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Use the coordinatewise partial order: $\omega \le \eta$ means $\omega (e)\le \eta (e)$ for every $e\in E$. An increasing event $A$ is upward closed: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

Under the independent Bernoulli distribution product measure ${\mathbb{P}}_p$ on these coordinates, the Harris-FKG inequality statesfor increasing events $A,B$. For $K\subseteq E$, let $[\omega {]}_K$ be the cylinder of configurations agreeing with $\omega$ on $K$. The disjoint occurrence of increasing events $A\square B$ consists of configurations for which there exist disjoint $K,L$ with $[\omega {]}_K\subseteq A$ and $[\omega {]}_L\subseteq B$. For increasing events, witnesses can be taken to prescribe only open coordinates. The BK inequality isPositive association rewards simultaneous occurrence; the disjoint-witness requirement in the BK inequality gives a bound in the opposite direction. The inequalities also apply to different independent Bernoulli parameters in different coordinates.

Regard the boxes as sets of lattice graph vertices. Write ${\tau }_p(x,y)={\mathbb{P}}_p(x\leftrightarrow y)$, and put $g_0=1$. On the event defining $g_n$, choose an open self-avoiding walk from the origin to its first visit to $\partial {\Lambda }_n$. Let $y$ be its first visit to $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow y$. Its remaining segment ends at a point $z$ withFor $m<n$, truncate this remaining segment on its first visit to $y+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges, so they are disjoint witnesses. For $m=n$, the second event is the sure event with empty witness. Thus the union bound, the BK inequality and translation invariance giveAn unrestricted connection event can depend on infinitely many edges. Apply the finite-coordinate BK inequality first to connections confined to growing boxes and take increasing limits; each occurrence has a finite path witness. This justifies its use here. The estimate is the weighted BK boundary-splitting estimate.

By summing cluster indicators and using Tonelli theorem, the percolation susceptibility isIf $\chi (p)<\infty$, then $S_m\to 0$. At $p=0$ all $g_k$ vanish, so any positive exponential rate works. Otherwise choose $m\ge 1$ with $a=S_m\in (0,1)$. Iterating the weighted BK boundary-splitting estimate gives, for $k=jm+r$ with $0\le r<m$,For $k\ge m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To include the finitely many smaller indices without a prefactor, note that $p<1$ under finite susceptibility andTherefore one may takeFor $k<m$ this follows from $\mu (p)k\le -\log g_1$, and for $k\ge m$ it follows from the iterated estimate. This proves exponential one-arm decay from finite susceptibility with exactly the requested unit prefactor.

For a probability distribution $m$ on $\mathbb{R}$, take the canonical sample space $\Omega ={\mathbb{R}}^{\mathbb{N}}$, the product of the Borel $\sigma$-algebras, and the product measure $\mathbb{P}=m^{\otimes \mathbb{N}}$. The coordinate functions $X_j(\omega )={\omega }_j$ have joint cylinder probabilitiesThe Kolmogorov extension theorem supplies this measure from the consistent finite-dimensional distributions, and uniqueness follows from the generating cylinder sets. Thus the coordinates are independent and identically distributed; the same construction works for a given standard Borel state space.

The one-sided Bernoulli shift is $T({\omega }_1,{\omega }_2,\dots )=({\omega }_2,{\omega }_3,\dots )$. It is measurable and preserves all cylinder probabilities. The uniqueness theorem for probability measures extends this to the product $\sigma$-algebra, proving measure preservation.

To prove ergodicity, let $A$ be invariant modulo null sets and approximate it in probability by an event $C$ depending on only the first $r$ coordinates, with $\mathbb{P}(A△C)<ϵ$. Such approximations follow from the fact that finite-coordinate cylinder events generate the product $\sigma$-algebra. For $n\ge r$, independence gives $\mathbb{P}(C\cap T^{-n}C)=\mathbb{P}(C{)}^2$. Invariance and measure preservation giveAlso $|\mathbb{P}(A{)}^2-\mathbb{P}(C{)}^2|\le 2ϵ$. Letting $ϵ\downarrow 0$ proves $\mathbb{P}(A)=\mathbb{P}(A{)}^2$, so $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is exactly ergodicity.

