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The 2-core is what remains of a finite simple graph after repeatedly removing every vertex with fewer than two surviving graph neighbours. It strips away dangling trees while keeping the part that can support itself under this local rule. Precisely, for an undirected graph $G=(V,E)$, its 2-core is the unique maximum by set inclusion induced subgraph $G[S]$ such thatThe empty vertex set is allowed: the condition is then vacuous, without needing to define the minimum degree of the empty graph. "Maximum by inclusion" means that $S$ contains every other vertex set satisfying the condition. It is stronger than merely saying that one chosen set cannot be enlarged: the 2-core simultaneously contains all possible qualifying induced subgraphs. In this particular family, every inclusion-maximal qualifying set does equal that unique maximum; the union argument in why the 2-core is unique explains why. ^[ref]

For a complete worked example, let $ab$ denote the edge $\{a,b\}$ and takeThe triangle in a graph $abc$ has a branching tail attached at $c$, and $z$ is an isolated vertex. In alphabetical order the initial degrees of vertices are $(2,2,3,3,2,1,1,0)$. Deleting $f$ turns $e$ into a leaf of a graph; deleting $g$ leaves $d$ with degree of a vertex two; deleting $z$ changes no other degree of a vertex. Now delete $e$, which makes $d$ a leaf of a graph, and finally delete $d$. Exactly the three edges of $abc$ remain: $C_2(G)=G[\{a,b,c\}]$.

A forest has empty 2-core: every nonempty finite forest has an isolated vertex or a leaf of a graph, so deletion cannot stop at a nonempty remainder. In particular, an isolated edge, an isolated vertex, a path graph, and the empty graph all have empty 2-core. Every cycle graph is its own 2-core. Two disjoint triangles in a graph also form their own 2-core, so a 2-core need not be a connected graph; attaching a tree to either triangle changes neither surviving triangle.

A polynomial-time many-one reduction represents each variable by an adjacent literal pair and each clause by a triangle in a graph. Connect each occurrence vertex to its corresponding literal vertex. A vertex cover of size $n+2m$ must choose one literal vertex per variable and two vertices per clause, and its omitted clause vertex forces a true literal. Conversely a satisfying Boolean valuation supplies such a cover. This establishes NP-hardness of the vertex-cover decision problem from 3-SAT.

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.

The chromatic index $\chi \prime (G)$, also called the edge chromatic number, is the least number of labels in a proper edge colouring of a finite undirected graph without self-loops. Equivalently, it is the smallest number of matchings in a graph whose set partition covers all the edges. We take $\chi \prime (G)=0$ when there are no edges. For an arbitrary edge colouring, one colour always suffices whenever an edge exists, so properness is essential to this definition. The chromatic number $\chi (G)$ instead colours vertices.^[ref]

To establish an exact chromatic index, supply both a proper edge colouring and a reason that fewer colours cannot work. Every edge at a fixed vertex must have its own colour, giving the lower bound $\chi \prime (G)\ge \Delta (G)$, where $\Delta (G)$ is the maximum degree of a graph. The bound need not be attained: it counts a single resource's workload, while conflicts can also accumulate around a graph cycle.

For the triangle in a graph $K_3$ with edges $ab,bc,ca$, every pair of edges shares a vertex, so all three require different labels. The assignment $c(ab)=1,c(bc)=2,c(ca)=3$ works, proving $\chi \prime (K_3)=3$ although $\Delta (K_3)=2$. In its line graph, those three edges become a three-vertex clique. Treating this as three pair-resource jobs makes the obstruction tangible: each resource has only two jobs, but every pair of jobs conflicts, so no slot can contain two jobs.

For the square cycle graph $C_4$ with edges $ab,bc,cd,da$, use $c(ab)=c(cd)=1$ and $c(bc)=c(da)=2$. Each colour class is a matching in a graph, and each vertex has one edge of each colour. Since $\Delta (C_4)=2$, this proves $\chi \prime (C_4)=2$. More generally, alternation closes on any even cycle graph $C_{2m}$, $m\ge 2$. On an odd cycle $C_{2m+1}$, $m\ge 1$, two-colour alternation would make the last and first edges share a colour at their common vertex; a third colour on the last edge suffices. Thus even cycle graphs have chromatic index two and odd cycle graphs have chromatic index three. These exact examples show why maximum degree of a graph alone does not determine the answer.

