In this paper, we find the polynomials, indices and average distance for Schultz and modified Schultz of vertex identification chain for 4-cycle and 4- cycle complete.
Given the importance of topological evidence, which can be deduced from the polynomial by finding the derivative with respect to a specific variable and then compensating for this variable by one value, therefore we have in this paper found the polynomial of Schultz and modified Schultz for a chain of special graphs which is the 4-cycle and 4-cycle complete by identification symmetrical vertices.
In this paper, we can refer to the basic concepts in graph theory and topological indices to the references [
Introduced Schultz index by Schultz in 1989 [
S c ( G ) = ∑ { u , v } ⊆ V ( G ) ( d e g v + d e g u ) d ( u , v ) . (1.1)
S c ∗ ( G ) = ∑ { u , v } ⊆ V ( G ) ( d e g v ⋅ d e g u ) d ( u , v ) . (1.2)
The Schultz and modified Schultz polynomials are have defined respectively as:
S c ( G ; x ) = ∑ { u , v } ⊆ V ( G ) ( d e g v + d e g u ) x d ( u , v ) . (1.3)
S c ∗ ( G ; x ) = ∑ { u , v } ⊆ V ( G ) ( d e g v ⋅ d e g u ) x d ( u , v ) . (1.4)
From these polynomials can obtain:
1) The Schultz index:
S c ( G ) = d d x ( S c ( G ; x ) ) | x = 1 . (1.5)
2) The modified Schultz index:
S c * ( G ) = d d x ( S c * ( G ; x ) ) | x = 1 . (1.6)
3) The average distance of Schultz:
S c ¯ ( G ) = 2 S c ( G ) / p ( G ) ( p ( G ) − 1 ) (1.7)
4) The average distance of modified Schultz:
S c * ¯ ( G ) = 2 S c * ( G ) / p ( G ) ( p ( G ) − 1 ) . (1.8)
where d d x is represent the derivative w.r.t. x and p ( G ) is the order of G.
There are many recent papers on polynomials and indices for Schultz and modified Schultz, see to references [
Let D k ( r , h ) be the set of all ( u , v ) of G which distance between u and v is k such that d e g u = r and d e g v = h and | D k ( G ) | is the number of pairs ( u , v ) of G that are distance k “ D ( G , k ) ”.
From clearly that ∑ k = 1 d i a m ( G ) | D k ( G ) | = p ( G ) ( p ( G ) − 1 ) / 2 .
Let { G 1 , G 2 , ⋯ , G n } be a set of pairwise disjoint graphs with vertices u i , v i ∈ V ( G i ) , i = 1 , 2 , ⋯ , n , n ≥ 2 , then the vertex-identification chain graph C v ( G 1 , G 2 , ⋯ , G n ) ≡ C v ( G 1 , G 2 , ⋯ , G n : v 1 ⋅ u 2 ; v 2 ⋅ u 3 ; ⋯ ; v n − 1 ⋅ u n ) of { G i } i = 1 n with respect to the vertices { v i , u i + 1 } i = 1 n − 1 is the graph obtained from the graphs G 1 , G 2 , ⋯ , G n by identifying the vertex v i with the vertex u i + 1 for all i = 1 , 2 , ⋯ , n − 1 . (See
Some Properties of Graph C v ( G 1 , G 2 , ⋯ , G n ) :
1) p ( C v ( G 1 , G 2 , ⋯ , G n ) ) = ∑ i = 1 n p ( G i ) − ( n − 1 ) .
2) q ( C v ( G 1 , G 2 , ⋯ , G n ) ) = ∑ i = 1 n q ( G i ) .
3) n ≤ d i a m ( C v ( G 1 , G 2 , ⋯ , G n ) ) ≤ ∑ i = 1 n d i a m ( G i ) .
The equality of lower bound is satisfied at K 2 but the upper bound is satisfied at path graph.