The useful form is Talagrand's convex distance inequality. Let $({\Omega }_i,{\mu }_i)$ be standard probability spaces, let $\Omega ={\prod }_{i=1}^n{\Omega }_i$ carry the product measure $\mu ={⨂}_i{\mu }_i$, and let $A\subset \Omega$ be a measurable event with $\mu (A)>0$. For $x\in \Omega$, form the mismatch vectorsThe Talagrand convex distance is the distance from zero to their closed convex hull:Equivalently, by the Hahn-Banach separation theorem for two convex sets,Nonnegative weights suffice because every mismatch vector has nonnegative coordinates. Talagrand's convex distance inequality statesIn particular, Markov's inequality gives the product-space concentration inequalityThe same conclusion is interpreted with outer probabilities if a distance set is not measurable; the continuous geometric function used below has measurable level sets. A median $M$ of a real random variable $Z$ means that $\mathbb{P}(Z\le M)\ge 1/2$ and $\mathbb{P}(Z\ge M)\ge 1/2$. Applying the displayed concentration inequality to suitable level sets will control deviations around a median.

In bond percolation on an undirected graph $G=(V,E)$, a configuration is $\omega \in \{0,1{\}}^E$. The edge $e$ is open when ${\omega }_e=1$ and closed when ${\omega }_e=0$. Under ${\mathbb{P}}_p$, the coordinates are independent and identically distributed random variables with the Bernoulli distribution of parameter $p$. For a countable edge set, use the product sigma-algebra and the product measureThe open edges form a random subgraph with the original graph vertices. Its connected components of a graph are the percolation clusters. Write $x\leftrightarrow y$ when an open graph path joins $x$ to $y$; a zero-length graph path is allowed, so $x\leftrightarrow x$ always holds. The randomness is in the independently open bonds; all vertices remain present.

Order configurations coordinatewise: $\omega \le {\omega }^{\prime }$ means ${\omega }_e\le {\omega }_e^{\prime }$ for every edge. An increasing event $A$ is a measurable set such that $\omega \in A$ and $\omega \le {\omega }^{\prime }$ imply ${\omega }^{\prime }\in A$. Opening additional edges cannot destroy an increasing event. The Harris-FKG inequality, the product measure version of the FKG inequality, statesfor two increasing events. Equivalently, bounded coordinatewise increasing random variables $f,g$ satisfy ${\mathbb{E}}_p(fg)\ge {\mathbb{E}}_{p}f{\mathbb{E}}_{p}g$. The event inequality also holds for two decreasing events, by taking complements.

For disjoint occurrence of increasing events, a finite open edge set $K$ witnesses $A$ in $\omega$ if every configuration agreeing with $\omega$ on $K$ belongs to $A$. Since $A$ is an increasing event and all edges of $K$ are open, this says that the configuration with exactly $K$ open already forces $A$. Define $A\square B$ to consist of configurations admitting disjoint finite witnesses $K,L$ for $A,B$. For events depending on finitely many edges, this is the usual disjoint-occurrence definition; it also applies to finite-connection events on the infinite graph. The van den Berg-Kesten inequality statesOn a countable graph, its finite-witness version follows by taking increasing unions over finite edge sets. The Harris-FKG inequality concerns simultaneous occurrence, whereas the van den Berg-Kesten inequality requires separate certificates using disjoint coordinates.

Use the probability space $([0,1{]}^{{\mathbb{Z}}^d},\mathcal{F},\mathbb{P})$ with the product measure of the uniform distribution on $[0,1]$. Its coordinate random variables $U_v$ are independent and identically distributed random variables. Define simultaneously for every parameterFor $r\le s$, the indicator function inequality $X_r(v)\le X_s(v)$ holds pointwise at every graph vertex. For fixed $p$, each coordinate has the Bernoulli distribution with parameter $p$, and the coordinates are independent because they are functions of distinct independent $U_v$. Thus every marginal configuration is exactly the required site percolation model. This is a continuum-indexed family of processes, all on one probability space, and is the standard monotone coupling of Bernoulli percolation.

