Given a 1‐ipm μ, denote by Gμ the associated random graph model. Let ℳ1,⩾p(G) denote the collection of 1‐ipms μ on G for which each edge is included in Gμ with probability at least p. For G=Z2, Balister and Bollobás asked for the value of the least p⋆ such that for all p > p⋆ and all μ∈ℳ1,⩾p(G), Gμ almost surely contains an infinite component. In this paper, we significantly improve previous lower bounds on p⋆. We also determine the 1‐independent critical probability for the emergence of long paths on the line and ladder lattices. Finally, for finite graphs G we study fG1, (p), the infimum over all μ∈ℳ1,⩾p(G) of the probability that Gμ is connected. We determine fG1, (p) exactly when G is a path, a complete graph and a cycle of length at most 5.A probability measure  Spanning subgraphs of G are called configurations. In a configuration H, an edge is said to be open if it belongs to H, and closed otherwise. A bond percolation model on the host graph G is a probability measure μ on the spanning subgraphs of G, that is, on the space of configurations. Given such a measure, we denote the corresponding random graph model by Gμ, and refer to it as the μ‐random graph or μ‐random configuration.Let μ where the states (open or closed) of edges in subsets F1, F2 of E in a μ‐random configuration are independent provided that the edges in F1 and F2 are “sufficiently far apart.” To make this more precise, we make use of the following definition.In this paper, we study bond percolation models Definition 1F1, F2 ⊆ E are k‐distant if F1 ∩ F2 = ∅ and the shortest path of G from an edge in F1 to an edge in F2 contains at least k edges. A bond percolation model μ on G is k‐independent if for any pair  of k‐distant edge sets, the intersections Gμ∩F1 and Gμ∩F2 are independent random variables.Two edge sets μ is 0‐independent if each edge of Gμ is open at random independently of all the others, that is, μ can be viewed as a product of Bernoulli measures on the edges of G. A well‐studied 0‐independent model is the Erdős‐Rényi random graph Gn, p, where the host graph is G = Kn, the complete graph on n vertices, and where μ, known as the p‐random measure, sets each edge to be open with probability p, independently of all the others.So for example site percolation on the square integer lattice. In this case the host graph is the square integer lattice Z2 (where two vertices are joined by an edge if they lie at distance 1 apart), and the measure μ=μsite(θ) is obtained by switching each vertex of Z2on at random with probability θ, independently of all the others, and by setting an edge to be open if and only if both of its endpoints are switched on. Site percolation measures may be defined more generally on any host graph in the natural way.In this paper, we focus on the next strongest notion of independence, namely 1‐independence. Measures that are 1‐independent have the property that events determined by vertex‐disjoint edge sets are independent. For many 1‐independent models, the randomness can be thought to “reside in the vertices.” An important example of a 1‐independent model is that of v ∈ V(G) a state Sv at random, and set an edge uv to be open if and only if f = 1, for some deterministic function f (which may depend on u and v). We refer to such measures as vertex‐based measures , it is well‐known that a graph G may support many 1‐independent measures which cannot be realized as vertex‐based measures or as general “block factors,” see for instance G, it is not feasible to generate or simulate the collection of 1‐independent measures of G.An important point to note is that while all 0‐independent bond percolation models are a product of Bernoulli measures on the edges of Definition 2μ on a host graph G, the (lower)‐edge‐probability of μ is Given a bond percolation model p‐random measure has edge‐probability p, while a site percolation measure with parameter θ has edge‐probability θ2. The collection of k‐independent bond percolation models μ on a graph G with edge‐probability d(μ)⩾p is denoted by ℳk,⩾p(G).So for instance a Remark 3μ∈ℳk,⩾p(G), we may readily produce a measure μ˜∈ℳk,⩾p(G) such that μ˜({eis open})=p for all e ∈ E(G) via random sparsification: independently delete each edge e of Gμ with probability p/μ({eis open})∈. The resulting bond percolation model on G is clearly k‐independent and has the property that each edge is open with probability exactly p; the corresponding bond percolation measure μ˜ thus has the required properties.Given a measure 1.2Percolation theory is the study of random subgraphs of infinite graphs. Since its inception in Oxford in the 1950s, it has blossomed into a rich theory and has been the subject of several monographs G, and let μ be a 0‐independent bond percolation model on G. We say that percolation occurs in a configuration H on G if H contains an infinite connected component of open edges. By Kolmogorov's zero‐one law, for G and μ as above, percolation is a tail event whose μ‐probability is either zero or one. This allows one to thus define the Harris critical probability pc0, (G) for 0‐independent percolation: In the most fundamental instance of this problem, consider an infinite, locally finite connected graph Problem 4Given an infinite, locally finite connected graph G, determine pc0, (G).p0,c(Z2) to be 1/2.One of the cornerstones of percolation theory—and indeed one of the triumphs of twentieth century probability theory—is the Harris‐Kesten theorem, which established the value of Letμbe the p‐random measure onZ2. Then. Z2 if the assumption of 0‐independence is weakened to k‐independence. In particular, how much can local dependencies between the edges postpone the global phenomenon of percolation?In this paper, we focus on the question of what happens to the Harris critical probability in Definition 5G be an infinite, locally finite connected graph and let k∈N0. The Harris critical probability for k‐independent percolationG is defined to be: Let Problem 6Determinep1,c(Z2).Problem renormalisation, which entails reducing a 0‐independent model to a 1‐independent one (possibly on a different host graph), trading in some dependency for a boost in edge‐probabilities. Renormalisation arguments feature in many proofs in percolation theory; a powerful and particularly effective version of such arguments was developed by Balister, Bollobás, and Walters A standard technique in percolation is Z2  and Monte‐Carlo simulations to estimate the probabilities of bounded events, has been applied to give rigorous confidence intervals for critical probabilities/intensities in a wide variety of settings: various models of continuum percolation Z3 Their method, which relies on comparisons with 1‐independent models on k‐independent bond percolation models for k⩾1 are as natural an object of study as the more widely studied 0‐independent ones.From a more practical standpoint, many of the real‐world structures motivating the study of percolation theory exhibit short‐range interactions and local dependencies. For example a subunit within a polymer will interact and affect the state of nearby subunits, but perhaps not of distant ones. Similarly, the position or state of an atom within a crystalline network may have a significant influence on nearby atoms, while long‐range interactions may be weaker. Within a social network, we would again expect individuals to exert some influence in esthetic tastes or political opinions, say, on their circle of acquaintance, and also expect that influence to fade once we move outside that circle. This suggests that many very different 1‐independent models with edge‐probability p, and they tend to be harder to study than 0‐independent ones due to the extra dependencies between edges. In particular simulations are often of no avail to formulate conjectures or to get an intuition for 1‐independent models in general. Moreover, while the theoretical motivation outlined above is probabilistic in nature, the problem of determining a critical constant like p1,c(Z2) is extremal in nature—one has to determine what the worst possible 