# A Conjecture on Super Edge-Magic Total Labeling of 4-Cycle Books

Authors: Mudin Simanihuruk; Tri Atmojo Kusmayadi; Baki Swita; Minsi Romala; Friska Damanik
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

A graphG is called cycle books B4,m,2 if G consists of m cycles C4 with a common path P2. Figueroa-Centeno, Ichishima, and Muntaner-Batle conjecture that the graph B4,m,2 is super edge-magic total if and only if m is even or m≡5mod8. In this article, we prove this conjecture for m ≥ 36 and m  = 0 mod (2).


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## Body

## 1. Introduction

For undefined terms and notations in this study, we follow Chartrand, Lesniak, and Peng [1]. Let G be a graph with VG and EG be a set of vertices and edges, respectively. A graph G is called a p,qgraph if G has p and q number of vertices and edges, respectively. Kotzig and Rosa [2] defined that an edge-magic total labeling of G is a bijective function f: VG∪EG⟶ {1, 2, ..., p + q}, such that f(w) + f(wz) + f(z) = k for any edge wz∈EG. Moreover, a super edge-magic total labeling is an edge-magic total labeling f, such that fVG = {1, 2, ..., p}.The notion of edge-magic total labeling of a graph is generalized to edge-antimagic total labeling of graphs. Letα≥0 and β≥0 be integers. Let W=wxy:wxy=fx+fxy+fy,xyEG. If W forms an arithmetic sequence starting from α with common difference β, then G is called α,β−edge antimagic total labeling. Moreover, if fVG = {1, 2, ..., p}, then G is called superα,β−edge antimagic total labeling. Notice that when β = 0, the α,β−edge antimagic total labeling of G is the usual edge-magic total labeling of G with f(w) + f(wz) + f(z) =  α for any edge wz∈EG.One of the most popular problems in the theory of graph labeling is super edge-magic total labeling of tree. Enomoto et al. [3] proposed the following conjecture.Tree conjecture [3]: every tree is superα,0−edge antimagic total labeling.The tree conjecture is still an open problem; however, some authors proved that tree conjecture is true for some classes of tree. For example, Bhatti, Javaid and Hussain [4] and Raheem et al. [5] proved that tree conjecture is true for subdivision of caterpillar. Javaid, Bhatti, and Aslam [6] proved that tree conjecture is true for subdivision of stars. Other authors who studied tree conjecture can be found in Gallian [7].Another popular problem in the theory of graph labeling is super edge-magic total labeling of a cycle book. A cycle book graph is constructed from some cycles either with the same or different order. Letm≥1 be any positive integer and let Ca be the cycles of order a. A graph G is called a-cycle books BCa,m,2 if G consists of m cycles Ca with a common path P2. For m=1, we define BCa,m,2 to be a cycle Ca. From now on, the graphs, a-cycle books BCa,m,2 is denoted by Ba,m,2.Marr and Wallis ([8], Research problem 2.7, p.39) proposed the following problem.Problem 1.
Are all graphsBa,m,2 edge-magic total (super edge-magic total)?
GraphBa,m,2 is constructed from some cycles Ca of the same order. Swita et al. [9] contructed a graph from some cycles with different orders. A graph a,b−cycle bookBa,m,b,n,t is constructed from some cycles Ca and Cb with a common path Pt, a path of order t with m,n,a,b, and t as the positive integers.Problem 2.
Are all graphsBa,m,b,n,2 edge-magic total (super edge-magic total)?
Both Problems1 and 2 are interesting problems for at least the following two reasons. First reason is the solutions of Problems 1 and 2 that can be used to construct the secret sharing scheme in information technology. Reddy and Basha [10] and Imron et al. [11] used edge-magic total labeling of catepilar graphs to construct the secret sharing scheme. Baskoro, Simanjuntak, and Adithia [12] used edge-magic total labeling of star graphs to construct the secret sharing scheme.
