Closed-Form Formulas for Kirchhoff Index

JOSÉ LUIS PALACIOSDepartamento de Cómputo Científico y Estadística, Universidad Simón Bolívar, Caracas, Venezuela

Received 5 July 2000; revised 8 August 2000; accepted 17 August 2000

ABSTRACT: We find closed-form expressions for the resistance, or Kirchhoff index, ofcertain connected graphs using Foster’s theorems, random walks, and the superpositionprinciple. c© 2001 John Wiley & Sons, Inc. Int J Quantum Chem 81: 135–140, 2001

Key words: resistance distance; effective resistance; commute times; geodetic graphs;distance-transitive graphs

Introduction

T he resistance, or Kirchhoff index, R(G) of a con-nected undirected graph G = (V, E), with

vertex set V = {1, 2, . . . , N} and edge set E, was de-fined by Klein and Randic [1] as

R(G) =∑i<j

Rij, (1)

where Rij is the effective resistance between ver-tices i and j as computed with Ohm’s law when allthe edges of G are considered to be unit resistors.The quantity Rij is shown in [1] to be a distance func-tion on the set of vertices, and it is introduced alongwith the index R(G) as an alternative to the usualgraph-theoretic distance and the indices this clas-sic distance generates. For the relationships amongthe different indices and their uses in chemistry, thereader is referred to [2 – 5].

The computational effort needed to obtain R(G)is essentially that of inverting an N × N matrix (see[1, 5, 6]). It makes sense, then, to find a closed-formformula or near-closed-form formulas for R(G) that

E-mail: jopala@eece.unm.edu.

avoid this computational complexity. In that regard,we gave in [6] the formulas for the Kirchhoff indexsummarized in the first three propositions.

Let d(a, b) denote the (usual graph-theoretic) dis-tance between vertices a and b. Assume G is distancetransitive, that is, for any vertices u, v, w, z such thatd(u, v) = d(w, z) there is a graph automorphism α

(i.e., a bijection in the vertex set that preserves ad-jacencies) such that α(u) = w and α(v) = z. ThenRij = Rk whenever d(i, j) = k, or in words: theeffective resistance between two vertices dependsonly on the distance between the vertices. More-over, let n(k), k ≥ 1 denote the cardinality of the set{(i, j) : d(i, j) = k}, then we have:

Proposition 1.

R(G) =∑

d(i,j) = 1

Rij +∑

d(i,j) = 2

Rij +∑

d(i,j) = 3

Rij + · · ·

= R1n(1)+ R2n(2)+ R3n(3)+ · · · , (2)

where

n(k) = ∣∣{j : d(i, j) = k}∣∣N

2. (3)

Also, let M = max{d(i, j) : i, j ∈ V}, that is, let M bethe maximal distance between pairs of vertices in G,

International Journal of Quantum Chemistry, Vol. 81, 135–140 (2001)c© 2001 John Wiley & Sons, Inc.

PALACIOS

usually called the diameter of G; it may happen thatfor a given vertex i ∈ G there is a single vertex jat distance M, or it may happen that there is morethan one sharing that property. In the first case wesay that G has opposite vertices at distance M (thisoccurs, e.g., in all Platonic graphs except the tetra-hedron, and all even cycles, but not in odd cycles orthe Petersen graph). Then we have:

Proposition 2. If G is a distance-transitive graphwith opposite vertices at distance M, then

RM = 1|E| + RM−1.

Proposition 3. If G is an N vertex, distance-transitive graph with opposite vertices at distance 2,then

R(G) = N − 1+ (N − δ − 1)Nδ

, (4)

where δ is the degree (number of neighbors) of a ver-tex.

The purpose of this note is to use a few propertiesof electric networks and random walks on graphs inorder to provide closed-form formulas for R(G) ina variety of scenarios that extend or generalize thethree propositions above.

