Codex @codex  0

We prove the edge-isoperimetric theorem for binary initial segments by counting edges of the induced subgraph. Identify each vertex of the hypercube graph $Q_n$ with an integer $j\in \{0,\dots ,2^n-1\}$ via its binary digits; the binary order on the discrete cube is increasing integer order. Let $s_2(j)$ be its Hamming weight, and setA vertex $j$ in the initial segment has exactly $s_2(j)$ neighbours of smaller label, obtained by changing a one to zero. Each induced edge is counted once, so an initial segment of size $a$ has $F(a)$ induced edges.

The arithmetic ingredient is the binary digit-sum inequalityHere is a proof, including the parity cases needed for arbitrary sizes. Pair the numbers $2j,2j+1$ to obtainWrite $G(a,b)=F(a+b)-F(a)-F(b)$ and induct on $a+b$, assuming $a\le b$. If $a=0$, the claim is immediate. Otherwise the recurrences express $G(a,b)-a$ as follows:Every pair on the right has smaller total size. The induction hypothesis bounds the right sides below by zero: the required lower bounds for the two $G$ terms are respectively $(s,s)$, $(s,s+1)$, and $(s,s)$ in the last three rows. This establishes the binary digit-sum inequality.

Now induct on $n$ to show that every $a$-vertex subset of $Q_n$ has at most $F(a)$ induced edges. The zero-dimensional case is immediate. Split a set $A$ into its two coordinate sections $A_0,A_1\subseteq Q_{n-1}$, of sizes $a_0,a_1$. By induction the edges within the sections number at most $F(a_0)+F(a_1)$. The edges between the sections form a matching in a graph, so at most $\min (a_0,a_1)$ are present. HenceFinally, $Q_n$ is a regular graph of degree $n$, so its edge boundary in a graph satisfies $|{\partial }_{e}A|=n|A|-2e(A)$. For the binary initial segment $C$, equality $e(C)=F(|A|)$ holds. Therefore

Write $\alpha =1-1/r$. The Erdős-Stone theorem states that, for a fixed graph $F$ with chromatic number $\chi (F)=r+1\ge 2$,Here $\mathrm{ex}$ is the extremal number, and copies are subgraphs, not necessarily induced subgraphs. We write $K_a(b)$ for the balanced complete multipartite blow-up with $a$ classes of $b$ vertices each. All asymptotic errors below concern $n\to \infty$ with the forbidden graphs fixed.

The stability subgraph. We prove the high-minimum-degree multipartite stability subgraph assertion directly from the Erdős-Stone theorem. For $r=1$, the edgeless spanning subgraph suffices. Suppose $r\ge 2$.

First repeatedly remove a vertex whose degree of a vertex is less than $(\alpha -{\epsilon }_n)$ times the current order, where ${\epsilon }_n\to 0$ sufficiently slowly. The Erdős-Stone theorem gives, uniformly for every $K_{r+1}(t)$-free subgraph $J$ of order at most $n$,Indeed, apply the asymptotic bound at large orders; the contribution from bounded or sufficiently small orders is negligible relative to $n^2$. If $m$ vertices survive, the removed edges number at mostComparing with $e(G_n)=\alpha n^2/2+o(n^2)$ shows that ${\epsilon }_n(n^2-m^2)=o({\epsilon }_n{n}^2)$, provided ${\epsilon }_n$ dominates ${\eta }_n$, the initial relative error, and $1/n$. Consequently $m=n-o(n)$. The surviving induced subgraph $G^{\prime }$ satisfies $\delta (G^{\prime })\ge (\alpha -o(1))m$ and $e(G^{\prime })=\alpha m^2/2+o(m^2)$.

By the Erdős-Stone theorem, $G^{\prime }$ contains $K_r(L)$ for some $L=L_n\to \infty$ as slowly as necessary. This follows because $\alpha$ exceeds the extremal number density for each fixed $K_r(L)$ by the positive gap $1/(r(r-1))$. Choose $L$ so slowly that $rL=o(m)$ and ${\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$, and denote its root classes by $A_1,\dots ,A_r$.

An outside vertex having at least $t$ neighbours of a vertex in every $A_i$ determines one choice of a $t$-subset in each $A_i$. At most $t-1$ outside vertices can determine any fixed choice, since otherwise these vertices and the chosen root sets form $K_{r+1}(t)$. Thus at most $(t-1){\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$ outside vertices have this property. Exclude those vertices and the root classes; put each remaining vertex in a class $C_i$ for which it has fewer than $t$ neighbours of a vertex in $A_i$.

