Metric and ultrametric inequalities for resistances in directed graphs
aa r X i v : . [ m a t h . C O ] S e p Metric and ultrametric inequalities for resistancesin directed graphs.
Vladimir Gurvich ∗ October 1, 2020
Abstract
Consider an electrical circuit G each directed edge e of which is a semi-conductor with a monomial conductance function y ∗ e = f e ( y e ) = y se /µ re if y e ≥ y ∗ e = 0 if y e ≤
0. Here y e is the potential difference (voltage), y ∗ e is the current in e , and µ e is the resistance of e ; furthermore, r and s are two strictly positive real parameters common for all edges. In partic-ular, case r = s = 1 corresponds to the Ohm law, while r = , s = 1 maybe interpreted as the square law of resistance typical for hydraulics andgas dynamics.We will show that for every ordered pair of nodes a, b of the circuit, theeffective resistance µ a,b is well-defined. In other words, any two-pole net-work with poles a and b can be effectively replaced by two oppositelydirected edges, from a to b of resistance µ a,b and from b to a of resistance µ b,a .Furthermore, for every three nodes a, b, c the inequality µ s/ra,c + µ s/rc,b ≥ µ s/ra,b holds, in which the equality is achieved if and only if every directed pathfrom a to b contains c .Some limit values of parameters s and r correspond to classic triangleinequalities. Namely,(i) the length/time of a shortest directed path,(ii) the inverse width of a bottleneck path, and(iii) the inverse capacity (maximum flow per unit time)between any ordered pair of terminals a and b are assigned to:(i) r = s → ∞ , (ii) r = 1 , s → ∞ , (iii) r → , s = 1, respectively.These results generalize ones obtained in 1987 for the isotropic monomialcircuits, modelled by undirected graphs. In this special case resistancedistances form a metric space, while in general only a quasi-metric one:symmetry, µ a,b = µ b,a is lost.In linear symmetric case these results are known from 1960-s and weregeneralized to linear the non-symmetric case in 2016-th.MSC classes: 11J83, 90C25, 94C15, 94C99 ∗ National Research University Higher School of Economics (HSE) Moscow Russia;e-mail: [email protected] and [email protected] Introduction
Two-Pole Circuits
We consider a circuit modeled by a directed graph (digraph) G = ( V, E )in which each directed edge e ∈ E is a semiconductor with the monomialconductivity law y ∗ e = f e ( y e ) = y re /µ se if y e ≥ y ∗ e = 0 if y e ≤
0. Here y e is the voltage, or potentialdifference, y ∗ e ≥ µ e is the resistance of e , while r and s aretwo strictly positive real parameters, the same for all e ∈ E .In particular, the case r = 1 corresponds to Ohm’s law, while r = is the so-called square law of resistance typical for hydraulics and gasdynamics. In the first case y e is the drop of potential (voltage) and y ∗ e isthe current; in the second case y e is the drop of pressure and y ∗ e is the flow.Parameter s , in contrast to r, looks redundant, yet, it plays an importantrole helping to interpret some limit cases.Given a circuit G = ( V, E ), let us fix an ordered pair of nodes a, b ∈ V .We will show that the obtained two-pole circuit ( G, a, b ) satisfies the samemonomial conductivity law. Let y ∗ a,b denote the total current that comesfrom a into b and y a,b the drop of potential (voltage) between a and b . Itwill be shown that y ∗ a,b = f a,b ( y a,b ) = y ra,b /µ sa,b when y a,b ≥
0, and there exists a directed path from a to b in G . If there isno such path then y ∗ a,b = 0 for any y a,b ≥
0; in this case we set µ a,b = + ∞ .Also, by convention, we set y ∗ a,b = 0 when y a,b < a = b . In the lattercase y a,b = 0 always holds and we set µ a,b = 0, by convention.In other words, each two-pole circuit ( G, a, b ) can be effectively re-placed by two oppositely directed edges: from a to b of resistance µ a,b and from b to a of resistance µ b,a . Both numbers are 0 when a = b . Main inequality
For arbitrary three nodes a, b, c ∈ G , we will prove inequality µ s/ra,b ≤ µ s/ra,c + µ s/rc,b (1)Furthermore, the inequality in (1) is strict if and only if there exists adirected path from a to b that does not contain c .Obviously, equality holds if at least one of the three considered resis-tances equals 0 or + ∞ , that is, if at least two of the considered threenodes coincide, or at least one of three directed paths, from a to c , from c to b , or from a to b fails to exist. In the latter case, + ∞ = + ∞ byconvention.Clearly, if s ≥ r then (1) implies the standard metric inequality: µ a,b ≤ µ a,c + µ c,b (2)Thus, a circuit can be viewed as a quasi-metric space in which thedistance from a to b is the effective resistance µ a,b . Note that equality µ a,b = µ b,a holds only in symmetric case, but may fail in general. uasi-metric and quasi-ultrametric spaces corre-sponding to asymptotics of parameters r and s Playing with parameters r and s , one can get several interesting(but well-known) examples. Let r = r ( t ) and s = s ( t ) depend on a realparameter t . Then, these two functions define a curve in the positivequadrant r ≥ , s ≥
0. For the next four limit transitions, as t → ∞ , forall pairs of poles a, b ∈ V , the limits µ a,b = lim t →∞ µ a,b ( t ) exist and canbe interpreted as follows: • (i) Effective resistance of an Ohm semiconductor circuit from pole a to pole b ; s ( t ) = r ( t ) ≡ , or more generally, s ( t ) → r ( t ) → • (ii) Standard length (travel time or cost) of a shortest route fromterminal a to terminal b in a circuit of one-way roads; s ( t ) = r ( t ) →∞ , or more generally, s ( t ) → ∞ and s ( t ) /r ( t ) → • (iii) The inverse width of a widest bottleneck path from terminal a to terminal b in a circuit of one way-roads; s ( t ) → ∞ and r ( t ) ≡
