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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]

Here a loopless multigraph has $v$ vertices, $e$ edges, and at most $\mu$ parallel edges per pair. In a minimum-crossing drawing, crossings between incident edges can be removed by exchanging their initial segments, so every crossing involves four distinct vertices. For each parallel class choose an edge with probability $1/\mu$ per edge, with the remaining probability choosing none. Retain vertices independently with probability $p$. The sampled simple graph has expected edge count $p^{2}e/\mu$, expected vertex count $pv$, and expected crossing count $p^4\mathrm{cr}(G)/{\mu }^2$. Deleting one edge per crossing and using the planar graph edge bound gives $p^4\mathrm{cr}(G)/{\mu }^2\ge p^{2}e/\mu -3pv$. Choose $p=4\mu v/e$.

The complete graph $K_5$ has $n=5$ and $m=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, exceeding the planar graph edge bound $3\cdot 5-6=9$. Therefore it is not a planar graph. The complete bipartite graph $K_{3,3}$ has $n=6$ and $m=3\cdot 3=9$. The general bound allows twelve edges, but the bipartite planar graph edge bound allows only eight, proving nonplanarity.

These are necessary bounds, not a complete planarity test. For an original counterexample, subdivide every edge of $K_{3,3}$ once. The resulting simple graph is a connected graph with $n=15$, $m=18$, and is bipartite: put all original vertices in one part and all nine new vertices in the other. Both inequalities hold, $18\le 39$ and $18\le 26$. Nevertheless, a crossing-free graph drawing would let us trace each two-edge graph path as one curve, producing an impossible embedding of $K_{3,3}$. The graph subdivision preserves this obstruction. Kuratowski's theorem supplies a stronger characterization than these counts.

Let $t$ be the maximum number of positive Euclidean distances from one of the $n$ points to the others. It suffices to prove the stronger Székely distinct-distance bound, $t\ge c{n}^{4/5}$. Here $n\ge 2$; enlarge constants to handle bounded $n$.

First prove the crossing lemma for multigraphs. In an optimal drawing of a loopless multigraph with $v$ vertices, $e$ edges and maximum parallel multiplicity $\mu$, crossings of incident edges can be removed by interchanging initial segments, so each remaining crossing involves four distinct vertices. In each parallel class choose one of its edges with probability $1/\mu$ per edge, selecting none with the remaining probability. Independently keep each vertex with probability $p$. The resulting simple graph has expected edge count $p^{2}e/\mu$, expected vertex count $pv$, and expected crossings $p^{4}X/{\mu }^2$, where $X$ is the original crossing count. Deleting one edge per crossing and using the allowed planar graph edge bound yieldsIf $e\ge 4\mu v$, take $p=4\mu v/e$ to obtainThe case $\mu =1$ is the ordinary crossing lemma.

We also need the rich line bound. Given $L$ lines, connect consecutive incidence points on each line by straight edges. This simple graph has at least $I-L$ edges, where $I$ is the total number of incidences, and at most $L^2$ crossings. The crossing lemma gives $I=O((nL{)}^{2/3}+n+L)$, proving the Szemerédi–Trotter theorem here rather than assuming it. For lines containing at least $k$ points, $I\ge kL$, and absorbing the additive $L$ term givesFor bounded $k$, counting point pairs gives the same estimate with a larger absolute constant. Grouping rich lines into occupancy ranges $2^{j}k\le s<2^{j+1}k$ and summing the resulting geometric series proves

Now form the circular-arc graph for distinct distances. Around every point draw each positive-distance circle centred there; there are at most $nt$ such circles. A circle with at least three of the original points contributes one edge between each consecutive pair around it. Discard the circles with at most two points. Since the total number of centre-to-point incidences is $n(n-1)$, the remaining loopless multigraph hasIts circular-arc drawing has $O(n^2{t}^2)$ crossings: two distinct circles have at most two intersections, and small perturbations remove tangencies and multiple crossings without changing this order.

The apparent difficulty is unbounded parallel multiplicity. The rich-bisector deletion bound controls it. Every centre supporting an arc between fixed endpoints lies on their perpendicular bisector. Distinct parallel arcs come from distinct centres, since their common endpoints determine the radius once the centre is fixed. Thus multiplicity at least $k$ makes that line contain at least $k$ of the points. For a line with $s$ centres, each of its at most $st$ circles has at most two consecutive-point arcs symmetric about that line, one at each end of its diameter along the line. The occupancy sum above therefore bounds the number of high-multiplicity arcs by

If $t\ge n^{4/5}$ there is nothing to prove. Otherwise $t\log n=o(n)$ and $t=o(n)$. Choose $k=\lceil K\sqrt{t}\rceil$ with a sufficiently large absolute constant $K$. The first deletion term is at most a small fixed fraction of $n^2$ and the second is $o(n^2)$. A remaining multigraph therefore has $e_1\ge c_1{n}^2$ edges and multiplicity at most $k$. For large $n$, $e_1\ge 4kn$, so the crossing lemma for multigraphs givesRearranging,The whole distinct-distance set contains those distances from the selected point, proving the requested bound. The circular-arc method and the stronger one-point conclusion are due to Székely.

Let $G$ have $n$ vertices and $e$ edges. Its complement graph $\overline{G}$ hasedges. If both graphs were planar, the planar graph edge bound would giveEquivalently, $n^2-13n+24\le 0$, whose larger root is $(13+\sqrt{73})/2<11$. Thus an integer $n\ge 11$ is impossible, proving that

Now suppose that $G$ and $\overline{G}$ are both bipartite. In a bipartition $V(G)=A\bigsqcup B$, every pair within $A$ and every pair within $B$ is adjacent in $\overline{G}$. Since a bipartite graph contains no triangle, $|A|,|B|\le 2$, and hence $|V(G)|\le 4$. Equality is attained by the path $P_4$, whose complement is again a copy of $P_4$. Therefore the maximum order is

For each crossing pair, delete one of its two edges. At most $t(G)$ distinct edges are deleted, and the graph left behind has a crossing-free drawing. The planar graph edge bound therefore givessoThis is the linear estimate underlying the crossing lemma.