INTEGER-MAGIC SPECTRA OF AMALGAMATIONS OF STARS AND CYCLES. Sin-Min Lee San Jose State University San Jose, CA

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1 INTEGER-MAGIC SPECTRA OF AMALGAMATIONS OF STARS AND CYCLES Sin-Min Lee San Jose State Universit San Jose, CA 9592 Ebrahim Salehi Department of Mathematical Sciences Universit of Nevada Las Vegas Las Vegas, NV Abstract. For an positive integer k, a graph G = (V, E) is said to be Z k -magic if there eists a labeling l : E(G) Z k {0} such that the induced verte set labeling l + : V (G) Z k defined b l + (v) = { l(uv) : uv E(G) } is a constant map. For a given graph G, the set of all h Z + for which G is Z h -magic is called the integer-magic spectrum of G and is denoted b IM(G). In this paper, we will determine the integer-magic spectra of the graphs which are formed b the amalgamation of stars and ccles. In particular, we will provide eamples of graphs that for a given n > 2, the are not h-magic for all values of 2 h n.. Introduction For an abelian group A, written additivel, an mapping l : E(G) A {0} is called a labeling. Given a labeling on the edge set of G one can introduce a verte set labeling l + : V (G) A as follows: l + (v) = { l(uv) : uv E(G) }. A graph G is said to be A-magic if there is a labeling l : E(G) A {0} such that for each verte v, the sum of the labels of the edges incident with v are all equal to the same constant; that is, l + (v) = c for some fied c A. We will call < G, l > an A-magic graph with sum c. In general, a graph G ma admit more than one labeling to become an A-magic graph; for ARS COMBINATORIA 67 (2003), 99-22

2 eample, if A > 2 and l : E(G) A {0} is a magical labeling of G with sum c, then λ : E(G) A {0}, the inverse labeling of l, defined b λ(uv) = l(uv) will provide another magical labeling of G with sum c. The original concept of A-magic graph is due to J. Sedlacek [8, 9], who defined it to be a graph with a real-valued edge labeling such that () distinct edges have distinct nonnegative labels; and (2) the sum of the labels of the edges incident to a particular verte is the same for all vertices. Given a graph G, the problem of deciding whether G admits a magic labeling is equivalent to the problem of deciding whether a set of linear homogeneous Diophantine equations has a solution [20]. At present, given an abelian group, no general efficient algorithm is known for finding magic labelings for general graphs. When A = Z, the Z-magic graphs were considered in Stanle [20, 2], he pointed out that the theor of magic labeling can be put into the more general contet of linear homogeneous diophantine equations. When the group is Z k, we shall refer to the Z k -magic graph as k-magic. Graphs which are k-magic had been studied in [2, 5, 6, 9, 0, 2, 4]. For convenience, we will use the notation -magic instead of Z-magic. Doob [2, 3, 4], also considered A-magic graphs where A is an abelian group. He determined which wheels are Z-magic. A graph G = (V, E) is called full magic [9,, 7, 22, 25, 26] if it is A-magic for ever abelian group A, and it is called non-magic if for ever abelian group A it is not A-magic. Also, a graph G is said to be N-magic if there eists a labeling l : E(G) N such that l + (v) is a constant, for ever v V (G). It is well-known that a graph G is N-magic if and onl if each edge of G is contained in a -factor (a perfect matching) or a {, 2}-factor [, 7, 26]. Berge [] called a graph regularisable if a regular multigraph could be obtained from G b adding edges parallel to the edges of G. In fact, a graph is regularisable if and onl if it is N-magic. For N-magic graphs, readers are referred to [7, 8, 9, 0, 2, 22, 23, 24]. The notion of Z-magic is weaker than N-magic. Figure shows a graph which is Z-magic but not N-magic. Observation.. For an n 3, the path of order n is non-magic. Observation.2. In an magical labeling of a ccle the edges should alternativel be labeled the same group elements. 200

