Research article

On the extremal cacti with minimum Sombor index

  • Received: 25 September 2023 Revised: 26 October 2023 Accepted: 01 November 2023 Published: 06 November 2023
  • MSC : 05C09, 05C90

  • Let H be a graph with edge set EH. The Sombor index and the reduced Sombor index of a graph H are defined as SO(H)=uvEHdH(u)2+dH(v)2 and SOred(H)=uvEH(dH(u)1)2+(dH(v)1)2, respectively. Where dH(u) and dH(v) are the degrees of the vertices u and v in H, respectively. A cactus is a connected graph in which any two cycles have at most one common vertex. Let C(n,k) be the class of cacti of order n with k cycles. In this paper, the lower bound for the Sombor index of the cacti in C(n,k) is obtained and the corresponding extremal cacti are characterized when n4k2 and k2. Moreover, the lower bound of the reduced Sombor index of cacti is obtained by similar approach.

    Citation: Qiaozhi Geng, Shengjie He, Rong-Xia Hao. On the extremal cacti with minimum Sombor index[J]. AIMS Mathematics, 2023, 8(12): 30059-30074. doi: 10.3934/math.20231537

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  • Let H be a graph with edge set EH. The Sombor index and the reduced Sombor index of a graph H are defined as SO(H)=uvEHdH(u)2+dH(v)2 and SOred(H)=uvEH(dH(u)1)2+(dH(v)1)2, respectively. Where dH(u) and dH(v) are the degrees of the vertices u and v in H, respectively. A cactus is a connected graph in which any two cycles have at most one common vertex. Let C(n,k) be the class of cacti of order n with k cycles. In this paper, the lower bound for the Sombor index of the cacti in C(n,k) is obtained and the corresponding extremal cacti are characterized when n4k2 and k2. Moreover, the lower bound of the reduced Sombor index of cacti is obtained by similar approach.



    Throughout this paper, we consider simple and undirected graphs. Let H=(VH,EH) be a graph, where VH and EH be the vertex set and the edge set of H, respectively. The degree of a vertex uVH, denoted by dH(u), is the number of edges which connected to u in H. A vertex u is called a pendantvertex if dH(u)=1. For an edge e=xyEH, e is a pendantedge of H if dH(x)=1 or dH(y)=1. For a vertex vVH and an edge xyEH, Hv and Hxy denote the graphs obtained from H by deleting the vertex v and the edge xy, respectively. If x and y are two vertices in VH and xyEH, H+xy is the graph obtained from H by adding an edge xy. For any vertex uVH, NH(u) denoted the neighborhood vertex set in H. The symbols δ(H) and Δ(H) represent the minimum degree and the maximum degree of H, respectively. Denote by Pn and Cn the path and the cycle with n vertices, respectively. One can refer to [1] for other notations and terminologies undefined in this paper.

    Topological indices of graphs have been widely studied in mathematical chemistry. The topological indices can be used in theoretical, medicinal and organic chemistry for studying the structure and physicochemical properties of chemical molecular. The Wiener index is the most well-known topological index which is introduced by the famous chemist Harry Wiener for investigating boiling points of alkanes [2].

    Recently, two new degree-based graph topological indices, named Sombor index and reduced Sombor index, are introduced by Gutman [3]. The Sombor index and the reduced Sombor index of a graph H are defined, respectively, as

    SO(H)=e=uvEHdH(u)2+dH(v)2

    and

    SOred(H)=e=uvEH(dH(u)1)2+(dH(v)1)2.

    Nowdays, the study on the Sombor index and the reduced Sombor index of various graphs has been a hot topic in graph theory. Alidadi et al. [4] investigated the minimum Sombor index of the unicyclic graphs with given diameter. Zhou et al. [5] characterized the extremal trees and unicyclic graphs with minimum Sombor index among the trees and unicyclic graphs with given order and maximum degree. The lower and upper bounds of the Sombor index of the trees in terms of order, independence number and the number of pendant vertices were given by Das and Gutman in [6], and the corresponding extremal trees were characterized. Li et al. [7] characterized the extremal graphs with respect to the Sombor index among all the n-order trees with a given diameter. The maximum and minimum Sombor indices of trees with fixed domination number were presented by Sun and Du in [8]. Cruz et al. [9] discussed the Sombor index of chemical graphs, chemical trees and hexagonal systems and characterized the extremal graphs. The upper bound for the Sombor index among all molecular trees with given order was obtained by Deng et al. in [10]. Ülker et al. [11] studied the relations between energy and Sombor index of a graph in terms of its degrees. Horoldagva and Xu [12] investigated the lower and upper bounds for the Sombor index of the the graphs with given girth. Liu et al. [13] studied the reduced Sombor index of the graphs with given graph parameters, obtained the expected values of reduced Sombor index in random polyphenyl chain, and applied the reduced Sombor index to graph spectrum and energy problems.

