aa r X i v : . [ m a t h . P R ] M a y Electric network for non-reversible Markov chains
M´arton Bal´azs ∗ ´Aron Folly † September 17, 2018
Abstract
We give an analogy between non-reversible Markov chains and elec-tric networks much in the flavour of the classical reversible results orig-inating from Kakutani, and later Kem´eny-Snell-Knapp and Kelly. Non-reversibility is made possible by a voltage multiplier – a new electroniccomponent. We prove that absorption probabilities, escape probabilities,expected number of jumps over edges and commute times can be com-puted from electrical properties of the network as in the classical case.The central quantity is still the effective resistance, which we do have inour networks despite the fact that individual parts cannot be replaced bya simple resistor. We rewrite a recent non-reversible result of Gaudilli`ere-Landim about the Dirichlet and Thomson principles into the electricallanguage. We also give a few tools that can help in reducing and solvingthe network. The subtlety of our network is, however, that the classicalRayleigh monotonicity is lost.
Keywords:
Non-reversible Markov chains; Electric networks; Effective resistance;Absorption probability; Commute time
MSC:
Random walks or, more generally, reversible Markov chains have a strong connectionto electric resistor networks. Our knowledge of this analogy started with the workof Kakutani [8], Doob [3], Kem´eny, Snell and Knapp [9], Nash-Williams [11]. Sincethen the field became a foundational part of the theory of reversible Markov chains,we refer the readers to Doyle and Snell [5], Telcs [13], Lyons and Peres [10], Chandra,Raghavan, Ruzzo, Smolensky and Tiwari [2] as a few references in the huge literature.Among several results, escape probabilities, transience-recurrence problems, commuteand mixing times have been successfully investigated with the use of this analogy. Twofundamental tools were the Thomson (or Dirichlet) energy minimum principles, andRayleigh’s monotonicity law. The former say that under given boundary conditions, ∗ University of Bristol; part of this work was done while the author was affiliated withthe Institute of Mathematics, Budapest University of Technology and Economics; theMTA-BME Stochastics Research Group, and the Alfr´ed R´enyi Institute of Mathematics. [email protected] ; research partially supported by the Hungarian Scientific ResearchFund (OTKA) grants K60708, F67729, K100473, K109684, and the Bolyai Scholarship of theHungarian Academy of Sciences. † Department of Mathematics, Ludwig-Maximilians-Universit¨at M¨unchen; part of this workwas done while the author was affiliated with the Institute of Mathematics, Budapest Univer-sity of Technology and Economics; [email protected] he physical current (or voltage, resp.) minimises the power losses on the resistors. Asa consequence, Rayleigh’s monotonicity law states that the effective resistance of thenetwork is increasing in any of its individual resistances.The resistor is a symmetric component, and this fact has fundamentally restrictedapplications to the family of reversible Markov chains. Much less is known therefore inthe non-reversible case. The Thomson and Dirichlet principles have been establishedby Doyle [4] and Gaudilli`ere-Landim [7], and re-proved in an elementary way by Slowik[12]. As an application, Gaudilli`ere and Landim also prove recurrence theorems insome non-reversible systems. These studies use notions like energy , potential and conductance , but a genuine electric network is not featured behind these ideas.In this note we build a full electrical framework behind non-reversible Markovchains. The basic idea is to replace the single resistor by a non-symmetric electricalcomponent. This new part consists of traditional resistors and a new voltage-multiplierunit, which we will just call amplifier in short. As shown below, this unit is verydirectly linked to “how much a jump is non-reversible” in the Markov chain. Inparticular the amplifier becomes trivial and the network reduces to the classical resistorcircuit if the chain is reversible. Also, reversing a chain w.r.t. its stationary distributionwill simply have the effect of reversing the amplifiers.With this new component, many of the classical analogies work out flawlessly. Thestarting point is, as in the reversible case, to make voltages in the network directlyrelated to absorption probabilities of the Markov chain. The electric current also has aprobabilistic interpretation. Our first observation is that despite the fact that individ-ual components are more complicated than a single resistor, relevant networks can bereplaced by a single effective resistance between any to vertices (or even subsets on twodifferent constant potentials). We derive that the effective conductance (reciprocal ofthe effective resistance) is equal to what people call capacity in the theory of Markovchains. We show how symmetry properties and other simple observations regardingthe capacity and escape probabilities follow from the electrical point of view. Thebeautiful observation of Chandra, Raghavan, Ruzzo, Smolensky and Tiwari [2] thatconnects commute times and effective resistance also generalises without problems tothe non-reversible setting. We remark here that a nice mapping of states of non-reversible Markov chains to Euclidean space, based on commuting times, had beenworked out earlier by Doyle and Steiner [6].Problems start when we look at Rayleigh’s monotonicity principle. In its simplenaive form monotonicity is just not true in our networks – this we demonstrate with acounterexample. What can possibly come as a replacement is a question for the future.As a possible first step towards answering this, we rewrite the Dirichlet and Thomsonprinciples by Gaudilli`ere-Landim [7] and Slowik [12] into the electrical language. Allterms are then assigned an electrical-energetic meaning, but this has not helped ourintuition enough to come up with a sensible way of establishing monotonicity.Finally, we give a bit of further insight to the behaviour of our electric networksby showing how series and parallel substitutions work. In connection with the lack of(naive) monotonicity, it turns out that delta-star transformations, being essential inthe theory of resistor networks, cannot hold in general for our case. We begin with describing the electric component that we can later use in our analogywith irreversible Markov chains. The schematic picture we use is as follows: xy / R xy / ∗ λ yx i xy u x u y This unit is thought of as being connected to neighbouring vertices x and y of agraph. These vertices are on respective electric potentials u x and u y , which inducesa current i xy through the unit from vertex x to vertex y . We will always consider i ·· as an antisymmetric quantity in the sense that i xy = − i yx . Our unit consists of threecomponents: • an