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For a finite connected graph with at least one vertex, embedded as a plane graph, let $n$ count vertices, $m$ count edges, and $f$ count faces of a plane graph, including the outer face. ThenThe hypotheses refer to an actual embedding; regions in a crossing-filled graph drawing do not supply its $f$.^[ref]

In the plane graph $K_4$ pictured above, three bounded faces of a plane graph and the outer face give $4-6+4=2$. A square cycle graph gives $4-4+2=2$, while the graph path $A-B-C$ gives $3-2+1=2$. Changing a crossing into a new vertex changes the graph: the square-and-diagonals drawing then has five vertices, eight edges, and five faces of a plane graph, giving $5-8+5=2$.

A proof outline deletes edges on graph cycles. Each deletion preserves the connected graph and merges two faces of a plane graph, leaving $n-m+f$ unchanged. Eventually a tree remains, with $m=n-1$ and $f=1$. Adding the point at infinity identifies the plane picture with a sphere; the value two agrees with its Euler characteristic and the cell-count formula for Euler characteristic.

A simple random walk on a cycle graph of $n\ge 3$ vertices has expected value of its cover time equal to $n(n-1)/2$, from any starting vertex. Lift the walk to the integers using the same $\pm 1$ increments. Its contiguous visited interval covers all residues precisely when it reaches length $n$. Sum the expected time to expand a random-walk range, $1+2+\cdots +(n-1)$, using linearity of expectation; no independence of expansion times is needed.

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 coloring assigns labels called colours to elements of a graph, subject to specified constraints. Here we focus on proper vertex colouring of a simple graph $G=(V,E)$: it is an undirected graph with a finite set of vertices, and a colouring with at most $k$ colours is a functionThe colour labels need not be visual colours: they can represent time slots or resources. Unused labels are allowed. Properness concerns graph adjacency, so moving vertices in a drawing does not change whether the graph coloring is valid.^[ref]

Each colour class $c^{-1}(i)$ is an independent set: no edge has both endpoints in it. Conversely, a set partition of $V$ into independent sets gives a proper graph coloring by assigning one colour to each part. Thus graph coloring groups mutually compatible objects, and its chromatic number asks for the fewest such groups.

For an illustrative scheduling model, let each vertex represent an activity of equal duration and each edge represent a pair that cannot occur in the same time slot. A proper graph coloring is then a schedule respecting those pairwise conflicts. If the conflicts form a cycle graph $C_4$ with edges $12,23,34,41$, activities $1,3$ can share one slot and $2,4$ another. This model assumes interchangeable slots and includes only the pairwise conflicts encoded by the graph.^[ref]

The word "colouring" also occurs in edge coloring, which requires different colours on edges sharing an endpoint. The chromatic number discussed here always refers to proper vertex colouring.

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]

The Grötzsch graph is the Mycielskian of the five-vertex cycle graph. Take original edges $v_i{v}_{i+1}$, with indices interpreted modulo five; add $u_i{v}_{i-1},u_i{v}_{i+1}$ and $w{u}_i$ for every $i$. There are eleven vertices and twenty edges: five original, ten between originals and Mycielski shadow vertices, and five from the Mycielski apex.

Give $v_1,\dots ,v_5$ colours $(1,2,1,2,3)$, give each $u_i$ the colour of $v_i$, and give $w$ colour 4. This is a proper vertex colouring. The Mycielskian chromatic increase proves that three colours are impossible, so the chromatic number is exactly four. The Mycielskian triangle preservation proves that this is a triangle-free graph. In particular its clique number is only two, even though its chromatic number is four.

The Mycielski apex has an independent set as its graph neighbourhood, yet all three old colours are needed there in any three-colour proper vertex colouring of the subgraph obtained by deleting it. Thus a locally easy group of mutually nonadjacent neighbours can carry globally incompatible colour requirements. Absence of triangles in a graph alone cannot bound the chromatic number.

