I was going thorough this paper: (P=BPP unless E has sub-exponential circuits: https://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/IW97/proc.pdf) and I am somewhat struggling with some basics from the paper that I quote below:

Definition 1: Let $m$ and $l$ be positive integers. Let $f:\{{0,1}\}^m→\{{0,1}\}^l$. $S(f)$, the worst-case circuit complexity of $f$, is the minimum number of gates for a circuit $C$ with $C(x)=f(x)$ for every $x∈\{{0,1}\}^m$.

2.2 Hardness versus Randomness: For this section only, we use $f,g,h$ to denote a Boolean function on strings of all lengths, and by $f_n,g_n,h_n$ their segments on n-bit strings. Also, let $E=DTIME(2^{O(n)})$.

Theorem 2: If there is a Boolean function $f∈E$ with $S(f_n )=2^{Ω(n)}$ then $P=BPP$.

Query 1: I am assuming the boolean function is given by its truth table of size $2^m.l$ with all possible entries. I am unclear what it means by this truth table (denoted by $f$) $f∈E/DTIME(2^{O(n)})$? How do you assign the truth table its runtime without having some circuit represent it? I am clear about the fact that $S(f)$ represents the size(i.e. the number of gates in) the smallest possible circuit that represents/calculates $f$.

Query 2: Can someone please explain and simplify the theorem ("if...then") part? I am clear about the complexity classes $P$ and $BPP$, but I am struggling with the condition?

  • $\begingroup$ A function maps binary strings (of arbitrary length) to binary strings. A Boolean function is the special case where the output is a single bit. The restriction of $f$ to inputs of length $n$ is denoted $f_n$. $\endgroup$ Nov 8, 2021 at 22:26
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    $\begingroup$ This paper is not a good starting point for complexity theory. I suggest starting with something simpler. $\endgroup$ Nov 8, 2021 at 22:27

1 Answer 1


The function $f$ is a Boolean function. It maps an input of arbitrary length to a single bit. Its signature is $$f\colon \{0,1\}^* \to \{0,1\}.$$ We denote its restriction to $n$-bit inputs by $f_n$, whose signature is $$f_n\colon \{0,1\}^n \to \{0,1\}.$$ For example, the parity function returns the parity of its inputs: $$ f(x_1,\ldots,x_m) = x_1 \oplus \cdots \oplus x_m, $$ where $m \geq 0$ is arbitrary. The restriction for inputs of length $n$ is $$ f_n(x_1,\ldots,x_n) = x_1 \oplus \cdots \oplus x_n. $$ Here the input length is fixed.

The main theorem states that P=BPP if there is some function $f$ such that

  1. There is a constant $C > 0$ and a Turing machine computing $f$ which runs in time $C2^{Cn}$ on inputs of length $n$.
  2. There is a constant $c > 0$ such that for all $n$, any circuit computing $f_n$ has size at least $c2^{cn}$.
  • $\begingroup$ So effectively (very crudely speaking as I understand) we have to demonstrate a sort of equivalence b/w the worst case runtime of a TM and circuit size of the equivalent minimum circuit for at least one exponential function? $\endgroup$
    – J.Doe
    Nov 9, 2021 at 9:31
  • $\begingroup$ Right. That’s a very believable claim, which is why we believe that P=BPP. $\endgroup$ Nov 9, 2021 at 10:28

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