Random Walks and Electric Networks

An extremely useful tool and viewpoint for the study of random walks is Kirchhoff’s theory of electric networks. Our treatment here roughly follows, we also refer the reader for an in-depth comprehensive study.


Harmonic Functions and Voltages
Let G = (V , E) be a finite network. In physics classes it is taught that when we impose specific voltages at fixed vertices a and z, then current flows through the network according to certain laws (such as the series and parallel laws). An immediate consequence of these laws is that the function from V to R giving the voltage at each vertex is harmonic at each x ∈ V \ {a, z}.

Definition 2.2 A function
1 π x y:y∼x c xy h(y) where π x := y:y∼x c xy . (2.1) Instead of starting with the physical laws and proving that voltage is harmonic, we now take the axiomatically equivalent approach of defining voltage to be a harmonic function and deriving the laws as corollaries. Definition 2.3 Given a network G = (V , E) and two distinct vertices a, z ∈ V , a voltage is a function h : V → R that is harmonic at any x ∈ V \ {a, z}.
We will show in Claim 2.8 and Corollary 2.7 that for any α, β ∈ R, there is a unique voltage h such that h(a) = α and h(z) = β (this assertion is true only when the network is finite).
Claim 2.4 If h 1 , h 2 are harmonic at x then so is any linear combination of h 1 , h 2 .
Proof Leth = αh 1 + βh 2 for some α, β ∈ R. It holds that h(x) = αh 1 (x) + βh 2 (x) = 1 π x y:y∼x c xy αh 1 (y) + 1 π x y:y∼x c xy βh 2 (y) We present an alternative proof of existence based on the random walk on the network. Consider the Markov chain {X n } on the state space V with transition probabilities p xy := P(X t +1 = y | X t = x) = c xy π x . (2.2) This Markov chain is a weighted random walk (note that if c xy are all 1 then the described chain is the so-called simple random walk). We write P x and E x for the probability and expectation, respectively, conditioned on X 0 = x. For a vertex x, define the hitting time of x by Proof 2 We will find a voltage g satisfying g(a) = 0 and g(z) = 1 by setting g(x) = P x (τ z < τ a ).
Indeed, g is harmonic at x = a, z, since by the law of total probability and the Markov property we have g(x) = 1 π x y:y∼x c xy P x (τ z < τ a | X 1 = y) = 1 π x y:y∼x c xy P y (τ z < τ a ) = 1 π x y:y∼x c xy g(y).
By Claim 2.4, h is a voltage, and clearly h(a) = α and h(z) = β, concluding the proof.
This proof justifies the equality between simple random walk probabilities and voltages that was discussed at the start of this chapter: since the function x → P x (τ z < τ a ) is harmonic on V \ {a, z} and takes values 0, 1 at a, z respectively, it must be equal to the voltage at x when voltages 0, 1 are imposed at a, z.

Claim 2.9 If h is a voltage with h(a) ≤ h(z), then h(a) ≤ h(x) ≤ h(z) for all
Furthermore, if h(a) < h(z) and x ∈ V \ {a, z} is a vertex such that x is in the connected component of z in the graph G\{a}, and x is in the connected component of a in the graph G \ {z}, then h(a) < h(x) < h(z).
Proof This follows directly from the construction of h in Proof 2 of Claim 2.8 and the uniqueness statement of Corollary 2.7. Alternatively, one can argue as in the proof of Claim 2.6 that if M = max x h(x) and m = min x h(x), then the sets A = {x ∈ V : h(x) = M} and B = {x ∈ V : h(x) = m} must each contain at least one element of {a, z}.
To prove the second assertion, we note that by Claim 2.8 and Corollary 2.7 it is enough to check when h is the voltage with boundary values h(a) = 0 and h(z) = 1. In this case, the condition on x guarantees that the probabilities that the random walk started at x visits a before z or visits z before a are positive. By proof 2 of Claim 2.8 we find that h(x) ∈ (0, 1).

Flows and Currents
For a graph G = (V , E), denote by E the set of edges of G, each endowed with the two possible orientations. That is, (x, y) ∈ E iff {x, y} ∈ E (and in that case, (y, x) ∈ E as well).

