aa r X i v : . [ m a t h . C O ] A p r EFFECTIVE RESISTANCES AND KIRCHHOFF INDEX OF PRISMGRAPHS
ZUBEYIR CINKIR
Abstract.
We explicitly compute the effective resistances between any two vertices ofa prism graph by using circuit reductions and our earlier findings on a ladder graph.As an application, we derived a closed form formula for the Kirchhoff index of a prismgraph. We show as a byproduct that an explicit sum formula involving trigonometricfunctions hold by comparing our formula for the Kirchhoff index and previously knownresults in the literature. We also expressed our formulas in terms of certain generalizedFibonacci numbers. Introduction
A prism graph Y n is a planar graph that looks like Figure 1.
Prismgraph Y .a circular ladder with n rungs. Figure 1 illustrates Y . Y n has 2 n vertices and 3 n edges. Each of its edgeshas length 1, so the total length of Y n is ℓ ( Y n ) := 3 n .We give explicit formulas for the effective resis-tances between any two vertices of Y n . We consider Y n as an electrical network in which we set the resis-tances along edges as the corresponding edge lengths.If we connect two ladder graphs possibly of differentvertex numbers by adding four edges to their end ver-tices, we obtain a prism graph. We apply circuit reductions to each of those ladder graphsby keeping their four end vertices. This gives us a circuit reduction of the prism graph Y n .The reduced Y n will have 8 vertices. Thanks to knowing the resistance values on a laddergraph [3], we can determine the resistance values between the vertices of the reduced Y n by utilizing the discrete Laplacian and its pseudo inverse of this reduced graph.Let us define the sequence G n by the following recurrence relation G n +2 = 4 G n +1 − G n , if n ≥
2, and G = 0, G = 1 . We showed that the following equalities hold for Kirchhoff index of Y n (see Theorem 3.1and Equation (17) below), where n is a positive integer: Kf ( Y n ) = n ( n − n G n G n − G n = n ( n − n √ h − (2 − √ n − i . Similarly, for any positive integer n , we showed that the following identities of trigono-metric sum hold (see Theorem 3.2 and Equation (17) below): n − X k =0
11 + 2 sin ( kπn ) = 2 nG n G n − G n = n √ h − (2 − √ n − i . The resistance values on Wheel and Fan graphs (in [1]) and Ladder graphs (in [3] and[4]) are expressed in terms of generalized Fibonacci numbers. Our findings for resistancevalues on a Prism graph are analogues of those results.
Key words and phrases.
Prism graph, effective resistance, Kirchhoff index, circuit reduction. Y Y Figure 2. Resistances between any pairs of vertices in Y n A ladder graph L n is a planar graph that looks like a ladder with n rungs. It has 2 n vertices and 3 n − L n is ℓ ( L n ) := 3 n −
2. We obtain the prism graph Y n from L n by adding two edges connectingthe end vertices on the same side.If we delete an edge that is a rung in Y n , and then contract two edges that are on eachside of that rung, we obtain the prism graph Y n − . If we apply this process to 3-prismgraph Y , we obtain the graph on the right in Figure 2. We call it 2-prism graph Y .Similarly, if we apply this process to Y , we obtain the graph on the left in Figure 2. Wecall it 1-prism graph Y . These graphs are the natural extension of prism graphs to thecases n = 1 ,
2. We see that our formulas for resistance values, Kirchhoff index as wellas the spanning tree formulas are also valid for these two cases (see Theorem 3.1 andEquations (10), (16) and (17) below).For a prism graph Y n , we label the vertices on the right and left as { q , q , · · · , q n } and { p , p , · · · , p n } , respectively. This is illustrated in Figure 3, where 2 ≤ i ≤ n . Wewant to find the value of r ( p, q ) for any two vertices p and q of Y n , where r ( x, y ) is theresistance function on Y n . We implement the following strategy to do this: • Consider Y n as the union of the ladder graphs L n − i +1 and L i − as illustrated inFigure 4. • Apply circuit reductions on each of these ladder graphs by keeping the four endvertices as illustrated in Figure 5. • Use our earlier results on a ladder graph in [3] to find the resistances between theend points of the reduced ladder graphs. Note that certain resistance values areequal to each other because of the symmetries in the ladder graphs. • So far we obtain the circuit reduction of Y n by keeping its 8 vertices p n , q n , p , q , p i , q i , p i − and q i − . This is illustrated in Figure 6. Find the Moore-Penroseinverse L + of the discrete Laplacian matrix L of this reduced Y n . They are 8 × • Use L + and Lemma 2.1 to find out the resistances between the 8 vertices of Y n .Again note that there are symmetries in Y n and that i is an arbitrary value in { , , . . . , n } .Symmetries in Y n gives the following identities of resistances: r ( p , p i ) = r ( q , q i ) and r ( p , q i ) = r ( q , p i ) , for each i ∈ { , . . . , n } . (1)We recall that the resistance values can be expressed in terms of the entries of thepseudo inverse of the discrete Laplacian matrix. FFECTIVE RESISTANCES OF PRISM GRAPHS 3 p p i - q i - q i q n p n p i q Y n Figure 3. p p i - q i - q i q n p n p i q L i - L n - i + Figure 4.