1‐independent model is with respect to percolation—and calls for tools from the separate area of extremal combinatorics.Despite the motivation outlined above, 1‐independent models remain poorly understood. To quote Balister and Bollobás from their 2012 paper: “1‐independent percolation models have become a key tool in establishing bounds on critical probabilities […]. Given this, it is perhaps surprising that some of the most basic questions about 1‐independent models are open.” There are in fact some natural explanations for this state of affairs. As remarked on in the previous subsection, there are In this paper, we continue Balister and Bollobás's investigation into the many open problems and questions about and on these measures. Before we present our contributions to the topic, we first recall below previous work on 1‐independent percolation.1.3k‐independent models by 0‐independent ones were given by Liggett, Schonmann and Stacey p1,c(Z2)<1. Balister, Bollobás and Walters Z2 with edge‐probability at least 0.8639, the origin has a strictly positive chance of belonging to an infinite open component. This remains to this day the best upper bound on p1,c(Z2). In a different direction, Balister and Bollobás p1,c(G)⩾12 for any infinite, locally finite connected graph G. In the special case of the square integer lattice Z2, they recalled a simple construction due to Newman which gives θsite is the critical value of the θ‐parameter for site percolation, that is, the infimum of θ∈ such that switching vertices of Z2 on independently at random with probability θ almost surely yields an infinite connected component of on vertices. Plugging in the known rigorous bounds for 0.556⩽θsite⩽0.679492 p1,c(Z2)⩾0.5062, while using the nonrigorous estimate θsite≈0.592746 ⩾0.5172.Some general bounds for stochastic domination of k‐independent setting by Mathieu and Temmel With regards to other lattices, Balister and Bollobás completed a rigorous study of 1‐independent percolation models on infinite trees 1.4p1,c(Z2) with the following theorems.In this paper, we make a three‐fold contribution to the study of Problem Theorem 7For alld∈N⩾2, we have thatd = 2 given in θsite again denote the critical threshold for site percolation. Then the following holds.Theorem Theorem 8θsite⩾0.556 into Theorem p1,c(Z2)⩾0.531136, which does slightly worse than Theorem θsite≈0.592746 yields a significantly stronger lower bound of p1,c(Z2)⩾0.554974.Substituting the rigorous bound p1,c(Z2), and in particular to establish some 1‐independent analogs of the Russo‐Seymour‐Welsh (RSW) lemmas on the probability of crossing rectangles, we investigate the following problems. Let Pn denote the graph on the vertex set {1, 2, … n} with edges {12, 23, … , (n − 1)n}, that is, a path on n vertices. Given a connected graph G, denote by Pn × G the Cartesian product of Pn with G. A left‐right crossing of Pn × G is a path from a vertex in {1} × V(G) to a vertex in {n} × V(G). We define the crossing critical probability for 1‐independent percolation on Pn × G to be Pn × G, there is a strictly positive probability of being able to cross Pn × G from left to right.Second, motivated by efforts to improve the upper bounds on Problem 9Givenn∈Nand a finite, connected graph G, determine p1, ×(Pn × G).n → ∞ in Problem G be an infinite, locally finite connected graph. The long paths critical probability for 1‐independent percolation on G is G.Problem Problem 10Given an infinite, locally finite, connected graph G, determine pℓp1, (G).G consists of a vertex or an edge p1,ℓp(Z)=34, and(ii)p1,ℓp(Z×P2)=23.p1, ×(Pn × G) and p1,ℓp(Z×G) for a variety of graphs G. We summarize the latter, less technical, set of results below. Let Cn and Kn denote the cycle and the complete graph on n vertices respectively.Note that part (i) of Theorem Theorem 12We have that(i)0.5359…=4−23⩽p1,ℓp(Z×Cn)⩽p1,ℓp(Z×Pn)⩽23for alln⩾3;(ii)p1,ℓp(Z×K3)⩽116(13−5512814−2513+12814−2513)=0.63154… ;(iii)p1,ℓp(Z×C4)⩽(3−3)/2=0.63397… ;(iv)p1,ℓp(Z×C5)⩽0.63895… ;(v)0.5359…=4−23⩽limn→∞p1,ℓp(Z×Kn)⩽59=0.5555….Pn × G to the probability of a given copy of G being connected in that model. This motivated our third contribution to the study of 1‐independent models in this paper, namely an investigation into the connectivity of 1‐independent random graphs.A key ingredient in the proof of Theorems Definition 13G be a finite connected graph. For any p ∈ , we define the k‐independent connectivity function of G to be Let Problem 14Given a finite connected graph G, determine fG1, (p).G is a path, a complete graph or a cycle on at most 5 vertices.We resolve Problem Theorem 15Givenn∈N⩾2and p ∈ , letθ=θ(p):=1+4p−32andpn:=143−tan2πn+1. We have thatTheorem 16Givenn∈N⩾2and p ∈ , letθ=θ(p):=1+2p−12andpn:=12(1−tan2(π2n)). We have thatIn particular,Theorem 17For p ∈  we have thatTheorem 18For p ∈  we have thatℳk,⩽p(G) denote the collection of 1‐independent measures μ on G such that supe∈E(G)μ{eis open}⩽p. Set We also consider the opposite problem to Problem Problem 19Given a finite connected graph G, determine FG1, (p).G is a path, a complete graph or a cycle on at most 5 vertices.We resolve Problem Theorem 20For alln∈Nwithn⩾2,F1,Pn(p)=p⌊n2⌋.Theorem 21For alln∈Nwithn⩾2,F1,Kn(p)=1−f1,Kn(1−p).Theorem 22Theorem 23μ on Kn, Pn, C4 and C5—i.e., the range of values μ({connected}) can take if every edge is open with probability p. In Figure G with plots of fG1, (p), FG1, (p) and f0,G(p):=μGpis connected, where Gp is the 0‐independent model on G obtained by setting each edge of G to be open with probability exactly p, independently at random.Together, Theorems 1.5Our first set of results, Theorems Z×G. This result is used in Sections In Section fG1, (p) and FG1, (p) when G is a path, a complete graph or a short cycle. We apply these results in Section In Sections 1.6N for the set of natural numbers {1, 2, …}, N0 for the set N∪{0}, and N⩾k for the set of natural numbers greater than or equal to k.We write n]: = {1, 2, … n}. Given a set A, we write Ar) where V = V(G) and E = E(G) ⊆ V(G)(2) denote the vertex set and edge set of G respectively. Given a subset A ⊆ G, we denote by G[A] the subgraph of G induced by A. We also write N(A) for the set of vertices in G adjacent to at least one vertex in A.We set ][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Blue.If v is a vertex in Tk, where k ≡ 1[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Red with probability q/2 and color it Blue otherwise.If v is a vertex in Tk, where k ≡ 2[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Red with probability q and put it in the Inwards state I otherwise.If v is a vertex in Tk, where k ≡ 3[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Red.If v is a vertex in Tk, where k ≡ 4[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Blue with probability q/2 and color it Red otherwise.If v is a vertex in Tk, where k ≡ 5[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 6, then we color v Blue with probability q and put it in the Inwards state I otherwise.If Let Tk + 3, Tk + 4, Tk + 5 are the same as those for Tk, Tk + 1, Tk + 2 respectively, except with red and blue interchanged. See Figure T0, T1, T2 and T3 when d = 2. Now suppose that e = {v1, v2} is an edge in Zd. First we say that the edge e is open if either both v1 and v2 are Blue or both v1 and v2 are Red. We also say the edge e is open if, for some k, we have that v1 ∈ Tk, v2 ∈ Tk + 1, and v2 is in state I. In all other cases we say that the edge e is closed. It is clear that this gives a 1‐independent measure on Zd as it is vertex‐based, and it is also easy to check that every edge is present with probability at least 4−23.Note that the rules for μ, and let G:=Zd. We claim that in Gμ, for all k ≡ 0[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 3, there is no path of open edges from Tk to Tk + 3. Suppose this is not the case, and P is some path of open edges from a vertex in Tk to Tk + 3. We first note that P cannot include a vertex in state I, as such a vertex would be in Tk + 2 and would only be adjacent to a single edge. Thus every