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For a finite family with a dependency graph of events, if $0\le x_i<1$ satisfy $\mathbb{P}(E_i)\le x_i{\prod }_{j\in N(i)}(1-x_j)$, then $\mathbb{P}({\bigcap }_i{E}_i^c)\ge {\prod }_i(1-x_i)>0$. Choosing equal $x_i$ gives the familiar symmetric criterion $eq(d+1)\le 1$ when event probabilities are at most $q$ and the maximum dependency degree is $d$.

For bad events $A_i$, a correlation graph is sufficient for the Lopsided Lovász local lemma when, for every set $S$ of nonneighbors and every positive-probability conditioning event, $\mathbb{P}(A_i\mid {\bigcap }_{j\in S}{A}_j^c)\le \mathbb{P}(A_i)$. A dependency graph of events satisfies this with equality. The condition concerns simultaneous avoidance, not just signs of pairwise correlations.

Let $B_v$ be independent Bernoulli random variables on a rooted $d$-ary tree, and put $A_v=\{B_v=1,B_w=0\text{ for every child }w\}$. Events at nonadjacent vertices use disjoint variables, giving a dependency graph of events of maximum degree $d+1$. Assign leaf success probability $p$ and successive probabilities $r_{j+1}=p/(1-r_j{)}^d$. If $p>{\max }_{0\le r\le 1}r(1-r{)}^d=p_d$, the recursion eventually reaches one. Set the final root success probability to one. Each event has probability at most $p$, but following success vertices down from the root must reach an occurring event. Thus the bad events cover the sample space. This gives an upper bound for uniform avoidance thresholds without assuming a converse to the Lovász local lemma.

A dependency graph of events joins potentially dependent events. Each event must be independent of the sigma-algebra generated by all its nonneighbors, a joint condition stronger than pairwise independence. In a product probability space, joining events whose sets of underlying independent variables overlap gives such a graph. This supplies the hypothesis of the Lovász local lemma.

Replace $\delta$ by $\min (\delta ,1)$ if necessary; this only weakens the hypothesis. We prove the result for finite graphs using a random proper partial graph coloring, then a deterministic completion. All constants below may depend on this fixed positive $\delta$.

First record the extension of a partial colouring by neighbour-colour savings. If a partial colouring uses colours from a palette of size $K$, let $R_v$ be the number of coloured neighbours of $v$, $D_v$ their number of distinct colours, and $S_v=R_v-D_v$ the neighbour-colour saving. There are $K-D_v$ available colours and $d(v)-R_v$ uncoloured neighbours. Thusat every vertex suffices for greedy completion: each subsequently coloured neighbour removes at most one available colour while reducing the uncoloured-neighbour count by one.

Call $v$ high-degree if $d(v)\ge (1-\delta /8)\Delta$. Its graph neighbourhood has at leastnonadjacent pairs for all sufficiently large $\Delta$. Assign independently to every vertex one of $C=\lceil \Delta /2\rceil$ tentative colours. Erase the colour of every vertex having a neighbour with the same tentative colour. The remaining colouring is proper; erasure is based on the original assignments and is not an iterative cascade.

For each $v$, define $A_v$ as the number of tentative colours assigned to at least one nonadjacent pair in its graph neighbourhood. Let $B_v$ count those same colours for which at least one neighbour of $v$ loses its colour. Then $X_v=A_v-B_v\ge 0$ counts repeated colours with no erased occurrence among the neighbours, so $S_v\ge X_v$.

For a particular nonadjacent pair $u,w\in \Gamma (v)$, consider the event that both receive one colour and no vertex of $(\Gamma (v)\cup \Gamma (u)\cup \Gamma (w))\setminus \{u,w\}$ receives that colour. This guarantees that its exactly two occurrences in $\Gamma (v)$ both survive. Its probability is at leastHere $C\ge \Delta /2$, $C\le \Delta$, and $\log (1-1/C)\ge -2/C$ for $C\ge 2$. Different pairs cannot be counted by the same colour in this event, since each event specifies exactly two occurrences in the graph neighbourhood. Thereforefor every high-degree vertex. This is the expected colour savings from sparse neighbourhoods.