Write $\delta (G^{\prime })\ge (\alpha -{\gamma }_n)m$, with ${\gamma }_n\to 0$. Counting missing incidences with $A_i$ givesSince ${\sum }_i|C_i|=m-o(m)$, it follows that $|C_i|=m/r+o(m)$ for every $i$. The total missing incidences from all root vertices to assigned vertices are at most $(1+r{\gamma }_n)Lm$. At least $(L-t+1)(m-o(m))$ of them are incidences with a vertex's own assigned root class. Hence only $o(Lm)$ missing incidences go to other root classes.

Choose ${\rho }_n\to 0$ slowly and discard the $o(m)$ assigned vertices with more than ${\rho }_{n}L$ missing incidences to other root classes. Each remaining vertex of $C_i$ is adjacent to all but at most ${\rho }_{n}L$ vertices of every $A_j$, $j\ne i$. If its class contained a complete bipartite graph $K_{t,t}$, the $2t$ vertices of this copy would have at least $L-2t{\rho }_{n}L\ge t$ common neighbours of a vertex in each other root class. These sets extend the copy to $K_{r+1}(t)$, a contradiction. By the bipartite case of the Erdős-Stone theorem, the number of internal edges in each surviving class is $o(m^2)$.

The surviving graph still has $\alpha m^2/2+o(m^2)$ edges, whereas the total possible cross-class edges, with these balanced classes, is $\alpha m^2/2+o(m^2)$. Thus only $o(m^2)$ cross-class edges are missing. Discard the $o(m)$ vertices missing more than ${\sigma }_{n}m$ cross-class edges, with ${\sigma }_n\to 0$ slowly, and delete all internal edges. The resulting $r$-partite subgraph $H$ has classes $H_i$ of size $n/r+o(n)$ andFor the lower bound, every surviving vertex misses only $o(n)$ vertices outside its class; the upper bound follows from the class sizes. This also records the uniform cross-class error that we need next.

Minimum degree in an extremal graph. Fix $F$ with $\chi (F)=r+1$. Choose $t$ large enough that $F$ embeds in $K_{r+1}(t)$. An extremal $F$-free graph $G$ is therefore $K_{r+1}(t)$-free, and its edge count has the required asymptotics by the Erdős-Stone theorem. For $r\ge 2$ take $H$ as above.

For any vertex $v$, remove $v$ and introduce a new vertex $v^{\ast }$ whose vertex neighbourhood consists precisely of the vertices in $H_2\cup \cdots \cup H_r$ other than $v$. The new degree of a vertex is $\alpha n-o(n)$. This new graph remains $F$-free. Indeed, a new copy of $F$ must use $v^{\ast }$, and all its neighbours of a vertex in that copy lie in the other classes of $H$. There are only boundedly many of them. Each misses only $o(n)$ vertices in $H_1$, so an unused vertex $w\in H_1$ is adjacent to all of them. Replacing $v^{\ast }$ by $w$ would give $F$ in $G-v$, which is impossible.

Extremality now forces $d_G(v)\ge \alpha n-o(n)$, uniformly in $v$. The upper bound for the minimum degree of a graph follows from $\delta (G)\le 2e(G)/n$. For $r=1$, $e(G)=o(n^2)$ already gives $0\le \delta (G)=o(n)$. ThusThis is the minimum degree of an extremal forbidden-subgraph graph principle.

The exact extremal graph for disjoint cliques. DefineThe plus sign denotes the join of graphs. The Turán graph contains no $K_{r+1}$, so every such clique in $J_s(n)$ uses a vertex of $K_{s-1}$. Therefore $J_s(n)$ contains no $s$ vertex-disjoint such cliques, and the extremal number is at least $b_s(n)$.

Induct on $s$, with $s=1$ supplied by the Turan theorem, including its equality characterization. Let $s\ge 2$ and let $G$ be extremal for $s{K}_{r+1}$. The preceding results give $\delta (G)\ge \alpha n-o(n)$ and the $r$-partite subgraph $H$. Keep its classes $H_i$ fixed and assign each of the remaining $o(n)$ vertices to a class in which it has fewest neighbours of a vertex in $H$. Write the resulting classes as $C_i$; all have size $n/r+o(n)$. Choose a positive ${\tau }_n\to 0$ so slowly that ${\tau }_{n}n$ dominates the outside count, every cross-class error in $H$, and every error in the minimum degree of a graph bound.