    A cactus is a connected graph that any block is either a cut edge or a cycle. It is also a graph in which any two cycles have at most one common vertex. A cycle in a cactus is called pendantcycle if all but one vertex of this cycle have degree 2, a cycle C in a cactus is called interalcycle if C is not a pendant cycle. Let C(n,k) be the class of cacti of order n with k cycles.

    It is routine to check that C(n,0) is the set of trees and C(n,1) is the set of unicyclic graphs. Gutman investigated the Sombor index of trees in [3] and proved that SO(H)22n for any tree H with n vertices. Cruz and Rada [14] proved that SO(H)22n for any unicyclic graph H with n vertices.

    Recently, Wu, An and Wu [15] established the lower bound for the Sombor index of the cacti in C(n,k) and characterized the corresponding extremal cacti when n6k4 and k2. In this paper, the lower bound for the Sombor index of a cactus in C(n,k) is obtained and the corresponding extremal cacti are characterized when n4k2 and k2 which improves the result of Wu et al. [15]. Moreover, it is also shown that our approach is valid for the reduced Sombor index of the cacti in C(n,k). The following Theorems 1.1 and 1.2 are our main results.

    Theorem 1.1. Let H be a cactus in C(n,k) with n4k2 and k2. Then,

    SO(H)8n+213k+(52213)k2+213102

    with equality holds if and only if H˜C(n,k) (where ˜C(n,k) is a subset of C(n,k), and the definition of ˜C(n,k) is introduced in Section 4).

    Theorem 1.2. Let H be a cactus in C(n,k) with n4k2 and k2. Then,

    SOred(H)2n+(25+2)k+(3225)k2+2572

    with equality holds if and only if H˜C(n,k) (where ˜C(n,k) is a subset of C(n,k), and the definition of ˜C(n,k) is introduced in Section 4).

    The rest of this paper is organized as follows. In Sections 2 and 3, it is proved that the minimum and maximum degrees of the cacti in C(n,k) (n4k2 and k2) with minimum Sombor index (and reduced Sombor index) are 2 and 3, respectively. In Section 4, the proofs of Theorems 1.1 and 1.2 are presented.

    In this section, the minimum degree of the cacti in C(n,k) (n4k2 and k2) with minimum Sombor index and reduced Sombor index is discussed. Let TH be the graph obtained from a graph H in C(n,k) by contracting each cycle of H into a vertex (called a cyclic vertex). Let P=v1v2vl be a path in H. If dH(v1)3, dH(vl)=1 and dH(vi)=2 for 2il1, then we call P is a pendantpath of H.

    Lemma 2.1. [16] Let x and y be two nonnegative integers and z be a fixed nonnegative integer. Then the function (x+z)2+y2x2+y2 is increasing with x for fixed y and decreasing with y for fixed x.

    Lemma 2.2. Let H be a cactus in C(n,k) with k2. If P=v1v2vl and P=u1u2us are two different pendant paths of H with dH(v1)3 and dH(u1)3. Let H=Hv1v2+usv2. Then SO(H)>SO(H) and SOred(H)>SOred(H).

    Proof. Let dH(v1)=t (t3) and NH(v1)={v2,w1,w2,,wt1}. By the conditions, one has dH(v1)=t1, dH(us)=1, dH(us)=2 and dH(v)=dH(v) for any vertex vVH{v1,us}. We divide this problem into two cases.

    Case 1: l>2.

    By the definition of Sombor index, one has that

    SO(H)SO(H)=t1i=1dH(v1)2+dH(wi)2+dH(v1)2+dH(v2)2+dH(us1)2+dH(us)2t1i=1dH(v1)2+dH(wi)2dH(v2)2+dH(us)2dH(us1)2+dH(us)2=t1i=1t2+dH(wi)2+t2+22+dH(us1)2+12t1i=1(t1)2+dH(wi)222+22dH(us1)2+22t2+2222+22+dH(us1)2+12dH(us1)2+22138+dH(us1)2+12dH(us1)2+22.

