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Research article

Edge irregular reflexive labeling for the r-th power of the path

  • Received: 05 May 2021 Accepted: 13 July 2021 Published: 19 July 2021
  • MSC : 05C12, 05C78, 05C90

  • Let G(V,E) be a graph, where V(G) is the vertex set and E(G) is the edge set. Let k be a natural number, a total k-labeling φ:V(G)E(G){0,1,2,3,...,k} is called an edge irregular reflexive k-labeling if the vertices of G are labeled with the set of even numbers from {0,1,2,3,...,k} and the edges of G are labeled with numbers from {1,2,3,...,k} in such a way for every two different edges xy and xy their weights φ(x)+φ(xy)+φ(y) and φ(x)+φ(xy)+φ(y) are distinct. The reflexive edge strength of G, res(G), is defined as the minimum k for which G has an edge irregular reflexive k-labeling. In this paper, we determine the exact value of the reflexive edge strength for the r-th power of the path Pn, where r2, nr+4.

    Citation: Mohamed Basher. Edge irregular reflexive labeling for the r-th power of the path[J]. AIMS Mathematics, 2021, 6(10): 10405-10430. doi: 10.3934/math.2021604

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  • Let G(V,E) be a graph, where V(G) is the vertex set and E(G) is the edge set. Let k be a natural number, a total k-labeling φ:V(G)E(G){0,1,2,3,...,k} is called an edge irregular reflexive k-labeling if the vertices of G are labeled with the set of even numbers from {0,1,2,3,...,k} and the edges of G are labeled with numbers from {1,2,3,...,k} in such a way for every two different edges xy and xy their weights φ(x)+φ(xy)+φ(y) and φ(x)+φ(xy)+φ(y) are distinct. The reflexive edge strength of G, res(G), is defined as the minimum k for which G has an edge irregular reflexive k-labeling. In this paper, we determine the exact value of the reflexive edge strength for the r-th power of the path Pn, where r2, nr+4.



    Throughout this paper we consider G a connected, simple, and undirected graph, where V and E are denote to sets of vertices and edges of G with cardinalities |V| and |E|, respectively. Chartrand et al. presented in [12] the edge k-labeling of graph G, φ:E(G){1,2,3,...,k} such that the aggregate of the labels of edges incident with a vertex is different for all the vertices of G. Such labelings were referred to as irregular assignments and irregular strength, s(G), of a graph G is known as the smallest k for which G has an irregular assignment using labels at most k. In [16] Lahel gives a comprehensive over view of the strength of irregularity. For some studies on irregularity strength see papers by Amar and Tongi [5], Dimitiz et al. [13], Gyárfás [14], and Nierhoff [17]. In [7] Bača et al. motivated the concept of irregular strength and started to investigate the total edge irregularity strength of a graph. An edge irregular total k-labeling of the graph G is a labeling φ:V(G)E(G){1,2,3,...,k} such that for every two distinct edges xy, xy of G the edge weights wtφ(xy)=φ(x)+φ(xy)+φ(y)wtφ(xy)=φ(x)+φ(xy)+φ(y). The total edge irregularity strength, tes(G) is defined as the smallest k for which G has an edge irregular total k-labeling. Some interesting studies on the total edge irregularity strength can be seen in [1,2,3,4,8,9]. Further, Ryan, Munasinghe, and Tanna in [18] introduced the concept of the edge irregular reflexive. For a graph G(V,E) they define an edge labeling φe:E(G){1,2,3,...,ke} and a vertex labeling φv:V(G){0,2,4,...,2kv}, then defined the labeling φ by:

    φ(x)={φv(x),        ifxV(G)φe(x),        ifxE(G)

    is a total k-labeling, where k=max{ke,2kv}. Moreover, if for every two different edges xy and xy of G one has wtφ(xy)wtφ(xy), where wtφ(xy)=φe(xy)+φv(x)+φv(y), then the total k-labeling φ is called an edge irregular reflexive labeling of graph G. The reflexive edge strength, res(G), is defined as the smallest k for which G has an edge irregular reflexive k-labeling. For more research on reflexive edge strength see [6,10,11,15,20,21,22]. In this paper, we estimate the exact value of the reflexive edge strength for the r-th power of the path Pn, where r2, nr+4.

    Definition 1.1. ([19]) The r-th power of a graph G, denoted by Gr, is a graph with the same vertex set of G such that adding edges between the vertices which are at distance at most r, see Figure 1.

    Figure 1.  The 3-th power of P8.

    When we prove the result, we will often use the following lemma, which has been proved in [18].

    Lemma 1.1. ([18]). For every graph G,

    res(G){|E(G)|3,               if|E(G)|2,3(mod6),|E(G)|3+1,          if|E(G)|2,3(mod6).

    Furthermore, Bača et al. [10] proposed the following conjecture:

    Conjecture 1.1. ([10]) Consider the graph G, which has a maximum degree =(G). Hence:

    res(G)=max{+22,|E(G)|3+r}

    where r=1 for |E(G)|2,3 (mod 6), and zero otherwise.

    The r-th power of a path Pn denoted by Prn, n3,r2. Let us denote to the vertex set and edge set of Prn by V(Prn)={xi,1in} and E(Prn)=j=1r{xixi+j,1inj}. In the next theorem, we determine the reflexive edge strength of various powers of a path Pn.

    Theorem 1. For the r-th power of a path Pn, r2, nr+4.

    res(Prn)={r(2nr1)6,               ifr(2nr1)22,3(mod6),r(2nr1)6+1,          ifr(2nr1)22,3(mod6).

    Proof. Note that the r-th power of Pn has r(2nr1)2 edges. The lower bound for res of the r-th power of Pn is as follow from the Lemma 1. res(Prn)k=r(2nr1)6+1 if r(2nr1)22,3 (mod 6) and res(Prn)k=r(2nr1)6 if r(2nr1)22,3 (mod 6). Moreover, we prove that:

    res(Prn){r(2nr1)6,               ifr(2nr1)22,3(mod6),r(2nr1)6+1,          ifr(2nr1)22,3(mod6).

    Let k=max{2k+r(r+1)42r,3} and for nr+4, we recognise two cases.

    Case 1. When nk2r.

    Construct the total k-labeling φ of Prn in the following way:

    The corresponding labeling for P38 is illustrated in Figure 2. Otherwise we have the following labeling:

    φ(xi)={0,               1ik2rk4,       k+1ik+nk2k,               k+nk2+1in
    Figure 2.  A reflexive irregular 6-labeling of P38.