Definition 2.12 Given a voltage h, the current flow
In other words, the voltage difference across an edge is the product of the current flowing along the edge with the resistance of the edge. This is known as Ohm's law. According to this definition, the current flows from vertices with lower voltage to vertices with higher voltage. We will use this convention throughout, but the reader should be advised that some other sources use the opposite convention.
Claim 2.13 The current flow associated with a voltage is indeed a flow.
Proof The current flow is clearly antisymmetric by definition. To show that it satisfies the node law, observe that for x = a, z, since h is harmonic, Claim 2.14 The current flow associated with a voltage h satisfies Kirchhoff's cycle law, that is, for every directed cycle e 1 , . . . , e m , r i=1 r e i θ( e i ) = 0. Note that since θ satisfies the cycle law, the right hand side of (2.3) does not depend on the choice of the path, hence h(x) is well defined. Let x ∈ V , and consider a given path e 1 , . . . , e k from a to x (if x = a we take the empty path). To evaluate h(y) for y ∼ x, consider the path e 1 , . . . , e k , xy from a to y, so h(y) = h(x) + r xy θ(xy). It follows that h(y) − h(x) = r xy θ(xy), hence θ(xy) = c xy (h(y) − h(x)), meaning that θ is indeed the current flow associated with h.
Since θ(xy) = c xy (h(y) − h(x)) for any x ∼ y, the node law of immediately implies that h is a voltage. To show that h is unique up to an additive constant, suppose that g : V → R is another voltage such that r xy θ(xy) = g(y) − g(x). It follows that g(y) − h(y) = g(x) − h(x) for any x ∼ y. Since G is connected it follows that g − h is the constant function on V . where the first equality is due to antisymmetry, and the third equality is due to the node law. The claim follows again by antisymmetry.
Proof Letθ = θ 1 − θ 2 . According to Claim 2.11,θ is a flow. It also satisfies the cycle law, as for every cycle e 1 , . . . , e m , Observe in addition that θ = θ 1 − θ 2 = 0. Now, let h = hθ be the voltage defined in Claim 2.15, chosen so that h(a) = 0. Note that it is harmonic at a, since 1 π a x:x∼a c ax h(x) = 1 π a x:x∼a c ax (h(a) + r axθ (ax)) Similarly, using Claim 2.17 it is also harmonic at z. Since h is harmonic everywhere, it is constant by Claim 2.5, and thus h ≡ 0, henceθ ≡ 0 and so θ 1 = θ 2 .
This last claim prompts the following useful definition.

Definition 2.19
The unit current flow from a to z is the unique current flow from a to z of strength 1.

The Effective Resistance of a Network
Suppose we are given a voltage h on a network G with fixed vertices a and z.
Scaling h by a constant multiple causes the associated current flow to scale by the same multiple, while adding a constant to h does not change the current flow at all. Therefore, the strength of the current flow is proportional to the difference h(z) − h(a).
Claim 2.20 For every non-constant voltage h and a current flow θ corresponding to h, the ratio is a positive constant which does not depend on h.
Proof Let h 1 , h 2 be two non-constant voltages, and let θ 1 , θ 2 be their associated current flows. For i = 1, 2, leth i = h i / θ i and letθ i be the current flow associated withh i (note that since h i is non-constant θ i = 0). Thus, θ i = 1. By Claim 2.18 we getθ 1 =θ 2 and thereforeh 1 =h 2 + c for some constant c by Claim 2.15. It To see that this constant is positive, it is enough to check one particular choice of a voltage. By Claim 2.8, let h be the voltage with h(a) = 0 and h(z) = 1. By Claim 2.9 and since G is connected, we have that h(x) > 0 for at least one neighbor x of a. Thus, the corresponding current flow θ has θ > 0 making (2.4) positive.

Fig. 2.2
Examples for effective resistances of two networks with unit edge conductances. (a) For the voltage depicted, the voltage difference between a and z is 5, and the current flow's strength is 1, hence the effective resistance is 5/1 = 5. (b) For the voltage depicted, the voltage difference between a and z is 1, and the current flow's strength is 1, hence the effective resistance is 1/1 = 1 Claim 2.20 is the mathematical manifestation of Ohm's law which states that the voltage difference across an electric circuit is proportional to the current through it. The constant of proportionality is usually called the effective resistance of the circuit.

