Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 11H (Coding & Cryptography)

The parity-check matrix has rank $3$, so its kernel has dimension $7-3=4$ and size $16$; see linear code. Its columns are precisely the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$. Thus no weight-one or weight-two word is in the kernel, while three dependent columns give a weight-three word. Hence $d=3$.

By parity extension, each code keeps size $16$, gains length $8$, and has distance $4$. Extension is injective. A common extended word punctures to a word in $H\cap K$, so only the zero and all-one words occur; both extend to $00000000$ and $11111111$.

Write the three blocks as $u=a+x$, $v=b+x$, $w=a+b+x$. Over ${\mathbb{F}}_2$,Thus the parameters are unique and $|G|=|H^{\prime }{|}^2|K^{\prime }|=16^3=2^{12}$.

Every nonzero word has weight divisible by $4$. Weight $4$ is impossible: using $u+v=a+b$, $u+w=b$, $v+w=a$, and $u+v+w=x$, the distance-four properties force any nonzero one of these sums to have weight $4$; coordinate counting in total weight $4$ then forces two blocks to vanish, which would put a nontrivial word in $H^{\prime }\cap K^{\prime }$. That intersection contains only $0$ and $1$, and $1$ gives block weight at least $8$. Hence $d(G)\ge 8$, and taking $a\ne 0,b=x=0$ gives weight $2\mathrm{wt}(a)=8$, so $d(G)=8$.

Puncturing gives a binary code of length $23$, size $2^{12}$, and distance at least $7$. Radius-three balls are disjoint andThe Hamming bound is an equality, so $G^{-}$ is perfect.