    Furthermore, the labels of edges are defined as the following :

    φ(xixi+1)={i,                                            1ik1r(2kr1)22rk4+1,                i=k(r1)k4rk4+i,                k+1ik+nk21r(2kr1)2+rnk2k2rk4+1,                        i=k+nk2r(2nr1)2nk2(nk23)2n2k+i,                           k+nk2+1in1
    φ(xixi+2)={k+i1,                                 1ik2(r1)kr(r+1)22rk4+i+3,                         k1ik(r1)k4rk4++nk2+i1,                         k+1ik+nk22(r1)(k+nk2)r(r+1)22rk4k+i+3,                   k+nk21ik+nk2r(2nr1)2nk2(nk25)2n2k+i1,                          k+nk2+1in2

    For 3jr, there exist three subcases:

    Subcase 1.1. If nk2>r, hence the edges label as following:

    φ(xixi+j)={φ(xkj+1xk)+i,                     1ikjφ(xkxk+j1)k+j+i,          kj+1ikφ(xk+nk2j+1xk+nk2)k+i,                                    k+1ik+nk2jφ(xk+nk2xk+nk2+j1)nk2k+j+i,                 k+nk2j+1ik+nk2φ(xnj+1xn)nk2k+i,      k+nk2+1inj

    Subcase 1.2. If nk2=r, here for 3jr1 the edge labels are common the pervious subcase, and for j=r we define edge labels as follows:

    φ(xixi+r)={φ(xkr+1xk)+i,                                      1ikrφ(xkxk+r1)k+r+i,                         kr+1ikφ(xk+nk2xk+nk2+r1)k+i,            k+1ik+nk2.

    Subcase 1.3. If nk2=r1, for 3jr2 the edge labels are common the Subcase 1.1 now, we construct edge labels only for j=r1 and j=r as follows:

    φ(xixi+r1)={φ(xkr+2xk)+i,                          1ikr+1φ(xkxk+r2)k+r+i1,       kr+2ikφ(xk+nk2r+2xk+nk2)k+i,                                        k+1ik+nk2r+1φ(xk+nk2xk+nk2+r2)knk2+r+i1,               k+nk2r+2ik+nk21.
    φ(xixi+r)={φ(xkr+1xk)+i,                                  1ikrφ(xkxk+r1)k+r+i,                      kr+1ikφ(xk+nk2xk+nk2+r1)k+i,        k+1ik+nk21.

    An explanation of above labeling is depicted in Figure 3. Evidently the vertices of Prn labeled with even numbers. Hence we will compute the weights of edges under the labeling φ:

    Figure 3.  A reflexive irregular 10-labeling of P312.

    The edge set of Prn can be divided into five mutually separated subsets, As,1s5 as follows:

    For 1jr,

    A1={xixi+j:1ikj} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:kjik}: The set of all edges which have endpoints tagged with 0 and 2rk4,

    A3={xixi+j:kik+nk2j}: The set of all edges which have endpoints tagged with 2rk4,

    A4={xixi+j:k+nk2rik+nk2}: The set of all edges which have endpoints tagged with 2rk4 and k,

    A5={xixi+j:k+nk2+1inj}: The set of all edges which have endpoints tagged with k.

    Therefore, the edge weights of Prn under the labeling φ are the following:

    1. The edge weights of the set A1, get the consecutive numbers from the set {1,2,...,r(2kr1)2},

    2. The edge weights of the set A2, receive the consecutive numbers from the set {r(2kr1)2+1,...,rk},

    3. The edge weights of the set A3, get the consecutive numbers from the set {rk+1,...,r(2kr1)2+rnk2},

    4. The edge weights of the set A4, get the consecutive numbers from the set {r(2kr1)2+rnk2+1,...,r(2nr1)2nk2(nk21)2},

    5. Finally, the edge weights of the set A5, receive the consecutive numbers from the set {r(2nr1)2nk2(nk21)2+1,...,r(2nr1)2}.

    An explanation of above corresponding weights is depicted in Figure 4. It is easy to check that the weights of the edges are different numbers from the set {1,2,3,...,r(2nr1)2}.

    Figure 4.  The edge weights of P312.

    Case 2. When nk2<r.

    In this case we have three subcases.

    Subcase 2.1. If k<n2. We can define the total k-labeling φ of Prn as follows:

    The corresponding labelings for P37, P59 and P410 are illustrated in Figure 7, 8 and 9 respectively. Otherwise we have following labeling:

    φ(xi)={0,               1ik2k4,         k+1ink+1k,               nk+2in
    Figure 5.  A reflexive irregular 22-labeling of P713.
    Figure 6.  The edge weights of P713.
    Figure 7.  A reflexive irregular 6-labeling of P37.
    Figure 8.  A reflexive irregular 10-labeling of P59.
    Figure 9.  A reflexive irregular 10-labeling of P410.

    Moreover, to define edge labels for Prn we have two subcases:

    Subcase 2.1.1 If kr, then we construct the edge labels as follows:

    φ(xixi+1)={i,                                                          1ik1r(2kr1)22k4+1,                             i=k(r1)k4k4+i,                              k+1inkrk+(n2k+1)(n2k)22k4k+1,                                      i=nk+1r(2nr1)2(k1)(k4)2n2k+i,                                        nk+2in1

    For 1<j<n2k+1,

    φ(xixi+j)={φ(xkj+1xk)+i,                                     1ikjφ(xkxk+j1)k+i+j,                         kj+1ikφ(xnkj+1xnk)k+i,                        k+1inkjφ(xnkxnk+j1+knk+i,            nkj+1inkφ(xnj+1xn)kn+i,                         nk+1inj.
    φ(xixi+n2k+1)={φ(x3knxk)+i,                              1i3kn1φ(xkxnk)+n3k+i+1,         3knikφ(xnkx2n3k)k+i,                 k+1ink+1φ(x2kxn)n+k+i1,              nk+2i2k1. (2.1)
    φ(xixi+n2k+2)={φ(x3kn1xk)+i,                              1i3kn2φ(xkxnk1)+n3k+i+2,          3kn1ik1r(2kr1)2+(n2k+1)(2rn+2k)2k+1,                                               i=kφ(xnkx2n+3k)k+i,                      k+1ink+1φ(x2k1xn)n+k+i,                     nk+2i2k2

    Now, for n2k+3ir we have three subcases:

    Subcase 2.1.1.1 If kr+2, hence the labels of edges are defined as follows:

    φ(xixi+j)={φ(xkj+1xk)+i,                                 1ikjφ(xnkj+2xnk+1)k+j+i,           kj+1inkj+1φ(xkxk+j1)n+k+j+i1,       nkj+2ikφ(xnk+1xnk+j)k+i,                   k+1ink+1φ(xnj+1xn)n+k+i1,               nk+2inj. (2.2)

    Subcase 2.1.1.2 If k=r+1, thus the edges n2k+3jr1, xixi+j, 1inj are labeled by Eq (2.2) and for j=r the edges are labeled as follows:

    φ(xixi+r)={φ(xkr+1xk)+i,                                1ikrφ(xnkr+2xnk+1)k+r+i,           kr+1inkr+1φ(xkxk+r1)n+k+r+i1,        nkr+2ikφ(xnk+1xnk+r)k+i,                    k+1ink+1.