1, ormore generally, r ( t ) ≤ const , or even more generally s ( t ) /r ( t ) → ∞ . • (iv) The inverse capacity (maximum flow per unit time) from termi-nal a to terminal b in a one-way pipeline; s ( t ) ≡ r ( t ) → s ( t ) → , while r ( t ) → r s s and r . All four examples define quasi-metric spaces, since in all cases s ( t ) ≥ r ( t ) for any sufficiently large t and we assume that t → ∞ . Moreover, forthe last two examples the ultrametric inequality µ a,b ≤ max( µ a,c , µ c,b ) (3) olds for any three nodes a, b, c , because s ( t ) /r ( t ) → ∞ , as t → ∞ , incases (iii) and (iv).These four examples allow us to interpret s and r as some importantparameters of transportation problems.Parameter s can be viewed as a measure of divisibility of a transportedmaterial; s ( t ) → s ( t ) → ∞ for (ii) and (iii),because a car, a ship, or an individual travelling from a to b is indivisible.Ratio s/r can be viewed as a measure of subadditivity of the trans-portation cost; so s ( t ) /r ( t ) → s ( t ) /r ( t ) → ∞ for (iii) and (iv), because in these casesonly edges of the maximum cost (the width of a bottleneck) or capacityof a critical cut matter.Other values of parameters s and s/r , between 1 and ∞ , correspond toan intermediate divisibility of the transported material and subadditivityof the transportation cost, respectively.The metric inequality in case (ii) is obvious. Let µ a,c and µ c,b be thelengths (or the travel times) of the shortest directed paths p a,c from a to c and p c,b from c to b , respectively. Combining these two paths we obtaina walk p a,b from a to b . Thus, the triangle inequality 2 follows.The ultrametric inequality of case (iii) can be proven in a similar way.Let λ a,c and λ c,b be the largest width (or wejght) of an object that canbe transported from a to c and from c to b , respectively, and let p a,c )and p c,b ) be the corresponding transportation paths. Combining them weobtain a walk p a,b from a to b . Obviously, an object of width w can betransported along this walk,i.e., λ a,b ≥ w , whenever it can be transportedfrom a to c and from c to b , i.e., λ a,c ≥ w and λ c,b ≥ w . This implies theultrametric inequality (3).Note that in both above (“indivisible”) cases the inequalities are strictwhenever directed paths p a,c and p c,b intersect not only in c . Yet, we willshow that they may be strict in some other cases too.“Divisible” case (iv) requires a different approach. Given a circuit withthree fixed nodes a, c , and b , let λ a,c and λ c,b denote the capacities (thatis, the maximum feasible flows) from a to c and from c to b , respectively.Furthermore, let w be the value of a flow feasible in both cases, in otherwords, let inequalities λ a,c ≥ w and λ c,b ≥ w hold. Then, λ a,b ≥ w holds too, implying (3). However, to show this, one cannot just combine(sum up) two flows realizing λ a,c and λ c,b , because the resulting flow mayexceed capacities of some edges, thus becoming not feasible.Instead, inequality λ a,b ≥ w can be easily derived from the classic“Max Flow - Min Cut Theorem” [7]. According to it, each minimum( a − b )-cut C : V = V a ∪ V b (such that a ∈ V a , b ∈ V b , and V a ∩ V b = ∅ ) isof capacity λ a,b .If c ∈ V a then C is a ( c − b )-cut too and, hence, λ c,b ≤ λ a,b ;if c ∈ V b then C is a ( a − c )-cut too and, hence, λ a,c ≤ λ a,b . hus, min( λ a,c , λ c,b ) ≤ λ a,b , which is equivalent with (3). When (1) holds with equality
In this paper we will prove that (1) holds with equality if and only if eachdirected path from a to b contains c . (For the symmetric case this wasshown in [15, 9].) The statement holds for any strictly positive real r and s ; in particular, in case (i), when r = s = 1. Yet, for the asymptoticcases (ii, iii, iv) only ”if part” holds, while ”only if” one may fail. Threeexamples are as follows: • cases (ii) and (iii). Let G = ( V, E ) be the directed triangle in which V = { a, b, c } and E = { ( a, b ) , ( a, c ) , ( c, b ) } . Obviously, G contains adirected ( a − b )-path avoiding c (just the edge ( a, b )). Set µ ( a,c ) = µ ( c,b ) = 1 and µ ( a,b ) = 3. Then, in case (i) we have µ a,c = µ c,b = 1and µ a,b = 2. Thus, (2) holds with equality: 1 + 1 = 2. In case ( iii )we have µ a,c = µ c,b = µ a,b = 1. Indeed, edge ( a, b ) of the width λ a,b = 1 / ,
1) = 1. • cases (iv). Define digraph G = ( V, E ) by V = { a, b, c, k, ℓ } and E = { ( a, k ) , ( k, c ) , ( c, ℓ ) , ( ℓ, b ) , ( k, ℓ ) } . Again G contains a directed( a − b -path avoiding c ; it is given by vertex-sequence a, k, c, ℓ, b . Set µ e = 1 for all e ∈ E . Then again µ a,c = µ c,b = µ a,b = 1, since edge( k, ℓ ) of capacity λ k,l = 1 is not needed for transportation, and (3)holds with equality: max(1 ,
1) = 1.
Known special cases of the main inequality
Our main result (1) generalizes some well (or maybe, not so well) knowninequalities obtained earlier for the following special cases.
Symmetric case.
A digraph G is called symmetric if its edges aresplit into pairs of oppositely directed edges e ′ = ( v ′ , v ′′ ) , e ′′ = ( v ′′ , v ′ ).Respectively, a circuit is called symmetric if its graph is symmetric and µ e ′ = µ e ′′ for each pair e ′ , e ′′ introduced above. In this case one canreplace each such pair e ′ , e ′′ by a non-directed edge e , thus, replacing thedigraph of the circuit by a non-directed graph. For this case, the maininequality (1) was shown in [13]; see also [9, 10, 11] for more details.The equality holds in (1) if and only if every path from a to b contains c . First it was shown in Section 16.9 of [15]; see also [9]). [It was alsoshown in Section 16.9 that the monomial conductance law is the onlyonly when the effective resistance µ a,b of the two-pole circuit ( G, a, b ) is areal number. In general, it is a monotone non-decreasing function for anarbitrary monotone circuit [17].]Clearly, equality µ a,b = µ b,a holds in the isotropic (symmetric) case.Thus, resistance distances of symmetric circuits form a metric spaces. Yet,in anisotropic (non-symmetric) case the above equality may fail and weobtain only quasi-metric spaces, in general. Linear case , r = s = 1. In the symmetric linear case, the metricresistance inequality was discovered by Gerald E. Subak-Sharpe [32, 33]; ee also [16, 27, 31, 34, 35, 36, 37, 5, 6, 4] and preceding works [39, 28,26, 29, 30]. This result was rediscovered several times later.Let us notice that the proof of (1) given in [13, 9] for the isotropicmonomial conductance differs a lot from the proof of [32, 33] for thesymmetric linear case. In this paper we extend the first proof to theanisotropic case or, in other words, to digraphs. To make the presentationself-contained we copy here some parts of [9].For the linear non-symmetric case the quasi-metric inequality, alongwith many related results, was recently obtained in [38]. Continuum.
