Reset, then set p = 0.05 and n = 20. Predict that twenty fully independent bad-event indicators all vanish with probability 0.95²⁰, approximately 0.3585, although the union-bound lower bound is zero. Reset and set both p = 0.05 and n = 10 to compare. This lab shows a genuine zero-dependency special case; it has no dependency-graph control and cannot test a nonzero-degree local-lemma hypothesis.

A zero-dependency baselineThese 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 forA failed union-bound certificate can coexist with positive avoidance probability; graph-dependent extensions require their own proof.

Check your understanding

Which property must hold for a valid dependency graph in the usual local lemma?

Try it yourself

For six-uniform hyperedges, test the symmetric criterion when every edge intersects at most ten others. Repeat with eleven others. State exactly what the second calculation means.

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

Give me a hint

Use monochromatic probability 1/32 and include d+1 rather than d in the numerical test.

Compare with the solution

For d=10, the product is 11e/32≈0.934409<111e/32\approx0.934409<1, so proper two-coloring is certified. For d=11 it is 12e/32≈1.01936>112e/32\approx1.01936>1, so this sufficient symmetric condition fails. It does not show that the hypergraph cannot be two-colored, and it does not even show that the asymmetric local lemma would fail. A tighter dependency graph or a different argument could succeed. Both calculations assume the intersection bound refers to other edges and that every edge contains six distinct vertices, colored independently and fairly.

Take it with you

  • Dependency graphs require independence from collections of nonneighbors.
  • Local interaction degree can matter more than the total number of bad events.
  • The symmetric local-lemma inequality is sufficient, with failure meaning only that the test is inconclusive.

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