Subfield: Combinatorics


Question: In some country several pairs of cities are connected by direct two-way flights. It is possible to go from any city to any other by a sequence of flights. The distance between two cities is defined to be the least possible number of flights required to go from one of them to the other. It is known that for any city there are at most 100 cities at distance exactly three from it. Prove that there is no city such that more than 2550 other cities have distance exactly four from it.

Solution:
Let us denote by $d(a, b)$ the distance between the cities $a$ and $b$, and by

$$
S_{i}(a)=\{c: d(a, c)=i\}
$$

the set of cities at distance exactly $i$ from city $a$.

Assume that for some city $x$ the set $D=S_{4}(x)$ has size at least 2551 . Let $A=S_{1}(x)$. A subset $A^{\prime}$ of $A$ is said to be substantial, if every city in $D$ can be reached from $x$ with four flights while passing through some member of $A^{\prime}$; in other terms, every city in $D$ has distance 3 from some member of $A^{\prime}$, or $D \subseteq \bigcup_{a \in A^{\prime}} S_{3}(a)$. For instance, $A$ itself is substantial. Now let us fix some substantial subset $A^{*}$ of $A$ having the minimal cardinality $m=\left|A^{*}\right|$.

Since

$$
m(101-m) \leqslant 50 \cdot 51=2550
$$

there has to be a city $a \in A^{*}$ such that $\left|S_{3}(a) \cap D\right| \geqslant 102-m$. As $\left|S_{3}(a)\right| \leqslant 100$, we obtain that $S_{3}(a)$ may contain at most $100-(102-m)=m-2$ cities $c$ with $d(c, x) \leqslant 3$. Let us denote by $T=\left\{c \in S_{3}(a): d(x, c) \leqslant 3\right\}$ the set of all such cities, so $|T| \leqslant m-2$. Now, to get a contradiction, we will construct $m-1$ distinct elements in $T$, corresponding to $m-1$ elements of the set $A_{a}=A^{*} \backslash\{a\}$.

Firstly, due to the minimality of $A^{*}$, for each $y \in A_{a}$ there exists some city $d_{y} \in D$ which can only be reached with four flights from $x$ by passing through $y$. So, there is a way to get from $x$ to $d_{y}$ along $x-y-b_{y}-c_{y}-d_{y}$ for some cities $b_{y}$ and $c_{y}$; notice that $d\left(x, b_{y}\right)=2$ and $d\left(x, c_{y}\right)=3$ since this path has the minimal possible length.

Now we claim that all $2(m-1)$ cities of the form $b_{y}, c_{y}$ with $y \in A_{a}$ are distinct. Indeed, no $b_{y}$ may coincide with any $c_{z}$ since their distances from $x$ are different. On the other hand, if one had $b_{y}=b_{z}$ for $y \neq z$, then there would exist a path of length 4 from $x$ to $d_{z}$ via $y$, namely $x-y-b_{z}-c_{z}-d_{z}$; this is impossible by the choice of $d_{z}$. Similarly, $c_{y} \neq c_{z}$ for $y \neq z$.

So, it suffices to prove that for every $y \in A_{a}$, one of the cities $b_{y}$ and $c_{y}$ has distance 3 from $a$ (and thus belongs to $T$ ). For that, notice that $d(a, y) \leqslant 2$ due to the path $a-x-y$, while $d\left(a, d_{y}\right) \geqslant d\left(x, d_{y}\right)-d(x, a)=3$. Moreover, $d\left(a, d_{y}\right) \neq 3$ by the choice of $d_{y}$; thus $d\left(a, d_{y}\right)>3$. Finally, in the sequence $d(a, y), d\left(a, b_{y}\right), d\left(a, c_{y}\right), d\left(a, d_{y}\right)$ the neighboring terms differ by at most 1 , the first term is less than 3 , and the last one is greater than 3 ; thus there exists one which is equal to 3 , as required.