The Kullback-Leibler divergence of a product measure is the sum of the individual divergences, by expanding the logarithm of the likelihood ratio and taking its expected value. Put $q=1/2-t/\sqrt{n}$ and define $W(S)$ to be the unordered pairs lying within one group, while $C(S)$ is the set of unordered pairs crossing between groups. ThenThus choosing the printed $\partial S$ to mean $W(S)$ gives its order of the summands. If $\partial S$ denotes the usual graph cut $C(S)$, their two orders must be interchanged. Each unordered edge is counted once.

The inequality $\log x\le x-1$ givesSince $1/3<q<1/2$, each changed-edge divergence is at most $9t^2/(2n)$. Summing over at most $n(n-1)/2$ pairs gives $D_{\mathrm{KL}}(P_S\Vert P_{S^{\prime }})\le 9n{t}^2/4$. This is the information scale needed for Fano's inequality.

Use dimensionless interaction couplings in the exponent of the Boltzmann weight. In the following formulas $K,L$ denote the physical interaction energies divided by $k_{B}T$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.

Split the original square lattice into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained Ising spins by $s_e$ and eliminated Ising spins by ${\eta }_o$. For fixed retained Ising spins defineThe blocked Boltzmann weight in the checkerboard decimation of the square-lattice Ising model isUse the independent uniform product measure on Ising spins on $O$ to perform a cumulant expansion. Its one-spin mean vanishes, and $⟨{\eta }_o{\eta }_{o^{\prime }}{⟩}_0={\delta }_{o{o}^{\prime }}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.

Each eliminated site has four retained neighbours, soIts six distinct neighbour pairs comprise four corner pairs at distance $\sqrt{2}a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives spin inversion symmetry, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.

The primitive vectors of the retained square lattice are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt{2}$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt{2}$, the leading checkerboard Ising decimation recursion is thereforeThe constant part of the weight is also determined: $4(K^2/2)(N/2)=N{K}^2$, so the original partition function equalsto the same order in the logarithm of the weights. That normalization changes free energy but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed ${\beta }_T$ the relations read $K_{\mathrm{energy}}^{\prime }=L_{\mathrm{energy}}+2{\beta }_T{K}_{\mathrm{energy}}^2$ and $L_{\mathrm{energy}}^{\prime }={\beta }_T{K}_{\mathrm{energy}}^2$.

The fixed-point equations are $L_{\ast }=K_{\ast }^2$ and $K_{\ast }=3K_{\ast }^2$. Their only finite real solutions areThe origin is the high-temperature fixed point; its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete renormalization-group transformation there:The characteristic polynomial is ${\lambda }^2-(4/3)\lambda -2/3$, givingEigenvectors may be chosen as $(1,2/(3{\lambda }_{\pm }){)}^T$. The positive eigenvalue greater than one is the relevant direction of a fixed point; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function eigenvalues.

Let $t$ be the relevant thermal scaling coordinate, so $t^{\prime }={\lambda }_{+}t$ to first order. A correlation length measured in restored lattice units satisfies $\xi (t^{\prime })=\xi (t)/b$. If $\xi (t)\sim |t{|}^{-\nu }$, then ${\lambda }_{+}^{-\nu }=b^{-1}$. Hence the thermal exponent from a discrete renormalization map isThis is the value predicted by the stated low-order real-space renormalization group, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice Ising model.

Harris' inequality states that, for a product measure on $\{0,1{\}}^n$ with independent Bernoulli random variables of parameters $p_i\in [0,1]$, increasing events $E,F$ satisfyIn particular, the uniform probability measure on the Boolean lattice gives $|E\cap F|\ge |E||F|/2^n$. We give a direct mathematical induction proof, so no stronger correlation theorem is assumed.