If G i ≡ H p , for all 1 ≤ i ≤ n , where H p is a connected graph of order p, we denoted C v ( H p , H p , ⋯ , H p ) by C v ( H p ) n .
From
| + × | d e g u i = 2 | d e g v i = 2 | d e g w 1 = 2 | d e g w h = 4 | d e g w p + 1 = 2 |
|---|---|---|---|---|---|
| d e g u j = 2 | 4 4 | 4 4 | 4 4 | 6 8 | 4 4 |
| d e g v j = 2 | 4 4 | 4 4 | 4 4 | 6 8 | 4 4 |
| d e g w 1 = 2 | 4 4 | 4 4 | 6 8 | 4 4 | |
| d e g w k = 4 | 6 8 | 6 8 | 6 8 | 8 16 | 6 8 |
| d e g w p + 1 = 2 | 4 4 | 4 4 | 4 4 | 6 8 |
Theorem 2.1.1: For p ≥ 2 , then we have:
1) S c ( C v ( C 4 ) p ; x ) = 8 ( 3 p − 1 ) x + 4 ( 7 p − 5 ) x 2 + 4 ∑ k = 3 2 p ( 6 p − 3 k + 1 ) x k .
2) S c * ( C v ( C 4 ) p ; x ) = 16 ( 2 p − 1 ) x + 4 ( 9 p − 8 ) x 2 + 16 ∑ k = 3 2 p − 1 ( 2 p − k ) x k + 4 x 2 p .
Proof: For all p ≥ 4 and every two vertices u , v ∈ V ( C v ( C 4 ) p ) , there is
d ( u , v ) = k , 1 ≤ k ≤ 2 p , then obviously, ∑ i = 1 2 p | D i | = 3 p ( 3 p + 1 ) 2 . We will have
six partitions for proof:
P1. If d ( u , v ) = 1 , then | D 1 | = 4 p = q ( C v ( C 4 ) p ) and we have two subsets of it:
P1.1. | D 1 ( 2 , 2 ) | = | { ( u 1 , w 1 ) , ( u p , w p + 1 ) , ( v 1 , w 1 ) , ( v 1 , w p ) } | = 4 .
P1.2. | D 1 ( 2 , 4 ) | = | { ( u i − 1 , w i ) , ( v i − 1 , w i ) , ( u i , w i ) , ( v i , w i ) : 2 ≤ i ≤ p } | = 4 ( p − 1 ) .
P2. If d ( u , v ) = 2 , then | D 2 | = 6 p − 4 and we have three subsets of it:
P2.1. | D 2 ( 2 , 2 ) | = | { ( u i , u i + 1 ) , ( u i , v i + 1 ) , ( v i , v i + 1 ) , ( v i , u i + 1 ) : 1 ≤ i ≤ p − 1 } ∪ { ( u i , v i ) : 1 ≤ i ≤ p } | = 5 p − 4.
P2.2. | D 2 ( 2 , 4 ) | = | { ( w 1 , w 2 ) , ( w p + 1 , w p ) } | = 2 .
P2.3. | D 2 ( 4 , 4 ) | = | { ( w i , w i + 1 ) : 2 ≤ i ≤ p − 1 } | = p − 2 .
P3. If d ( u , v ) = k , then | D k | = 9 p + 1 2 k + 3 , and we have:
1) If k is an odd, 3 ≤ k ≤ 2 p − 3 , then | D k | = 4 p − 2 k + 2 , and we have two subsets of it:
P3.1. | D k ( 2 , 2 ) | = | { ( w 1 , u k − k − 1 2 ) , ( w 1 , v k − k − 1 2 ) , ( w 1 , u k − k − 1 2 ) , ( w 1 , v k − k − 1 2 ) } | = 4 .
P3.2. | D k ( 2 , 4 ) | = | { ( u i + k − 1 2 , w i ) , ( v i + k − 1 2 , w i ) , ( u i − 1 , w i + k − 1 2 ) , ( v i − 1 , w i + k − 1 2 ) : 2 ≤ i ≤ p − k − 1 2 } | = 4 ( p − k − 1 2 − 1 ) .
