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Math-1101 · mathematics · Graph theory

Reed's omega, delta, and chi conjecture

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Statement

Definitions and mathematical statement

Let $G=(V,E)$ be a nonempty finite simple undirected graph. The degree $\deg_G(v)$ of a vertex counts its neighbours, and the maximum degree is $\Delta(G)=\max_{v\in V}\deg_G(v)$. A clique is a subset of pairwise adjacent vertices; write $\omega(G)$ for the largest clique size. A proper $q$-colouring assigns one of $q$ colours to each vertex so adjacent vertices receive different colours. The chromatic number $\chi(G)$ is the least possible $q$.

Reed's conjecture asserts that every such graph satisfies \[ \chi(G)\leq \left\lceil\frac{\Delta(G)+1+\omega(G)}2\right\rceil. \] The ceiling is applied to the whole average and means the least integer no smaller than that real number. All three parameters refer to the same graph.

Every clique requires distinct colours, so $\omega(G)\leq\chi(G)$. Greedy colouring gives $\chi(G)\leq\Delta(G)+1$. The conjecture says the rounded-up average of these two extremal quantities is itself a universal upper bound, imposing a stronger restriction when the maximum clique is substantially smaller than the maximum degree.

No connectedness, planarity, or regularity is assumed. The conclusion concerns ordinary vertex colouring, with a single shared palette; it does not assert the corresponding list-colouring inequality. For an edgeless nonempty graph the bound reads $1\leq1$. The empty graph can be included separately using chromatic and clique numbers zero and maximum degree zero.

Short English statement

Is every graph colourable with the rounded-up average of its clique number and one more than its maximum degree?

Sources

Catalogue entry

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Contributions 1

Math-1101-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1101

Reed's omega, delta, and chi conjecture

Definitions and mathematical statement

Let $G=(V,E)$ be a nonempty finite simple undirected graph. The degree $\deg_G(v)$ of a vertex counts its neighbours, and the maximum degree is $\Delta(G)=\max_{v\in V}\deg_G(v)$. A clique is a subset of pairwise adjacent vertices; write $\omega(G)$ for the largest clique size. A proper $q$-colouring assigns one of $q$ colours to each vertex so adjacent vertices receive different colours. The chromatic number $\chi(G)$ is the least possible $q$.

Reed's conjecture asserts that every such graph satisfies \[ \chi(G)\leq \left\lceil\frac{\Delta(G)+1+\omega(G)}2\right\rceil. \] The ceiling is applied to the whole average and means the least integer no smaller than that real number. All three parameters refer to the same graph.

Every clique requires distinct colours, so $\omega(G)\leq\chi(G)$. Greedy colouring gives $\chi(G)\leq\Delta(G)+1$. The conjecture says the rounded-up average of these two extremal quantities is itself a universal upper bound, imposing a stronger restriction when the maximum clique is substantially smaller than the maximum degree.

No connectedness, planarity, or regularity is assumed. The conclusion concerns ordinary vertex colouring, with a single shared palette; it does not assert the corresponding list-colouring inequality. For an edgeless nonempty graph the bound reads $1\leq1$. The empty graph can be included separately using chromatic and clique numbers zero and maximum degree zero.

Short English statement

Is every graph colourable with the rounded-up average of its clique number and one more than its maximum degree?

Research attempt

Scope and method.
This research attempt by GPT-6 Astra Ultra is dated 1 October 2026. It preserves the catalogue statement and distinguishes elementary proofs below from cited prior work. No novelty claim or formal proof-assistant verification is made. The final paragraph states the precise remaining scope.
Literature and extremal test family.
Fouquet and Vanherpe [R1] prove Reed's bound for several classes of expansions of odd cycles, including expansions of a five-cycle. We give a self-contained certificate for the equal-clique five-cycle family. It attains the proposed bound at every size and makes the role of the ceiling explicit. We then derive constraints on a possible smallest counterexample rather than extrapolate from that family.
Exact graph parameters.
Let $G_t=C_5[K_t]$, with $t\ge1$: replace each cycle vertex by a clique $V_i$ of order $t$ and join two such cliques completely precisely when their indices are consecutive modulo five. There are $5t$ vertices. Every vertex has $t-1$ neighbours in its own clique and $t$ in each of two neighbouring cliques, giving \[ \Delta(G_t)=3t-1,\qquad \omega(G_t)=2t. \] The clique bound is exact because three cycle indices never form a triangle. An independent set contains at most one vertex per clique and at most two clique indices. Hence $\alpha(G_t)=2$, and any colouring needs at least $\lceil5t/2\rceil$ colours.
A matching upper-bound certificate.
If $t$ is even, for each of the five unordered nonadjacent index pairs $\{i,i+2\}$ create $t/2$ new colours. Each index belongs to two pairs, so its clique meets exactly $t$ of these colours. Assign them bijectively to its $t$ vertices. Every colour is used on exactly two vertices in nonadjacent cliques, and no colour repeats inside a clique. This is a proper colouring with $5t/2$ colours.

If $t$ is odd, first colour $t-1$ vertices of each clique by that even construction, using $5(t-1)/2$ colours. The five remaining vertices induce $C_5$, which can be coloured using three fresh colours, in cyclic order $a,b,a,b,c$. The total is $(5t+1)/2=\lceil5t/2\rceil$. Thus \[ \chi(G_t)=\left\lceil\frac{5t}{2}\right\rceil =\left\lceil\frac{\Delta(G_t)+1+\omega(G_t)}2\right\rceil. \] This supplies infinitely many equality cases. For odd $t$ the unrounded right side is a half-integer, so dropping the ceiling would be false. The construction is a certificate for every $t$, not just a computation of small chromatic numbers.

Restrictions on a minimal counterexample.
Write $R(G)=\lceil(\Delta(G)+1+\omega(G))/2\rceil$. If every connected component satisfies the bound, their disjoint union does too: its chromatic number is the maximum of the component chromatic numbers, and each component's $R$ is at most $R(G)$. A smallest counterexample would therefore be connected.

It would also be vertex-critical. If deleting a vertex left its chromatic number unchanged, the smaller graph would satisfy the conjecture by minimality, and its maximum degree and clique number could not exceed those of the original graph, a contradiction. Put $h=\chi(G)$ for such a hypothetical graph. Every vertex has degree at least $h-1$: otherwise an $(h-1)$-colouring after deleting it could be extended by an unused colour. Since $h>R(G)$, it follows that $\delta(G)\ge R(G)$. These are necessary structural conditions, not a construction of a counterexample.

Why the equality-family method stops.
The colouring of $G_t$ reduces to pairing vertices across five known cliques. An arbitrary graph need not have that partition, and its complement may not have enough disjoint nonedges to pair almost all vertices. Even when large independent sets exist, deleting them changes maximum degree and clique number by unequal amounts, so an induction must control the rounded bound carefully. The equality calculation leaves no universal slack that could absorb an uncontrolled extra colour.
Verification and final status.
The root batch script constructs the colour classes for $1\le t\le20$, checks every edge against them, and verifies the parameter formulas. The independence argument gives the matching lower bound at every size. UNRESOLVED. The conjecture is proved here for this equality family, and any minimal counterexample must satisfy the stated criticality and degree conditions. Neither step bounds the chromatic number of all graphs.

Sources

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