ordinary resistor of R xy / > • a voltage amplifier of parameter λ yx > • another ordinary resistor of R xy / R ·· is considered as a sym-metric quantity: R xy = R yx . The new element is the voltage amplifier of parameter λ yx . It has the following characteristics: • the current that flows into it on one end agrees with the current that comes outon the other end; • the potential, measured with respect to Ground, from its left end (closer to x )to its right end (closer to y ) gets multiplied by the positive real parameter λ yx .As the definition naturally suggests, the parameter λ ·· is log-antisymmetric: we alwaysassume λ yx = 1 /λ xy . We will also allow the graph to have loops (edges connecting avertex to itself) from some vertex x to x , in which case we require λ xx = 1.According to the above, we now follow the potential (with respect to Ground) fromleft to right in the above unit. First, according to Ohm’s law, a drop of i xy · R xy / λ yx . Finally, a second drop by i xy · R xy / u y = (cid:16) u x − i xy R xy (cid:17) · λ yx − i xy R xy , or i xy = 2 C xy λ yx · ( λ yx u x − u y )with the introduction of the (Ohmic) conductance C xy = C yx = 1 /R xy . Notice thatthe case λ yx = 1 reduces our unit to the classical single resistor of value R xy . Noticealso that currents are automatically zero along loops: i xx = 0 whenever x is a vertexwith a loop.We write z ∼ x for neighbouring vertices z and x in the graph. This includes x ∼ x for vertices x with a loop. For later use we introduce(2.2) γ xy : = p λ xy = 1 γ yx , D xy : = 2 γ xy λ xy C xy = D yx , D x : = X z ∼ x D xz γ zx . The symmetry of D ·· follows from that of C ·· and log-antisymmetry of γ ·· and λ ·· .With these quantities we rewrite the above as(2.3) i xy = D xy · ( γ yx u x − γ xy u y ) . We emphasise that the voltage amplifier is not a natural object. Sophisticatedengineering would be required to build a black box with this characteristics, and thisblack box would require an outer energy source (or energy absorber) for its operation.We do not consider this energy source (or absorber) as part of our network. .1 Two alternative parts Two alternative units will facilitate calculations in our networks. Using these is notrequired for any of the later arguments, but simplifies matters. Consider R xy / R xy / ∗ λ yx i xy u x u y R pr yx ∗ λ pr yx i xy u x u y R se yx ∗ λ se yx i xy u x u y which are the original unit, the primer unit and the secunder unit, respectively. Theprimer and secunder units are built of the same types of elements as before. Repeatingthe arguments we see for the latter two cases u y = ( u x − i xy R pr yx ) · λ pr yx and u y = u x λ se yx − i xy R se yx . Comparing this with (2.1) we conclude that these three units behave in a completelyidentical way under the choices(2.4) λ pr yx = λ se yx = λ yx , R pr yx = R xy λ yx + 12 λ yx , R se yx = R xy λ yx + 12which we will assume whenever we write the pr or se indexed quantities. Notice thatthe primer and secunder resistances are not symmetric quantities anymore. An electric network for our purposes consists of our units placed along the edges of afinite, connected graph G = ( V, E ). We allow G to have loops as well. Suppose that asubset W of the vertices is taken to fixed potentials, U x , x ∈ W . The only requirementwe make is that W is non-empty. We show below that there exists a unique solutionof the network with these boundary values that is, a unique set of currents i ·· with(2.5) X y ∼ x i xy = 0 ∀ x / ∈ W, and voltages u x with u x = U x ∀ x ∈ W that satisfy (2.1) for all x ∼ y . We start with uniqueness. Proposition 2.1.
Given the graph G and the boundary set W , fix the boundary con-dition ( U x ) x ∈ W , and suppose that we have two solutions u ′· , u · and i ′·· , i ·· with thisboundary condition. Then u ′ ≡ u and i ′ ≡ i .Proof. As (2.1) is linear, the difference of two solutions is yet another solution. There-fore u ′· − u · and i ′·· − i ·· is another solution with boundary condition 0 for all x ∈ W .Define now the set ⊕ ⊂ V − W of vertices where u ′ − u is positive. If this set isnonempty, then any edge that connects it with the rest of V sees an outflowing cur-rent by (2.1). But this contradicts (2.5) (summed up for x ∈ ⊕ ). A similar argumentshows that there are no vertices of negative potential either, thus u ′ − u ≡
0. Then by(2.1) it follows that i ′ − i ≡ e proceed by existence of solutions. Call the incoming current to a vertex x ∈ Wi x :(2.6) i x : = X y ∼ x i xy , this is zero for all x / ∈ W . Lemma 2.2.
Given the graph G and the boundary set W , suppose we have a solutionfor all boundary conditions ( U x ) x ∈ W . Then for any x ∈ W , i x is an affine increasingfunction of U x when keeping all other boundary voltages U y , x = y ∈ W constant.Proof. Fix x ∈ W , consider U ′ x > U x , U ′ y = U y for all x = y ∈ W and the corre-sponding solutions u ′· , u · and i ′·· , i ·· . Then u ′· − u · and i ′·· − i ·· is another solution withboundary condition U ′ x − U x > x , and 0 for all x = y ∈ W . Again looking at thecurrent out of the set ⊕ of vertices with positive potential it is clear that the incomingcurrent i ′ x − i x is strictly positive in this setting. Now, multiplying all of u ′ − u , i ′ − i by any factor β is yet another solution. It follows that i ′ x − i x is a positive constantmultiple of U ′ x − U x and the proof is complete. Proposition 2.3.
There is a solution for any set ∅ 6 = W ∈ V and boundary condition ( U x ) x ∈ W .Proof. We will call the set V − W the free vertices, and perform an induction on itssize n = | V − W | . When n = 0, all potentials are fixed, and the currents are simplycomputed by (2.1). Suppose now that the statement is true for n , and consider a set W with | V − W | = n + 1. Pick any vertex x ∈ | V − W | . Fixing all boundary values U y for y ∈ W and also the value U x , we only have n free vertices, and we know by theinduction hypothesis that we have a solution. We also know by the above lemma thatthe incoming current i x is an affine function of U x . Therefore there exists a particularvalue U x with the corresponding incoming current i x = 0, and a solution u · , i ·· thatgoes with the boundary condition U x for x and U y for y ∈ W . This will be a solutionwith boundary condition U y for y ∈ W only, and the induction step is complete. In this section we make a connection of electric networks, built of our units, to Markovchains. The novelty is that the chain does not need to be reversible. For the followingproposition two non-intersecting subsets, A and B of the vertex set V are supposedto be connected to constant external potentials U A and U B : u a ≡ U A ∀ a ∈ A and u b ≡ U B ∀ b ∈ B. All other vertices are free: they just connect to neighbouring ones via our units. Thestarting point is
Proposition 3.1.
For every x / ∈ A ∪ B , we have (3.1) u x = X y ∼ x u y D xy γ xy D x . Proof.
