3.9 Excursion
3.9.7 Electrical networks
In this section we review the basic concepts in electrical networks and the connection of these concepts to random walks on graphs. It turns out that many different problems related to random walks are in analogy to problems in electrical networks. These analogy problems are sometimes more intuitive and easier to understand. The Dirichlet problem has an interesting analogy in electrical networks, which is discussed in Section 3.9.7.
Flow on a graph
A flow on a graph is a function i= (iuv) defined on the graph edges such that iuv =−ivu. The net in-flow of a vertex v is defined as −→
iv =P
u∼vivu. The net out-flow of a vertex is defined accordingly←−
iv =P
u∼viuv =−−→
iv. It is easy to verify that the total net in(out)-flow of a graphGis 0,P
v∈V
−
→iv = 0. A vertexsis called a source if−→
is >0, and a vertextis called a sink if←−
it >0. We call the remaining nodes with in(out)-flow zero as internal nodes.
An interesting set of flow functions are the ones that run from a single source vertex s to a single sink vertext. Every other nodev in the graph has the flow conservation property:
P
u∼vivu= 0. We call such a flow as as-t flow. The total in-flow of a network is zero, which shows that −→
it = −−→
is. When −→
is = 1, the flow is called a unit s-t flow. If we denote the vertex-edge incidence matrix of the graph byQ, every units-t flow satisfiesQi= (es−et).
Electrical flow in a network
From a graph G, build an electrical network where each edge e has resistance re = 1/we. Connect the nodes to a 1-Amper current source and the sink node t to the ground (set its voltage to 0). Then we will have a units-t flow in the network, and the out-flow of vertext is 1. The units-t electrical flow i∗ has a unique property that distinguishes it from the set of all units-t flows: We can assign a voltage(potential) functionv:V →Rsuch that Ohm’s law holds for every edge:
v(a)−v(b) =iabrab.
Proposition 3.19 Assumeiis a s-t flow and a voltage(potential) functionvon nodes satisfies the Ohm’s law on every edge. Thenv is harmonic onV \ {s, t}.
Remark 3.20 Ifvis harmonic on vertexx, it satisfies both∆v(x) = 0andLv(x) = 0. This is because∆ =D−1L. Note that the condition∆v(x) =g(x)is equivalent toLv(x) =dxg(x).
Another characteristic of an electrical flow is that it minimizes the power dissipated in the network
Pow(i) := X
e=(a,b)∈E
v(a)−v(b)
iab, (3.21)
or equivalently
Pow(i) = X
e=(a,b)∈E
rabi2ab= X
e=(a,b)∈E
v(a)−v(b)2 rab
. (3.22)
It is easy to show that these two characteristics are equivalent: if a units-t flowiminimizes the power, then we can assign a voltage functionv such that the Ohm’s law holds. If for a units-t flow i, we could find a corresponding voltage function v, then the flowi minimizes the power dissipated in the network.
Effective resistance: The effective resistance (or resistance distance) R(s, t) between two verticessandtin the network is the resistance measured betweensandt. We can replace the network betweensand t by a single resistor with resistanceR(s, t). The resistance distance can be computed by minimizing the energy dissipated in the network (Equation 3.21) from a unit flow:
R(s, t) = minn X
e∈E
rei2e
i= (ie)e∈E unit flow fromstoto
. (3.23)
We can intuitively check the correctness of this formula. A unit electrical flow froms tot is the units-t flow that minimizes the power dissipation Pow(i). Now replace the network with an equivalent resistor betweensandtwith resistanceR(s, t). Then we have
Pow(i) =R(s, t)−→
is2=R(s, t).
The resistance distance can also derived from the energy dissipation function in Equation 3.22:
C(s, t) = minn X
e=(u,v)
(v(a)−v(b))2 re
v(s)−v(t) = 1o
(3.24)
Then R(s, t) = 1/C(s, t). Again we can intuitively check the correctness of this formula.
Replace the network with an equivalent resistor between sand t with resistanceR(s, t). As we have a unit voltage drop,
C(s, t) = Pow(i) = (v(s)−v(t))2/R(s, t) = 1/R(s, t).
