Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 3I / b

Solution

Over

F_{2}

X^{7} - 1

X^{7} - 1 = (X + 1) (X^{3} + X + 1) (X^{3} + X^{2} + 1) .

The two cubic factors have no root in

F_{2}

, so they are irreducible, and they are reciprocal to one another. Put

f = X^{3} + X + 1, f^{*} = X^{3} + X^{2} + 1.

Every cyclic code corresponds to one monic divisor of

X^{7} - 1

. The complete list is

g (X) 1 X + 1 f f^{*} (X + 1) f (X + 1) f^{*} f f^{*} = 1 + X + \dots + X^{6} (X + 1) f f^{*} = X^{7} - 1 parameters [7, 7, 1] [7, 6, 2] [7, 4, 3] [7, 4, 3] [7, 3, 4] [7, 3, 4] [7, 1, 7] [7, 0] code whole-space linear code even-weight binary code Hamming code reversed Hamming code binary simplex code reversed binary simplex code binary repetition code zero code .

Thus the two cubic generators give the two cyclic coordinate orientations of Hamming's

[7, 4, 3]

code. Their duals have generators

(X + 1) f

and

(X + 1) f^{*}

and are the

[7, 3, 4]

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