Probability Statistics

Acta Numerica 1997: Volume 6 by Arieh Iserles

By Arieh Iserles

Acta Numerica surveys every year crucial advancements in numerical research. the topics and authors, selected through a exclusive overseas panel, offer a survey of articles amazing of their caliber and breadth. This quantity contains articles on multivariate integration; numerical research of semiconductor units; quickly transforms in utilized arithmetic; complexity concerns in numerical research.

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If and only if x1 = x3 + x4 , x2 = 2x3 + x4 . 3), one gets d(B) ={(0, 0, 0, 0) , (1, 1, 0, 1) , (2, 2, 0, 2) , (1, 2, 1, 0) , (2, 0, 1, 1) , (0, 1, 1, 2) , (2, 1, 2, 0) , (0, 2, 2, 1) , (1, 0, 2, 2) }. 4) b ∈ R(B), where, as before, R(·) denotes the row space of a matrix. Since the k × n matrix B over GF (s) has full row rank, the cardinality of R(B) is sk . Since pencils are nonnull vectors and pencils with proportional entries are identical, it follows that there are (sk − 1)/(s − 1) defining pencils.

Wn ), the Hamming distance dist(u, w) = wt(u − w) is the number of j’s with wj = uj . The minimum distance of a code C is the smallest Hamming distance between any two distinct codewords of C. Let Ki (C) be the number of codewords of weight i in C. Then (K1 (C), K2 (C), . ) is called the weight distribution of C. , the smallest i > 0 such that Ki (C) > 0. For convenience, a linear code is denoted by [n, k, d; s] if its minimum distance is d. Continuing the previous interpretation, the minimum distance of a linear code C is mathematically equivalent to the resolution of the corresponding sn−k design.

Thus any treatment contrast belonging to the pencil (1, 1) is orthogonal to any treatment contrast belonging to the pencil (1, 2) . The ideas implicit in the above example will now be formalized. First note that the pencils (1, 2) and (2, 1) considered in parts (i) and (ii) of the example are proportional to each other, in the sense that (1, 2) = 2(2, 1) over GF (3). In general, consider any two pencils b and b∗ such that b∗ = λb for some λ(= 0) ∈ GF (s). 2), x ∈ Vj (b) ⇔ b x = αj ⇔ b∗ x = λαj .

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