Foster’s Theorems

On any connected N-vertex graph the followingrelations hold: ∑

i∼j

Rij = N − 1, (5)

where i ∼ j means d(i, j) = 1, and∑ Rij

δij= N − 2. (6)

The summation in (6) runs over all pairs of adjacentedges, Rij is the effective resistance measured acrossthe endpoints i, j of those adjacent edges, and δij isthe degree of their common vertex.

The original proof of (5) can be found in [7]; thatof (6) in [8]. Several rediscoveries of these facts canbe found in [1, 9 – 11].

In [6] we used (5) in a simple proof (see [12] foran earlier proof) of a general bound for the Kirchhoffindex. Thus:

R(G) =∑i∼j

Rij +∑i6∼j

Rij ≥∑i∼j

Rij = N − 1 = R(KN).

In the previous line, the second equality is (5) ap-plied to G, and the third equality is (5) applied tothe complete graph KN.

An obvious modification of the above in the caseof 2-diameter graphs yields

R(G) =∑i∼j

Rij +∑

d(i,j) = 2

Rij = N − 1+∑

d(i,j) = 2

Rij.

The idea now is to link∑

d(i,j)= 2 Rij with (6) in orderto obtain closed-form formulas for R(G).

Geodetic Graphs

The summation in (6) is tricky: It runs over allpairs of adjacent edges, and the endpoints of theseadjacent edges may be at distance 1 or 2. If we re-quire that the graph have no triangles, then the firstpossibility disappears. Also, if we ask that the graphbe regular with common degree δ, then δij = δ forall i, j, and thus (6) becomes∑

Rij = δ(N − 2), (7)

where the summation in (7) now runs over allpairs of adjacent vertices, whose endpoints satisfyd(i, j) = 2. Still, the summation does not run over allpairs i, j of vertices such that d(i, j) = 2; in order tohave that, we need that any pair of nonneighbor-ing vertices have a unique common neighbor. Thisis exactly the definition (in the case of diameter 2)of a geodetic graph (see [13]; for an arbitrary di-ameter, the definition requires that any two verticesbe joined by a unique shortest path). Thus we haveproved the following:

Proposition 4. If G is a geodetic δ-regular graphwithout triangles, then∑

d(i,j)= 2

Rij = δ(N − 2).

If in addition the diameter of G is 2, then

R(G) = N − 1+ δ(N − 2).

One may question the usefulness of Proposi-tion 4. After all, how many such graphs are there?A partial answer to this question can be foundin [13], where a whole section is dedicated to geo-detic graphs of diameter 2. Of these, there are threeknown examples of strongly regular graphs satisfy-ing the hypotheses of Proposition 4: the pentagon[N = 5, δ = 2, R(G) = 10], the Petersen graph[N = 10, δ = 3, R(G) = 33], and the Hoffman–Singleton graph [N = 50, δ = 7, R(G) = 385].

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In addition, there are infinitely many feasible pa-rameter sets (N, δ, λ,µ), where λ is the number ofcommon neighbors of two adjacent vertices (our in-terest is that λ = 0 in order not to have triangles)and where µ is the number of common neighbors oftwo nonadjacent vertices (our interest is that µ = 1in order to be geodetic), so there is room for furtherresearch along these lines.

Symmetric Graphs

In the previous section we tried to give a closedform to the whole summation

∑d(i,j)= 2 Rij without

knowing the individual (possibly different) valuesof each summand Rij. A different strategy, which infact can be traced back to the original study [8], isto force all these summands to be equal. This can beachieved if we require the conditions under whichwe obtained (7), that is, no triangles and regularity,together with the extra condition that for any ver-tices u, v, w, z such that d(u, v) = d(w, z) = 2 thereis a graph automorphism α such that α(u) = w andα(v) = z. This is one of the conditions in the defini-tion of distance transitivity above, and it guaranteesthat Rij is constant over all i, j such that d(i, j) = 2.Then we can solve for Rij in (7) on account of the factthat there are δ(δ − 1)N/2 adjacent pairs of edges,yielding the following:

Proposition 5. If G is δ-regular, has no triangles,and (“symmetry condition”) for every u, v, w, z suchthat d(u, v) = d(w, z) = 2 there is an automorphism α

such that α(u) = w and α(v) = z, then

Rij = 2(N− 2)N(δ − 1)

, (8)

whenever d(i, j) = 2. If, in addition, the diameterof G is 2, then

R(G) = N − 1+ (N− 2)(N− δ − 1)δ − 1

. (9)

The only unchecked claim is (9), which followsfrom (8) and the fact that there are N/2(N− δ − 1)pairs of vertices at distance 2 in the 2-diametergraph.