    It is routine to check that dH(us1)dH(us1)2 if s=2 and dH(us1)=dH(us1)=2 if s>2. Thus, by Lemma 2.1, we obtain that

    dH(us1)2+12dH(us1)2+2222+1222+22.

    Then SO(H)SO(H)138+58>0.

    By a similar calculation method, we get

    SOred(H)SOred(H)=t1i=1(dH(v1)1)2+(dH(wi)1)2+(dH(v1)1)2+(dH(v2)1)2+(dH(us1)1)2+(dH(us)1)2t1i=1(dH(v1)1)2+(dH(wi)1)2(dH(v2)1)2+(dH(us)1)2(dH(us1)1)2+(dH(us)1)2(t1)2+1212+12+(dH(us1)1)2+02(dH(us1)1)2+1252+(dH(us1)1)2+02(dH(us1)1)2+1252+12>0.

    Case 2: l=2.

    According to the definition of Sombor index, we have

    SO(H)SO(H)=t1i=1dH(v1)2+dH(wi)2+dH(v1)2+dH(v2)2+dH(us1)2+dH(us)2t1i=1dH(v1)2+dH(wi)2dH(v2)2+dH(us)2dH(us1)2+dH(us)2=t1i=1t2+dH(wi)2+t2+12+dH(us1)2+12t1i=1(t1)2+dH(wi)212+22dH(us1)2+22t2+1212+22+dH(us1)2+12dH(us1)2+22105+dH(us1)2+12dH(us1)2+22105+58>0.

    In a similar manner, we deduce that

    SOred(H)SOred(H)=t1i=1(dH(v1)1)2+(dH(wi)1)2+(dH(v1)1)2+(dH(v2)1)2+(dH(us1)1)2+(dH(us)1)2t1i=1(dH(v1)1)2+(dH(wi)1)2(dH(v2)1)2+(dH(us)1)2(dH(us1)1)2+(dH(us)1)2(t1)2+0202+12+(dH(us1)1)2+02(dH(us1)1)2+1241+12>0.

    These complete the proof.

    Lemma 2.3. Let H be a cactus in C(n,k) with k2. If there is at most one pendant path in H, then there exists an edge u1u2EH on some cycle of H such that dH(u1)=dH(u2)=2.

    Proof. By the fact that H is a cactus, then TH is a connected tree. Thus there exists at least two pendant vertices in H. By the condition that there is at most one pendant path in H, then there exists at least one pendant vertex which is cyclic vertex in TH. So there exists at least one pendant cycle in H. By the definition of pendant cycle, the result follows.

    Lemma 2.4. Let H be a cactus in C(n,k) (k2) with an edge u1u2EH on some cycle of H such that dH(u1)=dH(u2)=2. Let P=v1v2vl be a pendant path of H with dH(v1)3 and H=Hv1v2u1u2+u1v2+u2vl. Then SO(H)>SO(H) and SOred(H)>SOred(H).

    Proof. Let dH(v1)=t (t3) and NH(v1)={v2,w1,w2,,wt1}. By the conditions, one has dH(v1)=t1, dH(vl)=1, dH(vl)=2 and dH(v)=dH(v) for any vertex vVH{v1,vl}. We divide this problem into two cases.

    Case 1: l>2.

    By the definition of Sombor index, one has that

    SO(H)SO(H)=t1i=1dH(v1)2+dH(wi)2+dH(v1)2+dH(v2)2+dH(u1)2+dH(u2)2+dH(vl1)2+dH(vl)2t1i=1dH(v1)2+dH(wi)2dH(u1)2+dH(v2)2dH(u2)2+dH(vl)2dH(vl1)2+dH(vl)2=t1i=1t2+dH(wi)2+t2+22+22+22+22+12t1i=1(t1)2+dH(wi)222+2222+2222+22t2+22+22+12222+2213+528>0.

    The corresponding result for reduced Sombor index is the following:

    SOred(H)SOred(H)=t1i=1(dH(v1)21)+(dH(wi)1)2+(dH(v1)1)2+(dH(v2)1)2+(dH(u1)1)2+(dH(u2)1)2+(dH(vl1)1)2+(dH(vl)1)2t1i=1(dH(v1)1)2+(dH(wi)1)2(dH(u1)1)2+(dH(v2)1)2(dH(u2)1)2+(dH(vl)1)2(dH(vl1)1)2+(dH(vl)1)2(t1)2+12+12+02212+125+122>0.

