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New classes of reverse super edge magic graphs

  • A reverse edge magic (REM) labeling of a graph G(V,E) with p vertices and q edges is a bijection f:V(G)E(G){1,2,,p+q} such that k=f(uv){f(u)+f(v)} is a constant k for any edge uvE(G). A REM labeling f is called reverse super edge magic (RSEM) labeling if f(V(G))={1,2,3,4,5,,v} and f(E(G))={v+1,v+2,v+3,v+4,v+5,,v+e}. In this paper, we find some new classes of RSEM labeling and the investigation of the connection between the RSEM labeling and different classes of labeling.

    Citation: Kotte Amaranadha Reddy, S Sharief Basha. New classes of reverse super edge magic graphs[J]. AIMS Mathematics, 2022, 7(3): 3590-3602. doi: 10.3934/math.2022198

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  • A reverse edge magic (REM) labeling of a graph G(V,E) with p vertices and q edges is a bijection f:V(G)E(G){1,2,,p+q} such that k=f(uv){f(u)+f(v)} is a constant k for any edge uvE(G). A REM labeling f is called reverse super edge magic (RSEM) labeling if f(V(G))={1,2,3,4,5,,v} and f(E(G))={v+1,v+2,v+3,v+4,v+5,,v+e}. In this paper, we find some new classes of RSEM labeling and the investigation of the connection between the RSEM labeling and different classes of labeling.



    The edge magic labelings of graphs were introduced by Kotzig and A. Rosa [1] and they called also magic valuations of graphs. In [2], The super edge magic labelings of a graph then the idea of edge magic labelings is proved by H. Enomoto et al. In [8], R. M. Figueroa Centeno et al. proved all caterpillars are super edge magic also verified that mK1,n,m and m,n are positive integers with the super edge magic is odd. In [4], M. Figueroa Centeno et al. defined that the forest PmK1,n,m4 each positive integer n1. All trees are edge magic is verified by G. Ringel and A. Llado [8]. H. Enomoto et al. proposed in [2] a more difficult hypothesis: that every tree is super edge magic. All the lobsters are gracefully demonstrated by J. C. Bermond [7].

    If G be the (super) 2-regular edge magic graph with n positive integers, then G¯Kn is (super) edge magic and therefore for every two integers m3 and m1, then n crown Cm¯Kn is super edge magic these results proved by R. Figueroa Centeno et al. [6]. V. Yegnanarayanan [3] demonstrated that the graph is obtained through edge magic for t2 also introduced new pendant edges of the outermost C3 in Pt×C3 at each vertex. In [10], The total graph T(Pn) is harmonious is obtained by R. Balakrishnan and R. Sampath kumar. In [2], H. Enomoto et al. is obtained that the complete bipartite graph Km,n is super edge-magic iff m=1 or n=1. R. Balakrishnan et al. [11] obtained that the harmonious iff n is even and the graph K2+2K2 is magic iff n=3. In [9], K. Kathiresan proved that the subdivision graph S(Ln) obtained by subdividing every edge of G exactly one is graceful. In [5], V. Yegnanarayanan introduced several other variations of magic labelings and discuss what are called vertex-magic and vertex-antimagic of (1,1),(1,0) and (0,1) graphs. Also, discussed edge-magic and edge-antimagic of (1,0) and (0,1) graphs. Finally, exhibited such magic, anti-magic labelings for a number of classes of graphs and derived several general results governing these graphs.

    A reverse edge magic (REM) labeling of a graph G(V,E) with p vertices and q edges is a bijection f:V(G)E(G){1,2,,p+q} such that k=f(uv){f(u)+f(v)} is a constant k for any edge uvE(G). A REM labeling f is called reverse super edge magic (RSEM) labeling if f(V(G))={1,2,3,4,5,,v} and f(E(G))={v+1,v+2,v+3,v+4,v+5,,v+e}. In this paper, we find some new classes of RSEM labeling and the investigation of the connection between the RSEM labeling and different classes of labeling.

