1. Distance magic and group distance magic graphs Report on work in progress (and a few dead ends) Dalibor Froncek University of Minnesota Duluth

3. Incomplete round robin tournaments

4. What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team.

5. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options:

6. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options: All teams have the same strength of schedule

7. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options: All teams have the same strength of schedule The tournaments mimics the complete tournament (strongest team has easiest schedule)

8. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options: All teams have the same strength of schedule The tournaments mimics the complete tournament (strongest team has easiest schedule) All teams have the same chance of winning (weakest team has easiest schedule)

9. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options: All teams have the same strength of schedule The tournaments mimics the complete tournament (strongest team has easiest schedule) All teams have the same chance of winning (weakest team has easiest schedule)

10. Incomplete round robin tournaments What if we have 8 teams but not enough time to play a complete tournament? Say we can play only 4−6 games per team. What games we choose, if the teams are ranked according to their strength? We have several options: All teams have the same strength of schedule The tournaments mimics the complete tournament (strongest team has easiest schedule) All teams have the same chance of winning (weakest team has easiest schedule)

11. Distance magic labeling Distance magic vertex labeling of a graph G with n vertices: A bijection μ from the vertex set of G to {1, 2,…, n} such that sum w(x) of labels of the neighbors of each vertex (called the weight of x) is equal to the same constant m.

12. Distance magic labeling Distance magic vertex labeling of a graph G with n vertices: A bijection μ from the vertex set of G to {1, 2,…, n} such that sum w(x) of labels of the neighbors of each vertex (called the weight of x) is equal to the same constant m.

13. Distance magic labeling Distance magic vertex labeling of a graph G with n vertices: A bijection μ from the vertex set of G to {1, 2,…, n} such that sum w(x) of labels of the neighbors of each vertex (called the weight of x) is equal to the same constant m.

14. Distance magic labeling Distance magic vertex labeling of a graph G with n vertices: A bijection μ from the vertex set of G to {1, 2,…, n} such that sum w(x) of labels of the neighbors of each vertex (called the weight of x) is equal to the same constant m.

15. n-dimensional hypercube Q n Q n Vertex set: All binary n-tuples Edge set: Two vertices adjacent iff they differ in exactly one position

16. n-dimensional hypercube Q n Q n Vertex set: All binary n-tuples Edge set: Two vertices adjacent iff they differ in exactly one position

17. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2

18. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2

19. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2

20. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2

21. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2 or Q n-2  C 4

22. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2 or Q n-2  C 4

23. n-dimensional hypercube Q n Recursive definition of Q n Take two copies of Q n-1 and join the corresponding vertices by an edge This is the cartesian product Q n-1  K 2 or Q n-2  C 4

24. Distance magic labeling of Q n Remark: For k-regular graphs we can use for labels the set {0, 1, 2,…, n–1} instead of {1, 2,…, n}

25. Distance magic labeling of Q n Remark: For k-regular graphs we can use for labels the set {0, 1, 2,…, n–1} instead of {1, 2,…, n}

26. Distance magic labeling of Q n Remark: For k-regular graphs we can use for labels the set {0, 1, 2,…, n–1} instead of {1, 2,…, n} and define μ' (x) = μ(x) – 1

27. Distance magic labeling of Q n Remark: For k-regular graphs we can use for labels the set {0, 1, 2,…, n–1} instead of {1, 2,…, n} and define μ' (x) = μ(x) – 1 Then we have m ' = m – k

28. Distance magic labeling of Q n

29. Small cases: n = 1NO

30. Distance magic labeling of Q n Small cases: n = 1NO n = 2YES

31. Distance magic labeling of Q n Small cases: n = 1NO n = 2YES n = 3NO

32. Distance magic labeling of Q n Small cases: n = 1NO n = 2YES n = 3NO Conjecture?

33. Distance magic labeling of Q n Small cases: n = 1NO n = 2YES n = 3NO Conjecture? Q n is distance magic if and only if n is even. ;-)

34. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Conjecture? Q n is distance magic if and only if n is even.

35. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Conjecture? Q n is distance magic if and only if n is even. 

36. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Conjecture: (Acharya, Rao, Singh, Parameswaran) Q n is not distance magic for any n except n = 2.

37. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Conjecture: (Acharya, Rao, Singh, Parameswaran) Q n is not distance magic for any n except n = 2. :-O

38. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Conjecture: (Acharya, Rao, Singh, Parameswaran) Q n is not distance magic for any n except n = 2. :-O (what?)

39. Distance magic labeling of Q n Theorem 1: There is no r-regular DM graph for r odd. Theorem 2: Q n is not distance magic for n = 4k. Conjecture: (Acharya, Rao, Singh, Parameswaran) Q n is not distance magic for any n except n = 2.

40. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0

41. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i

42. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i

43. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m W 5 = 56m

44. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m W 5 = 56m

45. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m = 4S 4 + 6S 2 and 6S 2 = 56m – 4S 4 W 5 = 56m = 4S 4 + 6S 6 and 6S 6 = 56m – 4S 4 Hence, S 2 = S 6.

46. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m = 4S 4 + 6S 2 and 6S 2 = 56m – 4S 4 W 5 = 56m = 4S 4 + 6S 6 and 6S 6 = 56m – 4S 4 Hence, S 2 = S 6. Similarly, W 1 = 8m = 2S 2 + 8S 0 and 8S 0 = 8m – 2S 2 W 7 = 8m = 2S 6 + 8S 8 and 8S 8 = 8m – 2S 6

47. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m = 4S 4 + 6S 2 and 6S 2 = 56m – 4S 4 W 5 = 56m = 4S 4 + 6S 6 and 6S 6 = 56m – 4S 4 Hence, S 2 = S 6. Similarly, W 1 = 8m = 2S 2 + 8S 0 and 8S 0 = 8m – 2S 2 W 7 = 8m = 2S 6 + 8S 8 and 8S 8 = 8m – 2S 6 But S 2 = S 6 and then S 0 = S 8.

48. Theorem 2: Q n is not distance magic for n = 4k. Proof: (By example for n = 8.) Level i = all vertices at distance i from the vertex labeled 0 S i = sum of vertex labels at level i W i = sum of vertex weights at level i W 3 = 56m = 4S 4 + 6S 2 and 6S 2 = 56m – 4S 4 W 5 = 56m = 4S 4 + 6S 6 and 6S 6 = 56m – 4S 4 Hence, S 2 = S 6. Similarly, W 1 = 8m = 2S 2 + 8S 0 and 8S 0 = 8m – 2S 2 W 7 = 8m = 2S 6 + 8S 8 and 8S 8 = 8m – 2S 6 But S 2 = S 6 and then S 0 = S 8. Level 0 and level 8 contain just one vertex each (0 and x, resp.), we have µ(x) = 0, which is a contradiction.

49. Theorem 2*: Q n is not distance magic for n = 4k+2.

50. Theorem 2*: Q n is not distance magic for n = 4k+2, n ≥ 6. Proved independently by (C. Barrientos, E. Krop, C. Raridan) and DF

51. Theorem 2*: Q n is not distance magic for n = 4k+2, n ≥ 6. Proved independently by (C. Barrientos, E. Krop, C. Raridan) and DF (Un)fortunately, the proof was WRONG!

52. Theorem 2*: Q n is not distance magic for n = 4k+2, n ≥ 6. Proved independently by (C. Barrientos, E. Krop, C. Raridan) and DF (Un)fortunately, the proof was WRONG!

53. Theorem 2*: Q n is not distance magic for n = 4k+2, n ≥ 6. Proved independently by (C. Barrientos, E. Krop, C. Raridan) and DF (Un)fortunately, the proof was WRONG!

54. Theorem 2*: Q n is not distance magic for n = 4k+2, n ≥ 6. Proved independently by (C. Barrientos, E. Krop, C. Raridan) and DF (Un)fortunately, the proof was WRONG!

55. Distance magic labeling of Q 6

62. 6?!