vertex of P is either Blue or Red. However, as one end vertex of P is Blue and the other end vertex is Red, and there are no open edges with different colored end vertices, we have that such a path P cannot exist. As a result, every component of Gμ is sandwiched between some Tk − 3 and Tk + 3, where k ≡ 0[[spimspace="0.3em"]][[spimspace="0.3em"]] mod[[spimspace="0.3em"]] 3, and so is of finite size. Thus we have that p1,c(Zd)⩾4−23.Call this measure G, and a vertex set A ⊆ V(G), let A¯ be the closure of A under 2‐neighbor bootstrap percolation on G. That is, let A¯:=∪i⩾0Ai, where A0: = A and for i⩾1G has the finite 2‐percolation property if, for every finite set A ⊆ V(G), we have that A¯ is finite.The construction in Theorem Corollary 24If G has the finite 2‐percolation property, thenp1,c(G)⩾4−23.V(G) in the following way: pick any vertex v and set T0: = {v}. For k⩾1 let w ∈ Tk, then w is only adjacent to vertices in Tk − 1, Tk and Tk + 1. Moreover, w is adjacent to at most one vertex in Tk − 1—this is the crucial property needed for our construction. Since G has the finite 2‐percolation property, each Tk is finite. Thus we can use the Tk to construct a nonpercolating 1‐ipm on G in the exact same fashion as done for Zd in Theorem I are still dead ends, being incident to a unique edge), which in turn shows that p1,c(G)⩾4−23.Partition p1,c3,4,6,4⩾0.52981682. As this is less than 4−23, we have that our construction gives the (rigorous) improvement of p1,c3,4,6,4⩾4−23.An example of a lattice with the finite 2‐percolation property is the lattice , where here we are using the lattice notation of Grünbaum and Shephard Proof of Theorem 8ε > 0 sufficiently small so that q:=θsite(Z2)−ε is strictly larger than 1/4. For each vertex v∈Z2, we assign to it one of three states: On, L or D, and we do this independently for every vertex. We assign v to the On state with probability q, we assign it to the L state with probability 12(1−q), and else we assign it to the D state with probability 12(1−q).Fix e is open if both of its vertices are in the On state. If a vertex is in state L, then the edge adjacent and to the left of it is open. Similarly, if a vertex is in state D, then the edge adjacent and down from it is open. All other edges are closed. See Figure We now describe which edges are open based on the states of the vertices. We first say that the edge Z2 as it is vertex‐based, and every edge is present with probability q2+12(1−q). Call this measure μ and let G:=Z2. We will show that every component of Gμ has finite size. We begin by first proving an auxiliary lemma. Let t∈, and let us define another 1‐independent measure on Z2, which we call the left‐down measure with parameter t. In the left‐down measure, each vertex of Z2 is assigned to one of three states: Off, L or D, and we do this independently for every vertex. For each vertex v∈Z2, we assign it to state L with probability t, we assign it to state D with probability t, and we assign it to state Off with probability 1 − 2t. As above, if a vertex is in state L, then the edge adjacent and to the left of it is open, while if a vertex is in state D, then the edge adjacent and down from it is open. All other edges are closed. We use νt to denote the left‐down measure with parameter t.It is easy to see that this is a 1‐independent measure on Lemma 25If0⩽t⩽38, then all components inGνtare finite almost surely.z:=1−1−2t. As 0⩽t⩽38 we have that 0⩽z⩽12. We start by taking a random subgraph of Z2 where every edge is open with probability z, independently of all other edges. We then further modify it as follows. For each vertex v =  we look at the state of the edge e1 from v to the vertex , and the state of the edge e2 from v to the vertex . If at least one of e1 or e2 is closed, we do not change anything. However, if both e1 and e2 are open, with probability 12 we close the edge e1, and otherwise we close the edge e2. We do this independently for every vertex v of Z2.Let νt, the left‐down measure with parameter t. Indeed, to each vertex v =  as above we may assign a state Off if both the edge e1(v) to the vertex to the left of v and the edge e2(v) to the vertex below v are closed, a state L if e1(v) is open and a state D if e2(v) is open. The probabilities of these three states are (1 − z)2 = 1 − 2t, t and t respectively, and since the vertex states depend only on the pairwise disjoint edge sets {e1(v),e2(v)}v∈Z2, they are independent of one another just as in the νt measure.It is easy to see that this is an equivalent formulation of νt to the 0‐independent bond percolation measure ξ on Z2 with edge‐probability z. In this coupling we have that if an edge e is open in Gνt, then it is also open in Gξ. As z⩽0.5 we have that all components in Gξ are finite by the Harris‐Kesten theorem, and so we also have that all of the components in Gνt are finite too.Thus we have coupled 0⩽t<12, then almost surely all components in Gνt are finite. We make no use of this stronger result in this paper, so we omit its proof. It is also clear that when t=12, every vertex in Gνt is part of an infinite path consisting solely of steps to the left or steps downwards, and so percolation occurs in Gνt at this point.By considering an appropriate branching process it is possible to prove the stronger result that if μ, where every vertex is in state On, L or D. Recall that our aim is to show that all components have finite size in Gμ. Consider removing all vertices in state L or D, and also any edges adjacent to these vertices. What is left will be a collection of components consisting only of edges between vertices in the On state, which we call the On‐sections. The black edges in Figure q<θsite(Z2), we have that almost surely every On‐section is finite. Similarly, consider removing all edges in the On‐sections. What is left will be a collection of edges adjacent to vertices in the L or D states. We call these components the LD‐sections; the dashed red edges in Figure LD‐sections. As each vertex is in state L with probability 12(1−q)⩽38 and in state D with the same probability, Lemma LD‐section is finite.Let us return to our original 1‐independent measure v in state L orient the open edge to the left of it away from v, while for each vertex v in state D orient the open edge below it away from v. This gives a partial orientation of the open edges of Gμ, in which every vertex in state L or D has exactly one edge oriented away from it, and vertices in state On have no outgoing edge. Furthermore, if v1 is a vertex in the On state and v2 is a vertex in the L or D state, then the edge between them is oriented from v2 to v1. Since the LD‐sections are almost surely finite, this implies the LD sections under this orientation consist of directed trees, each of which is oriented from the leaves to a unique root, which is in the On state. In particular, every LD‐section attaches to at most one On‐section. As such, almost surely every component in Gμ consists of at most one On‐section, and a finite number of finite LD‐sections attached to it. Thus almost surely every component in Gμ is finite.For each vertex 3G be a finite connected graph. Set v(G): = |V(G)|. Recall that for any 1‐independent bond percolation measure μ∈ℳ1,⩾p(G), we have μ(Gμis connected)⩾f1,G(p).Let Theorem 26If p satisfiesfor someα∈. For any n∈N, the restriction of μ to [n] × V(G) is a measure from ℳ1,⩾p(Pn×G), and clearly all such measures can be obtained in this way. Furthermore, for every n∈N, the restriction of μ to {n} × V(G) is a measure from ℳ1,⩾p(G), and in particular the subgraph of (Z×G)μ induced by {n} × V(G) is connected with probability at least fG1, (p).Consider an arbitrary measure μ‐random graph (Z×G)μ. For n⩾1 let Yn be the event that [n] × V(G) induces a connected subgraph. For n⩾2, let Xn be the event that [n − 1] × V(G) induces a connected subgraph and at least one vertex in {n} × V(G) is connected to a vertex in {n − 1} × V(G). For n = 1, set X1 to be the trivially satisfied event occurring with probability 1. For n⩾1, let Vn be the event that {n} × V(G) induces a connected subgraph, and for n⩾2 let Hn be the event that at least one of the edges from {n − 1} × V(G) to {n} × V(G) is present.We consider the Xn = Yn − 1 ∩ Hn and Xn ∩ Vn ⊆ Yn. From here, we obtain the following inclusions:Xn + 1)c ∩ Yn = (Hn + 1)c ∩ Yn,(Yn ∩ Yn − 1 ⊇ (Vn ∩ Xn) ∩ Yn − 1, andYn)c ∩ Xn ⊆ (Vn)c ∩ Xn. and (b) we have, We begin by establishing two inductive relations for the sequences xn⩽αf1,G(p) for all n. Indeed x1 = 0, and if xn⩽αf1,G(p), then by Second, using (c), Finally, we have that G, fG1, (p) is a nondecreasing function of p with fG1, (1) = 1. Thus the function )2 is also nondecreasing in p and attains a maximum value of 1 at p = 1. On the other hand, the function 4(1 − p)v(G) is strictly decreasing in p and is equal to 4 at p = 0. Thus there exists a unique solution p⋆ = p⋆(G) in the interval  to the equation For any finite connected graph Theorem Corollary 27Let G be a finite connected graph. Let p⋆ = p⋆(G) be as above. Thenα=1/2.Apply Theorem 4In this section we prove a lemma that we shall use in Sections Lemma 28Let G be a finite graph, and let𝒬:={QH(θ):H⊆G}be a set of polynomials with real coefficients, indexed by subgraphs of G. Givenθ∈C, letμθbe the following function from subgraphs of G toC:Suppose there exists a nontrivial intervalI⊆Rsuch that, for allθ∈I, the functionμθdefines a 1‐ipm on G. Suppose further that there exists a setX⊆Csuch that, for allθ∈Xand all H ⊆ G,μθ(H)is a nonnegative real number. Thenμθis a 1‐ipm on G for allθ∈X.μθ is a measure on G for all θ∈X. As μθ(H) is a nonnegative real number for all θ∈X and all H ⊆ G, all that is left to prove is that θ with real coefficients, and is equal to zero for all θ in the interval I. By the fact that a nonzero polynomial over any field has only finitely many roots, the polynomial is identically zero and so θ.We start by proving that μθ is a 1‐ipm on G for all θ∈X. To do this we must show that the following holds true for all θ∈X, for all A, B ⊆ V(G) such that A and B are disjoint, and all G1 and G2 such that G1 is a subgraph of G[A] while G2 is a subgraph of G[B]: θ with real coefficients—the left hand side, for example, can be written as μθ is a 1‐ipm on G for all θ∈I, we have that these two polynomials agree on I, and so must be the same polynomial. Thus We now show that 5n∈N⩾2 and p ∈ , we let θ=θ(p):=1+4p−32 and pn:=143−tan2πn+1. Let gn(θ):=∑j=0nθj(1−θ)n−j.In this section we prove Theorem νp∈ℳ1,⩾p(Pn) as follows. Let us start with the case p⩾34. For each vertex of Pn, we set it to state 0 with probability θ, and set it to state 1 otherwise, and we do this independently for every vertex. Recall that for each j ∈ [n] we write Sj for the state of vertex j; in this construction, the states are independent and identically distributed random variables. We set the edge {j, j + 1} to be open if Sj⩽Sj+1, and closed otherwise. Thus, as p=θ+(1−θ)2, we have that each edge is open with probability p. Moreover (Pn)νp will be connected if and only if there exists some j ∈ [n + 1] such that Sk = 0 for all k < j, while Sk = 1 for all k⩾j. Therefore (Pn)νp is connected with probability gn(θ). As this construction is vertex‐based, it is clear that it is 1‐independent.We begin by constructing a measure p<34 we have that θ is a complex number, and so the above construction is no longer valid. However, as discussed in Section p ∈ . For each subgraph G of Pn, set QG(θ) to be the polynomial νp((Pn)νp=G) for all θ∈. The following claim, together with Lemma νp is a 1‐ipm on Pn for all p ∈ .When Claim 29p∈ such {j, j + 1} is not an edge, and so Sj = 1 while Sj + 1 = 0. Note that if j⩾2, then the edge {j − 1, j} is present in G regardless of the state of vertex j − 1. Similarly, if j⩽n−2, then the edge {j + 1, j + 2} is present in G regardless of the state of vertex j + 2. If we write G1:=G{1,…,j−1} and G2:=G{j+2,…,n}, then we have that QG1θp and QG2θp are positive real numbers for all p∈ by q1 = 0 and by the recurrence relation qk=min1−p1−qk−1,1 for k ≥ 2, which corresponds exactly to the equality case in inequality We now prove that this construction is optimal with respect to the connectivity function. Note that the following proof involves essentially following the proof of Theorem Proof of Theorem 15f1,Pn(p)⩾0 for all p, and so all that remains to show is that f1,Pn(p)⩾gnθp for all p ∈ .The above construction discussed shows that μ∈ℳ1,⩾p(Pn). For k ∈ [n], let Xk be the event that the subgraph of (Pn)μ induced by the vertex set [k] is connected, and let Hk be the event that the edge {k − 1, k} is not present in (Pn)μ. Applying random sparsification as in Remark k, the event Hk occurs with probability exactly 1 − p.Let q2μ:=μ(X2)c=1−p, and for k > 2 let qkμ:=μXkc|Xk−1. We have that Let μ((Pn)μ=Pn)=∏j=2n(1−qnμ). Thus to show that the previous construction is optimal with respect to the connectivity function it is enough to show that equality holds for inequalities μ=νp. In the measure νp, we have that every edge is present with probability exactly p, thus νp(Hk)=1−p and so equality holds in Note that θ(p), and so it is sufficient to show that equality holds for p⩾34, as that will show they are the same polynomial . Suppose that the event (Hk ∩ Xk − 2) occurs. As Hk has occurred we have that Sk − 1 = 1 while Sk = 0. As Sk − 1 = 1, we have that edge {k − 2, k − 1} is open, regardless of Sk − 2. Thus, as Xk − 2 has occurred we also have that Xk − 1 has occurred. Therefore (Hk ∩ Xk − 1) has also occurred, and so we are done.Both the left and right hand sides of μ∈ℳ1,⩾p(Pn): μ=νp. This leads us to another way to define gnθp: let g1θp:=1, g2θp:=p, and for all n⩾3 we have that We remark in similar fashion to the above proof that the following holds for any Proof of Theorem 11f1,P1(p)=1 into equation p⋆(P1)=34 and apply Corollary p1,ℓp(Z)⩽34.(i). For the upper bound we plug p<34 be fixed. As the sequence (pn)n∈N is monotone increasing and tends to 3/4 as n → ∞, there exists N∈N such that p < pN. We showed in Theorem νpN∈ℳ1,⩾pN(PN) such that the probability PNνpN is connected is equal to zero.For the lower bound, let ν∈ℳ1,⩾p(Z). For each i∈Z, we let the subgraphs Zν[(i(N−1)+[N])] on horizontal shifts of PN by i(N − 1) be independent identically distributed random variables with distribution given by νpN. This gives rise to a 1‐independent model ν on Z with edge‐probability at least p (in fact at least pN). Furthermore, all connected components of Zν have size at most 2(N − 1) − 1. In particular, p1,ℓp(Z)⩾p. Since p<34 was chosen arbitrarily, this gives the required lower bound p1,ℓp(Z)⩾34.We use this measure to create a measure 6Z×P2 with edge‐probability close to 2/3 for which with probability 1 there are no open left‐right crossings. The idea of this construction is due to Walters and the second author In this section we construct a family of 1‐ipms on segments of the ladder V(PN × P2) as [N] × [2]. As in the case of the line lattice, we independently assign to each vertex  a random state Sn, y)n∈N,(rn)n∈N,(sn)n∈N are suitably chosen sequences of real numbers, ensuring that the Sn, y) ∈ [N] × [2] as follows:n ∈ [N − 1] and y ∈ [2], the horizontal edge {, } is open in Gμ if and only if S⩽S,for each n ∈ [N], the vertical edge {, } is open in Gμ if and only if S−S))=0.for each Let us begin by giving an outline of our construction. We write the vertex set n, 1), } to be open can be rephrased as if and only if either Sn, 1)n∈N,(rn)n∈N,(sn)n∈N and taking N sufficiently large, one can ensure that in addition μ satisfies d(μ)≥p and μ(∃open left‐right crossing)=0. In particular, with this construction we prove the following result.Clearly the bond percolation measure Theorem 30Fixp∈12,23. Then there existsN∈Nsuch that for alln⩾N,p:=23−ε, with ε∈0,16. We start by defining the sequences pnn∈N, rnn∈N and snn∈N iteratively as follows. We set p1 = r1 = 1 and s1 = 0. Then for n∈N, we let Fix Lemma 31The following hold for alln∈N:(i)pn, rn ∈ ,(ii)sn ∈ ,(iii)pn+1⩽pn,(iv)rn+1⩽rn,(v)rn+1+sn+1⩽rn+sn.n. By definition of our sequences, p1=r1=1⩾p=p2=r2, s1 = 0, and 0<s2=(2p−1)(1−p)p<1−p=r1+s1−r2, and thus (i)‐(v) all hold in the base case n = 1.We prove the lemma by induction on n⩽N, for some N⩾1. Since pN and rN + sN both lie in , the definition of pN + 1 and rN + 1 implies these also both lie in . This establishes (i) for n = N + 1. By construction, sN+1⩾0, and by the inductive hypotheses (ii) and (v), we have n = N + 1.Suppose now (i)‐(v) hold for all pN + 2 = 0, then pN+2⩽pN+1 trivially holds (since pN+1⩾0 by (i)). On the other hand, suppose pN+2=1−1−prN+1+sN+1>0. Then we have rN + 1 + sN + 1 > 1 − p, which by our inductive hypothesis (v) implies rN+sN⩾rN+1+sN+1>1−p. The definition of pN + 1 then implies n = N + 1. Arguing in exactly the same way (using the inductive hypothesis (iii) instead of (v)), we obtain that rN+2⩽rN+1. Hence (iv) holds for n = N + 1.If n = N + 1, which is the most delicate part of the induction. We begin by recording two useful facts, the second of which we shall reuse later.Finally we consider (v) for Claim 32pN + 2 = 0 or rN + 2 = 0, then sN + 2 = 0.If pN + 2 = 0, then by construction sN + 2 = 0 and so we are done. If rN + 2 = 0, then by construction pN + 1 ≤ 1 − p, which by our inductive hypothesis (iii) implies pN + 2 ≤ 1 − p and hence sN+2=max1−2rN+2+rN+2−(1−p)PN+2,0=max1−1−ppN+2,0=0.If Claim 33pN + 2 and rN + 2 are both strictly positive, then for all i ∈ {2, … , N + 1}, we have si > 0.If i ∈ {2, … , N + 1}. By our inductive hypotheses (iii) and (iv)  and since i⩾2, we have 0<pN+2⩽pi⩽p2=p and 0<rN+2⩽ri⩽r2=p. Since ri + 1 > 0, we in fact have pi > 1 − p. We also have that y ∈ , the function x ↦ f is a nonincreasing function of x. Thus if ri⩾1/2, we have y ∈ , the function x ↦ f is strictly increasing in x. Therefore if ri < 1/2, we have f > 0, and thus si=max1pif,0>0.Fix sN + 2 = 0, then (v) follows immediately from (iv). Thus we may assume that sN + 2 > 0, whence by Claim pN + 2 > 0 and rN + 2 > 0. By Claim pi, ri and si are all strictly positive for i ∈ {2, 3 … , N + 2}. By definition of our sequences we thus have for all i ∈ [N + 1] that i ∈ {2, … , N + 1} that: With these results in hand, we return to the proof of (v). If Claim 34sN + 2 > 0, for all integers i ∈ [N + 1] we have Under our assumption that p1 = 1 and p2 = p, our claim holds for i = 1. Suppose it holds for some i⩽N. Then by rearranging terms, we have pi + 2 given by i + 1 as well.Since i ∈ [N + 1], we can write ri + 1 + si + 1 as a function of pi: It follows from Claim pi > (1 − p) , the expression above is an increasing function of pi. By our inductive hypothesis (iii) that pN+1⩽pN it follows that rN+2+sN+2⩽rN+1+sN+1 and we have verified that (v) holds for n = N + 1.For p=23−ε, for some fixed ε∈0,16.Recall that Lemma 35We have that pn = rn = sn = 0 for alln⩾Nε, where Nε: = ⌈2ε−1⌉.m ∈ [Nε − 1] such that rm = 0. Then sm = 0, so pm + 1 = 0 and sm + 1 = 0 by construction and rm + 1 = 0 by Lemma pn = rn = sn = 0 for all n∈N⩾m+1, as required.Suppose first that there exists rn > 0 for all n ∈ [Nε − 1] and there exists some m ∈ [Nε − 2] such that pm⩽1−p. Then rm + 1 = 0, and thus by the argument above, we have that pn = rn = sn = 0 for all n∈N⩾m+2, as required.Suppose instead that pn > 1 − p and rn > 0 both hold for all n ∈ [Nε − 2]. By Claim sn > 0 for all n ∈ {2, … , Nε − 3}. This allows us in turn to apply Claim n in this interval and to deduce that p1 = 1. As such, it follows from inequality pn⩽1−(n−1)3ε4 for all n ∈ [Nε − 2]. In particular, as Nε = ⌈2ε−1⌉ and ε∈0,16, we have Finally, suppose N = Nε be the integer constant whose existence is given by Lemma Gμ on the graph G = PN × P2 from independent random assignments of states Sn, y) in V(G) = [N] × [2], as described at the beginning of this section.Now let Sn, y) ∈ [N] × [2], and so μ is a well‐defined 1‐ipm. We recall here for the reader's convenience the state‐based rules governing which edges are open in Gμ:n ∈ [N − 1] and y ∈ [2], the horizontal edge {, } is open if and only if S⩽S,for each n ∈ [N], the vertical edge {, } is open if and only if either Sn, 1)⩾p.We have that n, y) ∈ [N − 1] × [2], consider the horizontal edge {, }, . If n + y is even, then by definition of rn + 1, n + y is odd, then by definition of pn + 1, n, 1), }, n ∈ [N], we have pn = 0, then rn−1⩽1−p by definition of pn, whence rn⩽1−p by Lemma 1−rn⩾p. On the other hand if pn ≠ 0, then by definition of sn the expression above is at least p. Thus each horizontal edge and each vertical edge is open in Gμ with probability at least p, and d(μ)⩾p as claimed.For  = S = 2. Furthermore, by Lemma N, pN = rN = sN = 0, whence SN, 1), } of 𝒫 is traversed from left to right.Let i ∈ [ℓ], we have Si ∈ {1, 2}. Indeed, by construction Sv0=2. Suppose there exists some 1⩽i<ℓ such that Svj∈{1,2} for all j < i. If Svi=2, then the edge vivi + 1 can be open in Gμ only if Svi+1∈{1,2}. What is more, Svi+1 can be equal to 1 if and only if vivi + 1 is a vertical edge. On the other hand, suppose Svi=1. Then vi − 1vi was a vertical edge , and so vi + 1 = vi + . But then vivi + 1 open in Gμ implies Svi+1=2. Thus for every vertex vi of 𝒫, we have that Svi is indeed in state 1 or 2.We claim that for all vℓ ∉ {N} × [2] ( = 0). Thus there is no open path in Gμ from {1} × [2] to [N] × [2].This implies in particular that μ is an element of ℳ1,⩾p(PN×P2) for which n⩾N, we may extend μ to an element μ′∈ℳ1,⩾p(Pn×P2) by letting every edge in Pn × P2 ∖ PN × P2 be open independently at random with probability p. In this way we obtain a 1‐independent bond percolation measure μ′ on Pn × P2 with edge‐probability p for which there almost surely are no open left‐right crossings of Pn × P2, giving the required lower bound on p1, ×(Pn × P2).Thus We conclude this section by proving Theorem Proof of Theorem 11P2  is f1,P2(p)=p. Thus the constant p⋆(P2) defined by equation x2 = 4(1 − x)2, namely p⋆(P2)=23. By Corollary p1,ℓp(Z×P2)⩽23.(ii). Trivially, the 1‐independent connectivity function of the path on 2 vertices p∈12,23. In the proof of Theorem N∈N and μ∈ℳ1,⩾p(PN×P2) such that(i)μ(∃open left‐right crossing)=0;(ii)μ{,}and{,}are open=1.For the lower bound, fix ν∈ℳ1,⩾p(Z×P2). Let G:=Z×P2. For each i∈Z, we let the subgraphs Gν(i(N−1)+[N])×[2] on horizontal shifts of the ladder PN × P2 by i(N − 1) be independent identically distributed random variables with distribution given by μ. Thanks to property (ii) recorded above, the random subgraphs agree on the vertical rungs {1 + i(N − 1)} × P2 of the ladder, and this gives rise to a bona fide 1‐independent model ν on Z×P2 with edge‐probability p. Furthermore, property (i) implies all connected components in Gν have size at most 4(N − 1) − 2 = 4N − 6. In particular, p1,ℓp(Z×P2)⩾p. Since p<23 was chosen arbitrarily, this gives the required lower bound p1,ℓp(Z×P2)⩾23.We use this measure to create a measure 7n∈N⩾2 and p ∈ , we let θ=θ(p):=1+2p−12 and pn:=121−tan2π2n. Let gn(θ):=θn+(1−θ)n.In this section we will prove Theorem 7.1νp∈ℳ1,⩾p(Kn) that shows f1,Kn(p)⩽gn(θ) for p⩾12. We call this measure the Red‐Blue construction. We think of Kn as the complete graph on vertex set [n], and we color each vertex Red with probability θ and color it Blue otherwise, and we do this independently for all vertices. The edge {i, j} ∈ [n](2) is open if and only if i and j have the same color. As p=θ2+(1−θ)2, we have that each edge is present in (Kn)νp with probability p. Note that (Kn)νp will either be either a disjoint union of two cliques, in which case it is disconnected, or the complete graph Kn, in which case it is connected. This latter case occurs if and only if every vertex receives the same color, and so the probability that (Kn)νp is connected is equal to gn(θ). As this construction is vertex‐based, it is clear that it is 1‐independent.Before proving Theorem p<12 then θ is a complex number, and so the Red‐Blue construction is no longer valid. However, as discussed in Section p ∈ . Given j ∈ {0, 1, … , n}, let j = 0 or j = n we have that gn,0(θ) and gn,n(θ) are each equal to gn(θ), and so we just write the latter instead. Given some A ⊆ [n], let HA be the disjoint union of a clique on A with a clique on [n] ∖ A. Note that when A = ∅ or [n] we have that HA is equal to Kn] ∖ A, let