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Research article

Higher-order Randić index and isomorphism of double starlike trees

  • For an integer h0, the hth order Randić index for a simple graph G is defined as Rh(G)=π1v1(π)v2(π)vh+1(π), where π extends over all paths of length h in G and vi(π) denotes the degree of the i-th vertex of the path π. In this paper, we showed that the hth order Randić index Rh(T) of a double starlike tree T (a tree with two vertices of degrees m1,m2>2) is completely determined by its branches of length h. As a consequence, we proved that the double starlike trees with equal h-Randić index are isomorphic, except for some special values for m1 and m2.

    Citation: Zhenhua Su, Zikai Tang, Hanyuan Deng. Higher-order Randić index and isomorphism of double starlike trees[J]. AIMS Mathematics, 2023, 8(12): 31186-31197. doi: 10.3934/math.20231596

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  • For an integer h0, the hth order Randić index for a simple graph G is defined as Rh(G)=π1v1(π)v2(π)vh+1(π), where π extends over all paths of length h in G and vi(π) denotes the degree of the i-th vertex of the path π. In this paper, we showed that the hth order Randić index Rh(T) of a double starlike tree T (a tree with two vertices of degrees m1,m2>2) is completely determined by its branches of length h. As a consequence, we proved that the double starlike trees with equal h-Randić index are isomorphic, except for some special values for m1 and m2.



    In mathematical chemistry, a topological index is a numeric quantity derived mathematically in a direct and unambiguous manner from the structural graph of a molecule, which is used to characterize some properties of a molecular graph. It is also a graph invariant since isomorphic graphs have the same value for a given topological index.

    Many topological indices have been developed through the years and related successfully to physicochemical properties of organic molecules [1,2]. In order to define the concept of branching in molecular species [3,4], Randić [5] introduced in 1975 a topological index–the connectivity index (now called the Randić index), defined for a simple graph G as

    R1(G)=π1v1(π)v2(π),

    where π extends over all paths of length 1 and vi(π) denotes the degree of the i-th vertex of the path π. Note that for a path with (n+1) vertices and n edges, both of the vertex and edge can appear exactly once. It has become one of the most widely used and most successful in applications to physical and chemical properties [6]. For a review of historical details and a further bibliography on the chemical applications of the Randić index see [7,8,9,10].

    For an integer h0, the connectivity index of order h (also called h-Randić index) [11] is defined as

    Rh(G)=π1v1(π)v2(π)vh+1(π),

    where π extends over all paths of length h in G and vi(π) denotes the degree of the i-th vertex of the path π.

    Higher-order Randić indices are of great interest in the theory of molecular graph theory and some of its mathematical properties have been reported in [12]. Examples of non-isomorphic trees T and T such that Rh(T)=Rh(T) for all h0 exist [13]. However, it will not occur in starlike trees, i.e. trees that have a unique vertex of a degree greater than 2. In fact, Rada and Araujo [14] proved that starlike trees which have equal h-connectivity index for h0 are isomorphic.

    Very recently, by a relation on trees with respect to edge division vectors, Song and Huang [15] found some sufficient conditions to determine whether some trees with the same topological index value are isomorphic or not. Further, several classes of trees, including balanced starlike trees and double star Sp,q, are uniquely determined by edge division vectors.

    This leads naturally to determine h-Randić index of a double starlike tree, i.e., a tree that has only two vertices of a degree greater than 2. In this paper, we will show that for every integer h0, the higher-order Randić Rh(T) of a double starlike tree T is completely determined by its branches of length h. As a consequence, we show that almost all the double starlike trees that have equal h-Randić index for all h0 are isomorphic.

    Let G be a graph. The maximum vertex degree and the number of i-vertices (an i-vertex is a vertex of degree i) will be denoted by Δ(G) and ki(G), respectively. For an integer m2, a starlike tree T is a tree for which k1(T)=Δ(T)=m. The set of all starlike trees on n vertices with the maximal degree m is denoted by Ωn,m.

    For two integers m1,m22, a double starlike tree T is a tree with an edge u0v0 such that the components of T{u0v0} are two starlike trees Tu0 and Tv0 with their maximal degrees m11 and m21, respectively, where dT(u0)=m1 and dT(v0)=m2. If n1=|V(Tu0)| and n2=|V(Tv0)|, then n1+n2=n=|V(T)|. A double starlike tree is displayed as in Figure 1. The set of all double starlike trees on n vertices with two adjacent branching vertices of degrees m1 and m2 are denoted by Ωn,m1,m2. Specifically, if one of the values m1 and m2 is 2, then the double starlike tree is a starlike tree.

    Figure 1.  The double starlike tree.