Definition 2.21
The number defined in (2.4) is called the effective resistance between a and z in the network, and is denoted R eff (a ↔ z). We call its reciprocal the effective conductance between a and z and is denoted For examples of computing the effective resistances of networks, see Fig. 2

.2.
Notation In most cases we write R eff (a ↔ z) and suppress the notation of which network we are working on. However, when it is important to us what the network is, we will write R eff (a ↔ z; G) for the effective resistance in the network G with unit edge conductances and R eff (a ↔ z; (G, {r e })) for the effective resistance in the network G with edge resistances {r e } e∈E . Furthermore, given disjoint subsets A and Z of vertices in a graph G, we write R eff (A ↔ Z) for the effective resistance between a and z in the network obtained from the original network by identifying all the vertices of A into a single vertex a, and all the vertices of Z into a single vertex z.
Probabilistic Interpretation For a vertex x we write τ + x for the stopping time where X t is the weighted random walk on the network, as defined in (2.2). Note that .
Proof Consider the voltage h satisfying h(a) = 0 and h(z) = 1, and let θ be the current flow associated with h. Due to uniqueness of h (Corollary 2.7) we have that .
Network Simplifications Sometimes a network can be replaced by a simpler network, without changing the effective resistance between a pair of vertices.
Claim 2.23 (Parallel Law) Conductances add in parallel. Suppose e 1 , e 2 are parallel edges between a pair of vertices, with conductances c 1 and c 2 , respectively. If we replace them with a single edge e with conductance c 1 + c 2 , then the effective resistance between a and z is unchanged.
A demonstration of the parallel law appears in Fig. 2.3.
Proof Let G be the graph where e 1 and e 2 are replaced with e with conductance c 1 + c 2 . Then it is immediate that if h is any voltage function on G, then it remains a voltage function on the network G . The claim follows.

Claim 2.24 (Series Law)
Resistances add in series. Suppose that u ∈ {a, z} is a vertex of degree 2 and that e 1 = (u, v 1 ) and e 2 = (u, v 2 ) are the two edges touching u with edge resistances r 1 and r 2 , respectively. If we erase u and replace e 1 and e 2 by a single edge e = (v 1 , v 2 ) of resistance r 1 + r 2 , then the effective resistance between a and z is unchanged.
The series law is depicted in Fig. 2.4.
Proof Denote by G the graph in which u is erased and e 1 and e 2 are replaced by a single edge (v 1 , v 2 ) of resistance r 1 + r 2 . Let θ be a current flow from a to z in G, and define a flow θ from a to z in G by putting θ (e) = θ(e) for any e = e 1 , e 2 and θ (  Thus θ satisfies the node law at any x ∈ {a, z} and θ = θ . Furthermore, since θ satisfies the cycle law, so does θ . We conclude θ is a current flow of the same strength as θ and the voltage difference they induce is the same.
The operation of gluing a subset of vertices S ⊂ V consists of identifying the vertices of S into a single vertex and keeping all the edges and their conductances. In this process we may generate parallel edges or loops. Proof This is immediate since the voltage on the glued graph is still harmonic.
Example: Spherically Symmetric Tree Let be a spherically symmetric tree, that is, a rooted tree where all vertices at the same distance from the root have the same number of children. Denote by ρ the root of the tree, and let {d n } n∈N be a sequence of positive integers. Every vertex at distance n from the root ρ has d n children. Denote by n the set of all vertices of height n. We would like to calculate R eff (ρ ↔ n ). Due to the tree's symmetry, all vertices at the same level have the same voltage and therefore by Claim 2.25 we can identify them. Our simplified network has now one vertex for each level, denoted by {v i } i∈N (where ρ = v 0 ), with | n+1 | edges between v n and v n+1 . Using the parallel law (Claim 2.23), we can reduce each set of | n | edges to a single edge with resistance 1 | n | , then, using the series law (Claim 2.24) we get By Claim 2.22 we learn that where τ n is the hitting time of n for the random walk on . Observe that P ρ τ n < τ + ρ for all n = P ρ (X t never returns to ρ) , so by (2.6) we reach an interesting dichotomy. If ∞ i=1 1 d 1 ···d i = ∞, then the random walker returns to ρ with probability 1, and hence returns to ρ infinitely often almost surely. If ∞ i=1 1 d 1 ···d i < ∞, then with positive probability the walker never returns to ρ, and hence visits ρ only finitely many times almost surely.
The former graph is called a recurrent graph and the latter is called transient. We will get back to this dichotomy in Sect. 2.5.

