Consider an n-dimensional cube and its complement graph, represented by G and…

Concept: For a simple graph G on m vertices, its complement H satisfies |E(G)| + |E(H)| = C(m, 2), the total number of edges in the complete graph Km on the same m vertices. Also, by the handshaking theorem, a d-regular graph on m vertices has exactly (d × m) / 2 edges — the degree and the vertex count alone fix the edge count.

Application:

1. The n-dimensional hypercube Qn (graph G) has m = 2n vertices, and every vertex has degree n (it differs from exactly n neighbours, one per coordinate).

2. For n = 11: the vertex count is m = 211, and the degree is 11. By the handshaking theorem, |E(G)| = (11 × 211) / 2 = 11 × 210.

3. The complete graph on the same m = 211 vertices has every vertex of degree m − 1 = 211 − 1 = 2047, so |E(Km)| = (2047 × 211) / 2 = 2047 × 210.

4. The complement's edge count follows from the concept above: |E(H)| = |E(Km)| − |E(G)| = (2047 − 11) × 210 = 2036 × 210.

5. Matching this to the given form y × 210, we get y = 2036.

Cross-check: Since the hypercube Qn is regular, its complement H is also regular, with degree (m − 1) − n = 2047 − 11 = 2036 at every vertex. Applying the handshaking theorem directly to H: |E(H)| = (2036 × 211) / 2 = 2036 × 210 — the same value, independently confirming the result.

Answer: y = 2036.