    Subcase 2.1.1.3 If k=r, then the edges xixi+j, n2k+3jr2, 1inj are labeled by Eq (2.2) and for j=r1 and j=r the edges are labeled as follows:

    φ(xixi+r1)={φ(x2xk)+1,                                     i=1φ(xn2k+3xnk+1)+i1,                 2in2k+2φ(xkx2k2)n+2k+i2,            n2k+3ikφ(xnk+1xn1)k+i,                     k+1ink+1.
    φ(xixi+r)={φ(xn2k+2xnk+1)+i,                         1in2k+1φ(xkx2k1)n+2k+i1,              n2k+2ikφ(xnk+1xn)k+i,                             k+1ink.

    Subcase 2.1.2 If k=r1, hence we can defined the edge labels as follows:

    φ(xixi+1)={i,                                                       1ik1k(k1)22k4+1,                              i=kk24k4+i,                                    k+1inkn(n+1)+2k(3k2n)+222k4k,              i=nk+1k(nk+1)2k+i3,                  nk+2in1.

    For j=n2k+1, we used the Eq (2.1) to label the edges and for j=n2k+2 the edge labels given by:

    φ(xixi+n2k+2)={φ(x3kn1xk)+i,                            1i3kn2φ(xkxnk1)+n3k+i+2,         3kn1ik1k(6n7k1)n(n1)+22k,                      i=kφ(xnkx2n+3k)k+i,                    k+1ink+1φ(x2k1xn)n+k+i1,              nk+2i2k2.

    Further, the edges for n2k+3jr3 are labeled by Eq (2.2) and for j=r2, j=r1, and j=r the edges are labeled as follows:

    φ(xixi+r2)={φ(x2xk)+1,                                      i=1φ(xn2k+3xnk+1)+i1,                   2in2k+2φ(xkx2k2)n+2k+i2,             n2k+3ikφ(xnk+1xn1)k+i,                     k+1ink+1.
    φ(xixi+r1)={φ(xnkr+3xnk+1)+i,                   1in2k+1φ(xkx2k1)n+2k+i1,          n2k+2ikφ(xnk+1xn)k+i,                      k+1ink.
    φ(xixi+r)={φ(xn2k+1xnk+1)+i,                          1in2kφ(xkx2k)+i,                                       n2k+1ik.

    An explanation of above labeling is depicted in Figure 5.

    Also in this case the vertices are labeled with even number. Now we will estimate the weights of edges under the labeling φ:

    We split the edge set of Prn into six mutually separated subsets, As, 1s6 as follows: In Subcase 2.1.1.1, and Subcase 2.1.1.2 we have:

    A1={xixi+j:1jr,1ikj} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:1jn2k+1,kj+1ik}{xixi+j:n2k+2jr,kj+1inkj+1}: The set of all edges which have endpoints tagged with 0 and 2k4,

    A3={xixi+j:1jn2k,k+1in2kj+1}: The set of all edges which have endpoints tagged with 2k4,

    A4={xixi+j:1jn2k+1,nkj+2ink+1}{xixi+j:n2k+2jr,k+1ink+1}: The set of all edges which have endpoints tagged with 2k4 and k,

    A5={xixi+j:1jk2,nk+2inj}: The set of all edges which have endpoints tagged with k,

    A6={xi,xi+j,n2k+2jr,nkj+2ik}: The set of all edges which have endpoints tagged with 0 and k.

    In Subcase 2.1.1.3, A1={xixi+j:1jr1,1ikj} and other subsets As, 2s6 as in the Subcase 2.1.1.1, and in Subcase 2.1.1.2, A1={xixi+j:1jk1,1ikj}, A2={xixi+j:1jn2k+1,kj+1ik}{xixi+j:n2k+2jk,kj+1inkj+1}{x1xnk+1} and other subsets As, 3s6 as in the Subcase 2.1.1.1. Therefore, we obtain the edge weights for the Subcases 2.1.1.1, 2.1.1.2, and 2.1.1.3 as follows:

    1. The edges of the set A1, obtain weights from the set of sequential integers {1,2,...,r(2kr1)2},

    2. The edges of the set A2, obtain weights from the set of sequential integers {r(2kr1)2+1,...,r(2kr1)2+(n2k+1)(2rn+2k)2},

    3. The edges of the set A3, obtain weights from the set of sequential integers {rk+1,...,rk+(n2k+1)(n2k)2},

    4. The edges of the set A4, obtain weights from the set of sequential integers {rk+(n2k+1)(n2k)2+1,...,r(2nr1)2(k1)(k4)2+k+1)},

    5. The edges of the set A5, get weights from the set of sequential integers {r(2nr1)2(k1)(k4)2+k+2,...,r(2nr1)2},

    6. Finally, the edges of the set A6, get weights from the set of sequential integers {r(2kr1)2+(n2k+1)(2rn+2k)2+1,...,rk}.

    An explanation of above corresponding weights is depicted in Figure 6.

    In the Subcase 2.1.2, the edge weights are obtained as follows :

    1. The edges of the set A1, get weights from the set of sequential integers {1,2,...,k(k1)2},

    2. The edges of the set A2, get weights from the set of sequential integers {k(k1)2+1,...,k(6n7k1)n(n1)+221)},

    3. The edges of the set A3, admit weights from the set of sequential integers {k2+k+1,...,n(n+1)+2k(3k2n)+221},

    4. The edges of the set A4, admit weights from the set of sequential integers {n(n+1)+2k(3k2n)+22,...,k(nk)+n2},

    5. The edges of the set A5, receive weights from the set of sequential integers {k(nk)+n1,...,r(2nr1)2},

    6. Finally, The edges of the set A6, receive weights from the set of sequential integers {k(6n7k1)n(n1)+22,...,k2+k}.

    It is not hard to see that the weights of edges are distinct numbers from the set {1,2,3,...,(r(2nr1)2}.

    Subcase 2.2. If k=n2. Define the total k-labeling φ of Prn as follows:

    The corresponding labeling for P48 is illustrated in Figure 10. Otherwise we have following labeling:

    φ(xi)={0,                             1ik12(k1)(k2)4,            kik+1k,                              k+2in
    Figure 10.  A reflexive irregular 8-labeling of P48.

    Now we define the edge labels as follows:

    For 1j3 we have two subcases.