The Commute Time Identity
The following lemma shows that the effective resistance between a and z is proportional to the expected time it takes the random walk starting at a to visit z and then return to a, in other words, the expected commute time between a and z. We will use this lemma only in Chap. 6 so the impatient reader may skip this section and return to it later.

Lemma 2.26 (Commute Time Identity) Let G = (V , E) be a finite network and a = z two vertices. Then
Proof We denote by G z : V × V → R the so-called Green function G z (a, x) = E a [number of visits to x before z] and note that It is straightforward to show that the function ν(x) = G z (a, x)/π x is harmonic in V \ {a, z}. Also, we have that G z (a, z) = 0 and G z (a, a) = 1 P a (τ z <τ a ) = π a R eff (a ↔ z). Thus, ν is a voltage function with boundary conditions ν(z) = 0 and ν(a) = R eff (a ↔ z) which satisfies Similarly, the same analysis for E z [τ a ] yields the same result, with the voltage function η which has boundary conditions η(z) = R eff (a ↔ z) and η(a) = 0. Therefore, η(x) = ν(a)−ν(x) for all x ∈ V since both sides are harmonic functions in V \ {a, z} that receive the same boundary values. This implies that

Energy
So far we have seen how to compute the effective resistance of a network via harmonic functions and current flows. However, in typical situations it is hard to find a flow satisfying the circle law. Luckily, an extremely useful property of the effective resistance is that it can be represented by a variational problem. Our intuition from highschool physics suggests that the energy of the unit current flow is minimal among all unit flows from a to z. The notion of energy can be made precise and will allow us to obtain valuable monotonicity properties. For instance, removing any edge from an electric network can only increase its effective resistance. Hence, any recurrent graph remains recurrent after removing any subset of edges from it. Two variational problems govern the effective resistance, Thomson's principle, which is used to bound the effective resistance from above, and Dirichlet's principle, used to bound it from below.

Definition 2.27
The energy of a flow θ from a to z, denoted by E(θ ), is defined to be E(θ ) := 1 2 e∈ E r e θ( e) 2 = e∈E θ(e) 2 r e .
Note that in the second sum we sum over undirected edges, but since θ(xy) 2 = θ(yx) 2 , this is well defined.

Theorem 2.28 (Thomson's Principle)
R eff (a ↔ z) = inf{E(θ ) : θ = 1, θ is a flow from a to z} and the unique minimizer is the unit current flow.
Proof First, we will show that the energy of the unit current flow is the effective resistance. Let I be the unit current flow, and h the corresponding (Claim 2.15) voltage function. Observe that in the second term of the right hand side, for every x = a, z the sum over all y ∼ x is 0 due to the node law, hence the entire term equals 1 2 (h(a) − h(z)). From antisymmetry of I , the first term on the right hand side equals − 1 2 (h(a) − h(z)), hence the right hand side equals altogether h(z) − h(a) = R eff (a ↔ z).
We will now show that every other flow J with J = 1 has E(J ) ≥ E(I ). Let J be such flow and write J = I + (J − I ). Set θ = J − I and note that θ = 0. We have where the last inequality follows from the same reasoning as before. We conclude that E(J ) ≥ E(I ) as required and that equality holds if and only if E(θ ) = 0, that is, if and only if J = I . This inequality is preserved while taking infimum over all flows with strength 1. Applying Theorem 2.28 finishes the proof.

Corollary 2.30 Gluing vertices cannot increase the effective resistance between a and z.
Proof Denote by G the original network and by G the network obtained from gluing a subset of vertices. Then every flow θ on G (viewed as a function on the edges) is a flow on G . Hence the infimum in Theorem 2.28 taken over flows in G is taken over a larger subset of flows.

Definition 2.31
The energy of a function h : Compare the following lemma with Thomson's principle (Theorem 2.28).