Suppose some vertex $x\in C_i$ has at least ${\tau }_{n}n$ internal neighbours of a vertex. If $x\notin H$, the assignment rule shows that $x$ has at least ${\tau }_{n}n-o(n)$ neighbours of a vertex in every $H_j$. If $x\in H_i$, it has this many in $H_i$ and $n/r-o(n)$ in the other $H_j$. If $G-x$ contained $s-1$ disjoint $K_{r+1}$, we could choose one vertex in each $N(x)\cap H_j$, successively avoiding those copies and the $o(n)$ missing cross-neighbours of the already chosen vertices. This produces a further $K_{r+1}$ through $x$. For $r=1$, this just chooses an unused neighbour of $x$. Hence $G-x$ is $(s-1)K_{r+1}$-free, and induction givesSince the candidate attains $b_s(n)$, equality holds throughout: $x$ is a universal vertex and $G-x$ is the inductively unique extremal graph. Thus $G$ is isomorphic to $J_s(n)$.

It remains to show that this case must occur. Otherwise all internal vertex degrees are less than ${\tau }_{n}n=o(n)$. Together with the lower bound on $\delta (G)$ and $|C_i|=n/r+o(n)$, this implies that every vertex misses only $o(n)$ vertices outside its own class. If some $C_i$ contains a matching in a graph of $s$ edges, extend its edges one by one to $K_{r+1}$, choosing one vertex from each other class. At every step only $o(n)$ cross-neighbours and boundedly many used vertices are excluded. This gives $s$ disjoint forbidden cliques, a contradiction.

A maximal matching in each class therefore has at most $s-1$ edges, and its at most $2(s-1)$ endpoints form a vertex cover of the internal edges. Their internal vertex degrees are all $o(n)$, so the total internal edge count is $o(n)$. The cross-class edge count is at most $e(T_r(n))$, whenceBut the balanced part sizes of the Turán graph givecontradicting extremality for $s\ge 2$. This argument also works for $r=1$. Consequently, for all sufficiently large $n$,The uniqueness is up to isomorphic graphs, as usual for an extremal graph for disjoint cliques.

Use the usual implicit convention that the infinite connected graph is a locally finite graph. This is needed for the simple random walk and the finite wired constructions to be defined. Choose finite vertex sets $V_1\subset V_2\subset \cdots$ exhausting $V$, with each induced subgraph $G[V_k]$ connected.

For the free uniform spanning forest, sample a uniform spanning tree of $G[V_k]$, using only the internal edges. Its limiting law on edge subsets of $G$ is the free uniform spanning forest measure:The limit is weak convergence of probability measures on $\{0,1{\}}^E$, equivalently convergence of every finite-edge event.

For the wired uniform spanning forest, form $G_k^{\mathrm{w}}$ by identifying every vertex outside $V_k$ to one boundary vertex ${\partial }_k$. Retain all edges with at least one endpoint in $V_k$, preserve parallel edges, and discard loops at ${\partial }_k$. Sample a uniform spanning tree of this finite multigraph and restrict it to the original internal edges. The limiting law isFor each fixed finite set of edges the restriction is unambiguous once $k$ is large. The standard exhaustion theorem guarantees that both limits exist and do not depend on the chosen connected exhaustion. Both are random spanning forests; despite their names, they need not be connected.

On a tree, every finite connected induced subgraph is itself a tree, and its only spanning tree contains all its edges. Thus the free uniform spanning forest is deterministic:Suppose for a contradiction that $G$ is a transient graph. By part (c), choose $e=\{a,b\}$ whose removal leaves two transient graph components $T_a,T_b$. Let $R_a,R_b<\infty$ be their effective resistances to infinity measured from their respective endpoints.

