Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 11G / a

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- Solution a

Solution

Delete any

d - 1

coordinates. The resulting puncturing map

π : C ⟶ F_{2}^{n - d + 1}

is injective: if

π (x) = π (y)

for distinct codewords, then the nonzero codeword

x - y

is supported on at most

d - 1

coordinates, contradicting the minimum distance. Thus

2^{k} = ∣ C ∣ \leq 2^{n - d + 1},

which is the Singleton bound

n \geq d + k - 1 .

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