3 Figure. The graph P 3 P 3 is Z-magic but is not N-magic. Proof. Let l : E(C n ) A be a magic labeling and e, e 2, e 3, e 4 be four consecutive edges of C n. Then l(e )+l(e 2 ) = l(e 2 )+l(e 3 ) and l(e 2 )+l(e 3 ) = l(e 3 )+l(e 4 ). Which implies that l(e ) = l(e 3 ) and l(e 2 ) = l(e 4 ). Observation.3. C 2n, the ccle of order 2n, with a pendant edge is non-magic. Proof. B the Observation.2, it is enough to prove the statement for C 4. As illustrated in the Figure 2, in an labeling, the sum of labels of the edges incident with verte v needs to be equal to the sum of labels of the edges incident with u; that is, + + = + = = 0, a contradiction. u v Figure 2. The ccle C 4 with an edge pendant is non-magic. In this paper, we will denote the set of positive integers b N, and for an k > 0, kn = { kn : n N }, also k + N = { k + n : n N }. For a given graph G the set of all positive integers h for which G is Z h -magic (or simpl h-magic) is called the integer-magic spectrum of G and is denoted b IM(G). Since an regular graph is full magic, then it is h-magic for all positive integers h 2; therefore, IM(G) = N. For more general results on integer-magic spectrum of graphs, the reader is referred to [3, 5, 6]. A graph G with a fied verte u will be denoted b the ordered pair (G, u). Given two ordered pairs (G, u) and (H, v), one can form a new graph b amalgamation: form the disjoint union of G and H and identif u and v. The resulting graph will be denoted b (G, u) (H, v). 20

4 u v (G,u) (H,v) (G,u)o(H,v) Figure 3. Amalgamation Construction. For convenience the complete bipartite graph K(, m), known as star with m leaves, will be denoted b ST (m). Given a star ST (m) and a ccle C n, depending on whether we identif the center of the star with a verte of C n or identif an end-verte of the star with a verte of C n, the amalgamation of these two graphs will result in two non-isomorphic graphs. We will denote the first one b ST (m)#c n, and the latter b ST (m)@c n. Figure 4 illustrates the two different amalgamations of ST (3) and C 4. ST(3)#C 4 ST(3)@C 4 Figure 4. Two different amalgamations of ST (3) and C Integer-Magic Spectrum of ST (m)#c n. B the Observation.2, in an magic labeling of C n, labels of the edges alternates. One immediate consequence of this fact is that, the graph ST (m)#c 3 (or ST (m)#c 4 ) is h-magic if and onl if the graph ST (m)#c 2k+ (or ST (m)#c 2k ) is h-magic. As a result, we will onl concentrate on ccles C 3 and C 4. Also, we observe that since {, 2} is a subset of the degree set of these tpes of graphs, ST (m)#c n can not be 2-magic. Theorem 2.. IM( ST ()#C 4 ) =. Proof. This is a direct result of the Observation.3. Theorem 2.2. IM( ST (2)#C 4 ) = 2 + 2N. 202

5 u w r- r- r r Figure 5. General labeling of ST (m)#c 4. Here = +. Proof. We observe that ST (2)#C 4 is h-magic if and onl if 2 h and h > 2. Because, as illustrated in the Figure 5, we need to have l + (w) = l + (u) or 2 0 (mod h), and this means is an order two element of the group Z h. Therefore, 2 h. On the other hand, if h = 2r, then the selection of =, = r, = r will work. Theorem 2.3. For an positive integer m 2, the graph ST (m)#c 4 is h-magic if and onl if h > 2 and gcd(m, h) >. Therefore, IM( ST (m)#c 4 ) = (2 + N) { h N : gcd(m, h) = }. Proof. In ST (m)#c 4 there are m pendant edges attached to the verte w and the equation l + (w) = l + (u) implies that (2.) m 0 (mod h). If gcd(m, h) =, then the equation 2. will be equivalent to 0 (mod h), which does not provided a non-ero solution. Now suppose gcd(m, h) = δ >. If δ = h, then we use = = and = 2 as our labels, with l + 2. Suppose < δ < h, and let h = δr (r 2). Then the selection =, = r, and naturall = r will work with l + r. Corollar 2.4. If p is a prime number bigger than 2, then IM( ST (p)#c 4 ) = pn. Corollar 2.5. If m = p α pα 2 2 pα k k is an odd positive integer, then k IM( ST (m)#c 4 ) = (p i N ). If m = 2 β p α pα 2 2 pα k k is even, then i= IM( ST (m)#c 4 ) = (2 + 2N ) Theorem 2.6. IM( ST ()#C 3 ) = 2 + 2N. 203 k (p i N ). i=