    Case 2: l=2.

    From the definition of Sombor index, we have

    SO(H)SO(H)=t1i=1dH(v1)2+dH(wi)2+dH(v1)2+dH(v2)2+dH(u1)2+dH(u2)2t1i=1dH(v1)2+dH(wi)2dH(u1)2+dH(v2)2dH(u2)2+dH(v2)2=t1i=1t2+dH(wi)2+t2+12+22+22t1i=1(t1)2+dH(wi)222+2222+22t2+12+22+22222+22108>0.

    By a similar calculation method, one obtains that

    SOred(H)SOred(H)=t1i=1(dH(v1)1)2+(dH(wi)1)2+(dH(v1)1)2+(dH(v2)1)2+(dH(u1)1)2+(dH(u2)1)2t1i=1(dH(v1)1)2+(dH(wi)1)2(dH(u1)1)2+(dH(v2)1)2(dH(u2)1)2+(dH(v2)1)2(t1)2+02+12+12212+1242>0.

    These end the proof.

    Lemma 2.5. Let H be a cactus in C(n,k) (k2) with minimum Sombor index. Then δ(H)=2.

    Proof. If H contains no pendant edge, by Lemma 2.3, there exists at least one vertex with degree 2, the result follows.

    If H contains pendant edges, from Lemma 2.2, H contains just one pendant edge. By Lemmas 2.3 and 2.4, there exists a cactus H in C(n,k) such that SO(H)>SO(H), which contradicts to the condition that H is a cactus with minimum Sombor index.

    This ends the proof.

    By a similar proof with Lemma 2.5, the following Corollary 2.6 can be obtained immediately.

    Corollary 2.6. Let H be a cactus in C(n,k) (k2) with minimum reduced Sombor index. Then δ(H)=2.

    In this section, the maximum degree of the cacti in C(n,k) (n4k2 and k2) with minimum Sombor index and reduced Sombor index is discussed.

    Lemma 3.1. Let H be a cactus in C(n,k) (k2) with minimum Sombor index. Then there does not exist a path u1u2ul (l3) in H such that dH(u1)3,dH(ul)3 and dH(ui)=2(i=2,,l1), where u1 and ul are not adjacent.

    Proof. Suppose to the contrary that there exists a path u1u2ul (l3) in H such that dH(u1)3,dH(ul)3 and dH(ui)=2(i=2,,l1), where u1 and ul are not adjacent. By Lemma 2.5, each end block of H is a cycle and there exists at least one edge e=v1v2 with dH(v1)=dH(v2)=2 on some end block of H. Let H=Hu1u2ul1ulv1v2+v1u2+v2ul1+u1ul. It is routine to check that dH(u)=dH(u) for each vertex u of H. Therefore,

    SO(H)SO(H)=dH(u1)2+dH(u2)2+dH(ul1)2+dH(ul)2+dH(v1)2+dH(v2)2dH(v1)2+dH(u2)2dH(v2)2+dH(ul1)2dH(u1)2+dH(ul)2=dH(u1)2+22+22+dH(ul)2+22+2222+2222+22dH(u1)2+dH(ul)2=dH(u1)2+22+22+dH(ul)222+22dH(u1)2+dH(ul)2=(dH(u1)2+2222+22)(dH(u1)2+dH(ul)222+dH(ul)2)=(dH(u1)2+2222+22)(dH(u1)2+dH(ul)222+dH(ul)2).

    Note that dH(u1)3 and dH(ul)>2, by Lemma 2.1, one has that

    (dH(u1)2+2222+22)(dH(u1)2+dH(ul)222+dH(ul)2)>0.

    Thus, SO(H)SO(H)>0 which contradicts to the fact that H has minimum Sombor index.

    This completes the proof.

    The corresponding result for reduced Sombor index is the following Lemma 3.2.

    Lemma 3.2. Let H be a cactus in C(n,k) (k2) with minimum reduced Sombor index. Then there does not exist a path u1u2ul(l3) in H such that dH(u1)3,dH(ul)3 and dH(ui)=2(i=2,,l1), where u1 and ul are not adjacent.

    From Lemmas 3.1 and 3.2, the following Corollary 3.3 can be obtained immediately.