Lemma 2.32 (Dirichlet's Principle)
Let G be a finite network with source a and sink z. Then Proof The infimum is obtained when h is the harmonic function taking 0 and 1 at a, z respectively. The reason is that if there exists v = a, z with then we can change the value of h at v to be the right hand side of (2.7) and the energy will only decrease. One way to see this is that if X is a random variable with a second moment, then the value E(X) minimizes the function Let h be that harmonic function and let I be its current flow, so I (xy) = c xy (h(y) − h(x)). WriteÎ = R eff (a ↔ z) · I , so Î = 1. By Thomson's principle,

Infinite Graphs
Let G = (V , E) be an infinite connected graph with edge resistances {r e } e∈E . We assume henceforth that this network is locally finite, that is, for any vertex x ∈ V we have y:y∼x c xy < ∞. Let {G n } be a sequence of finite subgraphs of G such that n∈N G n = G and G n ⊂ G n+1 ; we call such a sequence an exhaustive sequence of G. Identify all vertices of G \ G n with a single vertex z n .
Claim 2.33 Given an exhaustive sequence {G n } of G, the limit lim n→∞ R eff (a ↔ z n ; G n ∪ {z n }) (2.8)

exists.
Proof The graph G n ∪ {z n } can be obtained from G n+1 ∪ {z n+1 } by gluing the vertices in G n+1 \ G n with z n+1 and labeling the new vertex z n . By Corollary 2.30, the effective resistance R eff (a ↔ z n ; G n ∪ {z n }) is increasing in n.
Claim 2.34 The limit in (2.8) does not depend on the choice of exhaustive sequence {G n }.
Proof Indeed, let {G n } and {G n } be two exhaustive sequences of G. We can find subsequences {i k } k≥1 and {j k } k≥1 such that . .} is itself an exhaustive sequence of G, the limit of effective resistances for this sequence exists and equals the limits of effective resistances for the subsequences {G i k } and {G j k }. In turn, these are equal to the limits of effective resistances for the original sequences {G n } and {G n }, respectively.

Definition 2.35
In an infinite network, the effective resistance from a vertex a and ∞ is We are now able to address the question of recurrence versus transience of a graph systematically. Recall the definition of τ + x in (2.5). In an infinite network we define τ + a = ∞ when there is no time t ≥ 1 such that X t = a. Definition 2.36 A network (G, {r e } e∈E ) is called recurrent if P a (τ + a = ∞) = 0, that is, if the probability of the random walker started at a never returning to a is 0. Otherwise, it is called transient .
Observe that since G is connected, if P a (τ + a = ∞) = 0 for one vertex a, then it holds for all vertices in the network. As we have seen, if n is large enough so that a ∈ G n , then R eff (a ↔ z n ; G n ∪ {z n }) = 1 π a · P a τ G\G n < τ + a .

Definition 2.37
Let G be an infinite network. A function θ : E(G) → R is a flow from a to ∞ if it is anti-symmetric and satisfies the node law on each vertex v = a.
The following follows easily from Theorem 2.28, we omit the proof.

Theorem 2.38 (Thomson's Principle for Infinite Networks) Let G be an infinite network, then
R eff (a ↔ ∞) = inf{E(θ ) : θ is a flow from a to ∞ of strength 1}.

Corollary 2.39
Let G be an infinite graph. The following are equivalent: There exists a vertex a ∈ V such that R eff (a ↔ ∞) < ∞. Hence all vertices satisfy this. 3. There exists a vertex a ∈ V and a unit flow θ from a to ∞ with E(θ ) < ∞. Hence all vertices satisfy this.
We will now develop a useful method for bounding effective resistances from below. This will lead us to a popular sufficient criterion for recurrence in Corollary 2.43.
Definition 2.40 A cutset ⊆ E(G) separating a from z is a set of edges such that every path from a to z must use an edge from .
Claim 2.41 Let θ be a flow from a to z in a finite network, and let a cutset separating a from z. Then e∈ |θ(e)| ≥ θ .
Proof Denote by Z the set of vertices separated from a by . Denote by G the network where Z is identified to a single vertex x and all edges having both endpoints in Z are removed. Now, the restriction of θ to the edges of the new network is a flow from a to x. By Claim 2.17, we have y:y∼x θ(yx) = θ . Also, all edges incident to x must be in , since otherwise x is not separated from a by . Therefore Also, since n is a cutset, the flow passing through n is at least θ , Consider now an infinite network G = (V , E). We say that ⊂ E is a cutset separating a from ∞ if any infinite simple path from a must intersect .  { n } is a collection of disjoint cutsets separating 0 from ∞. Also, | n | = 4(2n + 1) and therefore n e∈ n c e −1 = ∞. We deduce by Corollary 2.43 that Z 2 is recurrent.

