$\require{mathtools}
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% Environments
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% Greek
\newcommand{\eps}{\epsilon}
\newcommand{\veps}{\varepsilon}
\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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\newcommand{\gs}{\greatersim}
% .. miscer
\newcommand{\q}{\quad}
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\newcommand{\heart}{\heartsuit}
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% Delimiters
% (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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% Logic
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% ... named tensors
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% ... information theory
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%%% SPECIALIZED COMPUTER SCIENCE %%%
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% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
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\newcommand{\PSPACE}{\mathsf{PSPACE}}
\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
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% .. circuits
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% .. resources
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% .. keywords
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% .. functions (small caps sadly doesn't work)
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% In text
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% .. bold
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% .. calligraphic
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% .. typewriter
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The principle
Suppose you want to show that no good randomized algorithm/protocol/whatever exists for some problem. It is sufficent (and necessary) to show an input distribution that is hard for all deterministic algorithms/protocols/whatever.
More precisely,
\[\min_{\CA} \max_x \Pr_{A \sim \CA}[\text{$A$ fails on $x$}] \geq \max_{\CX} \min_{A} \Pr_{x \sim \CX}[\text{$A$ fails on $x$}].\]
Put another way, if you can find a distribution $\CX$ on which any deterministic algorithm $A$ fails on a $> \eps$ fraction of the inputs, this shows that any randomized algorithm $\CA$ must fail on some input $x$ with at least $> \eps$ probability.
The proof
Expressions with mins and maxes are confusing, so we’ll instead consider games between Algie (the algorithm) and Imperia (the input). Algie is trying to minimize the probability of a failure, while Imperia is trying to maximize it.
We will consider a series of 4 games, and show that the changes only make the situation better for Algie and worse for Imperia.
- Game 1: $\min_\CA \max_x \Pr_{A \sim \CA}[A\text{ fails on }x]$.
First, Algie chooses a distribution $\CA$ over deterministic algorithms (aka a randomized algorithm), then Imperia chooses a single input $x$.
- Game 2: $\min_\CA \max_\CX \Pr_{A \sim \CA, x \sim \CX}[A\text{ fails on }x]$.
Instead of choosing a single input $x$, Imperia gets to choose a distribution $\CX$ over inputs. But since she’s going second, she knows $\CA$, so she can figure out which $x$ gives her the best outcome and put all her weight on it. Therefore the outcome doesn’t change.
- Game 3: $\max_\CX \min_\CA \Pr_{A \sim \CA, x \sim \CX}[A\text{ fails on }x]$.
The order of the players is swapped. Intuitively, it’s always better to go second, because you have more information. Therefore, this can only make the outcome better for Algie. The other direction (which we don’t need) is by von Neumann’s minimax theorem.
- Game 4: $\max_\CX \min_A \Pr_{x \sim \CX}[A\text{ fails on }x]$.
Instead of choosing a distribution $\CA$, Algie must choose a single algorithm $A$. But again, since Algie goes second, he doesn’t care.
Since Game 4 is better for Algie, if we show that Game 4 is tough for him, so is Game 1.
#to-write the other direction allows $A$ to assume that there is a prior $\CX$ over $x$ (in case where $A$’s goal is actually to that it doesn’t have direct access to $x$)