The second reason is both Problems1 and 2 provide a challenging problem for the researchers, since they are open problems. Swita et al. [9] proved Problem 2 for a=7 or a=4x−1 for any integer x. MacDougall and Wallis [13] proved Problem 2 for m=n=1 that a graph Ba,1,b,1,2 is a super edge-magic total labeling. Let l=mina,b−3. Notice that l is a chord of cycle Ca+b−2. Thus, l is a chord of graph Ba,1,b,1,2. Using Kotzig array, Singgih [14, 15] proposed a new method to construct an edge-magic total labeling (super edge-magic) of graph cycle Ca+b−22r+1h with 2r+1hz chords, each of length l=a,b−3, from an edge-magic total labeling (super edge-magic) of graph Ba,1,b,1,2, where hand z are the positive integers.
Berkman, Parnas and Roditty [16], Enomoto et al. [3], Kotzig and Rosa [2], and Godbold and Slater [17] are among others authors that discuss Problem 1 for m=1. For m≥2, Problem 1 is an open problem; however, some authors provided a partial solution. Figueroa-Centeno, Ichishima, and Muntaner-Batle [18] proved that B4,m,2 is an edge-magic total. Furthermore, they proved that B[(4, m), 2] is not super edge-magic total for m≡1,3,7mod8 and m = 4, but B[(4, m), 2] is super edge-magic total for m = 2, 5, 6, 8, 10, 11. Moreover, they made the following conjecture.
Cycle book conjecture [18]: the graph B[(4, m), 2] is super edge-magic total if and only if m is even or m≡5mod8.
Gallian [7] reported that Yuansheng et al. [19] proved this conjecture for m is even in Ars Combinatoria, 93 (2009) 431–438. A study [20] contains the abstract of Yuansheng et al. [19] and claims that Yuansheng et al. proved the cycle book conjecture is true for m is even. The study [19] is the same as that of Gallian [7]. We trace this reference, and we find that this reference is neither in the table of contents of Ars Combinatoria, 93 (2009), nor in the table of contents Ars Combinatoria from 1995 up to 1999. Hence, we assume that the article of Yuansheng et al. is unpublished. Therefore, it is reasonable to publish this article. Thus, this study proves the cycle book conjecture for m ≥ 36 and m = 0 mod (2). The solution of cycle book conjecture is available from the author for 12 ≤ m ≤ 34 and m = 0 mod (2).


## 2. Preliminary Notes

In this section, we provide some previous results on super edge-magic total labeling of a graph. Figuero-Centeno, Ichisma, and Mutaner-Batle [18] proved some necessary conditions for super edge-magic total labeling of a graph. We need them to prove the main results of this study. First, we define some notations in the following definition.Definition 1.
LetG be a graph B4,m,2,X=um+2,Y=vm+2, Z=u1,u2,…,um/2+1,um/2+3,…,um+1, and W=v1,v2,…,vm/2+1,vm/2+3,…,vm+1. We define the vertex set VG=X∪Y∪Z∪W and the edge set EG=x,y:x∈X,y∈Y∪x,y:x∈Z,y∈W∪x,y:x∈X,y∈Z∪x,y:x∈Y,y∈W.
The element ofX∪Y and the edge um/2+2,vm/2+2 are called the common vertices and common edge of the m copies of Ca, respectively.
The graph B[(4,m), 2] in Definition1 is shown in Figure 1 and the graph Ba,m,b,n,2 is shown in Figure 2.Figure 1
GraphB4,m,2.Figure 2
GraphBa,m,b,n,2.Theorem 1 (see [18]).
LetG be a graph, such that VG=p and EG=q. Then, G is super edge-magic total if and only if there exists a bijective function f:VG⟶1,2,…,p, such that the set S=fw+fz:f or any edge wz∈EG consists of q consecutive integers. In such a case, f extends to a super edge-magic total labeling of G with magic constant k=p+q+s, where s=minS and S=k−p+1,k−p+2,…,k−p+q.Theorem 2 (see [18]).
LetG be a graph, such that VG=p and EG=q and f be a super edge-magic total labeling of G. Let S=fw+fz:wz∈EG and s=minS. Then, ∑v∈VGfvdegv=qs+q−1q/2. In particular, 2∑v∈VGfvdeg v≡0modq.Theorem 3 (see [18]).
LetG be a graph B4,m,2, such that S=fw+fz:wz∈EG and s=minS. If G is super edge-magic total labeling, then s=m/2+3.The following theorem is derived from the proof of Theorem3 in [3]. For self-contained of this article, we rewrite the proof again.Theorem 4 (see [18]).