    Subcase 2.2.1 If (k1)(k+2)2k,

    φ(xixi+1)={i,                                                                       1ik2(k1)(k2)22(k1)(k2)4+1,                           i=k1k(4rk1)r(r+1)24(k1)(k2)4+1,                  i=kk(4rk1)r(r+1)22(k1)(k2)4k+2,            i=k+1r(2nr1)2k2k+422k+i,                             k+2in1
    φ(xixi+2)={k+i2,                                                            1ik3k25k+222(k1)(k2)4+4+i,                           k2ik1k(4rk3)r(r+1)22(k1)(k2)4+i+3,                 kik+1(1+r)nr(r+1)2k2+k+822k+i,                       k+2in2
    φ(xixi+3)={2k+i5,                                                1ik4k25k+222(k1)(k2)4+7+i,                  k3ik2k2+k22k+1,                                           i=k1k(4rk3)r(r+1)22(k1)(k2)4k+i+5,                                                   kik+1(2+r)nr(r+1)2k2+3k+1422k+i,            k+2in3

    Subcase 2.2.2 If (k1)(k+2)2<k,

    φ(xixi+1)={i,                                                              1ik2(k1)(k2)22(k1)(k2)4+1,                   i=k1(rk)(k1)+(2kr1)(r2)24(k1)(k2)4+k+1,                                i=k(rk)(k1)+(2kr1)(r2)22(k1)(k2)4+2,                                       i=k+1(rk+2)(k1)+(2kr1)(r2)2kk+i,                                                k+2in1
    φ(xixi+2)={k+i2,                                                  1ik3(k1)(k2)22(k1)(k2)4k+i+4,                                                  k2ik1(rk)(k1)+(2kr1)(r2)24(k1)(k2)4+kk+i+3,                     kik+1(rk+2)(k1)+(2kr1)(r2)2++n2kk+i2,                                   k+2in2
    φ(xixi+3)={2k+i5,                                             1ik4(k1)(k2)22(k1)(k2)4k+i+7,                                            k3ik21,                                                           i=k1(rk)(k1)+(2kr1)(r2)22(k1)(k2)4k+i+5,                    kik+1(rk+2)(k1)+(2kr1)(r2)2++2n3kk+i5,                             k+2in3

    For 4jk2 (if k6),

    φ(xixi+j)={φ(xkjxk1)+i,                                1ikj1φ(xkj2xk+1)k+j+i+1,           kjikj+1φ(xk1xk+j2)k+j+i,                  kj+1ik1φ(xk+1xk+j)k+i+1,                     kik+1φ(xnj1xn)k+i1,                      k+2inj.
    φ(xixi+k1)={φ(x3xk+1)+i,                               1i2φ(xk1x2k3)+i1,                     3ik1φ(xk+1x2k1)k+i+1,               kik+1.
    φ(xixi+k)={φ(x2xk+1)+1,                             i=1φ(xk1x2k2)+i1,                  2ik1φ(xk+1x2k)k+i+1,              kik+1.
    φ(xixi+k+1)=φ(xk1x2k1)+i,              kink1.

    For k+2j2rk+1,

    φ(xixi+j)=φ(xnj+1xn)+i,              1inj.

    Thus the vertices are labeled with even numbers. Now we will calculate the weights of edges under the labeling φ:

    Like in the previous case the edge set of Prn can be partied into six mutually separated subsets As,1i6 as follows:

    A1={xixi+j:1jk2,1ikj1} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:1j2,kjik1}{xixi+j:3jk1,kjikj+1}{x1xk+1}: The set of all edges which have endpoints tagged with 0 and 2(k1)(k2)4,

    A3={xkxk+1}: The set of only one edge which has endpoints 2(k1)(k2)4,

    A4={xkxk+j,xk+1xk+j+1:3jn1,}{xk+1xk+2,xkxn}: The set of all edges which have endpoints tagged with 2k4 and k,

    A5={xixi+j:1jk2,k+2inj}: The set of all edges which have endpoints tagged with k,

    A6={xixi+j,3jr,kj+2ik1}: The set of all edges which have endpoints tagged with 0 and k.

    Accordingly we obtain the edge weights as follows:

    In the Subcase 2.2.1,

    1. The edge weights of the set A1, admit the successive numbers from the set {1,2,...,(k1)(k2)2},

    2. The edge weights of the set A2, receive the successive numbers from the set {(k1)(k2)2+1,...,k(k+1)22},

    3. The edge weight of the set A3, admits the number {k(4rk1)r(r+1)2+1};

    4. The edge weights of the set A4, admit the successive numbers from the set {k(4rk1)r(r+1)2+2,...,r(2nr1)2k23k2},

    5. The edge weights of the set A5, receive the successive numbers from the set r(2nr1)2k23k2+1,...,r(2nr1)2},

    6. Finally, the edge weights of the set A6, receive the successive numbers from the set {k2+k22+1,...,k(4rk1)r(r+1)2}.

    In the Subcase 2.2.2,

    1. The edge weights of the set A1, admit the successive numbers from the set {1,2,...,(k1)(k2)2},

    2. The edge weights of the set A2, receive the successive numbers from the set {(k1)(k2)2+1,...,(k1)(k+2)2},

    3. The edge weight of the set A3, admits the number {(rk)(k1)+(2kr1)(r2)2+k+1},

    4. The edge weights of the set A4, admit the successive numbers from the set {(rk)(k1)+(2kr1)(r2)2+k,...,(rk+2)(k1)+(2kr1)(r2)2+k+1},

    5. The edge weights of the set A5, receive the successive numbers from the set {(rk+2)(k1)+(2kr1)(r2)2+k+2,...,(2rk+2)(k1)2+(2kr1)(r2)2+k+1},

    6. The edge weights of the set A6, receive the successive numbers from the set {k+1,...,(rk)(k1)+(2kr1)(r2)2+k}.

    Hence the edge weights in the Subcases 2.2.1 and 2.2.2 are distinct numbers from the sets {1,2,3,...,r(2nr1)2} and {1,2,3,...,(2rk+2)(k1)2+(2kr1)(r2)2+k+1} respectively.

    Subcase 2.3 If k>n2, then we have the following subcases:

    Subcase 2.3.1 If n is odd, we construct the total k-labeling φ of Prn as follows:

    φ(xi)={0,                                                            1ink12(nk1)(nk2)4,                                       nkin+122(nk1)(nk2)4+2(nk1)(2kn+3)4,       n+32ik+1k,                                                              k+2in.