6 Proof. For graph ST ()#C 3 to be h-magic, as illustrated in the Figure 6, we need to have 2+ = or 2 0 (mod h). That is, is a non-ero element of Z h with order 2. Therefore, 2 h. On the other hand, if h = 2r, then the selections = r, = will work; that is, the graph ST ()#C 3 has h-magic. r r+ r Figure 6. ST ()#C 3 and ST (2)#C 3. Theorem 2.7. IM( ST (2)#C 3 ) = N {2, 3}. Proof. For the graph ST (2)#C 3 to be h-magic, as illustrated in the Figure 6, we need to have = or (mod h). Here = +, therefore we need (mod h), and this equation does not provide non-ero solutions for h = 3. If h > 3, then the selections =, = h 3 will work. This means the graph is h-magic for all h 4. Furthermore, ST (2)#C 3 is Z-magic; because, the labels =, = 3 will work with l + 2. w u Figure 7. A tpical labeling of ST (m)#c 3. Here, = +. Note that when m >, the graph ST (m)#c 3 is Z-magic; because, the labels = m, = +m, and = 2 will work with l + 2. In addition, an magical labeling of ST (m)#c 3, as illustrated in the Figure 7, uses three non-ero group elements,, and = +. In particular, we need to have l + (u) = l + (w), which implies (2.2) (m ) (mod h), or equivalentl (2.3) (m + ) + (m ) 0 (mod h). 204

7 Theorem 2.8. The graph ST (m)#c 3 is 3-magic if and onl if 3 m. Proof. If m = 3k, the labeling = = will work with l + 2 (mod 3). If m = 3k +, the equation 2.3 will become (3k + 2) + 3k 0 (mod 3), which is equivalent to 2 0 (mod 3), or 0, which is not an acceptable answer. If m = 3k + 2, the equation 2.3 will become (3k + 3) + (3k + ) 0 (mod 3), which is equivalent to 0, not an acceptable answer. Theorem 2.9. For ever h > m +, the graph ST (m)#c 3 is h-magic. Proof. If h > m +, then we choose = h (m ), = m +, and naturall = 2. We notice that,, are non-ero elements of Z h and l + (w) = m + 2 = 2m + 2(h m + ) = 2h (mod h). Theorem 2.0. If m > 2, then the graph ST (m)#c 3 is m-magic. Proof. The labeling = = works with l + 2. Theorem 2.. The graph ST (m)#c 3 is h-magic for all even positive integers h 4. Proof. To prove the theorem it is enough to show that the equation 2.2 has two distinct non-ero solutions for and in Z h. Let h = 2 α µ, where µ is an odd number, and consider the equations (2.4) (m ) (mod 2 α ), (2.5) (m ) (mod µ). Given an, the equation 2.5 has the solution 2 (m ) for, where 2 is the multiplicative inverse of 2 in Z µ. For the equation 2.4 we consider two cases: Case. When m = 2k + is odd, then the equation 2.4 becomes 2k (mod 2 α ) or k + 0 (mod 2 α ), and for an we will have k (mod 2 α ). Therefore, k (mod 2 α µ), and for an non-ero Z h, one can choose either k or k + 2 α µ (mod h) for the solution of the equation 2.2. Case 2. When m = 2k is even, then has to be even. Let = 2ξ. Equation 2.4 becomes (m )ξ + 0 (mod 2 α ), and for an ξ we will have (m )ξ (mod 2 α ). Therefore, (m )ξ (mod 2 α µ), and for an non-ero = 2ξ Z h, one can choose either (m )ξ or (m )ξ + 2 α µ (mod 2 α ) for the solution of the equation

