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Suppose the only distinct eigenvalues are $d$ and $\theta$. As in part (c), $A$ acts by $d$ on $\mathrm{span}\{\mathbf{e}\}$ and by $\theta$ on ${\mathbf{e}}^{\perp }$, so its spectral decomposition isEvery diagonal entry of an adjacency matrix is zero, givingThus every off-diagonal entry is the same number $(d-\theta )/n=-\theta$. Since these entries belong to $\{0,1\}$ and the graph is connected with at least two vertices, that common value must be one. Hence $A=J-I$ and $G$ is the complete graph $K_n$.

Conversely, $K_n$ has adjacency matrix $J-I$, with eigenvalue $n-1$ on $\mathrm{span}\{\mathbf{e}\}$ and eigenvalue $-1$ on ${\mathbf{e}}^{\perp }$. Therefore the requested graphs are preciselyas stated by the connected regular graph with two adjacency eigenvalues characterization.

A 3-connected graph contains a cycle, and any cycle is a complete graph subdivision $T{K}_3$.

For $T{K}_4$, choose a cycle $C$ for which some $C$-bridge has at least three attachment vertices $a,b,c$; such a choice is always possible in a 3-connected graph. Indeed, every component outside a cycle has at least three attachments, since one or two would form a vertex cut, and if a chosen cycle is spanning, a chord can be used to choose a smaller cycle and expose the remaining bridge. In a minimal connected part of that bridge joining $a,b,c$, suppress degree-two vertices. The result has a branch vertex $w$ joined to $a,b,c$ by three paths whose interiors are mutually disjoint and avoid $C$. The three arcs of $C$ between $a,b,c$, together with these three paths, form a subdivision of the complete graph on branch vertices $w,a,b,c$. Hence

A 3-connected graph need not contain a $T{K}_5$. The complete graph $K_4$ itself is 3-connected and has only four vertices, so it cannot contain a graph subdivision of $K_5$. Equally, any planar 3-connected graph is a counterexample because a subdivision of the complete graph $K_5$ would be nonplanar.

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

Let $X$ count copies of the complete graph $K_4$ in $G$. Its expected value is

Write $X$ as a sum of indicator random variable. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows thatThus $\mathrm{var}(X)/{\mu }^2\to 0$. By the second moment method, $X/\mu \to 1$ in probability, so $\mathbb{P}(X\ge 100)\to 1$. This is the sparse clique-count concentration estimate.

Factor the proposed polynomial:In particular,By the two-colourability criterion for bipartite graphs, every finite bipartite graph has at least one proper two-colouring, so its chromatic polynomial is positive at $2$. Therefore $f$ cannot be the chromatic polynomial of a bipartite graph.

It is nevertheless a chromatic polynomial. Start with the complete graph $K_3$, whose vertices may be coloured inways, and attach one leaf to any vertex. After the triangle is coloured, the leaf has $t-1$ available colours. The chromatic polynomial after attaching a leaf therefore gives

Take the infinite familyEach complete graph is connected and has graph diameter one. If $A_n$ is its adjacency matrix, thenand direct multiplication gives the adjacency-matrix quadratic relation for a complete graphConsequentlyso $I,A_n,A_n^2$, which are precisely the powers through $d(G_n)+1=2$, are linearly dependent for every $n\ge 2$.

Letbe the terminal set, and let $C$ be the vertex set of the given complete graph $K_{2k}$. Any vertex set $S$ meeting every $T$-$C$ path has size at least $2k$: if $|S|<2k$, choose $t\in T\setminus S$ and $c\in C\setminus S$. Since $G$ is $2k$-connected, $G-S$ is connected, so it contains a $t$-$c$ path avoiding $S$, a contradiction.

The Set version of Menger theorem therefore gives $2k$ pairwise vertex-disjoint $T$-$C$ paths. Truncate them at their first and last vertices in $T\cup C$. Since both sets have $2k$ vertices, every terminal is the endpoint of one path and the other endpoints are distinct vertices of $C$. Denote the path from a terminal $z$ to its clique endpoint by $R_z$, and call that endpoint $c_z$.

For each $i$, the clique contains the edge $c_{x_i}{c}_{y_i}$. Concatenatinggives an $x_i$-$y_i$ path in a graph. The linkage paths are mutually vertex-disjoint, their clique endpoints are all distinct, and the joining clique edges pair those endpoints without introducing a new vertex. The resulting paths $P_1,\dots ,P_k$ are therefore mutually vertex-disjoint. In particular, under the stated clique hypothesis, the $2k$-connected graph is $k$-linked for these terminals.

In a dense graph with many common neighbours for every pair of vertices, choose disjoint random small sets. A set is good if few vertices have no neighbour in it. The Markov inequality controls the number of bad sets, and random sampling makes almost every pair of good sets adjacent. Add unused common neighbours to connect each set and to repair the remaining missing adjacencies. If the number of used vertices stays below every common-neighbour count, this constructs disjoint branch sets of a graph minor representing a large complete graph.

The ratio convention for cutoff for Markov chains requires $t_{\mathrm{mix}}^{(n)}(\epsilon )/t_{\mathrm{mix}}^{(n)}(1-\epsilon )\to 1$ for every fixed $0<\epsilon <1/2$, with positive denominators eventually. Equivalently, any two fixed levels have asymptotically equal mixing times. This convention permits a bounded-time jump; definitions which additionally require diverging mixing times exclude such a jump. For nonlazy simple random walk on the complete graph $K_n$, every fixed-level mixing time is eventually one.

Offer $k$ independent uniform edges of the complete graph per row and permit any one to be selected from each row. If a chosen graph has no graph component of order at least $2n/3$, the balanced component cut provides an empty cut whose sides both have at least $n/3$ vertices. Each candidate edge crosses that cut with probability at least $4/9$, so a row can avoid crossing with probability at most $1-(4/9{)}^k$. The union bound over at most $2^n$ cuts gives failure probability at most $2^n[1-(4/9{)}^k{]}^{cn}$. Any fixed integer $c$ with $c[-\log (1-(4/9{)}^k)]>\log 2$ suffices. The guarantee is simultaneous even for choices made after seeing every offered edge.

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.