For every finite simple graph, Vizing's theorem givesHere $\Delta (G)$ is the maximum degree of a graph, taken as zero when there are no vertices. A bipartite graph has the stronger equality $\chi \prime (G)=\Delta (G)$. The six-job complete bipartite graph above attains this lower bound, while the triangle in a graph attains the extra-colour alternative. The simple-graph restriction in the upper bound matters.^[ref]

The bipartite graph equality still holds for finite loopless multigraphs, with parallel edges counted separately in each vertex degree. For instance, $r$ parallel jobs between two resources need exactly $r$ slots: every pair conflicts, and $\Delta =\chi \prime =r$. For a general finite loopless multigraph, however, the simple graph upper bound can fail. Put two parallel edges on each side of a triangle in a graph. All six edges pairwise share a vertex, so each needs its own colour; every vertex has degree of a vertex four. Consequently $\chi \prime =6>\Delta +1=5$. A multiplicity-aware bound is $\chi \prime (G)\le \Delta (G)+\mu (G)$, where $\mu (G)$ is the largest number of parallel edges between a pair of vertices. The doubled triangle in a graph has $\mu =2$ and attains that bound. These statements exclude self-loops; the two-distinct-resource scheduling model excludes them too.^[ref]

For historical orientation, Graph Theory by Reinhard Diestel (Colouring preview, §5.3, p. 132) dates the bipartite equality to 1916 and Vizing's theorem to 1964. These are textbook attributions; the original papers have not been inspected here. The results state existence of optimal or nearly optimal proper edge colourings, rather than saying that arbitrary greedy choices will find them. The list-chromatic index asks the further question of success for every sufficiently large family of allowed-colour lists.

A closed walk is a walk with equal endpoint vertices, $v_m=v_0$. Under our convention this includes the zero-length walk $(v)$; returning after traversing at least one edge requires $m>0$. Closure imposes no other restriction on repeated vertices or edges. In a directed graph, a closed directed walk must also follow every directed edge in its permitted direction.^[ref]

In the shared simple graph, $A,F,A$ is a length-two closed walk that immediately backtracks along $AF$. It is not a trail, since $AF$ is used twice, and it is not a graph cycle. The length-six closed walk $A,B,C,A,D,E,A$ is a trail, since its six edges are distinct, but is not a graph cycle: it returns to $A$ midway through the journey. The length-three closed walk $A,B,C,A$ is a graph cycle, with all vertices distinct apart from the required endpoint repetition. In a simple graph that is an undirected graph, a graph cycle has length at least three; the two-step backtrack is excluded.^[ref]

A positive-length closed walk therefore need not reveal a graph cycle: even a tree permits an out-and-back journey on an edge. A positive-length closed trail in a simple graph does contain a graph cycle. For a proof outline, choose a shortest positive-length consecutive portion of the trail whose endpoint vertices coincide. It cannot repeat an internal vertex, since that would give a shorter such portion; it cannot have length one or two because self-loops and repeated edges are excluded. It is consequently a graph cycle. Cutting it out leaves another closed trail, possibly of length zero, so repetition partitions the original edges into graph cycles. In our six-step example these are the two triangles in a graph. A closed trail using every edge would be an Euler circuit; here $AF$ is omitted.

Counting return journeys makes closed walks useful even when they repeat edges. For a finite simple graph with adjacency matrix $M$, the walk count from powers of an adjacency matrix gives $(M^k{)}_{uv}$ length-$k$ walks from $u$ to $v$, for every nonnegative integer $k$. A proof outline is that matrix multiplication appends one allowed step, summing over the previous endpoint; $M^0$ is the identity matrix, recording the unique zero-length walk at each vertex. ConsequentlyThe trace counts closed walks with a specified starting vertex and traversal order, not distinct graph cycles. These formulas use adjacency entries, not arbitrary travel costs in a weighted graph.^[ref]

In our example, $(M^2{)}_{AA}=5$: the five length-two closed walks are $A,B,A$, $A,C,A$, $A,D,A$, $A,E,A$, and $A,F,A$. Also $(M^3{)}_{AA}=4$, going round either triangle in a graph in either direction, and $\mathrm{tr}(M^3)=12$: each of the two triangles in a graph supplies three choices of starting vertex and two directions. These exact counts for the constructed graph illustrate why return counts and graph cycle counts differ.