It would be natural to conjecture that the above approach can be devel-oped not only for the discrete circuits but for continuum as well: inequality(1) and its corollaries should hold in this case too. Sooner or later, thiswill become the subject of a separate research. The same four subcasesappear: linear and monomial, isotropic and anisotropic.
Conductance law
Let e be a semi-conductor with the monomial conductivity law y ∗ e = f e ( y e ) = λ se y re = y re µ se if y e ≥ y e ≤ . (4)Here y e is the voltage or potential difference , y ∗ e current , λ e conductance ,and µ e = λ − e resistance of e ; furthermore, r and s are two strictly positivereal parameters independent of e . Obviously, the monomial function f e is continuous, strictly monotone increasing when y e ≥
0, and taking allnon-negative real values.
Main variables and related equations
A semi-conductor circuit is modeled by a weighted digraph G = ( V, E, µ )in which weights of the edges are their positive resistances µ e , e ∈ E .Let us introduce the following four groups of real variables; two foreach v ∈ V and e ∈ E : potential x v ; difference of potentials, or voltage y e ; current y ∗ e ; sum of currents, or flux x ∗ v .The above variables are not independent. By (4), current y ∗ e dependson voltage y e . Furthermore, the voltage (respectively, flux) is a linerfunction of the potentials (respectively, of the currents). These functionsare defined by the node-edge incidence function of the digraph G :inc( v, e ) = +1 , if node v is the beginning of e ; − , if node v is the end of e ;0 , if v ; and e are not incident . (5) e will assume that the next two systems of linear equations alwayshold: y e = X v ∈ V inc( v, e ) x v ; (6) x ∗ v = X e ∈ E inc( v, e ) y ∗ e . (7)Let us notice that equation (6) for a directed edge e = ( v ′ , v ′′ ) can bereduced to y e = x v ′ − x v ′′ .We say that the first Kirchhoff law holds for a node v if x ∗ v = 0.Let us introduce four vectors, one for each group of variables: x = ( x v | v ∈ V ) , x ∗ = ( x ∗ v | v ∈ V ) , y = ( y e | e ∈ E ) , y ∗ = ( y ∗ e | e ∈ E ) ,x, x ∗ ∈ R n ; y, y ∗ ∈ R m , where n = | V | and m = | E | are the numbers of nodes and edges of thedigraph G = ( V, E ). Let A = A G be the edge-node m × n incidencematrix of graph G , that is, A ( v, e ) = inc ( v, e ) for all v ∈ V and e ∈ E .Equations (6) and (7) can be rewritten in this matrix notation as y = Ax and x ∗ = A T y ∗ , respectively.It is both obvious and well known that these two equations imply thefollowing chain of identities:( x, x ∗ ) = X v ∈ V x v x ∗ v = X e ∈ E y e y ∗ e = ( y, y ∗ ) . Recall that y ∗ is uniquely defined by y according to the conductancelaw (4). Thus, given vector x , the remaining three vectors y , y ∗ , and x ∗ are uniquely defined by x (6,4,7). This triple will not change if we add anarbitrary real constant c to all coordinates of x , while multiplying x by c will result in multiplying y by c and y ∗ , x ∗ by c r . More precisely, thefollowing scaling property cllearly holds. Lemma 1.
For any positive constant c , two quadruples ( x, y, y ∗ , x ∗ ) and ( cx, cy, c r y ∗ , c r x ∗ ) can satisfy all equations of (6,7,4) only simultaneously. Two-pole boundary conditions
In general theory of monotone circuits, one can consider arbitrary mono-tone functions: a non-decreasing one y ∗ e = f e ( y e ) for each e ∈ E and anon-increasing one x ∗ v = g v ( x v ) for each v ∈ V ; see [8, 17, 24, 25] and also[14, 15, 20, 21, 23].In case of the two-pole circuits we restrict ourselves by the monomialconductance law (4). Fix an ordered pair of poles a, b , potentials x a = x a , x b = x b , (8)in them, and require the first Kirchhoff law for any other node: ∗ v = 0 , v ∈ V \ { a, b } . (9)By convention, y ∗ a,b = 0 if a = b or y a,b = x a − x b ≤