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 set selects mutually compatible objects when an edge of a graph represents a conflict. For example, if vertices are activities and edges join activities that cannot occur together, an independent set is a collection that can share one time slot. Finding the largest such collection is measured by the independence number.^[ref]

In a finite simple graph $G=(V,E)$, an independent set is a subset $I\subseteq V$ with no edge joining two selected vertices:The empty set and every singleton set are independent sets. In the cycle graph $C_5$ drawn under vertex cover, $\{0,2\}$ is an independent set, but $\{0,4\}$ is not: the closing edge $40$ joins those two vertices. Choosing nonconsecutive labels is insufficient unless the wraparound edge is checked.

An independent set of maximum size has cardinality equal to the independence number. Being unable to add another vertex only makes an independent set inclusion-maximal. In the star graph $K_{1,3}$, the centre alone is inclusion-maximal, since it is adjacent to every leaf, but the three leaves form a larger independent set. This distinction mirrors the difference between an inclusion-minimal vertex cover and a minimum vertex cover.

Every independent set $I$ has a vertex cover $V\setminus I$ as its complement of a set; a largest independent set has a minimum vertex cover as its complement. In every proper vertex colouring, each colour class is an independent set, so graph coloring partitions the vertices into mutually compatible groups.

The Graph Laplacian converts local differences along edges into a linear map. Throughout this section, $G=(V,E)$ is a finite undirected graph with no self-loops or repeated edges, and each edge has weight one. In a fixed order of vertices, let $A$ be the adjacency matrix and $D$ the degree matrix. The combinatorial Graph Laplacian isThus the value at a vertex is compared with its neighbours. Every row sums to zero, so $L1=0$: a constant vector produces no differences. This definition and the associated Graph Laplacian quadratic form are developed in Spielman's authored survey.^[ref]

For the three-vertex path graph $A-B-C$, the vertex degrees are $1,2,1$, and in that orderDirect multiplication gives three eigenvectors and their eigenvalues:The constant eigenvector has eigenvalue zero; the other two describe progressively stronger neighbour disagreement. Dividing these eigenvectors by $\sqrt{3},\sqrt{2},\sqrt{6}$, respectively, gives an orthonormal basis. The four-vertex cycle graph is already worked out under circulant matrix, where Fourier diagonalization of a circulant matrix gives its eigenvalues.

Every formula here uses $L=D-A$ without degree normalization. For comparison, Chung's treatment uses the degree-scaled matrix $D^{-1/2}L{D}^{-1/2}$, with a convention for isolated vertices; its eigenvalues must not be substituted into the examples here.^[ref]

The Mycielski construction replaces a graph $G$ with a graph $\mu (G)$ having one shadow vertex for each original vertex and one further vertex adjacent to every shadow. It satisfiesand preserves triangle-freeness. Iterating it from the five-cycle gives triangle-free graphs of arbitrarily large chromatic number.

Here $G$ is a finite simple graph with at least one vertex; it need not have an edge. The Mycielskian separates two ways of needing many colours: a large clique forces distinct colours immediately, but a triangle-free graph can impose incompatible choices through its whole pattern of graph adjacency. The Mycielski construction increases that global constraint while preserving the absence of triangles in a graph.

Write the vertex set of $G$ as $\{v_1,\dots ,v_n\}$, with $n\ge 1$. Keep these vertices, add distinct Mycielski shadow vertices $u_1,\dots ,u_n$, and add a Mycielski apex $w$. The vertex set and complete list of edges areThere are no other edges: the Mycielski shadow vertices form an independent set, $w$ has no original vertex as a neighbour, and $u_i$ is not adjacent to $v_i$. Each Mycielski shadow vertex copies its original vertex's graph neighbourhood within the original vertex set, not its position in the drawing. This is the construction in the inspected research paper.^[ref]

For $n$ original vertices and $m$ original edges, direct counting gives $2n+1$ vertices and $3m+n$ edges. The original vertices induce a copy of $G$, so no original graph adjacency is lost. A drawing crossing is not an extra vertex.

Start with the complete graph $K_2$, whose only edge is $v_1{v}_2$, abbreviating $\{v_1,v_2\}$. Its Mycielskian has exactly the edges $v_1{v}_2,u_1{v}_2,u_2{v}_1,w{u}_1,w{u}_2$. Reading them as $v_1,v_2,u_1,w,u_2,v_1$ gives the cycle graph $C_5$. A proper vertex colouring gives $v_1,u_1$ colour 1, $v_2,u_2$ colour 2, and $w$ colour 3. Two colours cannot close an odd cycle, so its chromatic number is exactly three.

