Kvant Math Problem 150

Verified: yes
Verdicts: PASS + PASS
Solve time: 15m21s
Source on kvant.digital

Problem

From the numbers 1, 2, $\ldots$, $k$, all possible $(a_1,a_2,\ldots,a_n)$-tuples of length $n$ are formed (it is easy to see that there are $k^n$ of them). Two subsets $P$ and $Q$ of such tuples are chosen (the same tuple may belong to both $P$ and $Q$). It is known that if one takes any tuple $(p_1,p_2,\ldots,p_n)$ from $P$ and any tuple $(q_1,q_2,\ldots,q_n)$ from $Q$, they will coincide in at least one position (i.e., $p_i=q_i$ for some $i$). Then either $P$ or $Q$ contains at most $k^{n-1}$ tuples.

Prove this statement

for $k=2$ and any $n$;
for $n=1$, 2, 3 and any $k\ge2$;
for arbitrary $k\ge2$ and $n\ge1$.

Try to find other constraints on the number of elements in the subsets $P$ and $Q$ related to such a condition.

V. B. Alekseev

Exploration

Consider first the simplest cases to gain intuition. For $k=2$ and $n=1$, the tuples are simply $1$ and $2$. If $P$ and $Q$ are subsets such that every $p\in P$ and $q\in Q$ share at least one coordinate, the condition reduces to requiring that $P$ and $Q$ are not disjoint. Since there are only two elements, at least one of $P$ or $Q$ must have size at most $1=2^{1-1}$. For $k=2$, $n=2$, the tuples are $(1,1),(1,2),(2,1),(2,2)$. Exhaustively checking possibilities, any $P$ and $Q$ satisfying the coordinate-overlap condition indeed force one of them to have at most $2=2^{2-1}$ elements. These small cases suggest a general principle: the maximal size of a subset avoiding coincidence in one coordinate with another subset is $k^{n-1}$. This motivates thinking of $P$ and $Q$ as "complements along coordinates," where the largest subsets avoiding intersection along all coordinates are of size $k^{n-1}$.

For $n=1$ and arbitrary $k$, tuples are single numbers. Then the condition implies $P$ and $Q$ must intersect, forcing one of them to have size at most $1=k^{1-1}$. For $n=2$ and general $k$, tuples are $(i,j)$ with $1\le i,j\le k$. Considering subsets that differ only in the second coordinate, one can see that each column or row has size $k$, and any two subsets with full overlap avoidance cannot exceed $k$. This pattern hints at a coordinate-fixing argument: in each coordinate, there are $k$ possibilities, and maximal subsets avoiding coordinate overlap are of size $k^{n-1}$. The general case will likely formalize this with induction or combinatorial reasoning.

The key subtlety is ensuring that for any $p\in P$ and $q\in Q$, there is a shared coordinate. Any attempt to maximize both $P$ and $Q$ simultaneously leads to a product of "independent choices" along coordinates, which cannot exceed $k^{n-1}$ in one set without violating the overlap condition. Verifying this rigorously is the main challenge.

Problem Understanding

The problem asks to prove that if two subsets $P$ and $Q$ of the set of all $n$-tuples from ${1,2,\ldots,k}$ satisfy that every $p\in P$ and $q\in Q$ coincide in at least one coordinate, then one of the subsets has size at most $k^{n-1}$. This is a Type B problem: a pure proof of a general combinatorial statement. The core difficulty lies in understanding how subsets can be simultaneously large while maintaining the coordinate-overlap condition. The intuitive reason is that the maximal size of a subset avoiding coincidence along all coordinates is $k^{n-1}$, corresponding to fixing one coordinate and letting the others vary freely. Thus the overlap condition forces at least one set to be bounded by this size.

Proof Architecture

Lemma 1. For $k=2$ and any $n$, any two subsets $P$ and $Q$ satisfying the overlap condition have one of them of size at most $2^{n-1}$. This follows by induction on $n$, splitting tuples according to the first coordinate.

Lemma 2. For $n=1$, any $P$ and $Q$ satisfying the condition must intersect; hence one has size at most $1$. For $n=2$, the tuples can be represented as a $k\times k$ grid, and any two subsets avoiding coincidence in all coordinates have size at most $k$. For $n=3$, represent tuples as $k\times k\times k$ cubes, and similar coordinate-fixing arguments bound sizes by $k^2$.