0. So, w.l.o.g. wecan assume that a = b and x a ≥ x b . Remark 1.
By Lemma 1, it would be sufficient to replace (8) by x a = 1 and x b = 0 . It would be also possible to replace it by x ∗ a = x ∗ a (or x ∗ a = 1 ).Then, x ∗ b = − x ∗ a (resp., x ∗ b = − ) will automatically hold, by (9). We call a vector x = x ( a, b ) a solution of the two-pole circuit ( G, a, b )if the corresponding quadruple ( x, y, y ∗ , x ∗ ) satisfies all equations (4 - 9).In [9] the monomial symmetric case was considered and it was shownthat there exists a unique solution x = x ( a, b ) whenever a and b belongto the same connected component of G .Yet, this claim cannot be extended directly to digraphs. For example,let ( G, a, b ) be a directed ( a, b )- k -path a = v , v , . . . , v k = b from a to b .Then potential x is unique if x a ≥ x b , but otherwise, when x a < x b , anynon-decreasing x a = x ≤ x v ≤ . . . ≤ x v k = x b will be a solution, withno current, that is, y ∗ e = 0 for all e = ( v j − , v j ) for j = 1 , . . . , k .In the directed case we will prove that y ∗ (rather than x ) is the samein all solutions. Furthermore, let G + be the subgraph of G defined by alldirected edges e ∈ E such that y ∗ e >
0. Then in all solutions potentials x are uniquely defined on vertices of G ′ . Existence of a solution
We will apply Method of Successive Approximation (MSA) increasingpotentials of some nodes, one by one in a certain order.Obviously, when we increase x v (keeping all remaining potentials x u unchanged) the corresponding flux x ∗ v is non-decreasing; furthermore, it isstrictly increasing if and only if G contains an edge ( v, u ) with x v ≥ x u oran edge ( u, v ) with x v ≤ x u . Respectively, x ∗ u in any other node u ∈ V \{ v } is non-increasing; furthermore, it is strictly decreasing if and only if ( v, u )is an edge and x v ≥ x u or ( u, v ) is an edge and x v ≤ x u .Let us set x a = x a and x v = x b for all nodes v ∈ V \ { a } , including b . In the course of iterations, potentials x a = x a and x b = x b will remainunchanged, while the all other potentials x v , on the nodes from W = V \ { a, b } , will be recomputed by MSA as follows.Order arbitrarily the nodes of W and consider them one by one in thisorder repeating cyclically. If x ∗ v = 0, skip this node and go to the nextone. If x ∗ v <
0, increase x v until x ∗ v becomes 0. The latter is possible,since, by (4), f e is continuous and y ∗ e → + ∞ as y e → + ∞ . Let us noticethat x ∗ v may remain 0 for some time, but we stop increasing x v the firstmoment when x ∗ v becomes 0, and proceed to the next node.The following claims can be easily proven together by induction on thenumber of iterations. • (i) For any node v ∈ W its potential x v is monotone non-decreasingand it remains bounded by x a from above. Hence, it tends to a limit x v between x a and x b . (ii) These limit potentials solve the two-pole circuit ( G, a, b ). • (iii) Fluxes x ∗ v remain non-positive for all v ∈ V \ { a } . In contrast, x ∗ a remains non-negative. Furthermore, x ∗ a and x ∗ b are monotonenon-increasing. • (iv) For the limit values of potentials and fluxes we have:If G contains no directed path from a to b then x ∗ v = 0 for all v ∈ V ,including a and b . In this case x v = x a if G contains a directed pathfrom a to v , otherwise x v = 0. If G contains a directed path from a to b then x ∗ a >
0. (Respectively, x ∗ b = − x ∗ a < x ∗ v = 0 forall v ∈ W , in accordance with (9.) • (v) The limit potentials x v take only values x a and x b if and only if G contains no directed path from a to b or every such path consistsof only one edge.Existence of a solution is implied by (ii). Remark 2.
A very similar monotone potential reduction (pumping) al-gorithm for stochastic games with perfect information was suggested in[1, 2].
Uniqueness of the solution
A solution x of a two-pole circuit ( G, a, b ) may be not unique. Supposethat G contains an induced directed path P from u to v of length greaterthan 1 and that ( G, a, b ) has a solution x with x u < x v . Then everymonotone non-decreasing sequence of potentials on P is feasible. Notice,however, that the current along P will be zero for any such sequence.We will demonstrate that, in general, vector of currents y ∗ (and, hence, x ∗ too) is unique for all solutions of ( G, a, b ). It follows directly from anold classical result relating solutions of an arbitrary monotone circuit witha pair of dual problems of convex programming [8, 17, 24, 25]; see also[20, 21, 22, 23, 13, 14, 15].First, note that, by Lemma 1, we can replace the boundary conditions(8) by x ∗ a = x ∗ a , x ∗ b = − x ∗ a , (10)and recall that x ∗ v = 0 for all v ∈ W = V \ { a, b } by (9).The Joule-Lenz heat on e is defined by the current y ∗ e ≥ F ∗ e ( y ∗ e ) = Z f − e ( y ∗ e ) dy ∗ e = µ s/re /r y ∗ /re , (11)which is a strictly convex function of y ∗ e ≥
0. Furthermore, the total heatdissipated in the circuit is additive: F ∗ ( y ∗ ) = X e ∈ E F ∗ e ( y ∗ e ) . (12)It is a strictly convex function of y ∗ defined on the positive ortant, y ∗ ≥ G, a, b ) is equivalent with minimizing dissipation F ∗ ( y ∗ ) subject to thefollowing constraints ∗ a = x ∗ a , ( x ∗ b = − x ∗ a ) , x ∗ v = 0 ∀ v ∈ V \ { a, b } , and y ∗ ≥ . (13)Since x ∗ = Ay ∗ , we obtain the minimization problem for a strictlyconvex function of y ∗ subject to linear constraints on y ∗ . It is knownfrom calculus that solution y ∗ is unique in this case. Remark 3.
In the non-directed (isotropic) case the above equivalence iswell-known in physics as the minimum dissipation principle. It is applica-ble for arbitrary “bounddary conditions” not only to the two-pole circuits.
Thus, all solutions of (
G, a, b ) have the same current vector y ∗ . Thisimplies the uniqueness of the flux vector x ∗ = A ∗ y ∗ as well.Let us denote by G + = ( V + , E + ) the subgraph of G formed by theedges e ∈ E with positive currents, y ∗ >
0. The following properties of G + are obvious: • (j) Digraph G contains a directed path from a to b if and only if G + is not empty. • (jj) In the latter case it contains the poles, a, b ∈ V + , and at leastone directed path from a to b , but not necessarily all such pathes.Yet, any vertex or edge of G + belongs to such a path. • (jjj) Potentials are strictly decreasing on each edge of G + and, hence,it has no directed cycles. Conductance functions of two-pole networks
Given a two-pole network (
G, a, b ), let us define the potential drop andcurrent from a to b as y a,b = x a − x b , y ∗ a,b = x ∗ a = − x ∗ b . If there is no directed path from a to b in G , let us set µ a,b = + ∞ ,since in this case y ∗ a,b = 0 for any y a,b . Otherwise, by Lemma 1, y ∗ a,b depends on y a,b as in (4): y ∗ a,b = f a,b ( y a,b ) = λ sa,b y ra,b = y ra,b µ sa,b if y a,b ≥ y a,b ≤ . (14)Two strictly positive real values λ a,b and µ a,b = λ − a,b are called con-ductance and, respectively, resistance of ( G, a, b ). Remark 4.