    Definition 1. Let a be the path Pn,1in and T1 be a caterpillar obtained by position one end vertex at each vertex. Let T be the lobster created by linking a copy of P2 at each end vertex bi of 1in.

    The accompanying outcomes on trees give support to the conjecture that all trees are RSEM.

    Lemma 1. A graph G with p vertices and q edges is RSEM iff a bijective function f:V(G){1,2,...,p} so that the set S={f(x)+f(y):xyE(G)} contains q number of successive numbers. In such a case, f spreads to a RSEM labeling of the graph G with reverse magic constant k=p+qs, where s=max(S) and S={(p+1)k,(p+2)k,(p+3)k,...,(p+q)k}.

    Theorem 1. If m is odd. Then 3stat Sm,3 is RSEM.

    Proof. Let m is odd. Assume m be the degrees of vertex x in Sm,3 and 3 is the length of ith path of xuiviwi for 1im.

    The paths are RSEM and since S1,3P4, when m=1 the outcome is true. Assume that m is an odd number and m>3. Assume n=3m+1.

    Define the vertex labeling, f:V(Sm,3){1,2,3,4,5,...,n} such that

    f(x)=n+23f(u1)=n
    f(u2i)=2ifor1in46f(u2i+1)=n+53+2i1for1in46f(v2i)=n+53+2i1for1in46f(v2i+1)=2i+1for0in46
    f(w1)=2n+13
    f(w2i)=5n26i+1for1in46f(w2i+1)=nifor1in46.

    Note that

    S={f(x)+f(y):xyE(Sm,3),m3isodd}={4n+23,4n13,...,n+83,n+113},

    is one set of n1 successive integers. Accordingly, by using the Lemma 1, f extend to a RSEM labeling of Sm,3 with valence. k=p+qs=n+n1+n+83=2n53, when m3 is odd.

    Example 1. Figure 1 shows the RSEM labeling of the lobster T with n=13.

    Figure 1.  The RSEM labeling of the lobster T with n=13.

    Theorem 2. The lobster T characterized above is RSEM for total positive numbers n>3.

    Proof. We consider two cases.

    Case 1: If n is even.

    Let Ci denotes that the termination vertex of T at bi,1in.

    Characterize a vertex with labeling f:V(T){1,2,3,...,3n} such that

    f(ai)={i,ifiiseven,1in2n+i,ifiisodd,1in
    f(bi)={i,ifiiseven,1in2n+i,ifiisodd,1in
    f(c1)=2nf(c2i+3)=3n2(1+i)for0in42f(cn2i)=3n2+ifor0in22.

    Therefore,

    S={f(x)+f(y):xyE(T),niseven,n3}={2n2,2n1,2n,2n+1,...,5n4}.

    Accordingly, by using the Lemma 1, f extend to a RSEM labeling of T with valence k=p+qs=n+3.

    Example 2. Figure 2 shows the RSEM labeling of the lobster T with n=11.

    Figure 2.  The RSEM labeling of the lobster T with n=11.

    Case 2: If n is odd.

    Define the vertex labeling f:V(T){1,2,3,...,3n} here

    f(ai)={i,ifiiseven,1inn+i,ifiisodd,1in
    f(bi)={i,ifiiseven,1inn+i,ifiisodd,1in
    f(c1)=3nf(c2i+1)=3nifor1in12f(cn2i+1)=2n+ifor1in12.

    Since,

    S={f(x)+f(y):xyE(T),nisodd,n3}={n+2,n+3,2n,...,4n,}.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of T with valence k=p+qs=2n1.

    Example 3. Figure 3 shows the RSEM labeling of the lobster T with n=9.

    Figure 3.  The RSEM labeling of the lobster T with n=9.

    Definition 2. Let {a1k1,n1,a2k1,n2,...,apk1,np} be a family of stars where aik1,ni, ai denotes the isomorphic disjoint copies of k1,ni for 1ip and ai1. Let k1,n1 and vijk be the end vertices of Hij be the jth the isomorphic, k=1,2,...,ni if one end vertex of each star which is adjacent to a vertex w adjoin. Thus the trees subsequently defined by Ha1+a2++apw. These kinds of trees are prefers to as the banana tree.

