Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2315 https://internationalpubls.com A Study on Perfect Rings Dominating Energy of Graphs Prema M1, Ruby Selestina M2 and Purushothama S3 1Research Scholar, Department of Mathematics, Yuvaraja’s College, University of Mysore, Mysuru 2 Associate Professor, Department of Mathematics, Yuvaraja’s College, University of Mysore, Mysuru 3Associate Professor, Department of Mathematics, MIT Mysore, Mysuru Email: mpremamallaiah@gmail.com ruby.salestina@gmail.com psmandya@gmail.com Article History: Received: 12-01-2025 Revised: 15-02-2025 Accepted: 01-03-2025 Abstract: A novel parameter called perfect rings domination has been developed in this study. The corresponding perfect rings dominating matrix is also generated. The energy of graphs is the sum of absolute value of their spectrum. This study investigates the spectrum and energy of certain family of graphs as well as ϑ-obrazom of graphs corresponding to this matrix. Moreover upper and lower bound are also established. Keywords: PR Domination, PRD spectrum, PRD Energy 1. Introduction Graphs with no isolated vertex are considered for this study. Graph energy is a significant topological index in the domain of chemical graph theory which can be used in chemistry. The absicth of graph energy was debuted by I. Gutman [9] in 1978 as summation of modulus values of spectrum of G, corresponds to adjacency matrix. A vast study of utilization on graph energy was pursued by I. Gutman and Balakrishnan [2, 10]. The consummation of energy and the equel with bounds can be beholded in vigous scrutinize of graph energy [12,13]. The necessary properties and vital chemical utilizations were evaluated in the molecular orbital theory of conjugated molecules [3-6]. For graph theoretic parlance one may refer Harary [11]. Let S ⊆ V. S is a dominating set if all v ∈ V − S has a neighbor in S, minimum cardinality among such sets is called a minimum dominating set. A dominating set S is perfect if all v in G is dominated by strictly one element of S [13]. A dominating set S is rings domination if each v in V – S is adjacent to minimum two elements in V – S [1]. A dominating set A is a perfect rings dominating (PRD) set if (i) every v ∈ G is dominated by strictly one element of A (ii) ∀ v ∈ V\A, |N(v)⋂(V\A)| ≥ 2. PRD set with minimum cardinality is the minimum PRD set of G and notate minimum PRD number by ‘𝔭’. This paper scrutinize the perfect rings dominating spectrum (say 𝔓-spectrum) and perfect rings dominating energy (say 𝔓𝔈) of few classes of graphs as well as derive some bounds on 𝔓𝔈. 1.1 Lemma [14]: Let B = [ B0 B1 B1 B0 ] be a symmetric block matrix has order 2 with B0 and B1 are square matrices of same order. Then spectrum of B is the union of spectrum of B0 + B1 and B0 − B1. 