Source: USATST-2018-P4
Category: Extremal Problems

Problem:
Let $n$ be a positive integer and let $S \subseteq \{0,1\}^n$ be a set of binary strings of length $n$. Given an odd number $x_1, \dots, x_{2k+1} \in S$ of binary strings (not necessarily distinct), their *majority* is defined as the binary string $y \in \{0,1\}^n$ for which the $i$\textsuperscript{th} bit of $y$ is the most common bit among the $i$\textsuperscript{th} bits of $x_1$, \dots, $x_{2k+1}$. (For example, if $n=4$ the majority of $0000$, $0000$, $1101$, $1100$, $0101$ is $0100$.) Suppose that for some positive integer $k$, $S$ has the property $P_k$ that the majority of any $2k+1$ binary strings in $S$ (possibly with repetition) is also in $S$. Prove that $S$ has the same property $P_k$ for all positive integers $k$.

Solution:
Let $M$ denote the majority function (of any length).

**First solution (induction).** We prove all $P_k$ are equivalent by induction on $n \ge 2$, with the base case $n = 2$ being easy to check by hand. (The case $n=1$ is also vacuous; however, the inductive step is not able to go from $n=1$ to $n=2$.)

For the inductive step, we proceed by contradiction; assume $S$ satisfies $P_{\ell}$, but not $P_{k}$, so there exist $x_1, \dots, x_{2k+1} \in S$ whose majority $y = M(x_1, \dots, x_{2k+1})$ is not in $S$. We contend that:

**Claim:** Let $y_i$ be the string which differs from $y$ only in the $i$\textsuperscript{th} bit. Then $y_i \in S$.

*Proof.* For a string $s \in S$ we let $\hat s$ denote the string $s$ with the $i$\textsuperscript{th} bit deleted (hence with $n-1$ bits). Now let
\[
T = \left\{ \hat s \mid s \in S \right\}.
\]
Since $S$ satisfies $P_\ell$, so does $T$; thus by the induction hypothesis on $n$, $T$ satisfies $P_{k}$. Consequently, $T \ni M(\hat{x}_1, \dots, \hat{x}_{2k+1}) = \hat y$. Thus there exists $s \in S$ such that $\hat s = \hat y$. This implies $s = y$ or $s = y_i$. But since we assumed $y \notin S$ it follows $y_i \in S$ instead.

Now take any $2\ell+1$ copies of the $y_i$, about equally often (i.e., the number of times any two $y_i$ are taken differs by at most $1$). We see the majority of these is $y$ itself, contradiction.

**Second solution (circuit construction).** Note that $P_k \implies P_1$ for any $k$, since
\[
M( \underbrace{a,\dots,a}_k, \underbrace{b,\dots,b}_k, c ) = M(a,b,c)
\]
for any $a$, $b$, $c$. We will now prove $P_1 + P_k \implies P_{k+1}$ for any $k$, which will prove the result. Actually, we will show that the majority of any $2k+3$ strings $x_1$, \dots, $x_{2k+3}$ can be expressed by $3$-majorities and $(2k+1)$-majorities. WLOG assume that $M(x_1, \dots, x_{2k+3}) = 0\dots0$, and let $\odot$ denote binary AND.

**Claim:** We have $M(x_1, x_2, M(x_3, \dots, x_{2k+3})) = x_1 \odot x_2$.

[Proof continues with circuit construction arguments]


Grading guidelines:
(Partial) 1. Showed that for any k, property P_k implies P_1 (e.g. by padding M(a,b,c) as M(a,…,a,b,…,b,c) with k copies). 2. Set up the “delete one coordinate” reduction: defined \(\hat s\) by deleting the i-th bit, defined \(T=\{\hat s:s\in S\}\), and correctly observed that if S has P_k then T has P_k (majority commutes with coordinate deletion). 3. In a contradiction setup (S has P_\ell but fails P_k with majority y\notin S), proved the key claim that for each i the neighbor \(y_i\) obtained by flipping the i-th bit lies in S (using the reduction to T and induction on n, or any other valid argument producing all \(y_i\in S\)). (Almost) 1. Presented a complete contradiction argument via coordinate deletion and induction on n, and derived \(y_i\in S\) for all i, but did not fully justify the final multiset choice of \(2\ell+1\) copies of the \(y_i\) whose majority is exactly y (e.g. the “almost equally often” counting step is not checked). 2. Used the circuit/closure approach: proved P_k\(\Rightarrow\)P_1 and established (or nearly established) the recursive step P_1 + P_k \(\Rightarrow\) P_{k+1} by expressing a (2k+3)-majority using 3-majorities and (2k+1)-majorities, but left one of the circuit identities/boolean-case checks (or the final induction on k) with a small gap. 3. Solution is almost complete, but made minor mistakes which are not negligible.