    For any two vertices u,w in T, we use [u,w] to represent the shortest path connecting u and w, and keep in mind that the number of edges in [u,w] is called the distance. Let TΩn,m1,m2. If {u1,u2,,um11} is the set of 1-vertices in Tu0, an l-branch of Tu0 is a path [u0,ui] of Tu0 such that d(u0,ui)=l. There is a similar definition for Tv0. We denote by rl(T) and sl(T) the numbers of l-branches in Tu0 and Tv0, respectively (see Figure 1).

    Based on the notations introduced above, we clearly have the following relations:

     r1+r2++rt1=m11,s1+s2++st2=m21, r1+2r2++t1rt1=n11,s1+2s2++t2st2=n21,n1+n2=n,

    where t1 and t2 represent the length of the longest branch in Tu0 and Tv0, respectively.

    In this section, we will determine the higher-order Randić index of double starlike trees.

    Theorem 3.1. Let TΩn,m1,m2, where m1m2. Then

    R0(T)=(m1+m2)(112)+1m1+1m2+n22.

    As a consequence, if TΩn,m1,m2, then R0(T)=R0(T) if and only if m1=m1 and m2=m2.

    Proof.It is not difficult to obtain that k1(T)=m1+m22, k2(T)=nm1m2 and km1(T)=km2(T)=1. Hence,

    R0(T)=(m1+m22)1+(nm1m2)12+1m1+1m2
    =(m1+m2)(112)+1m1+1m2+n22.

    Let f(m1,m2)=(m1+m2)(112)+1m1+1m2. Clearly, m1+m2N+, and 1m1+1m2N+. If R0(T)=R0(T), then we have

    m1+m2=m1+m2. (3.1)
    1m1+1m2=1m1+1m2. (3.2)

    In the following, we show that m1=m1 and m2=m2. Assume that m1m1 and m2m2, thus m1m1m2m2. For convenience, let a=m1, b=m2, c=m1 and d=m2. Without loss of generality, assume b<d<c<a. From equations (3.1) and (3.2), it can be concluded that

    acdb=acdb,acdb=d+ba+c.

    Therefore,

    acdb=d+ba+c.

    On the other hand, according to b<d<c<a, we have ac>bd and a+c>b+d, i.e.,

    acdb>d+ba+c,

    a contradiction. Therefore, the assumption is not valid and the proof is done.

    In the following, the set of all rational numbers is denoted by Q and the set of all irrational numbers is RQ.

    Theorem 3.2. Let TΩn,m1,m2, where m1m2. Then

    R1(T)=(1m112m1+1212)r1+(1m212m2+1212)s1
    +[m1+m222+m112m1+m212m2+1m1m2(m1+m2)+n2+1].

    Consequently, for TΩn,m1,m2, where m1=m1 and m2=m2, if R1(T)=R1(T), then

    (i) r1=r1 and s1=s1 when m1=2 or m2=2;

    (ii) r1=r1 and s1=s1, except for the case m1,m2RQ and 2m1,2m2Q, when m1,m23.

    Proof. Note that 12m11i=r1+1d(u0,ui)+12m21i=s1+1d(v0,vi)=12(nr1s12). Keeping the notation of R1(T), we have

    R1(T)=r1i=11m1+m11i=r1+1[12+12(d(u0,ui)2)+12m1]
    +s1i=11m2+m21i=s1+1[12+12(d(v0,vi)2)+12m2]+1m1m2
    =(1m112m1+1212)r1+(1m212m2+1212)s1+λ(n,m1,m2),

    where λ(n,m1,m2)=[m1+m222+m112m1+m212m2+1m1m2(m1+m2)+n2+1].

    For convenience, let a=(1m112m1+1212), and b=(1m212m2+1212).

    (i) If one of m1 and m2 is 2, assume that m1=2. Then R1(T)=ar1+bs1+λ(n,m1,m2). We can immediately deduce s1=s1 from the condition R1(T)=R1(T). In this case, we have r1=r1=1 and a=0.

    (ii) If m1,m23, from R1(T)=R1(T), we obtain that ar1+bs1=ar1+bs1. If r1=r1(or s1=s1), then we can obtain that s1=s1 (or r1=r1), the theorem is true. In the following, we will deduce a contradiction under the assumption of r1r1 and s1s1. Denote

    ab=s1s1r1r1=t. (3.3)

    Let's discuss in three cases according to whether m1 and m2 are rational numbers.

    Case 1. m1=cQ and m2=dQ.

    From Equation (3.3), it can be concluded that

    ab=1c1c2+12121d1d2+1212=t,

    and we get t=d(c+2)c(d+2)=d(c+1)c(d+1). Furthermore, we have c=d, i.e., m1=m2, which contradicts with m1m2.