    Theorem 3. The banana tree Ha1+a2+apw corresponding to the family of stars {a1k1,n1,a2k1,n2,...,apk1,np},1<n1<n2<...<np,p2 and a1+a2+...+aini,i=1,2,...,p is RSEM.

    Proof. Conider the family of stars {a1k1,n1,a2k1,n2,...,apk1,np}. Let k1,ni,i=1,2,...,p, is Hij be the jth the isomorphic copy. Assume Hij is the end-vertices of vijk,k=1,2,3,4,...,ni and uij be the Hij is center.

    Let the new vertex be w which is adjacent to one end vertex vijβij from every star Hij of the family where βij=a0+a1+...+ai1+j and a0=0. The new tree obtained is denoted by Ha1+a2+apw. and has a1(n1+1)+a2(n2+1)+...+ap(np+1) vertices and a1n1+a2n2+...+apnp+(a1+a2+...+ap) edges.

    Let p=a1(n1+1)+a2(n2+1)+...+ap(np+1).

    Define a vertex labeling f:V(Ha1+a2+apw){1,2,...,p1}, such that

    f(v1jk)=(j1)n1+kfor1ja1,1kn1.f(vijk)=f(vi1ai1ni1)+(j1)ni+k,for2ip,1ja1,1kn1.f(w)=f(vpapnp)+1.f(uij)=f(w)+(a0+a1+...+ai1+j),2ip,1ja1.

    Note that, S={a1n1+a2n2+...+apnp+2,a2n2+...+apnp+3,...,2(a1n1+a2n2+...+apnp)+(a1+...+ap)+1}.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of Ha1+a2+apw with valence k=p+qs=a0+a1+...+ap.

    Definition 3. Consider the graph G(t,m)=P1×C2m+1 where x have t vertices (t2) is an odd cycle when the cartesian product of the path. Consider a new graph G(t,m,n) by defining the new pendant edges n at each vertex of the furthest odd numbered cycle in G(t,m).

    Theorem 4. The graph G(t,m,n) is RSEM, for t2 and m2.

    Proof. Let C2m+1 be the fixed vertex of innermost of v11 and we will collect the different types of vertices v12,v13,...,v1(2m+1) in clock-wise. For 2it, let vi1 be the ith copy of C2m+1 vertex was end to end to the vertex v(i1)(2m+1) in the (i1)th copy of C2m+1 and take the other is adjacent to the vertex vijk is the outermost C2m+1 for 1kn and 1j(2m+1).

    Define the vertex marking f:V(G(t,m,n)){1,2,...,(2m+1)(t+n)} such that

    f(vij)={(i1)(2m+1)+j+12,ifjisodd,(i1)(2m+1)+m+j+22,ifjiseven,

    for 1it and 1j(2m+1), f(vijk)=(2m+1)(t+k1)+(2m+2j)for1j(2m+1)and1kn.

    Note that

    S={f(x)+f(y):xyE(G(t,m,n)),t2,m2}={m+2,m+3,...,(m+1)+(2m+1)(2t+n1)}

    is a set of all consecutive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of G(t,m,n) with valence k=p+qs=(2m+3)n+(2t+m),fort2andm2.

    Example 4. Figure 4 shows the RSEM labeling of the graph G(3,2,2) with n=22.

    Figure 4.  The RSEM labeling of the graph G(3,2,2) with n=22.

    Theorem 5. The graph CnP2 is RSEM for all odd n3.

    Proof. Let n=2m+13, here n is an odd integer. Let v1,v2,...,vn be the vertices of the cycle Cn. Now cnP2 is the graph defined by the attaching P2 to every vertex of Cn. Let the rim vertices vi of Cn in CnP2 is adjacent to the vertices ai,bi,1in. The graph CnP2 has 3n vertices and 4n edges. Consider a labeling of vertex f:V(CnPn{1,2,...,3n}) such that

    f(ai)={i+12,ifiisodd,m+i+22,ifiiseven,
    f(ai)=2n+1ifor1inf(b2i)=2n+ifor1imf(b1)=2n+m+1f(b2i+1)=2n+m+1+ifor1im.