63. Distance magic labeling of Q 6 6?! “Isn’t it three too many, Anton Pavlovich?”

64. Distance magic labeling of Q 6 Reduce Q 6 to Q 3

65. Distance magic labeling of Q 6

70. Vertex labels in 4 horizontal levels 1312101116172322 6362565734353736 235431302425 4849555445444243 5859616039383233 89151421201819 5352505140414746 760126272928

71. Distance magic labeling of Q 6 Vertex labels in 4 horizontal levelsSum of 4 neighbors in the same horizontal level 1312101116172322134 118 6362565734353736134 118 235431302425118 134 4849555445444243118 134 5859616039383233134 118 89151421201819134 118 5352505140414746118 134 760126272928118 134

72. Distance magic labeling of Q 6 Sum of 2 neighbors in the same columnSum of 4 neighbors in the same horizontal level 55 71 134 118 55 71 134 118 71 55 118 134 71 55 118 134 55 71 134 118 55 71 134 118 71 55 118 134 71 55 118 134

73. Distance magic labeling of Q 6 Sum of 2 neighbors in the same columnSum of 4 neighbors in the same horizontal level 55 71 134 118 55 71 134 118 71 55 118 134 71 55 118 134 55 71 134 118 55 71 134 118 71 55 118 134 71 55 118 134 m = 55 + 134 = 71 + 118 = 189

74. Distance magic labeling of Q 4k+2 Problem: Find a distance magic labeling of Q 4k+2 for k >1

75. Distance magic labeling of Q 4k+2 Problem: Find a distance magic labeling of Q 4k+2 for k >1 Theorem (P. Gregor): There is a distance magic labeling of Q 4k+2 for k >1

76. Distance magic labeling of Q 4k+2 Proof: Based on some matrix application and a group distance labeling. Theorem (P. Gregor): There is a distance magic labeling of Q 4k+2 for k >1

77. Distance magic labeling of Q 4k+2 Proof: Based on some matrix application and a group distance labeling. What? Theorem (P. Gregor): There is a distance magic labeling of Q 4k+2 for k >1

78. Distance magic labeling of Q 4k+2 Proof: Based on some matrix application and a group distance labeling. What? Group distance magic labeling? Theorem (P. Gregor): There is a distance magic labeling of Q 4k+2 for k >1

80. Distance magic labeling of Q 4 revisited We know that distance magic labeling of Q 4 does not exist.

81. Distance magic labeling of Q 4 revisited We know that distance magic labeling of Q 4 does not exist. Couldn’t we find a less restrictive labeling?

82. Distance magic labeling of Q 4 revisited Definition: Let G be a graph with n vertices and G a group with n elements. We call a bijection g: V(G)  G G -distance magic labeling if w(x) = m for some m in G. We know that distance magic labeling of Q 4 does not exist. Couldn’t we find a less restrictive labeling?

83. Distance magic labeling of Q 4 revisited Definition: Let G be a graph with n vertices and G a group with n elements. We call a bijection g: V(G)  G G -distance magic labeling if w(x) = m for some m in G. We know that distance magic labeling of Q 4 does not exist. Couldn’t we find a less restrictive labeling? Maybe Z 16 -distance magic labeling?

84. Z n -distance magic labeling of Q 2k+1

85. Theorem: There is no r-regular distance magic graph for r odd.

86. Z n -distance magic labeling of Q 2k+1 Theorem: There is no r-regular distance magic graph for r odd. Theorem: (Sylwia Cichacz) There is no r-regular Z n -distance magic graph for r odd.

87. G - distance magic labeling of Q 2k+1 Theorem: There is no r-regular distance magic graph for r odd. Theorem: (Sylwia Cichacz) There is no r-regular Z n -distance magic graph for r odd. Theorem: There is no r-regular G -distance magic graph for r odd and any Abelian group G.

88. Z 16 -distance magic labeling of Q 4 Definition: Let G be a graph with n vertices and G a group with n elements. We call a bijection g: V(G)  G G -distance magic labeling if w(x) = m for some m in G. We know that distance magic labeling of Q 4 does not exist. Couldn’t we find a less restrictive labeling? Maybe Z 16 -distance magic labeling?

89. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 13121011 151489 2354 0176

90. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 0176 111312101113 915148915 423542 601760 13121011

91. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x)

92. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x) 6

93. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x) 6 6

94. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x) 6 6 6

95. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x) 6666 6666 6666 6666

96. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x)

97. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? Define g i (x) = g(x) + i 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x)

98. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? Define g i (x) = g(x) + i Then w i (x) = w(x) + 4i 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x)

99. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? Define g i (x) = g(x) + i Then w i (x) = w(x) + 4i Say i = 1 0176 111312101113 915148915 423542 601760 13121011 w(x)w(x)

100. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? Define g i (x) = g(x) + i Then w i (x) = w(x) + 4i Say i = 1 1287 121413111214 100159100 534653 712871 14131112 w(x)w(x)

101. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 How about another constant? Define g i (x) = g(x) + i Then w i (x) = w(x) + 4i Say i = 1. Then m = 10. 1287 121413111214 100159100 534653 712871 14131112 w(x)w(x) 10

102. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 Using i = 0, 1, 2, 3, we can do it for m = 2, 6, 10, 14 Can we do it for other constants than 2, 6, 10, 14? 1287 121413111214 100159100 534653 712871 14131112 w(x)w(x) 10

103. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 Using i = 0, 1, 2, 3, we can do it for m = 2, 6, 10, 14 Can we do it for other constants than 2, 6, 10, 14? No. 1287 121413111214 100159100 534653 712871 14131112 w(x)w(x) 10

104. Z 16 -distance magic labeling of Q 4 Z 16 -distance magic labeling with m = 6 Using i = 0, 1, 2, 3, we can do it for m = 2, 6, 10, 14 Can we do it for other constants than 2, 6, 10, 14? No. For 4-regular graphs with 0 (mod 4) vertices, m is always 2 (mod 4). 1287 121413111214 100159100 534653 712871 14131112 w(x)w(x) 10

105. Z n -distance magic labeling of Q 2k ? Z 16 -distance magic labeling of Q 4 with m = 2, 6, 10, 14 Recall that Q 2 allows a Z 4 -distance magic labeling Problem: Does there exist a Z n -distance magic labeling of Q 2k for n = 2 2k = 4 k and k > 2? Or in general a G -distance magic labeling of Q 4k ?

106. Z n -distance magic labeling of Q 2k ? Z 16 -distance magic labeling of Q 4 with m = 2, 6, 10, 14 Recall that Q 2 allows a Z 4 -distance magic labeling Problem: Does there exist a Z n -distance magic labeling of Q 2k for n = 2 2k = 4 k and k > 2? Or in general a G -distance magic labeling of Q 4k ?

107. So is there a G -distance magic labeling of Q 2k with k ≥ 3 for some group G ?

108. Z 2 ×Z 2 ×…×Z 2 -distance magic labeling of Q 2k Observation: There exists a G -distance magic labeling of Q 2k for k ≥ 1 and G = Z 2 ×Z 2 ×…×Z 2.

109. Z 2 ×Z 2 ×…×Z 2 -distance magic labeling of Q 2k Observation: There exists a G -distance magic labeling of Q 2k for k ≥ 1 and G = Z 2 ×Z 2 ×…×Z 2. Proof: Take the “natural” labeling. The neighbors of x = x 1 x 2 …x 2k are x i = x 1 x 2 …x i–1 (x i +1) x i+1 …x 2k for i = 1,2,…,2k.

110. Z 2 ×Z 2 ×…×Z 2 -distance magic labeling of Q 2k Observation: There exists a G -distance magic labeling of Q 2k for k ≥ 1 and G = Z 2 ×Z 2 ×…×Z 2. Proof: Take the “natural” labeling. The neighbors of x = x 1 x 2 …x 2k are x i = x 1 x 2 …x i–1 (x i +1) x i+1 …x 2k for i = 1,2,…,2k. Then x 1 +x 2 = = (2x 1 +1) (2x 2 +1)(2x 3 )…(2x 2k ) = = 1100…0