2) If k is an even, 4 ≤ k ≤ 2 p − 4 , then | D k | = 5 p + 5 2 k + 1 , and we have three
subsets of it:
P3.3. | D k ( 2 , 2 ) | = | { ( u i , u i + k 2 ) , ( u i , v i + k 2 ) , ( v i , v i + k 2 ) , ( v i , u i + k 2 ) : 1 ≤ i ≤ p − k 2 } | = 4 ( p − k 2 ) .
P3.4. | D k ( 2 , 4 ) | = | { ( w 1 , w 1 + k 2 ) , ( w p + 1 , w p − k 2 + 1 ) } | = 2 .
P3.5. | D k ( 4 , 4 ) | = | { ( w i , w i + k 2 ) : 2 ≤ i ≤ p − k 2 } | = p − k 2 − 1 .
P4. If d ( u , v ) = 2 p − 2 , then | D 2 p − 2 | = 6 , and we have two subsets of it:
P4.1. | D 2 p − 2 ( 2 , 2 ) | = | { ( u 1 , u p ) , ( v 1 , v p ) , ( u 1 , v p ) , ( v 1 , u p ) } | = 4 .
P4.2. | D 2 p − 2 ( 2 , 4 ) | = | { ( w 1 , w p ) , ( w p + 1 , w 2 ) } | = 2 .
P5. If d ( u , v ) = 2 p − 1 , then | D 2 p − 1 | = 4 , and we have one subset of it:
| D 2 p − 1 ( 2 , 2 ) | = | { ( w 1 , u p ) , ( w 1 , v p ) , ( w p + 1 , u 1 ) , ( w p + 1 , v 1 ) } | = 4 .
P6. If d ( u , v ) = 2 p , then | D 2 p | = 1 , and we have one subset of it:
| D 2 p ( 2 , 2 ) | = | { ( w 1 , w p + 1 ) } | = 1 .
Hence, from P1 to P6 and
S c ( C v ( C 4 ) p ; x ) = { 4 ( 4 ) + 6 ( 4 ( p − 1 ) ) } x + { 4 ( 5 p − 4 ) + 6 ( 2 ) + 8 ( p − 2 ) } x 2 + { ∑ k = 3 2 p − 3 { 4 ( 4 ) + 6 ( 4 ( p − k − 1 2 − 1 ) ) } x k , k is an odd ∑ k = 4 2 p − 4 { 4 ( 4 ( p − k 2 ) ) + 6 ( 2 ) + 8 ( p − k 2 − 1 ) } x k , k is an even + { 4 ( 4 ) + 6 ( 2 ) } x 2 p − 2 + { 4 ( 4 ) } x 2 p − 1 + { 4 ( 1 ) } x 2 p = 8 ( 3 p − 1 ) x + 4 ( 7 p − 5 ) x 2 + 4 ∑ k = 3 2 p ( 6 p − 3 k + 1 ) x k .
And,
S c * ( C v ( C 4 ) p ; x ) = { 4 ( 4 ) + 8 ( 4 ( p − 1 ) ) } x + { 4 ( 5 p − 4 ) + 8 ( 2 ) + 16 ( p − 2 ) } x 2 + { ∑ k = 3 2 p − 3 { 4 ( 4 ) + 8 ( 4 ( p − k − 1 2 − 1 ) ) } x k , k is an odd , ∑ k = 4 2 p − 4 { 4 ( 4 ( p − k 2 ) ) + 8 ( 2 ) + 16 ( p − k 2 − 1 ) } x k , k is an even + { 4 ( 4 ) + 8 ( 2 ) } x 2 p − 2 + { 4 ( 4 ) } x 2 p − 1 + { 4 ( 1 ) } x 2 p = 16 ( 2 p − 1 ) x + 4 ( 9 p − 8 ) x 2 + 16 ∑ k = 3 2 p − 1 ( 2 p − k ) x k + 4 x 2 p .