    Now, to define the edge labeling we have to consider the following two subcases:

    Subcase 2.3.1.1 If 2kn=1,

    φ(xixi+1)={i,                                                 1ik3(k2)(k3)22(k2)(k3)4+1,          i=k2
    φ(xk1xk)={r(2nr1)2(k2)(k+3)24(k2)(k3)42,                            if4(k2)(k3)4k(k2)(k+3)24(k2)(k3)4+1,             if4(k2)(k3)4<k.
    φ(xixi+1)={r(2nr1)2(k2)(k+7)24(k2)(k3)41,                                  i=kr(2nr1)2(k2)(k+3)22(k2)(k3)4k+1,                            i=k+1r(2nr1)2k23k+822k+i,                   k+2in1.
    φ(xixi+2)={k3+i,                                             1ik4k27k+1622(k2)(k3)4+i,                k3ik2r(2nr1)2(k2)(k+7)24(k2)(k3)4,                                     i=k1r(2nr1)2(k2)(k+3)22(k2)(k3)4k+1,                        i=kr(2nr1)2(k2)(k+3)22(k2)(k3)4k+2,                          i=k+1(2nr)(r+1)2k2k622k+i,                  k+2in2.
    φ(xixi+3)={k+i7,                                                  1ik5k27k+2222(k2)(k3)4+i,                       k4ik3(k2)(k3)22(k2)(k3)4+1,                    i=k2r(2nr1)2(k2)(k+5)22(k2)(k3)4k+i+3,                            k1ikr(2nr1)2(k2)(k+3)22(k2)(k3)4k+3,                              i=k+1(2nr)(r+1)2k2+k+182+n2k+i,                 k+2in3.
    φ(xixi+4)={k+i12,                                                 1ik6k27k+2822(k2)(k3)4+i,                      k5ik4(k2)(k3)22(k2)(k3)4+2,                   i=k3.
    φ(xk2xk+2)={(k2)(k+3)2k+1,               if4(k2)(k3)4k(k2)(k+3)2k+2,               if4(k2)(k3)4 <k.
    φ(xixi+4)={r(2nr1)2(k2)(k+5)22(k2)(k3)4k+i+5,                                                k1ikr(2nr1)2(k2)(k+3)22(k2)(k3)4k+4,                                                    i=k+1(2nr)(r+1)2k2+3k+262+2n2k+i,          k+2in4.

    For 5jk3 (if k8),

    φ(xixi+j)={φ(xkj1xk2)+i,                                1ikj2φ(xkj+1xk)k+j+i+2,                kj1ikjφ(xkj+2xk+1)+1,                                i=kj+1φ(xk2xk+j3)k+j+i1,              kj+2ik2φ(xkxk+j1)k+i+2,                      k1ikφ(xk+1xk+j)+1,                                  i=k+1φ(xnj+1xn)k+i1,                      k+2inj.
    φ(xixi+k2)={φ(x3xk)+i,                        1i2φ(x4xk+1)+1,                     i=3φ(xk2x2k5)3,               4ik2φ(xkx2k3)k+i+2,      k1ikφ(xk+1x2k2)+1,               i=k+1.
    φ(xixi+k1)={φ(x2xk)+1,                          i=1φ(x3xk+1)+1,                         i=2φ(xk2x2k4)+i2,               3ik2φ(xkx2k2)k+i+2,          k1ik.
    φ(xixi+k)={φ(x2xk+1)+1,                        i=1φ(xk2x2k3)+1,                  2ik2φ(xkx2k1)+1,                      i=k1.
    φ(xixi+k+1)=φ(xk2x2k2)+i,                         1ink1.

    For k+2jr,

    φ(xixi+j)=φ(xnj+1xn)+i,                         1inj.

    An explanation of above labeling is depicted in Figure 11. In this case, we split edge set of the Prn into nine mutually separated subsets as follows:

    Figure 11.  A reflexive irregular 24-labeling of P913.

    A1={xixi+j:1jk3,1ikj2} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:2jk2,kjikj+1}{xk2xk1,x1xk}: The set of all edges which have endpoints tagged with 0 and 2(k2)(k3)4,

    A3={xk1xk}: The set of only one edge which has endpoints labeled with 2(k2)(k3)4,

    A4={xk1xk+1,xk1xk+1}: The set of two edges which has endpoints labeled with 2(k2)(k3)4 and 2k+2(k2)(k3)44,

    A5={xixi+j:4jk+1,kj+2ik2}{xixi+j:k+1jr,1inj}: The set of all edges which have endpoints tagged with 0 and k,

    A6={xk1xk+j1,xkxk+j:3jk1}{x2xk+2,xk1xn}: The set of all edges which have endpoints tagged with 2(k2)(k3)4 and k,

    A7={xk+1xk+j:1jk1}: The set of all edges which have endpoints tagged with 2k+2(k2)(k3)44 and k,

    A8={xkj+1xk+1:3jk}: The set of all edges which have endpoints tagged with 0 and 2k+2(k2)(k3)44,

    A9={xixi+j:1jk3,k+2inj}: The set of all edges which have endpoints tagged with k.

    Observe that under the total k-labeling φ the edge (edges):

    1. from the set A1, receive the weights from the successive numbers {1,2,...,(k2)(k3)2},

    2. from the set A2, receive the weights from the successive numbers {(k2)(k3)2+1,...,(k1)(k+2)2},

    3. from the set A3, receives the weight {r(2nr1)2(k2)(k+3)22} if 4(k2)(k3)4k or receives the weight {(k2)(k+3)2+1} if 4(k2)(k3)4<k,

    4. from the set A4, admit the two weights {r(2nr1)2(k2)(k+3)21,r(2nr1)2(k2)(k+3)2},

    5. from the set A5, receive the weights from the successive numbers {(k1)(k+2)2+1,...,r(2nr1)2(k2)(k+3)23}, if 4(k2)(k3)4k or receive the weights from the set {(k2)(k+3)2+2,...,r(2nr1)2(k2)(k+3)2}, if 4(k2)(k3)4<k,

    6. from the set A6, admit the weights from the successive numbers {r(2nr1)2(k2)(k+3)2+1,...,r(2nr1)2(k2)(k+1)2},

    7. from the set A7, receive the weights from the successive numbers {r(2nr1)2(k2)(k1)2+1,...,r(2nr1)2(k2)(k3)2},

    8. from the set A8, admit the weights from the successive numbers {(k2)(k+1)2+1,...,(k2)(k+3)2},

    9. from the set A9, admit the weights from the successive numbers {r(2nr1)2(k2)(k3)2,...,r(2nr1)2}.

    An explanation of above corresponding weights is depicted in Figure 12.

    Figure 12.  The edge weights of P913.

    Subcase 2.3.1.2 If 2kn1, hence we define the edge labeling as follows:

    φ(xixi+1)={i,                                                                  1ink2(nk1)(nk2)22(nk1)(nk2)4+1,            i=nk1.