It is shown in Section 6.9 of [15] that among all monotoneconductance laws the monomial one is the only case when resistance of atwo-pole network is a real number; in other words, up to a real factor, thesame function f describes the conductances f e ( y e ) and f a,b ( y a,b ) . onotonicity of effective resistances and Braess’Paradox Given a two-pole circuit (
G, a, b ), where G = ( V, E, µ ), let us fix an edge e ∈ E , replace the resistance µ e by a smaller one, µ ′ e ≤ µ e , and denoteby G ′ = ( V, E, µ ′ ) the obtained circuit.Of course, the total resistance will not increase either, that is, µ ′ a,b ≤ µ a,b will hold. Yet, how to prove this ”intuitively obvious” statement?Somewhat surprisingly, the simplest way is to apply the the minimumdissipation principle again; see, for example, [19],Let y ∗ and y ∗ ′ be the (unique) current vectors that solve ( G, a, b ) and( G ′ , a, b ), respectively. Since µ ′ e ≤ µ e , inequality F ∗ ′ e ( y ∗ e ) ≤ F ∗ e ( y ∗ e )is implied by (11). Furthermore, F ∗ ′ e ( y e ) = F ∗ e ( y e ) for all other e ∈ E ,distinct from e . Hence, F ∗ ′ ( y ∗ ) ≤ F ∗ ( y ∗ ) holds by (12). As we know, allsolutions of ( G ′ , a, b ) have the same vector of currents y ∗ ′ , which may differfrom ( y ∗ ) and, by the minimum dissipation principle, we have F ∗ ′ ( y ∗ ′ ) ≤ F ∗ ′ ( y ∗ ). From this, by transitivity, we conclude that F ∗ ′ ( y ∗ ′ ) ≤ F ∗ ( y ∗ )and, by (11,12), conclude that µ ′ a,b ≤ µ a,b holds.In particular, when an edge e is eliminated from G , its finite resistance µ e is replaced by µ ′ e = + ∞ . It was just shown that, by this operation,the effective resistance is not reduced, that is, µ a,b ≤ µ ′ a,b holds.In general, for monotone circuits [8, 17] the conductance function ofits edge may be an arbitrary, not necessarily monomial, monotone non-decreasing function: y ∗ e = f e ( y e ) if y e ≥ y e ≤ , Then, by results of [17], the conductance law of a two-pole network(
G, a, b ) is represented by a similar formula: x ∗ a = − x ∗ b = y ∗ a,b = f a,b ( y a,b ) if y a,b ≥ y a,b ≤ f a,b is a monotone non-decreasing function too.When we reduce conductance function f e ( y e ) of an edge e ∈ E , theeffective conductance function f a,b may increase for some (certainly, notfor all) values of its argument y a,b = x a − x b . This phenomenon is knownas Braess paradox [3].The above monotonicity principle implies that this paradox is not pos-sible for circuits with the monomial conductance law provided parameters r and s are the same for all edges e ∈ E . Indeed, in this case resistance µ a,b between the poles is a real number; moreover, it is a monotone function ofresistances µ e of edges e ∈ E , as it was shown above. Yet, the paradox canappear for monomial circuits in which parameter r = r e depends on e , orwhen some edges have non-monomial monotone conductance functions. Here we prove our main result generalizing the triangle inequality of [13, 9]from graphs to digraphs as follows. heorem 1. Given a weighted digraph G = ( V, E, µ ) with strictly positiveweights-resistances ( µ e | e ∈ E ) , three arbitrary nodes a, b, c ∈ V , andstrictly positive real parameters r and s , inequality (1) holds: µ s/ra,b ≤ µ s/ra,c + µ s/rc,b . Moreover, it holds with equality if and only if node c belongsto every directed path from a to b in G .Proof. W.l.o.g. we can assume that G contains directed paths from a to c and from c to b . Indeed, otherwise µ a,c or µ c,b is + ∞ and there is nothingto prove. By this assumption, G contains a directed walk from a to b passing through c . Hence, G also contains a directed path from a to b ,but the latter may avoid c .Anyway, there exists a (not necessarily unique) solution of ( G, a, b ) forany fixed potentials x a , x b in the poles a, b . If x a ≤ x b then y ∗ a.b = 0, bydefinition, and again there is nothing to prove. Thus, w.l.o.g. we assumethat x a > x b (Moreover, we could assume w.l.o.g. that x a = 1 and x b = 0,but will not do this.)We make use of the same arguments as in subsection ”Existence of asolution”. Consider a solution x = x ( a, b ) constructed there and denoteby x c the obtained potential in c . By construction, x a ≥ x c ≥ x b andat least one of these two inequalities is strict. (Actually, such inequalitieshold for any solution x = x ( a, b ), but one chosen x c will be enough forour purposes.)Now let us consider the two-pole circuit ( G, a, c ) and fix in it x a = x a and x c = x c , standardly requiring the first Kirchhoff law, x ∗ v = 0 for allother vertices v ∈ W = V \ a, c , including v = b . Lemma 2.
