1. The Towers of Riga Chiu Chang Mathematics Education Foundation President Wen-Hsien SUN

2. There are twelve ancient towers in Riga. The base of each consists of three regular hexagons each sharing one side with each of the other two. On these regular hexagons are right prisms of varying heights, as indicated by the numbers in Figure 1 below.

3. 1 1 1 1 1 2 3 1 4 2 2 2 6 2 5 5 3 6 6 4 5 5 5 5 6 4 6 6 6 6 2 1 3 2 1 5 Figure 1 A C B E F D I J H K L G

4. A legend says that the gods used magic to gather these twelve towers to the center of the town, and stored them in a large hexagonal box with a flat top and bottom and no spaces in between.

5. The pieces are named by, starting at the shortest section, naming the height of each section in counterclockwise order. For example, the piece on the left would be referred to as (2, 5, 6). Naming the Pieces 6 2 5

6. If there are two identical sections that are shortest, they are not separated, for example the bottom right piece would be referred to as (1, 1, 2) Naming the Pieces 1 1 2

7. It is decided that the twelve towers are to be combined into a modern museum of height 7, with the pattern of the base as shown in either diagram of Figure 2. The pink central hexagon represents the elevator shaft. 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

8. Figure 2

9. Clearly, six of the towers will have to be upside down, so that their flat bases form part of the flat top of the museum. Assuming that the practical difficulties can be overcome, the natural question is whether the towers actually fit together. In other words, is there a mathematical solution?

10. If both the top and the bottom of the apartment block follow the same one of the two patterns in Figure 2, then the twelve towers must form six complementary pairs, such as the (1,1,1) and the (6,6,6). 7 7 7 7 7 7

11. 6 6 6 1 1 1 6 6 6 1 1 1 1 1 1 6 6 6 + 7 7 7 =

12. However, we can see quickly that this is not the case. For instance, we have a (4,6,6) but no (1,1,3).

13. 6 4 1 3 6 1 + 6 4 6 1 3 1 6 4 6 1 3 1 7 7 7 =

14. 1 1 1 1 1 2 3 1 4 2 2 2 6 2 5 5 3 6 6 4 5 5 5 5 6 4 6 6 6 6 2 1 3 2 1 5 A C B E F D I J H K L G 1 3 1

15. Thus we may assume that the base of the apartment block follows the pattern on the left of Figure 2, and the top follows that on the right.

16. 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

17. This implies that the twelve towers form six overlapping pairs, one right side up and the other upside down. Two of the hexagonal prisms of the latter are resting on two of those of the former, so that the two vertical neighbors have a combined height of 7.

18. 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

19. 6 4 6 1 1 1 1 1 1 1 1 1 + 1 1 1 6 4 6 6 4 6 6 4 6 1 1 1 A 1. A+J 2. A+L 120° 1 1 1 6 6 6 1 1 1 6 6 6 1 1 1 6 6 6 + 1 1 1 60° 1. 2. = 7 1 6 7 = 7 4 1 7

20. 5 2 6 1 1 2 B 2. B+H 1. B+G 1 1 2 1. + 6 2 5 60° 2 1 1 180° 2 1 1 2 1 1 60° 5 2 6 5 2 6 2 1 1 1 1 2 2. + 6 5 3 60° 6 3 5 6 3 5 1 1 2 1 1 2 6 3 5 1 1 2 = 2 7 1 7 = 1 3 7 7 180°

21. 5 4 6 4. B+J 1 1 2 B 3. B+I 2. B+H 1. B+G 3. 5 6 4 1 1 2 + 60° 2 1 1 180° 2 1 1 2 1 1 60° 2 1 1 5 4 6 5 4 6 + 2 1 1 1 2 1 6 4 6 4 6 6 4 6 6 4 6 1 2 6 1 1 2 1 4. 180° 2 1 1 120° = 7 4 1 7 = 7 2 4 7

