Graph Theory |

Retained-Set Descent for Diagonal Ramsey Numbers

Sun, 13 Sep 2026 00:00:00 +0000

1 Introduction

Let $R(k,\ell)$ be the least integer $N$ such that every red–blue coloring of $K_{N}$ contains a red $K_{k}$ or a blue $K_{\ell}\text{.}$ Ramsey’s theorem [Ram] guarantees that these numbers are finite. The Erdős–Szekeres recurrence [ES] gives $R(k,\ell)\leq\binom{k+\ell-2}{k-1}\text{,}$ and hence $R(k,k)\leq 4^{k}\text{.}$ In the other direction, Erdős’s random-coloring argument [Er] gives a lower bound of order $k2^{k/2}\text{;}$ Spencer [Spe] improved its leading constant using the Lovász local lemma.

Thomason [Tho] improved the classical upper bound by a polynomial factor using quasirandomness: a coloring close to the inductive bound must have degrees and small-subgraph counts close to those of a random coloring. Conlon [Con] developed this approach to obtain $R(k,k)\leq 4^{k}\exp(-c(\log k)^{2}/\log\log k)\text{.}$ Sah [Sah] strengthened the estimate to $R(k,k)\leq 4^{k}\exp(-c(\log k)^{2})\text{.}$ Each bound holds for a suitable absolute constant $c>0$ and all sufficiently large $k\text{.}$ These results gave successively larger subexponential improvements while retaining exponential base $4\text{.}$

Campos, Griffiths, Morris, and Sahasrabudhe [CGMS] obtained the first improvement in the exponential base, proving $R(k,k)\leq(4-\varepsilon)^{k}$ for an absolute $\varepsilon>0$ and all sufficiently large $k\text{.}$ Their book algorithm works with two vertex sets and controls the red density between them. Gupta, Ndiaye, Norin, and Wei [GNNW] replaced this algorithm by a candidate induction, obtained stronger off-diagonal estimates, and optimized the resulting parameters to give $R(k,k)\leq 3.8^{k+o(k)}\text{.}$ Their candidate induction and reuse of improved Ramsey bounds in this optimization are the starting points of our work.

We improve the diagonal bound to $3.69507^{k}$ for all sufficiently large $k\text{.}$ Our descent applies density bounds inside a vertex set $X$ while keeping a second set $Y$ fixed for a final inequality in $|X|^{w}|Y|\text{,}$ with $w>0\text{.}$ Each resulting bound can be used inside $X$ in a later descent. We also ask how small an exponent finitely many such repetitions can give from a fixed initial bound.

We call this initial bound the source. It has the form $\log R(a,b)\leq af(b/a)+o(a)$ for integers $1\leq b\leq a\text{,}$ where $f$ is a fixed concave function on $[0,1]$ and the error is uniform in $b\text{.}$ Section 2 specifies $f\text{,}$ and Appendix B proves the bound.

A conditional bound at density $p\in(0,1)$ and ratio $s\in(0,1]$ applies to colorings with at least a fraction $p$ of their edges red. An exponent $z$ means that, for some fixed $C\geq 1$ and all sufficiently large $k\text{,}$ order at least $C\mathrm{e}^{zk}$ guarantees a red $K_{k}$ or a blue $K_{\lfloor sk\rfloor}\text{.}$ We use families of such bounds on closed intervals of $s\text{,}$ with continuous exponents and common constants and thresholds.

Starting from the conditional bounds supplied by the source, we apply the descent rule using bounds proved at earlier steps. We may also restrict ratio intervals, increase a density threshold or an exponent, and combine bounds at the same density on a finite closed cover of a ratio interval. A finite proof uses only finitely many of these operations; Definition 4.1 gives the precise rules.

Let $U_{\rm fin}(p,s)$ be the infimum of the exponents obtained at density $p$ and ratio $s$ by such finite proofs. In the diagonal case $(p,s)=(1/2,1)\text{,}$ we construct a finite proof with rational exponent $z_{+}$ and a rational lower bound $z_{-}$ for all finite proofs. Their exact values are given in Section 6.

Theorem 1.

For this source and these finite proof rules,

$$1.305<z_{-}\leq U_{\rm fin}(1/2,1)\leq z_{+}<1.307.$$

(1)

The exponent $z_{+}$ has a finite proof. For all sufficiently large integers $k\text{,}$

$$R(k,k)\leq\left\lceil 150003\exp(z_{+}k)\right\rceil,$$

(2)

and hence $R(k,k)\leq 3.69507^{k}\text{.}$

For the Ramsey bound, choose the denser color as red and apply the conditional bound at $(p,s)=(1/2,1)\text{.}$

1.1 Proof Outline

We use disjoint sets $X,Y\text{,}$ keeping $Y$ fixed and requiring each vertex of $X$ to have at least a fixed positive fraction of $Y$ as red neighbors. The red target $k$ and the blue target $\ell$ in $Y$ stay fixed while the blue target $t$ in $X$ decreases. When $X$ is large enough for a conditional bound at density $q\text{,}$ we either finish at red density at least $q$ or descend to a blue neighborhood of size greater than $(1-q)(|X|-1)\text{.}$ Restricting $X$ to this neighborhood preserves its red degrees into $Y$ and reduces $t$ by one: a blue $K_{t-1}$ there extends through the chosen vertex.

Write the required size of $X$ at $s=t/k$ as $\exp(kg(s))\text{,}$ with $g$ piecewise affine. On a cell of slope $d\text{,}$ the threshold falls by a factor $\mathrm{e}^{-d}$ per step, so we take $d>-\log(1-q)$ to allow for the neighborhood loss. The profile $g$ must also exceed the exponents of the bounds used inside $X\text{.}$ With slopes and calls fixed, optimizing its vertical shift under these conditions and the initial and terminal inequalities gives the cost in Equation 12.

Reusing these bounds and iterating from the source gives a decreasing sequence of exponents whose limit is a greatest fixed point (Theorem 3). Strict inequalities persist on neighborhoods of ratios; compactness selects finitely many neighborhoods and hence a common finite depth for a whole descent. Hence every exponent strictly above the limit has a finite proof, uniformly near its target ratio.

For the lower bound, a concave polygon $F$ assigns exponent $F(s)$ at densities up to $1-\mathrm{e}^{-d}$ on a segment of slope $d\text{.}$ Where a descent using bounds that respect this estimate has $g<F\text{,}$ its call density must exceed $1-\mathrm{e}^{-d}\text{,}$ forcing the slope of $g$ above $d\text{.}$ Thus $g-F$ decreases as the descent moves to smaller ratios. We choose $F$ so that the terminal weighted inequality cannot hold below it. The supporting-line formula reduces this requirement to one-variable inequalities at its vertices (Theorem 4); the certificates in Section 6 complete the proof of Theorem 1.

The Lean 4 formalization and the Python code for exact certificate checks are available in the accompanying repository.¹¹ 1

 <a href="https://github.com/sichen-wang/diagonal-ramsey-numbers" title="" class="ltx_ref ltx_url ltx_font_typewriter">https://github.com/sichen-wang/diagonal-ramsey-numbers</a></span></span></span></p>

2 Source Bounds

We now describe the source function and express its bound through pairs $(x,y)\text{.}$ These pairs supply the weighted inequality and the initial conditional bounds used in the descent.

All logarithms are natural. Let

$$h(s)=(1+s)\log(1+s)-s\log s,\qquad h(0)=0.$$

The classical Ramsey bound gives $\log R(k,\ell)\leq kh(\ell/k)$ for $1\leq\ell\leq k$ [ES].

Our fixed source is

$$f(s)=h(s)+s\mathrm{e}^{-s}P(s)\qquad(0\leq s\leq 1),$$

(3)

where the piecewise cubic $P$ is specified by the rational partition points, values, and derivatives in data/source.json. This source satisfies

$$\log R(a,b)\leq af(b/a)+o(a)\qquad(1\leq b\leq a),$$

(4)

with an error uniform in $b\text{.}$ Appendix B proves this bound and the properties of $f$ used below.

2.1 Source Pairs and Supporting Lines

The weighted inequality uses bounds of the form $R(a,b)\leq x^{-a}y^{-b}\text{.}$ We say that $(x,y)\in(0,1)^{2}$ is an admissible Ramsey pair if this bound holds for all sufficiently large $a+b\text{.}$ Color symmetry in Equation 4 leads to the extension

$$\widehat{f}(t)=\begin{cases}f(t),&0\leq t\leq 1,\\ tf(1/t),&t\geq 1.\end{cases}$$

The source exponent for arbitrary positive targets $a,b$ is then $b\widehat{f}(a/b)\text{.}$ We call $(x,y)\in(0,1)^{2}$ a source pair if $b\widehat{f}(a/b)\leq-a\log x-b\log y$ for all $a,b>0\text{,}$ and write $\mathcal{B}_{f}$ for their set. Every interior point of $\mathcal{B}_{f}$ is admissible: choose a slightly larger pair still in $\mathcal{B}_{f}$ and apply Equation 4 with color symmetry. The resulting error is $o(a+b)\text{,}$ which the strict coordinate gaps absorb.

Supporting lines give a convenient description of these pairs. The source verification shows that $f$ is continuous on $[0,1]$ and is $C^{1}$ on $(0,1]\text{,}$ where it is strictly concave and increasing. It also gives $f(0)=0$ and $2f^{\prime}(1)>f(1)\text{.}$ It follows that $\widehat{f}$ is increasing and concave. On each smooth piece with $t>1\text{,}$ $\widehat{f}^{\prime\prime}(t)=t^{-3}f^{\prime\prime}(1/t)<0\text{;}$ its first derivative is continuous at the reflected source partition points. At $t=1\text{,}$ the derivative drops from $f^{\prime}(1)$ to $f(1)-f^{\prime}(1)\text{.}$ Strict concavity and $f(0)=0$ give $f(s)-sf^{\prime}(s)>0\text{,}$ so the reflected branch is increasing. The expression in Equation 3 gives $\widehat{f}^{\prime}(0+)=\infty\text{,}$ $\widehat{f}^{\prime}(\infty)=0\text{,}$ and $\widehat{f}(t)/t\to 0$ as $t\to\infty\text{.}$

For $A>0\text{,}$ define the intercept of the supporting line with slope $A$ by

$$b(A)=\sup_{t\geq 0}\{\widehat{f}(t)-At\},\qquad\widehat{f}(t)=\inf_{A>0}\{At+b(A)\}\quad(t>0).$$

(5)

For fixed $A>0\text{,}$ $\widehat{f}(t)-At\to-\infty$ as $t\to\infty\text{,}$ while $f(t)/t\to\infty$ as $t\downarrow 0$ makes this expression positive for small positive $t\text{.}$ By continuity, the supremum is positive and attained at some $t>0\text{.}$ At each $t>0\text{,}$ concavity supplies a supporting slope $A>0\text{:}$ use the derivative, or any slope between the one-sided derivatives at $t=1\text{.}$ For this slope $b(A)=\widehat{f}(t)-At\text{,}$ proving the second identity.

Writing $x=\mathrm{e}^{-A}$ and $y=\mathrm{e}^{-B}\text{,}$ the source-pair condition becomes $B\geq b(A)\text{,}$ so

$$\mathcal{B}_{f}=\{(\mathrm{e}^{-A},\mathrm{e}^{-B}):A>0,\ B\geq b(A)\}.$$

The finite convex function $b$ is continuous on $A>0\text{,}$ so interior source pairs correspond to $B>b(A)\text{.}$ Taking $(a,b)=(1,s)$ in the support inequality gives

$$A+sb(A)\geq f(s)\qquad(0<s\leq 1).$$

(6)

2.2 The Weighted Candidate Bound

Write $N_{\chi}(v)$ for the neighborhood of $v$ in color $\chi\in\{R,B\}$ and $\deg_{\chi}(v,Z)=|N_{\chi}(v)\cap Z|\text{.}$ For disjoint nonempty vertex sets $X,Y\text{,}$ let $e_{R}(X,Y)$ be the number of red edges between them and write $d_{R}(X,Y)=e_{R}(X,Y)/(|X||Y|)\text{.}$ Following [GNNW], a candidate is an ordered pair $(X,Y)$ of such sets. We say that a candidate is $(k,\ell,t)$-good if $X\cup Y$ contains a red $K_{k}\text{,}$ or $X$ contains a blue $K_{t}\text{,}$ or $Y$ contains a blue $K_{\ell}\text{.}$ The target in the retained set $Y$ is $\ell\text{,}$ while the target $t$ in $X$ may decrease.

Lemma 2.1 (Weighted Candidate Bound).

Fix $w>0$ and $\mu,p,x,y\in(0,1)$ with $(x,y)\in\operatorname{int}\mathcal{B}_{f}$ and

$$x<(1-\mu)^{w}p^{1/(1-\mu)}.$$

For all sufficiently large $\ell\text{,}$ uniformly in positive integers $k,t\text{,}$

$$d_{R}(X,Y)\geq p,\qquad|X|^{w}|Y|\geq x^{-k}y^{-\ell}\mu^{-wt}$$

imply that $(X,Y)$ is $(k,\ell,t)$-good.

Proof.

Choose a slightly larger pair $(\widehat{x},\widehat{y})$ still in $\operatorname{int}\mathcal{B}_{f}\text{.}$ Once $\ell$ is large, admissibility guarantees a red $K_{a}$ or a blue $K_{\ell}$ in every vertex set of size at least $\widehat{x}^{-a}\widehat{y}^{-\ell}\text{,}$ uniformly for $a\geq 1\text{.}$ Applying Lemma A.1 with $C=1$ proves the claim. ∎

2.3 Conditional Bounds

To reuse a bound during a descent, we need it for an interval of blue-to-red target ratios, with a single size constant and a single threshold for $k\text{.}$

Definition 2.2 (Conditional Bound).

For $p\in(0,1)\text{,}$ a compact interval $I\subset(0,1]\text{,}$ and a continuous function $L:I\to(0,\infty)\text{,}$ we say that the conditional bound $\mathsf{D}(p,L,I)$ holds if some $C\geq 1,K$ have the following property uniformly for $r\in I$ and integers $k\geq K\text{:}$ a coloring of red density at least $p$ and order at least $C\mathrm{e}^{kL(r)}$ contains a red $K_{k}$ or a blue $K_{\lfloor rk\rfloor}\text{.}$ We choose $K$ so that all blue targets are positive. For a constant exponent $z\text{,}$ the notation $\mathsf{D}(p,z,I)$ means $L\equiv z\text{.}$

We call $(w,\mu,x,y)$ a strict control at density $p$ when it satisfies the parameter conditions of Lemma 2.1. Write $A=-\log x\text{,}$ $B=-\log y\text{,}$ and $\beta=-\log\mu\text{.}$ Apply the weighted candidate bound with $t=\ell$ to a balanced partition whose red cross-density is at least $p\text{;}$ such a partition exists by averaging. Both set sizes are a constant fraction of the graph’s order, so the corresponding exponent is

$$L(r)=\frac{A+rB+wr\beta}{1+w}.$$

(7)

Choose $y_{1}>y$ with $(x,y_{1})\in\operatorname{int}\mathcal{B}_{f}\text{.}$ For $\ell=\lfloor rk\rfloor$ and a host of order $N\geq\mathrm{e}^{kL(r)}\text{,}$ the partition satisfies

$$|X|^{w}|Y|\geq\frac{x^{-k}y^{-\ell}\mu^{-w\ell}}{3^{w+1}}\geq x^{-k}y_{1}^{-\ell}\mu^{-w\ell}.$$

The first inequality uses $kr\geq\ell$ and $B+w\beta>0\text{;}$ the second holds once $(y_{1}/y)^{\ell}\geq 3^{w+1}\text{.}$ Applying Lemma 2.1 at $(x,y_{1})$ proves $\mathsf{D}(p,L,I)$ with $C=1$ on every compact $I\subset(0,1]\text{.}$ The threshold is uniform in $r\in I\text{,}$ since $\ell\geq k\min I-1\text{.}$ These weighted source bounds are the starting points of our finite derivations, and we call them source leaves.

3 Retained-Set Descent and Its Cost

We now allow the conditional bound used in $X$ to change as its blue target decreases. Throughout the descent we keep $\deg_{R}(v,Y)\geq\pi|Y|$ for every $v\in X\text{,}$ for a fixed $\pi\in(0,1)\text{.}$ This gives cross-density at least $\pi$ after every restriction of $X\text{.}$ A positive function $g(s)$ specifies the required size $\exp(kg(s))$ of $X$ at target $t=sk\text{.}$ To use a bound $\mathsf{D}(q,L,I)\text{,}$ we require $g>L$ to absorb its size constant; the slope condition below absorbs the loss in a blue-neighborhood step.

Definition 3.1 (Route).

