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An acyclic orientation is an orientation of a graph whose resulting directed graph has no directed cycle, hence is a Directed acyclic graph. A directed cycle must follow every arrow around its closed route. A cycle in a graph in the skeleton of a directed graph only requires the corresponding undirected edges; their arrows need not agree around the route.

For the same triangle in a graph with vertex set $\{A,B,C\}$ and edges $AB,BC,AC$, compareBoth have an undirected cycle in a graph in their skeleton of a directed graph. Only $D_2$ has a directed cycle. In $D_1$, $C$ has no outgoing directed edge, so a route cannot continue from $C$ back to $A$.

Thus an acyclic orientation does not require its underlying undirected graph to be a forest. Every orientation of a graph of a forest is acyclic, because a directed cycle would give an undirected cycle in a graph. Conversely, every finite simple graph, including one with many cycles in a graph, admits at least one acyclic orientation by the following ordering construction. On this triangle in a graph, six of the eight orientations of a graph are acyclic; the other two follow the directed cycle in opposite senses.

For a finite simple graph $G$ and a matching in a graph $M\subseteq E(G)$, $M$ is a maximum cardinality matching if and only if $G$ has no augmenting path in a matching relative to $M$. The graph need not be a bipartite graph, and the assertion concerns cardinality, not edge benefits.^[ref]

The criterion turns a global comparison with every competing matching in a graph into a search for one repair. Starting from any matching in a graph, repeatedly find and toggle an augmenting path in a matching. Each step adds one selected edge, so finiteness gives termination after at most $\lfloor |V(G)|/2\rfloor -|M|$ steps; absence of any further augmenting path in a matching certifies maximum cardinality. This is an optimality criterion, not by itself a bound on the work needed to find each graph path.

For a short proof outline, an augmenting path in a matching immediately contradicts maximum cardinality. Conversely, if another matching in a graph $N$ has $|N|>|M|$, consider the subgraph whose set of edges is $M△N$. Its vertices have degree of a vertex at most two; its nontrivial connected components are graph paths and even cycles in a graph, alternating between the two matchings in a graph. A cycle in a graph contributes equal numbers from each. Hence some graph path contributes one more edge from $N$ than from $M$ and starts and ends with edges from $N$. Its endpoints are unmatched vertices in a matching relative to $M$, yielding the required augmenting path in a matching.

A 2-core can contain bridges in a graph, and even vertices that lie on no cycle in a graph. Consider the dumbbell graph, also called a barbell in this example, made of two disjoint triangles in a graph joined by a graph path with two internal vertices:Here $c,d$ have degree of a vertex three, and every other vertex has degree of a vertex two. No deletion is possible, so the whole graph is its 2-core. Yet $cx,xy,yd$ are bridges in a graph: deleting any one separates the two triangles. None belongs to a cycle in a graph, and $x,y$ belong to no cycle in a graph either. Keeping only edges on graph cycles would incorrectly discard the entire joining graph path.

Every cycle in a graph does survive pruning, because each of its vertices keeps its two cycle graph neighbours. The barbell shows that the converse fails: the 2-core can be larger than the union of all graph cycles. Its condition is local minimum degree, not protection against a single connection failing.

A biconnected component is a different object: a maximal connected subgraph with no cut vertex of its own. Under the block convention used here, the barbell has five biconnected components: its two triangles and its three individual joining edges, each with its endpoints. Its cut vertices are $c,x,y,d$. A 2-connected component with at least three vertices must remain connected after any single vertex is deleted; the whole barbell fails that test. An informal phrase such as "biconnected core" therefore needs its own definition and must not be substituted for 2-core. Even an isolated edge is a block under this convention, although its 2-core is empty.

The usefulness of the 2-core is this precise separation of local persistence from global robustness. In a connected graph with nonempty 2-core, the deleted vertices form a forest, with each of its connected components of a graph attached by exactly one edge to the 2-core. Otherwise a deleted graph cycle, or a graph path joining two surviving attachment points, would itself have survived. Stripping these attached trees exposes a smaller structure for studying graph cycles, graph excess and the graph kernel, while preserving the joining bridges in a graph that organize that structure.

A chordal graph is a finite simple graph with no induced cycle of length at least four. Equivalently, every cycle in a graph with at least four vertices has a chord: an edge joining two nonconsecutive vertices of that graph cycle. The word "induced" matters: the induced subgraph on the chosen vertices must contain exactly the cycle's edges, with no chords. Triangles are allowed. This definition concerns adjacency, independently of how the graph is drawn.