Prove more generally that $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$ for real coordinatewise increasing functions $f,g$ on the finite cube. For $n=0$ both are constant and equality holds. For the inductive step let $p=p_n$, and let $f_j,g_j$ be the restrictions obtained by setting the last coordinate equal to $j\in \{0,1\}$. Write $a_j=\mathbb{E}[f_j]$, $b_j=\mathbb{E}[g_j]$, using the product measure on the other coordinates. Monotonicity gives $a_0\le a_1$, $b_0\le b_1$. By the inductive hypothesis and conditional expectation,The nonnegative correction term also handles $p=0$ and $p=1$. Taking indicator functions $f=1_E$, $g=1_F$ proves Harris' inequality.

For later use, an increasing event $E$ and a decreasing event $G$ have negative correlation of increasing and decreasing events. The complement $G^c$ is an increasing event, so

Use the nearest-neighbour lattice graph with graph vertex set ${\mathbb{Z}}^d$: an unordered pair $\{x,y\}$ is an edge precisely when $\Vert x-y{\Vert }_1=1$. A bond configuration is $\omega \in \{0,1{\}}^{E_d}$; under the product measure ${\mathbb{P}}_p$, its edge coordinates are independent Bernoulli distribution with success probability $p$. Edges with coordinate one are open. The open cluster $C(0)$ is the connected component of a graph reachable from the origin through open edges.

Define the percolation probability and percolation critical probability byEquivalently, $p_c(d)=\sup \{p:{\theta }_d(p)=0\}$. The monotone coupling of Bernoulli percolation proves the equivalence: assign independent uniform distribution $U_e$ to the edges and open $e$ at parameter $p$ when $U_e\le p$. Connection events and ${\theta }_d$ then increase with $p$. For $d\ge 1$, ${\theta }_d(0)=0$ and ${\theta }_d(1)=1$, so the defining set is nonempty. The definition does not settle what happens at $p=p_c(d)$.

By translation invariance all roots have the same probability. Since the graph vertex set is countable, ${\theta }_d(p)=0$ implies that no graph vertex lies in an infinite percolation cluster almost surely. Conversely a positive root probability gives positive probability of an infinite percolation cluster, and translation ergodicity of Bernoulli percolation then makes existence an almost-sure event. In question 2(b), $\theta$ means ${\theta }_2$.

Use the probability-system convention $\mu (X)=1$. The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The limit is integrable and has the same integral as $f$. On a probability space the convergence also holds in $L^1$, as in the allowed mean ergodic theorem. If $T$ is an ergodic transformation, $\mathcal{I}$ is trivial modulo null sets, giving

The integer multiplication map on the circle preserves Lebesgue measure: for any integrable $g$ on $[0,1)$,To prove the ergodicity of integer multiplication on the circle, suppose $u\in L^2(m)$ satisfies $u\circ T_K=u$. Let $c_j$ be its Fourier coefficients in the Fourier basis $e_j(x)=e^{2\pi ijx}$. Since $e_j\circ T_K=e_{Kj}$, the Fourier coefficients of $u\circ T_K$ at index $j$ are zero if $K$ does not divide $j$, and are $c_{j/K}$ otherwise. This identity holds for all $L^2$ functions by approximation with trigonometric polynomials and the isometry $u\mapsto u\circ T_K$. Invariance givesEvery nonzero integer $j$ can be divided by $K$ only finitely often. Thus $c_j=0$ for every $j\ne 0$, and completeness of the Fourier basis makes $u$ constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so

A normal number in base $K$ has every word $w=(w_1,\dots ,w_r)$ of $r$ base-$K$ digits occurring with limiting overlapping frequency $K^{-r}$. Use the expansion that is not eventually equal to $K-1$ when there are two expansions. The word $w$ corresponds to the half-open intervalA word starting at position $n+1$ occurs exactly when $T_K^{n}x\in I_w$. The Birkhoff ergodic theorem, applied to $1_{I_w}$, gives frequency $m(I_w)=K^{-r}$ almost everywhere. There are countably many pairs $(K,w)$, so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base-$K$ intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.