We need simultaneous concentration, not just a positive expectation. A single tentative-colour change can affect $A_v$ or $B_v$ only in the old and new colour categories, so each function changes by at most two. A witness for $A_v\ge s$ fixes two vertices for each of $s$ colours. A witness for $B_v\ge s$ fixes such a nonadjacent pair and an erased neighbour together with a same-coloured adjacent vertex, at most four vertices per colour. Thus both are certifiable functions with Lipschitz constant $2$ and certificate parameter at most $4$. Both are bounded by $C\le \Delta$.

To see exactly what concentration follows, let an integer-valued $Z$ change by at most $L$ per coordinate and have certificates of size at most $rb$. For an input with $Z\ge b$, fix such a certificate $J$. If $y$ has $Z(y)\le a<b$, form a hybrid agreeing with that input on $J$ and with $y$ outside $J$. The hybrid has value at least $b$, so at least $(b-a)/L$ certificate positions differ from $y$. Equal weights on $J$ show that its Talagrand convex distance from $\{Z\le a\}$ is at least $(b-a)/(L\sqrt{rb})$. Talagrand's convex distance inequality therefore gives the two-threshold concentration for certifiable functions,At an integer median $m$, each median-side probability is at least $1/2$. Taking thresholds $m-t,m$ or $m,m+t$ and using $0\le Z\le \Delta$ givesfor $L=2,r=4$. Summing these tails gives $|\mathbb{E}Z-m|=O(\sqrt{\Delta })$. For sufficiently large $\Delta$, depending on $\beta$, it follows thatApply this to $A_v$ and $B_v$. Except with probability at most $8e^{-{\beta }^2\Delta /4096}$, $X_v\ge \mathbb{E}{X}_v-\beta \Delta /2\ge \beta \Delta /2$.

A bad event at $v$ depends only on tentative colours at distance at most two. It is independent of the entire collection of bad events at distance greater than four. Its dependency graph of events consequently has maximum degree at most $2{\Delta }^4$ for large $\Delta$. The Lovász local lemma now makes all high-degree vertices good simultaneously: exponential decay beats this polynomial dependency bound.

The needed local-lemma step can be checked directly. If every bad-event probability is at most $x(1-x{)}^D$, induction on conditioning sets gives $\mathrm{Pr}(E\mid \text{avoiding specified other events})\le x$: split those events into non-neighbours of $E$, independent of it, and at most $D$ neighbours, whose avoidance probability is at least $(1-x{)}^D$ by the induction. Multiplying the resulting conditional avoidance probabilities makes total avoidance positive. Taking $x=1/(D+1)$ works whenever $ep(D+1)\le 1$, since $(1-1/(D+1){)}^D\ge e^{-1}$. Our bound satisfies this criterion once $\Delta \ge {\Delta }_0(\delta )$.

Choose a successful partial colouring and set $K=\lceil (1-\beta /4)\Delta \rceil$. This contains the original palette. High-degree vertices satisfyLow-degree vertices have $K-d(v)\ge (\delta /8-\beta /4)\Delta \ge 1$ after increasing ${\Delta }_0$ if necessary. Greedy completion therefore gives $\chi (H)\le K$. Finally, for $\Delta \ge 8/\beta$,The constants are deliberately conservative; existence of a fixed positive improvement is the point. The same proof uses an external maximum degree bound, so if infinite locally finite graphs are allowed, every finite subgraph has this uniform colouring bound and compactness extends the colouring to the whole graph.

