附加代价约束下笛卡尔积集的解码固定速率熵编码矢量量化

A. Khandani
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引用次数: 0

摘要

考虑由K=| a |个元素组成的离散点集a。非负成本c(A)与每个元素A /spl /A相关联。A的n次笛卡尔积缩进为{A}/sup n/。n倍元素的代价a=(a/sub 0/,…), a/下标n-1/)/spl isin/{a}/sup n/等于:c(a)=/spl Sigma//下标i/ c(a/下标i/)。作者选择n-fold元素的子集S/sub //spl isin/{a}/sup n/,其代价小于或等于给定值c/sub max/。它们称A为组成子集。它们考虑另一个n元组的集合X/下标n/作为输入集合。定义每个x=(x/下标0/,…)之间的非负距离, x / sub n - 1 /) / spl型号/ x / an /,和每个s = (s /子0 /…, s/下标n-1/)/spl isin/ s/下标n/。x/下标i/与s/下标i/之间的距离记为d(x/下标i/,s/下标i/)。x和s之间的距离等于d(x,s)=/spl Sigma//下标i/d(x/下标i/, s/下标i/)对元素x/spl isin/ x/ sub n/进行解码是为了找到与x距离最小的元素s/spl isin/ s/ sub n/。该解码问题的一个主要应用是固定速率熵编码矢量量化,其中A是矢量量化器的重构向量集合,代价相当于自信息。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Decoding of a Cartesian product set with a constraint on an additive cost; fixed-rate entropy-coded vector quantization
The authors consider a discrete set of points A composed of K=|A| elements. A non-negative cost c(a) is associated with each element a/spl isin/A. The n-fold Cartesian product of A is dented as {A}/sup n/. The cost of an n-fold element a=(a/sub 0/, ..., a/sub n-1/)/spl isin/{A}/sup n/ is equal to: c(a)=/spl Sigma//sub i/ c(a/sub i/). The authors select a subset of the n-fold elements, S/sub n//spl isin/{A}/sup n/, with a cost less than or equal to a given value c/sub max/. They refer to A as the constituent subset. They consider another set of n-tuples X/sub n/ denoted as the input set. A non-negative distance is defined between each x=(x/sub 0/, ..., x/sub n-1/)/spl isin/X/sub n/, and each s=(s/sub 0/, ..., s/sub n-1/)/spl isin/S/sub n/. The distance between x/sub i/ and s/sub i/ is denoted as d(x/sub i/,s/sub i/). The distance between x and s is equal to: d(x,s)=/spl Sigma//sub i/d(x/sub i/, s/sub i/). Decoding of an element x/spl isin/X/sub n/ is to find the element s/spl isin/S/sub n/ which has the minimum distance to x. A major application of this decoding problem is in fixed-rate entropy-coded vector quantization where A is the set of reconstruction vectors of a vector quantizer and cost is equivalent to self-information.<>
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