By simple, we can calculate:
1) S c ( C v ( C 4 ) 2 ; x ) = 40 x + 36 x 2 + 16 x 3 + 4 x 4 .
S c * ( C v ( C 4 ) 2 ; x ) = 48 x + 40 x 2 + 16 x 3 + 4 x 4 .
2) S c ( C v ( C 4 ) 3 ; x ) = 64 x + 64 x 2 + 40 x 3 + 28 x 4 + 16 x 5 + 4 x 6 .
S c * ( C v ( C 4 ) 3 ; x ) = 80 x + 76 x 2 + 48 x 3 + 32 x 4 + 16 x 5 + 4 x 6 . ∎
Corollary 2.1.2: For p ≥ 2 , then we have:
1) S c ( C v ( C 4 ) p ) = 8 p ( 2 p 2 + p + 1 ) .
2) S c * ( C v ( C 4 ) p ) = 32 3 p ( 2 p 2 + 1 ) . ∎
Corollary 2.1.3: For p ≥ 2 , then we have:
1) S c ¯ ( C v ( C 4 ) p ) = 16 27 ( 6 p + 1 + 8 / ( 3 p + 1 ) ) .
2) S c * ¯ ( C v ( C 4 ) p ) = 128 81 ( 3 p − 1 + 11 / 2 ( 3 p + 1 ) ) . ∎
From
| × + | d e g u i = 3 | d e g v i = 3 | d e g w 1 = 3 | d e g w h = 6 | d e g w p + 1 = 3 |
|---|---|---|---|---|---|
| d e g u j = 3 | 9 6 | 9 6 | 9 6 | 18 9 | 9 6 |
| d e g v j = 3 | 9 6 | 9 6 | 9 6 | 18 9 | 9 6 |
| d e g w 1 = 3 | 9 6 | 9 6 | 18 9 | 9 6 | |
| d e g w k = 6 | 18 9 | 18 9 | 18 9 | 36 12 | 18 9 |
| d e g w p + 1 = 3 | 9 6 | 9 6 | 9 6 | 18 9 |
Theorem 2.2.1: for p ≥ 3 , then we have:
1) S c ( C v ( K 4 ) p ; x ) = 18 ( 3 p − 1 ) x + 18 ∑ k = 2 p ( 4 p − 4 k + 3 ) x k .
2) S c * ( C v ( K 4 ) p ; x ) = 9 ( 13 p − 8 ) x + 72 ∑ k = 2 p − 1 ( 2 p − 2 k + 1 ) x k + 81 x p .
Proof: For all p ≥ 4 , and every vertex u , v ∈ V ( C v ( K 4 ) p ) there is
d ( u , v ) = k , 1 ≤ k ≤ p , then obviously, ∑ i = 1 2 p | D i | = 3 p ( 3 p + 1 ) 2 . We will have
four partitions for proof:
P1. If d ( u , v ) = 1 , then | D 1 | = 6 p = q ( C v ( K 4 ) p ) and we have three subsets of it:
P1.1. | D 1 ( 3 , 3 ) | = | { ( u i , v i ) : 1 ≤ i ≤ p } ∪ { ( u 1 , w 1 ) , ( v 1 , w 1 ) , ( u p , w p + 1 ) , ( v p , w p + 1 ) } | = p + 4.
P1.2. | D 1 ( 3 , 6 ) | = | { ( u i , w i ) , ( u i , w i + 1 ) , ( v i , w i ) , ( v i , w i + 1 ) : 2 ≤ i ≤ p − 1 } ∪ { ( u p , w p ) , ( u 1 , w 2 ) , ( v p , w p ) , ( v 1 , w 2 ) , ( w 1 , w 2 ) , ( w p , w p + 1 ) } | = 4 p − 2.
P1.3. | D 1 ( 6 , 6 ) | = | { ( w i , w i + 1 ) : 2 ≤ i ≤ p − 1 } | = p − 2 .