A route consists of a partition $0<\theta=u_{0}<\cdots<u_{m}=\lambda\leq 1$ and, on each closed cell $J_{j}=[u_{j-1},u_{j}]\text{,}$ a proved call $\mathsf{D}(q_{j},L_{j},I_{j})$ with $J_{j}\subseteq I_{j}\text{.}$ It also specifies slopes

$$d_{j}>c(q_{j}):=-\log(1-q_{j})$$

and a continuous positive function $g\text{,}$ constant on $[0,\theta]$ and affine with slope $d_{j}$ on $J_{j}\text{,}$ such that $g>L_{j}$ on every closed cell.

The descent starts at ratio $\lambda$ and stops at $\theta\text{.}$ Write $d(t)$ for the piecewise constant slope on $[\theta,\lambda]\text{,}$ taking the right-hand slope at $\theta$ and internal partition points and the left-hand slope at $\lambda\text{.}$

Theorem 2 (Retained-Set Rule).

Fix a route as above, an outer density $p\text{,}$ and a row density $0<\pi<p<1\text{.}$ Choose a strict control at density $\pi$ as the route’s terminal parameters. If

$$z>g(\lambda),\qquad z+wg(\theta)>A+\lambda B+w\theta\beta,$$

(8)

then $\mathsf{D}(p,z,\{\lambda\})$ holds with

$$C=\frac{3(1-\pi)}{p-\pi}.$$

(9)

The conclusion is uniform over truncations and vertical shifts of one fixed route when the new endpoint, output exponent, and shift depend continuously on a parameter in a compact set; the stopping ratio, slopes, densities, calls, and terminal controls stay fixed. At endpoint $r\text{,}$ retain the closed cells $[u_{j-1},\min\{u_{j},r\}]$ with $u_{j-1}<r$ and require positive uniform margins in $g>L_{j}$ and in Equation 8. Such a family may include $r=\theta\text{,}$ with no internal calls.

Proof.

If $\lambda=\theta\text{,}$ the size tests give

$$(1+w)z>z+wg(\theta)>A+\theta B+w\theta\beta.$$

The source leaf at density $\pi$ proves the conclusion at density $p>\pi\text{,}$ with constant $1\leq C\text{.}$

Suppose $\theta<\lambda\text{,}$ and take a graph of red density at least $p$ and order $N\geq C\mathrm{e}^{kz}\text{.}$ Averaging gives a balanced partition $X_{0},Y$ with cross-density at least $p$ and both parts of size at least $N/3\text{.}$ Keep in $X_{0}$ only the vertices with at least $\pi|Y|$ red neighbors in $Y\text{.}$ The surviving set $X$ satisfies

$$(p-\pi)|X_{0}||Y|\leq e_{R}(X,Y)-\pi|X||Y|\leq(1-\pi)|X||Y|.$$

The choice of $C$ gives $|X|,|Y|\geq\mathrm{e}^{kz}\text{.}$ Every nonempty subset of $X$ has red cross-density at least $\pi$ with this fixed $Y\text{.}$

Put $n_{0}=\lfloor\theta k\rfloor$ and $\ell=\lfloor\lambda k\rfloor\text{,}$ taking $k$ large enough that $n_{0}\geq 1\text{.}$ To track the loss, define for integers $0\leq t\leq\ell$

$$\Gamma_{k}(t)=kg(\theta)+\sum_{j=n_{0}+1}^{t}d(j/k)\quad(t>n_{0}),\qquad\Gamma_{k}(t)=kg(\theta)\quad(0\leq t\leq n_{0}).$$

For $V=d_{1}+\sum_{j<m}|d_{j+1}-d_{j}|\text{,}$

$$\Gamma_{k}(t)-\Gamma_{k}(t-1)=d(t/k)\quad(t>n_{0}),\qquad|\Gamma_{k}(t)-kg(t/k)|\leq V.$$

(10)

For the second bound, express $d$ as its initial step at $\theta$ and its subsequent jumps. For each unit step, the mesh count differs from $k$ times its integral by at most one, even when a jump lies on the mesh. Summing the absolute jump sizes gives $V\text{.}$

Let $C_{j}$ be the call constants and put

$$\delta=\min_{j}\min_{J_{j}}(g-L_{j})>0,\qquad\eta=\min_{j}(1-q_{j}-\mathrm{e}^{-d_{j}})>0.$$

Choose $k$ so that $k\delta>V+\max_{j}\log C_{j}$ and $\eta\mathrm{e}^{kg(\theta)}>1\text{,}$ and so that all call and candidate thresholds hold. We prove by induction on $t=1,\ldots,\ell$ that $(Z,Y)$ is $(k,\ell,t)$-good whenever $Z\subseteq X$ and $|Z|\geq\mathrm{e}^{\Gamma_{k}(t)}\text{.}$

For $t\leq n_{0}\text{,}$ the terminal test gives

$$\log(|Z|^{w}|Y|)\geq k[wg(\theta)+z]>k(A+\lambda B+w\theta\beta)\geq kA+\ell B+wt\beta.$$

Since $d_{R}(Z,Y)\geq\pi\text{,}$ Lemma 2.1 applies.

For $t>n_{0}\text{,}$ put $s=t/k$ and use the cell with assigned slope $d_{j}\text{.}$ If the red density in $Z$ is at least $q_{j}\text{,}$ then $\log|Z|\geq kg(s)-V>kL_{j}(s)+\log C_{j}\text{.}$ The call gives a red $K_{k}$ or a blue $K_{t}\text{.}$ Otherwise some vertex has a blue neighborhood $Z^{\prime}\subseteq Z$ with

$$|Z^{\prime}|>(\mathrm{e}^{-d_{j}}+\eta)(|Z|-1)>\mathrm{e}^{-d_{j}}|Z|\geq\mathrm{e}^{\Gamma_{k}(t-1)}.$$

The middle inequality follows from $\eta|Z|>1$ and $\mathrm{e}^{-d_{j}}+\eta<1\text{.}$ Induction applies to $(Z^{\prime},Y)\text{.}$ A blue $K_{t-1}$ extends through the chosen vertex; either of the other two outcomes already proves the claim.

Finally, Equation 10 and the endpoint test give $\Gamma_{k}(\ell)\leq kg(\lambda)+V<kz$ for large $k\text{.}$ The claim therefore applies to $Z=X$ at $t=\ell\text{.}$

For uniformity, take the minimum stopping height and the minima of the strict gaps over the compact parameter set, and the maxima of the finitely many call thresholds. The same $V$ bounds every truncated profile, since truncation only removes slope jumps. These choices give a common threshold for $k\text{;}$ at $\lambda=\theta$ use the fixed source control as above. ∎

3.1 The Cost of a Descent

Fix the partition, calls, slopes, and terminal controls, and vary the vertical position of $g\text{.}$ Put

$$S(s)=\int_{s}^{\lambda}d(t)\,dt,\qquad M=\max_{j}\max_{s\in J_{j}}\{L_{j}(s)+S(s)\}.$$

(11)

The quantity $S(s)$ is the cumulative logarithmic loss in descending from $\lambda$ to $s\text{.}$ Every $g$ with the specified slopes has the form $g(s)=H-S(s)$ on $[\theta,\lambda]\text{.}$ The internal conditions are exactly $H>M\text{;}$ they also make $g$ positive, since $M\geq L_{1}(\theta)+S(\theta)>S(\theta)\text{.}$ The conditions in Equation 8 now read

$$H<z,\qquad wH+z>A+\lambda B+w\theta\beta+wS(\theta).$$

There is such an $H>M$ exactly when

$$z>Z:=\max\left\{M,\frac{A+\lambda B+w\theta\beta+wS(\theta)}{1+w}\right\}.$$

(12)

Necessity follows from $z>H>M$ and $(1+w)z>z+wH\text{.}$ Conversely, if $z>Z\text{,}$ choose $H<z$ sufficiently close to $z$ to satisfy both inequalities. Thus $Z$ is the infimum output exponent over these vertical positions. The term $M$ accounts for every internal call together with the loss needed to reach it; the other term is the least exponent compatible with the weighted terminal inequality.

For one call $\mathsf{D}(q,L,[\theta,\lambda])$ with affine $L\text{,}$ choose a slope $d>c(q)\text{.}$ Here $S(s)=d(\lambda-s)\text{,}$ so $L+S$ is affine and its maximum occurs at an endpoint. Substitution in Equation 12 gives

$$Z=\max\left\{L(\lambda),\ L(\theta)+d(\lambda-\theta),\ \frac{A+\lambda B+w\theta\beta+wd(\lambda-\theta)}{1+w}\right\}.$$

The first two terms require enough vertices to use the call at both ends of the interval; the second includes all the loss incurred before reaching $\theta\text{.}$ The third requires enough vertices to apply the weighted bound at $\theta\text{.}$ Every $z>Z$ therefore gives a new conditional bound at $\lambda\text{.}$

3.2 Reuse on an Interval

To obtain bounds at smaller target ratios, keep the same calls and terminal controls. For $\theta\leq r\leq\lambda\text{,}$ restrict the profile $g$ to $[0,r]$ and allow a nonnegative vertical shift. Such a shift preserves the inequalities $g>L_{j}\text{.}$ The resulting infimum is

$$E(r)=\max\left\{g(r),\frac{A+rB+w\{\theta\beta+g(r)-g(\theta)\}}{1+w}\right\},\qquad\theta\leq r\leq\lambda.$$

(13)

For every $\varepsilon>0\text{,}$ the whole family $\mathsf{D}(p,E+\varepsilon,[\theta,\lambda])$ holds with the constant in Equation 9. To check strictness uniformly, use the shift $H_{r}=E(r)-g(r)+\varepsilon/2$ and output $z_{r}=E(r)+\varepsilon\text{.}$ The endpoint gap is $\varepsilon/2\text{,}$ and

$$z_{r}+w[g(\theta)+H_{r}]-A-rB-w\theta\beta\geq(1+w/2)\varepsilon.$$

Theorem 2 applies uniformly.

A source leaf is affine, and $E$ is the maximum of two affine functions on each affine piece of $g\text{.}$ When a new route uses one of these bounds as a call, refine its cell at the partition points of the called profile. On each resulting piece, adding the affine loss $S$ preserves convexity. The maximum in Equation 11 is therefore attained at an endpoint of a call cell or a partition point of the called profile inside that cell.

4 The Optimum over Finite Proofs

We keep the source $\mathcal{B}_{f}$ fixed and minimize over all finite repetitions of the retained-set rule.

Definition 4.1 (Finite Proof).

A finite proof is a finite derivation of conditional bounds using the following rules:

1.

Start with a weighted source bound from Equation 7.
2.

Apply Theorem 2 with previously derived conditional bounds as its calls, including its uniform families.
3.

Restrict to a closed ratio interval, increase the density, or increase the continuous exponent function pointwise.
4.

Combine finitely many bounds: on a finite closed cover of a compact ratio interval, use a continuous output that dominates a proved family at the same density on each cover interval.

Taking the maximum of the constants and thresholds validates the last rule. Let $U_{\rm fin}(p,s)$ be the infimum of the exponents obtained at $(p,s)$ by these finite proofs.

To compare these proofs across all finite depths, first let $U_{0}(p,s)$ be the infimum over source leaves alone. It is finite: take $w=1\text{,}$ $\mu=1/2\text{,}$ $A>\log 2-2\log p\text{,}$ and $B>b(A)\text{.}$ These strict controls give $0\leq U_{0}(p,s)<\infty\text{.}$

To treat all possible descents at once, let $u(p,s)$ specify the required exponent for a call at $(p,s)\text{.}$ Order these functions pointwise and consider the complete lattice $\mathcal{L}=\{u:(0,1)\times(0,1]\to\mathbb{R}:0\leq u\leq U_{0}\}\text{.}$

Define $T(u)(p,\lambda)$ as the infimum of $U_{0}(p,\lambda)$ and all $z>0$ admitted by the following data: a finite partition $0<\theta=u_{0}<\cdots<u_{m}=\lambda\text{,}$ densities $q_{j}\in(0,1)\text{,}$ slopes $d_{j}>c(q_{j})\text{,}$ and a continuous positive function $g\text{,}$ affine with slope $d_{j}$ on each $J_{j}$ and constant on $[0,\theta]\text{,}$ such that

$$g(s)>u(q_{j},s)\qquad(s\in J_{j}).$$

(14)

The terminal parameters form a strict control at some row density $0<\pi<p$ and satisfy Equation 8. All data are finite and fixed before $k$ is chosen.

The operator is monotone: decreasing the internal requirements can only add feasible routes. Its output is nonincreasing in the outer density $p\text{,}$ since any leaf or route valid at $p$ remains valid at a larger density. Set

$$U_{n+1}=T(U_{n}),\qquad U_{\infty}=\inf_{n\geq 0}U_{n}.$$

We say that $v\in\mathcal{L}$ is a lower bound preserved by the rules if $v\leq T(v)\text{.}$ In standard order terminology, such a function is a post-fixed point of $T\text{.}$ Equivalently, every leaf and every route using internal requirements $v$ has output at least $v\text{.}$ Here a proof on a relative neighborhood of $s$ means a conditional bound on a compact ratio interval containing that neighborhood.

Theorem 3 (Finite Proofs and Preserved Lower Bounds).

The sequence $U_{n}$ decreases to the greatest fixed point $U_{\infty}$ of $T$ in $\mathcal{L}\text{,}$ and

$$U_{\rm fin}(p,s)=U_{\infty}(p,s)=\sup_{\begin{subarray}{c}v\in\mathcal{L}\\ v\leq T(v)\end{subarray}}v(p,s).$$

(15)

Every $z>U_{\infty}(p,s)$ has a finite proof with constant output $z$ on a relative neighborhood of $s\text{,}$ with uniform constants.

The order comparison is the classical greatest-fixed-point principle [Tarski, Theorem 1]. Strict inequalities and compactness also identify the limit with finite conditional proofs, including the uniformity required for their reuse.

Proof.

Since $T(U_{0})\leq U_{0}\text{,}$ monotonicity gives $U_{n+1}\leq U_{n}\text{.}$ By induction on $n\text{,}$ we prove both that each ratio section $U_{n}(p,\cdot)$ is upper semicontinuous and that every $z>U_{n}(p,s)$ has a finite proof with constant output $z$ on a relative neighborhood of $s\text{.}$ Upper semicontinuity means that a strict inequality $U_{n}(p,s)<a$ persists near $s\text{.}$ At level zero, both conclusions follow by choosing a continuous source leaf strictly below $z\text{.}$

Suppose the claim holds at level $n\text{,}$ and let $z>U_{n+1}(p,\lambda)\text{.}$ Choose a leaf or a formal route against $U_{n}$ with output $z^{\prime}<z\text{.}$ A leaf again gives both conclusions by continuity. For a route, truncate or extend its last affine cell, keeping the density and slope. Upper semicontinuity preserves $g>U_{n}(q_{m},\cdot)$ near $\lambda\text{,}$ and the strict endpoint and terminal tests persist by continuity. Thus the same output $z^{\prime}$ is feasible for nearby endpoints, proving $U_{n+1}(p,r)<z$ there.

Choose a compact interval of endpoints within this range that contains a relative neighborhood of $\lambda\text{.}$ To replace the formal calls by finite proofs, consider the call cells up to the largest endpoint in this neighborhood. At each point $s$ of a cell with density $q_{j}\text{,}$ choose

$$U_{n}(q_{j},s)<z_{s}<g(s).$$

The induction hypothesis gives a finite proof with constant output $z_{s}$ near $s\text{.}$ Shrink this neighborhood until $z_{s}<g$ throughout. Compactness gives a finite subcover of each cell. Subdivide into closed subcells contained in these neighborhoods and use the corresponding proofs, with the original density and slope. The calls and their strict inequalities also hold at every shared endpoint. There are finitely many subcells, so their positive gaps have a common positive lower bound. The uniform part of Theorem 2 now gives a finite proof with constant output $z$ on the chosen endpoint neighborhood. This completes the induction.

We next show that $U_{\infty}$ is fixed. Monotonicity gives $T(U_{\infty})\leq U_{n+1}$ for every $n\text{,}$ hence $T(U_{\infty})\leq U_{\infty}\text{.}$ For the reverse inequality, take a route feasible against $U_{\infty}\text{.}$ On each closed cell $J_{j}\text{,}$ the sets

$$O_{n}=\{s\in J_{j}:U_{n}(q_{j},s)<g(s)\}$$

are relatively open, increase with $n\text{,}$ and cover $J_{j}\text{.}$ Compactness and nesting give a common $N$ with $O_{N}=J_{j}$ on every cell. The route is therefore feasible against $U_{N}\text{,}$ so its output is at least $U_{N+1}\geq U_{\infty}\text{.}$ The leaf alternative is also at least $U_{\infty}\text{.}$ Taking infima gives $T(U_{\infty})=U_{\infty}\text{.}$

If $v\in\mathcal{L}$ and $v\leq T(v)\text{,}$ then $v\leq U_{0}$ and

$$v\leq U_{n}\quad\Longrightarrow\quad v\leq T(v)\leq T(U_{n})=U_{n+1}.$$

Hence $v\leq U_{\infty}\text{.}$ As $U_{\infty}$ is itself fixed, it is the greatest post-fixed point.

