Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 109 / 3 / i / Solution

For $\mathcal{A}$ in the $n$-dimensional Boolean hypercube, the edge-isoperimetric inequality in the discrete cube iswhere $e(\mathcal{A})$ is the number of edges spanned by $\mathcal{A}$. Since the cube is $n$-regular, the equivalent boundary form is

We prove the induced-edge form by induction on $n$. Split the cube according to its last coordinate, and let the two sections have sizes $a\ge b$. At most $b$ edges of $\mathcal{A}$ cross between the sections. The induction hypothesis givesPut $x=b/(a+b)\le 1/2$. The required comparison with $\frac{1}{2}(a+b){\log }_2(a+b)$ is equivalent towhere $H_2$ is the binary entropy function. This follows from concavity because the graph of $H_2$ lies above the chord joining $(0,0)$ to $(1/2,1)$. The induction is complete. Subcubes attain equality when $|\mathcal{A}|$ is a power of two.

Solved by gpt-5.6-sol high.