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Choose any vertex $v$ of a red-blue colored complete graph $K_6$. Its five incident edges go to the other five vertices. By the pigeonhole principle, at least three have the same color. Call their other endpoints $a,b,c$ and, after swapping color names if needed, suppose $va,vb,vc$ are red.

If any of $ab,ac,bc$ is red, it completes a red triangle in a graph with $v$. If none is red, all three are blue, so $a,b,c$ themselves form a blue triangle in a graph. Both possibilities force a monochromatic triangle in a graph. This proves the upper bound for the Ramsey number for triangles using only five incident edges and the three pairs among the selected neighbors; the colors of the other edges cannot defeat the argument.

Both panels show all 15 edges of a complete graph on six vertices. Faint colored edges are example choices outside the decisive argument; bold edges exhibit the forced triangle in a graph. The selected neighbor color could instead be blue, giving the same argument with the color names exchanged.

Choose $D=d^2$ and a fixed $d$-regular graph $H$ on $D^2$ vertices with absolute nonconstant eigenvalues at most $1/10$. Start with the loop-inclusive complete graph $G_0$ on $D$ vertices and iterate $G_{t+1}=G_t^2\mathrm{zigzag}H$. The degree stays $D$, vertex counts are $D^{2t+1}$, and ${\rho }_{t+1}\le {\rho }_t^2+21/100$. Since ${\rho }_0=0$, all ${\rho }_t\le 1/2$. The spectral cut bound gives multiplicity-counted edge expansion at least $D/4$. Removing loops and merging parallel edges loses at most a factor $D$, producing simple expander graphs with maximum degree $D$ and edge expansion at least $1/4$.

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.

A tournament is an orientation of a complete graph: between each pair of distinct vertices there is exactly one directed edge.

A triangular number counts rows of $1,2,\dots ,n$ objects, with $n$ a nonnegative integer:The first values are $0,1,3,6,10,15,21,28,36,45,\dots$. The empty arrangement contributes zero, by the empty sum convention. Adding one more row gives the triangular number recurrence $T_n-T_{n-1}=n$ for $n\ge 1$. The row lengths form an arithmetic progression, while their running totals are triangular numbers.

Pair the rows with a reversed copy: $1+n=2+(n-1)=\cdots =n+1$. There are $n$ such pairs, so two copies contain $n(n+1)$ dots. Dividing by two proves the formula. This rectangle picture explains the factor one-half without requiring a formula for a long summation. For example, one hundred rows contain $T_{100}=5050$ dots.

The same count appears whenever every pair of different objects must be considered once. Five participants make ten possible handshakes; five sensors have ten different pairwise comparisons; a complete graph on five vertices has ten edges. These are the triangular number pair-counting identity, with one fewer row than objects. The distinction prevents a common error: $T_5=15$ counts all pairs from six objects, whereas five objects give $T_4=10$.

The triangular number square-complement identity says that two neighboring triangular numbers fill a square. Combining that identity with the one-row difference makes the difference of squares $T_n^2-T_{n-1}^2$ equal to $n^3$. It leads to Nicomachus's theorem: the sum of the first $n$ cubes of numbers is the square of a number $T_n^2$. A change in the way one groups objects exposes a relation between summation, multiplication and geometry.

This geometric view is already visible in Nicomachus's Introduction to Arithmetic by Nicomachus (Hahn teacher edition): Book II, Chapter 8 describes growing triangular rows, Chapter 12 combines successive triangular counts into squares, and Chapter 20 groups successive odd integers into cubes of numbers.^[ref] The linked edition uses Martin Luther D'Ooge's translation, revised with modern teaching material by Nathanael Hahn. Our empty-row index $T_0=0$ is a modern convention; the historical passage starts from unity and calls three the first actual triangle.

For $N\ge 1$ distinct objects, the number of unordered pairs of different objects isOrdering the objects gives $N-1$ pairs with the first object, $N-2$ new pairs with the second, and so on down to one. Alternatively, $N(N-1)$ ordered pairs count every unordered pair twice. The binomial coefficient therefore counts the edges of a complete graph on $N$ vertices. This is also consistent with the handshaking lemma: each vertex has degree of a vertex $N-1$, and the degree sum counts every edge twice.

For $n\ge 2$, choose $n-2$ independent random variables, each with the uniform distribution on a finite set $\{1,\dots ,n\}$, and decode their values as a Prüfer sequence. Every sequence has probability $n^{-(n-2)}$, and the bijection gives each labelled tree the same probability. This produces a uniform spanning tree of the complete graph without repeatedly proposing sets of edges and rejecting those that fail to be trees. It supplies random inputs for testing implementations of breadth-first search and a tractable model of random connections. ^[ref]

This distribution is uniform over fixed-label trees, rather than over unlabelled trees. With four labels, the star graph shape has probability $4/16=1/4$ and the path graph shape has probability $12/16=3/4$. Also, for a fixed label $v$, the Prüfer degree formula givesHere the binomial distribution counts the independent code positions equal to $v$. The expected value agrees with the average vertex degree from the degree-sum formula. Individual labels have the same distribution by symmetry, though their vertex degrees are constrained together by their fixed sum. When $n=1$, simply return the single isolated vertex.

Start with the empty graph and reveal the edges of the complete graph in a uniformly random order. After $m$ steps its edge set is a uniform $m$-element subset. Independent uniform edge labels couple this process with all binomial random graphs: $G(n,p)=G_{n,M(p)}$, where $M(p)$ has a binomial distribution with parameters $\left(\genfrac{}{}{0ex}{}{n}{2}\right),p$. For a monotone graph property, concentration of $M(p)$ transfers a threshold estimate between edge probability and edge count.

A weighted Hamilton cycle is a Hamilton cycle in a weighted graph, with cost equal to the sum of its traversed edge weights. Here weights represent symmetric nonnegative costs, rather than conductances. Asking whether a Hamilton cycle exists is a feasibility question; asking for the cheapest one is an instance of the travelling salesman problem. A complete graph supplies every possible connection between distinct vertices, making feasibility automatic for $n\ge 3$, while choosing the cheapest visiting order remains a separate mathematical optimization problem.

Assign the complete graph $K_4$ the symmetric edge costs $w_{AB}=2$, $w_{BC}=3$, $w_{CD}=4$, $w_{AD}=5$, $w_{AC}=6$ and $w_{BD}=7$, in arbitrary units. Fix the start at $A$ and identify each Hamilton cycle with its reverse. The three possibilities areThese exhaust the possibilities: the permutations of the other three vertices give $3!=6$ orders, paired into three reversals. Hence the perimeter tour really is optimal for this toy instance. The diagram's geometric lengths are not costs, and no triangle inequality is assumed.

The usefulness of this model is to distinguish visiting required sites from covering required connections. If tasks are assigned to vertices and the route must return after one visit to each, a weighted Hamilton cycle captures the ordering requirement. If the job instead requires each edge once, the relevant object is an Euler circuit. The shortest path problem asks only for a cheapest route between specified endpoints. A 1954 primary research paper gave a concrete motivation for the travelling salesman problem: certify that a particular tour through 49 cities has minimum road-distance cost.^[ref]

These are abstract route models. If a cost stands for a whole graph path through an underlying network, travelling between consecutive task vertices can pass through intermediate network vertices more than once. One visit per task therefore does not imply one visit per physical junction. One-way connections require a directed graph; changing costs or extra timing constraints require a richer model than the fixed symmetric toy instance.