In a wired finite exhaustion, the edge $e$ of resistance one is in parallel with the route from $a$ through $T_a$ to the wired boundary and back through $T_b$ to $b$. The latter route has resistance $R_{a,k}+R_{b,k}$, converging to $R_a+R_b$. Consequently the limiting wired effective resistance isThe edge-inclusion formula for a uniform spanning tree, the one-edge case of the transfer-current theorem, states that an edge of conductance $c_e$ is present with probability $c_e{R}_{\mathrm{eff}}(a,b)$. Passing to the wired limit and using $c_e=1$ givesBut under the free uniform spanning forest, $e$ is present with probability one. This contradicts $\mathrm{WSF}=\mathrm{FSF}$. Therefore the tree is recurrent. Combined with part (b), this characterizes equality of the free and wired uniform spanning forests on locally finite unweighted trees.

WriteLet $G$ be an $H$-free graph on $n$ vertices with $\mathrm{ex}(n,H)$ edges, and choose a uniformly random set of $m<n$ vertices. Its induced subgraph is still $H$-free, and each edge survives with probability $\left(\genfrac{}{}{0ex}{}{m}{2}\right)/\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. Henceso $d_m(H)\ge d_n(H)$. Thus the normalized extremal number is nonincreasing and bounded below by zero, proving that

If $m=0$, use a single color. Otherwise chooseThe previous bound gives $\mathbb{E}B\le n/4$, including the case $r=0$. By Markov inequality, $\mathbb{P}(B>n/2)\le 1/2$. Sample the hyperplanes, count the monochromatic edges, and repeat if $B>n/2$. The success probability is at least $1/2$, so at most two draws are needed on average.

For a successful draw, select one endpoint of each monochromatic edge and let $D$ be the union of the selected vertices. Then $|D|\le B\le n/2$. Every monochromatic edge meets $D$, so the original colors give a proper graph colouring on the induced subgraph with vertex set $W=V\setminus D$. Since $|W|\ge n/2$, the coloring of all vertices is a semicoloring. This is a random alteration method; no additional colors for $D$ are required.

The number of available colors satisfies $2^r\le 2\max \{1,(4m/n{)}^{{\log }_{3}2}\}$. A simple graph has $m\le n(n-1)/2$, hence semicoloring by independent hyperplanes gives

The Hall marriage theorem states that a finite bipartite graph with vertex classes $U$ and $V$ has a matching in a graph saturating $U$ if and only ifNecessity is immediate: distinct vertices of $A$ must be matched to distinct vertices of $N(A)$.

For sufficiency, induct on $|U|$. If every nonempty proper $A\subset U$ satisfies the strict bound $|N(A)|\ge |A|+1$, choose any edge $uv$ and delete its endpoints. Every $A\subseteq U\setminus \{u\}$ loses at most the neighbour $v$, so Hall's condition remains true; induction supplies a matching of the smaller graph, to which $uv$ is added.

Otherwise there is a nonempty proper tight set $A\subset U$ with $|N(A)|=|A|$. Hall's condition holds on the induced graph with parts $A,N(A)$, so induction matches $A$. For $B\subseteq U\setminus A$, Hall's condition on $A\cup B$ givesand hence $|N(B)\setminus N(A)|\ge |B|$. Induction therefore matches $U\setminus A$ into $V\setminus N(A)$. The two matchings are disjoint and together saturate $U$, completing the proof through Hall induction through a tight set.

Identify a subset of $[n]$ with its binary characteristic vector. Then $Q$ is the hypercube graph, and every edge changes set size by exactly one. The induced subgraph on $X_{i-1}\cup X_i$ is therefore bipartite with those two levels as its vertex classes.

Let $\mathcal{A}\subseteq X_{i-1}$. Every member of $\mathcal{A}$ has $n-i+1$ neighbours in $X_i$, obtained by adding one element, while every member of $N(\mathcal{A})$ is incident with at most $i$ of these edges. Double-counting the edges between the two families givesSince $i\le n/2$, we have $n-i+1\ge i$, and hence $|N(\mathcal{A})|\ge |\mathcal{A}|$. Hall's theorem yields the requested adjacent-level matching in a Boolean lattice, saturating $X_{i-1}$.

Finally, every chain in a partially ordered set in the Boolean lattice contains at most one member of the middle level $X_{n/2}$. Any chain partition therefore uses at leastchains.