Lemma 2.1. [9] , [5, Theorem A] Suppose G is a graph with the discrete Laplacian L andthe resistance function r ( x, y ) . For the pseudo inverse L + of L , we have r ( p, q ) = l + pp − l + pq + l + qq , for any two vertices p and q of G. Let K = 1 + k + m + s and S = 1 + a + b + c , where k , m , s , a , b and c arethe resistance values along the edges given in Figure 6. Considering the ordering of thevertices V = { p , p i − , p i , p n , q , q i − , q i , q n } , discrete Laplacian matrix L of the graph(reduced Y n ) given in Figure 6 is as follows: L = K − m − − s − k − m K − − k − s − S − a − b − c − − a S − c − b − s − k K − m − − k − s − m K − − b − c − S − a − c − b − − a S . ZUBEYIR CINKIR p n p q q q n p n p q n L n L n uu wwtt Figure 5. p q p n q n p i - q i - q i p i L n - i + L i - Y n m mkks saa b bcc Figure 6.
Then we can compute the Moore-Penrose inverse L + of L by using [7] with the followingformula (see [12, ch 10]):(2) L + = (cid:0) L −
18 J (cid:1) − + 18 J . where J is of size 8 × L + and Lemma 2.1 to obtain r ( p , p i − ) and r ( p , q i − ). In this way, we find that r ( p , p i − ) = PQ , where P = m (4 bc ( ms + k ( m + 2 s + ms )) + a ((2 + b )(2 + c ) ms + k (2(2 + b )(2 + c ) s + m (4 + 2 c +4 s + cs + b (2 + c + s )))) + 2 a (2 c ( ms + k ( m + 2 s + ms )) + b (2(1 + c ) ms + k (4(1 + c ) s + m (2 + 2 c + 2 s + cs ))))) and Q = 2( c (2 m + k (2 + m )) + a ((2 + c ) m + k (2 + c + m )))( b (2 s + m (2 + s )) + a ((2 + b ) s + m (2 + b + s ))). r ( p , q i − ) = RS , where R = k ( c (4 c (2 ms + k ( m + s + ms )) + b (4(2 + c ) ms + k (2(2 + c ) s + m (4 + 2 c + 4 s + cs )))) + a ( b (2 + c )(2(2 + c ) ms + k ((2 + c ) s + m (2 + c + 2 s ))) + c (4(2 + c ) ms + k (2(2 + c ) s + m (4 + 2 c + 4 s + cs ))))) and S =2( c (2 m + k (2 + m )) + a ((2 + c ) m + k (2 + c + m )))( c (2 s + k (2 + s )) + b ((2 + c ) s + k (2 + c + s ))). FFECTIVE RESISTANCES OF PRISM GRAPHS 5
We note that these resistance values can be expressed in a more compact form as below: r ( p , p i − ) = 12 aca + c + mkm + k + 1 aba + b + msm + s ,r ( p , q i − ) = 12 aca + c + mkm + k + 1 bcb + c + ksk + s . (3)To find the exact values of the resistances in Equation (3), we need the correspondingvalues from the circuit reduction of L n − i +1 and L i − . Thus, we turn our attention to thecircuit reduction of L n as in Figure 5. Let us apply circuit reductions to a ladder graph L n by keeping its vertices at its bottom and top so that we obtain a complete graph on4 vertices. Suppose p , q and p n , q n are the vertices at the bottom and top of the laddergraph. This is illustrated in Figure 5. Note that by the symmetry in L n , we have onlythree distinct edge lengths in the complete graph obtained. Let us consider the orderingof the vertices { p n , q n , p , q } . Using the notations in Figure 5, the Laplacian