A closed walk naturally models a delivery or inspection journey that returns to its depot. Repeated vertices let it combine several excursions; repeated edges let it serve a dead end and return. For an Euclidean travelling salesman tour, one can first construct a closed walk visiting all locations and then use the triangle inequality to shortcut repeated visits without increasing total length. That argument assumes direct travel between the retained locations is permitted, as in a complete graph weighted by Euclidean distance; an arbitrary road network need not contain those shortcut edges.

A complete graph is a simple graph in which every two distinct vertices are joined by an edge. The finite complete graph on $n$ vertices is denoted $K_n$. Here the edges are undirected, and self-loops are excluded. The usual pictures below use $n\ge 1$: $K_1$ is a single vertex, $K_2$ a single edge, and $K_3$ a triangle in a graph.^[ref]

Completeness describes all the possible pairs, rather than a drawing style. Different layouts can represent the same graph through a graph isomorphism. Crossing lines away from the drawn vertices do not add new vertices or edges.

There are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=n(n-1)/2$ unordered pairs of distinct vertices, so this is the edge count of a complete graph. Each vertex has vertex degree $n-1$. The degree-sum formula gives the same count: $2|E|=n(n-1)$. These formulas make a complete graph the upper extreme for edge counts among simple graphs of a fixed size.

This is useful when modelling pairwise relationships. If every pair of objects needs its own direct connection, ten vertices require $45$ edges, while twenty require $190$. Adding one more vertex to $K_n$ introduces $n$ further edges. This counts a specified direct-link model; it does not require every communication system to implement every relationship with a separate physical connection.

A proper vertex coloring of $K_n$ must give every vertex a different color, so its chromatic number is $n$. For $n\ge 1$, its independence number is $1$: two distinct vertices already have an edge between them. Thus complete graphs provide immediate test cases for graph coloring and independent sets.

Consider the following original five-vertex simple graph, where $AB$ denotes the undirected edge $\{A,B\}$:Its seven edges give initial expected value $q(\varnothing )=7/2=3.5$. Use vertex order $A,B,C,D,E$ and choose zero whenever the two child conditional expectations tie.

The complete decision ledger is below. A prefix lists fixed labels in vertex order; all later labels remain hypothetically independent fair bits. Each candidate entry is the conditional expectation of the final cut of a graph, including the contribution of unresolved edges.

For example, after $A=0,B=1$, edge $AB$ is fixed crossing and the other six edges are unresolved, giving $1+6/2=4$. Setting $C=0$ fixes $BC$ crossing and $AC$ noncrossing, leaving four unresolved edges: $2+4/2=4$. Setting $C=1$ instead fixes $AC$ crossing and $BC$ noncrossing, giving the same conditional expectation.

The final labels are $(0,1,0,0,1)$, so the two sides are $\{A,C,D\}$ and $\{B,E\}$. The five crossing edges are $AB,BC,DE,EA,BD$; $AC,CD$ do not cross. This exceeds $m/2=3.5$, and hence also the rounded-up guarantee of four edges.

It is not a maximum cut. Labels $(0,1,1,0,1)$ cross all edges except $BC$, giving six. Seven are impossible because the triangle in a graph $ABC$ cannot have all three edges crossing a two-sided cut of a graph. An independent enumeration of all $2^5=32$ complete label assignments gives the following distribution, counting an assignment and its label-swapped counterpart separately:

The weighted sum is $112$, so the average is $112/32=3.5$, agreeing with linearity of expectation. Choosing $C=0$ in the earlier tie discards the remaining size-six completion while preserving the average of four. The method protects an expected value guarantee; it need not preserve the best possible completion.