For the stronger triangle-free result, the appropriate brief route is semi-random palette refinement for triangle-free colouring. Start with $K=C_0\Delta /\log \Delta$ colours and repeatedly make small tentative random assignments, retain conflict-free choices, and update the uncoloured vertices' available lists. Triangle-freeness makes each graph neighbourhood independent. Track list size and average residual neighbour congestion per colour, prune unusually congested colours, and use concentration inequalities and the Lovász local lemma to preserve these invariants through the rounds. The ratio of residual congestion to list size decreases until a final completion using the Lovász local lemma or greedy coloring succeeds. Average control and pruning are important: four-cycles are allowed, so assuming tight uniform concentration for every individual colour load would be unjustified. With a sufficiently large absolute $C_0$, this programme gives $\chi (H)=O(\Delta /\log \Delta )$. The one-round fixed-fraction saving proved above alone does not establish that logarithmic improvement.

To show the order is best possible, let $d\to \infty$ through integers and sample $G(n,d/n)$ with $n=d^7$. A Chernoff bound and union bound give maximum degree at most $2d$ with probability tending to one. The expected triangle count is at most $d^3/6=o(n)$, so fewer than $n/2$ vertices need be deleted to remove all triangles with high probability. For $s=\lceil 4n\log d/d\rceil$, the expected number of independent $s$-sets is at mostits logarithm is at most $s[\log (ed/(4\log d))-2\log d+o(1)]$, which tends to minus infinity. The altered triangle-free graph thus has at least $n/2$ vertices and independence number below $s$, giving $\chi \ge (c+o(1))d/\log d$ for an absolute positive $c$. Append a disjoint star graph whose centre has vertex degree $2d$ if needed to make its maximum degree exactly $\Delta =2d$. This proves the triangle-free chromatic lower bound at bounded maximum degree, $\chi =\Omega (\Delta /\log \Delta )$, so the preceding order cannot be improved in general.

The asymmetric Lovász local lemma says the following. For finitely many bad events with a dependency graph of events, if $0\le x_i<1$ satisfythenHere independence from nonneighbors is independence from their whole generated system, not just pairwise independence.

Prove by induction on $|S|$ that $\mathrm{Pr}(A_i\mid {\bigcap }_{j\in S}{A}_j^c)\le x_i$ whenever $i\notin S$, with all conditioning probabilities positive. At $S=\varnothing$, the hypothesis gives $\mathrm{Pr}(A_i)\le x_i$. Split $S$ into its neighbor part $S_1$ and nonneighbor part $S_2$. ThenThe numerator uses joint nonneighbor independence. The denominator is exposed one event at a time; every conditional bad-event probability involves fewer than $|S|$ conditioning events, so induction supplies the product lower bound. The same product argument gives positivity of the conditioning events before using the fraction. Finally expose all complements in any order and multiply their conditional probabilities to obtain the claimed lower bound.

For maximum dependency degree $d\ge 1$ and event probabilities at most $p$, choose $x_i=1/(d+1)$. Since $(1+1/d{)}^d\le e$, the symmetric Lovász local lemma follows fromFor $d=0$, the events are jointly independent and avoidance has positive probability whenever $p<1$, so the displayed sufficient criterion also holds.

To obtain the local lemma Ramsey lower bound, colour the edges of $K_N$ independently with two equally likely colours. For every $k$-set $S$, the event that it is monochromatic has probability $p=2^{1-\left(\genfrac{}{}{0ex}{}{k}{2}\right)}$. Events with intersections of size at most one depend on disjoint edge trials. Counting potentially shared pairs givesThe Lovász local lemma excludes all monochromatic $K_k$ if $ep(d+1)\le 1$. For $N\le ck{2}^{k/2}$, Stirling's approximation and $\left(\genfrac{}{}{0ex}{}{N-2}{k-2}\right)\le N^{k-2}/(k-2)!$ give, uniformly for $c$ bounded above and away from zero,For a concrete approach to the limiting constant, choose $c_k=(\sqrt{2}/e)(1-3\log k/k)$ and take the integer part of $c_{k}k{2}^{k/2}$. Then $(e{c}_k/\sqrt{2}{)}^k\le k^{-3}$ for all sufficiently large $k$, so the criterion holds. Such a colouring shows $R(k)>N$, and thereforeThe undefined $o(i)$ printed in the PDF is read as the intended $o(1)$, with $k\to \infty$.