For the growth assertion, put $g=|f|$. For every $\epsilon >0$, the Tonelli theorem gives the useful summability boundSince $T$ is a measure-preserving transformation, $\mu (g\circ T^n>\epsilon n)=\mu (g>\epsilon n)$. The first Borel-Cantelli lemma shows that $g(T^{n}x)>\epsilon n$ occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for $\epsilon =1,1/2,1/3,\dots$ proves the linear growth bound for integrable observables, $f(T^{n}x)/n\to 0$. Multiplication by $n^{1-a}\le 1$ then gives

The threshold is sharp. For $0<a<1$, choose $p$ with $1<p<1/a$, and take the Bernoulli shift on $X=(0,1{)}^{\mathbb{Z}}$ with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define $f(x)=x_0^{-1/p}$; it is integrable becauseThe variables $f(T^{n}x)=x_n^{-1/p}$ are independent. For any fixed $M\ge 1$,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer $M$ even yields ${\mathrm{limsup}}_{n}f(T^{n}x)/n^a=\infty$. For $a\le 0$, the constant observable $f=1$ already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives

For a cost $c:X\times Y\to [0,+\infty ]$ that is a Borel measurable function, a transport map is a measurable $T:X\to Y$ whose pushforward measure satisfiesThe Monge optimal transport problem moves every source point to one destination:The Kantorovich optimal transport problem permits mass to split. Its admissible transport plans are the probability measures on $X\times Y$ with prescribed marginal distributions:Thus a transport plan is a coupling of probability distributions. The set $\Pi (\mu ,\nu )$ is never empty: it contains the product measure $\mu \otimes \nu$. Signed costs can also be used when their integrals are well defined, for example with an integrable lower bound of the form $a(x)+b(y)$.

On the Polish space $\mathbb{R}$, take the Dirac measuresEvery measurable map satisfies $T_{\#}{\delta }_0={\delta }_{T(0)}$, which cannot equal $\nu$. A transport map cannot split an atom of a measure, whereas the transport plan $\frac{1}{2}{\delta }_{(0,-1)}+\frac{1}{2}{\delta }_{(0,1)}$ can.

For $1\le p<\infty$, define the probability measures with finite $p$th absolute moment bywhere $x_0\in X$ is any fixed reference point. The condition is independent of the choice of $x_0$. The p-Wasserstein distance isThe infimum is over all transport plans, rather than only transport maps. The product measure gives a finite upper bound usingFor $p=1$ this is the Wasserstein distance with the metric of Euclidean space; for $p=2$ it is the second Wasserstein distance. “Bounded moment” here means a finite integral, and does not require $X$ to be bounded.

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.

On a probability space carrying an infinite IID sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega ),X_2(\omega ),\dots )$. Its pushforward measure $\mu$ satisfiesfor every cylinder set. Cylinder sets form a pi-system generating $\sigma (\mathcal{C})$, so the sigma-finite uniqueness theorem for measures proves uniqueness. This is the countable product measure $m^{\otimes \mathbb{N}}$.

For a cylinder $A=A_1\times \cdots \times A_N\times \mathbb{R}\times \cdots$,The same generating-class argument extends this equality to every measurable set, so $\theta$ is a measure-preserving transformation.

If $A={\theta }^{-1}A$, then $A={\theta }^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu (A)\in \{0,1\}$, proving that the shift is ergodic.

Expand the logarithm of the product measure and exchange the finite sums:where part c identifies the expression in braces.

For two increasing events of equal probability $a$ under a product measure, positive association of their complements gives $1-P(A\cup B)\ge (1-a{)}^2$. Hence $a\ge 1-\sqrt{1-P(A\cup B)}$. For $r$ equal-probability increasing events, the same reasoning gives $a\ge 1-(1-P({\bigcup }_j{A}_j){)}^{1/r}$.

For an event $A$ in a product of finitely many probability spaces with its product measure, take the closed convex hull of the mismatch vectors $(1_{x_i\ne y_i}{)}_i$, $y\in A$. The distance of this hull from zero is $d_T(x,A)$. Equivalently it is the supremum, over nonnegative coordinate weights of squared sum at most one, of their minimum mismatch sum. Averaging mismatch patterns is the feature that distinguishes this quantity from ordinary Hamming distance.