Every finite proof has output at least $U_{\infty}\text{,}$ by induction through its rules. This holds for leaves since $U_{0}\geq U_{\infty}\text{.}$ A route whose calls are at least $U_{\infty}$ is feasible against $U_{\infty}\text{,}$ so its output is at least $T(U_{\infty})=U_{\infty}\text{.}$ The case $\lambda=\theta$ reduces to a leaf. Restriction, increasing the exponent, and finite patching preserve the comparison. Increasing the density does too, since $U_{\infty}$ is nonincreasing in that density. Thus $U_{\rm fin}\geq U_{\infty}\text{.}$

Conversely, every $z>U_{\infty}(p,s)$ exceeds $U_{n}(p,s)$ for some finite $n\text{.}$ The local proof constructed above gives $U_{\rm fin}\leq U_{\infty}$ and the claimed uniformity. ∎

5 Concave Lower Bounds

We construct a function $v\leq T(v)\text{,}$ giving the lower estimate by Theorem 3. Its exponent will be a concave polygon: a continuous piecewise affine concave function $F:[0,1]\to\mathbb{R}\text{.}$ Write its partition points and values as

$$0=s_{0}<s_{1}<\cdots<s_{N}=1,\qquad F_{i}=F(s_{i}),\qquad F_{0}=0,$$

and suppose the slopes $d_{i}=(F_{i}-F_{i-1})/(s_{i}-s_{i-1})$ satisfy $d_{1}\geq\cdots\geq d_{N}>0\text{.}$ A call at density $q>1-\mathrm{e}^{-d_{i}}$ requires a route slope greater than $c(q)>d_{i}\text{.}$ This relation suggests using $1-\mathrm{e}^{-d_{i}}$ as the density up to which $F$ gives a lower bound. Define

$$p_{F}(s)=1-\mathrm{e}^{-d_{i}}\quad(s_{i-1}<s\leq s_{i}),\qquad v_{F}(p,s)=\begin{cases}F(s),&p\leq p_{F}(s),\\ 0,&p>p_{F}(s).\end{cases}$$

At each $s_{i}>0\text{,}$ $p_{F}$ uses the slope of the segment ending there.

Proposition 5.1 (A Preserved Lower Bound).

Suppose every strict terminal control of row density at most $p_{F}(\lambda)$ satisfies

$$F(\lambda)+wF(\theta)\leq A+\lambda B+w\theta\beta\qquad(0\leq\theta\leq\lambda\leq 1,\ \lambda>0).$$

(16)

Then $v_{F}\leq U_{\infty}\text{.}$ If $d_{N}\geq\log 2\text{,}$ in particular $F(1)\leq U_{\infty}(1/2,1)\text{.}$

Proof.

Taking $\theta=\lambda$ in Equation 16 shows that every source leaf at density $p\leq p_{F}(\lambda)$ has exponent at least $F(\lambda)\text{.}$ Thus $0\leq v_{F}\leq U_{0}\text{.}$

Consider a route feasible against $v_{F}$ at density $p\leq p_{F}(\lambda)\text{,}$ with output $z<F(\lambda)\text{.}$ Its endpoint satisfies $g(\lambda)<F(\lambda)\text{.}$ Refine the call partition at the points $s_{i}\text{.}$ On a resulting cell $[a,b]\text{,}$ let $\ell_{j}$ and $d_{i}$ be the slopes of $g$ and $F\text{.}$ If $g(b)<F(b)\text{,}$ the closed call condition forces $q_{j}>p_{F}(b)=1-\mathrm{e}^{-d_{i}}\text{,}$ since $p_{F}$ uses the left slope at $b\text{.}$ Consequently $\ell_{j}>c(q_{j})>d_{i}\text{,}$ and

$$g(a)-F(a)=g(b)-F(b)-(\ell_{j}-d_{i})(b-a)<0.$$

Starting at $\lambda$ and applying this implication backward through the finite partition gives $g(\theta)<F(\theta)\text{.}$ The row density satisfies $\pi<p\leq p_{F}(\lambda)\text{,}$ so Equation 16 now contradicts $z+wg(\theta)>A+\lambda B+w\theta\beta\text{.}$ Every route and leaf therefore has output at least $v_{F}\text{,}$ giving $v_{F}\leq T(v_{F})\text{.}$ Apply Theorem 3; the final assertion follows from $p_{F}(1)\geq 1/2\text{.}$ ∎

5.1 Elimination of the Terminal Controls

At each positive partition point, we maximize over the stopping ratio and combine the terminal inequalities with the supporting-line formula. This gives a test in the single parameter $\mu\text{.}$ For $0<\mu<1\text{,}$ write $\tau=-\log(1-\mu)$ and $\beta=-\log\mu\text{,}$ and define

$$\Psi(\lambda,z,\kappa,\mu)=\frac{1-\mu}{\kappa}\left[z-\lambda\widehat{f}\!\left(\frac{1-\kappa}{\lambda}\right)\right],\qquad\lambda>0,\quad 0<\kappa<1.$$

(17)

Theorem 4 (Finite Polygon Criterion).

Suppose $F$ is as above and

$$0<F_{i}<\min\{f(s_{i}),h(s_{i})\}\qquad(1\leq i\leq N).$$

(18)

For each $\mu\in(0,1)$ define

$$D_{i}(\beta)=\max_{0\leq j\leq i}(F_{j}-s_{j}\beta),\qquad\kappa_{i}=D_{i}(\beta)/\tau,\qquad P_{i}=-\log(1-\mathrm{e}^{-d_{i}}).$$

(19)

If, for every $i$ and every $\mu$ with $D_{i}(\beta)>0\text{,}$

$$\Psi(s_{i},F_{i},\kappa_{i},\mu)\leq P_{i},$$

(20)

then $v_{F}\leq U_{\infty}\text{.}$

Proof.

At endpoint $s_{i}\text{,}$ the function $F(\theta)-\theta\beta$ is affine between consecutive partition points, so

$$\max_{0\leq\theta\leq s_{i}}\{F(\theta)-\theta\beta\}=D_{i}(\beta).$$

Concavity of $h$ and Equation 18 put $F(\theta)$ strictly below $h(\theta)$ for $\theta>0\text{.}$ The minimum of $\tau+\theta\beta$ over $0<\mu<1$ occurs at $\mu=\theta/(1+\theta)$ and equals $h(\theta)\text{.}$ Thus

$$F(\theta)<h(\theta)\leq\tau+\theta\beta\qquad(\theta>0).$$

The term at $\theta=0$ is zero, so $0\leq D_{i}(\beta)<\tau\text{.}$

Fix a strict control at row density $\pi\leq 1-\mathrm{e}^{-d_{i}}$ and put $P_{\pi}=-\log\pi\geq P_{i}\text{.}$ If the terminal inequality fails at $\lambda=s_{i}$ for some stopping ratio, then

$$wD_{i}>A+s_{i}B-F_{i}>0,\qquad A>w\tau+\frac{P_{\pi}}{1-\mu}.$$

Here $A+s_{i}B-F_{i}>0$ because $A+s_{i}B\geq f(s_{i})>F_{i}\text{.}$ In particular $0<\kappa_{i}=D_{i}/\tau<1\text{.}$ Multiplying the second inequality by $\kappa_{i}$ and using the first gives

$$(1-\kappa_{i})A+s_{i}B<F_{i}-\frac{\kappa_{i}P_{\pi}}{1-\mu}.$$

Since $B>b(A)\text{,}$ the supporting-line formula Equation 5 gives

$$s_{i}\widehat{f}\!\left(\frac{1-\kappa_{i}}{s_{i}}\right)\leq(1-\kappa_{i})A+s_{i}B.$$

Combining these inequalities yields $P_{\pi}<\Psi(s_{i},F_{i},\kappa_{i},\mu)\leq P_{i}\text{,}$ a contradiction. Thus Equation 16 holds for every stopping ratio at each positive partition point.

Now fix controls at row density at most $1-\mathrm{e}^{-d_{i}}$ and let $\lambda\in[s_{i-1},s_{i}]\text{.}$ Since $F$ is affine on this interval,

$$\max_{0\leq\theta\leq\lambda}\{F(\theta)-\theta\beta\}=\max\{D_{i-1}(\beta),\ F(\lambda)-\lambda\beta\},$$

where $D_{0}=0\text{.}$ The largest terminal difference over all stopping ratios is therefore

$$F(\lambda)-A-\lambda B+w\max\{D_{i-1}(\beta),\ F(\lambda)-\lambda\beta\}.$$

This is convex in $\lambda\text{,}$ being affine plus a positive multiple of a maximum of two affine functions. It is nonpositive at $s_{i}$ by the inequality already proved there. The same holds at $s_{i-1}>0$ because the left slope of $F$ there is at least $d_{i}\text{;}$ at zero the value is $-A<0\text{.}$ Convexity gives Equation 16 throughout the interval. Apply Proposition 5.1. ∎

5.2 The Full Control Interval

For the supplied polygon, the remaining variable is $\tau\in(0,\infty)\text{,}$ with $\mu=1-\mathrm{e}^{-\tau}$ and $\beta(\tau)=-\log(1-\mathrm{e}^{-\tau})\text{.}$ We verify the criterion analytically near zero and infinity, leaving a compact interval for the finite computation. Let $L=1/10$ and $H=10\text{.}$ The supplied data satisfy

$$\beta(L)>d_{1},\qquad\beta(H)<F_{N}/2,\qquad F_{N}<H,\qquad 2H\mathrm{e}^{-H}<P_{1}.$$

(21)

For $\tau\leq L\text{,}$ concavity gives $F_{j}\leq d_{1}s_{j}<\beta s_{j}$ for $j>0\text{,}$ so $D_{i}=0\text{.}$ For $\tau\geq H\text{,}$ concavity gives $F_{i}/s_{i}\geq F_{N}$ and hence $D_{i}\geq F_{i}-s_{i}\beta\geq F_{i}/2\text{.}$ Also $D_{i}\leq F_{N}<\tau\text{.}$ The source term in Equation 17 is nonnegative, so

$$\Psi(s_{i},F_{i},\kappa_{i},\mu)\leq\mathrm{e}^{-\tau}\frac{F_{i}\tau}{D_{i}}\leq 2\tau\mathrm{e}^{-\tau}\leq 2H\mathrm{e}^{-H}<P_{1}\leq P_{i}.$$

Thus only the closed interval $[L,H]$ requires subdivision.

On each control cell, outward interval arithmetic first encloses $\beta\text{.}$ The function $D_{i}$ is nonincreasing in $\beta\text{,}$ so its two endpoint evaluations enclose the finite maximum in Equation 19. The source $\widehat{f}$ increases, so its endpoint values enclose the source term in Equation 17. A cell is accepted if it proves $D_{i}=0$ throughout, or a nonpositive numerator in Equation 17 wherever $D_{i}>0\text{.}$ Points with $D_{i}=0$ already satisfy the terminal condition, since $F_{i}<A+s_{i}B\text{.}$ Otherwise the checker requires a positive lower bound for $\kappa_{i}$ before division, and accepts the cell only if it proves $P_{i}-\Psi>0\text{.}$ All other cells are subdivided. The checker verifies an exact closed cover of $[L,H]$ for every positive partition point. Theorem 4 then covers all weights, source supports, stopping ratios, endpoint ratios, and finite depths.

6 The Diagonal Bounds

We now evaluate the two sides of Equation 15. The repository specifies the source, a finite upper proof, and a rational lower polygon, with the following complete interval covers.

Construction	Size	Checked Cells
Source profile	2,279 cubic pieces	12,319
Upper proof	2,932 routes; depth 27	1,065,279
Lower polygon	2,048 affine pieces	853,082

The source inequalities are evaluated at 224 fractional binary bits, and the upper and lower inequalities at both 224 and 320 bits. An independent integer-prefix computation reproduces every route height and loss from the checked primitive enclosures. The repository contains the four rational inputs in data/, the Python certificate checker in proof/, and the Lean formalization in lean/. The Python checker uses only the standard library. Running python proof/replay.py --workers 16 from the repository root writes the results and input and code hashes to proof/build/replay/. The root README.md gives the build and reproduction instructions for both projects.

6.1 Exact Interval Arithmetic

Transcendental quantities are enclosed by intervals whose endpoints are integers divided by a fixed power of two. Rational inputs and arithmetic operations are rounded outward. For logarithms, reduce to $y\in[1,2]$ and put $u=(y-1)/(y+1)\in[0,1/3]\text{.}$ Then

$$\log y=2\sum_{j=0}^{m-1}\frac{u^{2j+1}}{2j+1}+E_{m},\qquad 0\leq E_{m}\leq\frac{3u^{2m+1}}{2m+1}.$$

(22)

The tail is at most $2u^{2m+1}/((2m+1)(1-u^{2}))\text{.}$ The same formula encloses $\log 2$ for reversing the scaling.

For the exponential, put $S_{m}(z)=\sum_{j=0}^{m}z^{j}/j!$ for integers $m\geq 0\text{.}$ Successive terms after degree $m$ have ratio at most $z/(m+2)\text{,}$ so a geometric sum gives

$$S_{m}(z)<\mathrm{e}^{z}\leq S_{m}(z)+\frac{z^{m+1}}{(m+1)!}\frac{1}{1-z/(m+2)}\qquad(0<z<m+2).$$

(23)

At zero the sum is exact. The interval backend reduces a nonnegative argument to $0\leq y\leq 1/8\text{,}$ where this tail is at most $2y^{m+1}/(m+1)!\text{.}$ Repeated squaring reverses the scaling; negative arguments use outward reciprocals. A logarithm or division is accepted only after its domain is checked. For the final exponential bound, a separate rational calculation applies Equation 23 directly with $m=140$ and compares the upper enclosure with $3.69507\text{.}$

6.2 A Finite Upper Proof

The file data/upper.json contains an ordered list of routes. A call names either a weighted source bound or an earlier route with a terminal control. All 2,932 routes are reachable from the root. The density, height, and output margins are all $10^{-12}\text{;}$ rounding is upward to multiples of $10^{-12}\text{.}$ The source pairs use the inward factor $1-10^{-9}\text{,}$ and we put $\rho=1-2\cdot 10^{-5}\text{.}$

For each control, let $\pi$ be its leaf density when it defines a source bound, and $\rho p$ when it ends a route at outer density $p\text{.}$ The checker computes

$$x_{\rm cap}=(1-\mu)^{w}\pi^{1/(1-\mu)}$$

and an explicit boundary pair $(\xi,\zeta)$ with $\xi\geq x_{\rm cap}\text{.}$ It uses $x=(1-10^{-9})x_{\rm cap}$ and $y=(1-10^{-9})\zeta\text{.}$ Both coordinates are strictly below a supporting pair, and $x<x_{\rm cap}\text{,}$ so the control is strict. Source supports are given explicitly in the data and checked by interval inequalities. For a source leaf, the coefficients $A/(1+w)$ and $(B+w\beta)/(1+w)$ are rounded upward, giving a rational affine upper bound for Equation 7.

For a reused child with stored profile $g$ and stopping ratio $\theta\text{,}$ upward enclosures of its terminal parameters give rational coefficients

$$a^{+}\geq\frac{A+w\theta\beta}{1+w},\qquad b^{+}\geq\frac{B}{1+w},\qquad t^{+}\geq\frac{w}{1+w}.$$

Its called family is

$$\max\{g(s),\ a^{+}+b^{+}s+t^{+}(g(s)-g(\theta))\}+10^{-12}.$$

This dominates Equation 13 with a strict allowance; rounding $t^{+}$ upward is valid because $g(s)-g(\theta)\geq 0\text{.}$

With these calls established, the parent route can be evaluated. Each slope is an upward-rounded enclosure of $c(q_{j})+10^{-12}\text{,}$ so the sums defining $S$ are rational. The endpoint rule in Equation 11 gives $M$ by checking the parent cell endpoints and the child profiles’ partition points inside each cell. With $Q=10^{12}\text{,}$ the parent stores

$$H=\frac{\lceil Q(M+10^{-12})\rceil}{Q},\qquad g=H-S.$$

The checker also verifies the full partitions, density implications, child-domain containment, and strict ordering of child indices. Induction through the ordered list and Theorem 2 prove all its conditional families.