In [1] and [8] one can find proofs of (8) with morerestrictive hypotheses, in particular the assumptionof 2-transitivity (as defined in [14]), which requiresthat every pair of adjacent edges be mapped intoany other pair via a graph automorphism. This as-sumption obviously implies our “symmetry condi-tion” above, as well as the nonexistence of triangles:

If all pairs of adjacent edges can be mapped into oneanother via a graph automorphism, then either alltheir endpoints are at distance 1 (trivial case of thecomplete graph) or all of them are at distance 2, andtherefore, there are no triangles. In [15] it is shownthat (8) holds in the case of distance-regular graphswithout triangles.

Next we give an extension of (8) in case thereare triangles, based on a previous result on hittingtimes of random walks on graphs. The hitting time(also called the passage time) Tb of the vertex b isthe number of jumps the random walk on G needsto reach b, and the expected value of Tb when thewalk is started at the vertex a is denoted by EaTb.The book of Doyle and Snell [16] is an elegant andaccessible work explaining the relationship betweenelectrical networks and random walks on graphs.A fundamental relation between hitting times andeffective resistances, proven in [17], is

EaTb + EbTa = 2|E|Rab. (10)

In case EaTb = EbTa, then from (10) we can solve forRab thus:

Rab = 1|E|EaTb. (11)

In [18] we proved with a simple conditioning argu-ment that if G is an N-vertex regular edge-transitivegraph satisfying the “symmetry condition” above,then the hitting time between vertices at distance 2is a constant E0T(2), and

E0T(2) = N − 1+ N − δ − 1δ − a3 − 1

, (12)

where a3 is the number of triangles per edge, and δis the common degree of all vertices. Now, if we puttogether (11) and (12), we obtain the following:

Proposition 6. G is an N-vertex regular edge-transitive graph satisfying the symmetry condition,then, whenever d(i, j) = 2 we have

Rij =(

2δN

)δ(N − 2)− (N − 1)a3

δ − a3 − 1, (13)

where a3 is the number of triangles per edge. If inaddition G has diameter 2, then

R(G) = N − 1+ [δ(N − 2)− (N − 1)a3](N − δ − 1)δ(δ − a3 − 1)

.

(14)

Notice that if a3 = 0, then (13) and (14) become,respectively, (8) and (9). The octahedron is a graphsatisfying the hypotheses needed to apply both (4)

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PALACIOS

and (14); the reader may check that in both cases thecomputations yield R(G) = 13

2 .

At this point, one possible direction to general-ize a result such as Proposition 6 is to add extrahypotheses on automorphisms acting on pairs ofvertices at distances 3, 4, etc., in order to get for-mulas for the effective resistances at distances 3, 4,etc., in terms of numbers of triangles, squares, etc.,per edge. This line of work leads to very intricatetechnicalities (see [18]). Rather than adding thesehypotheses one at a time, we prefer to include themall at once, that is, we go back to the scenario whereG is distance transitive, as in Proposition 1, andfind closed-form expressions for R1, R2, . . . in termsof the so-called intersection numbers. This will bedone in the next section.

Distance-Regular Graphs

Any connected graph G can be represented witha level diagram as follows: Start off with a vertex v,this unique vertex makes up level 0; then the ver-tices u on level i are those such that d(v, u) = i. Nowlook at the edges of the graph: They occur eitherbetween vertices on adjacent levels or between ver-tices within the same level. Let M be the diameterof G, let w be a vertex on level i, and define

ai = number of vertices in level i adjacent to w,0 ≤ i ≤M,

bi = number of vertices in level i+ 1 adjacent to w,0 ≤ i ≤M− 1,

ci = number of vertices in level i− 1 adjacent to w,1 ≤ i ≤M.

