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Thread: optimized distance betwee vectors using their linear combinations ?

  1. #1
    Dec 2007

    optimized distance betwee vectors using their linear combinations ?


    given a set of fix vectors V = v_{1}, v_{2}, ..., v_{n} and a set of vectors A = \alpha_{1}, \alpha_{2}, ..., \alpha_{k} where all \alpha_{i} can be expressed as a convex combination of the vectors in V:

    <br />
a_{i} = \lambda_{i,1} \cdot v_{1} + \lambda_{i,2} \cdot v_{2} + ... + \lambda_{i,n} \cdot v_{n}<br />

    where  \lambda_{i,j} is the j'th coefficient of vector i.
    Also  \lambda_{i,1} + \lambda_{i,2}  + ... + \lambda_{i,n} = 1 and  \lambda_{i,j} >= 0

    Now the mission is: given a vector \alpha_{1} \in A, find the vector \alpha_{2} \in A which has the smallest distance to  \alpha_{1} using the squared euclid distance  \| \alpha_{1} - \alpha_{2} \|^{2} (and of course \alpha_{1} != \alpha_{2}).

    Suppose that the vectors in A are stored in their convex combination form and that the combination is not calculated yet.

    An easy but expensive way to solve the problem is:
    1) calculate the convex combinations for all vectors in A
    2) find \alpha_{2} by comparing the distance of \alpha_{1} to all vactors in A and picking the one with the smallest distance.

    Is there an alternative way to efficiently calculate this (from a practical + implementation point of view) ? Probably by respecting the fact that the vectors in V do not change and that the vectors in  A are expressed as convex combination.

    (We assume that all vectors are in  R^{n}.)

    kind regards,
    Last edited by sirandreus; June 5th 2008 at 04:35 AM.
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