The root has

$$p=\tfrac{1}{2},\qquad\pi=\tfrac{1}{2}-10^{-5},\qquad\theta=\tfrac{1}{4},\qquad\lambda=1,\qquad w=1,\qquad\mu=\tfrac{1}{4}.$$

It has 65 cells and uses the left source support in data/upper.json, at $t\approx 0.383\text{.}$ Exact reconstruction gives $g(1)>1.3\text{,}$ while the second term in Equation 13 is less than $1.2\text{.}$ Adding the output margin yields the constant conditional exponent

$$z_{+}=g(1)+10^{-12}=\frac{1306998938417}{10^{12}}.$$

Thus $U_{\infty}(1/2,1)\leq z_{+}\text{.}$

6.3 The Lower Polygon

The file data/lower.json specifies the 2,049 rational partition points and values of $F\text{.}$ Exact rational comparisons establish positive decreasing slopes and $F(0)=0\text{.}$ Interval comparisons establish Equation 18, the tail bounds in Equation 21, and $d_{N}>\log 2\text{.}$ The 853,082 closed control cells verify Equation 20. Theorem 4 and Proposition 5.1 therefore give

$$z_{-}:=F(1)=\frac{1305460275532}{10^{12}}\leq U_{\infty}(1/2,1).$$

The resulting interval for the finite-proof optimum has width less than $0.0016\text{.}$

Proof of Theorem 1.

The finite upper proof and the preserved polygon give Equation 1 by Theorem 3. In an arbitrary two-coloring, interchange the colors so that red has density at least $1/2$ and apply the root conditional bound. Its constant is

$$\frac{3(1-\pi)}{1/2-\pi}=150003,$$

which proves Equation 2. An independent rational Taylor enclosure gives $\mathrm{e}^{z_{+}}<3.69507\text{.}$ This strict gap absorbs the fixed factor and the ceiling for sufficiently large $k\text{.}$ ∎

References

[CGMS] M. Campos, S. Griffiths, R. Morris, and J. Sahasrabudhe, An exponential improvement for diagonal Ramsey, Annals of Mathematics 203 (2026), no. 3, 869–932. https://doi.org/10.4007/annals.2026.203.3.4.

[Con] D. Conlon, A new upper bound for diagonal Ramsey numbers, Annals of Mathematics 170 (2009), no. 2, 941–960. https://doi.org/10.4007/annals.2009.170.941.

[Er] P. Erdős, Some remarks on the theory of graphs, Bulletin of the American Mathematical Society 53 (1947), 292–294. https://doi.org/10.1090/S0002-9904-1947-08785-1.

[ES] P. Erdős and G. Szekeres, A combinatorial problem in geometry, Compositio Mathematica 2 (1935), 463–470. https://www.numdam.org/item/CM_1935__2__463_0/.

[GNNW] P. Gupta, Ndiamé Ndiaye, S. Norin, and L. Wei, Optimizing the CGMS upper bound on Ramsey numbers, arXiv:2407.19026v2, revised 29 August 2026. https://arxiv.org/abs/2407.19026v2.

[Ram] F. P. Ramsey, On a problem of formal logic, Proceedings of the London Mathematical Society (2) 30 (1930), 264–286. https://doi.org/10.1112/plms/s2-30.1.264.

[Sah] A. Sah, Diagonal Ramsey via effective quasirandomness, Duke Mathematical Journal 172 (2023), no. 3, 545–567. https://doi.org/10.1215/00127094-2022-0048.

[Spe] J. Spencer, Ramsey’s theorem—a new lower bound, Journal of Combinatorial Theory, Series A 18 (1975), 108–115. https://doi.org/10.1016/0097-3165(75)90071-0.

[Tarski] A. Tarski, A lattice-theoretical fixpoint theorem and its applications, Pacific Journal of Mathematics 5 (1955), 285–309. https://msp.org/pjm/1955/5-2/pjm-v5-n2-p11-s.pdf.

[Tho] A. Thomason, An upper bound for some Ramsey numbers, Journal of Graph Theory 12 (1988), no. 4, 509–517. https://doi.org/10.1002/jgt.3190120406.

Appendix A The Finite-Host Weighted Inequality

We prove the estimate used in Lemma 2.1. Its stopping assumption concerns only proper subsets of the host. This lets the same estimate prove the source bound by a smallest-counterexample argument in Appendix B.

Lemma A.1 (Finite-Host Weighted Candidate Bound).

Fix $w>0$ and parameters in $(0,1)$ satisfying

$$x_{0}<\widehat{x},\qquad y_{0}<\widehat{y},\qquad x_{0}<(1-\mu_{0})^{w}p^{1/(1-\mu_{0})}.$$

There is $L_{0}\text{,}$ depending only on these parameters, with the following property, uniformly in $C\geq 1$ and the finite host coloring $H\text{.}$ Fix $\ell\geq L_{0}\text{,}$ and suppose that for every integer $a\geq 1\text{,}$ every proper subset $Z\subsetneq V(H)$ of size

$$|Z|\geq C\widehat{x}^{-a}\widehat{y}^{-\ell}$$

(24)

contains a red $K_{a}$ or a blue $K_{\ell}\text{.}$ Then, for all $k,t\geq 1\text{,}$ every candidate in $H$ satisfying

$$d_{R}(X,Y)\geq p,\qquad|X|^{w}|Y|\geq C^{w+1}x_{0}^{-k}y_{0}^{-\ell}\mu_{0}^{-wt}$$

(25)

is $(k,\ell,t)$-good.

Proof.

We adapt the candidate induction of [GNNW, Lemma 12] to unequal weights. After removing rows with too few red neighbors, we either pass to a common blue neighborhood or choose a vertex whose red or blue child satisfies the induction hypothesis. A power of the excess red density controls the size loss in these steps.

Choose a fixed integer $r>\max\{1,w\}$ so large that $p^{1/r}>\mu_{0}$ and

$$x_{0}<(p^{1/r}-\mu_{0})^{r}(1-\mu_{0})^{w-r}.$$

The logarithm of the right side tends to $w\log(1-\mu_{0})+\log p/(1-\mu_{0})\text{.}$ The function

$$x_{0}^{1/r}(1-q)^{1-w/r}+\mu_{0}^{w/r}q^{1-w/r}$$

increases until $q=\mu_{0}/(\mu_{0}+x_{0}^{1/w})>\mu_{0}\text{,}$ and is less than $p^{1/r}$ at $q=\mu_{0}\text{.}$ By continuity choose

$$x_{0}<x<\widehat{x},\quad y_{0}<y<\widehat{y},\quad\mu_{0}<\mu<\beta<1,\quad 0<\delta<p/2,$$

and $\gamma>0$ so that, with $\bar{p}=p-\delta\text{,}$

$$\max_{0\leq q\leq\beta}\{x^{1/r}(1-q)^{1-w/r}+\mu^{w/r}q^{1-w/r}\}\leq(1-\gamma)\bar{p}^{1/r}.$$

(26)

All these choices precede $C,H,k,\ell,t\text{.}$

Put $n=k+t\text{,}$ $\delta_{n}=\delta/n\text{,}$ and induct on $n\text{,}$ for fixed sufficiently large $\ell\text{,}$ using the stronger hypothesis

$$d_{R}(X,Y)\geq p-\delta_{n},\qquad Q_{n}=(d_{R}(X,Y)-p+\delta_{n})^{r}|X|^{w}|Y|\geq C^{w+1}x^{-k}y^{-\ell}\mu^{-wt}.$$

(27)

After canceling $C^{w+1}\text{,}$ the conditions in Equation 25 imply this once $\ell$ is large, uniformly in $k,t\text{,}$ because

$$(\delta/n)^{r}(x/x_{0})^{k}(y/y_{0})^{\ell}(\mu/\mu_{0})^{wt}\geq 1.$$

The factors involving $k,t$ grow exponentially in $n\text{,}$ so their ratio to $n^{r}$ has a positive infimum; increasing $\ell$ makes the inequality hold. If $k=1$ or $t=1\text{,}$ any vertex of $X$ suffices. Henceforth $k,t\geq 2\text{,}$ and the density excess lies in $(0,1)\text{.}$

Regularize and stop if $Y$ is large. Delete rows of red degree less than $(p-\delta_{n})|Y|\text{.}$ The excess edge count $E=e_{R}(X,Y)-(p-\delta_{n})|X||Y|$ increases, and

$$Q_{n}=E^{r}|X|^{w-r}|Y|^{1-r}$$

does not decrease, since $r>w\text{.}$ Positive excess prevents deletion of the last row. Every nonempty $T\subseteq X$ now has $d_{R}(T,Y)\geq p-\delta_{n}\text{.}$ If $|Y|\geq C\widehat{x}^{-k}\widehat{y}^{-\ell}\text{,}$ apply the stopping assumption with $a=k\text{;}$ $Y$ is proper because $X\neq\varnothing\text{.}$ Otherwise $Q_{n}\leq|X|^{w}|Y|$ implies

$$|X|^{w}>C^{w}(\widehat{x}/x)^{k}(\widehat{y}/y)^{\ell}\mu^{-wt},\qquad|X|\geq C\exp(c(k+\ell+t))$$

(28)

for a fixed $c>0\text{.}$

Find a common blue neighborhood or a suitable vertex. Define

$$b=\left\lceil\frac{2r\log n+r\log(1/\delta)+w\log 2}{w\log(\beta/\mu)}\right\rceil,\qquad m=\lceil 10\beta^{-1}b^{2}\rceil,$$

and $W=\{v\in X:\deg_{B}(v,X)\geq\beta|X|\}\text{.}$ Put $\xi_{n}=\delta/(2n^{2})\text{.}$ We choose $L_{0}$ once so that

$$|X|\geq 5m^{2},\qquad R(k,m)<\bar{p}\xi_{n}|X|,\qquad\frac{2n^{2}}{\delta|X|}<\gamma.$$

To justify a uniform choice, note that $b=O(\log n)\text{,}$ $m=O((\log n)^{2})\text{,}$ and $R(k,m)\leq\binom{k+m-2}{m-1}\leq k^{m-1}\text{;}$ the last inequality counts nondecreasing tuples in $\{1,\ldots,k\}^{m-1}\text{.}$ The logarithm of each required lower bound on $|X|$ is therefore $O((\log n)^{3})\text{.}$ Subtracting $cn$ leaves a function bounded above in $n\text{,}$ so Equation 28 supplies all three bounds for one $L_{0}\text{,}$ independently of $C\geq 1,k,t\text{.}$ We use the common-neighborhood averaging argument of [GNNW, Lemma 9]. When $|W|\geq R(k,m)\text{,}$ either a red $K_{k}$ finishes the proof, or $W$ contains a blue $m$-clique $U\text{.}$ Put $N_{X}=|X|\text{.}$ The blue density $\sigma$ between $U$ and $X\setminus U$ satisfies

$$\sigma\geq\frac{\beta N_{X}-m}{N_{X}-m}=\beta-\frac{(1-\beta)m}{N_{X}-m}\geq\beta-\frac{1}{4m}.$$

Choose a uniform $b$-subset of $U\text{,}$ and let $T$ be its common blue neighborhood in $X\setminus U\text{.}$ Convexity of the piecewise-linear extension of $j\mapsto\binom{j}{b}$ on nonnegative integers gives

$$\mathbb{E}|T|\geq(N_{X}-m)\frac{\binom{\lfloor\sigma m\rfloor}{b}}{\binom{m}{b}}\geq(N_{X}-m)(\sigma-b/m)^{b}.$$

The second inequality follows factor by factor from the binomial product. Since $m\geq 10\beta^{-1}b^{2}\text{,}$

$$\sigma-b/m\geq\beta\left(1-\frac{1}{8b}\right)>0,\qquad N_{X}-m\geq\frac{4}{5}N_{X}.$$

Bernoulli’s inequality gives $\mathbb{E}|T|\geq(7/10)\beta^{b}N_{X}\text{.}$ Thus some blue $b$-clique $S\subseteq U$ has a disjoint common blue neighborhood $T$ with $|T|\geq\beta^{b}|X|/2\text{.}$

If $b\geq t\text{,}$ then $S$ already contains the required blue clique. For $b<t\text{,}$ we have

$$\delta_{n-b}-\delta_{n}\geq\delta/n^{2},\qquad Q_{n-b}(T,Y)\geq(\delta/n^{2})^{r}2^{-w}\beta^{wb}|X|^{w}|Y|\geq C^{w+1}x^{-k}y^{-\ell}\mu^{-w(t-b)}.$$

The last inequality uses $Q_{n}\leq|X|^{w}|Y|$ and $(\beta/\mu)^{wb}\geq 2^{w}(n^{2}/\delta)^{r}\text{,}$ which follows from the choice of $b\text{.}$ Induction applies; a blue $K_{t-b}$ in $T$ joins $S$ to form a blue $K_{t}\text{,}$ and either of the other two outcomes already suffices.

We may now assume $|W|<R(k,m)<\bar{p}\xi_{n}|X|\text{.}$ Put $Y_{v}=N_{R}(v)\cap Y\text{,}$ $d=d_{R}(X,Y)\text{,}$ and $E_{0}=e_{R}(X,Y)\text{.}$ Row regularization gives $|Y_{v}|\geq\bar{p}|Y|>0$ for every $v\in X\text{.}$ Counting by vertices in $Y$ and applying Cauchy–Schwarz gives

$$\sum_{v\in X}d_{R}(X,Y_{v})|Y_{v}|=\frac{1}{|X|}\sum_{y\in Y}\deg_{R}(y,X)^{2}\geq dE_{0}.$$

Removing the terms indexed by $W$ loses at most $|W||Y|\text{,}$ so

$$\sum_{v\notin W}d_{R}(X,Y_{v})|Y_{v}|\geq dE_{0}-|W||Y|>(d-\xi_{n})E_{0}\geq(d-\xi_{n})\sum_{v\notin W}|Y_{v}|.$$

Here $E_{0}\geq\bar{p}|X||Y|$ and $d-\xi_{n}>0\text{.}$ Hence some $v\notin W$ satisfies $d_{R}(X,Y_{v})\geq d-\xi_{n}\text{.}$

Pass to a red or blue child. Set $Y^{\prime}=Y_{v}\text{,}$ $X_{R}=N_{R}(v)\cap X\text{,}$ $X_{B}=N_{B}(v)\cap X\text{,}$ and

$$a=|X_{R}|/|X|,\quad q=|X_{B}|/|X|,\quad\alpha=d_{R}(X,Y^{\prime})-p+\delta_{n-1}.$$

Then $a+q\leq 1\text{,}$ $q\leq\beta\text{,}$ and $\alpha\geq d-p+\delta_{n}+\delta/(2n^{2})>0\text{.}$ For $\chi\in\{R,B\}\text{,}$ set $\alpha_{\chi}=d_{R}(X_{\chi},Y^{\prime})-p+\delta_{n-1}$ when $X_{\chi}\neq\varnothing\text{,}$ and set $\alpha_{\chi}=0$ otherwise. Write $u_{+}=\max\{u,0\}$ and

$$Q_{\chi}=(\alpha_{\chi})_{+}^{r}|X_{\chi}|^{w}|Y^{\prime}|.$$

Splitting $X$ into its two children and the root gives

$$1\leq a\frac{(\alpha_{R})_{+}}{\alpha}+q\frac{(\alpha_{B})_{+}}{\alpha}+\frac{1}{\alpha|X|}.$$

(29)

The root’s contribution is $(1-p+\delta_{n-1})/(\alpha|X|)\leq 1/(\alpha|X|)\text{.}$ We claim that

$$Q_{R}\geq xQ_{n}\qquad\text{or}\qquad Q_{B}\geq\mu^{w}Q_{n}.$$

If both inequalities failed, then $Q_{n}\leq\alpha^{r}|X|^{w}|Y|\text{,}$ $|Y^{\prime}|\geq\bar{p}|Y|\text{,}$ and Equation 29 would give

$$1\leq\bar{p}^{-1/r}\{x^{1/r}a^{1-w/r}+\mu^{w/r}q^{1-w/r}\}+\frac{1}{\alpha|X|}\leq 1-\gamma+\frac{2n^{2}}{\delta|X|}<1.$$

The second inequality uses $a\leq 1-q$ and Equation 26; the last is one of our three size bounds. Thus one child has a positive value of $Q_{n-1}$ meeting Equation 27, with $k$ reduced by one in the red child or $t$ reduced by one in the blue child. Induction applies. The root extends a red $K_{k-1}$ in $X_{R}\cup Y^{\prime}\text{,}$ or a blue $K_{t-1}$ in $X_{B}\text{,}$ respectively. Every other good outcome already gives a required clique. ∎

Corollary A.2 (Local Density Bound).

Under the stopping assumption of Lemma A.1, fix $w>0$ and $\mu,p,x,y\in(0,1)$ with $x<\widehat{x}\text{,}$ $y<\widehat{y}\text{,}$ and $x<(1-\mu)^{w}p^{1/(1-\mu)}\text{.}$ If a host has red density at least $p$ and order

$$N\geq Cx^{-k/(w+1)}(y\mu^{w})^{-\ell/(w+1)},$$

(30)

then it contains a red $K_{k}$ or a blue $K_{\ell}$ for all sufficiently large $\ell\text{,}$ with a threshold independent of $C\geq 1$ and $k\text{.}$

Proof.