    Corollary 3.3. Let H be a cactus in C(n,k) (k2) with minimum Sombor index or minimum redeced Sombor index. Then, the following results hold.

    (i) If u is a vertex of H with dH(u)=2, then u lies on some cycle of H.

    (ii) Let C be a cycle of H. Then, either C is an end block, or C contains exactly two adjacent vertices whose degrees are not 2, or no vertex of C with degree 2.

    Lemma 3.4. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum Sombor index. If Δ(H)4, then there exists a path v1v2v3v4 in H such that dH(v2)=dH(v3)=2 and v1v4.

    Proof. Let t=Δ(H) and ni be the number of vertices of H with degree i (i=1,2,,t). From Lemma 2.5, δ(H)=2. Then, we get

    n2+n3++nt=n (3.1)

    and

    2n2+3n3++tnt=2|EH|=2(n+k1). (3.2)

    From (3.1) and (3.2), one obtains that

    n3+2n4++(t2)nt=2k2

    and

    n2=nn3n4nt=n[n3+2n4++(t2)nt]+[n4+2n5++(t3)nt]4k2(2k2)+[n4+2n5++(t3)nt]=2k+[n4+2n5++(t3)nt].

    By the condition Δ(H)4, we have

    n4+2n5++(t3)nt1

    and

    n22k+1.

    From Corollary 3.3(i), each vertex with degree 2 lies on some cycle of H. Since there are exactly k cycles in H and n22k+1, there exists a cycle C in H which contains at least three vertices with degree 2. By Corollary 3.3(ii), C is an end block or C contains exactly two adjacent vertices whose degrees are not 2. Combining the fact that C contains at least three vertices with degree 2, the result holds.

    Lemma 3.5. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum Sombor index. Then Δ(H)=3.

    Proof. Suppose to the contrary that Δ(H)4. Let uVH be a vertex with dH(u)=Δ(H)=t and NH(u)={v1,v2,,vt}. It is routine to check that u is a cut vertex of H. Let H1,H2,,Hs (st) be the pairwise-vertex-disjoint connected components of Hu. By Lemma 3.4 and the condition that Δ(H)4, there exists a path P=w1w2w3w4 in H such that dH(w2)=dH(w3)=2 and w1w4. We divide this discussion into two cases.

    Case 1: u{w1,w4}.

    Without loss of generality, suppose that PHs. We claim that |VHi{v1,v2,,vt}|2 for each i=1,2,,s. Otherwise, one can suppose to the contrary that there exists some i such that |VHi{v1,v2,,vt}|3. Without loss of generality, suppose that {v1,v2,v3}VHi{v1,v2,,vt}. Then, there exist two different cycles C1 and C2 in H such that {v1,v2,u}VC1 and {v1,v3,u}VC2. Which contradicts to the definition of cactus that any two cycles have at most one common vertex.

    If |VHi{v1,v2,,vt}|=1 for each i=1,2,,s1, by t4, s3 and one can suppose that v1H1, v2H2. If there exists some j1,2,,s1 such that |VHj{v1,v2,,vt}|=2, by t4, s2 and one can suppose that v1,v2Hj.

    Let H=Huv1uv2w1w2w2w3+w1w3+v1w2+v2w2+uw2. Then dH(u)=t,dH(w2)=2,dH(u)=t1,dH(w2)=3 and dH(g)=dH(g) for each other vertex g of H. Thus,

    SO(H)SO(H)=ti=1dH(vi)2+dH(u)2+dH(w1)2+dH(w2)2+dH(w2)2+dH(w3)2ti=3dH(vi)2+dH(u)2dH(v1)2+dH(w2)2dH(v2)2+dH(w2)2dH(u)2+dH(w2)2dH(w1)2+dH(w3)2=ti=1dH(vi)2+t2+dH(w1)2+22+22+22ti=3dH(vi)2+(t1)2dH(v1)2+32dH(v2)2+32(t1)2+32dH(w1)2+22=ti=3[dH(vi)2+t2dH(vi)2+(t1)2]+2i=1[dH(vi)2+t2dH(vi)2+32]+22+22(t1)2+32ti=3[t2+t2t2+(t1)2]+2i=1[t2+t2t2+32]+22+22(t1)2+32=(t2)[t2+t2t2+(t1)2]+2[t2+t2t2+32]+8(t1)2+32=(t2)(2t1)t2+t2+t2+(t1)2+2(t29)t2+t2+t2+32+8(t1)2+32(t2)(2t1)22t+2(t29)22t+8(t1)2+32=2t5221622t+8(t1)2+32=2t122+4221622t(t1)2+322t122(t1)2+32.