Consider a vertex x / ∈ A ∪ B , and its neighbouring vertices y ∼ x , x = y . Wedemonstrate the situation for two neighbours with the picture xy / R xy / ∗ λ xy R xy ′ / R xy ′ / ∗ λ xy ′ u x u y u y ′ R se xy ∗ λ xy R se xy ′ ∗ λ xy ′ u x u y u y ′ where the second line shows an equivalent rewriting of the original setting. However,with this secunder representation our formula follows easily after realising that • the potentials on the x -side of the amplifiers are λ xy · u y for the respectivevertices y ; • from here the potential u x is computed using the well-known formula for avoltage divider (of conductances C se xy = 1 /R se xy ).Putting all that in formulas, we have u x = X y ∼ xy = x λ xy u y · C se xy P z ∼ xz = x C se xz = X y ∼ xy = x u y C xy λ xy λ xy +1 P z ∼ xz = x C xz λ xz +1 = P y ∼ x u y D xy γ xy − u x D xx D x − D xx with the use of (2.4), (2.2) and λ xx = 1. Rearranging finishes the proof.Let P ·· be the transition probabilities of an irreducible Markov chain on the finite,connected graph G . Throughout this manuscript we assume that P xy > x, y ) ∈ E is an edge of the graph, totally asymmetric steps are not handled by ourmethods (although we suspect that a meaningful limit could be worked out for thesecases). As usual, the graph has a loop on vertex x whenever P xx >
0. We now give arecipe of how to build an electric network for this chain so that the resulting voltagesand currents have the classical probabilistic interpretations (see e.g., Doyle-Snell [5]).This will be done regardless of whether the chain is reversible or irreversible. Theunique stationary distribution of the chain will be called µ · , and we now make thefollowing choices:(3.2) D xy : = p µ x · P xy · µ y · P yx ; γ xy : = s µ x · P xy µ y · P yx . Notice first that these choices are consistent with the respective symmetry and log-antisymmetry of D ·· and γ ·· . It is also clear that the conductances C ·· and the ampli-fying factors λ ·· can also be expressed with the help of the above quantities. Followingthe definition (2.2), we have(3.3) D x = X z ∼ x D xz γ zx = X z ∼ x µ z P zx = µ x . Recall the non-intersecting subsets A and B of the vertex set V , and define, for x ∈ V ,the first reaching times(3.4) τ A : = inf { t ≥ X ( t ) ∈ A } and similarly τ B of these sets by the Markov chain started from x . When x ∈ A ( B ),we define τ A ( τ B , respectively) to be 0. P x will stand for the probabilities associatedwith the chain started from x . For short, we will set h x : = P x { τ A < τ B } . heorem 3.2. Set up the electric network with the choices (3.2) , apply constantpotentials U A ≡ on vertices of the set A , U B ≡ on vertices of the set B , and makeno external connections to vertices of V − A − B . Then for every x ∈ V we have u x = h x .Proof. By definition we have h x = 1 for vertices x ∈ A , h x = 0 for vertices x ∈ B , andby a first step analysis of the Markov chain, h x = X y ∼ x h y P xy when x ∈ V − A − B . Next we find that by definition we have u x = 1 for vertices x ∈ A , u x = 0 for vertices x ∈ B , and by (3.1), (3.2) and (3.3), u x = X y ∼ x u y D xy γ xy D x = X y ∼ x u y P xy when x ∈ V − A − B . Thus h · and u · satisfy the same (well defined) equations withthe same boundary conditions, therefore they agree on all vertices.A nice consequence of the analogy is what happens to our electric network whenwe reverse our Markov chain. The reversed Markov chain has the same stationarydistribution µ x as the original one, and its transition probabilities becomeˆ P xy = P yx · µ y µ x . We will simply call the network that corresponds the reversed chain the reversednetwork , and its parameters will be marked by hats. They are(3.5) ˆ D xy = q µ x · ˆ P xy · µ y · ˆ P yx = p µ y · P yx · µ x · P xy = D xy ;ˆ γ xy = s µ x · ˆ P xy µ y · ˆ P yx = s µ y · P yx µ x · P xy = γ yx = 1 γ xy . This also implies ˆ C xy = C xy and ˆ λ xy = λ yx = 1 /λ xy for the reversed network, inother words reversing the Markov chain simply reverses the direction of our voltageamplifiers while keeps the resistance values intact . A Markov chain is reversible ifand only if the corresponding network has all its amplifiers with λ xy ≡
1. Indeed anamplifier of parameter 1 is just a plain wire, therefore this case reduces to the classicalreversible setting with ordinary resistors on the edges.
Notice that P ·· being a Markov transition probability imposes restrictions on our elec-tric network. From now on, P z ∼ x ∈ V will be our notation for double summation on allneighbouring vertices x and z in V . Theorem 3.3.
Suppose that we are given an electric network built of our componentson the edges of the finite connected graph G = ( V, E ) . There is an irreducible Markovchain of graph G and transition probabilities P ·· such that (3.2) holds if and only if wehave both X z ∼ x D xz γ xz = X z ∼ x D xz γ zx ( ∀ x ∈ V ) , and (3.6) X z ∼ x ∈ V D xz γ zx = 1 . (3.7) n this case (3.8) µ x = X z ∼ x D xz γ zx and P xy = D xy γ xy P z ∼ x D xz γ zx . Proof.
If (3.2) holds for a Markov transition probability P ·· , then the above formulasfollow from direct verification. Conversely, if (3.6) and (3.7) hold then we make thedefinition P xy = D xy γ xy P z ∼ x D xz γ zx , and notice that X y ∼ x P xy = X y ∼ x D xy γ xy P z ∼ x D xz γ zx = 1 ( ∀ x ∈ V ) , which shows that P ·· is a Markov transition probability matrix. It is irreducible bypositivity of our parameters, and its stationary distribution µ · is the unique vectorwith µ y = X x ∼ y µ x P xy = X x ∼ y µ x D xy γ xy P z ∼ x D xz γ zx ( ∀ y ∈ V ) and X x ∈ V µ x = 1 . Notice, however, that the vector (cid:0) P z ∼ x D xz γ zx (cid:1) x ∈ V satisfies the same properties: X x ∼ y X z ∼ x D xz γ zx P xy = X x ∼ y X z ∼ x D xz γ zx D xy γ xy P w ∼ x D xw γ wx = X x ∼ y D xy γ xy = X x ∼ y D yx γ xy by the symmetry of D ·· , and X z ∼ x ∈ V D xz γ zx = 1by (3.7). Therefore these two vectors agree. Remark 3.4.
The normalisation (3.7) is just an artificial choice, and is not essentialat all. Given an electric network, multiplying every resistor value by the same constant K while keeping the amplifiers unchanged will result in the same voltages everywherewith currents multiplied by /K . In particular, Theorem 3.2 holds true in this case. Remark 3.5.
The condition (3.6) is, on the other hand, very essential, and we willrefer to networks with this property as
Markovian . On a technical level it states thatwe can extend the definition (3.3) of D · by D x = X z ∼ x D xz γ zx = X z ∼ x D xz γ xz . Notice also that this implies (3.9) D x = X z ∼ x D xz γ zx + γ xz X z ∼ x C xz . The Markovian property also has a rather intuitive meaning: considering (3.1) it statesthat the constant potential u x ≡ U for all vertices is a valid solution of the (free)network. his was, of course, trivially true for the all-resistors networks that correspond toreversible Markov chains. Consider a set of connected resistors, and apply potential U on one of the vertices. Then all vertices will stay at potential U , with no currentflowing anywhere in the network. This is not at all straightforward with our generalisednetworks of resistors and amplifiers. Applying potential U on one of the vertices, theamplifiers will change voltages for different parts of the network, and this can keep upcurrents in the cycles of the graph G . The Markovian property is that, nevertheless,each vertex will still stay at the same potential U even if circular (that is, divergencefree) currents flow in the system.A classical result for Markov chains follows easily from the analogy. Corollary 3.6.