Choose $y<y_{1}<\widehat{y}\text{.}$ A uniformly chosen balanced partition has expected cross-density equal to the host density, so some disjoint $X,Y\text{,}$ each of size at least $N/3\text{,}$ satisfy $d_{R}(X,Y)\geq p\text{.}$ For large $\ell\text{,}$ $(y_{1}/y)^{\ell}\geq 3^{w+1}\text{.}$ The order bound therefore supplies Equation 25 at $(x,y_{1})$ with $t=\ell\text{,}$ and Lemma A.1 applies. ∎

Appendix B Admissibility of the Source

We first give a criterion for a concave Ramsey profile, then verify it for the specified function $f\text{.}$ For a continuous $F$ on $[0,1]$ with $F(0)=0\text{,}$ define

$$G_{F}(a,b)=\max(a,b)F\!\left(\frac{\min(a,b)}{\max(a,b)}\right),$$

$$\mathcal{B}_{F}=\{(x,y)\in(0,1)^{2}:G_{F}(a,b)\leq-a\log x-b\log y\text{ for all }a,b>0\}.$$

Thus $\mathcal{B}_{F}$ consists of the exponential bounds that dominate the profile in both coordinates. For $F=f\text{,}$ this agrees with the source pairs in Section 2, since $G_{f}(a,b)=b\widehat{f}(a/b)\text{.}$ In the source criterion, $X,Y$ denote scalar support parameters.

Lemma B.1 (Source Criterion).

Suppose $F$ is continuous on $[0,1]\text{,}$ $F(0)=0\text{,}$ is $C^{1}$ and concave on $(0,1]\text{,}$ and has $F^{\prime}>0\text{.}$ If for every $s\in(0,1]$ there are $w>0\text{,}$ $\mu,Y\in(0,1)$ with

$$p_{0}(s)=1-\mathrm{e}^{-F^{\prime}(s)},\quad X=(1-\mu)^{w}p_{0}(s)^{1/(1-\mu)},\quad(X,Y)\in\mathcal{B}_{F},$$

and

$$(w+1)F(s)+\log X+s\log Y+ws\log\mu>0,$$

(31)

then, for every $\varepsilon>0\text{,}$ some $C_{\varepsilon}\geq 1$ satisfies

$$R(a,b)\leq\left\lceil C_{\varepsilon}\exp\bigl(G_{F}(a,b)+\varepsilon(a+b)\bigr)\right\rceil\qquad(a,b\in\mathbb{Z}_{\geq 1}).$$

(32)

Proof.

Fix $\varepsilon>0\text{.}$ We first choose finitely many local bounds, then apply one of them to a smallest counterexample. Since $F(0)=0$ and $F^{\prime}>0\text{,}$ we have $F\geq 0\text{.}$ Choose $0<\alpha<1$ with $h(\alpha)<\varepsilon/2\text{;}$ the classical bound will handle smaller ratios. At each $s_{0}\in[\alpha,1]\text{,}$ choose $w,\mu,X,Y$ as in the hypothesis and put $\bar{x}=\mathrm{e}^{-\varepsilon}X\text{,}$ $\bar{y}=\mathrm{e}^{-\varepsilon}Y\text{.}$ Scaling the controls this way and replacing $F(s_{0})$ by $F(s_{0})+\varepsilon(1+s_{0})$ increases the slack by $w\varepsilon(1+s_{0})\text{.}$ We may therefore choose

$$0<x<\widehat{x}<\bar{x},\qquad 0<y<\widehat{y}<\bar{y},\qquad 0<p<p_{0}(s_{0}),$$

close enough to $(\bar{x},\bar{y},p_{0}(s_{0}))$ that $x<(1-\mu)^{w}p^{1/(1-\mu)}$ and

$$H(t):=-\frac{\log x+t\log(y\mu^{w})}{w+1},\qquad H(s_{0})<F(s_{0})+\varepsilon(1+s_{0}).$$

Continuity gives a neighborhood on which this inequality holds and $p_{0}(s)>p+2\eta_{s_{0}}$ for some $\eta_{s_{0}}>0\text{.}$ A finite subcover gives $\eta>0$ and a common threshold $L\geq 2$ for Corollary A.2, both independent of $C\text{.}$ At every $s\in[\alpha,1]\text{,}$ one selected local bound satisfies $H(s)<F(s)+\varepsilon(1+s)$ and $p_{0}(s)>p+2\eta\text{.}$ Every selected pair also satisfies

$$-a\log\widehat{x}-b\log\widehat{y}\geq G_{F}(a,b)+\varepsilon(a+b)\qquad(a,b>0),$$

(33)

by the support condition and the scaling by $\mathrm{e}^{-\varepsilon}\text{.}$

Choose an integer $K$ with $\alpha K\geq L\text{,}$ and then $C\geq\max\{3,2/\eta,4^{K}\}\text{.}$ Suppose there is a coloring with no required clique at an integer order

$$N\geq C\exp\bigl(G_{F}(a,b)+\varepsilon(a+b)\bigr).$$

Choose one of smallest order among all parameter pairs and colorings. Interchange the colors if needed and write its targets as $k\geq\ell\text{.}$ The classical bound [ES] excludes $k<K\text{.}$ It also excludes $\ell/k\leq\alpha\text{,}$ since

$$\log R(k,\ell)\leq kh(\ell/k)\leq kh(\alpha)<\varepsilon k/2\leq kF(\ell/k)+\varepsilon(k+\ell)+\log C.$$

Thus $s=\ell/k\in[\alpha,1]$ and $\ell\geq L\text{.}$

For every integer $a\geq 1\text{,}$ minimality and Equation 33 give a red $K_{a}$ or a blue $K_{\ell}$ in each proper subset satisfying Equation 24. Thus the stopping assumption of Corollary A.2 holds for the local bound selected at $s\text{.}$ If the host’s red density is at least $p_{0}(s)-\eta>p\text{,}$ then $N>C\mathrm{e}^{kH(s)}\text{,}$ and Corollary A.2 gives a required clique.

Otherwise its blue density exceeds $q=\mathrm{e}^{-F^{\prime}(s)}+\eta<1\text{.}$ A blue neighborhood $Z$ has size at least $q(N-1)>N\mathrm{e}^{-F^{\prime}(s)}\text{,}$ since $N\eta>1\text{.}$ Concavity gives

$$k\bigl(F(s)-F(s-1/k)\bigr)\geq F^{\prime}(s).$$

Consequently

$$|Z|>N\mathrm{e}^{-F^{\prime}(s)}\geq\mathrm{e}^{\varepsilon}C\exp\bigl(kF(s-1/k)+\varepsilon(k+\ell-1)\bigr).$$

The proper subset $Z$ therefore has order above the proposed $(k,\ell-1)$ threshold. Minimality gives a red $K_{k}$ or a blue $K_{\ell-1}\text{;}$ in the latter case, add the chosen vertex. Either outcome contradicts the choice of the host.

Hence Equation 32 holds. For $1\leq b\leq a\text{,}$ its logarithm is at most $aF(b/a)+2\varepsilon a+\log(2C_{\varepsilon})\text{.}$ Since $\varepsilon$ is arbitrary, the error is uniform in $b\text{.}$ ∎

B.1 The Piecewise Cubic Profile

We now verify the criterion for $f$ in Equation 3. The function $P$ has 2,279 rational cubic pieces. The file data/source.json gives the rational partition points, values, and derivatives of $P\text{.}$ On a cell $[a,b]\text{,}$ set $u=t-a\text{,}$ $\delta=b-a\text{,}$ and $\Delta=P(b)-P(a)\text{.}$ Its exact polynomial is

$$P(a+u)=P(a)+P^{\prime}(a)u+\frac{3\Delta/\delta-2P^{\prime}(a)-P^{\prime}(b)}{\delta}u^{2}+\frac{P^{\prime}(a)+P^{\prime}(b)-2\Delta/\delta}{\delta^{2}}u^{3}.$$

Exact endpoint identities ensure that $P$ is globally $C^{1}\text{.}$ For $t>0$ in the interior of a piece, with $j(t)=t\mathrm{e}^{-t}P(t)\text{,}$

$$j^{\prime\prime}(t)=\mathrm{e}^{-t}\bigl((t-2)P+(2-2t)P^{\prime}+tP^{\prime\prime}\bigr),\qquad f^{\prime\prime}(t)=-\frac{1}{t(1+t)}+j^{\prime\prime}(t).$$

The cover in data/source_cells.json proves $1-t(1+t)j^{\prime\prime}(t)>0$ on each closed piece, using that piece’s endpoint derivatives, together with $f^{\prime}(1)>0$ and $2f^{\prime}(1)>f(1)\text{.}$ Consequently $f(0)=0\text{,}$ $f$ is continuous on $[0,1]\text{,}$ $C^{1}$ on $(0,1]\text{,}$ strictly concave and increasing.

To describe the boundary of $\mathcal{B}_{f}\text{,}$ put $x_{f}(t)=\mathrm{e}^{-f^{\prime}(t)}$ and $y_{f}(t)=\mathrm{e}^{tf^{\prime}(t)-f(t)}\text{.}$ Their endpoint limits are

$$x_{f}(t)=\frac{t}{1+t}\mathrm{e}^{-j^{\prime}(t)}\longrightarrow 0,\qquad y_{f}(t)=\frac{1}{1+t}\mathrm{e}^{tj^{\prime}(t)-j(t)}\longrightarrow 1.$$

Strict concavity makes $x_{f}$ increasing and $y_{f}$ decreasing. Also $x_{f}(t)<y_{f}(t)\text{:}$ the function $(1+t)f^{\prime}(t)-f(t)$ decreases to its positive value at $t=1\text{.}$ Writing $x_{1}=x_{f}(1)$ and $y_{1}=y_{f}(1)\text{,}$ the upper boundary is

$$Y_{f}(x)=\begin{cases}y_{f}(t),&0<x<x_{1},\quad x_{f}(t)=x,\\ \mathrm{e}^{-f(1)}/x,&x_{1}\leq x\leq y_{1},\\ x_{f}(t),&y_{1}<x<1,\quad y_{f}(t)=x.\end{cases}$$

(34)

These are the supporting lines of $\widehat{f}$ on its two smooth branches and at its corner. Thus $b(A)=-\log Y_{f}(\mathrm{e}^{-A})\text{.}$

B.2 The Source Inequalities

The file data/source_cells.json supplies a cover of 12,319 closed cells and rational controls $r,w>0$ on each cell. Set $\mu=tr\text{,}$ take $X$ as in Lemma B.1, and choose $Y=Y_{f}(X)\text{.}$ With $\Delta_{f}$ denoting the left side of Equation 31, the cell inequalities give

$$\Delta_{f}(t)/t>10^{-8}\qquad(0<t\leq 1).$$

The cell inequalities also establish $0<\mu,p_{0}<1$ for positive $t\text{.}$ Since Lemma B.1 requires controls at each ratio, they may be chosen separately on each cell.

Near zero, we cancel the logarithmic terms before evaluating the slack. Define

$$L_{+}(u)=\frac{\log(1+u)}{u},\qquad L_{-}(u)=\frac{-\log(1-u)}{u},\qquad L_{+}(0)=L_{-}(0)=1,$$

$$U=(1+t)L_{+}(t)+\mathrm{e}^{-t}P(t),\quad u=\frac{\mathrm{e}^{-j^{\prime}(t)}}{1+t},\quad V=wrL_{-}(tr)+\frac{uL_{-}(tu)}{1-tr},$$

$$J(s)=L_{+}(s)+s\mathrm{e}^{-s}(P(s)-P^{\prime}(s)).$$

Then $f(t)/t=-\log t+U$ and $\log X=-tV\text{.}$ On the right supporting branch, write $X=y_{f}(t\sigma)\text{,}$ $Y=x_{f}(t\sigma)\text{.}$ Here $\log Y=\log t+\log\sigma-\log(1+t\sigma)-j^{\prime}(t\sigma)$ and $\log\mu=\log t+\log r\text{.}$ The terms in $\log t$ cancel in the normalized slack, giving

$$\begin{aligned} \displaystyle\sigma J(t\sigma) & \displaystyle=V, \\ \displaystyle\Delta_{f}(t)/t & \displaystyle=(w+1)U-V+\log\sigma-\log(1+t\sigma)-j^{\prime}(t\sigma)+w\log r. \end{aligned}$$

As $t\downarrow 0\text{,}$ $X\to 1$ implies $t\sigma\to 0\text{,}$ since $y_{f}$ is strictly decreasing with limit 1 only at zero. Thus $\sigma J(t\sigma)=V$ and $J(0)=1$ give the continuous extension $\sigma(0)=V(0)=wr+\mathrm{e}^{-P(0)}>0\text{.}$ On each cell near zero, the checker verifies the branch, a positive bracket for $\sigma\text{,}$ and $t\sigma\leq 1\text{.}$ The root is unique: $\sigma\mapsto\sigma J(t\sigma)=-\log y_{f}(t\sigma)/t$ is strictly increasing for $t>0\text{,}$ and is the identity at zero.

The verifier evaluates complete finite polynomial expansions. Source inverse brackets are accepted only after outward-rounded endpoint inequalities prove them. When $t$ and $t\sigma(t)$ lie in the interiors of source pieces, possibly different ones,

$$\sigma^{\prime}=\frac{V^{\prime}-\sigma^{2}J^{\prime}(t\sigma)}{J(t\sigma)+t\sigma J^{\prime}(t\sigma)},$$

with a checked positive denominator. The integral representations

$$L_{+}(u)=\int_{0}^{1}\frac{da}{1+ua},\qquad L_{-}(u)=\int_{0}^{1}\frac{da}{1-ua}$$

give endpoint enclosures for these functions and their derivatives, including at zero. Away from zero, the function $u\mapsto\log Y_{f}(\mathrm{e}^{u})\text{,}$ with $u<0\text{,}$ has derivative $-s\text{,}$ $-1\text{,}$ or $-1/s$ on the three branches of Equation 34, where $s$ is the supporting parameter. These derivatives agree at the branch junctions.

Together with Equations 22 and 23, these identities enclose each slack and its derivative. An enclosure at the cell center plus the derivative interval times the displacement encloses the slack throughout the cell. When a cell crosses a source partition point, enclosures from every adjoining piece are combined. The continuous compositions are locally Lipschitz, and the stated derivative bounds hold almost everywhere; integrating those bounds justifies the same enclosure across a junction. The complete cover verifies the hypotheses of Lemma B.1, which gives Equation 4 with an error uniform in $b\text{.}$

A Bound Below 2.8 for Tuza's Conjecture

Sat, 12 Sep 2026 00:00:00 +0000

1 Introduction

A triangle packing in a graph $G$ is a set of pairwise edge-disjoint triangles, and a triangle transversal is a set of edges meeting every triangle. Let $\nu(G)$ and $\tau(G)$ denote their maximum and minimum sizes. A transversal contains an edge of every packed triangle, and the edges of a maximum packing form a transversal, so $\nu(G)\leq\tau(G)\leq 3\nu(G)\text{.}$ Tuza [Tuz90] conjectured that $\tau(G)\leq 2\nu(G)$ for every graph $G\text{.}$ The constant $2$ would be best possible, as $K_{4}$ and $K_{5}$ show.

Haxell [Hax99] proved $\tau(G)\leq(66/23)\,\nu(G)\text{,}$ where $66/23\approx 2.870\text{,}$ and noted without proof that the argument can be pushed to $(1+\sqrt{481})/8\approx 2.866\text{.}$ Very recently, Yi [Yi26] lowered the constant to $(162+4\sqrt{3})/59\approx 2.863$ by proving $\tau\leq(1+\sqrt{3})\nu$ for every family of triangles that admits a red-blue edge coloring with one red edge per triangle, and observed that this route alone cannot bring the constant below $54/19\approx 2.842\text{.}$ Independently of Yi’s work, we prove the following bound, whose constant $165/59\approx 2.797$ crosses this barrier and lies below $2.8\text{.}$

Theorem 1.

Every finite simple graph $G$ satisfies

$$\tau(G)\leq\frac{165}{59}\nu(G).$$

Our starting point is the following exchange property. Let the edges of a graph be colored red and blue, and consider the family of all triangles with exactly one red edge. Fix a maximum packing in this family and delete the blue edges of its triangles. By maximality, every triangle of the family that survives meets the packing only in its red edge. Suppose now that a surviving triangle $uvy$ contains the red edge $uv$ of a packed triangle $uvx\text{,}$ and that $xy$ is a red edge not used by the packing. Then $uvy$ is the only surviving triangle containing $uv\text{.}$ Indeed, $uxy$ has exactly one red edge, and the family contains every such triangle. So if $uvz$ were a second surviving triangle containing $uv\text{,}$ then replacing $uvx$ by $uxy$ and $uvz$ would enlarge the packing (Lemma 2.1). A red edge that lies in a single surviving triangle can be omitted from any transversal containing the other two edges of that triangle. Two transversals of the family, built in Section 2, allow this omission. Combining the two resulting bounds gives $\tau\leq(8/3)\nu$ for every family of this kind, and every packed triangle whose red edge lies in no other triangle of the family lowers the bound by a further $2/3$ (Lemma 2.3).