For the cycle, independently choose one of the twelve vertices in each part. Ignore cycle edges whose endpoints lie in the same part, since both endpoints cannot be selected. Every other edge defines a bad event of probability $1/144$. It uses just the choices in its two endpoint parts. A part's twelve vertices meet at most twenty-four cycle edges, so a bad event has at most $48-1=47$ neighbors in a dependency graph of events. Since $e(47+1)/144=e/3<1$, the Lovász local lemma gives a selection with no selected adjacent pair. Thus an independent transversal of size $n$ exists, for every stated partition.

The general, or asymmetric Lovász local lemma, is as follows. Let $E_1,\dots ,E_m$ be events, with a dependency graph of events having neighbour sets $N(i)$. Each $E_i$ must be independent of the sigma-algebra generated by all the events indexed by its nonneighbours. If numbers $0\le x_i<1$ satisfythenFor completeness, the key inductive bound is $\mathbb{P}(E_i\mid {\bigcap }_{j\in S}{E}_j^c)\le x_i$ whenever $i\notin S$. Split $S$ into neighbours $S_1$ of $i$ and nonneighbours $S_2$. If $S_1$ is empty, the dependency condition gives the bound directly. Otherwise,The denominator bound follows by exposing the exclusions in $S_1$ one at a time and using the induction hypothesis with strictly fewer conditioning events. The same induction ensures that all conditioning events have positive probability. Finally multiply the successive conditional probabilities of avoiding $E_1,\dots ,E_m$ to obtain the stated lower bound.

For the symmetric form, suppose each event has probability at most $q$ and the dependency graph has maximum degree $D\ge 1$. Choose $x_i=1/(D+1)$. ThenThe final inequality follows from $(1+1/D{)}^D\le e$. Thus $eq(D+1)\le 1$ suffices for positive probability that no bad event occurs. If $D=0$, the dependency condition gives joint independence and the same criterion implies $q<1$, so the conclusion follows directly.

Apply this to a fair independent two-colouring of the hypergraph vertices, with $E_W$ the event that hypergraph edge $W$ is monochromatic. Its probability is $q=2^{1-r}$. Events on disjoint hypergraph edges use disjoint independent variables, giving a dependency graph by joining intersecting hypergraph edges. Each hypergraph edge meets at mostother hypergraph edges; repeated counting only enlarges this upper bound. The slightly looser bound $D+1\le r(\Delta +1)$ is enough for the printed condition. Indeed,The symmetric Lovász local lemma therefore gives a proper two-colouring. A hypergraph with no hypergraph edges is trivially colourable and needs no dependency calculation.

The claimed sufficient condition is false for the printed all-colours requirement. This is a polychromatic hypergraph colouring problem, which is stronger than merely avoiding monochromatic edges.

For a counterexample even with $k=3$, take $V=\{1,2,3,4\}$ and all four triples as edges. Every edge meets the other three, so $m=3$, andAny assignment of three colours to four vertices repeats a colour on a pair. Any triple containing that pair has at most two colours. Thus this uniform hypergraph satisfies the numerical hypothesis but has no required colouring. A single two-vertex edge would already provide a smaller counterexample if $k=2$ is allowed.

Here is the correlation graph of events argument that correctly applies to the printed colouring definition. Colour each vertex independently and uniformly with three colours, and let $B_E$ be the event that edge $E$ misses some colour. For $k\ge 1$, the inclusion-exclusion principle givesJoin two bad events when their edges intersect. This is a dependency graph of events, hence also a correlation graph of events, of maximum degree at most $m$: each $B_E$ depends only on colours on $E$, which are jointly independent of the colours on all disjoint edges. Precisely, the symmetric Lovász local lemma says that events of probability at most $q$ with such a graph have positive simultaneous avoidance probability whenever $eq(m+1)\le 1$. It follows from the asymmetric form $q\le x(1-x{)}^m$, taking $x=1/(m+1)$ for $m\ge 1$ and using $(m/(m+1){)}^m\ge e^{-1}$; the case $m=0$ is direct independence.

