If every pair of graphs in a family on has a graph intersection without an isolated vertex, the family has at most members. At each vertex, its possible graph neighbourhoods form an intersecting family, of size at most . Every edge occurs in two such neighbourhood projections. Shearer inequality bounds twice the full information entropy by the sum of their information entropies.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 13 2 iii Solution Created 2026-10-03 Updated 2026-10-06
The empty family satisfies the bound, so assume is nonempty and choose a graph according to the uniform distribution on a finite set . Let be its random vector of edge indicators, indexed by . Then its information entropy is .
For each vertex , put , so records the graph neighbourhood of . The possible graph neighbourhoods form an intersecting family of subsets of : the graph intersection of any two members of has no isolated vertex at . A subset and its complement cannot both occur. Pairing the subsets into complementary pairs gives at most possible graph neighbourhoods. The maximum entropy on a finite alphabet therefore givesThis argument applies for . For the assumed nonempty family cannot exist, since its only possible graph has an isolated vertex.
Each edge belongs to exactly two of the sets . To obtain the needed instance of Shearer inequality directly from the preceding compression argument, repeatedly apply a union-intersection compression to incomparable members of the multiset . Each such step strictly increases , byThe potential is bounded and integer-valued, so the process terminates in a chain under inclusion. Coordinate multiplicities stay equal to two, so, for , the terminal chain consists of two copies of and empty sets. The compression of an entropy sum now givesExponentiating proves the entropy bound for graphs with isolated-vertex-free intersections: