The Towers of Riga
Chiu Chang Mathematics Education Foundation
President Wen-Hsien SUN
• 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.
Figure 1
A
B
1
E
1
1
1
4
2
1
I
6
J
5
4
6
D
2
2
1
G
2
3
1
F
3
C
5
6
5
1
H
6
5
2
2
2
6
K
5
4
3
5
5
L
6
6
6
• 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.
Naming the Pieces
• 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).
6
5
2
Naming the Pieces
• 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)
2
1
1
• 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
7
7
7
7
7
7
7
7
7
7
7
7
7
4
7
7
7
7
Figure 2
• 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?
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
7
7
6
7
=
6
6
1
+
1
6
1
6
1
6
1
1
• However, we can see quickly that this is
not the case. For instance, we have a
(4,6,6) but no (1,1,3).
7
7
7
=
4
6
6
+
3
4
1
6
1
6
3
1
1
A
B
1
E
1
1
1
4
2
1
I
6
J
5
4
6
D
2
2
1
G
2
3
1
F
3
C
5
6
5
1
H
6
5
2
6
K
5
4
3
2
2
5
5
3
L
6
6
6
1
1
• 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.
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
4
7
7
7
7
• 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.
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
4
7
7
7
7
A
1
2. A+L
1. A+J
1
1
60°
60°
120°
1.
6
4
4
4
1
16
6
2.
66
6
1
1
1
+
7
=
4
6
1
1
1
7
1
6
6
1
1
1
+
1
6
1
6
1
6
1
1
1
7
1
6
=
7
1
B
1
1
60°
60°
1.
2
1
1
1
1. B+G 2. B+H
2
22
11
1
65
6
2.
5
+
60°
12
2
1
1
1
2
2
6
6
5
5
3
3
+
180°
1
2
1
1
5
6
2
=
7
1
7
1
6
2
1
180°
5
2
3
=
7
7
3
B
1
1
60°
60°
3.
2
1
1
1
1. B+G 2. B+H
3. B+I 4. B+J
2
22
11
1
56
6
+
120°
4.
5
120°
4
4
+
2
12
1
2
1
11
180°
1
2
1
5
6
=
4
4
180°
1
4
66
6
6
7
7
4
4
1
6
1
6
7
2
7
=
4
2
B
2
1
1
1. B+G 2. B+H
3. B+I 4. B+J 5. B+L
120°
5.
2
1
11
6
+
2
12
1
6
6
180°
6
1
6
1
6
7
2
=
6
7
2
C
1. C+G
3
2
2. C+J
1
120°
60°
60°
1.
5
6
6
23
3
1
23
2
1
2.
5
+
1
60°
13
2
1
3
23
2
2
2
+
1
180°
3
2
51
6
3
2
=
7
2
7
1
6
3
4
2
180°
2
6
=
66
6
6
4
4
7
7
6
C
1. C+G
3. C+I
3
2
1
120°
60°
3.(a)
23
2
+
1
2
1
6
3
4
60°
3.(b)
6
6
13
2
1
3
180°
5
=
2. C+J
4
2
5
5
4
23
3
1
23
2
1
7
7
60°
3
5
+
1
2
51
6
6
5
6
180°
3
4
=
5
4
4
7
7
4
5
D
2. D+G
1. D+I
2
1
60°
120°
120°
1.
2
1
2
1
5
5
2
+
6
6
5
4
5
4
60°
2.(a)
52
2
5
4 180°
5
5
6
6
5
+
2
2
1
1
1
6
2
251
15
7
=
7
4
5
5
2
5
1
5
180°6
2
=
7
7
2
5
D
2. D+G
1. D+I
2
1
60°
60°
2.(b)
2
2.(c)
51
5
2
6
5
+
2
1
2
1
6
5
180°
2
7
5
2
2
5
5
7
=
2
2
5
6
5
+
2
1
6
2
5
5
180°
2
5
11
5
2
74
1
=
6
7
1
E
4
3
E+J
120°
1
1
4
31
3
66
4
3
+
1
66
44
180°
6
1
6
3
4
4
=
6
7
7
4
F
2
2
F+K
2
5
5
5
2
+
2
5
52
5
2
2
2
=
5
7
7
2
A
B
C
D
E
5
1
1
1
1
+
J
6
L
1
6
6
6
G
I
J
L
H
2
6 5
4
6 6
4
G
I
6 6
6
5
6
3
J
+
6 5
2
6
G
+
5
2
6 5
4
6 6
4
J
6
+
6
4
6
I
5
4
2
1
1
+
2
2
2
1
6 5
4
3
2
+
4
6
3
2
F
K
5
5
5
A
B
C
D
E
5
1
1
1
1
+
6
1
6
6
I
L
H
6
2
6 5
4
G
+
6
5
2
G
+
5
2
6
5
6
3
J
6
+
6
4
6 6
6
I
5
4
6
I
5
4
2
1
1
+
2
2
2
1
6 5
4
3
2
+
G
L
3
2
F
K
5
5
5
A
B
C
D
E
5
1
1
1
1
+
L
6
3
2
6
6
H
6
6
5
G
+
6
5
2
G
+
5
2
J
6
+
6
4
3
6
I
5
4
6
I
5
4
2
1
1
+
2
2
2
1
+
4
3
2
1
F
K
5
5
5
A+L
(1, 1, 1)+(6, 6, 6)=(6, 1)
6
6
6
+
6
1
6
1
6
1
1
1
7
1
=
6
7
1
B+H
(1, 1, 2)+(3, 5, 6)=(3, 1)
120°
5
3
6
5
6
+
3
2
51
6
3
2
1
2
1
1
180°
7
1
=
7
3
1
E+J
(1, 3, 4)+(4, 6, 6)=(6, 4)
66
120°
66
44
1
4
31
3
+
6
1
6
3
4
43
180°
1
4
=
6
7
7
4
F+K
(2, 2, 2)+(5, 5, 5)=(5, 2)
5
5
5
2
+
2
5
52
2
2
2
5
=
5
7
7
2
Now we get the following accessories
(5, 2)
5
7
7
(6, 4)
2
6
7
7
(3, 1)
(6, 1)
7
7
4
1
7
1
7
6
3
D
C
5
G
6
5
3
1
6
5
4
2
2
I
1
2
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.
