Abstract
Suppose that a Boolean functionf is computed by a constant depth circuit with 2m AND-, OR-, and NOT-gates—andm majority-gates. We prove thatf is computed by a constant depth circuit with\(2^{m^{O(1)} }\) AND-, OR-, and NOT-gates—and a single majority-gate, which is at the root.
One consequence is that if a Boolean functionf is computed by an AC0 circuit plus polylog majority-gates, thenf is computed by a probabilistic perceptron having polylog order. Another consequence is that iff agrees with the parity function on three-fourths of all inputs, thenf cannot be computed by a constant depth circuit with AND-, OR-, and NOT-gates, and\(2^{n^{O(1)} } \) majority-gates.
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Beigel, R. When do extra majority gates help? Polylog (N) majority gates are equivalent to one. Comput Complexity 4, 314–324 (1994). https://doi.org/10.1007/BF01263420
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DOI: https://doi.org/10.1007/BF01263420