For the square cycle graph $C_4$, take vertex set $\{a,b,c,d\}$ and set of edges $\{ab,bc,cd,da\}$, where $ab$ abbreviates $\{a,b\}$. Its graph neighbourhoods are $N(a)=N(c)=\{b,d\}$ and $N(b)=N(d)=\{a,c\}$. The square is an induced cycle, so $C_4$ is not a chordal graph. Add the single edge $ac$: now $N(a)=\{b,c,d\}$, $N(b)=\{a,c\}$, $N(c)=\{a,b,d\}$, and $N(d)=\{a,c\}$. The only possible four-vertex graph cycle has chord $ac$, so this new graph is a chordal graph. Its maximal cliques are exactly $\{a,b,c\}$ and $\{a,c,d\}$.

Connect these two maximal cliques by a single edge labelled with their intersection $\{a,c\}$. The result is a junction tree, also called a clique tree: its nodes are maximal cliques, and the nodes containing any fixed original vertex induce a connected graph inside the tree. Here $a,c$ occur in both nodes, while $b,d$ each occur in just one. This is the running intersection property. More generally, a nonempty connected graph is a chordal graph exactly when its maximal cliques admit such a tree; a disconnected chordal graph can be represented by one such tree per connected component of a graph.^[ref]

A chordal graph need not be a planar graph: the complete graph $K_5$ is a chordal graph, since every pair of distinct vertices is adjacent, but its ten edges violate the planar graph edge bound $|E|\le 3|V|-6=9$. Conversely, even a plane drawing whose bounded faces are triangles can hide an induced cycle: a wheel graph with a four-vertex rim and a central vertex joined to the whole rim still has the rim as an induced cycle. Adding chords to obtain a chordal graph is called triangulation of an undirected graph; this is an adjacency condition, rather than a requirement about planar faces.

Why is this structure useful? The maximal cliques collect local interactions, and the running intersection property makes their overlaps sufficient for consistent communication. For random variables $x_a,x_b,x_c,x_d$ with finite state sets, a probabilistic graphical model of the square-plus-diagonal form can havewhere the two nonnegative clique potentials have a positive normalizing constant $Z$. Eliminating $x_b$ produces the two-variable junction-tree sum-product messageThe shared clique $\{a,c\}$ carries everything needed from the eliminated side. For $q$ possible states per random variable, this example uses two sums over $q^3$ configurations after reusing the message, rather than directly summing over $q^4$ configurations. The example's message is an exact finite sum, with no simulated data. Large maximal cliques still make exact inference expensive; chordal graphs provide a decomposition, not a universal guarantee of cheap inference.^[ref]

A Hamilton cycle, also called a Hamiltonian cycle, is a cycle in a graph containing every vertex of the graph. Here the convention is a finite simple graph whose edges are undirected, with $n=|V|\ge 3$. A traversal is a sequencewhere $v_0,\dots ,v_{n-1}$ are all the distinct vertices and $\{v_i,v_{i+1}\}\in E$ for $0\le i<n$. Thus "once" means once before the closing return: the start appears again at the end, and there are exactly $n$ traversed edges. The graph may contain additional unused edges.^[ref]

A Hamiltonian path visits all vertices once without requiring a return. Removing an edge from a Hamilton cycle gives a Hamiltonian path; a particular Hamiltonian path closes only if its endpoints are adjacent. An Euler circuit instead traverses every edge exactly once and can revisit vertices midway. The two requirements are independent, as these original labelled examples show.

For the complete graph $K_4$, use $V=\{A,B,C,D\}$ and $E=\{AB,AC,AD,BC,BD,CD\}$, where $AB$ abbreviates the unordered pair $\{A,B\}$. The route $A,B,C,D,A$ is a Hamilton cycle: it visits every vertex and uses four of the six edges, omitting $AC$ and $BD$. It is not an Euler circuit, and the graph admits no Euler circuit at all: every vertex degree is three, contradicting the Euler circuit criterion. The crossing of the drawn diagonals is not another vertex.

Conversely, take two triangles in a graph sharing only $X$, with $V=\{X,A,B,C,D\}$ and $E=\{XA,AB,BX,XC,CD,DX\}$. The route $X,A,B,X,C,D,X$ is an Euler circuit, using all six edges once; the vertex degrees are $4,2,2,2,2$. It cannot be a Hamilton cycle, since $X$ occurs midway as well as at the start and end. More strongly, no Hamilton cycle exists: $X$ is a cut vertex, and deleting it leaves the two separate edges $AB$ and $CD$. Nevertheless, $A,B,X,C,D$ is a Hamiltonian path, whose endpoints are not adjacent.

An induced cycle is a cycle in a graph whose vertices induce exactly its cycle edges, with no edge between nonconsecutive vertices. A triangle is an induced cycle of length three. A chordal graph forbids induced cycles of length at least four.

