$\require{mathtools}
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% Greek
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\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
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% Miscellaneous shortcuts
% .. over and under
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% .. relations
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% .. miscer
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% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% .... (use phantom to force at least the standard height of double bars)
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%% .. two-part
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% ..... (use phantom to force at least the standard height of double bar)
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% .. vanilla versions (is it within a constant?)
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% .. dotted versions (is it equal in the limit?)
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% .. log versions (is it equal up to log?)
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% .. operators
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% .. functions
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% .. analysis
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% .. sets
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% Complexity classes
% .. classical
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% .. probabilistic
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% .. circuits
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% .. resources
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% .. keywords
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Adapted from Section 1.6 of “Boolean function complexity” by Stasys Jukna.
We’re still far from proving super-polynomial lower bounds on circuit size for an explicit function. In fact, we haven’t even managed to show super-linear lower bounds yet (even within $\NC^1$)! But we can prove very crummy, linear lower bounds.
For now, we have:
- $3.1n-o(n)$ lower bounds for circuits over the basis with all fanin-2 gates;
- $5n - o(n)$ lower bounds when parity and its negations are removed from the basis (and thus in particular over $\and,\or,\neg$).
Idea
The basic idea of gate elimination is very simple: find a variable that you can fix to some value in such a way that the circuit loses $q$ gates in the process (assuming the circuit was large enough). For this to be possible, the circuit must have had $qn-O(1)$ gates to start with.
It’s interesting to think about whether this technique could give super-linear lower bounds: what would it take for a variable to eliminate more than a constant number of gates? This paper argues this is unlikely to work (but I’ve yet to read it).
Let’s go over two simple examples! As far as I know, both the $3.1n-o(n)$ result and the $5n-o(n)$ result are instantiations of the same idea, just with more elaborate case analysis.
Eliminating two general gates
We’ll first prove a lower bound over the basis of all fanin-2 gates. Note that in this basis, fanin-1 gates are useless (unless they’re the output gate): they can just be merged into the gates that read it. If you fix one of the inputs of a fanin-2 gate, that makes it functionally fanin-1, so you can eliminate it. So to eliminates many gates, it’s enough to find a variable that’s queried by many gates, and then fix it.
Let $\Th_k^n: \zo^n \to \zo$ be the threshold function: $\Th_k^n(x) = \One[x_1 + \cdots + x_n \geq k]$. We’ll show a $2n-O(1)$ lower bound for $\Th_2^n$ by eliminating two gates with just one variable. The main property we’ll use is that if you hardcode two variables $x_i$ and $x_j$, this can produce three different subfunctions depending on the value you give to $x_i+x_j$:
- if $x_i+x_j = 0$, then we get $\Th_2^{n-2}$;
- if $x_i + x_j = 1$, then we get $\Th_1^{n-2}$ (aka OR);
- if $x_i + x_j = 2$, then we get $\Th_0^{n-2}$ (aka the all-ones function).
Take the deepest gate $g$, and say it queries variables $x_i$ and $x_j$. These must be two distinct variables, because otherwise $g$ is effectively a fanin-1 gate, which can be eliminated for free. We’ll show that either $x_i$ or $x_j$ is queried somewhere else, and therefore we can fix that variable to $0$ to eliminate two gates and induce on the function $\Th_2^{n-1}$.
Assume for the sake of contradiction that $x_i$ and $x_j$ are queried only by $g$. Then $g$ is the only way for the circuit to know anything about $x_i$ and $x_j$. But fixing $x_i$ and $x_j$ can only make $g$ take two different values ($0$ or $1$), while (assuming $n\geq 3$) there are three possible subfunctions with $x_i$ and $x_j$ fixed: $\Th_2^{n-2}$, $\Th_1^{n-2}$, and $\Th_0^{n-2}$. So we get a contradiction.
Eliminating three $\and,\or$ gates
If we restrict the gates to just $\and,\or,\neg$ gates, then we get one more trick up our sleeve: we can fix any gate to a constant by setting one of its inputs to the right value. For example, if we set one of the inputs of an AND to $0$, the output will definitely be $0$. This then allows us to eliminate more gates downstream.
Here we’ll show that any $\and,\or,\neg$ circuit for the parity function on $n$ variables must have at least $3n-O(1)$ AND and OR gates. It turns out this is tight, because you can represent a fanin-2 parity gate $x \xor y$ as $(x \and \neg y) \or (\neg x \and y)$, which uses only three AND and OR gates.
Again, let $g$ be the deepest AND or OR gate, and say it queries variables $x_i$ and $x_j$. These must be distinct variables, otherwise the gate is useless. We claim that either $x_i$ or $x_j$ must also be queried by some other gate $g' \neq g$. Suppose they’re not: $x_i$ and $x_j$ are only queried by $g$. Then you can set $x_i$ in a way that fixes $g$ to a constant and thus make the output of the circuit independent of $x_j$. This is impossible if the circuit computes parity.
Without loss of generality, say $x_i$ is queried by $g'$. Note that gate $g$ must be read by at least one other gate $h$. If as above we set $x_i$ in a way sets $g$ to a constant, then this eliminates all three gates $g$, $g'$, and $h$ (since one of $h$’s inputs would get fixed). But $g'$ and $h$ might be the same gate! If that’s the case, then you should instead set $x_i$ to whatever value fixes $h$ (not $g$): this would eliminate $g$, $h$, and whichever gate reads $h$.