matrix M of the reduced graph can be given as follows: M = t + u + w − w − u − t − w t + u + w − t − u − u − t t + u + w − w − t − u − w t + u + w . Then we use symmetries in L n , compute the Moore-Penrose inverse M + of M and applyLemma 2.1 to write r L n ( p n , q n ) = r L n ( p , q ) = w ( wt + u ( w + 2 t ))2( u + w )( w + t ) = 12 (cid:18) wtw + t + uwu + w (cid:19) ,r L n ( p n , p ) = r L n ( q n , q ) = u (2 wt + u ( w + t ))2( u + w )( u + t ) = 12 (cid:18) uwu + w + utu + t (cid:19) ,r L n ( p n , q ) = r L n ( q n , p ) = t (2 uw + ( u + w ) t )2( u + t )( w + t ) = 12 (cid:18) utu + t + wtw + t (cid:19) . (4)Using Equation (4), we derive A n := wtw + t = r L n ( p n , q n ) + r L n ( p n , q ) − r L n ( p n , p ) ,B n := uwu + w = r L n ( p n , q n ) − r L n ( p n , q ) + r L n ( p n , p ) ,C n := utu + t = − r L n ( p n , q n ) + r L n ( p n , q ) + r L n ( p n , p ) . (5)As particular cases of Equation (5), we have aca + c = C n − i +1 , aba + b = B n − i +1 , bcb + c = A n − i +1 , (6) mkm + k = C i − , msm + s = B i − , ksk + s = A i − , (7)where we used the notations as in Equation (3) (and so as in Figure 6). In [3], we gaveexplicit formulas for the resistance values between any two vertices of a ladder graph. ZUBEYIR CINKIR
Next, we use [3, Equation 12 at page 959] to write A n = − − √ √ − (2 − √ n ,B n = − − √ √
31 + (2 − √ n ,C n = n − . (8)We use Equations (3), (6) and (7) to derive r ( p , p i ) = 12 C n − i + C i + 1 B n − i + B i ! ,r ( p , q i ) = 12 C n − i + C i + 1 A n − i + A i ! . (9)where 2 ≤ i ≤ n . Next, we use Equation (8) in Equation (10) and then work with [7] tosimplify the algebraic expressions as below: r ( p , p i ) = 1 + (cid:0) − √ (cid:1) n − (cid:0) − √ (cid:1) n − i +1 − (cid:0) − √ (cid:1) i − √ (cid:16) − (cid:0) − √ (cid:1) n (cid:17) + ( n − i + 1)( i − n ,r ( p , q i ) = 1 + (cid:0) − √ (cid:1) n + (cid:0) − √ (cid:1) n − i +1 + (cid:0) − √ (cid:1) i − √ (cid:16) − (cid:0) − √ (cid:1) n (cid:17) + ( n − i + 1)( i − n . (10)where 1 ≤ i ≤ n .By using the symmetries of the graph Y n , we note that for every i and j in { , , . . . , n } we have r ( p , p i ) = r ( p j , p k ) and r ( p , q i ) = r ( p j , q k ) , (11)where 1 ≤ k ≤ n and k ≡ j + i − n .Finally, the explicit values of r ( p, q ) between any two vertices p and q of the prismgraph Y n can be obtained by using Equations (1), (10) and (11).It follows from Equation (10) that r ( p , p i ) + r ( p , q i ) = 1 √ (cid:16) (cid:0) − √ (cid:1) n − (cid:0) − √ (cid:1) n (cid:17) + ( n − i + 1)( i − n . (12) 3. Kirchhoff Index of Y n In this section, we obtain an explicit formula for Kirchhoff index of Y n by using ourexplicit formulas derived in § Y n .Moreover, we obtain an interesting summation formula by combining our findings andwhat is known in the literature about Kirchhoff index of Y n . Theorem 3.1.