An odd cycle is a cycle in a graph with an odd number of edges.

A path-cycle matching is exactly an edge subset whose resulting subgraph has maximum degree at most two: every nontrivial connected component of a graph of this form is a graph path or a cycle in a graph. Thus the task is a bipartite b-matching problem with bound two at every vertex.

Find a bipartition $L,R$, orient every original edge from $L$ to $R$, and add a source $s$ and sink $t$. Give each arc $s\to u$, for $u\in L$, flow network edge capacity two; each original arc $u\to v$ flow network edge capacity one; and each arc $v\to t$, for $v\in R$, flow network edge capacity two. Compute an integral maximum flow and select the original edges whose flow is one. The integral max-flow theorem guarantees such an integral optimum.

By flow conservation, at most two selected edges touch each vertex, so they decompose into the required vertex-disjoint paths and cycles in a graph. Conversely, any permissible path-cycle matching gives a feasible flow: send one unit through its middle arc for each selected edge, with the outside arc amounts equal to the endpoint degrees. In both directions, the flow value equals the number of selected edges. Maximizing either quantity therefore maximizes the other.

The blue edges in this example form a four-cycle in a graph and a one-edge graph path; the dashed edge is unused. Their five units of flow are optimal: the two left vertices $a,b$ can each contribute at most two selected edges, and $c$ has only one incident edge.

The constructed flow network has $|V|+2$ vertices and $|E|+|V|$ arcs. A polynomial-time algorithm for maximum flow, such as the Edmonds–Karp algorithm, followed by a traversal of the selected subgraph, gives an efficient solution.

The first assertion is false. Give the root a first child that is a goal state of cost ten and another child leading to many cheaper non-goal search nodes. Depth-first search can take the first child and stop immediately. A-star search with the admissible heuristic $h=0$ behaves as uniform-cost search and expands the cheaper non-goal search nodes before selecting that goal. Thus an admissible heuristic does not make A-star search expand fewer nodes than every possible depth-first search ordering. The latter need not return an optimal solution.

The second assertion is true under the usual finite-branching factor assumption. Breadth-first search orders by search depth, not path cost. Every finite depth contains finitely many search nodes, so a goal at finite depth is eventually reached even if some steps cost zero. Zero-cost steps can affect cost optimality, but do not obstruct this completeness argument. An infinite branching factor would invalidate the argument irrespective of step costs.

The third assertion is true as a worst-case black-box requirement, with a finite optimum assumed. If outgoing actions are known only by expanding a state, an unexamined reachable branch can conceal a sufficiently negative-cost action followed by a route to a goal. Its total cost can beat the current solution by an arbitrarily large amount. Without additional lower bounds or structural information, finding one cheap solution cannot justify stopping; exhaustive examination can be necessary. This does not assert that every already-described irrelevant state must be visited on every input. Moreover, a reachable negative-cost cycle in a graph that can still reach a goal can make costs unbounded below, in which case no finite-cost optimal solution exists at all.

Cutset conditioning selects a cycle cutset $C$: deleting its vertices makes the binary constraint graph a forest. Enumerate assignments to $C$, reject those violating internal constraints, and filter each remaining constraint domain against the chosen cutset values. Solve the residual components by tree-structured constraint satisfaction. Stop at a successful assignment, or report failure after all cutset assignments fail. Completeness follows by considering the restriction of any full solution to $C$; soundness follows by checking internal, crossing and residual constraints. With maximum constraint domain size $q$ and $k$ cutset variables, a typical bound is $O(q^k(n-k)q^2)$ for the residual forest work, with cutset and crossing checks added explicitly.^[ref]

Here take $C=\{c,e\}$. The residual constraint graph has vertices $a,b,d$ and only edges $ab,ad$, a tree rooted at $a$. Choose the consistent cutset values $c=3,e=1$. Filtering against these assignments givesThe upward support pass from $b$ deletes $a=2$, because $b$ can only equal $2$. The pass from $d$ retains $a=1$, which has support $2$ or $4$. Assign $a=1$ at the root and then supported child values $b=2,d=2$. Together with $c=3,e=1$, this gives the solution in (c).

No singleton cycle cutset exists: deleting $a,b,d$ or $e$ leaves, respectively, a triangle on $bce,acd,abc$ or $abc$; deleting $c$ leaves the cycle in a graph $a-b-e-d-a$. Thus the two-element choice is also minimum, although the algorithm only requires a valid cutset.

Each vertex of a bipartite graph has a strict preference ordering of its neighbours; every neighbour is preferred to being unmatched. A stable matching in a bipartite graph is a matching in a graph with no blocking edge: an unused edge blocks if both endpoints prefer each other to their assigned partners, interpreting an unmatched endpoint as willing to accept any neighbour. Stability does not require a perfect matching in a graph.

