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Definitions. Let i) standard q-ary alphabet. ii) is the set of all q! permutations of q symbols. is a set of n elements. n -sequence. q -partition. For any. -complement. v). Number of positions i , i=1,2,…n where:.
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Definitions Let i) standard q-ary alphabet. ii) is the set of all q! permutations of q symbols. • is a set of n elements n-sequence q-partition
For any -complement v) Number of positions i, i=1,2,…n where: vi) Hamming Distance: • vii) Partition Distance
Applications • An individual is asked to split n objects up into parts, putting • seemingly similar objects into the same part. • On the base of preliminary testing of individuals with K known diagnoses one • can choose from a subset of q-partitions • corresponding to the K putative subtypes of the complex disease. • The patient’s result will be some partition . We then calculate the partition distance between and partitions in the K set of putative sub-types. The subtype(s) that are represented by the partitions that yield the smallest distance is possibly what the individual is suffering from.
Varshamov-Gilbert and Hamming bounds. Spheres The sphere centered at x of radius tis the set of codewords: The volume of any sphere (not in a partition code) of radius t is: Hamming Bound Suppose we have a code with minimum distance d=2t+1. Let M denote the maximal possible size of the code. Then around each codeword we will have a sphere of radius t. Since all spheres are disjoint then we must have:
STIRLING SET NUMBERS By definition a Stirling set number (Stirling number of the second kind is the number of ways of partitioning a set of n elements intok non-empty clusters. Denoted by: All possible q-partitions of an n set:
SPHERE VOLUME (PARTITION CODE) Space of un-ordered q-partitions endowed with partition distance is homogeneous. Partitions of distance k from 0 partition: If we have i>k then S(k,i)=0 Volume of Sphere:
Bounds for q-partition codes Hamming Bound Let M denote the maximal possible size of a q-partition code with, length n and minimum distance d= 2t+1 . Then we have that the Hamming bound is: