Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 15 (Percolation and Combinatorics)

Let $C_x^b$ and $C_x^s$ be the out-clusters in directed percolation, with the site percolation cluster empty when $x$ is closed. Write ${\theta }_b(p)={\mathbb{P}}_p(|C_x^b|=\infty )$ and ${\chi }_b(p)={\mathbb{E}}_p|C_x^b|$, and similarly for site percolation. The survival and susceptibility thresholds for directed percolation are the infima of the parameters at which these quantities become positive and infinite, respectively.

We construct a coupling of probability distributions that compares the whole clusters. First condition the site percolation root to be open. Explore its out-cluster, using a fixed order of the finitely many outgoing bonds at each vertex. When a previously untested vertex $v$ is first encountered along a bond $u\to v$ from an already reached vertex, use a fresh Bernoulli random variable of parameter $p$ both for the state of $v$ and for the state of this particular bond. If $v$ is open, add it to the exploration; if it is closed, mark it as tested and never add it. Assign fresh independent random variables of the same parameter to every other bond, including bonds leading to vertices already tested. The choice of the next unassigned bond depends only on previously revealed information. Its assigned coin is therefore independent of that information, so the resulting bond configuration has exactly the product bond percolation law. Parallel bonds cause no difficulty; loops may be discarded.

Every reached site has an open parent bond leading from a previously reached site. Consequently the conditional site percolation cluster is contained in the bond percolation cluster. The connected graph is countable because it is a locally finite graph; taking increasing finite stages of the exploration proves the same containment for infinite clusters. Removing the conditioning on the root givesThus positive site percolation survival implies positive bond percolation survival at the same parameter, and infinite percolation susceptibility for site percolation implies infinite percolation susceptibility for bond percolation. Taking the infima in the definitions proves both comparisons:

Use the incoming-edge coupling of site and bond percolation. Start with independent random variables indicating open bonds, each with probability $q$. For a vertex $v$ of incoming degree $d_v$, let $J_v$ indicate that at least one incoming bond is open. ThenThe indicators $J_v$ are independent: the sets of bonds entering different vertices are disjoint. Using an additional independent coin at each vertex, turn some zeros into ones so that the resulting indicators $S_v$ are independent with common probability $r$. Explicitly, when $J_v=0$, set $S_v=1$ with conditional probability $(r-\mathbb{P}(J_v=1))/(1-\mathbb{P}(J_v=1))$.

Every vertex other than $x$ in an open bond percolation out-cluster has an open incoming bond, and is consequently an open site. On $\{S_x=1\}$, the whole bond out-cluster is therefore contained in the site percolation out-cluster. The event of infinite bond reachability does not depend on bonds entering $x$: delete cycles from any finite open directed path starting at $x$. It is thus independent of $S_x$, including its additional coin. HenceFor every $q>p_H^b$ below $1$, monotonicity and the definition of the survival threshold give ${\theta }_b(q)>0$. Letting $q\downarrow p_H^b$ yields the quantitative boundIn particular, if $p_H^b<1$, the right-hand side is strictly less than $1$. If $p_H^b=1$, no separation from $1$ is implied or generally possible.

No. The parallel-bond ray with different site and bond thresholds supplies a locally finite graph counterexample. Take vertices $v_0,v_1,\dots$ and $n+1$ parallel directed edges from $v_n$ to $v_{n+1}$. Its underlying multigraph is connected and every individual degree is finite, but there is no uniform incoming-degree bound.

For bond percolation rooted at $v_0$, survival means that every successive group of parallel bonds contains at least one open bond. Independence givesFor $p>0$, the sum of the failure probabilities is ${\sum }_{n\ge 0}(1-p{)}^{n+1}=(1-p)/p<\infty$. Eventually these terms are at most $1/2$, and $\log (1-t)\ge -2t$ for $0\le t\le 1/2$. The tail of the infinite product is therefore bounded below by a positive exponential; its finitely many initial factors are positive as well. Thus ${\theta }_b(p)>0$ for every $p>0$.

For site percolation, however, every vertex of the unique underlying ray must be open. For $p<1$, the probability that its first $N$ vertices are all open is $p^N\to 0$. At $p=1$ the ray is open. Consequently

We use three precise inputs. The Harris-Kesten theorem says that independent bond percolation on the square lattice has percolation critical probability $p_c=1/2$. Sharpness of the percolation transition says that, for every $p<p_c$, there are $c,C>0$ such that ${\mathbb{P}}_p(0\leftrightarrow \partial [-R,R{]}^2)\le C{e}^{-cR}$ for all positive integers $R$. Finally, the BK inequality says that increasing connection events with disjoint bond witnesses have joint probability at most the product of their individual probabilities. All three statements concern the independent product law used here.

The sharpness bound implies finite percolation susceptibility. Indeed, with $\tau (u,v)={\mathbb{P}}_p(u\leftrightarrow v)$, summing over square shells givesA radius bound alone would give only a square-root exponential volume bound. To obtain the requested exponential volume bound, we prove the tree-graph moment bound for a percolation cluster:where $(-1)!!=1$.

Expand the left-hand side as a sum over ordered marked vertices $x_1,\dots ,x_k$ of the probability that $0,x_1,\dots ,x_k$ belong to one open cluster. Whenever that event occurs, a finite open tree connects the marked vertices. Delete unmarked leaves and suppress unmarked degree-two vertices. Treat a marked internal vertex as a leaf attached by a zero-length connection, and split any higher-degree branch into degree-three branch points, again with zero-length connections. This gives an abstract binary tree with $k+1$ labelled leaves, $k-1$ branch points and $2k-1$ connections. Its nonempty connecting paths use disjoint sets of open bonds. Coincident marked vertices are allowed: their zero-length connections have empty witnesses.

There are $(2k-3)!!$ such binary tree topologies. For example, the topology with two leaves is unique; inserting the next labelled leaf subdivides one of the $2k-1$ edges of a topology with $k+1$ leaves, giving the usual double-factorial recursion. For a fixed topology and fixed positions of all its vertices, the BK inequality bounds the disjoint connection event by the product of the corresponding $\tau$ factors. Sum over all $x_i$ and all branch-point positions. Keep the root fixed, and successively sum a leaf away. Translation invariance gives ${\sum }_y\tau (z,y)=\chi$ at every step. The resulting sum is ${\chi }^{2k-1}$. Summing the topologies proves the bound.

Since $(2k-3)!!\le 2^{k-1}(k-1)!$, expansion of the exponential, justified by monotone convergence, givesTake $t=1/(4{\chi }^2)$ and write $M=1+\log 2/(2\chi )$. Markov's inequality yields ${\mathbb{P}}_p(|C_0|\ge n)\le M{e}^{-tn}$. To remove the prefactor, choose an integer $N\ge 2$ with $tN\ge 2\log M$. For $n\ge N$ the bound is at most $e^{-tn/2}$. For $2\le n\le N$, useFor $0<p<1/2$, setting $a=\min \{t/2,-\log u/N\}>0$ therefore proves the bound with unit prefactor:At $p=0$, $|C_0|=1$ and any positive $a$ works.

Set $q=1-p<1/2$. Declare a bond of the planar dual graph open when its crossing primal bond is closed. These states have the independent dual bond percolation law with parameter $q$; the dual square lattice is a translate of the primal square lattice. Part (i) supplies an $a=a(q)>0$ such thatIn particular every dual cluster is finite almost surely.

Suppose the primal percolation cluster $C_0$ is finite. Fill its finite complementary holes. Every bond on the resulting outer boundary is closed, since an open bond leaving the cluster would reach another cluster vertex. The crossing dual bonds form a connected enclosing contour, possibly touching itself, contained in one dual cluster $K$. This is the dual contour bound for a finite planar cluster. If $m=|K|$, a connected set of $m$ square-lattice dual vertices has each coordinate width at most $m-1$. All vertices of $C_0$ lie inside the enclosing contour, soMoreover the contour surrounds $0$ and the diameter of $K$ is at most $m-1$ in the graph distance. Thus every vertex of $K$ lies in the square of radius $m+1$ about $0$. There are at most $(2m+3{)}^2$ possible dual vertices in that square, with a harmless enlargement if a boundary convention requires it.

On $\{n\le |C_0|<\infty \}$ there is therefore an integer $m\ge \lceil \sqrt{n}\rceil$ and a dual vertex in that square whose dual cluster has size exactly $m$. The enclosing contour has at least four vertices. A union bound and part (i) givefor constants $C_1,c_1>0$ depending only on $q$. The last estimate follows by bounding the polynomial factor by a constant times $e^{am/2}$ and summing a geometric series. This explains the square root: an enclosing dual cluster must have size at least the square root of the primal volume.

For $p<1$, the Harris-Kesten theorem and $p>p_c$ imply $\theta (p)>0$, so $v:={\mathbb{P}}_p(|C_0|<\infty )=1-\theta (p)<1$. Choose $N$ with $\log C_1\le (c_1/2)\sqrt{N}$. For $n\ge N$ use the preceding bound; for $1\le n\le N$ use the bound $v$. The positive constant $b=\min \{c_1/2,-\log v/\sqrt{N}\}$ gives the required conclusion:For $p=1$ the event is empty, so any $b>0$ works.

Independence or another ergodicity hypothesis is necessary. Read literally for an arbitrary translation-invariant site percolation measure, the assertion is false: the mixture assigning probability $1/2$ to the all-open configuration and $1/2$ to the all-closed configuration is translation-invariant, but the invariant event of an infinite percolation cluster has probability $1/2$. We prove the zero–one law for the invariant independent law used in part (ii); the proof also works for the stated finite-range independent alternative.

Let $\Gamma$ be the group of translations, or the specified group of graph automorphisms, and let $A$ be invariant under it, up to null sets. An event determined by the states in a finite vertex set is a cylinder set. For every $\epsilon >0$, approximation in the product sigma-algebra provides a cylinder set $B$, supported on a finite set $F$, with $\mathbb{P}(A△B)<\epsilon$. Choose $g\in \Gamma$ so that $F$ and $gF$ are disjoint. Then $B$ and $gB$ are independent events, have equal probabilities, and $\mathbb{P}(A△gB)<\epsilon$. HenceandLetting $\epsilon \downarrow 0$ proves the zero–one law $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is translation ergodicity of Bernoulli percolation.

For a quasi-transitive graph, the required displacement exists: put $F$ in a ball about a vertex $v$; its orbit is infinite, so an automorphism can move $v$, and therefore that whole ball, beyond any prescribed finite neighborhood. A $k$-independent site percolation law means that the sigma-algebras generated by vertex sets at graph distance greater than $k$ are independent. Move $F$ far enough that $\mathrm{dist}(F,gF)>k$ and the same proof applies. Thus finite-range independence does give the zero–one law, although it will not give uniqueness.

We first specify the standard hypotheses of percolation uniqueness on an amenable quasi-transitive graph. A locally finite graph has finite degree at every vertex; a quasi-transitive graph, or graph of finite type, has finitely many vertex orbits under the acting group $\Gamma$ of graph automorphisms. Here the site percolation law is invariant under $\Gamma$ and independent, with one parameter $p_j$ for each orbit. Initially suppose $0<p_j<1$. This gives the finite-energy property of Bernoulli percolation: conditional on all states outside any finite set, every assignment inside that set has positive probability. An amenable graph has finite sets $W_n$ with $|{\partial }^{+}{W}_n|/|W_n|\to 0$, where ${\partial }^{+}{W}_n$ denotes the external vertex boundary. Finite type bounds the degrees uniformly, so internal and external boundary conventions are equivalent here.

Let $N$ be the number of infinite percolation clusters, taking values in $\{0,1,2,\dots ,\infty \}$. Each event $\{N=j\}$ is invariant. Part (i), together with countability of these possibilities, shows that $N$ is an almost-sure constant.

Suppose first that this constant is a finite integer $j\ge 2$. Some finite connected set $F$ meets all $j$ infinite percolation clusters with positive probability, since increasing finite sets exhaust the graph. Make every vertex of $F$ open. This joins all the infinite clusters. Opening a finite set cannot produce a new infinite cluster from finite clusters: only finitely many clusters meet its finite boundary, and their finite union is finite. By the finite-energy property the modified event has positive probability, contradicting the almost-sure value $j$. Thus a finite $N$ must be $0$ or $1$.

Suppose instead that $N=\infty$ almost surely. Some finite connected set $F$ meets three distinct infinite percolation clusters with positive probability. Open all of $F$. Its vertices belong to one open cluster, and deleting $F$ from that cluster leaves at least three infinite components. This is an encounter box in percolation. We use boxes rather than assume that the local geometry permits a single trifurcation vertex in percolation.

Here is the boundary-counting contradiction in detail. Translates of $F$ have the same positive encounter probability. Finite type and bounded degree allow a deterministic packing of mutually disjoint such translates whose centers have positive lower density in the interiors of the Følner sets $W_n$: every vertex is within a fixed distance of one chosen orbit, and a bounded-radius greedy packing discards only a bounded number of possible centers per chosen center. Boundary layers of any fixed thickness have size $o(|W_n|)$. The expected number of encounter boxes wholly inside $W_n$ is consequently at least $c|W_n|$ for a fixed $c>0$ and all large $n$.

On the other hand, contract the open encounter boxes in each cluster to vertices and retain the part inside $W_n$. Each contracted encounter box separates at least three branches that run to infinity, and each branch must leave $W_n$. Choose a minimal forest connecting all these boundary exits. Every encounter junction has degree at least three in that forest: its three infinite branches remain in different components after deleting the junction, so none can be omitted. The leaves are boundary exits. For each nontrivial tree, the degree-sum identity gives at most the number of leaves minus two vertices of degree at least three. Thus the number of encounter boxes is at most a constant times $|{\partial }^{+}{W}_n|$, where the constant accounts for the bounded boundary degrees. Taking expected values contradicts amenability. This excludes $N=\infty$ and provesThere is a genuine endpoint qualification if different orbit parameters are allowed to be $0$ or $1$. Subdivide each horizontal edge of the square lattice once. Open every original vertex deterministically and close every new midpoint deterministically. This is an invariant independent law on an amenable graph that is also a quasi-transitive graph, but its infinite open clusters are the distinct vertical columns. Thus the finite-energy property must be included when interpreting a general vector $\mathbf{p}$. The homogeneous endpoint laws themselves cause no problem: they give zero or one infinite cluster.

The conclusion is false for $k$-independent laws, even with all site marginals equal to $1/2$. Here is a fully specified monochromatic line-graph percolation counterexample. Form a graph $B$ by replacing each vertex of ${\mathbb{Z}}^2$ by a complete graph on ten vertices and putting every possible edge between the blocks of two neighboring lattice vertices. Every vertex of $B$ has degree $9+4\cdot 10=49$. The graph is connected, periodic and amenable: a square of blocks has volume of order $n^2$ and boundary of order $n$. Assign independent fair red/blue colors to the vertices of $B$.

Call a block good when it contains both colors. Good-block indicators are independent, withThey percolate. To verify this without presupposing a site threshold, put $q=1-g=1/512$. A finite nearest-neighbor good-block cluster around a good origin is surrounded by a circuit of bad blocks in the matching square graph, where horizontal, vertical and diagonal neighbors are allowed. A surrounding circuit of length $\ell$ has an anchor within $[-\ell ,\ell {]}^2$. Overcounting its choices by $9{\ell }^2{8}^{\ell }$ gives the Peierls argumentThe sum bound follows directly from ${\sum }_{\ell \ge 1}{\ell }^2{r}^{\ell }=r(1+r)/(1-r{)}^3$, with $r=1/64$. An infinite good-block cluster therefore exists with positive probability, and by part (i) it exists almost surely. Inside each of its blocks the red vertices are connected to one another, as are the blue vertices; between adjacent good blocks the complete bipartite joining edges connect each color. Hence $B$ has both an infinite red component and an infinite blue component almost surely.

Now use site percolation on $\Gamma =L(B)$, the line graph of $B$: its vertices are the edges $\{u,v\}$ of $B$, two adjacent when they share an endpoint. Declare such a site open exactly when $u,v$ have the same color. Each site has open probability $1/2$. Sets of sites at graph distance greater than one involve disjoint sets of underlying color variables, so this is a translation-invariant $1$-independent law. The line graph is again connected, locally finite, amenable and of finite type. The infinite red and blue components of $B$ yield infinite open clusters in $L(B)$. An open path in $L(B)$ cannot change color at a shared endpoint, so those two clusters are distinct. Thus $N\ge 2$ almost surely. Local color constraints destroy the finite-energy property, precisely the ingredient missing from the uniqueness proof.

We use Zhang's argument, obtaining the critical nonpercolation conclusion from uniqueness rather than assuming it as part of the Harris-Kesten theorem. For independent bond percolation on ${\mathbb{Z}}^2$, apply part (ii) to independent site percolation on its line graph; open sites correspond to open bonds. The line graph is amenable and of finite type. Therefore an infinite bond cluster, if it exists, is almost surely unique. The same holds for the dual bond percolation model.

Assume for contradiction that $\theta (1/2)>0$. The planar dual graph is another square lattice, and its open bonds correspond to closed primal bonds, also with parameter $1/2$. Both models then have an infinite percolation cluster almost surely, by part (i). As a disc $D_R$ expands, the probability that it meets an infinite cluster tends to one, for each model. Choose $R$ avoiding all primal and dual vertices and all intersections of either graph with the four endpoints of the following arcs. Divide the boundary circle into four equal arcs $A_1,A_2,A_3,A_4$ in cyclic order. Let $L_i$ be the event of an infinite primal open ray whose last exit from $D_R$ is in $A_i$; define $L_i^{\ast }$ similarly for an infinite dual open ray. After that exit the entire ray stays outside the disc. The existence of a ray follows by finite branching in an infinite locally finite graph.

Each $L_i$ is increasing in the primal bond states, and their union has probability tending to one. Four-fold rotation symmetry makes their probabilities equal. The FKG inequality for their decreasing complements gives the square-root trick for positively associated eventsThe left side tends to zero, so $\mathbb{P}(L_i)\to 1$. Apply the same argument to the dual states to get $\mathbb{P}(L_i^{\ast })\to 1$. A union bound, requiring no independence between the models, now shows that for sufficiently large $R$ the alternating eventhas positive probability.

The alternating exterior-arm separation lemma says that the two primal rays on this event cannot be joined by a primal open path lying outside $D_R$. Indeed, such a joining path contains a simple crosscut of the exterior between the two primal exit points. Join that crosscut to a chord inside the disc. The Jordan curve theorem puts one of the alternating dual exit points on the bounded side. Its dual ray, which is restricted to the exterior of the disc and runs to infinity, would have to cross the open primal crosscut. A dual open bond cannot cross a primal open bond. This is impossible.

Let $F$ be the finite set of primal bonds whose drawn edges meet $D_R$, and let $E$ be the projection of the alternating-arm event onto the states outside $F$. Because $F$ is finite, this projection is a finite union of measurable events; $\mathbb{P}(E)>0$. Every configuration in $E$ admits an assignment on $F$ witnessing the original four arms. Now close all bonds of $F$. The infinite tails of the two primal rays beyond their first exterior vertices survive; any path joining the surviving tails would now avoid the disc. Appending their original exterior segments back to their exit points would give the forbidden exterior connection. Thus the modified configuration contains at least two infinite percolation clusters. By independence, or the finite-energy property of Bernoulli percolation, $\mathbb{P}(E\cap \{F\text{ all closed}\})=2^{-|F|}\mathbb{P}(E)>0$. This contradicts the uniqueness obtained from part (ii). Therefore

Let $G=\Lambda$ be the martini lattice and $G^{\ast }$ its planar dual graph. Write $c=p_c^b(G)$ and $c^{\ast }=p_c^b(G^{\ast })$. Both are periodic, connected, bounded-degree plane graphs with bounded face sizes. Dual bond percolation declares a dual bond open exactly when its crossing primal bond is closed, giving independent dual parameter $1-p$.

First, $c+c^{\ast }\le 1$. Suppose instead that the sum exceeds $1$, and choose $1-c^{\ast }<p<c$. Both the primal model at $p$ and the dual model at $1-p$ are strictly subcritical. Sharpness of the percolation transition gives exponential connection-radius decay on each periodic graph, uniformly over its finitely many vertex types. A crossing of a rectangle of side lengths proportional to $R$ starts at one of $O(R)$ boundary vertices and travels distance of order $R$. A union bound therefore shows that each crossing probability tends to zero. But planar duality for rectangle crossings says that exactly one of an open primal left-to-right crossing and an open dual top-to-bottom crossing occurs, using the complementary boundary convention. Their probabilities sum to $1$. This contradiction proves the first inequality.

For the opposite inequality, we need noncoexistence of primal and dual infinite percolation clusters. The four-fold version in Q3(iii) cannot be assumed without change: this decorated honeycomb lattice has three-fold rotation symmetry. Here is the necessary three-fold symmetric primal-dual noncoexistence argument.

Assume both models percolate at complementary parameters $p,1-p$. For a large disc, the probability of having no exterior infinite arm is at most $\epsilon$ in each model. Divide its boundary into six arcs by splitting each $120$-degree sector at the same relative angle. In the primal model let $a,b$ be the failure probabilities for the two alternating arc types. Rotation symmetry and the FKG inequality giveSlide the splitting point across one sector. Initially the first arc is empty; finally the second is empty. A jump occurs only when an arc gains or loses an edge-crossing point. An arm using that point requires its crossing bond to be open. The FKG inequality therefore ensures that a failure probability changes by at most a factor $(1-p{)}^{-1}$ at a single jump: for a newly added arm event $L_x$, $\mathbb{P}(L_x^c)\ge 1-p$ and $\mathbb{P}(L^c\cap L_x^c)\ge (1-p)\mathbb{P}(L^c)$. At the first reversal of the order of $a,b$, their ratio is consequently at most $(1-p{)}^{-2}$. At that splitting point,Thus every one of the six primal arcs has an exterior arm with probability tending to $1$. On the same six arcs the dual failure probabilities $a^{\ast },b^{\ast }$ satisfy $a^{\ast }{b}^{\ast }\le {\epsilon }^{1/3}$, so at least one dual arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Take two of those dual arcs separated by one arc, and the two intervening primal arcs. A union bound gives four alternating primal/dual infinite exterior arms with positive probability when $\epsilon$ is sufficiently small. The separation and finite-closure argument of Q3(iii), together with percolation uniqueness on an amenable quasi-transitive graph, gives a contradiction. This proves noncoexistence. The arc balancing is the extra step needed for three-fold symmetry; it is also developed in Bollobás and Riordan's original paper on dual lattices.

If $c+c^{\ast }<1$, choose $c<p<1-c^{\ast }$. Both complementary models would then percolate, contradicting noncoexistence. Hence $c+c^{\ast }\ge 1$. Combining the two inequalities proves the dual threshold relation:

Group the martini lattice into independent three-terminal cells. A cell has terminals $A,B,C$, an inner triangle with vertices $a,b,c$, and the three spokes $Aa,Bb,Cc$. All six bonds are independently open with parameter $p$. The cells arise from the vertices in the decorated class of the honeycomb lattice; they have disjoint bond sets and meet only at terminals. The terminal positions form a triangular lattice, with cells occupying one orientation of its triangular faces.

A cell induces one of five connection partitions of its terminals. Let $P_3(p)$ be the probability that all three terminals are connected, and $P_0(p)$ the probability that no pair is connected. Each of the three remaining pair partitions has equal probability by symmetry. The three-terminal cell partition duality gives a particularly useful exact criterion. Draw dual terminals on the three intervening boundary arcs. The dual cell has a triangle of these three terminals and a central vertex joined to all three: its outer bonds cross the primal spokes, and its inner spokes cross the primal triangle bonds. A primal connection separates the corresponding dual boundary arcs, and conversely a dual connection separates primal terminals. Consequently duality swaps the all-connected and all-separate partitions, and permutes the three pair partitions. For example, if $A,B$ connect and $C$ is isolated, the dual terminal between $A,B$ is isolated while the other two dual terminals connect. It follows that $P_3=P_0$ makes the entire symmetric cell partition law invariant under this duality.

The triangular arrangement of cells is itself self-dual: the dual terminals lie in the complementary triangular gaps, and reflection and translation identify their cell arrangement with the original one. The induced partitions of distinct cells are independent in both models. At a parameter satisfying $P_3=P_0$, the primal model at $p$ and the complementary dual bond percolation model at $1-p$ therefore have the same coarse connection law. Replacing a cell by its terminal partition preserves infinite connectivity, since the cells have uniformly bounded size. This gives a rigorous link between the finite calculation and the infinite model, rather than assuming that a symmetric-looking cell is critical.

For all three terminals to connect, all three spokes must be open and the inner triangle must be connected. The latter event requires two or three of its edges. ThereforeTo compute $P_0$, condition on the number of open spokes. Zero or one open spoke cannot connect a pair of terminals. If exactly two spokes are open, the two attached inner vertices must not connect: their direct triangle bond is closed, and the two-edge route through the third inner vertex is not fully open. These independent requirements have probability $(1-p)(1-p^2)$. If all three spokes are open, all three triangle bonds must be closed. HenceSubtracting and factoring givesThe second factor equals $1+p^2(p-1)(p-2)$ and is at least $1$ on $[0,1]$. Thus the unique equality parameter is $p_0=1/\sqrt{2}$.

Finally part (i) proves $c+c^{\ast }=1$. If $p_0>c$, the primal model percolates while $1-p_0<c^{\ast }$ makes the dual model subcritical, contradicting their identical coarse connection laws. If $p_0<c$, the same contradiction arises with the roles reversed. Therefore the critical probability isThe local partition transformation is the mechanism underlying this exact value; see Ziff's original cell–dual-cell paper.