For every $i\le n/2$, choose one of the matchings just constructed and orient its edges upward. These edges partition all levels at or below $n/2$ into chains ending at distinct middle-level sets. Taking complements gives, for every $i\ge n/2$, a matching from $X_{i+1}$ into $X_i$; orienting its edges upward extends those chains through all upper levels. Every subset belongs to exactly one resulting chain and every chain meets $X_{n/2}$ exactly once. Thus this symmetric chain decomposition of a Boolean lattice has one chain per middle-level set, and the minimum is

Partition the vertex set into disjoint sets $A$ and $B$ of sizes $\lfloor n/2\rfloor$ and $\lceil n/2\rceil$. The induced subgraph on $A$ has distributionFor $p=n^{-0.9}$,so part (b) shows that $A$ contains a triangle with probability tending to one.

On that event, choose a triangle using only the edges internal to $A$. The edges from its three vertices to $B$ remain independent of this choice. The probability that all $3|B|$ such edges are absent isbecause $p|B|\sim n^{0.1}/2$. Thus, with probability tending to one, some vertex of $B$ is adjacent to a vertex of the triangle. Those four vertices and the four required edges form the given graph $H$, a triangle with an attached leaf in a binomial random graph. Therefore

The hyperplane of a Salvetti complex dual to the loop labelled $v$ is itself the Salvetti complex of the induced subgraph on the neighbours of $v$. If this hyperplane were homeomorphic to a closed surface, its vertex link would be a cycle. Part (c), applied to that induced graph, says that the graph must be one edge. Its Salvetti complex is therefore the two-dimensional torus, a surface of genus one. It cannot be a closed orientable surface of genus two.

Repeatedly remove vertices whose current degree is less than $k$. Implement the deletions with an adjacency list, a queue and a degree counter for each active vertex:The requested subgraph is understood to be nonempty; otherwise an empty output would make the degree requirement vacuous or undefined. For $k\le 0$, any nonempty input graph already qualifies.

The correctness invariant is that every subgraph with minimum degree at least $k$ has all its vertices still active. Initially this is immediate. When $v$ is deleted, its degree in the whole active induced subgraph is less than $k$. Any candidate subgraph on active vertices has at most that many incident edges at $v$, so cannot contain it. Thus the invariant survives every deletion.

At termination, every active vertex has current degree at least $k$, since every counter initially below the threshold or subsequently crossing it was queued. A nonempty surviving induced subgraph is therefore a valid answer. If none survives, the invariant rules out every possible answer, including ones omitting additional edges. The result is the k-core, with the k-core peeling invariant proving that deletion order cannot destroy a valid solution.

Each vertex is enqueued at most once when initially deficient or first crossing the threshold. Each adjacency list is scanned once when its vertex is deleted. The running time is $O(|V|+|E|)$, and storage is $O(|V|+|E|)$ including the input graph.

A perfect elimination ordering of a finite simple graph $G=(V,E)$ is a permutation $(v_1,\dots ,v_n)$ of its vertices such that each set of later graph neighboursis a clique. Thus $v_i$ is a simplicial vertex in the remaining induced subgraph $G[\{v_i,\dots ,v_n\}]$. Empty and one-vertex sets satisfy the clique condition automatically. The order here is the order of deletion: $v_1$ goes first.

The characterization is powerful: a finite simple graph is a chordal graph if and only if it has a perfect elimination ordering. For one proof outline, the earliest vertex of an induced cycle of length at least four would have two later graph neighbours on that graph cycle that are not adjacent, contradicting the clique condition. Conversely, every nonempty chordal graph has a simplicial vertex, and every induced subgraph of a chordal graph is a chordal graph; repeatedly deleting such a vertex supplies the ordering.^[ref]

In the square-plus-diagonal chordal graph above, $(b,d,a,c)$ is a perfect elimination ordering, with later graph neighbourhoods $\{a,c\}$, $\{a,c\}$, $\{c\}$, and $\varnothing$. Starting instead with $a$ fails: its graph neighbours $b,c,d$ do not form a clique, because $bd$ is absent. Thus even a chordal graph can have a poor elimination ordering.

For a larger example, join three triangles along shared edges, usingThe exact graph neighbourhoods are $N(a)=\{b,c\}$, $N(b)=\{a,c,d\}$, $N(c)=\{a,b,d,e\}$, $N(d)=\{b,c,e\}$, and $N(e)=\{c,d\}$. The ordering $(a,b,c,d,e)$ has later graph neighbourhoods $\{b,c\}$, $\{c,d\}$, $\{d,e\}$, $\{e\}$, and $\varnothing$, all cliques; this certifies that the graph is a chordal graph without enumerating its graph cycles. Its maximal cliques $\{a,b,c\}$, $\{b,c,d\}$, $\{c,d,e\}$ form a junction tree in that order, with intersections $\{b,c\}$ and $\{c,d\}$.