A Hamiltonian graph must be a connected graph and have minimum degree $\delta (G)\ge 2$: each vertex has two distinct neighbours on its Hamilton cycle. It also has no cut vertex. Indeed, deleting any vertex from the Hamilton cycle leaves a graph path through every remaining vertex, so the remaining graph is connected. These are quick ways to rule out a Hamilton cycle.^[ref]

Even these conditions together are not sufficient: the complete bipartite graph $K_{2,3}$ is connected, has minimum degree two and no cut vertex, but has no Hamilton cycle. Every graph cycle in a bipartite graph alternates between its two parts, using equally many vertices from each; it cannot use all two and all three.

In the other direction, Dirac's theorem gives a sufficient condition: a finite simple graph with $n\ge 3$ and $\delta (G)\ge n/2$ has a Hamilton cycle. Failure of this inequality proves nothing by itself: the cycle graph $C_5$ has a Hamilton cycle despite $\delta (C_5)=2<5/2$.^[ref]

For a nonempty $r$-uniform set family $\mathcal{F}$, write its cardinality in its unique greedy binomial representationThe Kruskal-Katona theorem gives the exact minimum size of its lower shadow:Equality is attained by the colexicographic initial segment of the same cardinality. The empty family has empty shadow. In colexicographic order, $A$ precedes $B$ exactly when the largest point of $A△B$ belongs to $B$. The greedy expansion exists by choosing each top argument as large as possible; Pascal's identity makes the next residual less than $\left(\genfrac{}{}{0ex}{}{a_j}{j-1}\right)$, forcing the next top argument below $a_j$. This also proves uniqueness.

Here is a proof by UV-compression of the theorem, including the step that controls the lower shadow. For disjoint nonempty sets $U,V$ of equal size, let $C_{U,V}$ replace a member $A$ containing $V$ and avoiding $U$ by $A^{\prime }=(A\setminus V)\cup U$ only when $A^{\prime }$ is absent. Other members stay. This UV-compression preserves cardinality. Allow only $\max U<\max V$, and at each step choose a changing compression with $|U|=|V|$ as small as possible. Every smaller allowed UV-compression then fixes the current family.

For such a chosen UV-compression, we proveFirst let $S$ be a new shadow member, not in $\partial \mathcal{F}$. It has the form $A^{\prime }\setminus \{x\}$ for a moved member $A$. If $x\in U$, choose $y\in V$ with $y\ne \max V$ when $|V|>1$. The smaller UV-compression with sets $U\setminus \{x\},V\setminus \{y\}$ fixes $\mathcal{F}$, and therefore its target $S\cup \{y\}$ already belongs to $\mathcal{F}$. This would put $S$ in the old shadow, a contradiction. For $|V|=1$ the same target is simply $A$, giving the same contradiction without a compression. Hence $x\notin U$. Now $A\setminus \{x\}$ is an old shadow member whose compression target is $S$; since $S$ was absent, it appears in the compressed shadow.

It remains to check that a surviving old shadow member is not removed on the right side. Such a removal could occur only for $S$ containing $V$, avoiding $U$, with $S^{\prime }=(S\setminus V)\cup U$ absent from the old shadow. A witness $B=S\cup \{x\}$ in the compressed family must be a retained old member, since newly moved members avoid $V$. If $x\notin U$, retention of the eligible member $B$ forces its target $S^{\prime }\cup \{x\}$ to be in the old family, contrary to the absence of $S^{\prime }$ from its shadow. If $x\in U$, the smaller fixed compression just used forces $S^{\prime }\cup \{y\}$ into the old family; for $|U|=1$ this is $B$ itself. Again there is a contradiction. This proves the containment, and hence the chosen compression cannot increase lower shadow cardinality.