    For nkin12,

    φ(xixi+1)={r(2nr1)2(k2)(k+1)24(nk1)(nk2)4n+i,             if4(nk1)(nk2)4k(nk1)(3kn)24(nk1)(nk2)4+i,                  if4(nk1)(nk2)4<k
    φ(xixi+1)={r(2nr1)2(4k(n1)n(n4)3)84(nk1)(nk2)42(nk1)(2kn+3)4+1,            i=n+12r(2nr1)2(n24n+3)84(nk1)(nk2)44(nk1)(2kn+3)4n+32+i+1,                            n+32ikr(2nr1)2(nk1)(k1)22(nk1)(nk2)42(nk1)(2kn+3)4k+1,                                    i=k+1r(2nr1)2(nk1)(nk2)22kk+i1,              k+2in1.

    For 1<j2kn12,

    φ(xixi+j)={φ(xnkjxnk1)+i,                                1inkj1φ(xnk1xnk+j2)n+k++j+i+1,                                                  nkjink1φ(xn+12j+1xn+12)n+k+i+1,                  nkin+12jφ(xn+12j+1xn+12)n+12+j+i,                      n+12j+1in+12φ(xkj+2xk+1)n+32+i+1,                      n+32ikj+1φ(xk+1xk+j)k+j+i1,                      kj+2ik+1φ(xnj+1xn)k+i1,                            k+2inj.
    φ(xixi+2kn+12)={φ(xn32xnk1)+i,                             1i3n4k32φ(xnk1xn+12)3n4k32+i,              3n4k12ink1φ(x2n2k+22xn+12)+1,                            i=nkφ(xnk+1)n+k+i,                         nk+1in+12φ(xk+1x4kn+12)n+12+i,                    n+32ik+1φ(x3n2k+12xn)k+i1,                     k+2inj.
    φ(xixi+2kn+32)={φ(x3n4k32xnk1)+i,                       1i3n4k52φ(xnk1xn12)3n4k52+i,             3n4k32ink1φ(xn+12xk+1)n+k+i+1,                nkin124k(2k+5)12kn+3n(n4)+178++r(2nr1)22(nk1)(nk2)4k,        i=n+12φ(xk+1x4kn+32)n+12+i,                    n+32ik+1φ(x3n2k12xn)k+i1,                      k+2i3n2k32.
    φ(xixi+2kn+52)={φ(x3n4k52xnk1)+i,                         1i3n4k72φ(xnk1xn+12)3n4k52+i+1,          3n4k52ink2(nk1)(k+1)22(nk1)(nk2)42(nk1)(2kn+3)4+1                       i=nk1φ(xn12xk+1)n+k+i+1,                nkin32φ(xn+12xk+2)n32+i,                       n12in+12φ(xk+1x4kn+52)n+12+i,                  n+32ik+1φ(x3n2k32xn)k+i1,                   k+2i3n2k52.

    For 2kn+72j2kn+2,

    φ(xixi+j)={φ(xnkjxnk1)+i,                                1inkj1φ(xn+12j+1xn+12)n+k+j+i+1,           nkjin+12jφ(xnk1xnk+j2)n+12+j+i,              n+32jink1φ(xkj+2xk+1)n+k+i+1,                  nkikj+1φ(xn+12xn+12+j1)k+j+i1,              kj+2in+12φ(xk+1xk+j)n+12+i,                            n+32ik+1φ(xnj+1xn)k+i1,                            k+2inj.
    φ(xixi+2kn+3)={φ(x2n3k3xnk1)+i,              1i2n3k4φ(x3n4k32j+1xn+12)++3k2n+i+4,                      2n3k3i3n4k52φ(xnk1xk+1)++i3n4k52,                            3n4k32ink2
    φ(xnk1xk+2)={(nk1)(3kn+2)2k+1,              if   4(nk1)(nk2)4k(12nk8k212k+8n3n25)8k+1,                                     if   4(nk1)(nk2)4<k
    φ(xixi+2kn+3)={φ(xn+12x4kn+52)n+k+i+1,                 nkin+12φ(xk+1x3kn+3)n+12+i,                            n+32ik+1φ(x2n2k2xn)k+i1,                       k+2i2n2k3.

    For 2kn+4jnk2 (if 2n3k5>0),

    φ(xixi+j)={φ(xnkjxnk1)+i,                      1inkj1φ(xn+12j+1xn+12)n++k+j+i+1,                               1in+12jφ(xkj+2xk+1)n+12++j+i,                                           n+32jikj+1φ(xnk1xnk+j2)k+j+i1,                               kj+2ink1φ(xn+12xn+12+j1)+kn+i+1,                                      nkin+12φ(xk+12xk+j)n+12+i,                   n+32ik+1φ(xnj+1xn)k+i1,                  k+2inj.
    φ(xixi+nk1)={φ(x2kn+52xn+12)+i,                1i3kn+32φ(x2kn+3xk+1)2kn+52+i+1,                   2kn+52i4k2n+42φ(xnk1x2n2k3)2k+n+i2,                    2kn+3ink1φ(xn+12x3n2k32)n++k+i+1,                         nkin+12φ(xk+1xn1)n+12++k+i,                                n+32inj.

    For nkjn32,

    φ(xixi+j)={φ(xn+32jxn+12)+i,                                         1in+12jφ(xkj+2xk+1)n+12+j+i,                      n+32jikj+1φ(xnk1xnk+j2)k+j+i1,            kj+2ink1φ(xn+12xn12+j)n+k+i+1,                     nkin+12φ(xk+1xkj)n12+i,                                n+32inj.
    φ(xixi+n12)={φ(x2xn+12)+i,                                      i=1φ(x2kn+52xk+1)+i1,                        2i2kn+32φ(xnk1x3n2k52)2kn+32+i,          2kn+52ink1φ(xn+12xn1)n+k+i+1,                 nkin+12.

    For n+12jk,

    φ(xixi+j)={φ(xkj+2xk+1)+i,                                 1ikj+1φ(xnk1xnk+j2)k+j+i1,         kj+2ink1φ(xnj+1xn)n+k+i,                          nkinj.

    For k+1jr,

    φ(xixi+j)=φ(xnk1xnk+j2)+i,          1inj.