A Markov chain is reversible if and only if for every closed cycle x , x , x , . . . , x n = x in the graph G we have P x x · P x x · · · P x n − x = P x x n − · P x n − x n − · · · P x x . In particular, any Markov chain on a finite connected tree G is necessarily reversible.Proof. Rewriting the above formula and using (3.8) together with the symmetry of D ·· , we arrive to the equivalent statement γ x x · γ x x · · · γ x n − x = γ x x n − · γ x n − x n − · · · γ x x , or λ x x · λ x x · · · λ x n − x = 1 . This is of course trivially true in the reversible case where all of the amplifiers have λ xy = 1. For the other direction, assume now the above formula to hold, and turn itinto electrical language. It says that the total multiplication factor of the potentials isone along any closed cycle of the circuit. It follows that fixing one vertex at potential U , zero currents everywhere in the network is a solution. By uniqueness this is theonly solution. The network being Markovian on the other hand tells us that everyvertex has to be on potential U . With no currents the only way this can happen isthat all of the amplifiers have parameter one, and the chain is reversible.A similar argument works directly for the tree: since there are no cycles, no currentcan flow if only one vertex is fixed at potential U . The Markovian property again tellsus that every vertex will be on potential U which again means parameter one for allof the amplifiers and thus reversibility of the chain. In this section we make sense of effective resistance in our network, and give it aprobabilistic interpretation similar to that of the classical case. The setting is the oneof Proposition 3.1: two disjoint subsets A and B of the vertices are forced to be onconstant potentials U A and U B , respectively. Define the total incoming current to theset A (c.f. (2.6)) as(3.10) i A : = X x ∈ A i x = X y ∼ x ∈ A i xy . Notice that by conservation of currents, this agrees to the sum of currents of edgesacross the boundary of A , and also i A + i B = 0. The existence of the effective resistancebetween sets A and B means that the network between these sets can be replaced bya single resistor. This is not true for arbitrary configurations, since the amplifiers ingeneral push the characteristics away from that of a single resistor. It is, however, truefor networks that match a Markov chain, this is formulated in the next theorem: Theorem 3.7.
In a Markovian electric network, for any disjoint
A, B ⊂ V there isa constant R eff AB > such that U A − U B = R eff AB · i A ( ∀ U A , U B ∈ R ) . roof. The proof will again proceed along the lines of linearity. When U A = U B thenwe just have the Markovian solution with zero incoming currents, thus i A = 0 andeverything is trivial. Suppose that we are given arbitrary reals U A = U B , U ′ A = U ′ B . • We consider two solutions of our network: the one u · , i ·· that satisfies the givenboundary conditions u x ≡ U A for x ∈ A and u x ≡ U B for x ∈ B , and onethat comes from the Markovian property: u M x ≡ U B , i M x ≡ x , not to be mixed with currents i M xy of edges!) for all x ∈ V . Wethink of this latter one as a solution with boundary conditions u M x ≡ U B for all x ∈ A ∪ B . • The difference u − u M and i − i M of these two is yet another solution due tolinearity. It has boundary conditions u x − u Mx ≡ U A − U B on x ∈ A , u x − u Mx ≡ U B − U B = 0 on x ∈ B , and notice that the incoming current to the set A isstill i A − i M A = i A − i A . • Again by linearity every current and potential can be multiplied by the factor U ′ A − U ′ B U A − U B and we still have a valid system. This looks like( u · − u M · ) · U ′ A − U ′ B U A − U B , ( i ·· − i M ·· ) · U ′ A − U ′ B U A − U B and therefore has boundary conditions( U A − U B ) · U ′ A − U ′ B U A − U B = U ′ A − U ′ B on x ∈ A, and0 · U ′ A − U ′ B U A − U B = 0 on x ∈ B, with incoming current i A · U ′ A − U ′ B U A − U B on the set A . • Finally, add the Markovian solution with constant potential u M · ′ ≡ U ′ B every-where, and i M ·· ′ with zero incoming currents in all vertices. This results in( u · − u M · ) · U ′ A − U ′ B U A − U B + U ′ B , ( i ·· − i M ·· ) · U ′ A − U ′ B U A − U B + i M ′ and therefore has boundary conditions U ′ A − U ′ B + U ′ B = U ′ A on x ∈ A, and0 + U ′ B = U ′ B on x ∈ B, with incoming current i A · U ′ A − U ′ B U A − U B + 0 = i A · U ′ A − U ′ B U A − U B on the set A .We have thus produced the solution for the boundary conditions U ′ A and U ′ B , andconcluded that the corresponding incoming current to the set A is i ′ A = i A · U ′ A − U ′ B U A − U B . This is equivalent to the statement of the theorem, i.e., the ratio ( U A − U B ) /i A is aconstant for all boundary potentials U A and U B .A Markovian network has a further peculiar property: the effective resistance R eff AB stays the same if we reverse each of the amplifiers. Recall that the networkthen turns into that of the reversed Markov chain. To prove this property, we followSlowik’s argument [12], and first define the one-step Markov generator L on functions f : V → R : ( Lf ) x = X y ∼ x P xy ( f y − f x ) . ewriting this via (3.8) into electrical terms and then applying (3.6) we get − ( Lf ) x = X y ∼ x D xy γ xy D x ( f x − f y ) = 1 D x X y ∼ x D xy · ( γ yx f x − γ xy f y ) . This latter formula is meaningful in electrical terms, as soon as we imagine f · as apotential applied on vertices of the graph, and define the resulting currents(3.11) i fxy : = D xy · ( γ yx f x − γ xy f y )via (2.3). We thus see, c.f. (2.6), that − ( Lf ) x = i fx D x .i fx is the current we are required to pump in vertex x in order to maintain potential f x . The quantity E ( f ) : = X x ∈ V µ x f x · ( − Lf ) x = X x f x · i fx is referred to as the energy associated to the pair P ·· , µ · , and we now see that it is thetotal electric power we need to pump in the system in order to maintain potential f x at each vertex x . (As usual, we do not count the external energy sources (absorbers)required by the amplifiers to work.)With this preparation we now prove Proposition 3.8.
Reversing a Markovian network does not affect the effective resis-tance: [ R eff AB = R eff AB . Proof.
