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→History: The statement is inaccurate. See discussion at https://www.reddit.com/r/math/comments/dkka6g/superpolynomial_circuit_lower_bounds/ and survey paper at http://www.cs.columbia.edu/~rocco/Teaching/S17/6998/Boppana-Sipser.pdf . |
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==History==
Circuit complexity goes back to [[Claude Shannon|Shannon]] (1949), who proved that almost all Boolean functions on ''n'' variables require circuits of size Θ(2<sup>''n''</sup>/''n''). Despite this fact, complexity theorists have
The [[clique problem|''k''-clique problem]] is to decide whether a given graph on ''n'' vertices has a clique of size ''k''. For any particular choice of the constants ''n'' and ''k'', the graph can be encoded in binary using <math>{n \choose 2}</math> bits, which indicate for each possible edge whether it is present. Then the ''k''-clique problem is formalized as a function <math>f_k:\{0,1\}^{{n \choose 2}}\to\{0,1\}</math> such that <math>f_k</math> outputs 1 if and only if the graph encoded by the string contains a clique of size ''k''. This family of functions is monotone and can be computed by a family of circuits, but it has been shown that it cannot be computed by a polynomial-size family of monotone circuits (that is, circuits with AND and OR gates but without negation). The original result of [[Alexander Razborov|Razborov]] (1985) was later improved to an exponential-size lower bound by Alon and Boppana (1987). Rossman (2008) shows that constant-depth circuits with AND, OR, and NOT gates require size <math>\Omega(n^{k/4})</math> to solve the ''k''-clique problem even in the [[average-case complexity|average case]]. Moreover, there is a circuit of size <math>n^{k/4+O(1)}</math> that computes <math>f_k</math>.
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