An alternative method for finding the effective resistance R(s, t) is the following: Assume that the current 1 is injected into the vertexsand consequently goes out from vertext. The voltage vectorvsatisfies
Lv=es−et. (3.25)
This equation has many solutions, but they all differ by a constant vector. This can also verified using the fact that L has eigenvalue 0 with multiplicity 1 and constant eigenvector (see also Section 3.9.6). Ifv∗ satisfies Equation 3.25, thenR(s, t) =v∗(s)−v∗(t). One such solution isv∗ =L†(es−et), so
R(s, t) = (es−et)TL†(es−et). (3.26) Another way to find the solution of Equation 3.25 is by setting the voltage at nodetto 0. If we setA=V \ {t}, then the solution to Equation 3.25 also satisfiesLA,AvA= (es)A. As the matrixLA,A is non-singular, the last equation has a unique solutionv∗A=L−1A,A(es)A and
R(s, t) =vs∗−v∗t =v∗s=L−1A,A(s, s). (3.27)
Dirichlet problem and electrical networks
The Dirichlet problem has an interesting interpretation in the context of electrical networks.
Using the notation from Equation 3.13, the boundary function ¯f can be interpreted as con-necting each boundary vertexv∈∂Ato a voltage source ¯f(v). Also the functiongcan be seen as connecting each internal nodev∈A to a current source with in-flowg(v). The problem is then to find the induced voltages on nodes inA.
Proposition 3.21 We are given a resistive network G = {V, E}. The graph vertices are divided into two setsV =A∪∂A, where every vertex in∂Ais connected to at least one vertex in A. Each vertex v ∈ ∂A is connected to a voltage source f¯(v) and each vertex v ∈ A is connected to a current source with in-flowg(v). Then the voltages of nodes inAare determined by solving
(Lf(x) =g(x) x∈A
f(x) = ¯f(x) x∈∂A. (3.28)
This problem is equivalent to the Dirichlet problem (∆f(x) = ¯g(x) x∈A
f(x) = ¯f(x) x∈∂A (3.29)
withg¯=D−1A,Ag.
Proof. It simply follows from the Ohm’s law. 2
Resistance and commute time
The following theorem states the relation between the resistance distance and the commute time, first proved by Chandra et al. (1989).
Theorem 3.22 For any two vertexs andt, the commute time equals to C(s, t) = 2 vol(G)R(s, t).
The next proposition relates the resistance distance and the frequency of return to the source before hitting the sink vertex:
Proposition 3.23 Denote the number of times a random walk starting from s returns to s before visitingt byns,t. Then
• ns,t=R(s, t)×ds.
• ns,t×dt=nt,s×ds
Proof. Part a) By the mean of Proposition 3.21, the resistance distance can be written as R(s, t) =f(s)−f(t) wheref is an arbitrary solution of the Dirichlet problem with an empty boundary:
Lf(s) = 1 Lf(t) =−1
Lf(x) = 0 x∈V \ {s, t}
This equation has many solutions, all differ by a constant vector. We can fix to one of them by settingf(t) = 0. If we set t as a single boundary node, then the condition Lf(t) = −1 will hold automatically. The reason can be seen from electrical network interpretation: the total in-flow of the network is zero andLf(s) = 1. The vertextis the only vertex that has a non-zero out-flow, soLf(t) =−1 . This means that the resistance distance can be written as R(s, t) =f(s) wheref is the solution of following Dirichlet problem
(Lf(x) =es(x) x∈V \ {t}
f(t) = 0 or equivalently
(∆f(x) =d−1s es(x) ∈V \ {t}
f(t) = 0
Now using Proposition 3.17 shows thatR(s, t) =ns,t/ds.
Part b) If we repeat the procedure in Part a by setting f(s) = 0, we getR(s, t) = nt,s/dt. Putting together, we havens,t×dt=nt,s×ds which finishes the proof. 2 This proposition helps us in analyzing the recurrence of a random walk. Consider a random walk on an infinite graph that starts from vertexs. Connect all vertices outside a “ring” with (shortest path) distancerfromstogether and call itt. Asrgoes to infinity,ns,twill converge to the expected number of returns to the origin. This means that the effective resistance betweensand infinityReff(s) equals to the expected number of returns to the origin divided byds. On the other hand, Theorem 3.11 shows that a random walk is recurrent if and only if the expected number of returns to the origin is infinite. So we proved the following theorem:
Theorem 3.24 A random walk is recurrent if and only if the resistance between a vertex and infinity is infinite.