μp be the following function on subgraphs G of Kn: p∈12,1 this function matches the Red‐Blue construction given above, and so by defining νp((Kn)νp=G):=μp(G) for all subgraphs G ⊆ Kn, we obtain a measure νp which is a 1‐ipm defined without making reference to states of vertices. The following claim, together with Lemma νp is a 1‐ipm on Kn for all p ∈ .If Claim 38p∈pn,12 and all j ∈ {0, … , n} we have that gn,jθp is nonnegative real number.For all j = n. As p⩽12, we have that θ and 1−θ are complex conjugates, and so gn(θ(p)) is a real number for all p in this range. By writing θ=reiφ, where r:=1−p2 and φ:=arctan1−2p, we can write p∈ implies 0≤φ≤π2n, which in turn gives cos(nπ)≥0. By gnθp is a nonnegative real number for all p∈, which proves the claim when j = n. For general j ∈ {0, … , n}, we have that gn,jθp∈ for all p∈; at this stage we are using the fact that (pn)n⩾2 forms an increasing sequence, and so p⩾pn implies that p⩾ps for all s⩽n.Let us begin with the case gnθpn=0, we have that the probability (Kn)νpn is connected is equal to 0. As νpn∈ℳ1,⩾p(Kn) for all p⩽pn, we have that f1,Kn(p)=0 for all p⩽pn. We now prove that this construction is optimal with respect to the connectivity function.Note that as this proof shows that 7.2Proof of Theorem 16f1,Kn(p)⩾0 for all p, and so all that remains to show is that f1,Kn(p)⩾gn(θ) for p ∈ . We will prove this result by induction on n. The inequality is trivially true when n = 2, so let us assume that n > 2 and that the inequality is true for all cases from 2 up to n − 1. First, we note that gn(θ)=gj(θ)gn−j(θ)−gn,j(θ) for all j ∈ {0, 1, … , n}. Thus, if we multiply both sides of this equation by nj and sum over all j ∈ {0, 1, … , n}, we have that μ∈ℳ1,⩾p(Kn) and let C be the event that (Kn)μ is connected. Given A ⊆ [n], let XA be the event that (Kn)μ[A] and (Kn)μ[Ac] are each connected, where Ac = [n] ∖ A. Moreover, let YA be the event that (Kn)μ[A] and (Kn)μ[Ac] are each connected, and there are no edges between A and Ac in (Kn)μ. For all A ⊆ [n], we have that A = ∅ or A = [n], the above equation is trivially true due to the fact that C, X∅, Xn][ are all the same event. As μ is 1‐independent we have that if A is a nonempty proper subset of [n], then, by induction on n, we have (pn)n⩾2 forms an increasing sequence, and so p⩾pn implies that p⩾ps for all s⩽n. We are also using the fact that g1(θ)=1 for all θ∈. We proceed by summing n], and then applying C, Y∅ and Yn][ are all the same event to A ⊆ [n], the events YA and YAc are the same event, and so ∑A⊆[n]μ(YA)=2∑1∈A⊆[n]μ(YA). Moreover, the set {YA  : 1 ∈ A ⊆ [n]} consists of pairwise disjoint events, and so ∑1∈A⊆[n]μ(YA)⩽1. Thus The previous constructions discussed show that (2n−4)μ(C)⩾(2n−4)gn(θ). As n > 2, we have that μ(C)⩾gn(θ) and so we are done.We apply 7.3fk, G(p) analogously to fG1, (p) for k∈N0. For k = 0, f0,Kn(p) is exactly the probability that an instance of the Erdős‐Rényi random graph Gn, p contains a spanning tree. As far as we know, there is no nice closed form expression for this function.Clearly we can define f1,Kn(p) exactly, which is the other interesting case, as for k⩾2 the connectivity problem is trivial.In this section, we have computed Proposition 39For allk,n∈N⩾2, we have thatμ∈ℳk,⩾p(Kn). Since any subgraph of Kn with at least n2−(n−1) edges is connected, we can apply Markov's inequality to show that For the lower bound, consider G obtained as follows. Let x:=1−p2. With probability min, select a vertex i ∈ [n] = V(Kn) uniformly at random, and let G be the subgraph of Kn obtained by removing all edges incident with i. Otherwise, let G be the complete graph Kn. It is easy to check that G is a 2‐independent model with edge‐probability p and that G is connected if and only if G = Kn, an event which occurs with probability 1−min=max0,1−n(1−p)/2.For the upper bound, consider the random graph 88.1fG1, (p), for any graph G, as a (possibly nonlinear) program.In this subsection we describe how we can represent the problem of finding G on vertex set [n], let ℋ=ℋ(G) be the set of all labeled subgraphs of G. Throughout this section we treat these subgraphs as subsets of E(G), and always imagine them to be on the full vertex set [n]. For each labeled subgraph of G we write Given a graph μ:ℋ→R⩾0, we have μ∈ℳ1,⩾p(G) if and only if the following three conditions all hold:μ is a probability measure on labeled subgraphs of G,G is open in Gμ with probability at least p,Every edge of S,T∈ℋ such that S and T are supported on disjoint subsets of [n], μ(S)·μ(T)=μ(S∪T).Given nonempty Recall that for a function fG1, (p), and as randomly deleting edges cannot increase the probability of being connected, we may assume that in fact every edge of G is open in Gμ with probability exactly p (by applying random sparsification as in Remark ∑H∈ℋμ(Ĥ)=1,e ∈ E(G), we have that∑H∈ℋ1(e∈H)μ(Ĥ)=p,For all edges S,T∈ℋ, such that S and T are supported on disjoint subsets of [n], we have that For all nonempty As we are interested in determining A = A(G) be a matrix which has columns indexed by ℋ, and a row for each piece of information given by one of the above conditions. That is:A∅, H: = 1.We have a row for the empty set such that e ∈ G; the entry Ae,H:=1(e∈H);We have a row for S,T∈ℋ∖{∅} supported on disjoint subsets of [n]; the entry A{S,T},H:=1((S∪T)⊆H)−μ(T)·1(S⊆H).We have a row for each pair Let q = q(G) be a vector with indexing the same as the rows of A; let q∅: = 1, qe: = p for e ∈ G, and qS, T}{: = 0 for each pair S,T∈ℋ∖{∅} supported on disjoint subsets of [n]. Then a vector w, whose entries are indexed by ℋ, which satisfies wH⩾0 for all H∈ℋ, and also Aw = q corresponds precisely to a 1‐ipm μ on G with μ{eopen}=p for all edges e ∈ E(G).Let c be a vector indexed by ℋ defined by cH:=1(His connected). Just to make it clear, we say that H∈ℋ is connected if it contains a spanning tree of [n]. Then for a given value of p the vector w(p) satisfying Aw(p) = q and cTw minimal corresponds to a measure μ∈ℳ1,⩾p(G) such that μ(His connected)=f1,G(p).Let G is a graph on [5], and S and T are nonempty subgraphs of G supported on disjoint subsets of [5], then one of S and T must consist of precisely one edge of G. By choosing T to be this subgraph, we can always choose S and T for μ(T)=p. Thus for any choice of p, we can turn the problem of finding fG1, (p) into the following linear program: S and T such that μ(T) Observe that for any graph with five vertices or fewer, any partition of the graph into two parts has that one part must have at most two vertices in it. In particular, if  μ(T) in  is an una∗: p, for example using the software Maple, and the LPSolve function it contains. However we of course wish to find solutions for all values of p ∈ .The duality theorem states that the asymmetric dual problem has the same optimal solution A = (aij), w = (wj), c = (cj), q = (qi) and x = (xi) any solutions w and x must satisfy ∑jaijwj=qi, ∑iaijxi⩽cj, xi ≥ 0 and wi⩾0. Thus we have ∑iqixi=∑jcjwj and so the inequality must be an equality, that is j we either have wj = 0 or ∑iaijxi=cj. Thus in our attempt to obtain a function for all p, it seems reasonable to look at an optimal solution for one value of p and see which wj have been set to zero; assume for these indices that we always have wj = 0 and attempt to directly solve the equations that result from this. This motivates the following method:p to obtain a solution w(p) and a set J  : = {j ∈ [|w|] : wj(p) = 0}.Solve Aw)i = qi, wj = 0  : i ∈ [|w|], j ∈ J} to obtain functions of p for all wk, k ∈ [|w|], which we write as wk′(p).Solve the set of equations {(ATx)i = ci  : i ∈ [|w|] ∖ J} to obtain functions of p for all xk, k ∈ [|w|], which we write as xk′(p).Solve the set of equations {(w∗(p): =  cTw′(p) and x∗(p): =  qTx′(p).Write P ⊆  of values of p, check that (ATx′)i(p)⩽ci and wi′(p)⩾0, for all i ∈ [|w|].For a certain interval By writing P which works above, the conditions above ensure that the w′(p) and x′(p) obtained in this way are feasible solutions to w∗(p) = x∗(p), then by the duality theorem we have fG1, (p) = w∗(p). Furthermore, a