When do extra majority gates help?

Abstract

Suppose that f is computed by a constant depth circuit with 2m AND-, OR-, and NOT-gates, and m majority-gates. We prove that f is computed by a constant depth circuit with 2mo(1) AND-, OR-, and NOT-gates, and a single majority-gate, which is at the root.One consequence is that if f is computed by and AC0 circuit plus polylog majority-gates, then f is computed by a probabilistic perceptron having polylog order. Another consequence is that if f agrees with the parity function of three-fourths of all inputs, then f cannot be computed by a constant depth circuit with 2no(1) AND-, OR-, and NOT-gates, and no(1) majority-gates.