LetG be a graph B4,m,2 in Definition 1 and let um/2+2,vm/2+2 be the common edges of all cycles C4 in G. If G is super edge-magic total labeling, then fum/2+2+fvm/2+2=6ms+m2+2s−17m−12/2m−2.Proof.
LetG be a graph B4,m,2in Definition 1. Let VG=p,EG=q, and t=m/2+2. We first notice that p=2m+2 and q=3m+1. Moreover, degut=degvt=m+1,degu1=degv1=⋯=degut−1=degvt−1=degut+1=degvt+1=…,degum+1=degvm+1=2.
LetA = X∪Y and B = Z∪ W. By Theorem 2, we have m+1∑w∈Afw+2∑w∈Bfw=qs+q−1q/2 or m−1∑w∈Afw+2∑w∈Bfw=qs+q−1q/2. The last equality reduces to ∑w∈Afw+2∑i=12m+2i=qs+q−1q/2, since fw,fz∈1,2,…,2m+2, and fw≠fz. Thus, m−1∑w∈Afw+2m+22m+3=qs+q−1q/2. We substitute q=3m+1 to the last equation, and we have the following equation.(1)fut+fvt=6ms+m2+2s−17m−122m−2.
By the equation (1) and s=m/2+3 in Theorem 3, we conclude that fut+fvt=2m+3. Hence, the theorem.


## 3. Proof of Cycle Book Conjecture form Is Even

In this section, we prove that the cycle book conjecture is true ifm is even and m≥36.Theorem 5.
LetG be the graph B4,m,2 in Definition 1 withVG=p,EG=q, m is an even integer, m≥36., and let f be an edge-magic total labeling of G. Let S=fw+fs:f or any edge wz∈EG and s=minS. Then, G is a super edge-magic total if and only if(i)
fum/2+2+fvm/2+2=6ms+m2+2s−17m−12/2m−2(ii)
s=m/2+3(iii)
S is a set of q consecutive integers(iv)
k=p+q+sProof.
LetG be the graph B4,m,2 with VG=p,EG=q, and let G be a super edge-magic total. Note that p=2m+2 and q=3m+1. Let f be an edge-magic total labeling of G. If f is a super edge-magic total labeling of G, then the conditions (i) and (ii) follow from Theorems 4 and 3, respectively, and the conditions (iii) and (iv) follow from Theorem 1.
Letf satisfies the conditions (i), (ii), (iii), and (iv). By (i) and (ii), we conclude that(2)fum/2+2+fvm/2+2=2m+3.
The pair (m/2+2,3m/2+1) is one of the solutions of equation (2) with fum/2+2 =  m/2+2 and fvm/2+2=3m/2+1. By this solution, we define the bijection f1x, such that f1um/2+2=fum/2+2=m/2+2 and f1vm/2+2=fvm/2+2=3m/2+1 as follows.(3)f1ui=m2+2,i=m2+2,(4)f1vi=3m2+1,i=m2+2,(5)f1ui=i,i=1,2,…,m+1,(6)f1vi=3m2+3,i=1,(7)f1vi=m+2,i=2,(8)f1vi=3m2+2,i=3,(9)f1vi=2m+1,i=m+1.Case 1:m≡0mod4We define(10)f1vi=3m2+3−i+12,i=m−1,m−3,…,m2+5,m2+3,(11)f1vi=5m+84,i=m,(12)f1vi=3m2−i−62,i=m2,m2−2,…,8,6.