    Set f(t)=2t122(t1)2+32. Then f(t)=2(t1)(t1)2+32>21>0 for t4. This implies that

    f(t)f(4)=42122(41)2+32>0

    and

    SO(H)SO(H)2t122(t1)2+32>0.

    Which contradicts to the condition that H is a cactus with minimum Sombor index. Therefore, Δ(H)3.

    Case 2: u{w1,w4}.

    If u=w4, let H=Huv1uv2w1w2w2w3+w1w3+v1w2+v2w2+uw2. If u=w1, let H=Huv1uv2w4w3w2w3+w4w2+v1w3+v2w3+uw3. By a similar calculation method with Case 1, one has that SO(H)SO(H)>0. Which contradicts to the condition that H is a cactus with minimum Sombor index. Thus, Δ(H)3.

    On the other hand, by k2, there exists at least one vertex in H with degree 3. The result holds.

    By similar proof with Lemmas 3.4 and 3.5, the following Corollary 3.6 can be obtained immediately.

    Corollary 3.6. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum reduced Sombor index. Then Δ(H)=3.

    Let H be a cactus in C(n,k) (n4k2 and k2) with minimum reduced Sombor index. By Corollaries 2.6 and 3.6, we have 2dH(v)3 for each vertex v in H. Let Ei,j={uvEH|dH(u)=i,dH(v)=j} for i,j{2,3} and ei,j=|Ei,j|. Thus

    e2,2+e2,3+e3,3=n+k1. (4.1)

    Note that ni is the number of vertices of H with degree i (i{2,3}). It can be check that the degree sums of the vertices of degree 2 and degree 3, respectively, are

    2n2=2e2,2+e2,3

    and

    3n3=2e3,3+e2,3.

    By the fact n2+n3=n, one has that

    6e2,2+5e2,3+4e3,3=6n. (4.2)

    Combining (4.1) and (4.2), we have

    e2,2=n5k+5+e3,3 (4.3)

    and

    e2,3=6k62e3,3. (4.4)

    Lemma 4.1. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum reduced Sombor index. Then

    SOred(H)=2n+(6552)(k1)+(3225)e3,3.

    Proof. By the definition of the reduced Sombor index and the fact that 2dH(v)3 for each vertex v in H, combining (4.3) and (4.4), we get

    SOred(H)=e=uvEH(dH(u)1)2+(dH(v)1)2=2e2,2+5e2,3+8e3,3=2n+(6552)(k1)+(3225)e3,3.

    This completes the proof.

    By a similar proof with Lemma 4.1, the following Corollary 4.2 can be gotten directly.

    Corollary 4.2. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum Sombor index. Then

    SO(H)=8n+(613102)(k1)+(52213)e3,3.

    In the following, a new set of cacti, named ˜C(n,k), is introduced. Let ˜C(n,k) denote the set of the element H of C(n,k) with the following properties:

    (i) δ(H)=2 and Δ(H)=3.

    (ii) A vertex is cut vertex if and only if its degree is 3, and there are exactly 2k2 cut vertices.

    (iii) If k is even, there are k22 internal cycles and every internal cycle is triangle. Moreover, there is no vertex not belong to any cycle and the degree of each vertex on internal cycles is 3.

    (iv) If k is odd, there are k32 internal cycles, and each internal cycle is one of the following 3 kinds of cycles: (1) a 3-cycle whose vertices are all degree 3; (2) a 4-cycle whose vertices are all degree 3; (3) a cycle which contains exactly two adjacent 3-degree vertices. Moreover, there are b internal 4-cycles whose vertices are all degree 3, c2 cycles each of which contains exactly two adjacent 3-degree vertices, and t3 vertices with degree 3 which not belong to any cycle such that b+c2+t3=1.

    One element of ˜C(n,k) is shown in Figure 1 where k is even, and three elements of ˜C(n,k) are shown in Figure 2 where k is odd. Moreover, the graph of Type I in Figure 2 is an example graph with c2=1 and b=t3=0; the graph of Type II in Figure 2 is an example graph with t3=1 and c2=b=0; the graph of Type III in Figure 2 is an example graph with b=1 and c2=t3=0.

    Figure 1.  k is even.
    Figure 2.  k is odd.

