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Consider a parallelotope in R^n and some point "P" in R^n. What algorithms (except of brute force) can be suggested to find the closest vertex of paralleloptope to "P" ?

Is it NP ?

Parallelotope has 2^n vertex, not arbitrary 2^n point in R^n are vertex of paralleloptope, so clever algorithm should somehow use this additional information, while brute force search over 2^n points does not use.



after choosing origin in the center of parallelotope we can come to the following algebraic version of the problem: minimize over x_i = {-1,+1} the quadratic form: \sum a_ij x_i x_j - \sum x_i v_i

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2 Answers 2

This looks almost like the Max-CUT problem to me (you have minimize instead of maximize, but you can just flip the signs of the matrix $A$)

In general, you problem is an instance of a binary quadratic program, so it will be hard to solve. Have a look at some solvers on this webpage

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This is equivalent to the MIMO detection problem, which is NP hard. Here is a paper with a semidefinite relaxation: http://citeseerx.ist.psu.edu/viewdoc/download?doi=

Then, there exist some easy instances of the problem, if your matrix of the quadtratic form is negative semidefinite and rank deficient. Check these for example


http://www.telecom.tuc.gr/~karystinos/paper_TIT3.pdf (algorithm included)

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