We repeat Slowik’s arguments [12] in the electrical language. Take two func-tions f and g on V , and apply (3.6) in the first term and symmetry of the doublesummation and of D ·· in the second term below:(3.12) X x f x · i gx = X x ∈ V f x X y ∼ x D xy · ( γ yx g x − γ xy g y )= X x ∈ V f x g x X y ∼ x D xy γ yx − X y ∼ x ∈ V f x D xy γ xy g y = X x ∈ V f x g x X y ∼ x D xy γ xy − X y ∼ x ∈ V g x D xy γ yx f y = X x ∈ V g x X y ∼ x D xy · (ˆ γ yx f x − ˆ γ xy f y ) = X x g x · ˆ i fx . (This equation is the electrical way of saying that the adjoint of the generator is theone of the reversed process.) As before, fix the boundary conditions u x ≡ ≡ ˆ u x on x ∈ A and u x ≡ ≡ ˆ u x on x ∈ B for two scenarios: u · , i ·· of the original network andˆ u · , ˆ i ·· of the reversed one. This latter has all its amplifiers reversed, and it correspondsto the reversed Markov chain. We claim that in our situation i ux ≡ ≡ ˆ i ˆ ux on x / ∈ A ∪ B ,since these are free vertices. This, together with the common boundary condition forthe two networks implies E ( u ) = X x ∈ V u x · i ux = X x ∈ A u x · i ux = X x ∈ A ˆ u x · i ux = X x ∈ V ˆ u x · i ux = X x ∈ V u x · ˆ i ˆ ux = X x ∈ A u x · ˆ i ˆ ux = X x ∈ A ˆ u x · ˆ i ˆ ux = X x ∈ V ˆ u x · ˆ i ˆ ux = ˆ E (ˆ u ) . Rewriting the power we apply to maintain our boundary conditions gives C eff AB = ( U A − U B ) · C eff AB = E ( u ) = ˆ E (ˆ u ) = ( U A − U B ) · d C eff AB = d C eff AB . ext we introduce what is called the capacity in the theory of Markov chains, andshow that it has close connections to the effective resistance. We again assume that A, B ⊂ V are non-empty, disjoint, and follow Gaudilli`ere-Landim and Slowik [7, 12]by defining τ A : = inf { t > X ( t ) ∈ A } (c.f. (3.4)) and cap( A, B ) : = X x ∈ A µ x P x { τ B < τ A } . Proposition 3.9.
The above capacity is simply the effective conductance C eff AB =1 /R eff AB between the sets A and B .Proof. We use the analogy set up in Theorem 3.2 ascap(
A, B ) = X x ∈ A µ x X y ∼ x P xy · P y { τ B < τ A } = X y ∼ x ∈ A µ x P xy · (1 − P y { τ A < τ B } )= X y ∼ x ∈ A D xy γ xy (1 − u y ) = X y ∼ x ∈ A D xy ( γ yx · − γ xy u y )= X y ∼ x ∈ A D xy ( γ yx u x − γ xy u y ) = X y ∼ x ∈ A i xy = i A = U A · C eff AB = C eff AB . Along the way we also used (3.6), the fact that u x ≡ U A = 1 for all x ∈ A , and finally(2.3).It follows immediately that the capacity is a symmetric quantity in its two argu-ments A and B . The identity cap( A, B ) = d cap( B, A ) also follows from the previousproposition.
Remark 3.10.
Gaudilli`ere-Landim and Slowik [7, 12] also establish cap(
A, B ) = 12 X x ∼ y ∈ V µ x P s xy ( h x − h y ) , with the symmetrised transitions P s xy = ( P xy + ˆ P xy ) . (3.8) together with (3.5) and (2.2) gives (3.13) P s xy = D xy γ xy + D xy γ yx D x = C xy D x , and the capacity gets another interesting interpretation: in the setting of Theorem 3.2, (3.14) cap( A, B ) = 12 X x ∼ y ∈ V C xy ( u x − u y ) , the ohmic power loss on the resistors, should we apply the actual voltages u · on themwithout the amplifiers . It is important to note that this interpretation is non-physical:with the amplifiers the ohmic losses are not given by the above formula, without theamplifiers the voltages u · would be totally different. The capacity, being C eff AB is, however, equal to the total power ( U A − U B ) · C eff AB =(1 − · C eff AB we need to pump in the set A to keep it on potential U A = 1. e repeat the computation for (3.14) in the electrical language. First notice thatby (3.11) and (2.2),12 ( i uxy + ˆ i uxy ) = D xy · (cid:16) γ yx + γ xy u x − γ xy + γ yx u y (cid:17) = C xy ( u x − u y ) , fixing the voltages everywhere (!), the average of the current and the reversed currentis the one of the network without amplifiers. This is the starting point to expand theright hand-side of (3.14):12 X x ∼ y ∈ V C xy ( u x − u y ) = 14 X x ∼ y ∈ V ( u x − u y )( i uxy + ˆ i uxy )= 14 X x ∼ y ∈ V u x i uxy + 14 X x ∼ y ∈ V u x ˆ i uxy − X x ∼ y ∈ V u y i uxy − X x ∼ y ∈ V u y ˆ i uxy = 14 X x ∼ y ∈ V u x i uxy + 14 X x ∼ y ∈ V u x ˆ i uxy + 14 X x ∼ y ∈ V u y i uyx + 14 X x ∼ y ∈ V u y ˆ i uyx = 14 X x ∈ V u x i ux + 14 X x ∈ V u x ˆ i ux + 14 X y ∈ V u y i uy + 14 X y ∈ V u y ˆ i uy = 14 X x ∈ V u x i ux + 14 X x ∈ V u x i ux + 14 X y ∈ V u y i uy + 14 X y ∈ V u y i uy = X x ∈ V u x i ux . with the use of the adjoint identity (3.12). Now, as in the proof of Proposition 3.8,apply our usual boundary conditions, and the right hand-side becomes the total powerrequired to maintain the boundary conditions or, equivalently, C eff AB .Finally, we define the escape probability from the set A and show its connectionto the effective resistance, this goes exactly as in the reversible case. Suppose that theMarkov chain is started from its stationary distribution µ , conditioned on being in theset A . (When A = a is a singleton, this is just the unit mass on the vertex a .) Theescape probability is the chance that the chain reaches set B before its first return to A : P { τ B < τ A } = X x ∈ A µ x µ ( A ) P x { τ B < τ A } = cap( A, B ) µ ( A ) = C eff AB P z ∈ A D z = C eff AB P y ∼ z ∈ A C zy . The last step used (3.9). The right hand-side agrees word for word with the classicalreversible result, the starting point of elegant recurrence-transience proofs.By symmetrising a Markov chain we mean replacing its transition probabilities by P s ·· . Corollary 3.11.