8 As an application of the Theorem 2., we will show that ST (m)#c 3 is alwas 4-magic. Using the notation and the process of the theorem, we will consider two cases: Case. If m = 2k +, then we need k + 0 (mod 2) or k (mod 2). For non-ero = 3, we have two choices of either 3k (mod 4) or 3k + 2 (mod 4), which translates to either = or = 2. Case 2. If m = 2k, then we will deal with the equation (2k ) (mod 4), which implies that = 2ξ. With this consideration the equation becomes (2k )ξ + 0 (mod 2) or ξ (mod 2). Now for the onl choice of = 2 (or ξ = ) we will have two choices of either or 3 and the corresponding values of will be or 3, respectivel. Theorem 2.2. Let h 3 be an odd positive integer. Then ST (m)#c 3 is h-magic if and onl if h is not a divisor of m + and m. Proof. Suppose h is a divisor of m+ or m. Since gcd(m+, m ) = or 2 and h is odd, then h can onl be divisor of one of them. Without loss of generalit, we ma assume that h is an odd divisor of m+. As a result gcd(h, m ) =, and equation 2.3, (m+)+(m ) 0 (mod h), becomes 0 (mod h), that does not provide non-ero solution for. Conversel, assume that h is not a divisor of m and m +, and let m = hq + r. Then r ±; otherwise, one of m or m+ will be divisible b h. As a result, the selections of = d+ r and = +r are valid and will work with l + (w) = m +2 = 2(hq +r)+2(d+ r) 2 (mod h). We conclude the section b the following theorem, which is the natural consequence of the Theorems 2.7 through 2.2. This theorem will completel determine the integer-magic spectrum of ST (m)#c 3. Theorem 2.3. If m 2, then the integer-magic spectrum of the graph ST (m)#c 3 is N { d N : d = 2 or d is an odd divisor of m + or m }. As an application of this theorem, consider the graph ST (34)#C 23. To find its integer-magic spectrum it is enough to consider ST (34)#C 3, where m = 34. Now the odd divisors of m + = 35 and m = 33 are 3, 5, 7, 9, 5, 9, 27, 45, 33, and 35. Therefore, IM( ST (34)#C 23 ) = N { 2, 3, 5, 7, 9, 5, 9, 27, 45, 33, 35}. 206

9 3. Integer-Magic Spectrum of ST n. B the Observation.2, in an magic labeling of C n, the labels of the edges alternates. One immediate consequence of this fact is that, the graph ST (m)@c 3 (or ST (m)@c 4 )is h-magic if and onl if the graph ST (m)@c 2k+ (or ST (m)@c 2k ) is h-magic. As a result, we will onl concentrate on ccles C 3 and C 4. Also, we observe that since {, 2} is a subset of the degree set of these tpes of graphs, the can not be 2-magic. In general an magic labeling of ST (m)@c 3, as illustrated in the Figure 8, uses three non-ero distinct group elements,, and = +. Therefore, for an abelian group A, a necessar condition for ST (m)@c 3 to be A-magic is that A 4. Hence, for an m N, the graph ST (m)@c 3 is not 3-magic. Moreover, when m 4, then ST (m)@c 3 is Z-magic; because, the labels = m, = 3 m work with l + 2. Therefore, the integer-magic spectrum of these graphs will be contained in N {2, 3}. - w u Figure 8. A tpical magical labeling of ST (m)@c 3. Here = +. Theorem 3.. IM( ST (m)@c 4 ) = for ever m. Proof. This is a direct result of the Observation.3. We observe that ST ()@C 3 = ST ()#C 3, therefore IM( ST ()@C 3 ) = 2 + 2N. Also, b., ST (2)@C 3 is non-magic, or IM( ST (2)@C 3 ) =. To determine the integer-magic spectrum of ST (m)@c 3, from now on we will assume that m 3. Also, in an magic labeling of ST (m)@c 3 one needs to have l + (w) = l + (u) or (m )+ =, which implies (3.) (m 3) + (m ) 0 (mod h), or equivalentl ( = + ) (3.2) (m ) 2 0 (mod h), 207