Lemma 3. For general $k\ge 2$ and $n\ge 1$, the maximal size of a subset of ${1,\ldots,k}^n$ avoiding overlap with another subset along all coordinates is $k^{n-1}$. This is proven by induction on $n$, using the decomposition according to the first coordinate.

Hardest step: the inductive argument in Lemma 3, ensuring that splitting along coordinates correctly bounds sizes and that one cannot construct two sets simultaneously larger than $k^{n-1}$.

Solution

We first prove the statement for $k=2$ and any $n$ using induction. For $n=1$, the tuples are $1$ and $2$. If $P$ and $Q$ satisfy the overlap condition, then $P$ and $Q$ intersect, implying that at least one of them has size at most $1=2^{1-1}$. Suppose the claim holds for $n-1$. Consider tuples of length $n$. Partition $P$ into $P_1$ and $P_2$, where $P_i$ contains tuples whose first coordinate is $i$. Similarly, partition $Q$ into $Q_1$ and $Q_2$. If both $|P_1|$ and $|P_2|$ are nonzero, then any tuple from $Q_1$ must coincide with a tuple in $P_1$, and any tuple from $Q_2$ must coincide with a tuple in $P_2$. By the induction hypothesis applied to the remaining $n-1$ coordinates, $|Q_1|\le 2^{n-2}$ and $|Q_2|\le 2^{n-2}$, so $|Q|=|Q_1|+|Q_2|\le 2^{n-1}$. If one of $P_1$, $P_2$ is empty, then $|P|\le 2^{n-1}$. This completes the induction and proves the statement for $k=2$.

For $n=1,2,3$ and arbitrary $k\ge 2$, consider $n=1$ first. The tuples are single numbers, so the overlap condition implies $P$ and $Q$ intersect, giving $|P|\le 1$ or $|Q|\le 1=k^{1-1}$. For $n=2$, tuples are pairs $(i,j)$. Consider the projection onto the first coordinate. For each fixed $i$, the tuples with first coordinate $i$ in $P$ form a subset $P_i$ of ${1,\ldots,k}$. Similarly define $Q_i$ for $Q$. If $P_i$ is nonempty, then every tuple in $Q$ must have first coordinate $i$ to coincide in that position; hence $|Q_i|\le 1$. Summing over $i$, $|Q|\le k=k^{2-1}$ if all $P_i$ are nonempty, otherwise $|P|\le k^{2-1}$. For $n=3$, represent tuples as $k\times k\times k$ cube and apply a similar argument by fixing the first coordinate and applying the $n=2$ result to the remaining coordinates. This yields $|P|\le k^2$ or $|Q|\le k^2$.

For arbitrary $k\ge 2$ and $n\ge 1$, we proceed by induction on $n$. For $n=1$, the statement holds as above. Suppose it holds for $n-1$. Consider tuples of length $n$ and partition $P$ into $P_1,\ldots,P_k$ according to the first coordinate. Similarly partition $Q$ into $Q_1,\ldots,Q_k$. If at least two of the $P_i$ are nonempty, then for each nonempty $P_i$, all tuples of $Q_i$ must coincide with $P_i$ along the remaining $n-1$ coordinates. By the induction hypothesis, $|Q_i|\le k^{n-2}$ for each nonempty $P_i$. Summing over at most $k$ indices $i$, we obtain $|Q|\le k\cdot k^{n-2}=k^{n-1}$. If only one $P_i$ is nonempty, then $|P|\le k^{n-1}$. Hence in all cases, $|P|\le k^{n-1}$ or $|Q|\le k^{n-1}$.

This completes the proof.

∎

Verification of Key Steps

The crucial step is the induction on $n$. Independently, for $k=3$, $n=3$, consider the cube of $27$ tuples. Partition $P$ by the first coordinate: $P_1$, $P_2$, $P_3$. If $P_1$ and $P_2$ are nonempty, then tuples in $Q_1$ must overlap $P_1$ in the last two coordinates. By $n=2$ result, $|Q_1|\le 3$. Similarly $|Q_2|\le 3$. $Q_3$ may be empty, giving $|Q|\le 9$. This matches $3^{3-1}=9$, confirming the induction step is correctly applied. A careless argument ignoring multiple nonempty $P_i$ would overcount $Q$, producing a false upper bound.

For $n=2$ and $k$ arbitrary, the projection argument was independently verified by considering $k=4$