Families of this kind arise in Haxell’s construction. The construction fixes a maximum packing $P\text{,}$ calls the edges of its triangles old, and packs the remaining triangles in two further rounds: a maximum packing $A$ of triangles with one old edge, and, after the edges of $A$ are deleted, a packing $P^{\prime}$ with as many triangles with two old edges as possible. Color new edges red and old edges blue. After the deletion of the edges of $A\text{,}$ the triangles with two old edges form a family of the above kind, and so do the triangles that survive the further deletion of the old edges of $P^{\prime}\text{.}$ We build four transversals of $G$ from these packings, using books of triangles with a common edge, the bounds of Lemma 2.3, and a random bipartition of the vertex set. Each gives an inequality between $\tau(G)\text{,}$ $\nu(G)\text{,}$ and the sizes of the packings, and a linear combination of the four inequalities yields $165/59\text{.}$ Section 2 proves the exchange property and Lemma 2.3, and Section 3 carries out the construction. The complete proof has been formalized in Lean 4 with Mathlib.¹¹ 1

 <a href="https://github.com/sichen-wang/tuza-lean" title="" class="ltx_ref ltx_url ltx_font_typewriter">https://github.com/sichen-wang/tuza-lean</a></span></span></span></p>

2 Colored Triangle Families

Throughout, graphs are finite and simple, a triangle is identified with its set of three edges, and two triangles meet if they share an edge. For a family of triangles, a packing is a set of pairwise edge-disjoint members, a transversal is a set of edges meeting every member, and a packing is maximal if no member of the family can be added to it. Maximum packings are maximal. For a set $P$ of triangles, $E(P)$ is the union of their edge sets, and $H-X$ is the graph $H$ with the edge set $X$ deleted.

This section proves the three lemmas used in Section 3: the exchange property (Lemma 2.1), a covering lemma for books (Lemma 2.2), and the two colored bounds (Lemma 2.3). Let $H$ be a graph whose edges are colored red and blue, and let $\mathcal{F}(H)$ be the family of all triangles of $H$ with exactly one red edge. We write $\mathcal{F}$ for $\mathcal{F}(H)$ when $H$ is clear. For a packing $P$ in $\mathcal{F}(H)\text{,}$ let $R(P)$ and $B(P)$ be the sets of red and blue edges of its members, and let

$$\mathcal{F}_{P}=\mathcal{F}(H-B(P))$$

be the family of triangles with one red edge that survive the deletion of the blue edges of $P\text{.}$ If $P$ is maximum, every member of $\mathcal{F}_{P}$ has its red edge in $R(P)\text{,}$ since otherwise it could be added to $P\text{.}$ Hence it meets $E(P)$ exactly in its red edge, and it meets exactly one member of $P\text{.}$ We say that an edge is private in a family of triangles if exactly one member of the family contains it.

Lemma 2.1.

Let $P$ be a maximum packing in $\mathcal{F}(H)\text{,}$ let $uvx\in P$ have red edge $uv\text{,}$ and let $uvy\in\mathcal{F}_{P}\text{.}$ If $xy$ is a red edge outside $R(P)\text{,}$ then $uv$ is private in $\mathcal{F}_{P}\text{.}$

Proof.

Suppose that $uvz\in\mathcal{F}_{P}$ for some $z\neq y\text{.}$ Since $uz\notin B(P)$ while $ux\in B(P)\text{,}$ also $z\neq x\text{.}$ The triangle $uxy$ has blue edges $ux,uy$ and red edge $xy\text{,}$ so it belongs to $\mathcal{F}(H)\text{,}$ and it meets $E(P)$ only in $ux\text{,}$ because $uy\notin B(P)$ and $xy\notin R(P)\text{.}$ The triangle $uvz$ meets $E(P)$ only in $uv\text{.}$ As $z\notin\{x,y\}\text{,}$ the triangles $uxy$ and $uvz$ are edge-disjoint, so replacing $uvx$ by both of them gives a packing in $\mathcal{F}(H)$ larger than $P\text{.}$ ∎

The transversals below are built from books. A book with spine $uv$ is a set of triangles containing the edge $uv\text{.}$ Its members are its pages, and the union of their edge sets is its support. If a book has exactly two pages $uvx$ and $uvy$ and $xy$ is an edge, then $xy$ is the opposite edge of the book.

Lemma 2.2.

Let $\mathcal{B}$ be a set of books with at most three pages each and pairwise edge-disjoint supports, and let $\mathcal{T}$ be a family of triangles such that every selection of one page from each book is a maximal packing in $\mathcal{T}\text{.}$ Let $C$ consist of all edges of the one-page books and the spines of the other books. Then every member of $\mathcal{T}$ avoiding $C$ is $uxy$ or $vxy$ for some two-page book $\{uvx,uvy\}$ in $\mathcal{B}\text{.}$ Consequently, $C$ together with the opposite edges of the two-page books is a transversal of $\mathcal{T}$ of size at most $\sum_{B\in\mathcal{B}}(4-|B|)\text{.}$

Proof.

Let $T\in\mathcal{T}$ avoid $C\text{.}$ If every book had a page edge-disjoint from $T\text{,}$ choosing such a page in each book would give a maximal packing in $\mathcal{T}$ to which $T$ could still be added, a contradiction. So $T$ meets every page of some book $B\text{.}$ Since all edges of a one-page book lie in $C\text{,}$ $B$ has two or three pages, and $T$ avoids its spine $uv\text{,}$ so $T$ contains $uz$ or $vz$ for every page $uvz$ of $B\text{.}$ If $B$ had three pages, these would be three edges of $T$ with distinct ends outside $\{u,v\}\text{,}$ but a triangle has only three vertices. So $B=\{uvx,uvy\}\text{,}$ and $T$ contains one of $ux,vx$ and one of $uy,vy\text{.}$ Two edges of a triangle share a vertex, so these are $ux,uy$ or $vx,vy\text{,}$ and $T$ is $uxy$ or $vxy\text{.}$ Finally, a book with $k$ pages contributes at most $4-k$ edges to the transversal. ∎

Lemma 2.3.

Let $P$ be a maximum packing in $\mathcal{F}(H)$ and $Q$ a maximum packing in $\mathcal{F}_{P}\text{,}$ with $p=|P|$ and $q=|Q|\text{,}$ and let $u$ be the number of members of $P$ whose red edge is private in $\mathcal{F}(H)\text{.}$ Then

$$\begin{aligned} \displaystyle\tau(\mathcal{F}(H)) & \displaystyle\leq\tfrac{8}{3}p-\tfrac{2}{3}u, \\ \displaystyle\tau(\mathcal{F}(H)) & \displaystyle\leq\tfrac{9}{4}p+\tfrac{3}{2}q-\tfrac{1}{4}u. \end{aligned}$$

(1)(2)

Proof.

Let $r$ be the number of members of $Q$ whose red edge is private in $\mathcal{F}_{P}\text{.}$ We build two transversals of $\mathcal{F}\text{,}$ which give

$$\tau(\mathcal{F})\leq 3p-2q+r-u\qquad\text{and}\qquad\tau(\mathcal{F})\leq 2p+3q-r,$$

and then combine these bounds, once directly and once after applying Equation 1 to $\mathcal{F}_{P}\text{.}$

The first transversal comes from books. Each member of $Q$ shares its red edge with exactly one member of $P\text{,}$ and the two form a book with that red edge as spine. Each unpaired member of $P$ forms a one-page book. No member of $P$ is paired twice, as $Q$ is a packing, and the supports are edge-disjoint, as the blue edges of $Q$ avoid $B(P)\text{.}$ Every selection of one page per book is a packing in $\mathcal{F}$ of size $p=\nu(\mathcal{F})\text{,}$ hence maximal. A member of $P$ whose red edge is private in $\mathcal{F}$ is unpaired, since a page of $Q$ on the same spine would be a second member of $\mathcal{F}$ containing that edge.

Let $C$ consist of the $3(p-q)$ edges of the one-page books and the $q$ spines. By Lemma 2.2, a member of $\mathcal{F}$ avoiding $C$ is $uxy$ or $vxy$ for a two-page book $\{uvx,uvy\}$ with $uvx\in P$ and $uvy\in Q\text{.}$ The two edges it shares with the pages are blue, so $xy$ is red, and $xy\notin R(P)$ since $R(P)\subseteq C\text{.}$ By Lemma 2.1, $uv$ is private in $\mathcal{F}_{P}\text{,}$ so $uvy$ is one of the $r$ members of $Q$ with a private red edge. Adding the opposite edges of these at most $r$ books to $C$ therefore gives a transversal of $\mathcal{F}\text{.}$ Finally, delete the red edges of the $u$ unpaired members of $P$ whose red edge is private in $\mathcal{F}\text{:}$ the only member of $\mathcal{F}$ containing such an edge is that member of $P\text{,}$ which is met by its blue edges in $C\text{.}$ The transversal now has at most $3(p-q)+q+r-u$ edges, which is the first bound.

The second transversal starts from $B(P)\cup E(Q)\text{.}$ A member of $\mathcal{F}$ avoiding $B(P)$ lies in $\mathcal{F}_{P}$ and, as $Q$ is maximal there, meets $E(Q)\text{.}$ So $B(P)\cup E(Q)$ is a transversal of $\mathcal{F}\text{.}$ Delete the red edges of the $r$ members of $Q$ whose red edge is private in $\mathcal{F}_{P}\text{:}$ a member of $\mathcal{F}$ avoiding the remaining edges lies in $\mathcal{F}_{P}$ and contains a deleted edge, so it is the corresponding member of $Q\text{,}$ which is met by its blue edges. The remaining $2p+3q-r$ edges give the second bound.

Twice the first bound plus the second gives $3\tau(\mathcal{F})\leq 8p-q+r-2u\leq 8p-2u\text{,}$ as $r\leq q\text{,}$ which is Equation 1. Next, $Q$ is a maximum packing in $\mathcal{F}(H-B(P))=\mathcal{F}_{P}\text{,}$ and $r$ of its members have a red edge that is private in $\mathcal{F}_{P}\text{,}$ so Equation 1 for the graph $H-B(P)$ gives $\tau(\mathcal{F}_{P})\leq\tfrac{8}{3}q-\tfrac{2}{3}r\text{.}$ Since $B(P)$ meets every member of $\mathcal{F}\setminus\mathcal{F}_{P}\text{,}$

$$\tau(\mathcal{F})\leq 2p+\tfrac{8}{3}q-\tfrac{2}{3}r.$$

Three times this bound plus the first bound gives $4\tau(\mathcal{F})\leq 9p+6q-u-r\leq 9p+6q-u\text{,}$ which is Equation 2. ∎

3 The General Bound

Proof of Theorem 1.

We follow Haxell’s construction [Hax99]: the packings below and the first transversal are taken from it, and the other three transversals rest on the lemmas of Section 2. Choose a maximum triangle packing $P$ in $G\text{,}$ and put $n=|P|=\nu(G)$ and $O=E(P)\text{.}$ Call edges in $O$ old and all other edges new. A triangle has type $j$ if it contains exactly $j$ old edges. By maximality of $P$ there are no type-0 triangles.

Choose a maximum packing $A$ of type-1 triangles and put $a=|A|\text{.}$ The first transversal comes from books formed by $P$ and $A\text{.}$ Pair each member of $A$ with the member of $P$ containing its old edge. If two members of $A$ met the same member of $P\text{,}$ replacing it by both would enlarge $P\text{,}$ since a type-1 triangle meets $E(P)$ only in its old edge. These pairs and the remaining members of $P$ form $n$ books with $n+a$ pages. Their supports are edge-disjoint, since the new edges of $A$ avoid $O\text{,}$ and every selection of one page per book is a packing of size $n=\nu(G)\text{,}$ hence maximal. Lemma 2.2 gives a transversal with at most $3(n-a)+2a$ edges, that is,

$$\tau(G)\leq 3n-a.$$

(3)

Put $G^{\prime}=G-E(A)$ and $O^{\prime}=O\setminus E(A)\text{.}$ Every type-1 triangle meets $E(A)$ by maximality of $A\text{,}$ so every triangle of $G^{\prime}$ has type $2$ or $3\text{,}$ and $|O^{\prime}|=3n-a\text{.}$ Each remaining transversal of $G$ consists of $E(A)$ and a transversal of $G^{\prime}\text{.}$ Color the new edges of $G^{\prime}$ red and its old edges blue. Then $\mathcal{F}(G^{\prime})$ is the family $\mathcal{F}_{2}$ of type-2 triangles of $G^{\prime}\text{.}$

Choose a packing $P^{\prime}$ in $G^{\prime}$ that maximizes first its number of type-2 members and then its size $m\text{.}$ Let $B^{*}$ be the set of type-2 members of $P^{\prime}$ and put $b=|B^{*}|\text{.}$ Any packing in $\mathcal{F}_{2}$ is a packing in $G^{\prime}\text{,}$ so $B^{*}$ is a maximum packing in $\mathcal{F}_{2}\text{.}$ The new edges of $P^{\prime}$ are the new edges of the members of $B^{*}\text{,}$ one for each. Let $N$ be their set, so that $R(B^{*})=N$ and $|N|=b\text{.}$ Since $A\cup P^{\prime}$ is a packing in $G\text{,}$ we have $a+m\leq n\text{.}$ The choice of $P^{\prime}$ makes every packing in $G^{\prime}$ with $m$ members, $b$ of them of type $2\text{,}$ maximal in $G^{\prime}\text{:}$ adding a triangle would give a packing with at least $b$ type-2 members and $m+1$ members.

Let $B(P^{\prime})$ be the set of old edges of $P^{\prime}\text{,}$ and put

$$\mathcal{S}=\mathcal{F}(G^{\prime}-B(P^{\prime})),$$

the family of type-2 triangles of $G^{\prime}$ whose old edges avoid $E(P^{\prime})\text{.}$ Since $B(B^{*})\subseteq B(P^{\prime})\text{,}$ we have $\mathcal{S}\subseteq(\mathcal{F}_{2})_{B^{*}}=\mathcal{F}(G^{\prime}-B(B^{*}))\text{,}$ and since $B^{*}$ is a maximum packing in $\mathcal{F}_{2}\text{,}$ the red edge of every member of $\mathcal{S}$ lies in $R(B^{*})=N\text{.}$ Choose a maximum packing $Q$ in $\mathcal{S}\text{,}$ put $c=|Q|\text{,}$ and let $h$ be the number of members of $Q$ whose red edge is private in $\mathcal{S}\text{.}$

The second transversal combines books formed by $B^{*}$ and $Q$ with a random bipartition. Pair each member of $Q$ with the member of $B^{*}$ containing its red edge. No member of $B^{*}$ is paired twice, as $Q$ is a packing. These pairs and the unpaired members of $B^{*}$ form $b$ books, whose supports are edge-disjoint since the old edges of $Q$ avoid $E(P^{\prime})\text{.}$ Every selection of one page per book consists of $b$ type-2 triangles and is therefore a maximum packing in $\mathcal{F}_{2}\text{.}$ Let $W$ be the set of old edges of the one-page books, so $|W|=2(b-c)\text{.}$ The set $C$ of Lemma 2.2 for these books lies in $W\cup N\text{,}$ since spines are edges of $N$ and a one-page book has its old edges in $W$ and its new edge in $N\text{.}$

Assign each vertex of $G^{\prime}$ a bit, independently and uniformly at random, and call an edge crossing if its ends receive different bits. Select $W$ and every non-crossing edge of $O^{\prime}\text{,}$ then add the new edge of each triangle of $G^{\prime}$ not yet met. A type-3 triangle has two vertices with the same bit and hence a non-crossing old edge, so the selected edges meet every triangle of $G^{\prime}\text{.}$ Their expected number is $|W|+\frac{1}{2}(|O^{\prime}|-|W|)$ plus the expected number of added new edges, which we bound in two parts.