(13)f1vi=2m+3−i2,i=m−2,m−4,…,m2+6,m2+4,(14)f1vi=7m+124,i=m2+1.Case 1.1:m≡0mod12We define(15)f1vi=2m+3−i2,i=5,11,17,…,m2−7,m2−1,(16)f1vi=2m−i−32,i=7,13,19,…,m2−11,m2−5,(17)f1vi=2m−i−92,i=9,15,21,…,m2−9,m2−3,(18)f1vi=2m+2,i=4.Next, we will show thatS is a set of q consecutive integers with q=3m+1. Recall that X=um/2+2 and Y=vm/2+2.LetZ11=u1,u2,…,um/2+2+1 and S11=f1x+f1y:x∈X,y∈Z1. Let x∈X. By equations (3) and (5), we conclude that f1x+f1ui=m/2+2+i,i=1,2,…,m/2+1. Hence, S11=m/2+3,m/2+4,…,m+2,m+3.LetZ21=u2,W11=v2 and S21=f1x+f1y:x∈Z2,y∈W1. Let x∈Z2 and y∈W1. By equations (5) and (7), we conclude that f1x+f1y=2+m+2=m+4. Hence, S21=m+4.Let =Z31=um/2+3,um/2+4,…,um,um+1,Z41=X, and S31=f1x+f1y:x∈Z31,y∈Z41. Let x∈Z41 and ui∈Z31. By equations (3) and (5), we conclude that f1x+f1ui=m/2+2+i,i=m/2+3,m/2+4,…,m+1. Hence, S31=m+5,m+6,…,3m/2,3m/2+2,3m/2+3.LetZ51 = u1,u3, W21 = v1,v3, and S41=f1ui+f1vi:ui∈Z5,vi∈W2. By equations (5) and (6), we conclude that f1u1+f1v1=1+3m/2+3=3m/2+4. By equations (5) and (8), we conclude that f1u3+f1v3=3+3m/2+2=3m/2+5. Hence, S41=3m/2+4,3m/2+5.LetZ61=u6,u8,…,um/2−2,um/2,W31=v6,v8,…,vm/2−2,vm/2, and S51=f1ui+f1vi:ui∈Z61,vi∈W31,i=6,8,…,m/2−2,m/2. By equations (5) and (12), we conclude that f1ui+f1vi=i+3m/2−i−6/2,i=6,8,…,m/2. Hence, S51=3m/2+6,3m/2+7,3m/2+8,…,7m/4+2,7m/4+3.LetZ71 = um/2+3,um/2+5,….,um−3,um−1, W41 = vm/2+3,vm/2+5,…,um−3,um−1, and S61=f1ui+f1vi:ui∈Z71,vi∈W41,i=m/2+3,m/2+5,…,m−3,m−1. By equations (5) and (10), we conclude that f1ui+f1vi=i+3m/2+3−i+1/2,i=m/2+3,m/2+5,…,m−3,m−1. Hence, S61=7m/4+4,7m/4+5,…,2m+1,2m+2.LetZ81=X,W51=Y and S71=f1x+f1y:x∈Z8,y∈W5. By equations (3) and (4), we conclude that f1um/2+2+f1vm/2+2=m/2+2+3m/2+1=2m+3. Hence, S71=2m+3.LetZ91=ui:i≡5mod6=u5,u11,…,um/2−7,um/2−1, W61=vi:i≡5mod6 = v5,v11,…,vm/2−7,vm/2−1, and S81=f1ui+f1vi:ui∈Z91,vi∈W61. By equations (5) and (15), we conclude that f1ui+f1vi=i+2m+3−i/2,i=5,11,17,…,m/2−7,m/2−1. Hence, S81=2m+4,2m+7,…,9m/4−2,9m/4+1.LetZ101 = ui:i≡1mod6=u7,u13,…,um/2−11,um/2−5, W71 = vi:i≡1mod6=v7,v13,…,vm/2−11,vm/2−5, and S91=f1ui+f1vi:ui∈Z101,vi∈W7. By equations (5) and (16), we conclude that f1ui+f1vi=i+2m−I−3/2,i=7,13,19,…,m/2−11,m/2−5. Hence, S91=2m+5,2m+8,2m+11,…,9m/4−1.LetZ111 = u4, W81 = v4, and S101=f1x+f1y:x∈Z111,y∈W8. By equations (5) and (18), we conclude that f1u4+f1v4=2m+6. Hence, S101=2m+6.LetZ121=ui:i≡3 = mod6=u9,u15,…,um/2−11,um/2−9,um/2−3, W91 = vi:i≡3mod6=v9,v15,…,vm/2−15,vm/2−9,um/2−3, and S111=f1ui+fvi:ui∈Z121,vi∈W91. By equations (5) and (17), we conclude that f1ui+f1vi=i+2m−I−9/2,i=9,15,21,…,m/2−9,m/2−3. Hence, S111=2m+9,2m+12,2m+15,…,9m/4−3,9m/4,9m/4+3.LetZ131 = um, W101 = vm, and S121=f1x+f1y:x∈Z131,y∈W101. By equations (5) and (11), we conclude that f1um+f1vm=m+5m/4+2=9m/2+2. Hence, S121=9m/4+2.LetZ141 = um/2+1, W111 = vm/2+1, and S131=f1x+f1y:x∈Z141,y∈W111. By equations (5) and (14), we conclude that f1um/2+1+f1vm/2+1=m/2+1+7m/4+3=9m/4+4. Hence, S131=9m/4+4.LetZ151 = um/2+4,um/2+6,…,um−4,um−2, W121 = vm/2+4,vm/2+6,…,vm−4,vm−2, and S141=f1ui+f1vi:ui∈Z151,vi∈W121. By equations (5) and (13), we conclude that f1ui+f1vi=i+2m+3−i/2,i=m/2+4,m/2+6,…,m−4,m−2. Hence, S141=9m/4+5,9m/4+6,9m/4+7,…,5m/2,5m/2+1,5m/2+2.LetW131 = Y=vm/2+2, W141 = v2, and S151=f1x+f1y:x∈W131,y∈W141. By equations (4) and (7), we conclude that f1vm/2+2+f1v2=3m/2+1+m+2=5m/2+3. Hence, S151=5m/2+3.LetW151 = Y=vm/2+2,W161 = vm−1,vm−3,…,vm/2+5,vm/2+3, and S161=f1x+f1y:x∈W151,y∈W161. By equations (4) and (10), we conclude that f1vm/2+2+f1vi=3m/2+1+3m/2+3−i+1/2,i=m−1,m−3,m−5,…,m/2+5,m/2+3. Hence, S161=5m/2+4,5m/2+5,…,11m/4+1,11m/4+2.LetW171 = Y=vm/2+2, W181 = vm, and S171=f1x+f1y:x∈W171,y∈W181. By equations (4) and (11), we conclude that f1vm/2+2+f1vm=3m/2+1+5m/4+2=11m/4+3. Hence, S171=11m/4+3.LetW191 = Y=vm/2+2, W201 = vm/2,vm/2−2,…,v8,v6, and S161=f1x+f1y:x∈W19,y∈W201. By equations (4) and (12), we conclude that f1vm/2+2+f1vi=3m/2+1+3m/2−i−6/2,i=m/2,m/2−2,m/2−4,…,8,6. Hence, S181=11m/4+4,11m/4+5,…,3m,3m+1.LetZ161=um+1, W211 = vm+1, and S191=f1x+f1y:x∈Z161,y∈W211. By equations (5) and (9), we conclude that f1um+1+f1vm+1=m+1+2m+1=3m+2. Hence, S191=3m+2,.LetW221 = Y=vm/2+2, W231 = v3,v1, and S201=f1x+f1y:x∈W221,y∈W231. By equations (4) and (8), we conclude that f1vm/2+2+f1v3=3m/2+1+3m/2+2=3m+3. By equations (4) and (6), we conclude that f1vm/2+2+f1v1=3m/2+1+3m/2+3=3m+4. Hence, S201=3m+3,3m+4.LetW241 = Y=vm/2+2, W251 = vm−2,vm−4,…,vm/2+6,vm/2+4, and S211=f1x+f1y:x∈W241,y∈W251. By equations (4) and (13), we conclude that f1vm/2+2+f1vi=3m/2+1+2m+3−i/2,i=m−2,m−4,…., m/2+6,m/2+4. Hence, S211=3m+5,3m+6,…,13m/4+1,13m/4+2.LetW261 = Y=vm/2+2, W271 = vm/2−1,vm/2−7,…,v11,v5, and S221=f1x+f1y:x∈W261,y∈W271. By equations (4) and (15), we conclude that f1vm/2+2+f1vi=3m/2+1+2m+3−i/2,i=m/2−1,m/2−7,…,11,5. Hence, S221=13m/4+3,13m/4+6,…,7m/2−3,7m/2.LetW281 = Y=vm/2+2, W291 = vm/2+1, and S231=f1x+f1y:x∈W281,y∈W291. By equations (4) and (14), we conclude that f1vm/2+2+f1vm/2+1=3m/2+1+7m/4+3=13m/4+4. Hence, S231=13m/4+4.LetW301 = Y=vm/2+2,W311 = vm/2−5,vm/2−11,…,v13,v7, and S241=f1x+f1y:x∈W301,y∈W311. By equations (4) and (16), we conclude that f1vm/2+2+f1vi=3m/2+1+2m−i−3/2,i=m/2−5,m/2−11,…,13,7. Hence, S241=13m/4+5,13m/4+8,…,7m/2−4,7m/2−1.LetW321 = Y=vm/2+2,W331 = vm/2−3,vm/2−9,…,v15,v9, and S251=f1x+f1y:x∈W321,y∈W331. By equations (4) and (17), we conclude that f1vm/2+2+f1vi=3m/2+1+2m−i−9/2,i=m/2−3,m/2−9,…,15,9. Hence, S251=13m/4+7,13m/4+10,…,7m/2−2,7m/2+1.LetW341 = Y=vm/2+2,W351 = vm+1, and S261=f1x+f1y:x∈W341,y∈W351. By equations (4) and (9), we conclude that f1vm/2+2+f1vm+1=3m/2+1+2m+1=7m/2+2. Hence, S261=7m/2+2.LetW361 = Y=vm/2+2,W371 = v4, and S271=f1x+f1y:x∈W361,y∈W371. By equations (4) and (18), we conclude that f1vm/2+2+f1v4=3m/2+1+2m+2=7m/2+3. Hence, S271=7m/2+3.Next, we will show thatS81∪S91∪S101∪S111∪S121 consists of 3m/12 consecutive integers. Simple counting shows that S81=m/12, and S91=S111=m/12−1, S101 =  S121=1. We arrange the term of S81∪S91∪S101∪S111∪S121 in Table 1.We observe from Table1 that S81∪S91∪S101∪S111∪S121  =  2m+4,2m+5,2m+6,…,9m/4+1,9m/4+2,9m/4+3 consists of 3m/12 consecutive integers.In addition, we will show thatS221∪S251∪S231∪S241 consists of 3m/12−1 consecutive integers. Simple counting shows that S221 =  m/12, S241 = S251 =  m/13−1, and S23=1. By these information, we arrange the terms of S221∪S251∪S231∪S241 in Table 2.We observe from Table2 that S221∪S251∪S23∪S241 =  13m/4+3,13m/4+4,13m/4+5,…,7m/2−1,7m/2,7m/2+1 consists of 3m/12−1 consecutive integers. Let S =  S11∪S21∪⋯∪S261∪S271. We will show that S consists of q consecutive integers with q=3m+1.LetS11=m/2+3,m/2+4,…,m+2,m+3=T11,S21=m+4=T21,S31=m+5,m+6,…,3m/2+2,3m/2+3=T31,S41=3m/2+4,3m/2+5=T41,S51=3m/2+6,3m/2+7,…,7m/4+2,7m/4+3=T51,S61=7m/4+4,7m/4+5,…,2m+1,2m+2=T61,S71=2m+3=T71,S81∪S91∪S101∪S111∪S121=2m+4,2m+5,2m+6,…,9m/4+1,9m/4+2,9m/4+3=T81S131=9m/4+4=T91,S141=9m/4+5,9m/4+6,9m/4+7,,…,5m/2,5m/2+1,5m/2+2=T101,S151=5m/2+3=T111,S161=5m/2+4,5m/2+5,…,11m/4+1,11m/4+2=T121,S171=11m/4+3=T131,S181=11m/4+4,11m/4+5,…,3m,3m+1=T141,S191=3m+2=T151,S201=3m+3,3m+4=T161,S211=3m+5,3m+6,…,13m/4+1,13m/4+2=T171,S221∪S251∪S231∪S241=13m/4+3,13m/4+4,13m/4+5,…,7m/2−1,7m/2,7m/2+1=T181,S261=7m/2+2=T191,S271=7m/2+3=T201.Notice thatminTi1=maxTi−11+1,i=2,3,…,20. Moreover, Ti1 consists of consecutive integers. Simple counting shows that S consists of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Moreover, by iv, we have f1uv=k−f1u+f1v for all uv∈EG. Hence, the theorem in this case.Case 1.2:m≡4mod12We define(19)f1vi=2m+i−92,i=5,11,17,…,m2−9,m2−3,(20)f1vi=2m−i−32,i=7,13,19,…,m2−7,m2−1,(21)f1vi=2m−i−32,i=9,15,21,…,m2−11,m2−5.It can be proved in the same