For any positive integer n , we have Kf ( Y n ) = n ( n − n √ h − (2 − √ n − i . FFECTIVE RESISTANCES OF PRISM GRAPHS 7
Proof.
With the notation of vertices as in Figure 3 we have Kf ( Y n ) = 12 X p, q ∈ V ( Y n ) r ( p, q ) , by definition [13] . = n n X i =1 r ( p , p i ) + r ( p , q i ) , Equations (1) and (11).Then the result follows if we use first Equation (12) and do some algebra [7]. (cid:3)
Alternatively, we can express the Kirchhoff index formula in Theorem 3.1 as follows: Kf ( Y n ) = n ( n − − n √ (cid:0) n − √ (cid:1) .Kf ( Y n ) have rational values. For example, its values for 1 ≤ n ≤
10 are as follows:1, 11 /
3, 47 /
5, 58 /
3, 655 /
19, 279 /
5, 5985 /
71, 2540 /
21, 44193 / / Theorem 3.2.
For any positive integer n , we have n − X k =0
11 + 2 sin ( kπn ) = n √ h − (2 − √ n − i . Proof.
Prism graph Y n can be seen as the cartesian product P (cid:3) C n , where P is the pathgraph with 2 vertices and C n is the cycle graph with n vertices. Moreover, consideringthe Laplacian eigenvalues of P and C n we see that the Laplacian eigenvalues of Y n (see[8], [10] and [11]) are λ ij = 4 − iπ − jπn ) , where i = 0 , j = 0 , , . . . , n − . (13)We recall that [14, pg 644] n − X k =1 ( kπn ) = n − . (14)Now, we can express the Kirchhoff index via the eigenvalues of the discrete Laplacianmatrix of Y n [13]: Kf ( Y n ) = 2 n X λ ij =0 λ ij = n n − X k =1 − cos ( kπn ) + n n − X k =0 − cos ( kπn ) , by Equation (13) , = n n − X k =1
12 sin ( kπn ) + n n − X k =0
11 + 2 sin ( kπn ) , using 1 − cos ( 2 kπn ) = 2 sin ( kπn ) , = n ( n − n n − X k =0
11 + 2 sin ( kπn ) , by Equation (14) . (15)Then the proof is completed by combining Equation (15) and the result in Theorem 3.1. (cid:3) ZUBEYIR CINKIR Recursive Formulations
In this section, we give recursive formulas for the resistance values obtained in §
2, theKirchhoff index of Y n and the trigonometric formula given in Theorem 3.2. As we did in [3]for Ladder graph, we use the sequence of integers G n defined by the following recurrencerelation G n +2 = 4 G n +1 − G n , if n ≥
2, and G = 0, G = 1 . We have G n = (2 − √ − n − (2 − √ n √ , for each integer n ≥ . The sequence G n has various well-known properties [15]. For example, it gives the numberof spanning trees of L n [2], and the number of the spanning trees of the prism graph Y n [6] is given by n (cid:16) (2 + √ n + (2 − √ n − (cid:17) = n (cid:16) G n G n − (cid:17) . (16)Let g n = (2 − √ n = G n +1 − (2 −√ G n for any nonnegative integer n . Then, for any integer i ∈ { , , . . . , n } we have r ( p , p i ) = ( n − i + 1)( i − n + G n G n − G n − (cid:16) √ G n G n − G n (cid:17) ( g n − i +1 + g i − ) ,r ( p , q i ) = ( n − i + 1)( i − n + G n G n − G n + (cid:16) √ G n G n − G n (cid:17) ( g n − i +1 + g i − ) . Similarly, the other resistance values can be expressed in terms of G n by using the sym-metries in Y n like Equation (1) and Equation (11).Here are how we can express the results given in Theorem 3.1 and Theorem 3.2 in termsof G n : Kf ( Y n ) = n ( n − n G n G n − G n and n − X k =0
11 + 2 sin ( kπn ) = 2 nG n G n − G n . (17) Acknowledgements:
This work is supported by Abdullah Gul University Foundationof Turkey.
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Zubeyir Cinkir, Department of Industrial Engineering, Abdullah G¨ul University, Kay-seri, TURKEY.
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