10 Theorem 3.2. IM(ST 3 ) = 2 + 2N. + w + r+ w r+ - r- r Figure 9. IM(ST (3)@C 3 ) = 2 + 2N. Proof. When m = 3, the equation 3. will become 2 0 (mod h). This means that is a member of Z h, which has order 2, therefore 2 h and h is even. On the other hand, whenever h = 2r, the graph is h-magic, as illustrated in the Figure 9. Theorem 3.3. If m 4, then the graph ST (m)@c 3 is Z-magic. Proof. We observe that the choices of = m, = m+3 and = 2 will give us three distinct non-ero integers with l + (w) = (m ) + = 2(m ) m + 3 m + = 2. Theorem 3.4. For an abelian group A, if A 4, then ST (4)@C 3 is not A-magic. Furthermore, IM(ST (4)@C 3 ) = N {2, 3, 4}. + - w + + h-2 4 w h-2 h-2 h-3 h-3 Figure 0. IM(ST (4)@C 3 ) = N {2, 3, 4}. Proof. An magic labeling of this graph, as illustrated b the Figure 0, will require three distinct non-ero elements of the abelian group A; namel,,, and +. Also, in this case, equation

11 becomes + 3 = 0 (mod h). This implies that + 2 is another non-ero element of this group other than,, and +. Therefore, the group A must have at least five elements. Furthermore, as illustrated in the Figure 0, the graph is h-magic for ever h 5. Corollar 3.5. The graph ST (m)@c 3 is not 3-magic. Theorem 3.6. For an m 5 the graph ST (m)@c 3 is m-magic. Proof. The choices of = m, = 3, and = 2 will provide three distinct non-ero elements of Z m with l + (w) = 2. Theorem 3.7. If m is an odd positive integer, then the graph ST (2k + )@C 3 is 4-magic. Proof. We consider two cases: Case. If m = 4k+, then the choices of = 2, = 3, and = will work with l + (w) = 4k+ (mod 4). Case 2. If m = 4k + 3, then the choices of = 3, = 2, and = will work with l + (w) = (4k + 2) + 3 (mod 4). Theorem 3.8. If m = 2k + 5, then the integer-magic spectrum of ST (2k + )@C 3 is N {2} { d > : d is an odd divisor of one of m, m 2, or m 3 }. Proof. We will prove the theorem in five steps: Step. In this part we will show that for an h 2k the graph ST (2k + )@C 3 is h-magic. Because, the labeling of = k, = h + k, and naturall, = will work. Here we note that = h + 2k / 0 (mod h) and l + (w) = 2k + / (mod h). Step 2. If h is an divisor of m 2, then ST (m)@c 3 is not h-magic. Because, the equation 3.2, is equivalent to (m 2) + 0 (mod h), and since h is a divisor of m 2, we will have 0 (mod h), which is not an acceptable solution. Step 3. If h is an odd divisor of either m = 2k or m 3 = 2k 2, then ST (2k + )@C 3 is not h-magic. Because, the equation 3. becomes (2k 2) + 2k 0 (mod h) or (k ) + k 0 (mod h). Since gcd(k, k) =, without loss of generalit, we ma assume that d (k ) and gcd(d, k) =. As a result will get 0 (mod h), which is not an acceptable answer. 209