We say that a connected G is distance-regular if thenumbers ai, bi, ci (these are called the intersectionnumbers of G) depend only on i and not on the choiceof the level diagram or on the choice of w. Sinceai = δ − bi − ci for 0 ≤ i ≤ M, the parametersof interest are (b0, b1, . . . , bM−1; c1, c2, . . . , cM); this se-quence of numbers is called the intersection array ofthe graph. For a fixed v, let Li = |{z : d(v, z) = i}|be the size of the ith level, which is independent ofthe choice of v. Then the following is well known(see [14]):

Proposition 7. If G is distance-transitive, it is reg-ular, say of degree δ, and distance-regular, and theintersection numbers satisfy

(i) a0 = 0, b0 = δ, c1 = 1.

(ii) Li−1bi−1 = Lici, 1 ≤ i ≤M.(iii) 1 ≤ c2 ≤ c3 ≤ · · · ≤ cM.(iv) δ ≥ b1 ≥ b2 ≥ · · · ≥ bM−1.

As an illustration, for the dodecahedron the in-tersection array and the sequence (Li)0≤i≤M are, re-spectively, (3, 2, 1, 1, 1; 1, 1, 1, 2, 3) and (1, 3, 6, 6, 3, 1).Many familiar graphs are distance-regular: cycles,complete graphs, hypercubes, regular complete bi-partite graphs, and the Platonic graphs. Althoughdistance-regularity places no restrictions on the au-tomorphism group of the graph, most distance-regular graphs are also distance-transitive. A 26-vertex exception to this rule is described in [14].

Next we are going to find closed-form formu-las for the effective resistances Rij for arbitrary i, jin a distance-transitive graph. As mentioned in theIntroduction, the existence of the automorphismsguarantees that Rij depends only on the distanced(i, j), and in what follows we will use the obviousnotation R1, R2 . . . . The ideas follow closely thoseof [19], which were applied only to the case of thePlatonic graphs.

Given a distance-transitive graph, we feed a cur-rent I into a vertex v and allow the current to leaveat the remaining N − 1 vertices in equal portions ofmagnitude I/(N − 1). Then each “level,” as definedabove with respect to v, is in fact an equipotentialplane, and we can evaluate the potentials of these“planes” or “levels” as

Vi =i∑

j= 1

Ij

nj= I

i∑j= 1

1nj

(1− 1

N − 1

j−1∑k= 1

Lk

). (15)

Here, Ij denotes the total current flowing betweenthe equipotential planes j−1 and j, and it is equal tothe original current I minus all the current lost in thesuccessive equipotential planes up to j − 1; also, nj

denotes the number of resistors in parallel betweenthe equipotential planes j− 1 and j.

The effective resistance Ri between two verticesat distance i is found by superposition of the sit-uation described above with a similar situation inwhich the full current leaves the graph at one ver-tex w such that d(v, w) = i while it is fed into thegraph in N−1 equal portions of magnitude I/(N−1)each at the other vertices. The superposition entailsthat a current of magnitude I + I/(N− 1) entersthrough v and exits through w, no current enters orexits at any other vertex, and the voltage between vand w is doubled. This leads to(

I + IN − 1

)Ri = 2Vi. (16)

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Solving for Ri in (16) and using (15) yields

Ri = 2(N − 1)Vi

NI= 2

N

i∑j= 1

1nj

(N − 1−

j−1∑k= 1

Lk

).

Minor rearranging gives us then the following:

Proposition 8. For a distance-transitive graph withdiameter M

Ri = 2N

i∑j= 1

1nj

(N −

j−1∑k= 0

Lk

), (17)

for 1 ≤ i ≤M, where the number of edges nj is givenby either side of (ii) in Proposition 7:

nj = Lj−1bj−1 = Ljcj, (18)

and where Lk can be obtained from (iii) in Proposi-tion 7 as:

Lk = b0b1 · · · bk−1

c1c2 · · · ck. (19)

Thus (17), (18), and (19) give all effective resis-tances in terms of the intersection array of the graph.Biggs first obtained these results (see [15]) witha very similar method. Perhaps the exposition givenhere, in the same vein as that of [19] for Platonicgraphs, is more concise.