    Note that in this subcase the edge set of Prn can be partied into ten mutually separated subsets as follows:

    A1={xixi+j:1jnk2,1inkj1} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:1j2kn+32,nkjink1}{xixi+j:2kn+52jnk2,nkjin+12j}{xixi+j:nk1jn12,1in+12j}: The set of all edges which have endpoints tagged with 0 and 2(nk1)(nk2)4,

    A3={xixi+j:1j2kn+12,nkin+12j}: The set of all edges which have endpoints tagged with 2(nk1)(k2)4,

    A4={xixi+j:1j2kn+12,n+32jin+12}{xixi+2kn+32:nkin12}{xixi+j:2kn+52j2kn+1,nkikj+1}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4 and 2(nk1)(nk2)4+2(nk1)(2kn+3)4,

    A5={xixi+j:2kn+3jk+1,kj+2ink1}{xixi+j:k+2jr,1inj}: The set of all edges which have endpoints tagged with 0 and k,

    A6={xixi+j:2kn+32j2kn+2,kj+2in+12}{xixi+j:2kn+3jn12,nkin+12}{xixi+j:n+12jk,nkinj}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4 and k,

    A7={xixi+j:1j2kn+12,k+2ik+j+1}{xixi+j:2kn+32jnk1,n+32ik+1}{xixi+j:nkjn32,n+32inj}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4+2(nk1)(2kn+3)4 and k,

    A8={xixi+j:2kn+52j2kn+2,n+32jink1}{xixi+j:2kn+3jn+12,n+32jikj+1}{xixi+j:n+32jk,1ikj+1}: The set of all edges which have endpoints tagged with 0 and 2(nk1)(nk2)4+2(nk1)(2kn+3)4,

    A9={xixi+j:1j(2kn1)2,n+32ik+j1}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4+2(nk1)(2kn+3)4,

    A10={xixi+j:1jnk2,k+2inj}: The set of all edges which have endpoints tagged with k.

    It is obvious that under the total k-labeling φ the edge (edges):

    1. from the set A1, receive the weights from the successive numbers {1,2,...,(nk1)(nk2)2},

    2. from the set A2, receive the weights from the successive numbers {(nk1)(nk2)2+1,...,(nk1)(k+1)2},

    3. from the set A3, obtain the weights from the successive numbers {r(2nr1)2k(k+1)2+1,...,r(2nr1)2(4k(n1)n(n4)3)8} if 4(nk1)(nk2)4k or obtain the weights from the set {(nk1)(3kn+2)2+1,...,(12nk8k212k+8n3n25)8} if 4(nk1)(nk2)4<k,

    4. from the set A4, admit the weights from the successive numbers {r(2nr1)2(4k(n1)n(n4)3)8+1,...,r(2nr1)212kn4k(2k+5)3n(n4)98},

    5. from the set A5, admit the weights from the successive numbers {(nk1)(3kn+2)2+1,...,r(2nr1)2(4k(n1)n(n4)3)8}, if 4(nk1)(nk2)4k or admit the weights from the set {(12nk8k212k+8n3n25)8+1,...,r(2nr1)2(4k(n1)n(n4)3)8}, if 4(nk1)(nk2)4<k,

    6. from the set A6, admit the weights from the successive numbers {r(2nr1)212kn4k(2k+5)3n(n4)98+1,...,r(2nr1)2(n2+4n+3)8},

    7. from the set A7, obtain the weights from the successive numbers {r(2nr1)2(nk1)(k1)2+1,...,r(2nr1)2(nk1)(nk2)2},

    8. from the set A8, obtain the weights from the successive numbers {(nk1)(k+1)2+1,...,(nk1)(3kn+2)2},

    9. from the set A9, admit the weights from the successive numbers {r(2nr1)2(n2+4n+3)8+1,...,r(2nr1)2(nk1)(k1)2},

    10. from the set A10, admit the weights from the successive numbers {r(2nr1)2(nk1)(nk2)2+1,...,r(2nr1)2}.

    Subcase 2.3.2 If n is even, therefore we construct the total k-labeling φ of Prn as follows:

    φ(xi)={0,                                                            1ink12(nk1)(nk2)4,                                     nkin22(nk1)(nk2)4+2(nk1)(2kn+2)4,     n+22ik+1k,                                                             k+2in.
    φ(xixi+1)={i,                                                                   1ink2(nk1)(nk2)22(nk1)(nk2)4+1,            i=nk1(nk1)(3kn)24(nk1)(nk2)4+i,              nkin21r(2nr1)2n(4kn+2)84(nk1)(nk2)42(nk1)(2kn+2)4+1,                                 i=n2r(2nr1)2(2kn+2)(3n2k4)8k(nk1)24(nk1)(nk2)44(nk1)(2kn+3)4n2+i,                                                           n2+1ikr(2nr1)2k(nk1)22(nk1)(nk2)42(nk1)(2kn+2)4k+1,                             i=k+1r(2nr1)2(nk1)(nk2)22kk+i1,         k+2in1.

    For 2jkn2 (if kn21),

    φ(xixi+j)={φ(xnkjxnk1)+i,                            1inkj1φ(xnk1xnk+j2)n+k++j+i+1,                                            nkjink1φ(xn2j+1xn2)n+k+i+1,                 nkin2jφ(xn2xn2+j1)n2+j+i,                        n2j+1in2φ(xkj+2xk+1)n2+i,                        n2+1ikj+1φ(xk+1xk+j)k+j+i1,                kj+2ik+1φ(xnj+1xn)k+i1,                       k+2inj.
    φ(xixi+kn2+1)={φ(x3n22k1xnk1)+i,                   1i3n22k2φ(xnk1xnk)3n2+2k+i+2,                             3n22k1ink1φ(xn2xk)n+k+i+1,                  nkin2φ(xk+1x2kn2+1)n2+i,                  n2+1ik+1φ(x3n2kxn)k+i1,                   k+2i3n2k1.
    φ(xixi+kn2+2)={φ(x3n22k2xnk1)+i,                 1i3n22k3φ(xn2nk2xn2)3n2++2k+i+3,                                 3n22k2ink2k(nk1)22(nk1)(nk2)42(nk1)(2kn+2)4+1,                i=nk1φ(xn2xk+1)n+k+i+1,            nkin21r(2nr1)2(nk1)(3kn+2)22(nk1)(nk2)4k+1,             i=n2φ(xk+1x2kn2+2)n2+i,                n2+1ik+1φ(x3n2k1xn)k+i1,               k+2i3n2k2.

    For kn2+3j2kn+2,

    φ(xixi+j)={φ(xnkjxnk1)+i,                              1inkj1φ(xn2j+1xn2)n+k+j+i+1,              nkjin2jφ(xnk1xnk+j2)n2+j+i,                n2j+1ink1φ(xn2xn2+j1)n+k+i+1,                     nkin2φ(xk+1xk+j)n2+i,                               n2+1ik+1φ(xnj+1xn)k+i1,                          k+2inj.
    φ(xixi+2kn+3)={φ(x2n3k3xnk1)+i,               1i2n3k4φ(x3n22k2xn2)2n+3k++i+4,                                        2n3k3i3n22k3φ(xnk1xk+1)3n2+2k++i+3,                                        3n22k2ink24k(nk1)+(2kn+2)(3n2k4)8+k+1,                                        i=nk1φ(xn2x2kn2+2)n+k+i+1,       nkin2φ(xk+1x3kn+3)n2+i,               n2+1ik+1φ(x2n2k2xn)k+i1,             k+2i2n2k3.