A cut vertex, also called an articulation point, is a vertex $v$ of an undirected graph whose deletion increases the number of connected components of a graph. Write $k(G)$ for that count, including isolated vertices, and $G-v$ for the graph obtained by removing $v$ and every incident edge. Then the test is $k(G-v)>k(G)$, even when the original graph is disconnected. By contrast, deleting a bridge in a graph removes just one edge and keeps both endpoint vertices. These are different failure models. ^[ref]

First take two triangles in a graph, $ABC$ and $ADE$, sharing only the vertex $A$. This simple graph hasHere $XY$ abbreviates the unordered pair $\{X,Y\}$. Removing $A$ leaves the two connected components of a graph $\{B,C\}$ and $\{D,E\}$, so $k(G-A)=2>1=k(G)$. Removing any other vertex leaves a connected graph. Thus $A$ is the only cut vertex. Every edge lies on a triangle in a graph, so deleting any single edge still leaves a graph path between its endpoints along the other two edges. There are no bridges in a graph.

The generator computes the adjacency lists and checks every vertex deletion, giving the following complete table:

Now take two disjoint triangles in a graph, $ABC$ and $DEF$, joined by the edge $CD$:Deleting only $CD$ preserves all six vertices and leaves connected components of a graph $\{A,B,C\}$ and $\{D,E,F\}$: $CD$ is the only bridge in a graph. Deleting $C$ also removes $AC$ and $BC$, leaving connected components of a graph $\{A,B\}$ and $\{D,E,F\}$. Deleting $D$ instead leaves $\{A,B,C\}$ and $\{E,F\}$. Thus $C,D$ are exactly the cut vertices.

Again the adjacency lists and all vertex deletion counts are computed directly:

For a disconnected example, adjoin an isolated vertex $Z$ to the first graph. Now $k(G)=2$, deleting $A$ gives $k(G-A)=3$, and deleting $Z$ gives $k(G-Z)=1$. The original graph was already disconnected, but $A$ is still a cut vertex; $Z$ is not. The empty graph has count zero, so deleting the only vertex of a one-vertex graph does not create a cut vertex. Nor is either endpoint of a single isolated edge a cut vertex, although that edge is a bridge in a graph.

In a communications network, a cut vertex models a router whose failure separates surviving routers. The shared-$A$ example shows why having an alternative graph path around each individual edge does not protect against failure of the common router. An additional edge from $B$ to $D$ removes this particular vulnerability: after deleting $A$, $B,D$ still join the two sides. A bridge in a graph instead models a vulnerable link. This distinction tells us whether to add another cable between existing devices or a route avoiding an indispensable device; it describes connectivity, rather than traffic capacity or failure probabilities.

The beautiful structural consequence is that a graph splits into biconnected components joined at cut vertices. Within each nontrivial robust piece, a single vertex failure cannot separate the other vertices; between pieces, the shared cut vertices reveal the bottlenecks. Also, a Hamiltonian graph has no cut vertex: deleting any vertex from its Hamilton cycle leaves a Hamiltonian path through all remaining vertices.

Historically, Tarjan's June 1972 paper developed depth-first search as a tool for finding biconnected components in time linear in the numbers of vertices and edges. ^[ref] In his 1983 recollection, Tarjan credited Hopcroft with the original linear-time biconnected component algorithm during their collaboration beginning in summer 1970, and described the published version as a refinement. The historical lesson is how a carefully organized traversal turns many separate deletion experiments into one computation. ^[ref]

An edge colouring assigns a label called a colour to every edge of a graph: it is a function $c:E(G)\to \{1,\dots ,k\}$. Here this definition allows arbitrary assignments. Two edges sharing a vertex may receive the same colour. A proper edge colouring imposes the additional condition that such edges receive different colours. Some textbooks use "edge colouring" as shorthand for this proper version; check the convention before applying a theorem.^[ref]

The distinction matters in Ramsey theory. In a red-blue edge colouring of a complete graph, a triangle in a graph is monochromatic when its three edges have the same colour. These are arbitrary edge colourings: forbidding equal colours on edges sharing a vertex would forbid the very monochromatic triangles in a graph being sought. For example, assigning red to every edge of $K_3$ is a valid arbitrary edge colouring, although it is not a proper edge colouring. Colour labels can also stand for time slots: then proper edge colouring turns incompatible jobs into separate rounds.