    Case 2. m1=cQ and m2RQ.

    Subcase 2.1. 2m2=dQ.

    From Equation (3.3), it can be concluded that

    ab=1c1c2+12121m21d+1212=t.

    Thus, we get t=d(c+2)c(d2)=d(c+1)c(d2), i.e., c+1=c+2, a contradiction.

    Subcase 2.2. 2m2RQ.

    By Equation (3.3), we obtain

    ab=1c1c2+12121m212m2+1212=t,

    and we have t=c+2c=m2(c+1)(2m21)c, which implies 2(2+c)m2(2+c)=0, i.e., c=2 and m2=0, a contradiction.

    Case 3. m1RQ, m2RQ.

    Subcase 3.1. 2m1=cQ, 2m2RQ.

    By Equation (3.3), we obtain

    ab=1m11c+12121m212m2+1212=t.

    Then we have t=c2c=2m2(2m1)c(2m21), i.e., m1=2, which contradicts with m13.

    Subcase 3.2. 2m1RQ, 2m2RQ.

    By Equation (3.3), we obtain

    ab=1m112m1+12121m212m2+1212=t.

    Similarly, it can be concluded that t=1=1m112m11m212m2, i.e., m1=m2, which contradicts with m1m2.

    Therefore, taking into account the above situations, we obtain that r1=r1 and s1=s1, except for the case m1,m2RQ and 2m1,2m2Q.

    Next, We will extend this result to Rh(T), where h2.

    Let u0 and v0 be the m1-vertex and m2-vertex of T (see Figure 1). Let Ph represent the set of all paths of length h in T, N={0,1,2,} as the set of natural numbers and the function Ψ:PhNh+1 by Ψ(π)=(v1(π),,vh+1(π)), where vi(π) denotes the degree of the i-th vertex of the path π.

    We first consider all paths in Ph that contain u0 or v0 as an end-vertex, or do not contain u0 and v0. There are twelve possibilities for the images under Ψ of these paths:

     X1=(1,2,,2h1,m1),X1=(1,2,,2h1,m2), X2=(2,,2h,m1),X2=(2,,2h,m2), X3=(2,,2h+1),X3=(2,,2h+1), X4=(1,2,,2h),X4=(1,2,,2h), X5=(1,2,,2h2,m1,m2),X5=(1,2,,2h2,m2,m1), X6=(2,,2h1,m1,m2),X6=(2,,2h1,m2,m1).

    We then consider all paths in Ph that contain only one of u0 and v0, but not as an end-vertex. In this case the image of Ψ is as follows:

    Y1(a)=(1,2,,2a,m1,2,,2h1a),Y1(a)=(1,2,,2a,m2,2,,2h1a), where 0ah2.

    Y2(a)=(1,2,,2a,m1,2,,2h2a,1),Y2(a)=(1,2,,2a,m2,2,,2h2a,1), where 0ah21.

    Y3(a)=(2,,2a,m1,2,,2ha),Y3(a)=(2,,2a,m2,2,,2ha), where 1ah2.

    Finally, consider all paths in Ph that contain both v0 and u0, but not one as an end-vertex. In this case the image of Ψ is as follows:

    Z1(a)=(1,2,,2a,m1,m2,2,,2h2a),Z1(a)=(1,2,,2a,m2,m1,2,,2h2a), where 0ah3.

    Z2(a)=(1,2,,2a,m1,m2,2,,2h3a,1), where 0ah3.

    Z3(a)=(2,,2a,m1,m2,2,,2h1a), where 1ah2.

    Now, let's prove our main result.

    Theorem 3.3. Let TΩn,m1,m2, where m1m2. Then

     Rh(T)=(12h1m112hm1+12h+112h)rh+(12h1m212hm2 +12h+112h)sh+λ(h,n,m1,m2,r1,s1,,rh1,sh1),

    where λ(h,n,m1,m2,r1,s1,,rh1,sh1) is a real number determined by the values of h,n,m1, m2,,rh1,sh1.