    For 2kn+4jnk2 (if 2n3k5>0),

    φ(xixi+j)={φ(xnkjxnk1)+i,                     1inkj1φ(xn2j+1xn2)n+k+j++i+1,                                            nkjin2jφ(xkj+2xk+1)n2+j+i,               n2j+1ikj+1φ(xnk1xnk+j2)k++j+i1,                                      kj+2ink1φ(xn2xn2+j1)n+k+i+1,            nkin2φ(xk+1xk+j)n2+i,                         n2+1ik+1φ(xnj+1xn)k+i1,                   k+2inj.
    φ(xixi+nk1)={φ(xkn2+2xn2)n+k++i+1,                                     1ikn2+1φ(x2kn+3xk+1)n2+i,             kn2+2i2kn+2φ(xnk1xnk3)2k++n+i2,                               2kn+3ink1φ(xn2x3n2k2)n+k++i+1,                                    nkin2φ(xk+1xn1)n2+i,                 n2+1ik+1.

    For nkjn21,

    φ(xixi+j)={φ(xn2j+1xn2)+i,                              1in2jφ(xkj+2xk+1)n2+j+i,               n2j+1ikj+1φ(xnk1xnk+j2)k++j+i1,                                       kj+2ink1φ(xn2xn2+j1)n+k+i+1,             nkin2φ(xk+1xk+j)n2+i,                       n2+1inj.

    For n2jk,

    φ(xixi+j)={φ(xkj+2xk+1)+i,                                  1ikj+1φ(xnk1xnk+j2)k+j+i1,           kj+2ink1φ(xn2xn2+j1)n+k+i+1,                      nki nj.

    For k+1jr,

    φ(xixi+j)=φ(xnj+1xn)+i,                  1inj.

    As the pervious Subcase 2.3.1.2, the edge set can be divided into ten mutually separated subsets as follows:

    A1={xixi+j:1jnk2,1jnkj1} : The set of all edges which have endpoints tagged with 0,

    A2={xixi+j:1j2kn+22,nkjink1}{xixi+j:2kn+42jnk1,nkjin2j}{xixi+j:nkjn21,1in2j}: The set of all edges which have endpoints tagged with 0 and 2(nk1)(nk2)4,

    A3={xixi+j:1j2kn2,nkin2j}: The set of all edges which have endpoints tagged with 2(k2)(k3)4,

    A4={xixi+j:1j2kn+22,n+22jin2}{xixi+j:2kn+42j2kn+1,nkikj+1}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4 and 2(nk1)(nk2)4+2(nk1)(2kn+2)4,

    A5={xixi+j:2kn+3jk+1,kj+2ink1}{xixi+j:k+2jr,1inj}: The set of all edges which have endpoints tagged with 0 and k,

    A6={xixi+j:2kn+42j2kn+2,kj+2in2}{xixi+j:2kn+3jk,nkinj}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4 and k,

    A7={xixi+j:1j2kn+22,kj+2ik+1}{xixi+j:2kn+42jnk1,n2+1ik+1}{nkjn21,n2+1inj}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4+2(nk1)(2kn+2)4 and k,

    A8={xixi+j:2kn+42j2kn+2,n2j+1ink1}{xixi+j:2kn+3jn2,n2+1jikj+1}{xixi+j:n2+1jk,1ikj+1}: The set of all edges which have endpoints tagged with 0 and 2(nk1)(nk2)4+2(nk1)(2kn+2)4,

    A9={xixi+j:1j(2kn)2,n+32ik+j1}: The set of all edges which have endpoints tagged with 2(nk1)(nk2)4+2(nk1)(2kn+2)4,

    A10={xixi+j:1jnk2,k+2inj}: The set of all edges which have endpoints tagged with k.

    One can easily check that under the total k-labeling φ the edge (edges):

    1. from the set A1, get the weights from the successive numbers {1,2,...,(nk1)(nk2)2},

    2. from the set A2, get the weights from the successive numbers {(nk1)(nk2)2+1,...,k(nk1)2},

    3. from the set A3, get the weights from the successive numbers {(nk1)(3kn+2)2+1,...,4k(nk1)+(2kn+2)(3n2k4)8},

    4. from the set A4, get the weights from the successive numbers {r(2nr1)2n(4kn+2)8+1,...,r(2nr1)2(2kn+2)(3n2k4)8},

    5. from the set A5, admit the weights from the successive numbers\\ {4k(nk1)+(2kn+2)(3n2k4)8+1,...,r(2nr1)2n(4kn+2)8},

    6. from the set A6, admit the weights from the successive numbers {r(2nr1)2(nk1)(3kn+2)2+1,...,r(2nr1)2n(4kn+2)8},

    7. from the set A7, obtain the weights from the successive numbers {r(2nr1)2k(nk1)2+1,...,r(2nr1)2(nk1)(nk2)2},

    8. from the set A8, obtain the weights from the successive numbers {k(nk1)2+1,...,(nk1)(3kn+2)2},

    9. from the set A9, admit the weights from the successive numbers {r(2nr1)2(2kn+2)(3n2k4)8+1,...,r(2nr1)2(nk1)(3kn+2)2},

    10. from the set A10, admit the weights from the successive numbers {r(2nr1)2(nk1)(nk2)2+1,...,r(2nr1)2}.

    By direct computation we can verify that in the both subcases all vertices are labeled by even numbers and the edge weights of the edges are different numbers from the set {1,2,3,...,r(2nr1)2}. Hence we prove that in all cases vertices are even numbers and the labels of edges are less than or equal to k=r(2nr1)6+1 where r(2nr1)22,3 (mod 6) and less than or equal to k=r(2nr1)6 where r(2nr1)22,3 (mod 6). Thus the edge irregular reflexive strength of the r-th power of the path Prn:

    res(Prn)={r(2nr1)6, if r(2nr1)22,3(mod

    In this paper we discussed the edge irregular reflexive k-labeling of the r -th power of the path P_{n} , where r\geq2 , n\geq r+4 . Also, we computed the precise value of the reflexive edge strength of P^{r}_{n} , r\geq2 , n\geq r+4 . In the future, we would like to calculate the reflexive edge strength, res, for r-th power of other graphs.

    The authors declare that they have no competing interest.



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