12 Step 4. If h is an odd number that is not a divisor of an one of m = 2k, m 2 = 2k, m 3 = 2k 2, then ST (m)@c 3 is h-magic. Because, the labels k, k, and naturall = are three non-ero distinct elements of Z h, will work with l +. Note that 2k / 0 (mod h). Step 5. If 4 < h m = 2k is an even number, then ST (2k + )@C 3 is h-magic. Because, the equation 3.2 becomes 2k 2 0 (mod h), which is equivalent to k 0 (mod h/2). Now for an non-ero Z h, we have two choices for ; namel, k (mod h), or k + h 2 (mod h). Eamples 3.9. (a) IM( ST (5)@C 3 ) = N {2, 3}. Here, m = 5. We need to eclude 2 and the odd divisors of m = 4, m 2 = 3, and m 3 = 2. (b) IM( ST (7)@C 3 ) = N {2, 3, 5}. Here, m = 7. We need to eclude 2 and the odd divisors of m = 6, m 2 = 5, and m 3 = 4. (c) IM( ST (45)@C 3 ) = N {2, 3, 7,, 2, 43}. Here, m = 45. We need to eclude 2 and the odd divisors of m = 44, m 2 = 43, and m 3 = 42. (d) IM( ST (35)@C 95 ) = N {2, 3, 7,, 9, 33, 67, 33}. Theorem 3.0. If m = 2k 4, then the integer-magic spectrum of ST (m)@c 3 is N { h > : h (2m 4), or h (m ), or h (m 3) }. Proof. We will prove the theorem in three steps: Step. If h > 2m 4, then ST (m)@c 3 is h-magic. Because, the choices of = m, = h m+3, and naturall = + = 2, will work with l + (w) = 2(m ) + 4 2m = 2. Also, note that = 4 2m / 0 (mod h). Step 2. If h > is an divisor of 2m 4, then the graph ST (m)@c 3 is not h-magic. It is enough to prove this statement for h > 2. Let 2m 4 = hq, which implies that m = (m 3)+hq or m (m 3) (mod h). In this case the equation 3. becomes (m 3)( ) 0 (mod h). Since gcd(m 3, 2m 4) = or 2, we have gcd(h, m 3) =, as a result (mod h), which does not provide a valid labeling. Step 3. If h is an odd divisor of either m or m 3, then ST (m)@c 3 is not h-magic. Here we observe that since gcd(m 3, m ) = or 2 and h is odd, then h cannot divide both 20

13 m 3 and m. Without loss of generalit, we ma assume that h is a divisor of m 3 and gcd(h, m ) =. In this case, the equation 3. becomes 0 (mod h), which does not provide a valid solution. Eamples 3.. (a) IM( ST (4)@C 3 ) = N {2, 3, 4}. Here, m = 4, and we need to eclude all the divisors d > of 2m 4 = 4, m = 3, and m 3 =. (b) IM( ST (6)@C 3 ) = N {2, 3, 4, 5, 8}. Here, m = 6. We need to eclude all the divisors of 2m 4 = 8, m = 5, and m 3 = 3. (c) IM( ST (4)@C 3 ) = N {2, 3, 4, 6, 8,, 2, 3, 24}. Here, m = 4. We need to eclude all the divisors of 2m 4 = 24, m = 3, and m 3 =. (d) The integer-magic spectrum of ST (38)@C 95 is N {h N : 2 h 9} {2, 8, 24, 35, 36, 37, 72}. Theorem 3.2. Given an n 2, there is a graph G such that G is not h-magic for ever h = 2, 3,, n. Proof. As we observed in the 3., the graph ST (4)@C 3 is not h-magic for h = 2, 3, 4. So we ma assume that n 4. Let µ be the least common multiple of the numbers 2, 3,, n, and consider the graph G = ST (m)@c 3, where m = µ + 4. Note that here µ is divisible b 4, and so is µ + 4, 2 which implies that m is even. Also, an h = 2, 3,, n, is a divisor of µ = 2m 4. Therefore, the graph ST (m)@c 3 is not h-magic. We note that the number m, presented in the proof of the theorem 3.2, might not be the smallest possible answer. For eample, in 3., we realied that ST (38)@C 3 is not h-magic for all 2 h 9, while it is 0-magic. In this case, number 38 works, while the least common multiple of the numbers 2, 3,, 9 is µ = 2520, and the number provided b this theorem is m = 262. We conclude this paper b the following problems: Problem 3.3. For an positive integer n Z + find the smallest m N such that the graph ST (m)@c 3 is not h-magic for all 2 h < n. Problem 3.4. As eamined in 3., the graph ST (6)@C 3 has the propert that it is 6-magic but it is not h-magic for all 2 h 5. Find all m N such that ST (m)@c 3 is not h-magic, for all 2 h < m. 2

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