Past exam of the graph theory course of the University of Oxford / 2023 / 3

A graph is a k-connected graph if it has at least $k+1$ vertices and deleting any set of fewer than $k$ vertices, together with their incident edges, leaves a connected graph.

The vertex connectivity $\kappa (G)$ is the largest such $k$. Equivalently, for a graph with at least two vertices, it is the smallest number of vertices whose deletion either disconnects the graph or leaves a single vertex. The latter convention gives $\kappa (K_n)=n-1$ for a complete graph. A disconnected graph has vertex connectivity zero; one usually also sets $\kappa (K_1)=0$.

For distinct nonadjacent vertices $x,y$, the local connectivity isSuch a set $S$ is an $x$–$y$ vertex separator. By Menger's theorem, local connectivity also equals the maximum number of internally vertex-disjoint paths from $x$ to $y$. If no graph path exists already, the empty vertex separator gives zero.

The vertex separator version of Menger's theorem says that, for distinct nonadjacent vertices $x,y$ of a finite graph, the maximum number of internally vertex-disjoint paths from $x$ to $y$ equals the minimum size of an $x$–$y$ vertex separator.

Use the following version of the max-flow min-cut theorem: in a finite directed flow network with nonnegative integer flow network edge capacities, the maximum flow value equals the minimum cut capacity, and a maximum flow can be chosen to have integer values on every directed edge. The last clause is the integral max-flow theorem.

Here is the reduction, including the correspondence with vertex separators. Put $N=|V(G)|$. Perform vertex splitting on each vertex $v\ne x,y$, replacing it by $v_{\mathrm{in}},v_{\mathrm{out}}$ with a directed edge $v_{\mathrm{in}}\to v_{\mathrm{out}}$ of flow network edge capacity one. Keep $x$ as the source and $y$ as the sink. For each original undirected edge $uv$, put directed edges from the output of $u$ to the input of $v$ and from the output of $v$ to the input of $u$, each of flow network edge capacity $N$; an unsplit endpoint uses $x$ or $y$ itself.

A family of $p$ internally vertex-disjoint paths gives a feasible integral flow of value $p$, by sending one unit down each graph path. Conversely, decompose an integral flow into unit source-to-sink graph paths and directed cycles, discarding the directed cycles. This decomposition can be obtained by following positive-flow directed edges and subtracting one unit along a source-to-sink graph path or a directed cycle, repeating until the flow value is exhausted. A capacity-one split edge can belong to at most one unit graph path. The projected graph paths in $G$ are therefore internally vertex-disjoint paths. Thus the largest number of such graph paths equals the maximum flow value.

Deleting a vertex separator $S$ corresponds to removing its split edges, of total flow network edge capacity $|S|$. This destroys every source-to-sink graph path; taking the source-reachable vertices after removal gives a cut of a flow network of cut capacity at most $|S|$. Thus the minimum cut capacity is no greater than the minimum vertex separator size.

For the reverse inequality, deleting all $N-2$ internal vertices separates $x,y$, because they are nonadjacent. Hence a minimum cut has cut capacity at most $N-2<N$ and cannot contain a directed edge of flow network edge capacity $N$. Its outgoing edges are therefore split edges. Let $S$ be the original vertices associated with them. Any $x$–$y$ graph path avoiding $S$ would lift to a source-to-sink graph path crossing this cut of a flow network without using an outgoing edge, an impossibility. Hence $S$ is a vertex separator, with size exactly the cut capacity. The two minima agree, and the max-flow min-cut theorem proves Menger's theorem.

Add two new vertices $s,t$ to $G$, joining $s$ to every vertex of $U$ and $t$ to every vertex of $W$, with no other new edges. The vertices $s,t$ are nonadjacent. Suppose $S\subseteq V(G)$ has $|S|<k$. Since $G$ is a k-connected graph, $G-S$ is a connected graph. Both $U\setminus S$ and $W\setminus S$ are nonempty because each original set has cardinality $k$. Choose $u\in U\setminus S$ and $w\in W\setminus S$; a graph path between them in $G-S$ extends to an $s$–$t$ graph path avoiding $S$.

Consequently no fewer than $k$ vertices can form an $s$–$t$ vertex separator. By Menger's theorem, there are at least $k$ internally vertex-disjoint paths from $s$ to $t$. There cannot be more than $k$, since $s$ has only $k$ neighbours.

Remove $s,t$ from these $k$ graph paths. The resulting graph paths in $G$ start at distinct vertices of $U$, end at distinct vertices of $W$, and are pairwise vertex-disjoint paths: any shared vertex would have been an internal vertex common to two of the original graph paths. Because there are exactly $k$ graph paths, they use every vertex of $U$ and $W$ as an endpoint. None can pass through another terminal internally, since that terminal already occurs as an endpoint of one of the vertex-disjoint paths. Thus there are $k$ vertex-disjoint paths from $U$ to $W$, with some pairing of the endpoints. This is also an immediate application of the Set version of Menger theorem.

No. A counterexample works for every $k\ge 2$. Take the complete bipartite graph $G=K_{k,k}$ with vertex setsIt is a k-connected graph: deleting fewer than $k$ vertices leaves at least one vertex in each side, and the remaining complete bipartite graph is a connected graph.

Chooseand order these sets so that $u_1=b_1$ and $w_1=b_2$. Order the other members arbitrarily. These are disjoint sets of cardinality $k$, and every vertex of $G$ is one of the specified endpoints.

If the prescribed $k$ vertex-disjoint paths existed, no one of them could have an internal vertex, because every vertex must serve as the endpoint of one of the graph paths. Thus each graph path would have to be a single edge. But $b_1,b_2$ lie in the same side of the bipartite graph, so there is no edge $b_1{b}_2$. This contradicts the requirement on $P_1$.

For $k=2$ this is the four-vertex cycle graph, with opposite vertices paired. The stronger property that every prescribed pairing can be realized is the linked graph property; vertex connectivity alone does not guarantee it.

We verify the Hall condition for $U$. Let $S\subseteq U$. If $S$ is empty, then $|N(S)|=0=|S|$. Suppose $0<|S|\le n/2$. Choose $u\in S$. Its graph neighbourhood is contained in $N(S)$, so the minimum degree hypothesis givesIf $|S|>n/2$, suppose for a contradiction that some vertex $w\in W\setminus N(S)$ exists. All neighbours of $w$ lie in $U\setminus S$, socontradicting the minimum degree hypothesis. Thus $N(S)=W$ and $|N(S)|=n\ge |S|$.

Every subset $S\subseteq U$ satisfies the Hall condition. By Hall's marriage theorem, there is a matching in a graph saturating $U$. Since $|U|=|W|=n$, its $n$ edges also cover every vertex of $W$. It is therefore a complete matching, meaning a perfect matching. For $n=0$, the empty matching in a graph already has this property.