Symmetrising a Markov chain never increases the escape probabili-ties.Proof.
By symmetrising we keep the stationary distribution µ · = D · and the conduc-tances C ·· unchanged while the amplifiers all become trivial: λ ·· ≡
1. This can easilybe seen via (3.5) and (3.13). Denote the potentials that result our usual boundaryconditions U A ≡ U B ≡ u · in the original network and by u s · in the sym-metrised one. The classical Dirichlet principle for the reversible case tells us that u s · isthe potential that minimises the ohmic power losses in the resistors for the reversiblechain. Therefore 12 X x ∼ y ∈ V C xy ( u s x − u s y ) ≤ X x ∼ y ∈ V C xy ( u x − u y ) . ince the conductances agree for the two networks, the left hand-side is the capacityof the symmetrised network, while the right hand-side is the one of the original net-work. Dominance of the capacities then implies that of the escape probabilities as theconductances are not changed by symmetrising. In this section we give a probabilistic interpretation of the currents i ·· of the network.We take a singleton set A = { a } and another, arbitrary set B a . Start the Markovchain from a , and define v x as the expected number of visits to vertex x before thefirst hitting of the set B . Clearly v x ≡ x ∈ B . A last step analysis shows thatfor any a = x / ∈ B we have v x = X y ∼ x v y P yx = X y ∼ x v y ˆ P xy · µ x µ y , in other words v x /µ x is harmonic w.r.t. ˆ P . The boundary conditions are v a /µ a is fixed(to be determined later), and v x /µ x ≡ x ∈ B . Consider now the correspondingelectric network of parameters C ·· and ˆ λ ·· with these boundary conditions. It followsthat its potentials ˆ u x agree with v x /µ x for all x ∈ V , and its currents are (see (2.3))ˆ i xy = D xy · (cid:0) ˆ γ yx · v x µ x − ˆ γ xy · v y µ y (cid:1) = D xy · (cid:0) γ xy · v x µ x − γ yx · v y µ y (cid:1) = v x P xy − v y P xy . This latter is the expected number of jumps from x to y minus that from y to x beforeabsorption in B . It remains to fix the boundary term ˆ u a = v a /µ a . This is done bythe simple observation that the chain has to exit vertex a one more times than enterit, thus ˆ i a = X y ∼ a ˆ i ay = 1 . This fixes a unique potential ˆ u a = v a /µ a for boundary. Everything is analogue to theclassical reversible case, except that we need to use the reversed network. We have found that a classical result of Chandra, Raghavan, Ruzzo, Smolensky andTiwari [2] on effective resistance and commute times (or costs) can be extended easilyto the irreversible case. Work on commute times in this case has also been done byDoyle and Steiner [6]. Fix two vertices a = b of the graph and a cost function k xy onedges of the graph. Costs k xy and k yx can be different, we do not require any relationbetween these two. The expected cost of the chain from a to b is the expected priceto pay until the first hitting of the chain to vertex b if started from a . It is defined as H kab : = E a τ b X t =1 k X ( t − X ( t ) , where we (ab)used definition (3.4) (by writing τ b for τ { b } ). We consider an empty sumto be zero for the case H kaa = 0. In particular, for k ≡ H ab = E a ( τ b ). Define the expected commute cost K kab = H kab + H kba , thisbecomes the expected commute time for k ≡ Theorem 3.12.
The expected commute cost can be computed by K kab = R eff ab · D k , with D k : = P y ∼ x ∈ V D xy γ xy k xy . e spell out the case k ≡
1: the expected commute time is K ab = R eff ab · D = R eff ab · X y ∼ x ∈ V D xy γ xy = R eff ab · X x ∈ V D x = R eff ab · X z ∼ x ∈ V C xz via (3.9). This is the exact same formula as the one of [2] for the reversible case. Proof.
We start with a first step analysis and write, for any x = b ,(3.15) H kxb = X y ∼ x P xy ( k xy + H kyb ) = X y ∼ x D xy γ xy D x ( k xy + H kyb ) . In our electric network, impose boundary conditions u x = H kxb on each vertex x . Thisresults in currents i xy = D xy · ( γ yx H kxb − γ xy H kyb )according to (2.3), and, via (3.15), the necessity of pumping external currents i x = X y ∼ x i xy = X y ∼ x D xy γ yx H kxb − X y ∼ x D xy γ xy H kyb = D x H kxb − D x H kxb + X y ∼ x D xy γ xy k xy = X y ∼ x D xy γ xy k xy into each vertex x = b . By conservation of current, i b = X y ∼ b D by γ by k by − D k with D k : = P y ∼ x ∈ V D xy γ xy k xy .A second configuration we consider is u ′ x = H kxa on each vertex x , in a similarfashion this has external currents i ′ x = X y ∼ x D xy γ xy k xy for all x = a , and i ′ a = X y ∼ a D ay γ ay k ay − D k . Our equations being linear, the difference u − u ′ is also a solution of the network. Ithas potentials and external currents u a − u ′ a = H kab − H kaa = H kab and i a − i ′ a = X y ∼ a D ay γ ay k ay − X y ∼ a D ay γ ay k ay + D k = D k in vertex a,u b − u ′ b = H kbb − H kba = − H kba and i b − i ′ b = X y ∼ b D by γ by k by − D k − X y ∼ b D by γ by k by = − D k in vertex b,i x − i ′ x = X y ∼ x D xy γ xy k xy − X y ∼ x D xy γ xy k xy = 0 elsewhere . Therefore this combination only has boundary conditions at a and b , all other verticesare free. The effective conductance between a and b is given by C eff ab = i a − i ′ a u a − u ′ a − u b + u ′ b = D k H kab + H kba = D k K kab , which completes the proof. .6 A non-monotone example We have seen many nice properties of the network. The next step in the reversible caseis making use of Rayleigh’s monotonicity property: in the reversible case the effectiveresistance is a non-decreasing function of any of the individual resistances. Here weshow an example to demonstrate that this is not the case in the irreversible case, thenaive approach does not work. Resistance values below are in Ohms. / / ∗ / / / ∗ / / / ∗ / / ∗ / R a b We immediately rewrite this network to an equivalent form using the primer andsecunder alternatives: / ∗ / / ∗ / ∗ / ∗ / R a b First notice that a circular current of 4 Amperes in the positive direction and no currentthrough R gives a constant 9 Volts free solution, thus the network is Markovian for all R values. Therefore it has an effective resistance, and it is perhaps easiest to computeif we fix u a = 5 Volts and u b = 0. Then we just need to figure out currents in /
