A threshold function has the following meaning: if , the probability of a tends to zero; if , that probability tends to one. Behavior at a fixed constant multiple is not specified by this definition.
Let count the five-vertex subsets spanning in the Erdős-Rényi model . ThenIf , then , so Markov inequality gives .
Now let with . Then . Distinct copies have dependent indicators only if their vertex sets share at least two vertices. If they share vertices, they share edges; there are ordered such pairs, and their joint probability is . ConsequentlyPart (b) gives . ThereforeThis is the five-vertex instance of the fixed-size clique appearance threshold.
For the Erdős-Rényi model , choose the number of vertices and an edge probability . Visit every unordered pair of distinct vertices once and include its edge independently with probability . This construction gives a simple graph with expected edge count and mean vertex degree .
Increasing at fixed makes the graph denser, increases typical vertex degree and local clustering coefficient, and tends to shorten graph distances and improve connectivity. Increasing at fixed also increases the mean vertex degree; to retain a sparse comparison one instead decreases approximately as .
Its degree distribution is binomial distribution, approximately Poisson distribution in the sparse large- regime, so it is not a scale-free network. Connected realizations can have short average shortest path length, but their clustering coefficients are not elevated above the random baseline they themselves define. Thus they are not small-world networks under the high-clustering convention, although “small world” is sometimes used more loosely to mean only short distances.
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 4 17F Solution Created 2026-09-23 Updated 2026-09-25
In the Erdős-Rényi model , the vertex set is and every one of the possible edges is present independently with probability .
Chebyshev inequality states thatNow suppose . Monotonicity lets us use the largest allowed . Withthe supplied lower exponential estimate showsFor distinct vertices ,HenceandChebyshev with gives , so .
For connectivity, the lower-threshold result is immediate because a connected graph has no isolated vertex:For the upper threshold, if a graph is disconnected it has a component with vertex set of some size . In particular, every one of the edges from to its complement is absent. ThereforeChooseFor , the standard bound givesThe sum of these terms tends to zero. For , use and :Thus above the threshold. Together,This is the connectivity threshold in the Erdős-Rényi model.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 17H a Solution Created 2026-09-24 Updated 2026-09-25
Fix and sample from the Erdős-Rényi model with . Let count copies of the complete bipartite graph . By the expected subgraph count in the Erdős-Rényi model,
Put and let count independent set of size . If the chromatic number satisfies , some colour class has at least vertices, so . Moreover,because the positive part of the exponent is whereas has order . The first moment method therefore gives .
Using Markov inequality for and the union bound,for all sufficiently large . Hence at least one such contains no and has .
Past exam of the mathematics course of the University of Cambridge 2024 ia Paper 2 12F i Solution Created 2026-09-24 Updated 2026-09-25
The degree of counts the incident edges, each present independently with probability . Thus in the Erdős-Rényi model,
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 122 1 c Solution Created 2026-09-24 Updated 2026-09-25
The sharp Hamilton cycle threshold for the Erdős-Rényi model says thatonly above the window . The hypothesis therefore places above that window. The Hamiltonicity-to-pancyclicity sprinkling principle then says that three independent rounds contain every cycle graph , , with high probability: one round supplies a Hamilton cycle, while the other two supply the chords and short-cycle edges used to obtain all intermediate lengths.
The union of the three rounds has individual edge probabilityBy the standard monotone coupling, it is a subgraph of . Since being pancyclic is an increasing graph property, the required probability tends to one. If , interpret the latter parameter as , in which case the conclusion is immediate.