A new edge is added only if both old edges of its triangle are crossing, and then it is non-crossing. Each edge of $N$ is non-crossing with probability $\frac{1}{2}\text{,}$ so the expected number of added edges in $N$ is at most $\frac{b}{2}\text{.}$

Now let an edge $xy\notin N$ be added as the new edge of a type-2 triangle $T$ not yet met. Then $T$ contains no edge of $N$ and avoids $W\text{,}$ hence avoids $C\text{,}$ and Lemma 2.2 for these books and $\mathcal{F}_{2}$ shows that $T$ is $uxy$ or $vxy$ for a two-page book $\{uvx,uvy\}$ with $uvx\in B^{*}$ and $uvy\in Q\text{.}$ The two edges that $T$ shares with the pages are old, so its new edge $xy$ is the opposite edge of the book. Since $uvy\in\mathcal{S}\subseteq(\mathcal{F}_{2})_{B^{*}}$ and $xy$ is a red edge outside $R(B^{*})=N\text{,}$ Lemma 2.1 shows that $uv$ is private in $(\mathcal{F}_{2})_{B^{*}}\text{,}$ hence in $\mathcal{S}\text{,}$ so $uvy$ is one of the $h$ members of $Q$ with a private red edge. Such a book contributes an added edge only if $uxy$ or $vxy$ is not yet met, which requires $x$ and $y$ to receive the same bit and $u$ or $v$ a different one. As $u,v,x,y$ are distinct, this has probability $\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\text{.}$ Distinct added edges come from distinct books, as each book has one opposite edge, so the expected number of added edges outside $N$ is at most $\frac{3}{8}h\text{.}$ The expected total is at most

$$|W|+\frac{|O^{\prime}|-|W|}{2}+\frac{b}{2}+\frac{3}{8}h=\frac{|O^{\prime}|}{2}+\frac{3}{2}b-c+\frac{3}{8}h,$$

so some assignment of bits gives a transversal of $G^{\prime}$ of at most this size. Adding $E(A)$ and using $|O^{\prime}|=3n-a$ gives

$$\tau(G)\leq\tfrac{3}{2}n+\tfrac{5}{2}a+\tfrac{3}{2}b-c+\tfrac{3}{8}h.$$

(4)

For the third transversal, we add a third page to some of these books and include the remaining members of $P^{\prime}\text{.}$ Choose a maximum packing $Q_{1}$ in $\mathcal{S}_{Q}=\mathcal{F}(G^{\prime}-B(P^{\prime})-B(Q))$ and put $d=|Q_{1}|\text{.}$ The red edge of each member of $Q_{1}$ lies in $R(Q)\text{,}$ and distinct members of $Q_{1}$ have distinct red edges. Add each member of $Q_{1}$ as a third page to the book whose spine is its red edge, and add the members of $P^{\prime}\setminus B^{*}$ as one-page books. The old edges of $Q_{1}$ avoid $E(P^{\prime})\cup E(Q)\text{,}$ so the supports remain edge-disjoint. Every selection of one page per book is a packing in $G^{\prime}$ with $m$ members, $b$ of them of type $2\text{,}$ hence maximal in $G^{\prime}\text{.}$ These $m$ books have $m+c+d$ pages, so Lemma 2.2 gives a transversal of $G^{\prime}$ of size at most $3m-c-d\text{.}$ Adding $E(A)$ and using $a+m\leq n$ gives

$$\tau(G)\leq 3n-c-d.$$

(5)

The fourth transversal consists of $E(A)\text{,}$ the old edges of $P^{\prime}\text{,}$ and a transversal of $\mathcal{S}\text{.}$ The first two sets have at most $3a+3m-b\leq 3n-b$ edges. A triangle avoiding them lies in $G^{\prime}-B(P^{\prime})$ and is not of type $3\text{,}$ since a type-3 triangle avoiding $B(P^{\prime})$ would avoid $E(P^{\prime})$ and could be added to $P^{\prime}\text{.}$ So it belongs to $\mathcal{S}\text{.}$ Lemma 2.3 applied to $G^{\prime}-B(P^{\prime})\text{,}$ with $Q\text{,}$ $Q_{1}\text{,}$ $c\text{,}$ $d\text{,}$ $h$ in the roles of $P\text{,}$ $Q\text{,}$ $p\text{,}$ $q\text{,}$ $u\text{,}$ gives $\tau(\mathcal{S})\leq\frac{9}{4}c+\frac{3}{2}d-\frac{1}{4}h$ by Equation 2, so

$$\tau(G)\leq 3n-b+\tfrac{9}{4}c+\tfrac{3}{2}d-\tfrac{1}{4}h.$$

(6)

Multiply Equations 3, 4, 5 and 6 by $20,8,19,12\text{,}$ respectively, and add. The coefficients of $a,b,c,h$ cancel, giving

$$59\tau(G)\leq 165n-d\leq 165\nu(G).$$

Remark.

The weights are best possible for the system Equations 3, 4, 5 and 6. For $a=c=12n/59\text{,}$ $b=m=39n/59\text{,}$ and $d=h=0\text{,}$ which satisfy the relations $a+m\leq n\text{,}$ $d\leq c\leq b\leq m\text{,}$ and $h\leq c$ that hold in the proof, all four right-hand sides equal $165n/59\text{.}$ So no nonnegative combination of the four inequalities gives a smaller constant.

References

[Hax99] P. E. Haxell, Packing and covering triangles in graphs, Discrete Mathematics 195 (1999), 251–254. doi:10.1016/S0012-365X(98)00183-6.

[Tuz90] Z. Tuza, A conjecture on triangles of graphs, Graphs and Combinatorics 6 (1990), 373–380. doi:10.1007/BF01787705.

[Yi26] L. Yi, An improved upper bound for Tuza’s conjecture via 2-colorable triangle families, arXiv:2608.23010v1 (2026).

(带权)并查集

Sat, 24 Jun 2023 11:00:56 +0000

用途

并查集可以用来维护一类具有传递性的关系，维护形如将 $x$ 与 $y$ 所在集合合并和询问 $x$ 和 $y$ 是否在同一个集合的操作。

实现

查询祖先节点

对于普通并查集来说，有路径压缩和按秩合并等优化，路径压缩一般实现如下：

int Find (int x) {
 if (x != fa[x]) fa[x] = Find(fa[x]);
 return fa[x];
}

按秩合并则取点数小的连向点数大的即可。

而对于带边权的并查集，我们需要在路径压缩的时候考虑加上其父亲的权值以维护其到根的差值，有路径压缩优化的带权并查集一般实现如下：

int Find (int x) {
 if (x != fa[x]) {
 int mark = fa[x];
 fa[x] = Find(fa[x]), val[x] += val[mark];
 }
 return fa[x];
}

由于路径压缩后 $x$ 的父亲就是根节点了，所以需要记录其父亲到根的总权值并加到它的边权上。

另外，如果要查询某个点 $x$ 到根的权值和，由于一次路径压缩以后 $x$ 的父亲会变成根，而此时的 $val_{x}$ 就是路径和，所以可以先调用一次 Find(x)，此时的 $val_{x}$ 即为答案。

合并两个结点所在集合

对于普通并查集来说，两个集合合并直接将其中一个根节点接到另一个根节点上即可。

而对于带权并查集而言，我们必须求出需要连接的两个根节点之间连边的权值，而这个权值是可以直接确定的，假设现在要给 $x\to y$ 连一条长度为 $w$ 的边，设 $x$ 的祖先为 $Rx$，则 $val_{Rx}=val_{y}+w-val_{x}.$ 另外，如果 $x$ 和 $y$ 已经在同一个集合中，并且 $val_{Rx}+val_{x}-val_{y}\not = w$，则这条连边不合法。实现如下：

bool Merge (int x, int y, int w) {
 int Rx = Find(x), Ry = Find(y);
 if (Rx != Ry) fa[Rx] = Ry, val[Rx] = val[y] + w - val[x];
 return val[Rx] + val[x] - val[y] == w;
}

Code

带路径压缩和按秩合并优化的普通并查集 :

int Find (int x) {
 if (x != fa[x]) fa[x] = Find(fa[x]);
 return fa[x];
}

bool Merge (int x, int y) {
 int Rx = Find(x), Ry = Find(y);
 if (x != y) {
 if (pnum[Rx] < pnum[Ry])
 fa[Rx] = Ry, pnum[Ry] += pnum[Rx];
 else fa[Ry] = Rx, pnum[Rx] += pnum[Ry];
 }
 return Rx != Ry;
}

带路径压缩和按秩合并优化的带权并查集：

int Find (int x) {
 if (x != fa[x]) {
 int mark = fa[x];
 fa[x] = Find(fa[x]), val[x] += val[mark];
 }
 return fa[x];
}
bool Merge (int x, int y, int w) {
 int Rx = Find(x), Ry = Find(y);
 if (Rx == Ry) return (val[x] - val[y] == w);
 if (pnum[Rx] < pnum[Ry])
 fa[Rx] = Ry, pnum[Ry] += pnum[Rx], val[Rx] = val[y] + w - val[x];
 else fa[Ry] = Rx, pnum[Rx] += pnum[Ry], val[Ry] = val[x] - w - val[y];
 return true;
}

不带任何优化的并查集复杂度是 $\Theta(n^{2})$ 的。

带路径压缩的并查集时间复杂度 $\Theta(n \log n)$，期望复杂度 $\Theta(n \alpha(n))$。

再加上按秩合并可以做到严格 $\Theta(n \alpha(n))$。

2-SAT 问题

Sat, 24 Jun 2023 11:00:56 +0000

求解 2-SAT 问题

$2-\rm Sat$ 问题是求解某个 $\rm bool$ 方程组，方程组中的每个方程被描述为：$x_{i}$ 为真$/$假或 $x_{j}$ 为真$/$假。

我们可以对每个 $x_{i}$ 和 $\lnot x_{i}$ 分别建点，$x_{i}$ 为假看作 $\lnot x_{i}$ 为真，这样方程就仅有一种类型了，即若 $p$ 为真则 $q$ 为真。具体来说，建一些边 $u\to v$ 表示当 $u$ 为真的时候 $v$ 一定为真，比如有方程 $p$ 为真或 $q$ 为真时，建两条边 $p\to \lnot q$ 和 $q\to \lnot {p}$。

建完图后，显然同一个强连通分量中的点取值必然相同，那么如果 $a$ 和 $\lnot a$ 在同一个强连通分量里则该 $\rm bool$ 方程组无解。易知，存在 $a$ 和 $\lnot a$ 在同一个强连通分量中为该 $\rm bool$ 方程组无解的充要条件。

对于其它的点，每对 $a$ 和 $\lnot a$ 中有且仅有一个为真，我们令拓扑序大的那个为真。可以分情况证明其最优性：

若 $a$ 和 $\lnot a$ 在 $\rm DAG$ 的同一条链上，那么拓扑序大的必然为真，否则两个都为真与题意相悖。
若 $a$ 和 $\lnot a$ 在不同链上，那么两个点的取值其实无所谓，但一定要跟拓扑序相关。因为每条链上拓扑序的大小关系时严格的，用拓扑序规定某个点取值是否为真可以使为真的点在同一条链上，尽可能降低对其他点的影响。另外，为了保证跟上一种情况的统一性，仍选择令拓扑序大的点为真。

另一方面，我们发现 $\rm Tarjan$ 缩点所求出的 $\rm SCC$ 强连通分量数组的值负相关，可以理解为拓扑序更小的点更后入栈，也更先出栈编号。故最后对比 $a$ 和 $\lnot a$ 的 $\rm SCC$ 值即可判断它们谁为真谁为假。

经典例题

[LG6378] Riddle

对树上每个点建立一个 $\rm bool$ 变量，取值为 $1$ 表示这个点选。

使用 $\rm 2-Sat$ 的通用做法，建立为每个点的状态建立两个点 $u$ 和 $\lnot u$，分别对每个限制考虑：

每条边至少有一个关键点，则对于一条边 $u-v$ 若 $\lnot u$ 则 $v$，若 $\lnot v$ 则 $u$，故连接 $\lnot u\to v$ 和 $\lnot v\to u$。
每个部分最多有一个关键点，则如果 $u$ 被选了，那么同一个块中所有点都不能选，连接 $u\to \lnot v$。但是我们发现这样连边是 $\Theta(n^{2})$ 的，所以前缀和优化连边即可。

建好图以后直接跑 $\rm 2-Sat$，时间复杂度 $\Theta(n)$。

[POI2011] KON-Conspiracy

首先转化题意，这道题要求的是将一张无向图划分成一个团和一个独立集的方案数。

首先找出一种合法的方案是很容易的，是一个裸的 $\rm 2-Sat$ 问题。

考虑找出一种方案后统计方案数。首先有如下性质：

一个团中不可能取出一个以上的点加入独立集，因为两点之间必有连边。

一个独立集中不可能取出一个以上的点加入团，因为两点之间必无连边。

所以可以直接分别枚举团和独立集中的每个点，交换它们即可。

时间复杂度 $\rm \Theta(n^{2})$。

[北京省选集训 2019] 完美塔防

首先预处理哪些炮台会互相打到，若存在，则输出 IMPOSSIBLE.

然后发现有如下结论：

每个空地最多可以被两个炮台打到。因为每个空地最多在纵横两个方向上被打到，若还存在一个，则必定会打到其中一个炮台。

于是对每个炮台，它的纵横情况设为 $0/1$，每个空地会加入一个形如 $a\ or\ b$ 的条件，这就变成了经典的 $\rm 2-Sat$ 问题。

时间复杂度 $\Theta(Tn^{3})$。

[HNOI2010] 平面图判定

首先有平面图中的欧拉定理，设一张无向连通图点数为 $v$，边数为 $e$，面数为 $f$，则有：

\[ v-e+f=2 \]

考虑通过对 $e$ 进行归纳证明该式。

当 $e=0$ 时显然成立。

若当 $e'=e-1$ 时该式成立，我们要证 $v-e+f=1$ 成立只需证添加一条边会使得面数恰好增加 $1$。而对于这个问题，我们可以分类讨论，若这条边加在某个面的中间，那么其必将这个面分成两半，若这条边加在图形外面，那么会围出一块新的区域，也会使面数恰好增加 $1$。

故该式成立。

我们进一步探究平面图的性质。定义面的度数为与其相邻的面的数量，显然对于每条边都会有恰好两个面与其相邻，故所有面的度数之和为 $2e$。又因为每个面至少三条边，所以度数不小于 $3$，进而有 $3f$ 不大于所有面的度数之和，故有 $3f\leqslant 2e$，将其带入平面图中的欧拉定理可得：

\[ e\leqslant 3v-6 \]

于是不满足该式的无向连通图均不为平面图，进而可以将 $m$ 限制到与 $n$ 同数量级。

我们发现题目给定了哈密顿回路，于是考虑以这个环作为中心考虑问题。

我么发现对于任意一条边，它要么从环内连接，要么从环外连接，并且如果有两条边 $u-v$ 和 $u'-v'$，且满足 $u\lt u'\lt v\lt v'$，那么它们不能连在同一面。于是可以通过这种关系构建 $\rm 2-Sat$ 模型。

时间复杂度 $\Theta(Tn)$。

Prufer 序列

Sat, 24 Jun 2023 11:00:56 +0000

Prufer 序列的定义

$\rm Prufer$ 序列可以将一颗结点数为 $n$ 的有标号无根树用一个长度为 $n-2$，值域为 $[1,n]$ 的数列唯一表示，即有标号无根树和 $\rm Prufer$ 序列呈双射关系。

具体来说，有标号无根树的构建方式如下：

选择一个编号最小的叶子节点，并将其删除。
将这个叶子节点所连接的点的编号加入 $\rm Prufer$ 序列中。
重复以上步骤 $n-2$ 次，直到树上只剩下 $2$ 个点。

这是某一颗点数为 $n$ 的有标号无根树的 $\rm Prufer$ 序列的构造过程：

快速求解 Prufer 序列

由其过程，显然有一个使用堆优化的 $\Theta(n\log n)$ 做法，但 $\rm Prufer$ 序列其实可以线性构造。

记录所有点的度数和一个指针 $p$ 指向编号最小的叶子节点，进行以下操作：

将 $p$ 指向的结点删除，检查是否出现了新的叶结点。
如果产生了新的叶结点，记其编号为 $x$。若 $x\gt p$，则不做任何操作，否则将其删除，并检查删除 $x$ 以后是否出现了新的叶结点，重复这一步操作直到没有产生新的叶结点或新产生的结点编号大于 $p$。
让 $p$ 自增直到遇见下一个叶结点为止。
重复以上操作直到结点数为 $2$ 可以得到这棵树的 $\rm Prufer$ 序列。

可以发现，在算法流程中，每条边只在其连接的外层结点被删除时被遍历过一次，并且 $p$ 单调递增，树中的节点数单调递减，所以每个点也只被遍历过一次。

总时间复杂度 $\Theta(n)$。

快速还原 Prufer 序列

显然还是有一个堆优化的 $\Theta(n\log n)$ 做法。

从前到后枚举 $\rm Prufer$ 序列中的每一个数，维护不在序列中的结点编号的最小值。显然不在序列中的是叶子节点，并且其最小值为最后一个加入的，于是将其与当前枚举的序列中的数连边。连边后若序列中已没有枚举的这个数，那么将其踢出序列作为新的叶子结点。