An edge contraction merges the two endpoints of an existing edge. For a finite simple graph $G$ and $e=uv$, replace $u,v$ by a new vertex $w$. Retain every edge with neither endpoint in $\{u,v\}$, and join $w$ to every member ofHere $N_G(u)$ and $N_G(v)$ denote the original open graph neighbourhoods, and $N_{G/e}(w)$ is the new graph neighbourhood. Remove the self-loop arising from $uv$ and merge duplicate edges, so the result $G/e$ is again a simple graph. This is not deletion of $uv$, which keeps both original vertices.^[ref]

The left example contracts $uv$ in the cycle graph $C_4$ to give $\{wa,ab,bw\}$. Its endpoints have no common neighbours, so four edges become three. The right example contracts $uv$ in the triangle in a graph with edges $uv,ua,va$. Its endpoints have the single common neighbour $a$, so the two edges $ua,va$ become one edge $wa$: three edges become one.

For any such edge contraction, the exact counts areThe contracted edge accounts for one loss. Each common neighbour $z$ contributes exactly one further loss, since $uz,vz$ become a single edge $wz$. All other edges correspond one-to-one to surviving edges. In particular the new degree of a vertex isThese formulas depend on simplifying to a simple graph; a multigraph convention that retains duplicate edges has different counts. In the six-vertex branch set of a graph minor diagram, contracting $12$ loses one edge, contracting the resulting $34$ loses two because of its common neighbour $A$, and contracting $56$ loses one. Hence its seven edges become the three edges of $K_3$: $7-1-2-1=3$.

An edge contraction can increase a degree of a vertex even while decreasing both total counts. For example, contracting $uv$ in the tree with edges $uv,ua,ub,vc,vd$ merges two vertices of degree of a vertex three into one of degree of a vertex four. This is the mechanism behind the distinction between a graph minor and a topological minor.

Place five labeled vertices around a pentagon. In the complete graph $K_5$, color the five perimeter edges red and the five diagonals blue. Writing $ij$ for the unordered pair $\{i,j\}$, the red pairs are $12,23,34,45,15$, and the blue pairs are $13,14,24,25,35$. This assigns a color to every pair, including every diagonal.

The red subgraph is the cycle graph $1,2,3,4,5,1$; the blue subgraph is the cycle graph $1,3,5,2,4,1$. Neither contains a triangle in a graph. Consequently, none of the ten triples has all three edges red or all three blue, proving the lower bound for the Ramsey number for triangles. Crossings in the drawing are not additional vertices.

A graph minor compresses connected regions of a graph into single vertices while retaining selected connections between them. For finite simple graphs, $H\preccurlyeq G$ means that some sequence of deleting vertices, deleting edges, and performing edge contractions produces $H$ up to graph isomorphism. Deleting a vertex also deletes all its incident edges; deleting an edge leaves its endpoints. The edge contraction convention here removes resulting self-loops and duplicate edges.^[ref]

For an exact small example, let $G$ be the cycle graph $C_4$ with vertices $u,v,a,b$ and edges $uv,va,ab,bu$. Contract $uv$ to a new vertex $w$. The surviving edges are $wa,ab,bw$, giving the complete graph $K_3$. Thus $K_3\preccurlyeq C_4$, although $C_4$ has no triangle in a graph as a subgraph. The branch sets of a graph minor are $\{u,v\},\{a\},\{b\}$, with the original edges $va,ab,bu$ witnessing all three required adjacencies. The first panel of the edge contraction diagram shows this operation.

A subgraph uses existing vertices and edges; a topological minor permits a graph subdivision of the target as a subgraph, so target edges become graph paths with mutually disjoint interiors and no target vertex in an interior. An ordinary graph minor permits whole connected branch sets of a graph minor in place of target vertices. Every topological minor is a graph minor, by contracting the replacement graph paths.^[ref]

The converse fails. Take the tree with edges $uv,ua,ub,vc,vd$. Contracting $uv$ gives the star graph $K_{1,4}$. Yet the original tree has maximum degree three, whereas every graph subdivision of $K_{1,4}$ retains a centre of degree of a vertex four. Hence $K_{1,4}$ is a graph minor but not a topological minor of this tree. The difference is that a connected branch set of a graph minor may gather connections incident to several original vertices.