As an application of (17) we will generalizePropositions 2 and 3. First notice that, by adding allvertices in all levels, we get

M∑j = 1

Lj = N. (20)

Also, (17) obviously yields the recurrence

Ri − Ri−1 = 2Lici

(1− 1

N

i−1∑k= 0

Lk

). (21)

In particular, for i =M, (21) and (20) yield

RM − RM−1 = 2LMCM

(1− 1

N

M−1∑k= 0

Lk

)= 2

cMN. (22)

Now if cM = δ, then the right hand side of (22) is1/|E| and we have proved the following:

Proposition 9. Propositions 2 and 3 hold if we re-place “with opposite vertices at distance M” with“cM = δ.”

Clearly, if the graph has opposite vertices, cM = δholds. The converse is not true in general: The linegraph of the Petersen graph has intersection array

{4, 2, 1; 1, 1, 4} (see [13]) and so cM = δ holds; how-ever, for this graph there are no opposite vertices;indeed LM = 2.

Next we give a more compact and explicit formof Proposition 1. Notice that formula (3) can be con-nected to the size of the levels through the equation

n(k) = N2

Lk, (23)

and therefore, (2), (17), and (23) yield the following:

Proposition 10. If G is distance-regular with diam-eter M, then

R(G) =M∑

i= 1

Li

i∑j= 1

1Ljcj

(N −

j−1∑k= 0

Lk

), (24)

where L0 = 1 and for 1 ≤ k ≤M:

Lk = b0b1 · · · bk−1

c1c2 · · · ck.

A singularly compact form of (17) occurs for thecase i = M in graphs with opposite vertices atdistance M. Under this hypothesis, the intersectionarray is palindromic (see [13]), that is: bi = cM−i for0 ≤ i ≤ M − 1. This implies, after some algebra,that nj = nM−j+1 for 1 ≤ j ≤ M and Lk = LM−k for0 ≤ k ≤M. Then, we have

M∑j= 1

j−1∑k= 0

Lk

nj=

M∑j= 1

j−1∑k= 0

LM−k

nM−j+1=

M∑r= 1

M−r∑k= 0

LM−k

nr

=M∑

r= 1

M∑s= r

Ls

nr=

M∑j= 1

M∑k= j

Lk

nj

= 12

(M∑

j= 1

j−1∑k= 0

Lk

nj+

M∑j= 1

M∑k= j

Lk

nj

)

= 12

M∑j= 1

M∑k= 0

Lk

nj= N

2

M∑j= 1

1nj

. (25)

Inserting (25) into (17) yields

RM =M∑

i= 1

1ni

, (26)

which is the expected electrical result: If we imaginea battery connected between the opposite verticesat distance M, then all vertices within each levelare shorted (they share the same potential), andthus the graph is turned into a linear circuit (wherethe resistance of the individual resistors is added)of M resistors r1, r2, . . . , rM, where ri is made up of ni

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PALACIOS

unit resistors in parallel (thus the resistance of ri

equals 1/ni).All results concerning distance-transitive graphs

can be extended to distance-regular graphs, as isnoted by Biggs in [15]: Equations (15) for the poten-tials involve only the intersection array; thus theseequations provide “potentials” for any distance-regular (not necessarily distance-transitive) graphwith that intersection array, and since these “poten-tials” provide a solution to Laplace’s equation onthe graph, by the uniqueness of the solution, theymust be the potentials of the distance-regular graph.

Along these lines, it seems plausible that in thepreceding section, the symmetry condition can bedispensed with in Proposition 5 and replaced inProposition 6 by a condition such as “the numberof triangles per edge is constant,” but we will notpursue this issue any further.

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