    Define

    S={I(x)+f(y):xyE(cnp2)}={m+2,m+3,m+4,...,m+4n+1}

    is a set of 4n successive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of CnP2 with valence k=p+qs=15n12, when n3 is odd number.

    Example 5. Figure 5 shows the RSEM labeling of the graph C5P2 with n=12.

    Figure 5.  The RSEM labeling of the graph C5P2 with n=12.

    Theorem 6. The graph CnP3 is RSEM for every odd n3.

    Proof. Let Cn be an odd cycle with n=2m+13 vertices. The cycle Cn with the vertices v1,v2,...,vn. Let the path of three vertices is P3. Now 4n vertices and 6n edges is a graph CnP3 is obtained by attaching P3.

    Consider a vertex labeling f:V(CnP3){1,2,...,4n} such that

    f(vi)={i+12,ifiisodd,m+i+22,ifiiseven.

    If n is odd, Then m even for fvalues and the rim vertices of fvalues is m+1 odd. Let us the label 3n vertices outside the rim of Cn in CnP3 as follows. Let u1,u2,...um, be the vertex degree two outside the rim, fvalues are 2m,2m2,...,4,2 is adjacent to the rim vertices respectively. Again let un+1,un+2,...un+m be the remaining vertices of degree two, whose fvalues are 2m,2m2,...,4,2 is adjacent to the rim vertices respectively. Let um+1,um+2,...,un be the vertex degree two outside the rim, whose fvalues are n,n2,...,3,1 is adjacent to the rim vertices respectively. Also, let un+m+1,un+m+2,...,u2n be the continuing vertex degree two outside the rim, whose fvalues are n,n2,...,3,1 is adjacent to the rim vertices respectively. Let u2n+1.u2n+2,...,u2n+m+1 be the vertices of degree three outside the rim whose fvalues are n,n2,...,3,1 is adjacent to the rim vertices respectively. Finally, let u2n+m+2,u2n+m+3,...,u3n be the vertices of degree three whose fvalues are 2m,2m2,...,4,2 is adjacent to the rim vertices respectively.

    Consider f(ui)=n+ifor1i3n.

    Note that

    S={f(x)+f(y):xyE(CnP3)}={m+2,m+3,..,m+6n+1}

    is a set of all consecutive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of CnP3 with valence k=p+qs=7n12.

    Example 6. Figure 6 shows the RSEM labeling of the graph C7P3 with n=24.

    Figure 6.  The RSEM labeling of the graph C7P3 with n=24.

    Definition 4. Let Ln denote the ladder graph Pn×P2 and LnK1 be the graph containing the connecting an edge at every vertex of Ln.

    Theorem 7. The graph LnK1 is RSEM for odd n.

    Proof. Let V((Ln)={u1,u2,...,un;v1,v2,...,vn} and E((Ln)={uiui+1,vivi+1,ujvj,1i(n1),1jn}.

    Let u1i and v1i be the vertices is adjacent to the ui and vi respectively in LnK1. Then V((LnK1)={ui,vi,u1i,v1i:1 leqin,}. and V((LnK1)={uiui+1,vivi+1,ujvj,uju1j,vjv1j,1i(n1),1jn}. The graph LnK1 has 4n vertices and 5n2 edges.

    Define f:V(LnK1){1,2,...,4n} is the vertex labeling where

    f(x)={4n+i+12,ifx=uiiisoddand1in5n+i+12,ifx=uiiisevenand1in3n+i2,ifx=viiisoddand1in2n+i2,ifx=viiisevenand1inn,ifx=v117n+12,ifx=v12i,ifx=v12i+1,1i(n12)n+2i12,ifx=v12i,2i(n12)7n+2i+12,ifx=u12i1,2i(n12)3n+1+i,ifx=u12i,1i(n32)n+12,ifx=u1n13n+1,ifx=u1n.