measure μ on the subgraphs of G which is extremal is given directly by w′(p). In the following subsection we give, as examples, two results which are proved using the above method.For the given interval 8.2In this subsection we prove Theorems Proof of Theorem 17C4 and p∈ an extremal construction is given by the measure μ, defined by C4 and p∈ an extremal construction is given by the measure μ, defined by For μ∈ℳ1,⩾p(C4), we have by 1‐independence that μ above , this gives a second and perhaps more insightful proof of Theorem f1,C5(p), we do not have a combinatorial proof, and our result relies solely on linear optimisation.We can in fact give a direct combinatorial proof of the lower bound in Theorem Proof of Theorem 18C5 and p∈ an extremal construction is given by the measure μ, defined by C5 and p∈ an extremal construction is given by the measure μ, defined by For 8.3f1,Cn(p) for n⩾6.We can use Markov's inequality to derive the following simple lower bound on Proposition 40Forn∈N, withn⩾6, and p ∈ , we havef1,Cn(p)⩾np−(n−2)2.n = 6.A small adjustment to this argument gives the following improvement for Proposition 41For p ∈  we have thatf1,C6(p)⩾−p3+3p2−1.Proof of Proposition 40μ∈ℳ1,⩾p(Cn). Note that Gμ is connected if and only if it has at most one closed edge. Thus by Markov's inequality, we have Let Proof of Proposition 41X be the number of closed edges in Gμ. Cyclically label the edges of C6 as e1, … , e6. Then by simple counting, Let 9f1,Kn(p) and hence prove Theorem In this section, we derive our results for maximizing connectivity in 1‐independent modes. First of all Theorem Proof of Theorem 21G on Kn with edge‐probability at least 1 − p, observe that the complement Gc of G in Kn is a 1‐independent model in which every edge is open with probability at most p. Furthermore, Gc is connected whenever G fails to be connected. This immediately implies νp we constructed to obtain the upper bound on f1,Kn(p) in the proof of Theorem νp‐random graph is connected if and only if its complement fails to be connected. This immediately implies that we have equality in Given a 1‐independent model F1,Pn(p).For paths, a simple construction achieves the obvious upper bound for Proof of Theorem 20μ∈ℳ1,⩽p(Pn), we have by 1‐independence that F1,Pn(p)⩽p⌊n2⌋. For the lower bound, we construct a 1‐ipm as follows. For each integer i: 1⩽i⩽n/2, we assign a state On to the vertex 2i with probability p, and a state Off otherwise, independently at random. Then set an edge of Pn to be open if one of its endpoints is in state On, and closed otherwise. This is easily seen to yield a 1‐ipm μ on Pn in which every edge is open with probability p, and for which For any measure F1,Pn(p)⩾p⌊n2⌋, as claimed.Thus Cn appears to be slightly more subtle. For the 4‐cycle, as in the previous section, we can give two proofs, one combinatorial and the other via linear optimisation.The case of cycles Proof of Theorem 17μ∈ℳ1,⩽p(C4), we have by 1‐independence that p)2 = 2p − p2, we obtain F1,C4(p).The theorem immediately follows from an application of the linear optimisation techniques from Section p. For p∈ consider the measure μ defined by μ∈ℳ1,⩽p(C4) and that μ({connected})=1−(1−p(2−p))=2p−p2, which is maximal for p in that range.For the lower bound, we give two different constructions, depending on the value of p∈, consider the measure μ defined by μ∈ℳ1,⩽p(C4) and that μ({connected})=μ{⩾3edges open}=2p2, which is maximal for p in that range.For Proof of Theorem 18We simply apply the linear optimisation method from Section p∈ an extremal construction is given by the measure μ, defined by p∈ an extremal construction is given by the measure μ, defined by p∈ an extremal construction is given by the measure μ, defined by For 10Combining Corollary Proof of Theorem 12Z×Cn has the finite 2‐percolation property. Thus, as described after the proof of Theorem p1,ℓp(Z×Cn)⩾4−23. For the upper bound in part (i), since the long paths critical probability is nondecreasing under the addition of edges, we have G = K3, C4, C5, we plug in the value of fG1, (p) in equation p⋆(G) and apply Corollary For the lower bound in part (i), we note that p1,ℓp(Z×Kn)n∈N is nonincreasing in  and hence tends to a limit as n → ∞. For the lower bound in (v), observe that for any n∈N the graph Z×Kn has the finite 2‐percolation property—indeed for any finite k, the closure of a copy of Pk × Kn under 2‐neighbor bootstrap percolation in Z×Kn is equal to itself. We construct a 1‐ipm μ on Z×Kn as in Corollary T0 = {0} × V(Kn) and hence Tk = ({k} × V(Kn)) ∪ ({− k} × V(Kn)). It is easily checked that μ‐almost surely, all components  in a μ‐random graph have length at most 5n. Since by construction d(μ)=4−23, this proves n∈N. For the upper bound, we perform some simple analysis. By solving a quadratic equation, we see that p∈. Then by Theorem p and all n sufficiently large, we have that p⋆(Kn) < p for all n sufficiently large, which by Corollary p1,ℓpZ×Kn<p.In part (v), we begin by noting that as we are considering an increasing nested sequence of graphs, the sequence 1111.1p1,c(Z2) can be improved. This problem is, we suspect, very hard in general. However, it may prove more tractable if we restrict our attention to a smaller family of measures.The most obvious open problem about 1‐independent percolation is of course whether the known lower and upper bounds on Definition 42G be a graph. A G‐partition is a partitioned set ⊔v∈V(G)Ωv, with nonempty parts indexed by the vertices of G. A G‐partite graph is a graph H on a G‐partition V(H)=⊔v∈V(G)Ωv whose edges are a subset of the union of the complete bipartite graphs ⊔uv∈E(G){ωuωv:ωu∈Ωu,ωv∈Ωv} corresponding to the edges of G.Let G‐partite graph H on a G‐partition ⊔v∈V(G)Ωv, we have a natural way of constructing 1‐independent bond percolation models: given a family X=Svv∈V(G) of independent random variables with Sv taking values in Ωv, the X)‐random subgraph obtained by setting uv to be open if and only if SuSv ∈ E(H).Given a Definition 43G be a graph. A measure μ∈ℳ1,⩾p(G) is said to be vertex‐based if there existG‐partition ⊔v∈V(G)Ωv,a G‐partite graph H, andan associated Sv)v ∈ V (G) with Sv taking values in Ωv,a collection of independent random variables ‐random subgraph H[X] has the same distribution as the μ‐random graph Gμ.such that the  denote the collection of all vertex‐based measures on G with edge‐probability at least p.Let Problem 44Determineinf{p∈:∀μ∈ℳvb,⩾p(Z2),μ({percolation})=1}.Ωv have bounded size.Vertex‐based measures arise naturally in renormalising arguments, and are thus a natural class of examples to consider. A special case of Problems Definition 45μ on a graph G is N‐uniformly bounded if it as in Definition v ∈ V(G), |Ωv|⩽N. Furthermore, a vertex‐based measure μ on a graph G is uniformly bounded if it is N‐uniformly bounded for some N∈N.A vertex‐based measure ℳN−ubvb,⩾p(G) and ℳubvb,⩾p(G) denote the collection of all vertex‐based measures on G with edge‐probability at least p that are N‐uniformly bounded and uniformly bounded respectively.Let Problem 46(i)ForN∈N, determine(ii)Determinep1,c(Z2), is arguably that of giving bounds on p1,ℓp(Z2) and closely related variants. Such problems, which correspond to new questions in extremal graph theory, are discussed in the subsections below. For these problems too we believe restrictions to the class of uniformly bounded vertex‐based 1‐ipms could be both fruitful and interesting in their own right.Finally, let us note that the second most obvious problem arising from our work, besides that of improving the bounds on 11.2Z2, it is natural to ask about bounds on pc1, (G) for some of the other commonly studied lattices in percolation theory.Beyond Problem 47Give good bounds on the value of pc1, (G) when G is one of the eleven Archimedean lattices in the plane or the d‐dimensional integer latticeZd.G is the triangular lattice or the honeycomb lattice , or the cubic integer lattice Z3 (which is important in applications). A challenge in all cases is finding constructions of nonpercolating 1‐independent measures with high edge‐probability—indeed, our