22. 6 6 5. B+L 4. B+J 1 1 2 B 3. B+I 2. B+H 1. B+G + 1 2 180° 120° 6 6 6 6 6 6 6 1 2 1 1 1 2 1 1 2 1 5. 2 1 1 = 7 6 2 7

23. 2 180° 2. C+J 1. C+G 2 1 3 C 6 2 5 + 1. 3 2 1 5 2 6 60° 3 5 2 6 5 2 6 2 3 1 2 3 1 2 3 1 1 2. + 3 2 1 6 4 6 60° 6 6 4 6 6 4 6 6 4 180° 2 3 1 1 2 3 120° 1 2 3 1 2 3 = 7 3 2 7 = 7 6 2 7

24. 2 3 5 4 6 6 5 4 2. C+J 3. C+I 1. C+G 2 1 3 C 3.(a) + 3 2 1 60° 180° 2 3 1 1 2 3 120° 1 2 3 1 2 3 5 6 4 6 5 4 6 5 4 3.(b) 180° + 3 2 1 60° 2 3 1 2 3 1 2 3 1 1 5 6 4 5 4 6 5 4 6 = = 4 3 7 7 7 2 5 7

25. 1. D+I 2 1 5 D + 1. 5 6 4 5 2 1 120° 6 4 5 6 4 5 1 2 5 60° 1 5 2 2. D+G 2.(a) 5 2 1 + 6 2 5 5 2 6 60° 5 2 6 1 2 5 120° 2 5 1 6 4 1 5 5 2 5 2 2 5 6 1 = = 4 7 5 7 7 5 2 7 180° 1 5 2 2 5 1

26. 2 6 2 5 1 5 2. D+G 1. D+I D 2 1 5 2.(b)2.(c) 5 2 1 + 6 2 5 60° 1 2 5 6 5 2 6 5 2 6 5 2 1 2 5 1 2 5 5 2 1 + 6 2 5 1 2 5 180° 6 2 5 60° 2 1 5 2 1 5 = = 1 6 4 7 7 7 5 2 7 180°

27. E+J 3 1 4 E 6 6 6 6 4 3 1 4 6 4 6 + 1 4 3 1 4 4 3 1 4 3 120° 180° 4 3 1 = 6 7 7 4

28. F+K 2 2 2 F 5 5 5 5 5 5 2 2 2 2 2 2 5 5 5 2 2 2 + = 5 7 7 2

29. ABCDEF ++++++ 1 1 1 1 1 2 3 1 4 2 2 2 2 1 3 2 1 5 5 5 5 K 6 4 6 J I G 6 2 5 6 4 5 6 4 6 I J G 6 6 6 L 5 3 6 H 6 4 6 6 6 6 J L 6 2 5 6 4 5 6 4 6 I J G 6 4 5 6 2 5

30. ABCDEF ++++++ 1 1 1 1 1 2 3 1 4 2 2 2 2 1 3 2 1 5 5 5 5 K 6 4 6 J 6 2 5 6 4 5 I G 6 6 6 L 5 3 6 H 6 6 6 L I G 6 4 5 6 2 5 I G 6 4 5 6 2 5