For existence, run the Gale-Shapley algorithm. Whenever a vertex in the first part is unmatched and has an untried neighbour, it proposes to its most preferred untried neighbour. A receiver keeps its preferred proposal and rejects the other, including any previously held proposal. At most $e(G)$ proposals occur, so the procedure stops with a matching in a graph. Suppose an edge $bg$ blocked this final matching in a graph. Since $b$ prefers $g$ to its final partner, or is unmatched, $b$ must have proposed to $g$. But a receiver's held proposal only improves, so $g$'s final partner is preferred to $b$ if $b$ was rejected. If $b$ was not rejected, the edge is in the matching in a graph. Both alternatives contradict blocking. Every finite bipartite preference instance therefore has a stable matching.

For the mutual-worst edge in stable matchings, let $M,N$ be stable matchings with $bg\in M$ and $bg\notin N$. We give the argument using alternating graph components including possible unmatched endpoints. If $b$ were unmatched in $N$, then $g$ would have to prefer its $N$-partner to $b$, since otherwise $bg$ would block $N$. Following the $N$-edge from $g$, stability of $M$ forces its other endpoint to have and prefer an $M$-partner. Stability of $N$ then forces the next vertex to have and prefer an $N$-partner. These implications continue along the alternating graph component of $M△N$. They cannot stop, because at a missing next edge the preceding edge would block the appropriate matching in a graph. They also cannot close into a cycle in a graph, since the graph component already has endpoint $b$. This contradiction proves that $b$ is matched in $N$. The same reasoning applies to $g$.

Now $b$ prefers its $N$-partner to $g$, because $g$ is its worst choice. Along the alternating graph component, stability of $M$ forces that partner to prefer its $M$-partner; stability of $N$ forces the next vertex in $b$'s part to prefer its $N$-partner, and so on. A graph path graph component is impossible by the same endpoint argument, so this is a cycle in a graph. Thus throughout this cycle in a graph the part containing $b$ prefers $N$ and the part containing $g$ prefers $M$. In particular $g$ must prefer $b$ to its $N$-partner. That contradicts $b$ being $g$'s worst choice. Hence an edge ranked last by both endpoints, if present in one stable matching, is present in every stable matching.

For the list edge coloring assertion, let $\Delta$ be the maximum degree. We first prove that the chromatic index is $\Delta$. Use induction on the number of edges, with a proper $\Delta$-edge coloring of $G-uv$. Some color $a$ is missing at $u$ and some color $b$ is missing at $v$. If $a=b$, color $uv$ with it. Otherwise the edges of colors $a,b$ form alternating graph paths and even cycles in a graph. The graph component containing $u$ cannot contain $v$: an alternating graph path from $u$, starting with color $b$, would end at $v$ with color $a$, have even length, and place $u,v$ in the same bipartition class. Swap $a,b$ in the graph component containing $u$. Now $b$ is missing at both ends of $uv$, so it can be colored $b$. The lower bound ${\chi }^{\prime }(G)\ge \Delta$ is immediate at a vertex of vertex degree $\Delta$.

Fix such a proper coloring $c:E(G)\to \{1,\dots ,\Delta \}$. Orient the line graph toward the smaller color at vertices in the first part of $G$, and toward the larger color at vertices in the second part. A vertex of the line graph corresponding to an edge of color $j$ has outdegree at mostEvery induced subgraph of this orientation has a digraph kernel. Indeed, restrict to its edge set in the original bipartite graph, give first-part vertices a preference for smaller colors and second-part vertices a preference for larger colors, and choose a stable matching. Its edges form an independent set in the line graph. Every edge outside the matching in a graph has an endpoint whose partner is preferred to it, by stability; thus it has an outgoing edge to the independent set. This proves that the orientation is a kernel-perfect directed graph.

Here is the kernel list-coloring lemma with its proof. If a kernel-perfect directed graph $D$ has lists satisfying $|L(v)|\ge d_D^{+}(v)+1$, choose any color $a$ appearing in a list and a digraph kernel $K$ of the induced subgraph on vertices whose lists contain $a$. Color $K$ with $a$, delete $K$, and delete $a$ from the remaining lists. Each remaining vertex that loses a color had an outgoing neighbour in $K$, so its outdegree decreases by at least one; the list inequality persists. Vertices not losing a color also retain the inequality. Induction on the number of vertices completes a proper list coloring.

