Subfield: Combinatorics


Question: The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.

Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{A B}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{B A}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{A B}=N_{B A}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

Solution:
We recall some graph-theoretical notions. Let $G$ be a finite graph, and let $V$ be the set of its vertices; fix two distinct vertices $s, t \in V$. An $(s, t)$-cut is a partition of $V$ into two parts $V=S \sqcup T$ such that $s \in S$ and $t \in T$. The cut-edges in the cut $(S, T)$ are the edges going from $S$ to $T$, and the size $e(S, T)$ of the cut is the number of cut-edges.

We will make use of the following theorem (which is a partial case of the Ford-Fulkerson "min-cut max-flow" theorem).

Theorem (Menger). Let $G$ be a directed graph, and let $s$ and $t$ be its distinct vertices. Then the maximal number of edge-disjoint paths from $s$ to $t$ is equal to the minimal size of an $(s, t)$-cut.

Back to the problem. Consider a directed graph $G$ whose vertices are the cities, and edges correspond to the roads. Then $N_{A B}$ is the maximal number of edge-disjoint paths from $A$ to $B$ in this graph; the number $N_{B A}$ is interpreted similarly.

As in the previous solution, denote by $a$ and $b$ the out-degrees of vertices $A$ and $B$, respectively. To solve the problem, we show that for any $(A, B)$-cut $\left(S_{A}, T_{A}\right)$ in our graph there exists a $(B, A)$-cut $\left(S_{B}, T_{B}\right)$ satisfying

$$
e\left(S_{B}, T_{B}\right)=e\left(S_{A}, T_{A}\right)+(b-a)
$$

This yields

$$
N_{B A} \leqslant N_{A B}+(b-a) ; \quad \text { similarly, we get } \quad N_{A B} \leqslant N_{B A}+(a-b),
$$

whence again $N_{B A}-N_{A B}=b-a$.

The construction is simple: we put $S_{B}=S_{A} \cup\{B\} \backslash\{A\}$ and hence $T_{B}=T_{A} \cup\{A\} \backslash\{B\}$. To show that it works, let $\mathrm{A}$ and $\mathrm{B}$ denote the sets of cut-edges in $\left(S_{A}, T_{A}\right)$ and $\left(S_{B}, T_{B}\right)$, respectively. Let $a_{s}$ and $a_{t}=a-a_{s}$ denote the numbers of edges going from $A$ to $S_{A}$ and $T_{A}$, respectively. Similarly, denote by $b_{s}$ and $b_{t}=b-b_{s}$ the numbers of edges going from $B$ to $S_{B}$ and $T_{B}$, respectively.

Notice that any edge incident to none of $A$ and $B$ either belongs to both $\mathrm{A}$ and $\mathrm{B}$, or belongs to none of them. Denote the number of such edges belonging to $A$ by $c$. The edges in $A$ which are not yet accounted for split into two categories: those going out from $A$ to $T_{A}$ (including $A \rightarrow B$ if it exists), and those going from $S_{A} \backslash\{A\}$ to $B$ - in other words, going from $S_{B}$ to $B$. The numbers of edges in the two categories are $a_{t}$ and $\left|S_{B}\right|-1-b_{s}$, respectively. Therefore,

$$
|\mathrm{A}|=c+a_{t}+\left(\left|S_{B}\right|-b_{s}-1\right) . \quad \text { Similarly, we get } \quad|\mathrm{B}|=c+b_{t}+\left(\left|S_{A}\right|-a_{s}-1\right)
$$

and hence

$$
|\mathrm{B}|-|\mathrm{A}|=\left(b_{t}+b_{s}\right)-\left(a_{t}+a_{s}\right)=b-a
$$

since $\left|S_{A}\right|=\left|S_{B}\right|$. This finishes the solution.