31. ABCDEF ++++++ 1 1 1 1 1 2 3 1 4 2 2 2 2 1 3 2 1 5 5 5 5 K 6 4 6 J 6 6 6 L 5 3 6 H I G 6 4 5 6 2 5 I G 6 4 5 6 2 5

32. 6 6 6 1 1 1 6 6 6 1 1 1 6 6 6 + 1 1 1 = 7 1 6 7 (1, 1, 1)+(6, 6, 6)=(6, 1) A+L

33. 1 1 2 5 3 6 + 5 3 6 2 1 1 5 3 6 2 1 1 5 3 6 2 1 1 = 3 7 7 1 (1, 1, 2)+(3, 5, 6)=(3, 1) 120° 180° B+H

34. 6 6 6 6 4 3 1 4 6 4 6 + 1 4 3 1 4 4 3 1 4 3 = 6 7 7 4 (1, 3, 4)+(4, 6, 6)=(6, 4) 120° 180° 4 3 1 E+J

35. 5 5 5 5 5 5 2 2 2 2 2 2 5 5 5 2 2 2 + = 5 7 7 2 (2, 2, 2)+(5, 5, 5)=(5, 2) F+K

36. 6 7 7 4 5 7 7 2 3 7 7 1 7 1 6 7 6 2 5 6 4 5 2 1 3 2 1 5 Now we get the following accessories (5, 2)(6, 4) (3, 1) (6, 1) C D G I

37. We identify (2,2,2)+(5,5,5), (1,3,4)+(4,6,6), (1,1,2)+(3,5,6) and (1,1,1)+(6,6,6) with the dominoes (2,5), (4,6), (1,3) and (1,6), respectively.

38. Each of the dominoes (4,6) and (1,3) links a (1,6) with a (1,6). 6 7 7 4 7 7 1 3 7 7 1 6 7 7 4 7 7 1 3 + + 7 1 6 7 6 7 7 4 7 7 6 7 7 1 6 7 7 1 1 7 7 1 = 1 6 7 7 7 7 7 7 7 7 60° 120°

39. In order to get the (2,5) into the domino ring, we need at least one domino in which one of the numbers is 2 or 5, and the other is 1 or 6. With this in mind, we now examine the remaining four towers. There are two possible couplings. 5 7 7 2 1 6 7 7 7 7 7 7 7 7

40. CD ++ 2 1 3 2 1 5 I G 6 4 5 6 2 5 I G 6 4 5 6 2 5 (i)C+I 、 D+G (ii)C+G 、 D+I

41. + + 7 1 6 7 7 5 2 7 7 3 4 7 7 1 6 7 7 5 2 7 (i) False C+I D+G 5 6 4 2 1 3 2 1 5 6 2 5

42. 7 2 3 7 7 5 4 7 (ii) C+G D+I + + 5 6 4 2 1 3 2 1 5 6 2 5 False

43. Each of the dominoes (4,6) and (1,6) links a (1,3) with a (3,4). [Method 1] 7 6 4 7 6 7 7 4 7 6 4 7 7 6 4 7 7 7 6 7 7 3 7 7 6 7 7 3 3 7 7 1 + + 7 1 6 7 = 7 7 6 1 7 7 3 1 4 7 7 7 7 7 7 7 7 3 180° 60° 7 1 6 7 180° 60° 7 7 1 3 180°

44. Each of the dominoes (4,6) and (1,6) links a (1,3) with a (3,4). [Method 2] 7 6 4 7 6 7 7 4 7 6 4 7 7 6 4 7 7 7 6 7 7 3 7 7 6 + + 3 7 7 1 7 1 6 7 7 7 6 1 7 7 3 7 7 3 1 = 4 7 7 7 7 7 7 7 7 3 180° 7 1 6 7 60° 180° 7 1 3 7 60°

45. In order to get the (2,5) into the domino ring, we need at least one domino in which one of the numbers is 2 or 5, and the other is 3 or 4. With this in mind, we now examine the remaining four towers. There are two possible couplings. 5 7 7 2 4 7 7 7 7 7 7 7 7 3 4 7 7 7 7 7 7 7 7 3

46. CD ++ 2 1 3 2 1 5 I G 6 4 5 6 2 5 I G 6 4 5 6 2 5 (i)C+I 、 D+G (ii)C+G 、 D+I

47. + + 7 1 6 7 7 5 2 7 7 3 4 7 7 1 6 7 7 5 2 7 (i) False C+I D+G 5 6 4 2 1 3 2 1 5 6 2 5

48. 7 2 3 7 7 5 4 7 (ii) C+G D+I + + 5 6 4 2 1 3 2 1 5 6 2 5 OK!