111. Z 2 ×Z 2 ×…×Z 2 -distance magic labeling of Q 2k Observation: There exists a G -distance magic labeling of Q 2k for k ≥ 1 and G = Z 2 ×Z 2 ×…×Z 2. Proof: Take the “natural” labeling. The neighbors of x = x 1 x 2 …x 2k are x i = x 1 x 2 …x i–1 (x i +1) x i+1 …x 2k for i = 1,2,…,2k. Then x 1 +x 2 = = (2x 1 +1) (2x 2 +1)(2x 3 )…(2x 2k ) = = 1100…0 Similarly we have x 2i–1 +x 2i = = (2x 1 )…(2x 2i–1 +1) (2x 2i +1) …(2x 2k ) = = 00…0110…0

112. Z 2 ×Z 2 ×…×Z 2 -distance magic labeling of Q 2k Observation: There exists a G -distance magic labeling of Q 2k for k ≥ 1 and G = Z 2 ×Z 2 ×…×Z 2. Proof: Take the “natural” labeling. The neighbors of x = x 1 x 2 …x 2k are x i = x 1 x 2 …x i–1 (x i +1) x i+1 …x 2k for i = 1,2,…,2k. Then x 1 +x 2 = = (2x 1 +1) (2x 2 +1)(2x 3 )…(2x 2k ) = = 1100…0 Similarly we have x 2i–1 +x 2i = = (2x 1 )…(2x 2i–1 +1) (2x 2i +1) …(2x 2k ) = = 00…0110…0 and the sum of all neighbors is 111…1.

113. Z n -distance magic labeling of Q 2k

114. Theorem: There exists a Z n -distance magic labeling of Q 2k for k ≥ 1 and n = 4 k.

115. Z n -distance magic labeling of Q 2k Theorem: There exists a Z n -distance magic labeling of Q 2k for k ≥ 1 and n = 4 k. Proof: Recursive construction.

116. Z n -distance magic labeling of Q 2k Theorem: There exists a Z n -distance magic labeling of Q 2k for k ≥ 1 and n = 4 k. Proof: Recursive construction. To label Q 2k+2 : take four copies of Q 2k multiply the labels by 4 or −4 in each copy, add to the label of the four images of a vertex x elements 0, 1, 2, 3 of Z 4n, resp.

117. Z n -distance magic labeling of Q 2k To label Q 2k+2 : take four copies of Q 2k multiply the labels by 4 or −4 in each copy, add to the label of the four images of a vertex x elements 0, 1, 2, 3 of Z 4n, resp.

118. Is there a G -distance magic labeling of Q 2k with k ≥ 3 for other groups G ?

119. Theorem (??): There exists a G -distance magic labeling of Q 2k for k ≥ 1 and any Abelian group G of order n = 4 k.

120. Is there a G -distance magic labeling of Q 2k with k ≥ 3 for other groups G ? Theorem (??): There exists a G -distance magic labeling of Q 2k for k ≥ 1 and any Abelian group G of order n = 4 k. Proof: Recursive construction similar to the previous one.

122. Z mk -distance magic labeling of C m □ C k Problem: Does there exist a Z mk -distance magic labeling of C m  C k ?

123. Z mk -distance magic labeling of C m □ C k Problem: Does there exist a Z mk -distance magic labeling of C m  C k ? Theorem: (Rao, Singh, Parameswaran) G =C m  C k is distance magic if and only if m = k = 4s + 2

124. Z mk -distance magic labeling of C m □ C k Problem: Does there exist a Z mk -distance magic labeling of C m  C k ? Theorem: (Rao, Singh, Parameswaran) G =C m  C k is distance magic if and only if m = k = 4s + 2 Theorem: There exists a Z mk -distance magic labeling of C m  C k if and only if mk is even.

125. Z mk -distance magic labeling of C m □ C k Theorem: There exists a Z mk -distance magic labeling of C m  C k if and only if mk is even.

