Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 3K iii Solution Created 2026-09-24 Updated 2026-09-25
For a chosen coordinate and symbol , the shortened code retains the words having that symbol in the chosen coordinate and then deletes that coordinate. For the last coordinate,At least one of the two coordinate fibres contains at least words. Choose that fibre and, if necessary, discard surplus words. Because all retained words agree in the deleted coordinate, their mutual distances do not change. Hence one can always obtainWithout discarding words, its size is that of the chosen fibre and its minimum distance is at least .
For the final calculation, the given code is the whole space , so its parity extension is the length-four even-weight binary code. In a binary symmetric channel, parity fails to notice a nonzero error exactly when an even number of bits flips. The possible error weights are therefore two and four. Their total probability isThis excludes the weight-zero event, since it is not an undetected error.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 3 3I Solution Created 2026-09-23 Updated 2026-09-25
Write for the all-one column vector. The parity-check extension of a linear code isThe defining map is linear, so is a binary linear code. If is a generator matrix and a parity-check matrix for , then one may take
The punctured code in, say, the last coordinate isCoordinate deletion is a linear map, so is linear. Its generator matrix is obtained by deleting the last column of . If the minimum distance is at least two, deletion is injective on , and hence has the same dimension as .
For its parity-check matrix, first use invertible row operations on to make its nonzero last column a coordinate vector. The last column is nonzero because otherwise the weight-one word supported there would belong to , contrary to . Delete the pivot row and the last column; the resulting matrix is a parity-check matrix for . Equivalently, is obtained by shortening the dual code at the deleted coordinate.
Finally, the shortened code in the last coordinate isIt is linear whenever is linear, because it is the puncture of the subspace .