P2. If d ( u , v ) = k , 2 ≤ k ≤ p − 2 , then | D k | = 9 ( p − k + 1 ) and we have three subsets of it:
P2.1
| D k ( 3 , 3 ) | = | { ( u i , u i + k − 1 ) , ( v i , v i + k − 1 ) , ( u i , v i + k − 1 ) , ( v i , u i + k − 1 ) : 1 ≤ i ≤ p − k + 1 } ∪ { ( u k , w 1 ) , ( u p − k + 1 , w p + 1 ) , ( v k , w 1 ) , ( v p − k + 1 , w p + 1 ) } | = 4 ( p − k + 2 ) .
P2.2.
| D k ( 3 , 6 ) | = | { ( u i , w i + k ) , ( u i + k − 1 , w i ) , ( v i , w i + k ) , ( v i + k − 1 , w i ) : 2 ≤ i ≤ p − k } ∪ { ( u 1 , w 1 + k ) , ( u p , w p − k + 1 ) , ( v 1 , w 1 + k ) , ( v p , w p − k + 1 ) , ( w 1 , w 1 + k ) , ( w p + 1 , w p − k + 1 ) } | = 2 ( 2 p − 2 k + 1 ) .
P2.3. | D k ( 6 , 6 ) | = | { ( w i , w i + k ) : 2 ≤ i ≤ p − k } | = p − k − 1 .
P3. If d ( u , v ) = p − 1 , then | D p − 1 | = 18 and we have two subsets of it:
P3.1. | D p − 1 ( 3 , 3 ) | = { { ( u i , u i + p − 2 ) , ( u i , v i + p − 2 ) , ( v i , v i + p − 2 ) , ( v i , u i + p − 2 ) : i = 1 , 2 } ∪ { ( u 2 , w p − 1 ) , ( u p − 1 , w 1 ) , ( v 2 , w p + 1 ) , ( v p − 1 , w 1 ) } | = 12.
P3.2. | D p − 1 ( 3 , 6 ) | = | { ( u 1 , w p ) , ( u p , w 2 ) , ( v p , w 2 ) , ( v 1 , w p ) , ( w 1 , w p ) , ( w 1 , w p ) } | = 6 .
P4. If d ( u , v ) = p , then | D p | = 9 , and we have one subset of it:
| D p ( 3 , 3 ) | = | { ( u 1 , u p ) , ( u 1 , w p + 1 ) , ( u 1 , v p ) , ( v 1 , w p + 1 ) , ( v 1 , v p ) , ( u p , w 1 ) , ( u p , v 1 ) , ( w 1 , w p + 1 ) , ( w 1 , v p ) } | = 9
Hence, from P1 to P4 and
S c ( C v ( K 4 ) p ; x ) = { 6 ( p + 4 ) + 9 ( 4 p − 2 ) + 12 ( p − 2 ) } x + ∑ k = 2 p − 2 { 6 ( 4 p − 4 k + 8 ) + 9 ( 4 p − 4 k + 2 ) + 12 ( p − k − 1 ) } x k + { 6 ( 12 ) + 9 ( 6 ) } x p − 1 + { 6 ( 9 ) } x p = 18 ( 3 p − 1 ) x + 18 ∑ k = 2 p ( 4 p − 4 k + 3 ) x k .
And,
S c * ( C v ( K 4 ) p ; x ) = { 9 ( p + 4 ) + 18 ( 4 p − 2 ) + 36 ( p − 2 ) } x + ∑ k = 2 p − 2 { 9 ( 4 p − 4 k + 8 ) + 18 ( 4 p − 4 k + 2 ) + 36 ( p − k − 1 ) } x k + { 9 ( 12 ) + 18 ( 6 ) } x p − 1 + { 9 ( 9 ) } x p = 9 ( 13 p − 8 ) x + 72 ∑ k = 2 p − 1 ( 2 p − 2 k + 1 ) x k + 81 x p .