arsenal of constructions for 1‐independent percolation problems is so sparse that any new construction could be of independent interest.This problem is particularly interesting when Zd+1 contains a copy of Zd, whence the sequence p1,c(Zd)d∈N is nonincreasing in  and converges to a limit. Balister and Bollobás asked for its value:In a different direction, we can observe that Problem 48 Determinelimd→∞p1,c(Zd).4−23.Note that by Theorem 11.3G be an infinite, locally finite connected graph, and v0 a fixed vertex of G. Given a bond percolation model μ on G, we let Cv0 denote the connected component of Gμ containing v0.Let μ is 0‐independent, then μ({percolation})=1 if and only if μ({|C0|=∞})>0. However this need not be true for a 1‐independent measure. Indeed, consider the 1‐ipm on Z2 obtained by taking the measure constructed in the proof of Theorem ℓ∞ ball of radius 3 around the origin and setting every other edge to be open independently at random with probability 4−23. Then in this model percolation occurs almost surely, but the origin is contained inside a component of order at most 28.If G for percolation to occur somewhere with probability 1 and for it to occur anywhere with strictly positive probability. Indeed, if p1,c(Z2) were strictly less than 3/4, then one could obtain examples of such a graph G by attaching a long path to the origin in Z2.Thus in principle there are different edge‐probability thresholds in 1‐independent percolation on a graph Temperley critical probability, which in 0‐independent percolation is the threshold pT at which E|Cv|=∞ for any vertex v (and every 0‐independent measure with edge‐probability >pT). In general this threshold is different from the Harris critical probability. Again for 1‐independent percolation we have that the threshold for some vertex v ∈ V(G) to satisfy E|Cv|=∞ and for the threshold for all vertices of G to satisfy this are different.Another critical edge‐probability of interest is the Problem 49Given an infinite, locally finite connected graph G, determine the following four critical probabilities:It follows from their definition that these four critical probabilities satisfy G with p1,H1(G)=12. For any p: 12<p<34, we have shown in Theorem N such that f1,PN(p)=0. Attach one end of a path of length N to an arbitrary vertex of G to form a graph G1, and let v denote the other end of the path. Then there exist 1‐ipm μ∈ℳ1,≥p(G1) such that with probability 1 the component of v in a μ‐random graph has order at most N, which is finite. Thus we have G2 obtained from the line lattice by attaching to each vertex i∈Z a collection of 2i|+2| leaves. Clearly, p1,H1(G2)=p1,H1(Z)=1. Now consider a 1‐ipm μ∈ℳ1,≥34(G2). By Theorem G = K1 and α=12, for any path P of length i in G2, the μ‐probability that all edges in P are open is at least 2i + 1)−(. Thus for any v0 ∈ V(G2), the expected size of |Cv0| is In general, these four critical probabilities are all different. Indeed, Balister and Bollobás showed in  What isp1,F(Z2)?11.4An obvious problem is to tighten the bounds in Theorem Problem 54Determinelimn→∞p1,ℓpZ×Kn.p1,ℓp(Z×Pn) is a nonincreasing function of n (since Z×Pn+1 contains Z×Pn as a subgraph). In this paper, we have given constructions showing that for all integers n⩾3, p1,ℓp(Z×Pn)n∈N tends to a limit in the interval  as n → ∞.In a similar vein, the sequence Problem 55DetermineZ2 .An in principle different but related problem is determining the value of the long paths critical probability in Problem 56Determinep1,ℓpZ2.k‐independent versions of the long paths critical probability. Defining pk,ℓpG mutatis mutandis, it is straightforward to adapt our arguments and constructions from Section Z by k‐independent ones.We can also ask for Theorem 57 For anyk∈N0, we havewith the convention that 00 = 1.k fixed, it is easy to construct a 3k‐ipm μ on Z2 with d(μ)=1−2k and no open path of length more than (2k + 1)2. Indeed, build a random graph model as follows:Z2 open;begin with all edges of ∈Z2, choose Hij ∈ [k + 1] uniformly at random and then for all j′: j(k + 1) − k ≤ j′ ≤ j(k + 1) + k, set the horizontal edge {(i(k + 1) + Hij − 1, j′), (i(k + 1) + Hij, j′) to be closed;independently for each ∈Z2, choose Vij ∈ [k + 1] uniformly at random and then for all i′: i(k + 1) − k ≤ i′ ≤ i(k + 1) + k, set the vertical edge { + Vij − 1),  + Vij) to be closed.independently for each Given k‐independent, has edge probability at least k2(k+1)2≥1−2k and that every connected component has order at most (2k + 1)2.It is easy to check that this random graph model is 3Corollary 58For any fixedk∈N,□[[spimspace="1em"]]limk→∞pk,ℓp(Z2)=1 (and in fact a similar construction shows this remains true in Zd).In particular we have G does not imply that for every ℓ∈N every vertex of G has a strictly positive probability of being part of a path of length at least ℓ. Thus we may actually define a second long paths critical probability, Finally, as in Section Problem 59Determinep1,ℓp2(Z2).p1,ℓp2(Z2)⩾4−23, and we know it is upper‐bounded by p1,H2(Z2)⩽0.8639. As in Section Our construction in the proof of Theorem n‐uniformly bounded Z2‐partite graph H with partition ⊔v∈Z2Ωv. A transversal subgraph of H is a subgraph of H induced by a set of distinct representatives S for the parts of H, that is, a set of vertices of H such that |S∩Ωv|=1 for all v∈Z2. The G‐partite density of H is Given an Question 60Suppose H is an n‐uniformly boundedZ2‐partite graph in which in every transversal subgraph the connected component containing the origin is…(a)of size at most C, for some constantC∈N.… (b)finite.… How large can dG(H) be?This question can be viewed as a problem from extremal multipartite graph theory. Plausibly some tools from that area, in particular the work of Bondy, Shen, Thomassé and Thomassen 11.5f1,Cn(p) for cycles Cn of length at most 5. It is natural to ask what happens for longer cycles.We determined in Sections Problem 61Determinef1,Cn(p)forn∈N⩾6.f1,C6(p) is nonlinear. Nevertheless, one can use software, such as Maple and its contained NLPSolve function, to try to estimate the answer. This suggests the following:f1,C6(p) becomes nonzero is approximately p = 0.59733;The threshold at which p just above this threshold, the best “asymmetric” (see the next subsection for a definition) measure is better than the best “symmetric” measure; e.g. at p = 0.62 we have f1,C6(0.62) is approximately 0.007, but is as high as 0.11 when restricted to “symmetric” measures.For As mentioned in Section More generally, one can ask what happens in cycles if we have higher dependency or if we try to maximize connectivity rather than minimize.Problem 62Determinefk,Cn(p)for all p ∈ ,k∈Nand integersn⩾k+2.Problem 63Determinefk,Cn(p)for all p ∈ ,k∈Nand integersn⩾k+2.Qn and in the n × n toroidal grid Cn × Cn. Progress on either of these would likely lead to progress on other problems in 1‐independent percolation as well.Beyond paths, cycles and complete graphs, the 1‐independent connectivity problem is perhaps most natural to study in the hypercube graph Problem 64Determinef1,Qn(p)for alln⩾3.Problem 65Determinef1,Cn×Cn(p)for alln⩾3.fG1, (p) can be required to have “nice” properties. For C4 and p ∈  another extremal construction for f1,C4(p) is given by the measure μ, defined by μ∈ℳ1,⩾p(G)symmetric if for any pair of labeled subgraphs S and T of G such that there exists an automorphism of G mapping S to T, then μ(Ŝ)=μ(T^). Note that the above measure is an example of a nonsymmetric extremal construction for f1,C4(p), whereas the measure given at the end of Section In a different direction, we can ask whether the extremal measures attaining Question 66For any G and any p ∈ , does there always exist a symmetric measureμ∈ℳ1,⩾p(G)which achieves fG1, (p)?fG1, (p) attained via our method in Section G is yes(see the appendix for a proof of this fact). Another natural question is when the extremal connectivity can be attained by vertex‐based measures.If the program for solving Question 67For which G and which p does there exist a vertex‐based measureμ∈ℳ1,⩾p(G)which achieves fG1, (p)?