My favorite example of: the probabilistic method

The “probabilistic method” is the art of applying probabilistic thinking to non-probabilistic problems. Applications of the probabilistic method often feel like magic. Here is my favorite example:

Theorem (Erdös, 1965). Call a set {X} sum-free if for all ${a, b \in X}$ , we have ${a + b \not\in X}$ . For any finite set {S} of positive integers, there is a sum-free subset ${X \subseteq S}$ of size {|X| > |S| / 3} .

This theorem seems to have nothing to do with probability. Yet Erdös’s proof relies on a short and beautiful probabilistic argument.

Proof. Erdös’s first insight is to treat the set {S} as living in ${\mathbb{Z}_p}$ for some large prime number {p} , rather than living in ${\mathbb{Z}}$ . Living in ${\mathbb{Z}_p}$ only makes the theorem harder to prove, since any set {X} that is sum-free in ${\mathbb{Z}_p}$ is also sum-free in ${\mathbb{Z}}$ . But, as we shall see, ${\mathbb{Z}_p}$ has several structural properties that Erdös beautifully exploits.

Let {p = 3k + 2} be a prime number satisfying {p > s} for all ${s \in S}$ . (The fact that such a prime exists is actually nontrivial, and uses the fact that there are infinitely many primes congruent to {2} mod {3} .) We’ll be using {p} and {k} throughout the rest of the proof.

Erdös begins his construction by defining a large sum-free set in ${\mathbb{Z}_p}$ which is not necessarily a subset of {S} :

$\displaystyle \text{MIDDLE-THIRD} = \{k + 1, k + 2, \ldots, 2k + 1\}.$

As the name suggests, MIDDLE-THIRD consists of what is roughly the middle third of ${\mathbb{Z}_p}$ . One can easily confirm that MIDDLE-THIRD is sum-free in ${\mathbb{Z}_p}$ .

Next, Erdös selects a random value ${r \in \{1, 2, \ldots, p - 1\}}$ , and examines the set

$\displaystyle X_r = S \cap (r \cdot \text{MIDDLE-THIRD}),$

In words, ${X_r}$ is the set of ${s \in S}$ that can be written as ${rx \pmod p}$ for some ${x \in \text{MIDDLE-THIRD}}$ . If we could show that ${X_r}$ is a sum-free set of size greater than {|S| / 3} , then the proof would be complete.

The good news is that ${X_r}$ is always sum-free, no matter the value of {r} . This is because, if we have {a + b = c} for some ${a, b, c \in X_r}$ , then we also have {a / r + b / r = c / r} , violating the sum-freeness of MIDDLE-THIRD. (Here we are implicitly exploiting a nontrivial property of ${\mathbb{Z}_p}$ , namely that it supports division!)

The bad news is that ${|X_r|}$ could potentially be too small. To prove the theorem, we need to show that there is some option for {r} that results in ${|X_r| > |S| / 3}$ .

Rather than directly finding a value of {r} for which ${|X_r| > |S| / 3}$ , Erdös instead proves a probabilistic statement: the expected value of ${|X_r|}$ for a randomly chosen {r} satisfies ${\mathbb{E}[|X_r|] > |S| / 3}$ . Of course, if ${\mathop{\mathbb E}[|X_r|] > |S| / 3}$ , then there must exist some {r} for which ${|X_r| > |S| / 3}$ . What’s neat is that we don’t actually construct {r} , we just prove it exists.

Our new challenge is to prove that ${\mathbb{E}[|X_r|] > |S| / 3}$ . One way to think of ${|X_r|}$ is as the number of ${x \in \{k + 1, k + 2, \ldots, 2k + 1\}}$ for which ${rx \in S}$ . Thus we can expand ${\mathbb{E}[|X_r|]}$ as

$\displaystyle \mathbb{E}[|X_r|] = \sum_{x = k + 1}^{2k + 1} \Pr[rx \in S].$

This is where Erdös really exploits the structure of ${\mathbb{Z}_p}$ . In ${\mathbb{Z}_p}$ , every non-zero element {x} is an additive generator, meaning that the multiples ${x, 2x, 3x, \ldots, (p - 1)x}$ of {x} are just the numbers ${1, 2, \ldots, p - 1}$ in some order. The result is that {rx} is just a random element of ${\{1, 2, \ldots, p - 1\}}$ , and thus

$\displaystyle \mathbb{E}[|X_r|] = \sum_{x = k + 1}^{2k + 1} \frac{|S|}{p - 1}$

$\displaystyle = (k + 1) \frac{|S|}{p - 1}$

$\displaystyle > |S| / 3.$

Putting the pieces together, we have shown that ${\mathbb{E}[|X_r|] > |S| / 3}$ , which implies that there is some {r} for which ${|X_r| > |S| / 3}$ . Since ${X_r}$ is sum-free, this completes the proof of the theorem.

Magic!

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