CF 102980B - Поврежденный пароль
Let $mathcal{A}$ be a set of $t$-combinations and let $ $$kappat N = min{ where $partial mathcal{A}$ is the set of all $(t-1)$-subsets obtained by deleting one element from a member of $mathcal{A}$.
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Solution
Solution
Let $\mathcal{A}$ be a set of $t$-combinations and let $|\mathcal{A}| = N$. The operator $\kappa_t N$ denotes the minimum possible size of the shadow of any family of $N$ $t$-combinations, that is
$$\kappa_t N = \min_{|\mathcal{A}| = N} |\partial \mathcal{A}|,$$
where $\partial \mathcal{A}$ is the set of all $(t-1)$-subsets obtained by deleting one element from a member of $\mathcal{A}$.
From the definition of $\partial$, each $t$-combination contributes exactly $t$ distinct $(t-1)$-subsets before identifications from overlaps. Hence for every $\mathcal{A}$,
$$|\partial \mathcal{A}| \le t |\mathcal{A}| = tN.$$
This bound is attained when $N=1$, since a single $t$-combination has exactly $t$ distinct $(t-1)$-subsets. Therefore,
$$\kappa_t 1 = t.$$
Consequently,
$$\kappa_t 1 - 1 = t - 1.$$
To determine whether any larger value can occur, consider $N \ge 2$. Any two distinct $t$-combinations share at least one $(t-1)$-subset only when they differ in exactly one element; in that case their shadows overlap in at least one element, reducing the total shadow size strictly below $tN$. Hence for $N \ge 2$,
$$\kappa_t N \le tN - 1,$$
since at least one overlap occurs in the shadow of any non-singleton family. This implies
$$\kappa_t N - N \le (t-1)N - 1,$$
which is strictly less than $t-1$ for all $N \ge 2$.
For $N=0$, $\kappa_t 0 = 0$ and the difference is $0$.
For $N=1$, the value is $t-1$, and this is the only case where the upper bound $t$ shadow size is achieved without overlap.
Thus the maximum of $\kappa_t N - N$ over all $N \ge 0$ is attained at $N=1$ and equals
$$\boxed{t-1}.$$
This completes the proof. ∎