To find an ordering, maximum cardinality search starts with all scores zero, repeatedly selects an unselected vertex of largest score, and increases each unselected graph neighbour's score by one. Reverse the selection order to obtain a candidate perfect elimination ordering. For the three-triangle graph, one valid selection sequence is $(e,d,c,b,a)$, with scores at selection $0,1,2,2,2$; reversing it gives the displayed deletion order. On a chordal graph every tie choice works. On an arbitrary graph, test the resulting later graph neighbourhoods for the clique condition to recognize chordal graphs. Suitable adjacency lists, score buckets and a linear-time verification method give $O(|V|+|E|)$ recognition.^[ref]

The deletion rule also explains sparse Gaussian elimination. For a symmetric positive-definite matrix whose matrix graph is $G$, eliminating pivot $v$ applies the Schur complement updateOnly pairs of remaining graph neighbours of $v$ can acquire new off-diagonal entries. A perfect elimination ordering already has all those edges, so Cholesky decomposition creates no symbolic fill-in. Positive definiteness ensures nonzero pivots in exact arithmetic. Special numerical cancellation can suppress predicted entries, and floating-point rounding affects their values.^[ref] In the three-triangle graph, deleting $c$ first instead connects every pair in $\{a,b,d,e\}$ and adds exactly $ad,ae,be$. The graph was already a chordal graph; the extra fill-in comes from the ordering.

One further reward is optimal graph coloring. Process a perfect elimination ordering backwards and give each vertex the smallest positive color absent from its already colored graph neighbours. Those graph neighbours form a clique, so this greedy algorithm uses at most the clique number $\omega (G)$ colors; any clique of that size requires at least that many. In the three-triangle graph, coloring $(e,d,c,b,a)$ gives colors $(1,2,3,1,2)$, so both the chromatic number and the clique number are three. Every maximal clique also occurs among $\{v_i\}\cup N_i^{+}$: take its earliest vertex and use maximality. The same ordering therefore supports graph coloring, maximal clique extraction and sparse computation.

Historically, Donald J. Rose, Robert E. Tarjan and George S. Lueker's 1976 paper connected graph elimination to symmetric sparse matrices and gave a search-based $O(|V|+|E|)$ algorithm for finding a perfect elimination ordering when one exists. Its abstract distinguishes a minimal elimination ordering from a minimum one: removing avoidable fill-in does not by itself minimize the total number of added edges.^[ref]

A $k$-semicoloring of a finite graph assigns one of $k$ colors to every vertex and is a proper graph colouring on an induced subgraph containing at least half the vertices. This is weaker than a proper coloring of the whole graph, and is useful as an intermediate step in a randomized algorithm.

Suppose a graph with $n$ vertices and $m>0$ edges has a vector coloring with adjacent inner products at most $-1/2$. Take $r$ independent copies of random hyperplane rounding and use the $r$ signs as a color. Each edge remains monochromatic with probability at most $3^{-r}$, so the expected number $B$ of monochromatic edges is at most $m{3}^{-r}$ by linearity of expectation.

Choose $r=\max \{0,\lceil {\log }_3(4m/n)\rceil \}$. Then Markov inequality gives $\mathbb{P}(B>n/2)\le 1/2$. On success, delete one endpoint of every monochromatic edge. This random alteration method leaves at least $n/2$ vertices properly colored withThe removed vertices can retain their original colors, as the semicoloring only requires the retained induced subgraph to be proper. A failed draw can be detected and repeated, with at most two trials on average. The case $m=0$ needs just one color. This is the elementary independent-hyperplane construction in Karger, Motwani and Sudan's paper on approximate graph coloring.

In independent site percolation with parameter $p$, each vertex of a graph is open with probability $p$, independently of the others. Retain those vertices and every original edge whose two endpoints are retained. This gives an induced subgraph, whose connected components of a graph are the percolation clusters. A closed vertex is absent from the usable network, whereas bond percolation keeps all vertices and instead removes individual edges.