    Note that S={f(x)+f(y):xyE(LnK1)}={3n+52,3n+72,...,13n12} is a set of alternative integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of LnK1 with valence k=p+qs=5n32, for all odd n. Accordingly, Lemma 1, if G is a RSEM labeling of (p,q) graph then q2p3.

    The next theorem gives a RSEM graph with q=2p3.

    Example 7. Figure 7 shows the RSEM labeling of the graph L5K1 with n=11.

    Figure 7.  The RSEM labeling of the graph L5K1 with n=11.

    Theorem 8. The total graph T(Pn) is RSEM for every integer n.

    Proof. Let Pn be the path u1,u2,...,un and ej be the edge uj,uj+1 for 1j(n1). Then the vertex and edge set of T(Pn) as denoted as V(T(Pn))={uj,ej:1jn,1j(n1).}

    Note that T(Pn) has 2n1 vertices and 4n5 edges, then q=2p3.

    Now define f:V(T(Pn)){1,2,...,(2n1)} as the vertex labeling such that

    f(ui)=2i1,for1inf(ei)=2i,for1i(n1).

    Since S={f(x)+f(y):xyE(T(Pn))}={3,4,...,(4n3)} is a set of successive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of T(Pn) with valence k=2n3.

    Example 8. Figure 8 shows the RSEM labeling of the graph T(P4) with n=5.

    Figure 8.  The RSEM labeling of the graph T(P4) with n=5.

    Theorem 9. The cycle graph Cn with a chord of distance 3 consisting two vertices is RSEM for each odd number n, where n>7.

    Proof: Let G be the graph and Cn is a chord with consisting two vertices of Cn(n7) at a distance 3. Let (G)={v1,v2,...,vn}, join the vertices v1 and vn2 as a chord for G so that d(v1,vn)=3. Note that G has n vertices and n+1 edges. Define f:V(G){1,2,...,n} the vertex labeling such that

    f(vi)={i+12, i is odd n+i+12, i is even. 

    Note that S={f(x)+f(y):xyE(G)}={n+12,n+32,...,3n+12} is a set of successive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of G with valence k=p+qs=n+12.

    Example 9. Figure 9 shows the RSEM labeling of the graph C7 with n=4.

    Figure 9.  The RSEM labeling of the graph C7 with n=4.

    Theorem 10. Let G1,G2,...,Gm be m disconnect and n cycles having vertex sets vi={vi1,vi2,...,vin},i=1,2,...,m here n is odd and n3. Let G the graph attained by connecting v1n to v2j,1jn and vkn to vk+1j,1jn,2j(m1). Then G is a RSEM graph.

    Proof. The graph G has containing mn vertices and n(2m1) edges.

    Consider f:V(G){1,2,...,mn} a vertex labeling such that

    f(v1i)={i+12,ifiisodd,1inn+1+i2,ifiiseven,1in

    f(vri)=5r4, if 2rm

    f(vri)={f(vr1)+i12,ifiisodd,1in,2rmf(vrn)+i2,ifiiseven,2in,2rm.

    It is easy to see that S={f(x)+f(y):xyE(G)}={n+32,n+52,...,n+10m7} is a set of n(2m1) successive integers.

    Accordingly, by using the Lemma 1,f extend to a RSEM labeling of G with valence k=p+qs=n(3m2)10m+7.

    Example 10. Figure 10 shows the RSEM labeling of the graph G1 with n=4.

    Figure 10.  The RSEM labeling of the graph G1 with n=4.

    Permitting to the outcome and argument we establish reverse edge magic valuation of the 3-star Sm,3 if m is odd, the lobster T characterized above is RSEM for all integers n>3, the banana tree Ha1+a2++apw. for t2 and m2 the graph G(t,m,n) the graphs CnP2,CnP3 for all odd m3, the graph LnK1 for odd n, the total graph T(Pn) for any positive integer n and the graph Cn is a cycle with a chord conncection two vertices at the distance of 3 units for all odd n,n>7.



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