• Each of the dominoes (4,6) and (1,3) links a
(1,6) with a (1,6).
6
7
7
120°
60°
4
7
7
+
77
1
1
+
33
6
7
7
4
7
7
1
7
1
7
7 7 6 1
6 1 7 7
6
=
7
7
7
7
7
1
7
7
7
• 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.
7
7
5
6
7
7
2
7
7
7
1
7
7
7
C
D
5
3
2
(i) C+I、D+G
2
1
1
+
(ii) C+G、D+I
6
G
6
5
2
6
I
+
5
4
G
2
6
I
5
5
4
(i)
6
3
+
2
1
7
6
D+G
C+I
7
5
5
4
2
1
7
1
5
7
7
7
7
2
6
7
6
+
2
1
7
3
5
4
False
5
7
2
(ii)
D+I
C+G
6
3
+
2
1
7
7
5
5
2
2
1
6
+
7
3
7
4
2
False
5
4
5
• Each of the dominoes (4,6) and (1,6) links a
(1,3) with a (3,4). [Method 1]
7
7
6
46
+
4
60°
7
180°
6
7
67
60°
61
+ 180°
3
1
1
7 7 7 71
7
7
1
3
3
1
180°
7
7
4
6
7
7
3
7
6
7
7
7
=
4
7
7
7
3
7
7
7
• Each of the dominoes (4,6) and (1,6) links a
(1,3) with a (3,4). [Method 2]
6
6
7
7
60°
4
+
4
7
60°
6 7
7 1
7
6 6
1
7
+
1
1
180°
7
6
7
6
7
7
3
7
7
4
=
180°
7
7
7
7
7
7
3
3
7
7
4
77
7 3
1
77
3
1
180°
• 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
7
7
7
7
7
7
3
7
7
7
7
4
4
7
7
7
3
7
7
7
C
D
5
3
2
(i) C+I、D+G
2
1
1
+
(ii) C+G、D+I
6
G
6
5
2
6
I
+
5
4
G
2
6
I
5
5
4
(i)
6
3
+
2
1
7
6
D+G
C+I
7
5
5
4
2
1
7
1
5
7
7
7
7
2
6
7
6
+
2
1
7
3
5
4
False
5
7
2
(ii)
D+I
C+G
6
3
+
2
1
7
7
5
5
2
2
1
6
+
7
3
7
4
2
OK!
5
4
5
7
7
4
7
7
7
3
Method 1
7
7
7
+
=
5
4
7
7
5
+
4
5
5
7 7
2
7
7
2
7
7
7
+
2
3
73
=
7
57
7
73
7
47
5
2
2
7
7
7
7
7
7
7
3
4
2
7
7
7
Method 1
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
7
7
4
7
7
7
7
Method 1
1
65
5
2
2
3
2
45
5
5
2
2
6
4
7
2
3
1
5
6
6
6
6
3
1
1
4
1
6
6
5
1
12
6
6
3
4
1
6
2
6
5
6
5
6
5
2
5
5
3
1
1
4
1
3
6
1
1
5
2
1
2
1
5
2
2
2
4
6
7
7
7
7
7
7
3
Method 2
7
7
7
4
+
5
7
7
+
4
5
2
37
77
+
2
73
2
=
7
7
7
7
7
7
3
7
7
7
7
3
7
7
7
4
5
7
57
7
2
4
2
75
77
2
Method 2
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
7
7
4
7
7
7
7
Method 2
6
6
6
1
1
1
6
6
5
3
1
1
4
2
4
6
7
3
1
62
1
5
3
1
6
45
5
2
2
2
5
5
5
2
2
4
6
3
6
6
2
1
6
6
5
5
5
6
1
5
3
5
2
3
4
1
1
1
5
1
6
1
2
2
2
1
6
2
4
2
5
6
6
3
4
1
6
2
6
5
6
5
6
5
2
5
5
3
1
4
6
3
6
6
2
1
6
6
5
5
5
6
1
5
3
5
2
1
4
1
3
6
1
1
5
2
1
2
1
2
5
2
2
4
6
3
4
1
1
1
5
1
6
1
2
2
2
1
6
2
4
2
5
J
C
L
K
H
D
E
G
F
J
I
C
L
K
H
A
A
B
B
D
E
G
I
F
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.
All Combination not higher than 6
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.
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.
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.
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
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.