59 18 / / R xy One way of proceeding is to write the equations for the voltage dividers in x and y .These are: u x = 1 · + 0 · + u y · R + + R = 10 R R + 18 + 65 R + 6 · u y ,u y = 25 · + 0 · + u x · R + + R = 50 R R + 18 + 65 R + 6 · u x . The solution is u x = R +7215 R +36 and u y = R +7215 R +36 , and the effective resistance is R eff ab = u a i a = u a − i b = u a u x + u y = 675 R + 1620350 R + 648 = 2714 + 12961225 R + 2268 , decreasing function of R . The situation reminds the authors to Braess’s paradox [1]. Having the direct monotonicity approach failed, we now give an insight to the Dirich-let and Thomson energy minimum principles for the irreversible case. These are thefundamental principles that enable one to derive Rayleigh’s monotonicity law in the re-versible case. The irreversible case was established in this form by Gaudilli`ere-Landimand Slowik [7, 12], below we simply give a translation of their results without proofinto the electrical language. In fact we stick to the notation of Slowik’s Proposition2.6 as closely as possible.As before, take two sets A ∩ B = ∅ of vertices, and define H A,B : = { u : V → R : u | A ≡ , u | B ≡ } , we think of such functions as voltages with respective boundary conditions on A and B . Set U A,B as the set of currents i ·· with zero external currents i x for x / ∈ ( A ∪ B )(c.f. (2.6)), and total incoming current i A = 0 = i B to the set A (and therefore to theset B as well), see (3.10). The next quantity to define is, for currents i ·· , D ( i ) = 12 X y ∼ x ∈ V µ x P s xy i xy = 12 X y ∼ x ∈ V R xy i xy , the ohmic power losses on the resistors, see (3.13). Finally, recall (3.11), and reversethe amplifiers in there to get ˆ i · . We can now state Proposition 3.13 (Dirichlet principle, Slowik [12] Proposition 2.6) . cap( A, B ) = min u ∈H A,B min i ∈ U A,B D (ˆ i u − i ) . The minimum is attained for u = ( u AB + ˆ u AB ) and i = ˆ i u − i s u AB , where u AB and ˆ u AB are the physical potentials in the network and in the reversed network under ourboundary conditions, respectively, and i s u AB is the current that would result under thepotential ( u AB ) · without the amplifiers. In words: find a potential function u with our boundary conditions (this resultsin currents ˆ i u in the reversed network) and a divergence free current i on the freevertices with total incoming flow i A = 0 such that the difference of these two currentsminimises the ohmic losses on the network. Then these ohmic losses sum up to thetotal physical power required to maintain the boundary conditions (this is the effectiveconductance C eff AB = cap( A, B ) since the boundary voltage difference U A − U B = 1).We emphasise that the minimisers u and i are non-physical, except in the reversiblecase when u becomes the physical voltage while i ≡ G A,B : = { u : V → R : u | A ≡ u | B ≡ } , and U A,B as the set of currents i ·· with zero external currents i x for x / ∈ ( A ∪ B ) (c.f.(2.6)), and total incoming current i A = 1 = − i B to the set A (and therefore -1 to theset B ), see (3.10). Proposition 3.14 (Thomson principle, Slowik [12] Proposition 2.6) . cap( A, B ) = max i ∈ U A,B max u ∈ G A,B D ( i − ˆ i u ) . The maximum is attained for u = ( u AB − ˆ u AB ) / cap( A, B ) , and i = i s u AB + ˆ i u ,where i s u AB is the current that would result under the potential ( u AB ) · without theamplifiers, except that it is normalised to have unit total inflow in A . n words: find a potential function u that vanishes in A and B (this results incurrents ˆ i u in the reversed network), and a unit flow i such that the difference of thesecurrents minimises the ohmic losses on the network. Then these ohmic losses sum upto the total physical power required to maintain a unit flow (the reciprocal is againthe effective conductance since the total current flow is one). Again, the minimisersare non-physical, except for the reversible case when u ≡ i is the physical unitcurrent flow.How these principles can be used towards monotonicity is a question left for futurework. In this last section we dive a little bit into “network-algebra” by showing how units inseries, parallel, star or delta configurations behave. Series and parallel are simple andnice situations:
Proposition 4.1.
Two of our units in series of respective parameters ( R, λ ) and ( Q, µ ) can be replaced by a single unit of parameters (cid:16) R ( λ + 1) µλµ + 1 + Q µ + 1 λµ + 1 , λµ (cid:17) . Proof.
We use the primer and secunder alternatives as R/ R/ ∗ λ Q/ Q/ ∗ µR pr ∗ λ Q se ∗ µR pr Q se ∗ λµR pr Q ′ pr ∗ λµS pr ∗ λµS/ S/ ∗ λµ It is obvious that the parameter of the voltage amplifier is λµ . Applying the transcrip-tion formula for each step, the resistance of the substitute element can be determined: S = S pr λµλµ + 1 = ( R pr + Q ′ pr ) 2 λµλµ + 1 = R pr λµλµ + 1 + Q se λµ + 1= R ( λ + 1) µλµ + 1 + Q µ + 1 λµ + 1 . Proposition 4.2.
Two of our units in parallel of respective parameters ( R, λ ) and ( Q, µ ) can be replaced by a single unit of parameters (cid:16) RQR + Q , Q ( µ + 1) Q ( µ + 1) + R ( λ + 1) · λ + R ( λ + 1) Q ( µ + 1) + R ( λ + 1) · µ (cid:17) . otice the classical parallel formula for the resistance, and the weighted averagefor the amplifier. Proof.
This case cannot be reduced with transformations into one single unit, but thealternative elements are still useful. Below are two equivalent circuits. Q/ Q/ ∗ µR/ R/ ∗ λu x u y Q se ∗ µ R se ∗ λu x u y The total current from x to y will be the sum of currents like in (2.1) for the top andthe bottom branches, therefore it will be in the same form. This proves that a singleunit can be used as a replacement. Its secunder alternative will be S se ∗ νu x u y It remains to determine the parameters S and ν . Assume first u x = 1 and zero totalcurrent that is, leave vertex y free. Then the secunder resistors act as a voltage divider,giving u y = λQ se + µR se Q se + R se . In the simple unit this agrees to the value ν of the amplifier, therefore ν = Q se λ + R se µQ se + R se = Qλ ( µ + 1) + Rµ ( λ + 1) Q ( µ + 1) + R ( λ + 1) . Next, when u x = 0, the amplifiers keep the potentials at zero, and the parallel formula S se = R se Q se R se + Q se follows. Returning to the original alternative, S = 2 ν + 1 · S se = 2 Q ( µ + 1) + R ( λ + 1)( R + Q )( λ + 1)( µ + 1) · R se Q se R se + Q se = RQR + Q .