The obtained currents in the circuits ( G, a, b ) and ( G, a, c ) satisfy inequality y ∗ a,b ≥ y ∗ a,c . Moreover, the equality holds if and only if c belongs to every directed path from a to b .Proof. First, recall that the current vectors (and, hence, the values of x ∗ ( a, b ) = y ∗ a,b and x ∗ ( a, c ) = y ∗ a,c too) are well-defined, that is, remainthe same for any solutions x ( a, b ) and x ( a, c ) of ( G, a, b ) and (
G, a, c ) withboundary conditions x a , x b and x a , x c , respectively.Again we apply MSA to compute x ( a, c ), yet, this time we take x ( a, b )as the initialization. Thus, in the beginning we have x ∗ a ( a, b ) = − x ∗ b ( a, b ) , x ∗ c ( a, b ) = 0and at the end we will have x ∗ a ( a, c ) = − x ∗ c ( a, c ) , x ∗ b ( a, c ) = 0.Potentials x a = x a and x c = x c satisfy the boundary conditions andwill stay unchanged in the course of iterations, while the remaining po-tentials x v on the nodes from W = V \ { a, c } will be determined by MSAas follows. Order arbitrarily the nodes of W and consider them one byone in this order repeating cyclically. If x ∗ v = 0, skip this node v and goto the next one. If x ∗ v <
0, increase x v until (the very first moment when) x ∗ v becomes 0. Then proceed with the next node. The following claimscan be easily proven together, by induction on the number of iterations. (i) Fluxes x ∗ a and x ∗ c are both monotone non-decreasing and remainnon-negative and non-positive, respectively. Moreover, x ∗ v remainnon-positive for all v ∈ V \ { a } . • (ii) All potentials x v are monotone non-decreasing and remain boundedby x a from above. Hence, x v tends to a limit x v ( a, c ) between x a and x v ( a, b ). • (iii) These limit potentials solve the two-pole circuit ( G, a, c ). • (iv) Since G contains directed pathes from a to c and from c to b ,for the limit values of the fluxes we have: x ∗ a ( a, c ) > , x ∗ c ( a, c ) = − x ∗ a ( a, c ), and x ∗ v ( a, c ) = 0 for all v ∈ V \ { a, c } , in accordancewith (9). • (v) x ∗ a ( a, b ) = y ∗ a,b ≥ y ∗ a,c = x ∗ a ( a, c )Existence of a solution is implied by (iii). The last inequality holds,because potential x a is constant, while all other potentials are not de-creasing. Hence, the flux from a cannot increase.If all directed pathes from a to b contain c then equality holds in (v).Indeed, in this case x v will not be changed by MSA for any vertex v that belongs to a path from a to c . Hence, the flux x ∗ a remains constant,resulting in y ∗ a,b = y ∗ a,c .Suppose conversely that G contains a directed path P from a to b avoiding c . Then, order the nodes of W = V \ { a, c } so that the nodesof P go first ordered from b to a . Obviously, in | P | − x ∗ a will be strictly reduced. Furthermore, x ∗ a is non-increasing, and its initialand limit values are y ∗ a,b and y ∗ a,c , respectively. Thus, y ∗ a,b > y ∗ a,c . Recallthat these two numbers are well-defined, although solutions of ( G, a, b )and (
G, a, c ) are not necessarily unique. This proves the lemma.
Remark 5.
The same arguments prove that inequality f a,b ( x a − x b ) = y ∗ a,b ≥ y ∗ a,c = f a,c ( x a − x c ) holds not only for monomial but for arbitrary monotone non-decreasingconductivity functions. In the exactly same way we can apply MSA to (
G, c, b ) again takinga solution of (
G, a, b ) as an initial approximation. Clearly this will resultin inequality y ∗ a,b ≥ y ∗ a,c , in which the equality holds if and only if c belongs to every path between a and b . Summarizing we obtain thefollowing statement: Proposition 1.
For an arbitrary weighted digraph G and nodes a, b, c in it, the inequality y ∗ a,b ≥ max ( y ∗ a,c , y ∗ c,b ) holds and the following fivestatements are equivalent: • (ac) y ∗ a,b = y ∗ a,c ; • (bc) y ∗ a,b = y ∗ c,b ; • (ab) y ∗ a,c = y ∗ c,b ; • (acb) every directed path from a to b contains c ; • (=) µ s/ra,b = µ s/ra,c + µ s/rc,b . or the rest of the proof of Theorem 1 we will need only elementary”high-school” transformations: y ∗ a,b = ( x a − x b ) r µ sa,b ≥ ( x a − x c ) r µ sa,c = y ∗ a,c ; y ∗ a,b = ( x a − x b ) r µ sa,b ≥ ( x c − x b ) r µ sc,b = y ∗ c,b , (15)which can be obviously rewritten as follows (cid:18) µ a,c µ a,b (cid:19) s/r ≥ x a − x c x a − x b ; (cid:18) µ c,b µ a,b (cid:19) s/r ≥ x c − x b x a − x b (16)Summing up these two inequalities we obtain (1).The above computations show that (1) holds with equality if and onlyif y ∗ a,b = y ∗ a,c and y ∗ a,b = y ∗ c,b . By Propositions 1 these two equations areequicallent and hold if and only if c belongs to each directed path from a to b . Parallel and series connection of edges
Let us consider two simplest two-pole circuits given in Figure 2. a be ′ e ′′ a c be ′ e ′′ Figure 2: Parallel and series connection. All edges are directed from left toright.
Proposition 2.
The resistances of these two circuits can be determined,respectively, from formulas µ − sa,b = ( µ − se ′ + µ − se ′′ ) and µ s/ra,b = ( µ s/re ′ + µ s/re ′′ ) . (17) Proof. If r = s = 1 then (17) turns into familiar high-school formulas.The general case is just a little more difficult. Without loss of generalitylet us assume that y a,b = x a − x b ≥ y ∗ a,b = f a,b ( y a,b ) = y ra,b µ sa,b = f e ′ ( y a,b )+ f e ′′ ( y a,b ) = y re ′ µ se ′ + y re ′′ µ se ′′ = y ra,b µ se ′ + y ra,b µ se ′′ . Let us compare the third and the last terms; dividing both by thenumerator y ra,b we arrive at (17). n case of the series connection, let us start with determining x c fromthe first Kirchhoff law: y ∗ a,b = f a,b ( y a,b ) = y ra,b µ sa,b = ( x a − x b ) r µ sa,b = y ∗ e ′ = f e ′ ( y e ′ ) = f e ′ ( x a − x c ) = ( x a − x c ) r µ se ′ = y ∗ e ′′ = f e ′′ ( y e ′′ ) = f e ′′ ( x c − x b ) = ( x c − x b ) r µ se ′′ . It is sufficient to compare the last and eighth terms to get x c = x b µ s/re ′ + x a µ s/re ′′ µ s/re ′ + µ s/re ′′ . Then, let compare the last and forth terms, substitute the obtained x c , and get (17).Now, let us consider the convolution µ ( t ) = ( µ te ′ + µ te ′′ ) /t ; it is wellknown and easy to see that µ ( t ) → max( µ e ′ , µ e ′′ ) , as t → + ∞ , and µ ( t ) → min( µ e ′ , µ e ′′ ) , as t → −∞ . (18) Main four examples of resistance distances
Let us fix a weighted digraph G = ( V, E, µ ) and two strictly positive realparameters r and s . As we proved, the obtained circuit can be viewed asa quasi-metric space in which the distance from a to b is defined as theeffective resistance µ a,b . As announced in the introduction, this model re-sults in several interesting examples of quasi-metric and quasi-ultrametricspaces. Yet, to arrive to them we should allow for r and s/r to take values0 and + ∞ . More accurately, let r = r ( t ) and s = s ( t ) depend on a realpositive parameter t , or in other words, these two functions define a curvein the positive quadrant s ≥ , r ≥ µ a,b ( t ) are well-defined for every two nodes a, b ∈ V and each t . We will show that, for the four limit transitions listedbelow, limits µ a,b ( t ) = lim t →∞ µ a,b ( t ), exist for all a, b ∈ V and can beinterpreted as follows: Example 1: the effective Ohm resistance of an electrical cir-cuit.