lines as previous proof of Case 1.1 thatS is a set of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Hence, the theorem in this case.Case 1.3:m≡8mod12We define(22)f1vi=2m+2,i=4,(23)f1vi=2m−1,i=5,(24)f1vi=2m−4,i=9,(25)f1vi=2m,i=7,(26)f1vi=2m−3,i=11,(27)f1vi=2m−i−92,i=13,19,…,m2−9,m2−3,(28)f1vi=2m−i−32,i=15,21,…,m2−7,m2−1,(29)f1vi=2m−i−32,i=17,23,…,m2−11,m2−5.It can be proved in the same lines as the previous proof of Case 1.1 thatS is a set of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Hence, the theorem in this case.Case 2:m≡2mod4(30)f1vi=3m2+3−i+12,i=m−1,m−3,…,m2+6,m2+4,(31)f1vi=5m+64,i=m,(32)f1vi=3m2−i−62,i=m2+1,m2−1,…,8,6,(33)f1vi=2m+3−i2,i=m−2,m−4,…,m2+5,m2+3,(34)f1vi=7m+144,i=m2.Case 2.1:m≡2mod12We define(35)f1vi=2m+3−i2,i=5,11,17,…,m2−8,m2−2,(36)f1vi=2m−i−32,i=7,13,19,…,m2−12,m2−6,(37)f1vi=2m−i−92,i=9,15,21,…,m2−10,m2−4,(38)f1vi=2m+2,i=4.It can be proved in the same lines as the previous proof of Case 1.1 thatS is a set of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Hence, the theorem in this case.Case 2.2:m≡6mod12We define(39)f1vi=2m−i−92,i=5,11,17,…,m2−10,m2−4,(40)f1vi=2m−i−32,i=7,13,19,…,m2−8,m2−2,(41)f1vi=2m−i−32,i=9,15,21,…,m2−12,m2−6,(42)f1vi=2m,i=4.It can be proved in the same lines as the previous proof of Case 1.1 thatS is a set of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Hence, the theorem in this case.Case 2.3
:m≡10mod12We define(43)f1vi=2m+2,i=4,(44)f1vi=2m−1,i=5,(45)f1vi=2m−4,i=9,(46)f1vi=2m,i=7,(47)f1vi=2m−3,i=11,(48)f1vi=2m−i−92,i=13,19,…,m2−10,m2−4,(49)f1vi=2m−i−32,i=15,21,…,m2−8,m2−2,(50)f1vi=2m−i−32,i=17,23,…,m2−12,m2−6.It can be proved in the same lines as previous proof of Case 1.1 thatS is a set of q consecutive integers with q=3m+1. Thus, by Theorem 1, we conclude that G is super edge-magic total. Hence, the theorem.Table 1
S81∪S91∪S101∪S111∪S121 consists of 3m/12 consecutive integers.
S81S91∪S121S111∪S1012m+42m+52m+6=S1012m+72m+82m+92m+102m+112m+12………9m/4−49m/4−49m/4−39m/4−29m/4−19m/49m/4+19m/4+2=S129m/4+3Table 2
S221∪S251∪S23∪S241 consists of 3m/12−1 consecutive integers.
S221S251∪S231S24113m/4+313m/4+4=S23113m/4+513m/4+613m/4+713m/4+813m/4+913m/4+1013m/4+11………7m/2−67m/2−57m/2−47m/2−37m/2−27m/2−17m/27m/2+1


## 4. Conclusion

We are able to prove the cycle book conjecture form is even, but we cannot prove it for m≡5mod8. Hence, the cycle book conjecture is an open problem.



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*Source: 8483926-2021-08-13.xml*