# Counting the Faces of Randomly-Projected Hypercubes and Orthants, with Applications

Discrete & Computational Geometry, Sep 2009

Let A be an n×N real-valued matrix with n<N; we count the number of k-faces f k (AQ) when Q is either the standard N-dimensional hypercube I N or else the positive orthant ℝ + N . To state results simply, consider a proportional-growth asymptotic, where for fixed δ,ρ in (0,1), we have a sequence of matrices $A_{n,N_{n}}$ and of integers k n with n/N n →δ and k n /n→ρ as n→∞. If each matrix $A_{n,N_{n}}$ has its columns in general position, then f k (AI N )/f k (I N ) tends to zero or one depending on whether ρ>min (0,2−δ −1) or ρ<min (0,2−δ −1). Also, if each $A_{n,N_{n}}$ is a random draw from a distribution which is invariant under right multiplication by signed permutations, then f k (Aℝ + N )/f k (ℝ + N ) tends almost surely to zero or one depending on whether ρ>min (0,2−δ −1) or ρ<min (0,2−δ −1). We make a variety of contrasts to related work on projections of the simplex and/or cross-polytope. These geometric face-counting results have implications for signal processing, information theory, inverse problems, and optimization. Indeed, face counting is related to conditions for uniqueness of solutions of underdetermined systems of linear equations. Below, let A be a fixed n×N matrix, n<N, with columns in general position. (a) Call a vector in ℝ + N k -sparse if it has at most k nonzeros. For such a k-sparse vector x 0, b=Ax 0 generates an underdetermined system b=Ax having k-sparse solution. Among inequality-constrained systems Ax=b, x≥0, having k-sparse solutions, the fraction having a unique nonnegative solution is f k (Aℝ + N )/f k (ℝ + N ).   (b) Call a vector in the hypercube I N k-simple if all entries except at most k are at the bounds 0 or 1. For such a k-simple vector x 0, b=Ax 0 generates an underdetermined system b=Ax with k-simple solution. Among inequality-constrained systems Ax=b, x∈I N , having k-simple solutions, the fraction having a unique hypercube-constrained solution is f k (AI N )/f k (I N ).

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David L. Donoho, Jared Tanner. Counting the Faces of Randomly-Projected Hypercubes and Orthants, with Applications, Discrete & Computational Geometry, 2009, 522-541, DOI: 10.1007/s00454-009-9221-z