Let a weighted digraph G = ( V, E, µ ) model an electrical circuit inwhich µ e is the resistance of a directed edge (semiconductor) e and r ( t ) = s ( t ) ≡
1, or more generally, r ( t ) → s ( t ) →
1, as t → + ∞ . Then, µ a,b is the effective Ohm resistance from a to b . For parallel and seriesconnection of two directed edges e ′ and e ′′ , as in Figure 2, we obtain,respectively, µ − a,b = µ − e ′ + µ − e ′′ and µ a,b = µ e ′ + µ e ′′ , which is knownfrom the high school. Example 2: the length of a shortest route.
Let a weighted digraph G = ( V, E, µ ) model a road network in which µ e is the length (milage, traveling time, or gas consumption) of a one-way oad e . Then, µ a,b can be viewed as the distance from a to b , that is,the length of a shortest directed path between them. In this case, forparallel and series connection of e ′ and e ′′ , we obtain, respectively, µ a,b =min( µ e ′ , µ e ′′ ) and µ a,b = µ e ′ + µ e ′′ . Hence, by (18), − s ( t ) → −∞ and s ( t ) ≡ r ( t ) for all t , as in Figure 1; or more generally, s ( t ) → ∞ and s ( t ) /r ( t ) →
1, as t → + ∞ . Example 3: the inverse width of a bottleneck route.
Now, let digraph G = ( V, E, µ ) model a system of one-way passages(rivers, canals, bridges, etc.), where the conductance λ e = µ − e is the ”width” of a passage e , that is, the maximum size (or tonnage) of a shipor a car that can pass e , yet. Then, the effective conductance λ a,b = µ − a,b is interpreted as the maximum width of a (bottleneck) path between a and b , that is, the maximum size (or tonnage) of a ship or a car that can stillpass between terminals a and b . In this case, λ a,b = max( λ e ′ , λ e ′′ ) forthe parallel connection and λ a,b = min( λ e ′ , λ e ′′ ) for the series connection.Hence, s ( t ) → ∞ and s ( t ) /r ( t ) → ∞ , as t → ∞ ; in particular, r mightbe bounded by a constant, r ( t ) ≤ const , or just r ( t ) ≡ t , as inFigure 1. Example 4: the inverse value of a maximal flow.
Finally, let digraph G = ( V, E, µ ) model a pipeline or transportationnetwork in which the conductance λ e = µ − e is the capacity of a one-waypipe or road e . Then, λ a,b = µ − a,b is the capacity of the whole two-polenetwork ( G, a, b ) from terminal a to b . (Standardly, the capacity is definedas the amount of material that can be transported through e , or from a to b in the whole circuit, per unit time.) In this case, λ a,b = λ e ′ + λ e ′′ for theparallel connection and λ a,b = min( λ e ′ , λ e ′′ ) for the series connection.Hence, − s ( t ) ≡ − s ( t ) /r ( t ) → ∞ , that is, s ( t ) ≡ r ( t ) →
0, asin Figure 1, or more generally, s ( t ) →
1, while r ( t ) →
0, as t → ∞ . Theorem 2.
In all four examples, the limits µ a,b = lim t → + ∞ µ a,b ( t ) existand equal the corresponding distances from a to b for all a, b ∈ V . In allfour cases these distances define quasi-metric spaces and in the last two -quasi-ultrametric spaces.Proof. (sketch) For Example 1 there is nothing to prove. Also, for theseries-parallel circuits the statement is obvious in all cases. It remansto consider Examples 2, 3, and 4 for general circuits. In each case ouranalysis will be based on the minimum dissipation principle ( ?? ).For simplicity, we will omit argument t in r, s, µ e , µa, b, λ e , λ a,b , y ∗ e , y a,b ,remembering, however, that all these variables depend on t , as indicatedin the definitions of Examples 2,3 and 4. Example 2 . In this case F e ( y ∗ e ) ∼ µ e y ∗ e , as t → + ∞ .Hence, “moving some current to a shorter directed path” reduces thetotal dissipation F ∗ ( y ∗ ) when t is large enough. More precisely, let p ′ = p ′ ( s, t ) and p ′′ ( s, t ) be two directed paths from s to t in G such that thefirst one is shorter, that is, µ ( p ′ ) = P e ∈ p ′ µ e < P e ∈ p ′′ µ e = µ ( p ′′ ). uppose that min( y ∗ e | e ∈ p ′ ∪ p ′′ ) = y ∗ >
0. Obviously, dissipation F ∗ ( y ∗ ) will we reduced by about ( µ ( p ′′ ) − µ ( p ′ )) y ∗ if we move the current y ∗ from p ′′ to p ′ , that is, we subtract y ∗ from y ∗ e for each e ∈ p ′′ and addit to y ∗ e for each e ∈ p ′ .Recall that for every fixed t > G, a, b ) hasa unique distribution of currents y ∗ , which minimizes the total dissipation F ∗ ( y ∗ ). Thus, the above observation implies that all currents will tendto the shortest directed paths from a to b , as t → + ∞ . Moreover, y ∗ e wellbecome just 0 for any e that does not belong to some shortest ( a − b )-pathif t is large enough.Clearly, by an arbitrary small perturbation of µ e , one can make thelengths of all directed paths from a to b distinct. After such perturbation,the shortest ( a − b )-path p in G becomes unique and all currents outsideof it become 0, that is, y ∗ e = 0 whenever e p , in particular, µ a,b becomesthe length of p when t is large enough. Example 3 . In this case F e ( y ∗ e ) ∼ µ se y ∗ e , where s = s ( t ) → + ∞ , as t → + ∞ . Thus, all currents will tend to the widest