Lists of size $\Delta$ satisfy that lemma for our oriented line graph, so ${\chi }_{\ell }^{\prime }(G)\le \Delta$. Identical lists show ${\chi }_{\ell }^{\prime }(G)\ge {\chi }^{\prime }(G)$. Consequently the Galvin theorem for bipartite list edge coloring givesThe empty graph is immediate; if parallel edges are allowed, orient their line-graph adjacency in both directions when the two endpoints prescribe opposite directions. The outdegree bound still holds, and the same stable matching and digraph kernel proof applies with preferences on incident edges.

A linked graph is $k$-linked if it has at least $2k$ vertices and, for every choice of distinct terminals $s_1,\dots ,s_k,t_1,\dots ,t_k$, it has pairwise vertex-disjoint graph paths joining $s_i$ to $t_i$ for all $i$. The pairing is prescribed, which is stronger than the unpaired conclusion of the Menger theorem.

We prove the rooted dense minor linkage step carefully. We first prove the rooted branch-set reduction. Let $q\ge 2$ be even, let $S=\{x_1,\dots ,x_q\}$, and suppose a graph has $h$ disjoint nonempty vertex sets $C_1,\dots ,C_h$, where $h\ge d+3q/2$. Suppose each $G[C_i]$ is connected or every one of its graph components meets $S$, and each $C_i$ is adjacent to all but at most $d$ of the other $C_j$ not meeting $S$. Assume also that no rooted separation of a graph of order below $q$ avoids $d+1$ of these sets. Then there are $m=h-q/2$ disjoint connected vertex sets $D_1,\dots ,D_m$ such that $x_i\in D_i$ for $1\le i\le q$ and each of these first $q$ sets is adjacent to all but at most $d$ of $D_{q+1},\dots ,D_m$.

Here adjacency of sets means existence of an edge between them. A rooted separation of a graph is $(A,B)$ with $A\cup B=V(G)$, $S\subseteq A$, no edges between $A\setminus B$ and $B\setminus A$, and order $|A\cap B|$. It avoids $C_i$ when $A\cap C_i=\varnothing$. We prove the reduction by induction, first on the vertex count and then on the edge count; any counterexample chosen minimal in this ordering will be impossible.

We may delete edges joining two vertices of $S$. They cannot affect rooted separations of a graph, since $S$ lies wholly on the first side. Within a branch set, their deletion can only split a graph component into graph components each containing an endpoint in $S$, so the branch-set hypothesis also persists. An isolated vertex outside $S$ and all branch sets can be deleted. An isolated vertex $v\in S$ is also impossible, since $(S,V(G)\setminus \{v\})$ has order $q-1$ and avoids at least $h-q\ge d+q/2\ge d+1$ sets. Thus in a minimal counterexample $S$ is independent, and every isolated vertex outside $S$ belongs to a singleton branch set.

Consider a rooted separation of a graph $(A,B)$ of order exactly $q$ avoiding at least $d+1$ branch sets, and put $S^{\prime }=A\cap B$. Restrict to $G^{\prime }=G[B]-E(G[S^{\prime }])$ and $C_i^{\prime }=C_i\cap B$. Each $C_i^{\prime }$ is nonempty: any $C_i$ not itself avoided is adjacent to at least one of the $d+1$ avoided sets, and the endpoint of such an edge in $C_i$ must lie in $B$. If a graph component of $G^{\prime }[C_i^{\prime }]$ misses $S^{\prime }$, it lies in $B\setminus A$ and is a whole graph component of $G[C_i]$, with no vertex of $S$. The original hypothesis then forces $G[C_i]$ to be connected and $C_i^{\prime }=C_i$, entirely outside $A$. Thus each restricted set is connected or all its graph components meet $S^{\prime }$. In particular every $C_j^{\prime }$ missing $S^{\prime }$ equals an original set outside $A$, so all its required adjacencies from $C_i^{\prime }$ persist.

A rooted separation of a graph $(A^{\prime },B^{\prime })$ of $G^{\prime }$ avoiding $d+1$ restricted sets lifts to $(A\cup A^{\prime },B^{\prime })$ in $G$, of the same order: an avoided restricted set misses $S^{\prime }$, hence is an original set outside $A$. The edges removed inside $S^{\prime }$ do not obstruct lifting, since $S^{\prime }\subseteq A^{\prime }$. This proves the required separation hypothesis for $G^{\prime }$. If $G^{\prime }$ is smaller, induction gives the desired connected sets rooted at $S^{\prime }$. Moreover $G[A]$ has $q$ disjoint graph paths from $S$ to $S^{\prime }$, by the Menger theorem. Indeed a separator of smaller order between these sets would give a rooted separation of a graph of $G$ of smaller order still avoiding those $d+1$ original sets. Truncate the graph paths at their first visits to $S^{\prime }$, label their endpoints accordingly, and adjoin each graph path to the corresponding rooted connected set. They meet $B$ only in their distinct endpoints, so this gives the desired sets for $G$, a contradiction. Consequently in a minimal counterexample every such separation of order $q$ has $B=V(G)$ and $A=S$, an independent set.