Notice that in both the series or parallel formulas the resulting resistances aremonotone increasing functions of the original ones. Not all networks can, however, bereduced using only series or parallel substitutions. The next step of transformationsfor classical resistor networks is the star-delta transformation. As we will see shortlyit is here where non-monotonicity issues begin. Our non-monotone example is alsoone that cannot be reduced using only series or parallel substitutions.In our case, star and delta look like / S / ∗ ν R / R / ∗ λ Q / Q / ∗ µ U x U y U z S ′ / S ′ / ∗ ν ′ R ′ / R ′ / ∗ λ ′ Q ′ / Q ′ / ∗ µ ′ U x U y U z where it is essential that the centre of star has no further connections. The questionis whether the parameters can be linked so that these two networks behave identicallyunder all scenarios. We start by rewriting the above into the equivalent secunderalternatives, and work with those thereafter. Any formulas can be rewritten into theoriginal parameters via (2.4), we avoid that for the sake of simplicity (well...). S s e ∗ ν R s e ∗ λ Q s e ∗ µ U x U y U z S ′ se ∗ ν ′ R ′ s e ∗ λ ′ Q ′ s e ∗ µ ′ U x U y U z With the notations of this picture, we have
Proposition 4.3.
Any star can be transformed into an equivalent delta, the parame-ters of which are given by S ′ se = R se S se + Q se S se + Q se R se λS se ,Q ′ se = R se S se + Q se S se + Q se R se νQ se ,R ′ se = R se S se + Q se S se + Q se R se µR se , and λ ′ = νµ , ν ′ = µλ , µ ′ = λν . Not every delta, however, can be transformed into a star.
Proposition 4.4.
A delta can be transformed into an equivalent star if and only if (4.1) λ ′ · ν ′ · µ ′ = 1 . ven in this case the resulting star is not unique: with any positive number α > , itcan have parameters S se = λ ′ / ν ′ / αR ′ se Q ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se ,Q se = ν ′ / µ ′ / αR ′ se S ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se ,R se = λ ′ / µ ′ / αQ ′ se S ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se , and λ = λ ′ / µ ′ / α , ν = λ ′ / ν ′ / α , µ = ν ′ / µ ′ / α .Proof of both star → delta and delta → star. We determine and compare the incomingcurrents on vertices x, y and z in the two networks. We start with star. The voltagesat the outer points of the resistances are µU x , λU y and νU z , thus the voltage in thecentre point is U = µU x Q se + λU y R se + νU z S se Q se + R se + S se = µU x R se S se + λU y Q se S se + νU z Q se R se R se S se + Q se S se + Q se R se . Therefore, the respective currents flowing from x , y and z into the centre point are i x = µU x − UQ se = µ ( S se + R se ) U x − λS se U y − νR se U z R se S se + Q se S se + Q se R se ,i y = λU y − UR se = λ ( S se + Q se ) U y − µS se U x − νQ se U z R se S se + Q se S se + Q se R se ,i z = νU z − US se = ν ( R se + Q se ) U z − µR se U x − λQ se U y R se S se + Q se S se + Q se R se . Next we turn to delta. The currents flowing on the edges are: i xy = ν ′ U x − U y S ′ se , i yz = µ ′ U y − U z Q ′ se , i zx = λ ′ U z − U x R ′ se . Thus, the currents flowing from x , y and z into the network can be written as i x = i xy − i zx = ( ν ′ R ′ se + S ′ se ) U x − R ′ se U y − λ ′ S ′ se U z R ′ se S ′ se ,i y = i yz − i xy = ( µ ′ S ′ se + Q ′ se ) U y − ν ′ Q ′ se U x − S ′ se U z Q ′ se S ′ se ,i z = i zx − i yz = ( λ ′ Q ′ se + R ′ se ) U z − Q ′ se U x − µ ′ R ′ se U y Q ′ se R ′ se . In a star / delta substitution the currents have to equal for all possible voltages U x , U y and U z . Hence, by comparing the coefficients of the voltages in the formulas for thecurrents, the connections between the quantities can be determined. It is subservientto consider first the coefficient of U y in the formula for i x , the coefficient of U z in theformula for i y and the coefficient of U x in the formula for i z :(4.2) 1 S ′ se = λS se R se S se + Q se S se + Q se R se , Q ′ se = νQ se R se S se + Q se S se + Q se R se , R ′ se = µR se R se S se + Q se S se + Q se R se . econd, consider the coefficient of U z in i x , the coefficient of U x in i y and the coefficientof U y in i z : λ ′ R ′ se = νR se R se S se + Q se S se + Q se R se ,ν ′ S ′ se = µS se R se S se + Q se S se + Q se R se ,µ ′ Q ′ se = λQ se R se S se + Q se S se + Q se R se . By dividing the corresponding equations in the two triplets of equations we get: λ ′ = νµ , ν ′ = µλ and µ ′ = λν . Finally, with these substitutions the coefficients of U x in i x , U y in i y and U z in i z alsomatch. This proves Proposition 4.3.Notice that for any star, in the substitute delta we have λ ′ · ν ′ · µ ′ = 1 and thatmultiplying the amplifiers in the star by a common constant does not change theparameters of the voltage amplifiers in the substitute delta. Therefore, the inversionof the previous formulas is possible only if (4.1) holds, and even in this case it isnot uniquely determined, a constant factor has to be chosen. This can be written as λνµ = α where α > λ = λ ′ / µ ′ / α, ν = λ ′ / ν ′ / α and µ = ν ′ / µ ′ / α. To invert the resistances (4.2) first note that1 µR ′ se · λS ′ se + 1 νQ ′ se · λS ′ se + 1 νQ ′ se · µR ′ se = 1 R se S se + Q se S se + Q se R se . Thus from (4.2) and from (4.1): S se = λS ′ se µR ′ se · λS ′ se + νQ ′ se · λS ′ se + νQ ′ se · µR ′ se = λ ′ / ν ′ / αR ′ se Q ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se ,Q se = νQ ′ se µR ′ se · λS ′ se + νQ ′ se · λS ′ se + νQ ′ se · µR ′ se = ν ′ / µ ′ / αR ′ se S ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se ,R se = µR ′ se µR ′ se · λS ′ se + νQ ′ se · λS ′ se + νQ ′ se · µR ′ se = λ ′ / µ ′ / αQ ′ se S ′ se λ ′ / ν ′ / Q ′ se + ν ′ / µ ′ / R ′ se + λ ′ / µ ′ / S ′ se . The condition (4.1) is that at any constant potential the delta has no circularcurrent by itself. This is rather restrictive, thus delta → star transformations cannotbe used to reduce a general network. After the lack of monotonicity, this is the secondserious drawback of our networks compared to the classical resistor-only case. cknowledgements The authors wish to thank Edward Crane, Nic Freeman, Alexandre Gaudilli`ere, Clau-dio Landim, G´abor Pete, Martin Slowik, Andr´as Telcs and B´alint T´oth for stimulatingdiscussions on this project.
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