advanced30 min studyLesson 622 in Mathematics

Supersaturation and counting subgraphs

Turn a fixed density excess into polynomially many copies and track labeled versus unlabeled subgraph counts.

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What you’ll understand

  • State supersaturation with a fixed forbidden graph and explicit quantifier order.
  • Double-count small witnesses without confusing embeddings with unlabelled copies.
  • Distinguish fixed-density excess from adding a bounded number of edges.

Crossing an extremal threshold guarantees at least one forbidden pattern. Supersaturation asks whether a substantial excess forces many such patterns. The scale of the excess is decisive: one extra edge and a fixed positive fraction of all possible edges are very different hypotheses. We again use finite simple undirected graphs and ordinary, noninduced copies. A copy of H means an injective adjacency-preserving embedding modulo automorphisms of H, so the convention does not count every labeling separately.

Read the quantifiers before the asymptotic notation

Fix a graph H with h vertices and at least one edge, and fix a positive real number epsilon. Supersaturation states that there are positive constants c and n-zero, depending on H and epsilon, such that every n-vertex graph with n at least n-zero and at least ex(n,H) plus epsilon times n squared edges contains at least c times n to the h copies of H. The constants are chosen before seeing the particular large graph. If the edge hypothesis is impossible, the implication is simply vacuous.

For a fixed clique, one can instead use a fixed positive density excess over its limiting Turán density. The two forms agree asymptotically after adjusting constants, but they should not be conflated in a finite calculation. If epsilon is allowed to shrink with n, the theorem does not supply a uniform constant c. Adding a single edge to a balanced complete bipartite graph creates only linearly many triangles, so it cannot justify a positive constant fraction of n cubed triangles.

∀ε>0 ∃c>0,n0 ∀n≥n0:e(G)≥ex⁡(n,H)+εn2 ⟹ NH(G)≥cnv(H)\forall\varepsilon>0\ \exists c>0,n_0\ \forall n\ge n_0:\quad e(G)\ge\operatorname{ex}(n,H)+\varepsilon n^2\ \Longrightarrow\ N_H(G)\ge c n^{v(H)}
H is fixed first. N_H counts unlabelled ordinary copies; c and n₀ depend on H and ε, not on G or n.↔ Scroll wide equations horizontally.

Find many small witnesses by averaging

The proof mechanism samples a fixed-size vertex subset. With its size chosen sufficiently large using H and epsilon, the expected number of sampled edges remains above the relevant small-graph threshold by a controlled margin. Since no sample can have arbitrarily many edges, a positive proportion of samples must actually cross that threshold. Each such sample contains H. This is a proof sketch: choosing the sample size rigorously uses convergence of normalized extremal numbers and must precede sending the host order to infinity.

Now count pairs consisting of a sampled vertex set and a copy of H lying inside it. A fixed copy occurs in exactly the binomial number of samples obtained by adding the remaining sample vertices outside that copy. Dividing the lower bound for witness-containing samples by this upper multiplicity gives a constant times n to the h copies. The argument needs no assumption that copies are edge-disjoint. Heavy overlap is permitted; the multiplicity calculation is what prevents overlap from invalidating the count.

NH(G)(n−hk−h)≥α(nk)⟹NH(G)≥α (nh)(kh)N_H(G)\binom{n-h}{k-h}\ge\alpha\binom nk\quad\Longrightarrow\quad N_H(G)\ge\alpha\,\frac{\binom nh}{\binom kh}
If at least an α proportion of k-subsets contain H, count copy–subset pairs. Each good subset contributes at least one pair; each fixed copy belongs to exactly the stated number of subsets.↔ Scroll wide equations horizontally.

Choose a counting convention that survives comparisons

An injective labeled embedding specifies where each named vertex of H goes. Each unlabelled copy gives exactly the number of automorphisms of H such embeddings. For a triangle that factor is six. Homomorphisms allow collisions between nonadjacent source vertices, so they are another count; for a fixed simple H, noninjective maps contribute at most a constant times n to the h minus one. That lower-order difference matters in exact finite examples but disappears after normalizing by n to the h.

Density alone does not guarantee the random-graph count for every pattern. A complete balanced bipartite graph has many edges and zero triangles, whereas an independent-edge model at similar edge density has many expected triangles. Supersaturation concerns crossing a forbidden-pattern threshold, not asserting independence. Later regularity and graphon methods explain which additional structural information controls counts. For now, state both the pattern being counted and the convention, then verify that every normalization in a proof uses those same choices.

One added edge creates many but not cubically many triangles

Start with K₅,₅ and add one edge inside its first part. Count triangles and compare with the total number of triples.

  1. The original bipartite graph has 25 edges and no triangles. After adding one internal edge, a triangle must use that edge because every other edge still crosses the bipartition.

  2. Its third vertex can be any of the five vertices in the opposite part. These choices give exactly five triangles, with no other possibility.

  3. There are C(10,3)=120 vertex triples. Thus five of those triples are triangles, and the count of labeled triangle embeddings is 6×5=30.

The result

There are exactly five unlabelled triangles. In the analogous Kₘ,ₘ construction there are m triangles, so one extra edge gives order n rather than order n³.

Explore the model

Interactive lab

Interactive lab · probability

A random benchmark is not a universal count

Pause, reset, then set success probability p=0.5 and trials n=3. Predict probability 1/8 at three successes, representing the event that all three possible edges on a fixed triple appear independently. Observe the exact binomial bar. This is a G(3,p) benchmark, not a graph enumeration or a supersaturation proof, and it cannot model the dependence in the added-edge construction.

A random benchmark is not a universal countThese 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 random benchmark can calibrate counts only after its independence assumptions are stated.

Check your understanding

In the fixed-density supersaturation statement, which dependence of the copy-count constant is allowed?

Try it yourself

In K₄,₆ add two distinct internal edges in the four-vertex part and no other edges. Count triangles, whether or not the two new edges share an endpoint.

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

Give me a hint

Every triangle must use an internal edge and a vertex from the opposite part. Check that two internal edges cannot complete a triangle by themselves.

Compare with the solution

Each internal edge has six possible third vertices in the opposite part, giving twelve triangles. These triangles are distinct even when the internal edges share an endpoint, because the triangle's unique internal edge identifies its family. There is no triangle entirely within the four-vertex part: only two internal edges were added, while three would be needed. The labeled embedding count is 72.

Take it with you

  • A fixed positive density excess forces a constant-times-nᵛ⁽ᴴ⁾ copy count for fixed H.
  • Double-counting controls overlap by tracking how often each copy is counted.
  • One-edge excess, fixed-density excess, and independent-edge expectation are different regimes.

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