    Proof. Clearly, Rh(T) is determined by the numbers |Ψ1(Xi)|, |Ψ1(Xi)|, |Ψ1(Yi(a))|, |Ψ1(Yi(a))|, |Ψ1(Zi(a))| and |Ψ1(Zi(a))|, where |Ψ1(W)| represents the number of elements in the inverse image of W under Ψ, i.e.,

     Rh(T)=|Ψ1(X1)|12h1m1+|Ψ1(X1)|12h1m2+|Ψ1(X2)|12hm1+|Ψ1(X2)|12hm2 +|Ψ1(X3)|12h+1+|Ψ1(X3)|12h+1+|Ψ1(X4)|12h+|Ψ1(X4)|12h +|Ψ1(X5)|12h2m1m2+|Ψ1(X5)|12h2m1m2+|Ψ1(X6)|12h1m1m2 +|Ψ1(X6)|12h1m1m2+h2a=0|Ψ1(Y1(a))|12h1m1+h2a=0|Ψ1(Y1(a))|12h1m2 +h21a=0|Ψ1(Y2(a))|12h2m1+h21a=0|Ψ1(Y2(a))|12h2m2+h2a=1|Ψ1(Y3(a))|12hm1 +h2a=1|Ψ1(Y3(a))|12hm2+h3a=0|Ψ1(Z1(a))|12h2m1m2+h3a=0|Ψ1(Z1(a))|12h2m1m2 +h3a=0|Ψ1(Z2(a))|12h3m1m2+h2a=1|Ψ1(Z3(a))|12h1m1m2.

    We can express |Ψ1(Xi)|, |Ψ1(Xi)|, |Ψ1(Yi(a))|, |Ψ1(Yi(a))|, |Ψ1(Zi(a))| and |Ψ1(Zi(a))| in terms of r1,s1,,rh,sh by a counting argument together with the reduction formulas as follows:

     |Ψ1(X1)|=rh,|Ψ1(X1)|=sh, |Ψ1(X2)|=rh+1++rt1=m11hi=1ri,|Ψ1(X2)|=m21hi=1si, |Ψ1(X3)|=rh+1+2rh+2++(th1)rt1=(h+1)t1i=h+1ri+t1i=h+1iri, =(h+1)[m11hi=1ri]+[n11hi=1iri], |Ψ1(X3)|=(h+1)[m21hi=1si]+[n21hi=1isi],wheren1+n2=n, |Ψ1(X4)|=m11hi=1ri,|Ψ1(X4)|=m21hi=1si, |Ψ1(X5)|=rh1,|Ψ1(X5)|=sh1, |Ψ1(X6)|=rh++rt1=m11h1i=1ri,|Ψ1(X6)|=m11h1i=1si,
    |Ψ1(Y1(a))|={ra+1(m11h1ai=1ri),if0a<h12,ra+1(m12h1ai=1ri),ifh12ah2.
    |Ψ1(Y1(a))|={sa+1(m21h1ai=1si),if0a<h12,sa+1(m22h1ai=1si),ifh12ah2.
    |Ψ1(Y2(a))|={ra+1rh1a,if0a<h21,12ra+1(ra+11),ifa=h21.
    |Ψ1(Y2(a))|={sa+1sh1a,if0a<h21,12sa+1(sa+11),ifa=h21.
    |Ψ1(Y3(a))|={[m11hai=1ri][m12ai=1ri],if1ah12,12[m11hai=1ri][m12ai=1ri],ifa=h2.
    |Ψ1(Y3(a))|={[m21hai=1si][m22ai=1si],if1ah12,12[m21hai=1si][m22ai=1si],ifa=h2.
     |Ψ1(Z1(a))|=ra+1(m21h2ai=1si),if0ah3, |Ψ1(Z1(a))|=sa+1(m11h2ai=1ri),if0ah3, |Ψ1(Z2(a))|=ra+1sh2a,if0ah3, |Ψ1(Z3(a))|=(m11ai=1ri)(m21h1ai=1si),if1ah2.

    We can see that |Ψ1(Xi)|, |Ψ1(Xi)|, |Ψ1(Yj(a))|, |Ψ1(Yj(a))|, |Ψ1(Zj(a))| and |Ψ1(Zj(a))| depend on the numbers h,m,r1,s1,,rh1,sh1 for all a and i=5,6 andj=1,2,3; while for i=1,2,3,4, |Ψ1(Xi)| and |Ψ1(Xi)| depend on the numbers h,m,r1,s1,,rh,sh. Hence, by grouping in a convenient way we can get

    Rh(T)=λ(h,n,m1,m2,r1,s1,,rh1,sh1)+μ(h,m1,m2,rh,sh),

    where λ(h,n,m1,m2,r1,s1,,rh1,sh1) is a real number determined by the values of h,n,m1, m2,r1,s1,,rh1,sh1, and

     μ(h,m1,m2,rh,sh)=(12h1m112hm1+12h+112h)rh +(12h1m212hm2+12h+112h)sh.

    Thus, the theorem holds.

