Extremal bipartite graphs and Zarankiewicz bounds
Count common neighborhoods to bound forbidden complete bipartite subgraphs, and explain why bipartite extremal problems need finer scales.
On this page
The question for today
What you’ll understand
- Specify which side of a bipartition contains each forbidden pattern part.
- Derive a K₂,ₜ-free edge bound by counting pairs of common neighbors.
- Translate a double-counting inequality into the balanced n-to-the-three-halves scale.
Forbidding a bipartite graph often forces edge density to vanish, leaving a finer question: how quickly can the number of edges still grow? Zarankiewicz problems isolate this issue in an already bipartite host. Here the host is finite, simple, with declared left and right vertex sets of sizes m and n. We forbid a complete pattern with specified numbers of vertices on each side. Specifying the sides prevents accidental symmetry assumptions when their sizes differ.
Count a pair and its common neighbors
Suppose the host contains no with its two vertices on the left and its t vertices on the right, where is fixed. Any pair of distinct left vertices then has at most t minus one common neighbors on the right. Count triples consisting of a left pair and a right vertex adjacent to both. Summing over right vertices gives a binomial coefficient of each right degree; summing over left pairs gives the forbidden-pattern upper bound.
The same triples are counted exactly once in both descriptions, so the inequality needs no randomness or regularity assumption. Degrees are measured only across the declared bipartition, and repeated vertices are not allowed in the pair. A four-cycle is precisely a in a bipartite graph. Thus a four-cycle-free bipartite graph has at most one common right neighbor for each left pair, a particularly transparent starting case.
Convexity turns degree data into an edge bound
Cauchy–Schwarz gives the sum of squared right degrees at least e squared divided by n. Expand each degree's binomial coefficient as half its square minus half its value. Combining with the preceding count yields a quadratic inequality for e. Solving it gives an explicit finite bound; a simpler expression follows by bounding its positive root above. When the two sides have comparable size, the leading term has order n to the three halves rather than n squared.
The general Kővári–Sós–Turán method counts s-element left subsets inside right neighborhoods. If each such subset has at most t minus one common neighbors, convexity yields an upper bound of order for fixed s and t. In the balanced case this becomes order . This is an upper bound, not a claim that a matching construction exists for every pair of forbidden part sizes. Sharp constants and exponents can require additional algebraic constructions or remain separate questions.
Constructions explain why the scale is meaningful
Incidence graphs connect point vertices to line vertices when a point lies on a line. If two distinct points determine at most one of the chosen lines, the graph has no four-cycle. A finite projective plane of order q gives q squared plus q plus one points and equally many lines, each of degree q plus one. Its edge count is therefore comparable to the three-halves power of a part size. This construction exists for prime-power q; no existence assertion for arbitrary integer q is needed.
There are two related extremal settings. A Zarankiewicz number concerns a bipartite host with specified side sizes and oriented forbidden parts. The ordinary extremal number ex(N,H) concerns any simple graph on N vertices. One can relate their orders of growth through bipartition arguments, but their finite definitions and constants differ. An average density calculation cannot replace the common-neighbor restriction: a graph with the same number of edges may concentrate many of them on a small complete bipartite block.
Work it through
A four-cycle-free incidence graph on fourteen vertices
Take the seven points and seven lines of the projective plane of order two. Each line contains three points, and each pair of points lies on one line. Count edges and test the K₂,₂ bound.
There are seven line vertices, each with degree three, so the incidence graph has 7×3=21 edges.
Each pair of points lies on only one line. A four-cycle would require two point vertices on two distinct common lines, which is impossible.
The common-neighbor sum is 7·C(3,2)=21, equal to C(7,2). In the quadratic inequality, e²/7−e=63−21=42 equals 7·6=42.
This graph has 21 edges and no four-cycle, meeting the degree-count inequality exactly for side sizes seven and seven.
The living notebook
Explore the model
Interactive lab · function
Inspect the convex degree penalty
Pause, reset, then set coefficient a=1 and position x=1 to read x²=1. For a fresh comparison, pause and reset, set a=1 and x=2, and read x²=4. Relate this convexity to the degree-square term in the double count. The continuous quadratic lab neither stores integer graph degrees nor checks common neighborhoods; the inequality connecting those values must be derived on paper.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forConvexity penalizes concentrating degree, while the forbidden-subgraph condition bounds the total common-neighbor count.
Pause & reason
Check your understanding
Your turn
Try it yourself
Can a K₂,₂-free bipartite graph with six vertices on each side have 19 edges? Use the degree-square inequality.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Insert m=n=6 and e=19 into e²/n−e≤m(m−1), keeping fractions exact.
Compare with the solution
The right side is 6·5=30. The left side would be 361/6−19=(361−114)/6=247/6, greater than 30. Thus 19 edges are impossible without a four-cycle. The quadratic upper root is (6+sqrt(756))/2, approximately 16.75, so this argument actually limits the integer edge count to at most 16; it does not prove that 16 is attainable.
Keep the idea
Take it with you
- A forbidden complete bipartite pattern bounds common neighborhoods of fixed-size subsets.
- Double counting plus convexity converts that restriction into an edge-count inequality.
- Bipartite-host definitions, ordinary extremal numbers, and sharp constructions must be distinguished.
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