Thus a valid replacement sufficient condition isor the simpler, stronger condition $3e(m+1)\le (3/2{)}^k$. Under either replacement, avoiding all $B_E$ gives exactly the requested polychromatic hypergraph colouring. For infinitely many vertices, finite subfamilies can be coloured and compactness of the product of three-point spaces extends this to all constraints; positive simultaneous avoidance probability is only asserted for finite families. The printed $3^k$ bound cannot be justified by substituting monochromatic-event probabilities for missing-colour probabilities.

Color each integer independently and uniformly with one of $k$ colors. For a translate $S+t$, let $E_t$ be the event that some color is missing. The union bound givesJoin two events in the dependency graph of events when their translates overlap. Nonneighbors depend on disjoint sets of independent colors, giving the required joint independence. Overlap occurs only if $u-t\in S-S$, so the maximum degree is at most $s(s-1)$.

With natural logarithms, $s\ge 6k\log k$ and $k\ge 20$ implyThe middle estimate uses that $s^2{e}^{-s/k}$ is decreasing for $s\ge 2k$, and the next uses $\log k\le \sqrt{k}$. Therefore the Lovász local lemma supplies a coloring avoiding every bad event in any specified finite family of translates.

To obtain one coloring for all translates, use the compactness extension of the Lovász local lemma. At level $N$, consider colorings of $[-N,N]\cap \mathbb{Z}$ satisfying every constraint whose translate is contained in that interval. There are finitely many constraints, and the preceding argument makes the level nonempty. Restrictions connect these colorings into a finitely branching tree. The König infinity lemma gives an infinite branch, hence a coloring of all integers. Every translate eventually lies in a level of this branch, so it contains every color.

Thus there exists a $k$-coloring of $\mathbb{Z}$ in which every translate of $S$ contains all $k$ colors. This is a polychromatic coloring of integer translates; the compactness step establishes existence of one simultaneous coloring.

An independence graph of events, also called a dependency graph of events, has one vertex for each event $E_i$, with $E_i$ independent of the entire family of events indexed by its nonneighbors. Precisely, $E_i$ is independent of the sigma-algebra generated by those events. Pairwise independence alone is insufficient.

The asymmetric Lovász local lemma says that if numbers $0\le x_i<1$ satisfythen $\mathbb{P}({\bigcap }_i{E}_i^c)\ge {\prod }_i(1-x_i)>0$ for a finite family. In the commonly used symmetric Lovász local lemma, if each probability is at most $q$ and the maximum graph degree is at most $d$, the sufficient condition is

For finite $S\subset \mathbb{Z}$, use vertices $\mathbb{Z}$ and hyperedges $S+t$. Independent uniform $k$-coloring gives missing-color probability at most $k{e}^{-|S|/k}$, and the dependency graph of events has degree at most $|S|(|S|-1)$. For $k\ge 20$ and $|S|\ge 6k\log k$, the symmetric Lovász local lemma condition holds. The compactness extension of the Lovász local lemma gives a coloring in which every translate contains every color.

A colouring is polychromatic with $r$ colours if every hypergraph edge contains a vertex of every colour. This is stronger than requiring every edge to be nonmonochromatic. Under independent uniform colours, a $k$-element edge misses at least one of three colours with probability $3(2/3{)}^k-3(1/3{)}^k$. If each edge meets at most $m$ other edges, the edge-intersection dependency graph of events and the Lovász local lemma therefore give a polychromatic three-colouring whenever $e(m+1)[3(2/3{)}^k-3(1/3{)}^k]\le 1$.