Remark 2.45
There are recurrent graphs for which there exists M < ∞ such that for every collection { n } of disjoint cutsets, n e∈ n c e −1 ≤ M. Therefore, the Nash-Williams inequality is not sharp. See Exercise 4 of this chapter.

Random Paths
We now present the method of random paths, which is one of the most useful methods for generating unit flows on a network and bounding their energy. In fact, it is possible to show that the electric flow can be represented by such a random path. Suppose G is a network with fixed vertices a, z and μ is a probability measure on the set of paths from a to z. Then θ is a flow from a to z with θ = 1.
Proof θ is antisymmetric since θ γ is antisymmetric for every γ , and it satisfies the node law since θ γ satisfies the node law. Similarly, the "strength" of θ γ (i.e., x∼a θ γ (ax)) is 1, hence θ = 1. An example of the use of this method is the following classical result.
Proof For R > 0 denote by B R = {(x, y, z) : x 2 + y 2 + z 2 ≤ R 2 } the ball of radius R in R 3 . Put V R = B R ∩ Z 3 and let ∂V R be the external vertex boundary of V R , that is, the set of vertices not in V R which belong to an edge with an endpoint in V R . We construct a random path μ from the origin 0 to ∂V R by choosing a uniform random point p in ∂B R = {(x, y, z) : x 2 + y 2 + z 2 = R 2 }, drawing a straight line between 0 and p in R 3 , considering the set of distance at most 10 in R 3 from the line, and then choosing (in some arbitrary fashion) a path in Z 3 which is contained inside this set. The non-optimal constant 10 was chosen in order to guarantee that such a discrete path exists for any point p ∈ ∂B R . By Claim 2.46, the measure μ corresponds to a flow from 0 to ∂V R . To estimate the energy of this flow, we note that if e is an edge at distance r ≤ R from the origin, then the probability that it is traversed by a path drawn by μ is O(r −2 ). Furthermore, there are O(r 2 ) such edges. Hence the energy of the flow is at most for some constant C < ∞ which does not depend on R. By Claim 2.46 and Theorem 2.28 we learn that R eff (0 ↔ ∂V R ) ≤ C for all R, and so by Corollary 2.39 we deduce that Z 3 is transient.

Exercises
1. Let G z (a, x) be the Green's function, that is, G z (a, x) = E a #visits to x before visiting z .
Show that the function h(x) = G z (a, x)/π(x) is a voltage. 2. Show that the effective resistance satisfies the triangle inequality. That is, for any three vertices x, y, z we have R eff (x ↔ z) ≤ R eff (x ↔ y) + R eff (y ↔ z) . (2.9) 3. Let a, z be two vertices of a finite network and let τ a , τ z be the first visit time to a and z, respectively, of the weighted random walk. Show that for any vertex x P x (τ a < τ z ) ≤ R eff (x ↔ {a, z}) R eff (x ↔ a) .
4. Consider the following tree T . At height n it has 2 n vertices (the root is at height n = 0) and if (v 1 , . . . , v 2 n ) are the vertices at level n we make it so that v k has 1 child at level n + 1 and if 1 ≤ k ≤ 2 n−1 and v k has 3 children at level n + 1 for all other k.
(a) Show that T is recurrent. (b) Show that for any disjoint edge cutsets n we have that n | n | −1 < ∞.
(So, the Nash-Williams criterion for recurrence is not sharp) 5. (a) Let G be a finite planar graph with two distinct vertices a = z such that a, z are on the outer face. Consider an embedding of G so that a is the left most point on the real axis and z is the right most point on the real axis. Split the outer face of G into two by adding the ray from a to −∞ and the ray from z to +∞. Consider the dual graph G * of G and write a * and z * for the two vertices corresponding to the split outer face of G. Assume that all edge resistances are 1. Show that R eff (a ↔ z; G) = 1 R eff (a * ↔ z * ; G * ) .
Is G recurrent or transient?
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