Now contract any edge whose endpoints do not belong to two different branch sets. Its endpoints cannot both lie in $S$, so the image of $S$ still has size $q$. The branch-set conditions persist. If the contraction created a forbidden rooted separation of a graph of order below $q$, its lift would have order below $q$, except possibly when the contracted vertex lies in its separator. In that case the lifted order increases by one and is at most $q$. Equality would put the contracted edge inside $A$, contradicting the just-proved independence of $A$ for an order-$q$ separation avoiding $d+1$ sets. Thus the separation hypothesis also persists. Applying induction in the contracted graph and expanding the contracted vertex would produce the required disjoint connected sets. Hence every edge in a minimal counterexample joins two different branch sets.

Every vertex outside the branch sets would be isolated, since every edge joins two branch sets; such an isolated vertex could be deleted. Thus the branch sets cover the graph. Each nonsingleton branch set is therefore an independent set, and all its vertices belong to $S$, because each of its singleton graph components must meet $S$. Let $C$ be the union of the nonsingleton branch sets, and write $c=|C|\le q$. There is a matching in a graph from $C$ into $V(G)\setminus S$. Otherwise the Hall marriage theorem gives $X\subseteq C$ with neighbourhood $Y\subseteq V(G)\setminus S$ satisfying $|Y|<|X|$. Since $S$ is independent, $(S\cup Y,V(G)\setminus X)$ is a rooted separation of a graph of order $q-|X|+|Y|<q$. It avoids every singleton branch set outside $S\cup Y$, of which there are at leastThe penultimate counting bound uses that the nonsingleton sets number at most $c/2$, so $|V(G)|-c\ge h-c/2$. This contradicts the separation hypothesis.

Use that matching in a graph to make a two-vertex connected set for each terminal in $C$; terminals outside $C$ remain singleton sets. There are at leastunused vertices, all singleton original branch sets. Choose $m-q$ of them for the remaining $D_i$. Every rooted $D_i$ contains a singleton original branch set: either its terminal, or its terminal's matched neighbour. That singleton misses at most $d$ of the selected nonterminal singleton sets. This constructs exactly the desired family, contradicting minimality and proving the branch-set reduction.

Return to the given $22k$-connected graph. By the allowed assumption it has a graph minor $H$ with $2\delta (H)\ge h+4k-1$, where $h=|H|$. Put $d=h-1-\delta (H)$. ThenTake a branch-set model of $H$ and the $2k$ prescribed terminals as $S$. The branch sets are connected and each misses at most $d$ of the others. The separation hypothesis holds with $q=2k$: a separation of order below $2k$ avoiding a branch set would separate a surviving terminal from that set, contrary even to $2k$-connectivity. Since $h\ge d+3k$, the reduction applies. It produces $m=h-k$ sets with $s_i\in D_i$, $t_i\in D_{k+i}$, each terminal set missing at most $d$ of the $m-2k$ nonterminal sets. For each pair $D_i,D_{k+i}$ at leastnonterminal sets are adjacent to both. Choose different such sets for the $k$ pairs successively. The connected graph property inside the three sets and the two joining edges give a graph path from $s_i$ to $t_i$; take a simple graph path within their union. Different graph paths use disjoint sets. We have provedIn fact the argument proves the stronger transfer statement with $2k$-connectivity and the stated dense graph minor.

For the negative assertion, it suffices to give a value of $k$ for which the proposed general implication fails. Take $k=2$ and the graph on cyclic indices $0,\dots ,7$ in which two indices are adjacent when their cyclic distance is one or two. This is the four-connected planar obstruction to two-linkage. Draw the even cycle in a graph $0,2,4,6$ as the outer square, the odd cycle in a graph $1,3,5,7$ as an inner square rotated by $45$ degrees of vertices, and join each odd vertex to the two neighbouring even vertices; the annulus is triangulated without crossings. The outer face has vertices $0,2,4,6$ in this order.

Deleting at most three vertices leaves the graph connected. In cyclic order, consecutive surviving vertices are joined unless separated by a run of at least two deleted vertices. Disconnection requires two distinct such runs, hence at least four deletions. Conversely deleting the four neighbours of a vertex isolates it, so the vertex connectivity is exactly four. Disjoint graph paths joining $0$ to $4$ and $2$ to $6$ cannot exist: both would lie in the closed disk complementary to the outer face, and their endpoints alternate on its boundary. A first such graph path is a crosscut of the disk; the Jordan curve theorem puts the other two endpoints on opposite sides, forcing the second graph path to meet it. ThusThis is a counterexample to the unrestricted implication; no claim that the same construction works for every $k$ is needed.