    Example 3.4. Let TΩ18,7,4 with r1(T)=3, r2(T)=2, r3(T)=1, s1(T)=2 and s4(T)=1, where m1=7,m2=4. In order to calculate R3(T), we first determine the number of paths of each type:

     |Ψ1(X1)|=r3=1,|Ψ1(X1)|=s3=0, |Ψ1(X2)|=m113i=1ri=0,|Ψ1(X2)|=1, |Ψ1(X3)|=(3+1)[m113i=1ri]+[n113i=1iri]=0,|Ψ1(X3)|=0, |Ψ1(X4)|=m113i=1ri=0,|Ψ1(X4)|=1, |Ψ1(X5)|=r2=2,|Ψ1(X5)|=s2=0, |Ψ1(X6)|=m112i=1ri=1,|Ψ1(X6)|=1, |Ψ1(Y1(0))|=r1(m112i=1ri)=3,|Ψ1(Y1(1))|=r2(m12r1)=4, |Ψ1(Y1(0))|=s1(m212i=1si)=2,|Ψ1(Y1(1))|=s2(m22s1)=0, |Ψ1(Y2(0))|=r1r2=6,|Ψ1(Y2(0))|=0, |Ψ1(Y3(1))|=[m112i=1ri][m12r1]=2,|Ψ1(Y3(1))|=0, |Ψ1(Z1(0))|=r1(m21s1)=3,|Ψ1(Z1(0))|=s1(m11r1)=6, |Ψ1(Z2(0))|=r1s1=6,|Ψ1(Z3(1))|=(m11r1)(m21s1)=3.

    Hence, we obtain from Theorem 3.3

     R3(T)=128+0116+0156+1132+0116+0116+018+18+2156 +0156+1112+1112+(3+4)128+(2+0)116+6114+018 +2156+0132+3156+6156+6128+31112 =12+122+142+(7+54)17+(6+132)114.

    Theorem 3.5. Let T,TΩn,m1,m2, where m1m2. Then T and T are isomorphic if and only if

    (i) Rh(T)=Rh(T) for all h0, where min{m1,m2}=2; or

    (ii) Rh(T)=Rh(T) for all h0, where m1,m23, except for the case m1,m2RQ and 2m1,2m2Q.

    Proof. The necessity is clear, we only need to prove sufficiency. Assume that T,TΩn,m1,m2. Since R0(T)=R0(T), by Theorem 3.1, m1=m1 and m2=m2. Now, from R1(T)=R1(T) and Theorem 3.2, we have r1=r1 and s1=s1, where min{m1,m2}=2, or where m1,m23, except for the case m1,m2RQ and 2m1,2m2Q. Next, applying Theorem 3.3 for h=2, we get

     (121m1122m1+123122)r2+(121m2122m2+123122)s2+λ =(121m1122m1+123122)r2+(121m2122m2+123122)s2+λ

    i.e.,

     (1m112m1+1212)r2+(1m212m2+1212)s2 =(1m112m1+1212)r2+(1m212m2+1212)s2

    By a proof process similar to Theorem 3.2, we have r2=r2 and s2=s2. Continuing this process by repeated use of Theorem 3.3 and Theorem 3.2, we can conclude that ri=ri and si=si for all iN. Therefore, T and T are isomorphic.

    Finally, let TΩn,m1,m2. If min{m1,m2}=2, then T is a starlike tree, i.e., TΩn,m, where m=max{m1,m2}. From Theorem 3.3 and Theorem 3.5, we have the following corollaries which are given in [14].

    Corollary 3.6. [14] Let TΩn,m. Then

     Rh(T)=(12h1m12hm+12h+112h)sh+λ(h,n,m,s1,,sh1),

    where λ(h,n,m,s1,,sh1) is a real number determined by the values of h,n,m,s1,,sh1.

    Corollary 3.7. [14] Let T,TΩn,m. Then T and T are isomorphic if and only if Rh(T)=Rh(T) for all h0.

    In this paper, we mainly investigated the hth order Randić index Rh(T) of the double starlike tree T, which is a tree with two vertices of degrees m1,m2>2. First, the formula of the hth order Randić index Rh(T) has been completely determined by its branches of length h. Second, it was taken that m1m2, and we have shown that almost all the double starlike trees TΩn,m1,m2 with equal h-Randić index for all h0 are isomorphic. In addition, some results of starlike trees have been obtained, which were given in [14].

    These results lead to a natural question, which we pose as a problem.

    Problem 3.8. Which index can determine isomorphism of double starlike trees TΩn,m1,m2, where m1=m2?

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

    This work is supported by the Department of Education of Hunan Province (22B0763), (22B0828), the National Natural Science Foundation of China (12201634), and the Natural Science Foundation of Hunan Province (2023JJ30070).

    The authors declare that they have no conflicts of interest.



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