The counts of vertices and edges never increase under these operations. However, the $C_4$ example shows that being a bipartite graph is not preserved: its graph minor $K_3$ contains an odd cycle. Its chromatic number also increases from two to three. A connected graph stays connected under an edge contraction, because the image of each graph path still connects its endpoints, but deletion can disconnect it. Thus one must check the particular graph property before treating it as preserved by taking graph minors.

Being a planar graph is preserved: in a crossing-free drawing, shrink a contracted edge and bring its incident edges into the merged vertex; deletion merely removes parts of the drawing. This makes a nonplanar graph minor a certificate that the original graph is nonplanar. More strongly, for a finite simple graph, being a planar graph is equivalent to having neither the complete graph $K_5$ nor the complete bipartite graph $K_{3,3}$ as a graph minor. Finding the corresponding branch sets of a graph minor settles nonplanarity even when no literal $K_5$ or $K_{3,3}$ subgraph is present. The converse is a substantial theorem, rather than a consequence of preservation alone.^[ref]

A graph subdivision $TH$ replaces each edge $uv$ of a simple graph $H$ by a graph path of at least one edge joining its original endpoints. The original vertices are branch vertices of a graph subdivision. The replacement graph paths have pairwise disjoint interiors, contain no branch vertices of a graph subdivision internally, and together make up the entire new graph. Leaving an edge unchanged is allowed.^[ref]

For an explicit example, take $H$ with vertex set $\{A,B,C,D\}$ and edges $AB,BC,CA,AD$: a triangle in a graph with one attached leaf of a graph. Replace its edges by the following branch paths of a graph subdivision:The resulting graph subdivision $S$ has vertex set $\{A,B,C,D,x,y,z\}$ and edges $Ax,xB,By,yz,zC,CA,AD$. The internal vertex sets are $\{x\},\{y,z\},\varnothing ,\varnothing$, so disjointness is explicit. The degrees of vertices of $A,B,C,D$ remain $3,2,2,1$, respectively; the three new internal vertices each have degree of a vertex two.

Adding $m$ internal vertices to a finite graph increases both its vertex count and its edge count by $m$. Here $m=3$, so four vertices and four edges become seven of each; the graph excess is unchanged. Each replacement graph path remains a route between the same junctions. Thus graph subdivision preserves whether a graph is a planar graph: insert intermediate vertices along a crossing-free drawing, or trace each replacement graph path as one edge to recover a drawing of the original graph. Travel times or capacities are separate modelling choices and are not determined by this adjacency construction.

The independence number is the largest cardinality of an independent set in a finite simple graph:It answers "how many mutually compatible objects can be selected at once?", while the chromatic number asks how many compatible groups are needed for all the vertices. Diestel uses this notation for a largest independent set.^[ref]

For the cycle graph $C_5$, the independent set $\{0,2\}$ shows that $\alpha (C_5)\ge 2$. Three selected vertices would require at least three unselected vertices separating them around the cycle graph, which has only five vertices. Thus $\alpha (C_5)=2$. There are five largest independent sets, obtained by rotating $\{0,2\}$ around the cycle graph, and five corresponding minimum vertex covers.

For the complete bipartite graph $K_{2,3}$, an independent set cannot contain a vertex from each side, since every such pair is joined by an edge. The larger side has three vertices, so $\alpha (K_{2,3})=3$. A triangle in a graph has independence number one. A graph with no edges and $n$ vertices has independence number $n$, including $\alpha (G)=0$ when $n=0$.

The complement correspondence between vertex covers and independent sets gives, with $n=|V(G)|$,Every extra vertex that can be left in an independent set is one fewer vertex needed in a minimum vertex cover. This identity holds for all finite simple graphs, including the triangle in a graph and the cycle graph $C_5$; it does not require the bipartite graph hypothesis of Kőnig's theorem for bipartite matching.

For a nonempty graph, every colour class of a proper vertex colouring contains at most $\alpha (G)$ vertices. Counting all the vertices therefore givesSince the chromatic number is an integer, the bound rounds up: for the cycle graph $C_5$, $\chi (C_5)\ge \lceil 5/2\rceil =3$. It is attained by assigning colour $A$ to $0,2$, colour $B$ to $1,3$, and colour $C$ to $4$. All five edges, including $40$, have differently coloured endpoints.

