- Conference Article
1
- 10.5555/2095116.2095206
Testing odd-cycle-freeness in Boolean functions
- Jan 17, 2012
- Arnab Bhattacharyya + 3 more +3
A function f: Fn2 → {0, 1} is odd-cycle-free if there are no x1,..., xk e Fn2 with k an odd integer such that f(x1) =... = f(xk) = 1 and x1 +... + xk = 0. We show that one can distinguish odd-cycle-free functions from those e-far from being odd-cycle-free by making poly(1/e) queries to an evaluation oracle. We give two proofs of this result, each shedding light on a different connection between testability of properties of Boolean functions and of dense graphs.The first problem we study is directly reducing testing linear-invariant properties of Boolean functions to testing associated graph properties. We show a black-box reduction from testing odd-cycle-freeness to testing bipartiteness of graphs. Such reductions have been shown previously (Kral-Serra-Vena, Israel J. Math 2011; Shapira, STOC 2009) for monotone linear-invariant properties defined by forbidding solutions to a finite number of equations. But for odd-cycle-freeness whose description involves an infinite number of forbidden equations, a reduction to graph property testing was not previously known. If one could show such a reduction more generally for any linear-invariant property closed under restrictions to subspaces, then it would likely lead to a characterization of the one-sided testable linear-invariant properties, an open problem raised by Sudan.The second issue we study is whether there is an efficient canonical tester for linear-invariant properties of Boolean functions. A canonical tester for linear-invariant properties operates by picking a random linear subspace and then checking if the restriction of the input function to the subspace satisfies a fixed property or not. The question is whether for every linear-invariant property, there is a canonical tester for which there is only a polynomial blowup from the optimal query complexity. We answer the question affirmatively for odd-cycle-freeness. The general question still remains open.
Read more