2. PERFECT RINGS DOMINATING ENERGY 2.1 Definition: ConsiderG = (p, q). Let A ⊆ V(G) be its minimum perfect rings dominating (PRD) set. Then the perfect rings dominating matrix of G corresponding to A is a matrix 𝔓A has order p, defined as 𝔓A = { 1 if vivj ∈ E 1 if i = j, vi ∈ A 0 otherwise mailto:mpremamallaiah@gmail.com mailto:ruby.salestina@gmail.com mailto:psmandya@gmail.com Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2316 https://internationalpubls.com The characteristic polynomial of 𝔓A is φ(𝔓A, λ) = det (𝔓A − λI). The 𝔓-spectrum of G is the eigenvalues of the matrix𝔓A. Let λ1, λ2, … λp be the spectrum of 𝔓A. Then the perfect rings dominating energy 𝔓𝔈 of G corresponding to A is defined as 𝔓𝔈A(G) = ∑ |λi| p i=1 . Let λ1, λ2, λ3, … λp be the spectrum of 𝔓A and they can be notated as SpecA(G) = { λ1 λ2 λ3 … λp m1 m2 m3 … mp } where mi is the algebraic multiplicity of eigenvaluesλi, for 1 ≤ i ≤ p. 2.2 Remark: Though all the minimum PRD sets are of same cardinality, the perfect rings dominating energy 𝔓𝔈(G) need not be same for all PRD set. 2.3 Remark: If G has a unique minimum PRD set, then 𝔓𝔈A(G) can be denoted as 𝔓𝔈(G). 3. PRD ENERGY OF GRAPHS 3.1 Theorem: For any complete graph Kp with p ≥ 3, 𝔓𝔈A(Kp) = p − 2 + √p(p − 2) + 5 Proof: Let V(Kp) = {v1, v2, . . . , vp}. Any arbitrary element of G will be a PRD-set. Consider A = {v1} as a PRD-set. Therefore the PRD-matrix has the structure 𝔓A(Kp) = ( 1 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) Where J and I represents the matrix of 1’s and identity matrix. And the corresponding characteristic polynomial is φ(Kp, λ) = det (𝔓A(Kp) − λI) ⟹ φ(Kp, λ) = (−1) p(λ + 1)p−2(λ2 − (p − 1)λ − 1) Then, SpecA(Kp) = { −1 (p − 2) + √p2 − 2p + 5 2 (p − 2) − √p2 − 2p + 5 2 p − 2 1 1 } Thus the PRD-energy is 𝔓𝔈A(Kp) = (p − 2)|−1| + | (p − 2) + √p2 − 2p + 5 2 | + | (p − 2) − √p2 − 2p + 5 2 | = p − 2 + √p2 − 2p + 5 𝔓𝔈A(Kp) = p − 2 + √p(p − 2) + 5 3.2 Theorem: For Kp,q with p, q ≥ 2, 𝔓𝔈A(Kp,q) = { √(p − 1)2 + 3+ √(p + 1)2 − 4 if p = q 2(1 + √pq) if p < q & if p > q Proof: Let V(Kp,q) = {v1, . . . , vp, v1′, . . . , vq′}. Consider A = {v1, v1′} as a PRD-set. Therefore the PRD-matrix has the structure 𝔓A(Kp,q) = ( 1 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) i) Suppose p = q, then 𝔓A(Kp,p) has the structure 𝔓A(Kp,p) = ( B0 B1 B1 B0 ) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2317 https://internationalpubls.com Where B0 = ( 1 O1×p−1 Op−1×1 Op−1 ) and B1 = (Jp), here J and O represents the matrix of 1’s and 0’s. Therefore by Lemma 1.1, SpecA(Kp,p) = Spec(B0 + B1)⋃ Spec(B0 − B1). Consider, B0 + B1: B0 + B1 = ( 2 J1×p−1 Jp−1×1 Jp−1 ) |(B0 + B1) − λI| = (−1) pλp−2(λ2 − (p + 1)λ + (p − 1)) Therefore, SpecA(B0 + B1) = { 0 (p + 1) + √(p − 1)2 + 3 2 (p + 1) − √(p − 1)2 + 3 2 p − 2 1 1 } Consider, B0 − B1: B0 − B1 = ( 0 −J1×p−1 −Jp−1×1 −Jp−1 ) |(B0 − B1) − λI| = (−1) pλp−2(λ2 + (p − 1)λ − (p − 1)) Therefore, SpecA(B0 − B1) = { 0 (1 − p) + √(p + 1)2 − 4 2 (1 − p) − √(p + 1)2 − 4 2 p − 2 1 1 } Hence SpecA(Kp,p) = { 0 (p + 1) + √(p − 1)2 + 3 2 (p + 1) − √(p − 1)2 + 3 2 2p − 4 1 1 (1 − p) + √(p + 1)2 − 4 2 (1 − p) − √(p + 1)2 − 4 2 1 1 } Now, 𝔓𝔈A(Kp,p) = | (p + 1) + √(p − 1)2 + 3 2 | + | (p + 1) − √(p − 1)2 + 3 2 | + | (1 − p) + √(p + 1)2 − 4 2 | + | (1 − p) − √(p + 1)2 − 4 2 | Thus, 𝔓𝔈A(Kp,p) = 2√(p − 1)2 + 3 2 + 2√(p + 1)2 − 4 2 𝔓𝔈A(Kp,p) = √(p − 1)2 + 3 +√(p + 1)2 − 4 ii) Suppose p > q, then the PRD matrix is of the form 𝔓A(Kp,q) = ( A B C D ) where A = ( 1 O1×p−1 Op−1×1 Op−1 ); B = (Jp×q) C = (Jq×p); D = ( 1 O1×q−1 Oq−1×1 Oq−1 ), here J and O represents the matrix of 1’s and 0’s. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2318 https://internationalpubls.com And the corresponding characteristic polynomial is, φ(Kp,q, λ) = det(𝔓A(Kp,q) − λI) φ(Kp,q, λ) = (−1) pλp+q−4 [λ4 − 2λ3 − (pq − 1)λ2 + (2pq − (p + q))λ − [pq − (p + q) + 1]] ≈ λp+q−4(λ − 1)[λ3 − λ2 − pqλ − (pq− (p + q))] φ(Kp,q, λ) ≈ λ p+q−4(λ − 1)2(λ2 − pq) iii) Suppose p > q, then the PRD matrix is of the form 𝔓A(Kp,q) = ( P Q R S ) where P = ( 1 O1×p−1 Op−1×1 Op−1 ); Q = (Jp×q) R = (Jq×p); S = ( 1 O1×q−1 Oq−1×1 Oq−1 ) here J and O represents the matrix of 1’s and 0’s. Therefore the corresponding characteristic polynomial is, φ(Kp,q, λ) = det(𝔓A(Kp,q) − λI) φ(Kp,q, λ) = (−1) pλp+q−4 [λ4 − 2λ3 − (pq − 1)λ2 + (2pq − (p + q))λ − [pq − (p + q) + 1]] ⟹φ(Kp,q, λ) ≈ λ p+q−4(λ − 1)2(λ2 − pq) For the both cases ii), iii) the characteristic polynomial remains same. Hence, SpecA(Kp,q) = { 0 1 √pq −√pq p + q − 4 2 1 1 } 𝔓𝔈A(Kp,p) = 2(1 + √pq) 3.3 Theorem: For a crown graph Sp 0 with p ≥ 2, 𝔓𝔈A(Sp 0) = 2(p − 2) + √p2 − 2p + 5 + √p2 + 2p − 3 Proof: Let V(Sp 0) = {v1, . . . , vp , v1′, . . . , vp′}, let A = {v1, v1′} be a PRD set. Then 𝔓A(Sp 0) has the form ( B0 B1 B1 B0 ), where B0 = ( 1 O1×p−1 Op−1×1 Op−1 ) and B1 = (Jp − Ip), here J, O and I represents the matrix of 1’s, 0’s and identity matrix. Therefore by Lemma 1.1, SpecA(Sp 0) = Spec(B0 + B1)⋃Spec(B0 − B1). Consider, B0 + B1: B0 + B1 = ( 1 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) |(B0 + B1) − λI| = (−1) p(λ + 1)p−2(λ2 − (p − 1)λ − 1) Therefore, SpecA(B0 + B1) = { −1 (p − 1) + √p2 − 2p + 5 2 (p − 1) − √p2 − 2p + 5 2 p − 2 1 1 } Consider, B0 − B1: B0 − B1 = ( 1 −J1×p−1 −Jp−1×1 −(Jp−1 − Ip−1) ) |(B0 − B1) − λI| = (−1) p(λ − 1)p−2(λ2 + (p − 3)λ − (2p − 3)) Therefore, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2319 https://internationalpubls.com SpecA(B0 − B1) = { 1 (3 − p) + √p2 + 2p − 3 2 (3 − p) − √p2 + 2p − 3 2 p − 2 1 1 } Hence SpecA(Sp 0) = { −1 1 (p − 1) + √p2 − 2p + 5 2 (p − 1) − √p2 − 2p + 5 2 p − 2 p − 2 1 1 (3 − p) + √p2 + 2p − 3 2 (3 − p) − √p2 + 2p − 3 2 1 1 } Now, 𝔓𝔈A(Sp 0) = (p − 2)|−1| + (p − 2)|1| + | (p − 1) + √p2 − 2p + 5 2 | + | (p − 1) − √p2 − 2p + 5 2 | + | (3 − p) + √p2 + 2p − 3 2 | + | (3 − p) − √p2 + 2p − 3 2 | Thus, 𝔓𝔈A(Sp 0) = 2(p − 2) + √p2 − 2p + 5 + √p2 + 2p − 3 3.4 Theorem: For a barbell graph Bp,p with p > 3, 𝔓𝔈A(Bp,p) = 3p − 4 + √(p − 2)2 + 8 Proof: Let the vertex set V(Bp,p) = {v1, . . . , vp, v1′, . . . , vp′}. Consider the PRD set A = {v1, v1′}. Then 𝔓A(Bp,p) has the form ( B0 B1 B1 B0 ), where B0 = ( 1 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) and B1 = ( 1 O1×p−1 Op−1×1 Op−1 ), here J and O represents the matrix of 1’s and 0’s. Therefore by Lemma 1.1, SpecA(Bp,p) = Spec(B0 + B1)⋃Spec(B0 − B1). Consider, B0 + B1: B0 + B1 = ( 2 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) |(B0 + B1) − λI| = (−1) p(λ + 1)p−2(λ2 − pλ + (p − 3)) Therefore, SpecA(B0 + B1) = { −1 p + √(p − 2)2 + 8 2 p − √(p − 2)2 + 8 2 p − 2 1 1 } Consider, B0 − B1: B0 − B1 = ( 0 J1×p−1 Jp−1×1 Jp−1 − Ip−1 ) |(B0 − B1) − λI| = (−1) p(λ − (p − 1))(λ + 1)p−1 Therefore, SpecA(B0 − B1) = { p − 1 −1 1 p − 1 } Hence Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2320 https://internationalpubls.com SpecA(Bp,p) = { −1 p − 1 p + √(p − 2)2 + 8 2 p − √(p − 2)2 + 8 2 2p − 3 1 1 1 Now, 𝔓𝔈A(Bp,p) = (2p − 3)|−1| + |p − 1| + | p + √(p − 2)2 + 8 2 | + | p − √(p − 2)2 + 8 2 | Thus, 𝔓𝔈A(Bp,p) = 3p − 4 + √(p − 2)2 + 8 3.5 Theorem: Let G be obtained by removing an edge ′e′ from Kp, p > 4. Then G has the spectrump − 1, -1 and 0 with multiplicities 1,p − 2 and 1 respectively. And hence 𝔓𝔈A(G) = 2p − 3. Proof: Let G = Kp − e. Let v1, vp be the non-adjacent vertices of G. Then the vertices other than v1, vp would be act as a PD-set. Since p ≥ 5, deg(v1) = deg(vp) = p − 2, any PD-set will satisfy the constrain of RD-set. Let A = {v2} is the PRD-set of G. Then the PRD matrix is, 𝔓A(G) = ( P Q R S ) where A = ( 0 1 1 1 ); B = ( J1×p−3 0 J1×p−3 1 ) C = ( Jp−3×1 Jp−3×1 0 1 ); D = (Jp−3 − Ip−3) where J and I represents the matrix of 1’s and identity matrix. Now, |𝔓A(G) − λI| = (−1) pλ(λ + 1)p−4(λ3 − (p − 3)λ2 − (2p − 3)λ − 2) ⟹φ(G,λ) ≈ (−1)pλ(λ + 1)p−4(λ3 − (p − 3)λ2 − (2p − 3)λ − (p − 1)) ⟹φ(G, λ) ≈ (−1)pλ(λ + 1)p−2(λ − (p − 1)) Therefore, SpecA(G) = { p − 1 −1 0 1 p − 2 1 } Thus, 𝔓𝔈A(G) = |p − 1| + (p − 2)|−1| = p − 1 + p − 2 𝔓𝔈A(G) = 2p − 3 4. PRD ENERGY OF LINE GRAPHS OF STAR GRAPHS 4.1 Theorem: For the line graph of a star graph L(K1,p−1) withp > 4, 𝔓𝔈A (L(K1,p−1)) = p − 3 + √(p − 2)2 + 4 Proof: Let E(K1,p−1) = {e1, . . . , ep−1}. And it’s known that L(K1,p−1) ≅ Kp−1. Therefore L(K1,p−1) and Kp−1 are co-spectral and hence equi-energetic. Thus 𝔓𝔈A (L(K1,p−1)) ≅ 𝔓𝔈A(Kp−1) = p − 3 +√(p − 2)2 + 4. Hence 𝔓𝔈A (L(K1,p−1)) = p − 3 + √(p − 2)2 + 4 4.2 Theorem: For the line graph of a double star graph L(Sp,p) with p > 3, 𝔓𝔈(L(Sp,p)) = 4p − 5 Proof: Let V(L(Sp,p)) = {e0, e1, . . . , ep−1, e1′, . . . , ep−1′}, where e0 is the bridge connecting two star graphs. Then the unique PRD set of L(Sp,p) is A = {e0}. Therefore 𝔓A(L(Sp,p)) = ( 1 J1×p−1 J1×p−1 Jp−1×1 Jp−1 − Ip−1 Op−1 Jp−1×1 Op−1 Jp−1 − Ip−1) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2321 https://internationalpubls.com where J, O and I represents the matrix of 1’s, 0’s and identity matrix. And the corresponding characteristic polynomial is, φ(L(Sp,p), λ) = |𝔓A(L(Sp,p)) − λI)| φ(L(Sp,p), λ) = −(λ + 1) 2p−3(λ − (p − 2))( λ − p) Hence SpecA(L(Sp,p)) = { 1 2 − p −p 2p − 3 1 1 } Now, 𝔓𝔈(L(Sp,p)) = (2p − 3)|−1| + |2 − p| + |−p| ⟹𝔓𝔈(L(Sp,p)) = 4p − 5 5. CHARACTERISTICS OF PRD SPECTRUM 5.1 Theorem: Consider graph G. If λ1, λ2, … λp are the 𝔓-spectrum of 𝔓A(G), then the following condition holds. (i) ∑ λi p i=1 = 𝔭. (ii) ∑ λi 2p i=1 = 2q + 𝔭. Proof: As sum of spectrum of 𝔓A(G) is same as its trace, ∑λi p i=1 =∑𝔭ii p i=1 = |A| = 𝔭 (i) As summation of squares of the spectrum of 𝔓A(G) is the trace of [𝔓A(G) 2], ∑λi 2 p i=1 =∑∑𝔭ij 𝔭ji p i=1 p i=1 =∑(𝔭ii) 2 p i=1 +∑𝔭ij 𝔭ji i≠j =∑(𝔭ii) 2 p i=1 + 2∑(𝔭ij) 2 i