49. 4 7 7 7 7 7 7 7 7 3 3 7 7 2 + + = 4 3 7 7 2 3 5 7 7 2 7 2 3 7 7 5 4 7 4 7 7 7 7 7 7 7 7 3 + 7 7 7 7 7 7 7 7 = 7 5 4 7 7 7 2 5 7 7 2 5 7 7 2 5 7 7 2 7 5 4 3 7 Method 1 7 7 7 5 4 7

50. 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

51. 1 6 1 6 4 3 3 4 6 1 1 6 5 2 1 6 1 6 1 3 4 2 2 2 1 2 1 1 1 1 5 6 4 6 2 5 6 6 6 2 1 5 3 2 1 6 4 6 6 3 5 5 5 5 7 2 2 1 5 5 5 6 2 5 2 5 2 2 5 4 3 6 1

52. 7 3 4 7 7 7 7 7 7 7 + + = 5 7 7 2 7 2 3 7 7 5 4 7 + 3 7 7 2 7 7 2 3 7 7 2 5 7 7 2 5 7 7 2 7 5 4 4 7 7 7 7 7 7 7 7 3 4 7 7 7 7 7 7 7 7 3 3 7 Method 2 7 7 5 7 7 2 7 5 4 7

53. 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7

54. 3 4 1 6 4 3 1 6 5 2 1 6 1 6 1 1 2 5 4 6 6 5 3 6 5 5 5 3 1 2 1 2 5 6 6 6 5 2 6 6 5 4 1 1 1 3 1 4 2 2 2 1 1 2 6 7 2 6 6 2 5 5 1 6 1 2 5 2 5 4 3 5 2

55. 4 6 6 5 3 6 5 5 5 3 1 2 1 2 5 6 6 6 5 2 6 6 5 4 1 1 1 3 1 4 2 2 2 1 1 2 1 3 4 2 2 2 1 2 1 1 1 1 5 6 4 6 2 5 6 6 6 2 1 5 3 2 1 6 4 6 6 3 5 5 5 5

56. A A B B C C D D E E F F G G H H I I J J K K L L

57. Remark It should be mentioned that uniqueness is up to symmetry. Had we assumed that the base of the apartment block follows the pattern on the right of Figure 2, and the top follows that on the left, we would obtain the solution in the right of last page. We do not consider these two solutions distinct.

58. All Combination not higher than 6

59. History The idea for the Tower of Riga was found in a 32-page pamphlet “ Jug of Diamonds ” (in Russian), published by the Ukrainian puzzlist Serhiy Grabarchuk in Uzhgorod, 1991. The towers there are all non-symmetric, but it has many solutions.

60. History The late Latvian mathematician and computing scientist Alberts Vanags found by hand a set of towers with only three solutions. Later, Atis Blumbergs, a retired Latvian physician, found by computer a set with a unique solution. This feat was duplicated by Marija Babiča, a master degree student at the University of Latvia. However, these sets contain many symmetric towers.

61. History Our set, found by Andris Cibulis, has only six symmetric towers. It was used as his Exchange Gift at the 25th International Puzzle Party in Helsinki in July, 2005. Afterwards, the right to manufacture it is granted to Chiu Chang Mathematics Books and Puzzles in Taipei, a company affiliated with Chiu Chang Mathematics Foundation.

62. History A power-point demonstration of the solution may be found on the Foundation's website: http://www.chiuchang.org.tw/download/catalo g/rigatower.ppt

63. History In 2006, Marija Babiča found the following set of towers containing only four symmetric ones and yet has a unique solution: (1,1,1), (3,3,3), (4,4,4), (6,6,6), (1,2,3), (1,3,5), (1,5,6), (1,6,4), (2,3,5), (2,5,4), (2,6,4) and (3,4,6). Using our approach, the reader should have little difficulty solving this version of the Towers of Riga.