Each change strictly decreases the integer potential ${\sum }_{A\in \mathcal{F}}{\sum }_{a\in A}{2}^a$. The procedure therefore terminates. Its terminal family is an initial segment of colexicographic order: otherwise take an absent earlier set $X$ and a present later set $Y$. The choice $U=X\setminus Y,V=Y\setminus X$ is an allowed changing compression. Thus UV-compression transforms $\mathcal{F}$ into the colexicographic initial segment of the same size without increasing its lower shadow.

Finally compute that shadow. The first block comprises all $r$-subsets of $[a_r]$ and contributes $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$ shadow members. The next block has maximum point $a_r+1$ and its remaining $(r-1)$ points form a smaller colexicographic initial segment. Shadow members omitting $a_r+1$ are already in the first block's shadow; those containing it are counted by the lower shadow of that smaller initial segment. Recursion through the binomial representation gives precisely the displayed sum. This completes the proof.

To deduce the Erdős-Ko-Rado theorem, let $1\le r\le n/2$ and let $\mathcal{F}$ be an intersecting family. Put $k=n-r$ and take the complement family $\mathcal{B}=\{[n]\setminus A:A\in \mathcal{F}\}$. Its rank-$r$ lower shadow, obtained by deleting $k-r$ points, is disjoint from $\mathcal{F}$: membership of $A$ in that shadow would give $A\cap C=\varnothing$ for some $C\in \mathcal{F}$. If $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, iterating the proved Kruskal-Katona theorem gives this shadow at least as many members as the corresponding colex shadow. The colex family contains every $k$-set of $[n-1]$ and a further $k$-set containing $n$. Its rank-$r$ shadow therefore contains every $r$-set of $[n-1]$ and at least one more $r$-set containing $n$. This applies also when $k=r$, with no shadow operation. The two disjoint families would consequently have total size greater thana contradiction. Thus $\boxed{|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)}$, attained by all $r$-sets containing a fixed point.

For nonisomorphic colex shadow minimizers of rank two, take $n=4$ and $\mathcal{F}=\{12,23,34,14\}$. Its shadow has four singletons, as does the initial four-set colex segment $\mathcal{I}=\{12,13,23,14\}$. The first family is the edge set of a cycle graph; its four point degrees are all two. The second has point degrees $3,2,2,1$, so no permutation makes them isomorphic.

For rank three, take $n=5$ andThe first lower shadow comprises the three pairs in $\{1,2,3\}$ and the six pairs joining that set to $\{4,5\}$. The second shadow comprises every pair on $[5]$ except $45$. Both therefore have size nine. Their point-degree multisets are respectively $\{4,4,4,3,3\}$ and $\{5,4,4,3,2\}$, so again the two families are not isomorphic. These examples can be embedded in larger ground sets by adding unused points.

The Brooks' theorem states that a finite connected simple graph with maximum vertex degree $\Delta$ has chromatic number at most $\Delta$, unless it is a complete graph or an odd cycle graph; those exceptions have chromatic number $\Delta +1$.

First, if the graph is not regular, choose a vertex $v$ of degree less than $\Delta$ and a spanning tree rooted at $v$. Order vertices with every child before its parent, and apply the greedy coloring algorithm. Every vertex except $v$ has an as-yet uncolored parent, so has at most $\Delta -1$ colored neighbors; $v$ also has fewer than $\Delta$ neighbors. Thus $\Delta$ colors suffice. This justifies reducing to the regular case.

Now let $G$ be regular, 3-connected and not complete. A shortest path between two nonadjacent vertices contains an induced two-edge path $a,v,b$ with $a,b$ nonadjacent. By 3-connectivity, $G-\{a,b\}$ is connected. Color $a,b$ with the same color. Choose a spanning tree of the remaining graph rooted at $v$ and color its vertices in child-before-parent order. Each vertex other than $v$ again has an uncolored parent and thus at most $\Delta -1$ already colored neighbors. At the last vertex $v$, the two neighbors $a,b$ share a color, so its $\Delta$ neighbors use at most $\Delta -1$ distinct colors. A color is therefore available at every step, proving $\boxed{\chi (G)\le \Delta }$ in the required 3-connected noncomplete case. A 3-connected graph has $\Delta \ge 3$, so the odd-cycle exception cannot arise here.