126. Z mk -distance magic labeling of C m □ C k Proof:

127. Z mk -distance magic labeling of C m □ C k Proof:

128. Z mk -distance magic labeling of C m □ C k Proof:

129. Z mk -distance magic labeling of C m □ C k Proof:

130. Z mk -distance magic labeling of C m □ C k Proof:

131. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k)

132. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 12 x 21 Σ x 23 x 32

133. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 12 x 21 Σ x 23 x 32 Σ x 34 x 43

134. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … x 12 x 21 Σ x 23 x 32 Σ x 34 x 43

135. Z mk -distance magic labeling of C m ×C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … but when l is odd …=x 32 + x 23 = x 54 + x 45 =… x 12 x 21 Σ x 23 x 32 Σ x 34 x 43

136. Z mk -distance magic labeling of C m ×C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … but when l is odd …=x 32 + x 23 = x 54 + x 45 =… so x ij + x i+1,j-1 = Σ/2 x 23 x 32 Σ/2

137. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … but when l is odd …=x 32 + x 23 = x 54 + x 45 =… so x ij + x i+1,j-1 = Σ/2 but x i,j+2 + x i+1,j-1 = Σ/2 x 23 Σ/2 x 34

138. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … but when l is odd …=x 32 + x 23 = x 54 + x 45 =… so x ij + x i+1,j-1 = Σ/2 but x i,j+2 + x i+1,j-1 = Σ/2 and x i,j+2 = x i,j x 23 x 32 Σ/2 x 34

139. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) x 21 + x 12 = x 43 + x 34 = … but when l is odd …=x 32 + x 23 = x 54 + x 45 =… so x ij + x i+1,j-1 = Σ/2 but x i,j+2 + x i+1,j-1 = Σ/2 and x i,j+2 = x i,j Contradiction! x 23 x 32 Σ/2 x 34

140. Z mk -distance magic labeling of C m □ C k Proof: Diagonal length is l = lcm(m,k) Hence, mk cannot be odd. x 23 x 32 Σ/2 x 34

141. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Then h = 2n/l. Fill half of a diagonal by H 0 h 2h2h

142. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Then h = 2n/l. Fill half of a diagonal by H, and the rest by coset H – 1 going “backwards.” 0 n–1 h –h–1 2h2h …

143. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Then h = 2n/l. Fill half of a diagonal by H, and the rest by coset H – 1 going “backwards.” 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

144. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Now fill other diagonals carefully by cosets H+i and H–i–1 to obtain the same sums, n–1 and h–1. 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

145. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Now fill other diagonals carefully by cosets H+i and H–i–1 to obtain the same sums, n–1 and h–1. Then Σ = h–2. 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

146. Z mk -distance magic labeling of C m □ C k Proof: When n=mk even, l is even. Find a subgroup of Z n of order l/2, say H =. Now fill other diagonals carefully by cosets H+i and H–i–1 to obtain the same sums, n–1 and h–1. Then Σ = h–2. 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

147. Z mk -distance magic labeling of C m □ C k Remark: For n=mk ≡ 0 (mod 4), we have always Σ ≡ 2 (mod 4). For n=mk ≡ 2 (mod 4), we have Σ ≡ 0 (mod 4). 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

148. Z mk -distance magic labeling of C m □C k Remark: For n=mk ≡ 0 (mod 4), we have always Σ ≡ 2 (mod 4). For n=mk ≡ 2 (mod 4), we have Σ ≡ 0 (mod 4). Theorem: There is no r-regular DM graph with n vertices for r  n  2 mod 4. 0n–1 h–1 hn–1 –h–1 h–1 2h2hn–1 … h–1

149. G -distance magic labeling of C m □C k Theorem: (Rao, Singh, Parameswaran) G =C m  C k is distance magic if and only if m = k = 4s + 2 Theorem: There exists a Z mk -distance magic labeling of C m  C k if and only if mk is even. Theorem (Sylwia): There exists a G -distance magic labeling of C m  C k if mk is even and G = Z l × A where l ≡ 0 mod lcm(m,k).

150. G -distance magic labeling of C m □C k Theorem: (Rao, Singh, Parameswaran) G =C m  C k is distance magic if and only if m = k = 4s + 2 Theorem: There exists a Z mk -distance magic labeling of C m  C k if and only if mk is even. Theorem (Sylwia): There exists a G -distance magic labeling of C m  C k if mk is even and G = Z l × A where l ≡ 0 mod lcm(m,k). Observation: There exists a G -distance magic labeling of C 12  C 36 by G = Z 6 × Z 6 × Z 6 × Z 2.

152. Thank you!