    Remark. In [15], Wu et al. defined a set of cacti C(n,k) which was different with the set ˜C(n,k) in this paper. Actually, when k is odd, the set C(n,k) in [15] contains two types of cacti meanwhile the set ˜C(n,k) in this paper contains three types of cacti. Furthermore, if k is odd, the two types cacti of C(n,k) in [15] are just the cacti of the Types I and II in this paper.

    Lemma 4.3. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum Sombor index. Then, e3,35k24 with equality holds if and only if H˜C(n,k).

    Proof. By Lemmas 2.5 and 3.5, we have 2dH(v)3 for each vertex v in H. Let c1 be the number of end blocks, c2 be the number of the cycles which contains exactly two adjacent vertices whose degrees are not 2. By Corollary 3.3(ii), there are c3=kc1c2 cycles containing no vertex with degree 2, and let them be C1,C2,,Cc3. Let li=|VCi| for i=1,2,,c3. Let t3 be the number of vertices which does not lie on any cycle of H.

    Let TH be the tree obtained by contracting each cycle of H into a vertex. Then |VTH|=k+t3=|ETH|+1, and the degree sum of all vertices in TH is

    3t3+2c2+c1+l1+l2++lc3=2(k+t31).

    Therefore,

    t3+c2+l1+l2++lc3+c1+c2=2k2 (4.5)

    and

    t3+c2+l1+l2++lc33c3+2c3+k=2k2.

    Thus,

    2c3=k2[t3+c2+c3i=1(li3)]. (4.6)

    On the other hand, by Corollary 3.3, one has that

    e3,3=(k+t31)+c2+l1+l2++lc3. (4.7)

    Combining (4.5), (4.6) and (4.7), we get

    e3,3=(k+t31)+c2+l1+l2++lc3=k1+2k2c1c2=2k3+c3=2k3+12{k2[t3+c2+c3i=1(li3)]}=5k2412[t3+c2+c3i=1(li3)].

    If k is even, we obtain

    e3,35k24

    with equality holds if and only if t3=c2=0 and li=3 for i=1,2,,c3, that is H˜C(n,k).

    If k is odd, we have

    e3,35k124

    with equality holds if and only if either t3=0,c2=1 and li=3 for i=1,2,,c3 (i.e., H is the graph of Type I in Figure 2), or t3=1,c2=0 and li=3 for i=1,2,,c3 (i.e., H is the graph of Type II in Figure 2), or t3=c2=0 and c3i=1(li3)=1 (i.e., H is the graph of Type III in Figure 2), that is H˜C(n,k).

    In a similar manner with Lemma 4.3, the following Corollary 4.4 can be deduced.

    Corollary 4.4. Let H be a cactus in C(n,k) (n4k2 and k2) with minimum reduced Sombor index. Then, e3,35k24 with equality holds if and only if H˜C(n,k).

    Proof of Theorem 1.1: From Corollary 4.2 and Lemma 4.3, one obtains that

    SO(H)=8e2,2+13e2,3+18e3,3=8n+(613102)(k1)+(52213)e3,38n+213k+(52213)k2+213102.

    Moreover, the equality holds if and only if H˜C(n,k).

    Proof of Theorem 1.2: From Lemma 4.1 and Corollary 4.4, we get

    SOred(H)=2e2,2+5e2,3+8e3,3=2n+(6552)(k1)+(3225)e3,32n+(25+2)k+(3225)k2+2572.

    Moreover, the equality holds if and only if H˜C(n,k).

    In this paper, the Sombor index and the reduced Sombor index on cacti with n vertices and k cycles are discussed. The minimum Sombor index on cacti with n vertices and k cycles (n4k2 and k2) is obtained and the corresponding extremal cacti are characterized which improves the result of Wu et al. [15]. Moreover, the minimum reduced Sombor index of cacti with n vertices and k cycles (n4k2 and k2) is obtained and the corresponding extremal cacti are characterized as well. For further study, it would be interesting to generalize the Theorems 1.1 and 1.2 to the case of 3k+1n4k3 and k2. Furthermore, it would be meaningful to study the Sombor index and the reduced Sombor index of other kinds of graphs.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This research is supported by Tianjin Education Commission Research Program Project (No.2022KJ007). The authors would like to thank the anonymous reviewers for their valuable and kind suggestions which greatly improved the original manuscript.

    The authors declare that they have no conflicts of interest.



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