By simply we can calculate:
S c ( C v ( K 4 ) 3 ; x ) = 144 x + 126 x 2 + 54 x 2 .
S c ( C v ( K 4 ) 3 ; x ) = 279 x + 216 x 2 + 81 x 2 .
This completes the proof. ∎
Remark:
1) S c ( C v ( K 4 ) 2 ; x ) = 90 x + 54 x 2 .
2) S c * ( C v ( K 4 ) 2 ; x ) = 162 x + 81 x 2 .
Corollary 2.2.2: For p ≥ 2 , then we have:
1) S c ( C v ( K 4 ) p ) = 3 p ( 4 p 2 + 9 p − 1 ) .
2) S c * ( C v ( K 4 ) p ) = 6 p ( 4 p 2 + 6 p − 1 ) . ∎
Corollary 2.2.3: For p ≥ 2 , then we have:
1) S c ¯ ( C v ( K 4 ) p ) = 2 9 ( 12 p + 23 − 32 / ( 3 p + 1 ) ) .
2) S c * ¯ ( C v ( K 4 ) p ) = 8 9 ( 6 p + 7 − 23 / 2 ( 3 p + 1 ) ) . ∎
A finite sequence ( a 1 , a 2 , ⋯ , a h ) of h positive integers is coefficients of polynomial P ( x ) = ∑ i = 1 h a i x i . Then:
1) The polynomial P ( x ) is called j-unimodal if, for some index j, a 1 ≤ a 2 ≤ ⋯ ≤ a j ≥ a j + 1 ≥ ⋯ ≥ a h and it is strictly j-unimodal if the inequality holds without equalities.
2) The polynomial P ( x ) is called monotonically increasing (or monotonically decreasing) if, a i ≤ a i + 1 or a i ≥ a i + 1 , respectively, for all 1 ≤ i ≤ h and it is strictly-increasing or strictly-decreasing respectively if the inequalities hold without equalities.
3) The polynomial P ( x ) is called palindromic if a i = a h − i + 1 , for all 1 ≤ i ≤ h and is called semi-palindromic if a j = a h − j + 1 , 1 + i ≤ j ≤ h − i and for all 1 ≤ i ≤ h − 2 .
4) The polynomial P ( x ) is called troubled if a i ≠ a i + 1 , for all 1 ≤ i ≤ h .
5) The polynomial P ( x ) is called equality if a i = a i + 1 , for all 1 ≤ i ≤ h and is called semi-equality if a i = a i + 1 for some values i.
The following
In this paper, we obtained the general formulas of polynomials for Schultz and modified Schultz polynomials of operation vertex identification chain for square and complete square graphs and indices of its. Also, we discuss some properties of the coefficients of these polynomials.
| Property 5 | Property 4 | Property 3 | Property 2 | Property 1 | Some Properties Polynomials of types graphs |
|---|---|---|---|---|---|
| Not satisfy | Satisfy | Not satisfy | Not satisfy | Satisfy at j = 2 | S c ( C v ( C 4 ) p ; x ) |
| Not satisfy | Satisfy | Not satisfy | Not satisfy | Satisfy at j = 2 | S c * ( C v ( C 4 ) p ; x ) |
| Not satisfy | Satisfy | Not satisfy | Not satisfy | Satisfy at j = 2 | S c ( C v ( K 4 ) p ; x ) |
| Not satisfy | Satisfy | Not satisfy | Not satisfy | Satisfy at j = 2 | S c * ( C v ( K 4 ) p ; x ) |
This paper is has supported from Master’s thesis for the student Mahmood M. A. from college computer sciences and mathematics, University of Mosul, Republic of Iraq.
The authors declare no conflicts of interest regarding the publication of this paper.
Abdullah, M.M. and Ali, A.M. (2020) Schultz and Modified Schultz Polynomials of Vertex Identification Chain for Square and Complete Square Graphs. Open Access Library Journal, 7: e6309. https://doi.org/10.4236/oalib.1106309