Counting lemmas and graph removal
Turn regular dense pairs into many embeddings and convert a sparse pattern count into a small deletion certificate.
On this page
The question for today
What you’ll understand
- Derive a triangle-count lower bound with explicit regularity and density conditions.
- State removal with deletion tolerance fixed before the permitted copy density.
- Explain the partition–clean–count argument and its parameter dependencies.
A regular partition is useful only when its coarse information forces actual patterns in the original graph. Counting lemmas provide that bridge. Removal lemmas then run the bridge backward: if very few patterns exist, a small number of edge deletions eliminates all of them. In this lesson the host is a finite simple undirected graph, triangles are unlabelled three-vertex copies, and any tripartite count uses three disjoint declared vertex sets.
Count triangles by extending typical vertices
Suppose A, B, and C are disjoint nonempty sets and each of their three pairs is epsilon-regular, with each density at least twice epsilon. All but at most epsilon times |A| vertices have at least density-minus-epsilon times |B| neighbors in B. Otherwise that exceptional subset and all of B would witness irregularity. The same assertion holds for C, so at least a one-minus-two-epsilon fraction of A is good for both neighbor sets.
For a good vertex, its B-neighborhood and C-neighborhood both satisfy the size thresholds needed to apply regularity of the pair B,C. Their crossing edge density is at least d(B,C) minus epsilon, so many choices close a triangle. Multiplying these three density-minus-epsilon factors and summing over good A vertices gives the displayed bound. Each triangle using one vertex from each declared part is counted once. The assumption that the densities dominate epsilon is essential to keep the neighborhoods above the regularity threshold.
Small counts permit small edits, with dependent tolerances
The triangle removal lemma says that for every positive deletion tolerance eta there exists a positive delta such that any n-vertex graph with at most delta n cubed triangles can be made triangle-free by deleting at most eta n squared edges. The small copy tolerance is chosen in response to the desired edit tolerance. It is not an assertion that delta equals eta, nor that one can delete a number of edges proportional to the observed triangle count with a uniform constant.
For a fixed graph H the analogous statement uses at most delta n to the v(H) copies and permits eta n squared edge deletions. Ordinary copies and induced copies require different statements: deleting edges cannot create an ordinary H, but it can create an induced H by removing an unwanted edge. The ordinary removal lemma therefore does not directly establish induced removal. A graph with few copies can still contain some; the conclusion concerns the existence of a nearby pattern-free graph, not that the original was already free.
Partition, clean, then force a contradiction
Choose eta first. Next choose a density threshold much smaller than eta, a regularity accuracy sufficiently smaller than that threshold, and a sufficiently large minimum partition size. The regularity lemma then supplies a bounded maximum number M of parts. Delete edges touching exceptional vertices, lying inside parts, crossing irregular pairs, or crossing pairs below the density threshold. With the choices made in that order, the total deletions fit the eta budget. This describes a proof architecture rather than every constant in its bookkeeping.
If a triangle survived, its three parts would form regular dense pairs. The counting lemma would then force at least a fixed constant times (n/M) cubed original triangles. Choose delta still smaller than that guaranteed coefficient, after M has been determined, to contradict the assumed small count. A sufficiently large host order handles rounding; finitely many smaller cases can be absorbed by decreasing delta. Reversing this order by choosing delta before the partition bound would leave the contradiction unjustified.
Work it through
A concrete regular-triplet counting coefficient
Suppose three disjoint parts each have N vertices, all three pair densities are 0.5, and each pair is 0.1-regular. Derive a triangle bound in terms of N and evaluate it conditionally at N=100.
The density requirement holds because 0.5 is at least 2×0.1. The typical-vertex factor is 1−2×0.1=0.8.
Each density-minus-error factor is 0.5−0.1=0.4. Their product is 0.4³=0.064.
Multiplying by 0.8 and N³ gives 0.0512N³ triangles. If the stated hypotheses hold with N=100, this evaluates to 51,200. The calculation does not construct such a graph at that particular order or assume independent edges.
Under the regularity hypotheses, at least 0.0512N³ triangles cross the three parts. The value 51,200 is a conditional substitution at N=100; pair densities alone do not justify either bound.
The living notebook
Explore the model
Interactive lab · probability
Compare a random count with a deterministic guarantee
Pause, reset, then set p=0.5 and trials n=3. Predict probability 0.125 at three successes, the independent-edge probability for a fixed triangle. Compare it with the worked lower-bound coefficient 0.0512. The exact binomial bars describe the independent benchmark; they do not test regularity or implement an edge-removal procedure.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
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 counting lemma gives a robust deterministic lower bound, which need not equal the random-model expectation.
Pause & reason
Check your understanding
Your turn
Try it yourself
Take q vertex-disjoint triangles and no other edges, so n=3q. Calculate the normalized triangle count and the minimum normalized deletions needed to eliminate all triangles.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Each component contains exactly one triangle and needs one edge deletion. Divide by n³ for copies and n² for edits.
Compare with the solution
There are q triangles, giving q/(3q)³=1/(27q²). Every triangle needs at least one of its own edges deleted, and the components share no edges, so exactly q deletions are necessary and sufficient. The normalized edit count is q/(3q)²=1/(9q). Both quantities tend to zero, consistently with removal. Their different rates also show why equating the two tolerances is unwarranted.
Keep the idea
Take it with you
- Regular dense pairs let typical-neighbor arguments count many crossing triangles.
- Removal converts small normalized copy count into small dense-scale edit distance.
- The required copy tolerance depends on all earlier cleaning and partition choices.
Read further
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