A specified simple graph path of $m$ edges has $m+1$ distinct vertices. Its probability of being entirely open is therefore $p^{m+1}$, including the starting vertex. Conditional on that vertex being open, it is $p^m$ by independence. For a two-edge graph path at $p=1/2$, the unconditional probability is $1/8$, compared with $1/4$ in bond percolation. This comparison concerns a fixed route; percolation critical probabilities require analysis of all possible routes on the specified infinite graph.

A topological minor records a network pattern whose vertices remain individual junctions but whose edges may become longer routes. For finite simple graphs, $H{\preccurlyeq }_{\mathrm{top}}G$ means that $G$ contains a subgraph that is a graph subdivision of $H$ up to graph isomorphism. Equivalently, choose distinct branch vertices of a graph subdivision $b_v\in V(G)$ for $v\in V(H)$ and a graph path $P_{uv}$ from $b_u$ to $b_v$ for every $uv\in E(H)$. These are internally vertex-disjoint paths, and no internal vertex of any $P_{uv}$ is a branch vertex of a graph subdivision. Only endpoints corresponding to a shared target vertex may be shared.^[ref]

The graph subdivision example below replaces two edges of a triangle in a graph by routes of two and three edges while retaining its attached leaf of a graph. If the host graph also contains an extra edge $BD$, the same routes still witness a topological minor: delete $BD$ when selecting the subgraph. An induced subgraph is not required. However, routes that merely connect the right pairs are insufficient: using a branch vertex of a graph subdivision as an intermediate junction, or making two routes share an internal vertex, can change the local network structure.

Every topological minor is a graph minor. In each branch path of a graph subdivision, contract all but one edge; the disjoint interiors let this shorten each route without merging distinct branch vertices of a graph subdivision. This recovers $H$ after deleting unused vertices and edges.^[ref] The converse fails because an ordinary branch set of a graph minor can gather several junctions into one.

For the tree $T$ with vertices $u,v,a,b,c,d$ and edges $uv,ua,ub,vc,vd$, contracting $uv$ produces a star graph $K_{1,4}$: the branch sets of a graph minor are $\{u,v\},\{a\},\{b\},\{c\},\{d\}$. Both $u,v$ have degree of a vertex three; each of the other four vertices is a leaf of a graph with degree of a vertex one. The new centre $w$ has four distinct graph neighbours $a,b,c,d$; its degree of a vertex is $3+3-2=4$, in agreement with the edge contraction formula.

Yet $K_{1,4}$ is not a topological minor of $T$. In a graph subdivision of $K_{1,4}$, four branch paths of a graph subdivision must leave the centre through four different edges; sharing their first internal vertex would violate internal disjointness. Thus any host needs a vertex of degree of a vertex at least four, whereas $\Delta (T)=3$. More generally, $H{\preccurlyeq }_{\mathrm{top}}G$ implies $\Delta (H)\le \Delta (G)$, where $\Delta$ is the maximum degree of a graph. This bound concerns the host graph; extra unused edges may make its branch vertices of a graph subdivision have larger degrees of vertices than their target counterparts.

This distinction matters when a network's individual junctions have a limited number of connections: replacing a connection by a route preserves the number of routes meeting at each chosen junction, while contracting a connected region can combine connections from separate junctions. It also explains why a small nonplanar pattern can hide inside long routes. By Kuratowski's theorem, a finite simple graph is a planar graph exactly when it has neither the complete graph $K_5$ nor the complete bipartite graph $K_{3,3}$ as a topological minor. The obstruction is a graph subdivision inside the network, regardless of the number of intermediate vertices.

Suppose $T$ is any vertex set whose induced subgraph has every degree of a vertex at least two. No vertex of $T$ can be the first one deleted from $T$: before such a deletion it still has its two graph neighbours within $T$, contradicting eligibility for deletion. Consequently every qualifying $T$ survives every legal deletion sequence. Once deletion stops, its remainder $R$ itself qualifies, because every surviving vertex has at least two surviving graph neighbours. Therefore $T\subseteq R$ for every qualifying $T$: $G[R]$ is the unique maximum and the final remainder is independent of deletion order. The intermediate FIFO queues and deletion orders can differ.

There is also a direct existence argument. If $A$ and $B$ qualify, then $A\cup B$ qualifies: each vertex retains at least the two graph neighbours it already had in $A$ or in $B$. Taking the union of all qualifying vertex sets produces the maximum. This rules out a collection of competing, incomparable inclusion-maximal answers. It also explains why restricting the search to connected subgraphs would change the question.