A topological graph is a CW complex with only zero-dimensional and one-dimensional cells: its vertices form the zero-skeleton, and each edge is a copy of an open interval attached by its two endpoints to vertices, possibly the same vertex. It has the weak topology of a CW complex. A subgraph is a CW subcomplex, obtained by choosing vertices and edges and retaining the endpoints of every chosen edge. Loops and multiple edges are allowed.

A spanning tree is a connected spanning subgraph containing no cycle in a graph. To construct one in any connected topological graph $\Gamma$, consider all acyclic spanning subgraphs, ordered by inclusion; the subgraph with every vertex and no edges is one. The union of a chain is still acyclic, since any cycle uses only finitely many edges and would already occur in a member of the chain. The Zorn lemma therefore gives a maximal such subgraph $\Delta$.

If $\Delta$ were disconnected, a finite edge path in $\Gamma$ between vertices of two of its components would contain an edge joining different components. Adding that edge creates no cycle, contradicting maximality. Thus $\Delta$ is a spanning tree. Fix a root vertex. Each point of $\Delta$ has a unique finite path to the root; give every edge length one and move the point a fraction $s$ of the way along that path at time $s\in [0,1]$. This defines a contraction of a topological space of $\Delta$. On each closed edge times $[0,1]$, the construction takes place in a finite tree and is continuous. Since the interval is a locally finite CW complex, the product with $\Delta$ has the corresponding weak topology of a CW complex, so these restrictions establish continuity globally. ConsequentlyThis argument includes infinite graphs and does not assume local finiteness of $\Gamma$.

If $p:\tilde{\Gamma }\to \Gamma$ is a covering map, take $p^{-1}({\Gamma }^0)$ as the vertices upstairs. Each component of the inverse image of an open edge is an open interval mapping homeomorphically to that edge, because an interval is simply connected. Its parametrization lifts along the closed edge to attach its endpoints to lifted vertices, by path lifting theorem. Small star neighborhoods at vertices are lifted homeomorphically by the covering map, and give exactly the neighborhoods of the resulting topological graph. Thus the cell structure has the topology of the original covering space. The total space may be disconnected, but every component is a topological graph.

The Nielsen–Schreier formula isFor a proof, realize $F_n$ as the fundamental group of the wedge sum of $n$ circles. Construct its connected covering graph with vertices the right cosets $Hg$, and an oriented edge labeled $x_i$ from $Hg$ to $Hg{x}_i$ for each free generator $x_i$. There is exactly one incoming and one outgoing edge of each label at every vertex, which establishes the local covering map property. Reading a word from $H$ ends at its coset, so the closed paths at $H$ give exactly the words in $H$. The induced map of fundamental groups is injective by the homotopy lifting property of a covering map, and its image is $H$.

For index $d$, this covering graph has $d$ vertices and $dn$ edges. A spanning tree in it has $d-1$ edges. The permitted rank formula for a finite connected graph now givesIn particular this proves both freeness and the index formula.

If $n\ge 2$ and $d>1$, then $1+d(n-1)>n$, so a proper finite-index subgroup cannot be isomorphic to $F_n$. For $n=1$, $F_1\cong \mathbb{Z}$, and $d\mathbb{Z}$ is a proper index-$d$ subgroup isomorphic to $\mathbb{Z}$ for every $d\ge 2$. Hence

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.

For $m\ge 1$, let $S_m$ be the set of vertices of $B(m)$ connected to $\partial B(m)$ by an open path in a graph using only edges of $B(m)$. Every infinite-cluster vertex of $B(m)$ belongs to $S_m$, by stopping an infinite open path in a graph at its first boundary visit. Hence the increasing eventhas probability at least $\theta (p)/2$ by part (a). Importantly, $E_m$ depends only on the edges with both endpoints in $B(m)$.

Open all $8m$ perimeter edges joining successive vertices of $\partial B(m)$. They form a cycle in a graph, and opening them joins every vertex of $S_m$ into one percolation cluster, containing the specified perimeter vertex $v_m=(m,0)$. The Harris-FKG inequality bounds the probability of this event together with $E_m$ below by $p^{8m}\theta (p)/2$.