An independent dominating set is simultaneously an independent set and a dominating set. Equivalently, it is an independent set maximal under inclusion: an outside vertex can be added while preserving independence exactly when it has no neighbour in the selected set. On a triangle in a graph, any singleton is an independent dominating set but no independent vertex cover exists. Independence and domination are directly expressible in first-order logic when a fixed bound on the number of selected vertices is given.

A pairwise intersecting family of two-element sets is contained in a star graph or in the three edges of a triangle in a graph. If two members are $\{a,b\},\{a,c\}$ and a third avoids $a$, the third must be $\{b,c\}$. Every member meeting all three belongs to that triangle.

The complete bipartite graph $K_{3,3}$ has vertex sets $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$, with every edge between $A$ and $B$ and none inside either part. Its six vertices support $3\cdot 3=9$ edges and no triangles in a graph: among any three vertices, two belong to the same independent set and have no joining edge.

Add the single edge $a_1{a}_2$. Each $b_j$ is a common neighbour of its endpoints, producing the triangle in a graph $a_1{a}_2{b}_j$. These are exactly the three new triangles in a graph, because every triangle in a graph must use the only edge inside a part. The new count is ten edges; nine was the last possible count without a triangle in a graph on six vertices.

In general, every simple graph with a finite set of $n\ge 0$ vertices that is triangle-free has at mostedges. This is the $r=2$ case of Turán's theorem, with forbidden clique $K_3$. Equality holds exactly for the balanced complete bipartite graph $K_{\lfloor n/2\rfloor ,\lceil n/2\rceil }$, up to graph isomorphism; empty parts are allowed for $n<2$.^[ref]

The floor function gives the exact integer bound. Splitting $n$ vertices into two independent sets of sizes $a$ and $n-a$, and putting in every cross edge, givesThe count is largest when $a$ is an integer nearest $n/2$. For $n=2k$ it is $k^2$; for $n=2k+1$ it is $k(k+1)$. Both equal $\lfloor n^2/4\rfloor$. This proves attainability for every $n$, including the empty graph.

A short proof of the upper bound uses degrees of vertices. In a triangle-free graph, the endpoints $u,v$ of an edge have disjoint vertex neighbourhoods: a common neighbour would complete a triangle in a graph. Thus $d(u)+d(v)\le n$. Summing over edges counts $d(v)$ once for each of the $d(v)$ edges incident to $v$, givingFor $n>0$, the Cauchy-Schwarz inequality and handshaking lemma yieldIf $m>0$, division gives $m\le n^2/4$; $m=0$ is immediate. Since $m$ is an integer, apply the floor function. The case $n=0$ is immediate too. This proves the bound; the full equality characterization follows from Turán's theorem.

The useful reversal is that more than $\lfloor n^2/4\rfloor$ edges force a triangle in a graph, whatever their placement. A global count certifies a local pattern. The extremizer is bipartite, but a triangle-free graph need not be bipartite: a five-vertex graph cycle has five edges, below the bound of six. Historically, Willem Mantel posed the triangle-exclusion question in 1907.^[ref]

A maximum cardinality matching is a matching in a graph with as many edges as any matching in a graph of the same graph. It has cardinality $\nu (G)$, the matching number, and need not be unique or be a perfect matching. For example, any single edge of a triangle in a graph is a maximum cardinality matching, but leaves one vertex unmatched. Berge's theorem characterizes these optima by the absence of an augmenting path in a matching.

A subset $A\subseteq X$ is monochromatic for a finite coloring $c:X\to C$ if there is a color $i\in C$ such thatEquivalently, the restriction of a function $c\upharpoonright A$ is a constant function, or $A$ lies in one colour class $c^{-1}(i)$. The word describes agreement of colors; it does not require $A$ to equal an entire colour class.

For example, color the integers by integer parity, using red for even and blue for odd. The set $\{2,4,8\}$ is monochromatic, whereas $\{2,3\}$ is not. If the colored objects are edges, a monochromatic triangle in a graph means that its three edges form a monochromatic set. In a coloring of $r$-element subsets of a set $S$, a homogeneous set for a colouring $H\subseteq S$ means that the collection of all $r$-element subsets of $H$ is monochromatic. For pairs, this is a complete graph on $H$ whose edges all have one color, rather than a coloring of the vertices in $H$.