考虑以与之前构造 $\rm Prufer$ 序列相同的方法构造线性做法。

记一个指针 $p$ 指向不在序列中的最小结点，如果将其连向序列中枚举的位置，并将枚举的位置删除。序列中没有这个结点了，那么如果被删除的结点比 $p$ 小，直接将其与现在序列中枚举的数连边后将其删除，重复操作。如果这个结点的值大于 $p$，那么不用管它，在后面它一定会被枚举到。

显然，这种算法的时间复杂度为 $\Theta(n)$。

Prufer 序列的应用

Cayley 公式

完全图 $K_{n}$ 有 $n^{n-2}$ 颗生成树。

证明方法很多，其中最简单的应该是利用 $\rm Prufer$ 序列。

考虑到任何一个长度为 $n-2$ 的 $\rm Prufer$ 序列唯一对应一颗大小为 $n$ 的有标号无根树，于是考虑计算长度为 $n-2$ 的 $\rm Prufer$ 序列的数量，又因为其值域为 $[1,n]$，故总方案数为 $n^{n-2}$。

限定度数的有标号无根树计数

求有 $n$ 个结点，且第 $i$ 个结点的度数为 $d_{i}$ 的有标号无根树的数量。

考虑到一个点的度数为其在 $\rm Prufer$ 序列中出现次数 $+1$，于是对于一个度数为 $d_{i}$ 的点，其在 $\rm Prufer$ 序列中的出现次数应为 $d_{i}-1$，故存在这样的有标号无根树当且仅当 $\sum_{i=1}^{n} d_{i}-1=n-2$，并且这样的树的数量为：

\[ \dbinom{n-2}{d_1-1,d_2-1,\cdots ,d_n-1}=\frac{(n-2)!}{(d_1-1)!(d_2-1)!\cdots (d_n-1)!} \]

差分约束

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差分约束是用来解形如 $x_{1}-x_{2}\leqslant k$ 的若干个方程的算法。

我们发现可以将方程移项成 $x_{1}\leqslant x_{2}+k$，解方程的时候考虑建图，从 $x_{2}$ 向 $x_{1}$ 连一条长度为 $k$ 的边，然后对这张图建一个超级源点，向所有点连一条长度为 $0$ 的边，最后跑最短/最长路得出的 $dis$ 就是每个变量的一个取值。

为什么这样做是对的呢？因为如果有一条 $u$ 到 $v$ 的长度为 $w$ 的边，那么跑完最短路以后 $dis_{v}$ 一定不大于 $dis_{u} + w$，恰好满足一个方程。那么将所有边连上以后再跑最短路就可以满足所有方程，最长路同理。

另外，如果有负环则无解。

时间复杂度 $\Theta(\rm SPFA)$。

需要注意的是，最短路求的是以起始点为基准的最大解，而最长路求的是以起始点为基准的最小解。

最小生成树

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Kruskal

最常用且大部分情况下效率最高的最小生成树算法。

将边按长度从小到大排序，每次贪心选择最短的一条边，若这条边连接的两个点还为连通，则将两个点所在的连通块用这条边连接，连通块用并查集维护。

证明: 使用归纳法，证明任何时候 Kruskal 算法选择的边集都被某棵 MST 所包含。基础: 对于算法刚开始时，显然成立（最小生成树存在）。归纳: 假设某时刻成立，当前边集为 $F$，令 $T$ 为这棵 MST，考虑下一条加入的边 $e$。如果 $e$ 属于 $T$，那么成立。否则，$T+e$ 一定存在一个环，考虑这个环上不属于 $F$ 的另一条边 $f$（一定只有一条）。首先，$f$ 的权值一定不会比 $e$ 小，不然 $f$ 会在 $e$ 之前被选取。然后，$f$ 的权值一定不会比 $e$ 大，不然 $T+e-f$ 就是一棵比 $T$ 还优的生成树了。所以，$T+e-f$ 包含了 $F$，并且也是一棵最小生成树，归纳成立。

时间复杂度 $\Theta(m\log m)$。

Kruskal 重构树

这是 Kruskal 算法除了求最小生成树以外的一个重要应用。

在使用 $\rm kruskal$ 最小生成树算法的时候会从小到大加入 $n-1$ 条边。现在仍按照这个顺序，每次加入一条边时将其变成一个点权等于边权的点，然后把它作为所连接两个连通块的父亲节点。

不难发现两个结点在 MST 上的最短路径上的边权最大值为它们在 $\rm kruskal$ 重构树上 $\rm LCA$ 的点权，并且从一个点 $u$ 出发只经过边权小于某值的边能够到达的点集为 $\rm Kruskal$ 重构树上某点的子树内所有的叶子结点。

Prim

在稠密图上复杂度最优的最小生成树算法。

与 Dijkstra 算法类似，选择一个点作为起点，每次选择到当前已选的点集中任意一个点距离最短的未选择的点并向其连边，再递归操作。

证明: 还是说明在每一步，都存在一棵最小生成树包含已选边集。基础: 只有一个结点的时候，显然成立。归纳: 如果某一步成立，当前边集为 $F$，属于 $T$ 这棵 MST，接下来要加入边 $e$。如果 $e$ 属于 $T$，那么成立。否则考虑 $T+e$ 中环上另一条可以加入当前边集的边 $f$。首先，$f$ 的权值一定不小于 $e$ 的权值，否则就会选择 $f$ 而不是 $e$ 了。然后，$f$ 的权值一定不大于 $e$ 的权值，否则 $T+e-f$ 就是一棵更小的生成树了。因此，$e$ 和 $f$ 的权值相等，$T+e-f$ 也是一棵最小生成树，且包含了 $F$。

时间复杂度 $\Theta[(n+m)\log n]$。

Boruvka

在大多数完全图上求最小生成树所使用的算法。

每次遍历所有的点，求出它当前属于哪个连通块。
遍历所有的边 $u-v$，如果 $u$ 和 $v$ 不在同一个连通块，就分别用 $u$ 和 $v$ 去更新它们所在连通块的最小出边。
将每个连通块的最小出边加入 MST 的边集。

由于每次连通块数量减半，所以时间复杂度 $\Theta(m\log n)$。

最短路

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最短路算法拓展技巧

Floyd 传递闭包

有时候我们需要维护一些具有传递性的关系，如相等，连通等。

初始条件往往是给定几组关系，要把所有的关系求出来。

可以把 Floyd 算法做一些调整：

\[ dis_{i,j}=dis_{i,j}~|~(dis_{i,k}~\mathbin{\&}~dis_{k,j}) \]

时间复杂度还是 $\Theta(n^{3})$，但由于位运算的特殊性，有些时候可以使用 bitset 优化到 $\Theta\left(\frac{n^{3}}{\omega}\right)$。

Johnson 重赋权

对于多源最短路，如果我们枚举一个点然后跑堆优化的 Dijkstra，那么复杂度是 $\Theta(nm \log n)$ 的，在图比较稀疏的情况下，这个复杂度要优于 Floyd 算法的 $\Theta(n^{3})$。

但是 Dijkstra 算法要求所有边权均非负，于是就有了重赋权的技巧。

具体来说，我们新建一个 $0$ 号点，并且从这个点出发向所有点连一条边权为 $0$ 的边，然后跑单源最短路(SPFA 或者 Bellman-Ford)。设距离数组为 $h$，接下来对于每条边 $(u, v)$，令

\[ w^{\prime}(u, v)=w(u, v)+h(u)-h(v) \]

这样所有的边权都变成非负的，我们就可以跑 Dijkstra 算法了。

接下来证明这样做为什么是对的：

首先由于 $h(v) \leqslant h(u)+w(u, v),$ 所以新图的边权一定非负。

设新图上的最短路径为 $d^{\prime}$，原图上的最短路径为 $d$。

\[ \begin{aligned} d^{\prime}(u, v) &= \min_{a_{1}, a_{2}, \ldots, a_{k}} w^{\prime}\left(u, a_{1}\right)+w^{\prime}\left(a_{1}, a_{2}\right)+\cdots+w^{\prime}\left(a_{k}, v\right) \\ &= \min_{a_{1}, a_{2}, \ldots, a_{k}} w\left(u, a_{1}\right)+\left(h(u)-h\left(a_{1}\right)\right)+w\left(a_{1}, a_{2}\right)+\left(h\left(a_{2}\right)-h\left(a_{1}\right)\right)+\cdots+w\left(a_{k}, v\right)+\left(h(v)-h\left(a_{k}\right)\right) \\ &= h(u)-h(v)+\min_{a_{1}, a_{2}, \ldots, a_{k}} w\left(u, a_{1}\right)+\cdots+w\left(a_{k}, v\right) \\ &= h(u)-h(v)+d(u, v) \end{aligned} \]

这个证明用到了势能分析的思想，如果想进一步了解话可以参考本站《「杂项」复杂度分析》一文。

最短路图

所谓最短路图，就是在求完从 $S$ 出发的单源最短路之后，只保留最短路上的边形成的有向图。

只需要在求的过程中维护一个 $pre_{u}$ 数组表示点 $u$ 的前驱即可。

很多最短路的变种都需要用这个算法，有时候考虑最短路图也有利于发现一些隐藏的性质。

经典例题

[LG1266] 速度限制

记 $dis_{i,j}$ 表示当前已经到 $i$，上一个点是 $j$ 的时间最短路。

本质上是个分层图最短路，直接跑 Dijkstra 就好了。

时间复杂度 $\Theta(nm\log n)$。

[ICPC-BJ2006] 狼抓兔子

首先有定义：

平面图是指的存在某种方式使得图上所有边平铺在平面上无交。

对偶图是指的将平面图上的每个区域看作一个点，两点连边当且仅当两区域相邻，边权为将两区域分隔开的边的权值。

有定理：

平面图两点间最小割等于对偶图两点间最短路。

于是在这道题中我们就可以将每个三角形区域看作一个点，相邻区域连边后跑从左下角到右上角的最短路。

时间复杂度 $\Theta(n^{2}\log n)$。

[GXOI/GZOI2019] 旅行者

我们发现可以对 $k$ 个点二进制分组，每次将 $k$ 个点分成两个集合 $S_{1}$ 和 $S_{2}$，建一个源点 $s$ 向所有 $S_{1}$ 中的结点连长度为 $0$ 的边，建一个汇点 $t$ 从所有 $S_{2}$ 中的结点开始向它连长度为 $0$，的边，求从 $s$ 到 $t$ 的最短路。

可以发现，若答案为两点 $x$ 和 $y$ 之间的最短路，那么一定有 $x\not= y$，这意味着它们至少在某一次二进制分组中被分到了不同的集合，所以这样一定可以找到两点间的最短路。

时间复杂度 $\Theta(Tn\log n\log k)$。

欧拉回路

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定义

欧拉路径：如果图中的一个路径包括每个边恰好一次，则该路径称为欧拉路径。
欧拉回路：首尾相接的欧拉路径被称为欧拉回路。

判定

由于每一条边都要经过恰好一次，因此对于除了起点和终点之外的任意一个节点，只要进来，一定要出去，故有如下结论：

一个无向图存在欧拉回路，当且仅当该图所有顶点度数都为偶数，且该图只有一个存在边的连通块。
一个无向图存在欧拉路径，当且仅当该图中奇点的数量为 $0$ 或 $2$ 且该图只有一个存在边的连通块。
一个有向图存在欧拉回路，当且仅当所有点的入度等于出度。
一个混合图存在欧拉回路，当且仅当存在一个对所有无向边定向的方案，使得所有点的入度等于出度。求解需要用到网络流相关算法。

求解

一般使用 $\rm dfs$ 算法求出一张图的欧拉回路。

给每一条边记一个 $\rm vis$ 数组，表示其是否被访问过，然后从一个点出发，遍历所有的边。

直接 $\rm dfs$ 的话可能会有一些点无法被遍历到，于是在记录答案的时候可以倒着记录，即当通过 $u\to v$ 这条边的时候，可以先将点 $v\ \rm dfs$ 完，再加入 $u\to v$ 这条边。这样做相当于如果碰到两个相邻的环，但只遍历了其中一个就走到头了，于是我们需要走完另一个环后将两个环拼起来。

由于只需要将图遍历一遍，所以时间复杂度是 $\Theta(n + m)$ 的。

参考代码

void dfs (int u) {
 for (int i = last[u]; i; i = Next[i]) {
 if (vis[i]) continue;
 vis[i] = true;
 int record = i;
 dfs(to[i]);
 Sta[++top] = record;
 }
}

缩点 & 割点 & 割边

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Tarjan 算法

强连通分量

若图 $G$ 中存在一个子图 $G'$，$G'$ 中任意两点均可互相到达，则称 $G'$ 为 $G$ 的一个强连通分量。若 $G'$ 不存在一个严格子图为强连通分量，则 $G'$ 为 $G$ 的极大强连通分量。

图的强连通分量可以考虑使用 $\rm Tarjan$ 算法求解。

$\rm Tarjan$ 算法的思想是这样的，如果一个点的 $\rm dfs$ 序等于他能访问到的最小的 $\rm dfs$ 序，那么在 $\rm dfs$ 树中，它的下方会出现一个强连通分量，且所有最小能访问到的 $\rm dfs$ 序等于它的点，属于这个强连通分量。

时间复杂度 $\Theta(n+m)$。

参考代码：

void Tarjan (int u) {
 dfn[u] = low[u] = ++tix, Sta[++top] = u;
 for (int i = last[u]; i; i = Next[i])
 if (!dfn[to[i]]) Tarjan(to[i]), low[u] = min(low[u], low[to[i]]);
 else if (!SCC[to[i]]) low[u] = min(low[u], dfn[to[i]]);
 if (dfn[u] == low[u]) {
 SCC[u] = ++id;
 while (Sta[top] != u) SCC[Sta[top--]] = id;
 top--;
 }
}

点/边双连通分量

如果在 $\rm dfs$ 树上，某个结点 $u$ 存在一个子节点 $v$ ，满足 $low_{v}\geqslant dfn_{u}$ ，则说明 $v$ 想到达 $u$ 的祖先结点必须经过 $u$，所以 $u$ 是一个割点。

当然，根节点不能通过这种方式来判断，应该看它的子节点数是否大于等于 $2$，如果是，则它为割点。

遍历的时候要记得判掉父子边。

和判断割点的方法几乎一模一样，唯一的区别是判断 $dfn_{u}\lt low_{v}$ 而不是 $dfn_{u}\leqslant low_{v}$。（如果从 $v$ 出发连 $u$ 都无法到达，那么 $(u, v)$ 就是一条割边。

判断割边的时候根节点不需要特殊考虑。

时间复杂度 $\Theta(n+m)$。

参考代码：

void Tarjan (int u, int rt) {
 dfn[u] = low[u] = ++tix;
 int son = 0;
 for (int i = last[u]; i; i = Next[i]) {
 if (!dfn[to[i]]) {
 son++, Tarjan(to[i], rt);
 low[u] = min(low[u], low[to[i]]);
 if (low[to[i]] >= dfn[u] && u != rt) cut[u] = true;
 }
 low[u] = min(low[u], dfn[to[i]]);
 }
 if (u == rt && son > 1) cut[u] = true;
}

经典例题

[国家集训队] 稳定婚姻

首先在夫妻之间连有向边 $B_{i}\to G_{i}$，然后在情人之间连有向边 $G_{i}\to B_{i}$。

可以发现，若 $G_{i}$ 和 $B_{i}$ 的婚姻情况发生了改变，那么一定存在一个从 $G_{i}$ 开始 $B_{i}$ 结束的环。

于是我们使用 $\rm Tarjan$ 算法求强连通分量，若 $B_{i}$ 与 $G_{i}$ 属于同一个强连通分量，那他们一定属于一个环，则他们是不安全的，否则就是安全的。

时间复杂度 $\Theta(n+m)$。

[ZJOI2007] 最大半连通子图

首先缩点，发现强连通分量内的点一定属于同一个极大半连通子图。

然后将 $\rm DAG$ 画出来发现最大半连通子图一定是某条从无入度结点到无出度结点的链，长度和方案数直接 $\rm DAG$ 上 $\rm DP$ 即可。

需要注意的是，在这题中，由于求的是导出子图，所以不能有重边。

时间复杂度 $\Theta(n+m)$。

[HNOI2012] 矿场搭建

先把个点和点双连通分量处理出来，对每个点双分开处理：

若该点双无割点，则需要建两个出口。
若该点双有一个割点，则需要在这个点双内建一个出口，且不能建在割点上。
若该点有两个以上割点，则不需要建出口，因为一定可以从某个割点跑出去。

方案数直接乘法原理计算即可。

时间复杂度 $\Theta(n+m)$。

[HAOI2010] 软件安装

可以发现这是一棵基环外向树，依题意环中结点必须选，剩下的构成一棵树，直接树上背包即可。

时间复杂度 $\Theta(nm)$。