bottleneck directedpaths from a to b , as t → + ∞ . Moreover, y ∗ e well become just 0 for any e that does not belong to such a path if t is large enough.Recall that the widest bottleneck directed ( a − b )-path p in G is definedas one maximizing min( µ e | e ∈ p ).Unlike the shortest ( a − b )-path the widest bottleneck one is ”typically”not unique in G . Let us refine slightly this concept and introduce the lexicographically widest bottleneck directed ( a, b )-path in G . To do so,consider all widest bottleneck directed ( a − b )-paths in G . Among themchoose those that maximize the second smallest width, etc. In severalsteps (at most | V | ) we will obtain the required path.Clearly, by an arbitrary small perturbation of µ e , one can make thewidths of all directed paths from a to b distinct. Under this condition, thelexicographically widest bottleneck directed ( a, b )-path p in G becomesunique, and all currents outside of it become 0, that is, y ∗ e = 0 whenever e p , in particular, µ a,b becomes the width of p when t is large enough. Example 4 . In this case, setting s = 1, we obtain F e ( y ∗ e ) ∼ ry ∗ e ( µ e y ∗ e ) r = ry ∗ e ( y ∗ e λ e ) r , where r = r ( t ) → t → + ∞ .Let us recall the concept of the so-called balanced flow introduced in[12] for the multi-pole circuits. Given a weighted digraph G = ( V, E )consider the following boundary conditions: x ∗ v = x ∗ v for all v ∈ V .A flow y ∗ is called satisfactory if it satisfies these conditions. We willassume that such a flow exists. In particular, P v ∈ V x ∗ = 0 must hold.Two-pole boundary conditions (13), considered in the present paper,form a special case of the multi-pole conditions. In this case a satisfactoryflow exists if and only if digraph G contains a directed path from a to b .Introduce resistances µ e (and conductances λ e = µ − e ) for all directededges e ∈ E , thus getting a weighted multi-pole circuit G = ( V, E, µ ).Among all satisfactory flows choose all that minimize max( y ∗ e λ e | e ∈ E );then among them choose all that minimize the second largest value of y ∗ e λ e , tc. For all e ∈ E order the ratios y ∗ e λ e non-decreasingly. A satisfactoryflow y ∗ realizing the lex-min over all such vectors is called balanced .Let us briefly recall the algorithm from [12] constructing a balancedflow in a multi-pole circuit. A cut in a digraph G = ( V, E ), is defined asan ordered pair C = ( V ′ , V ”) that partitions V properly, that is, V ′ = ∅ , V ′′ = ∅ , V ′ ∩ V ′′ = ∅ and V ′ ∪ V ′′ = V .We say that a directed edge e = ( u, w ) is in C if u ∈ V ′ and w ∈ V ′′ .Consider a multi-pole circuit defined by a weighted digraph G = ( V, E, µ )and boundary conditions x ∗ v = x ∗ v , v ∈ V ; fix a cut C in G . Its deficiency and capacity are defined by formulas: D ( C ) = P v ∈ V ′ x ∗ = − P v ∈ V ′′ x ∗ ; Λ( C ) = P e ∈ C λ e .Note that Λ( C ) ≥
0, by this definition, while D ( C ) may be negative.A multi-pole problem has no satisfactory vector if and only if thereexists a cut C such that R ( C ) = + ∞ , or in other words, such that D ( C ) > C ) = 0. In a two-pole circuit this happens if and only if there isno directed path from a to b . Note also that in the two-pole case we have D ( C ) = 0 whenever a, b ∈ V ′ or a, b ∈ V ′′ .Cut C is called critical if it realizes the maximum of the ratio R ( C ) = D ( C ) λ ( C ) . Choose such C and set y ∗ e = R ( C ) λ e for each e ∈ C . Then, y ∗ e λ e take the same value R ( C ) for all e ∈ C and we have P e ∈ C y ∗ e = λ ( C ).Reduce digraph G eliminating all edges of C from it. Recompute newmulti-pole boundary conditions in the obtained reduced digraph G ′ takinginto account flows y ∗ e on the deleted edges e ∈ C . Then, find a criticalcut C ′ in G ′ , and repeat. It is shown in [12] that the values R, R ′ , . . . aremonotone non-increasing. Hence, the ratio y ∗ e λ e takes the largest values on e ∈ C at the first stage. It is also shown in [12] that the balanced flow isunique.Recall that F e ( y ∗ e ) ∼ ( ry ∗ e )( y ∗ e λ e ) r , where r = r ( t ) → t → + ∞ .From this we conclude that a satisfactory flow y ∗ , minimizing the totaldissipation F ∗ ( y ∗ ), becomes a balanced flow when t is large enough.A satisfactory flow y ∗ is called feasible if y ∗ e ≤ λ e . Obviously, the(unique) balanced flow y ∗ is feasible whenever a latter exists. We sawthat in this case y ∗ is a solution of the multi-pole network problem, since y ∗ minimizes F ∗ ( y ∗ ) when t is large enough.In particular, this is true for the two-pole problems. In this case letus set x ∗ a = − x ∗ b = Λ( a, b ), where Λ( a, b ) is the capacity from a to b ofthe circuit G = ( V, E, µ ) with poles a and b . Obviously, on the first stageof the algorithm we obtain a cut C of the unit ratio R ( C ) = 1, that is, y ∗ e = λe for all e ∈ C . Thus, λ ( a, b ) → Λ( a, b ), as t → + ∞ Acknowledgements
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