Close all $8m+4$ edges in the edge boundary of $B(m)$. These edges are distinct from the internal edges already considered, so their states are independent of those events. The resulting percolation cluster of $v_m$ is finite, contained in $B(m)$, and has at least $|S_m|$ vertices. Therefore, by translating $v_m$ to the origin,For $n\ge 1$, choose $m=\lceil \sqrt{n/(2\theta (p))}\rceil$. Then the threshold is at least $n$, and $m\le (1+(2\theta (p){)}^{-1/2})\sqrt{n}$. Put $A=-\log p$ and $B=-\log (1-p)$; both are positive. The preceding lower bound is at least $e^{-c_1\sqrt{n}}$ with, for example,The construction modifies only order-$m$ boundary edges while trapping order-$m^2$ vertices, which explains the stretched exponential scale. Here $\mathbb{N}$ is interpreted as the positive integers; the proposed lower bound at $n=0$ would be false because $\theta (p)>0$.

Declare an edge of the planar dual graph open exactly when its crossed primal edge is closed. This dual bond percolation has parameter $q=1-p<1/2=p_c$, using the Harris-Kesten theorem. We use the standard exponential tail of subcritical cluster size: for some $K,a>0$, uniformly in the dual vertex $v$,This is decay of the number of vertices in the percolation cluster, not merely its radius.

On $\{|C|=n\}$, the geometric fact allowed in the paper supplies a simple dual cycle in a graph surrounding $C$, with length $\ell \ge \alpha \sqrt{n}$. Every crossed edge is in the edge boundary of $C$ and is therefore closed, so the dual cycle is open. Its $\ell$ distinct vertices lie in one dual percolation cluster of size at least $\ell$.

A dual cycle in a graph of length $\ell$ surrounding the origin has all its vertices within sup-norm distance $\ell +1$ of the origin: its coordinate spans are at most $\ell$, and the origin lies between each pair of extreme coordinates. There are at most $K_0(\ell +1{)}^2$ possible dual vertices there. The union bound over lengths and possible vertices yieldsfor some $K_1,b>0$, since a polynomial factor can be absorbed into a slower exponential decay.

To remove the constant prefactor for all $n\ge 1$, first choose $n_0$ so that the bound is at most $e^{-(b/2)\sqrt{n}}$ for $n\ge n_0$. For the remaining finitely many $n$, use ${\mathbb{P}}_p(|C|=n)\le 1-\theta (p)<1$ and chooseFor $1\le n<n_0$, the chosen $c_2$ ensures $e^{-c_2\sqrt{n}}\ge 1-\theta (p)$.

Suppose first that $G$ is bipartite. As one traverses any cycle in a graph, its vertices alternate between the two vertex classes. Returning to the initial class therefore requires an even number of edges, so $G$ contains no odd cycle.

Conversely, suppose that $G$ has no odd cycle. Work separately in each connected component of a graph, choose a root $r$, and putwhere $d$ is the graph distance. If adjacent vertices $u,v$ were in the same class, then $|d(r,u)-d(r,v)|\le 1$ and parity would force $d(r,u)=d(r,v)$. Choose a breadth-first spanning tree. The two tree paths from $r$ to $u$ and $v$ diverge at some last common vertex; their remaining segments have equal length, and adjoining the edge $uv$ gives an odd cycle. This contradiction shows that every edge joins $A$ to $B$, proving the odd-cycle characterization of bipartite graphs.

A path-cycle matching in a finite simple graph is a collection of mutually vertex-disjoint paths and cycles in a graph. Its used edges form a subgraph with every degree of a vertex at most two. Conversely, every finite connected component of a graph with maximum degree at most two is a graph path, a cycle in a graph, or an isolated vertex; following successive edges proves this classification. Discarding unused isolated vertices therefore converts any such subgraph into a path-cycle matching without changing its edge count. On a bipartite graph, maximizing the used edges is exactly bipartite b-matching with every $b_v=2$.

A pseudoforest is an undirected graph in which every connected component of a graph contains at most one cycle in a graph. Thus each component is a tree or a unicyclic component. A finite tree component has one fewer edge than vertices; a finite unicyclic component has equally many edges and vertices.

A finite nonempty connected graph whose edges are undirected admits a strong orientation if and only if it has no bridge in a graph. This statement holds for finite multigraphs as well as finite simple graphs; self-loops do not help connect distinct vertices. The single-vertex case is allowed, with length-zero directed paths.^[ref]

Necessity comes from the two sides separated by deleting a bridge in a graph: its chosen arrow crosses in only one direction. For sufficiency, the cited constructive proof uses depth-first search to find a cycle in a graph, orients it cyclically, contracts it, and repeats. Undoing the contractions replaces junctions by directed cycles, retaining two-way reachability. Contraction can produce multiple edges, so these must remain distinct. This is a construction outline; the input graph is finite and bridgeless, and remaining edges within contracted pieces can receive arbitrary directions.^[ref]