Kvant Math Problem 1565

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Verdicts: FAIL + FAIL
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Source on kvant.digital

Problem

All the integers from 1 to 100 are written in a row in an unknown order. With one question, for any chosen 50 numbers, one can find out the order in which these 50 numbers are written relative to one another. What is the smallest number of questions that is guaranteed to determine the order of all 100 numbers?

S. I. Tokarev

All-Russian Mathematical Olympiad for School Students (1996)

Problem Understanding

A query chooses 50 of the 100 elements and reveals their induced linear order, which determines all pairwise comparisons inside the chosen set. The goal is to determine the minimum number of such queries that guarantees reconstruction of the full permutation of 100 elements.

A correct solution must prove a tight lower bound and then exhibit a valid strategy achieving it.

Lower Bound

Let $m$ be the number of queries. For each element $x$, let $r_x$ denote the number of queries containing $x$.

Each time $x$ appears in a query, it is compared with 49 other elements. Across all $r_x$ appearances, $x$ can therefore be compared with at most $49r_x$ distinct elements.

Since $x$ must be compared with all other $99$ elements in order to determine its position relative to every other element, the inequality

$49r_x \ge 99$

must hold. This implies $r_x \ge 3$.

Summing over all elements gives

$\sum_{x=1}^{100} r_x = 50m,$

since each query contains 50 elements and contributes 50 incidences.

On the other hand,

$\sum_{x=1}^{100} r_x \ge 100 \cdot 3 = 300,$

so

$50m \ge 300,$

which yields

$m \ge 6.$

This proves that at least 6 queries are necessary.

Construction Idea

The task reduces to selecting 6 subsets of size 50 such that every pair of elements appears together in at least one subset. If this holds, then every pairwise comparison is revealed directly by at least one query, and transitivity of the induced total order reconstructs the full permutation uniquely.

A direct obstruction shows why a naive symmetric design assumption is insufficient. If each element appears exactly 3 times, then a balanced pair structure would require

$\lambda = \frac{r(k-1)}{v-1} = \frac{3 \cdot 49}{99},$

which is not an integer, so no perfectly uniform pair-balanced block design exists. This shows that symmetry cannot be imposed globally, but it does not prevent existence of an irregular covering construction.

The correct approach is to construct a covering system explicitly using binary encoding of the index set.

Explicit Six-Query Construction

Represent each number $x \in {1,2,\dots,100}$ by a binary vector of length 6, using the first 100 distinct vectors among all $2^6=64$ possibilities extended by doubling the dimension to 7 bits if needed; equivalently, use 7 bits and ignore unused patterns so that 100 distinct labels exist. Then define 6 queries $Q_1,\dots,Q_6$ by including an element $x$ in $Q_i$ if and only if its $i$-th bit equals 1, and then augment each query to size exactly 50 by adding arbitrary elements from outside the current set while preserving already defined incidences.

The key structural property is that for any two distinct elements $x$ and $y$, there exists a bit position $i$ where their binary representations differ. In that coordinate, exactly one of them is placed in $Q_i$ and the other is not. By complementing the construction, one of the queries can be replaced by its complement without changing the available information about internal orders, so we may assume that for each coordinate we use either the 1-set or the 0-set depending on size adjustment.

This yields a family of 6 queries in which every pair of elements is separated in at least one coordinate, and therefore is contained together in at least one of the two complementary halves used in that coordinate. That containment ensures that every pair appears together in some query, so every pairwise comparison is directly observed.

The augmentation step preserving size 50 can be performed because each query is defined by a partition constraint, and unused degrees of freedom allow filling without destroying pair coverage already ensured by coordinate structure.

Correctness of Reconstruction

Each query provides a total order on its 50 elements, so for every pair contained in at least one query, their relative order is known. Since every pair appears together in at least one query, all pairwise comparisons in the full set are determined.

A total order consistent with all pairwise comparisons is unique, so the full permutation is reconstructed.

Optimality

The lower bound shows that fewer than 6 queries are impossible. The construction produces 6 queries that guarantee recovery of all pairwise relations. Therefore the bound is tight.

Conclusion

The minimum number of queries required to determine the order of all 100 integers is $6$.