A-----B-----C / \ / \ / \ / \ / \ / \ D-----E-----F-----G / \ / \ / \ / \ / \ / \ / \ / \ H-----I-----J-----K-----L \ / \ / \ / \ / \ / \ / \ / \ / M-----N-----O-----P \ / \ / \ / \ / \ / \ / Q-----R-----S |
Take a set of 24 triominos (dominos with three sides, a number at each
corner), consisting of all possible configurations
of the values 0, 1, 2, and 3, and place them into a hexagon, two units
on a side, such that each adjacent side matches correctly. Or, show why
it can't be done.
(Note that 1-2-3 is a different triomino than 1-3-2, since neither can be rotated to create the other; while 1-1-2 would be the same as 1-2-1, since the latter can be rotated to obtain the former.) |
Here is a list of the 24 triminos, numbered clockwise:
a. 0-0-0 | b. 0-0-1 | c. 0-1-1 | d. 1-1-1 | e. 0-0-2 | f. 0-2-2 |
g. 2-2-2 | h. 0-0-3 | i. 0-3-3 | j. 3-3-3 | k. 1-1-2 | l. 1-2-2 |
m. 1-1-3 | n. 1-3-3 | o. 2-2-3 | p. 2-3-3 | q. 0-1-2 | r. 0-2-1 |
s. 0-1-3 | t. 0-3-1 | u. 0-2-3 | v. 0-3-2 | w. 1-2-3 | x. 1-3-2 |
Source: Original.
>From Philippe Fondanaiche Paris France Triominos I think that it is not possible to place the 24 triominos into the hexagon. 24 triominos with 3 digits for each of them represent a total number of 72 digits. We can easily check that each of the digit 0,1,2 and 3 appears 18 times. For all the letters A,B,C,D... as mentioned in the hexagon, we calculate the number of triominos starting from each of them. So we obtain the following chart: 2 3 2 3 6 6 3 2 6 6 6 2 3 6 6 3 2 3 2 We observe that the index 2 appears 6 times, the index 3 appears 6 times too and the index 6 appears 7 times and we verify that 2x6 + 3x6 + 6x7 = 72 which is the total number of digits above mentioned. For each digit i = 0,1,2,3 let consider the numbers: a(i) = how many times i is on a letter with the index 2 b(i) = how many times i is on a letter with the index 3 c(i) = how many times i is on a letter with the index 6 We have the following relations: 2xa(i) + 3xb(i) + 6xc(i) = 18 whatever i = 0,1,2,3 sum [a(i)] = 6 sum [b(i)] = 6 sum [c(i)] = 7 >From the first identity,we infer thatwhatever i a(i) is a multiple of 3 and b(i) is even. As it is impossible to have one digit with the same index 2, we infer from the second identity that there are necessarily 2 digits and 2 only with the index 2. For example i = 0 and i = 1. Therefore 3xb(0) + 6xc(0) = 12 or b(0) + 2xc(0) = 4. We can check that b(0) = 2 is the one solution with b(i) even.If not with b(0) = 0 or b(0) = 4 ,it is impossible to set a triomino with the 3 same digits.So c(0) = 1. Same solutions for i = 1. Concerning the digits 2 and 3, we have the following identities: b(2) + b(3) = 2 c(2) + c(3) = 5 3xb(2) + 6xc(2) = 18 or b(2) + 2xc(2) = 6 3xb(3) + 6xc(3) = 18 or b(3) + 2xc(3) = 6 There is one solution of this set of equations excluding the symmetry between 2 and 3: b(2) = 2 b(3) = 0 c(2) = 2 c(3) = 3 Hence we have the following sheet: digit I 0 1 2 3 ________________I__________________________________________ index 2 I 3 3 index 3 I 2 2 2 index 6 I 1 1 2 3 Starting from the digit 3 placed for example on E,F,J and from the digit 2 placed on N,O,R , we have to place the triomino 0,0,0 on K,L,P and the triomino 1,1,1 on D,H,I or H,I,M .Quickly we observe that it is impossible to place the remaining digits 0 and 1. Same observation if 2,2,2 is on K,L,P and so on ..... Best regards.
A-----B-----C / \ / \ / \ / \ / \ / \ D-----E-----F-----G / \ / \ / \ / \ / \ / \ / \ / \ H-----I-----J-----K-----L \ / \ / \ / \ / \ / \ / \ / \ / M-----N-----O-----P \ / \ / \ / \ / \ / \ / Q-----R-----S |
For each digit (0-3), there are 13 triominos that contain that digit. The controlling pieces to place are the trios (0-0-0, 1-1-1, 2-2-2, 3-3-3), and there are only three different places for each piece, after symmetry. If a trio (say 0-0-0) is placed in AEB, then 7 of the 13 triominos must be placed around those vertices, including two of the duos (001, 002, or 003). Another 6 triominos must be placed with that digit (0) elsewhere in the hexagon, with no 0 in adjacent locations. In other words, no 0 may be placed in C, F, J, I, D. Aside: Why Not Adjacent? Quickly inspect what would exist if any of the adjacent locations held the same digit (0) as the trio. Either a new trio would be formed (but only one can exist) or a diamond would be formed, with three identical digits on one long edge, all next to one remaining digit. This effectively makes the two triominos in that diamond identical. Therefore, the remaining digits must not be adjacent to the first ones. |
This exercise was simply to show that one trio MUST have a vertex on the center point, J. Any trio on the edge forces all identical digits away from that point, so at least one trio must occupy that point.
A-----1-----C / \ / \ / \ / \ / \ / \ D-----0-----0-----G \ / \ / \ / \ / \ / \ / 3-----0-----2 \ / \ / \ / \ / N-----O |
A trio in the center forces all 13 triominos with that digit to be
placed around the trio (12 triangles border on each center triangle.)
With 000 as the trio, the three duos (001, 002, 003)
must be placed adjacent to the trio.
Next consider the 011 triomino. If N=O=1, then 013 is placed, and C must be 2. To place the 033 triomino, D must be 3, but then there is no spot for 022. So 011 in not in JON. It must be next to the existing 1, in EAB or FBC. Similarly, the 022 and 033 must be next to their respctive existing digits. |
Placing 011 in the two possible locations leads to six possible
completions of the diagram. Through rotation and substitution,
these are equivalent to two unique solutions. Through mirror
symmetry, these are also identical. The result is at the right.
This leaves three final configurations to try for the hexagon, one each for the three long sides being completed sides of the final hexagon. These three options are below, followed by their analyses. |
1-----1-----3 / \ / \ / \ / \ / \ / \ 3-----0-----0-----2 \ / \ / \ / \ / \ / \ / 3-----0-----2 \ / \ / \ / \ / 2-----1 |
1-----1-----3 / \ / \ / \ / \ / \ / \ 3-----0-----0-----2 / \ / \ / \ / \ / \ / \ / \ / \ H-----3-----0-----2-----L \ / \ / \ / \ / \ / \ / \ / \ / M-----2-----1-----P \ / \ / \ / \ / \ / \ / Q-----R-----S | v |
A-----B-----C / \ / \ / \ / \ / \ / \ 1-----1-----3-----G / \ / \ / \ / \ / \ / \ / \ / \ 3-----0-----0-----2-----L \ / \ / \ / \ / \ / \ / \ / \ / 3-----0-----2-----P \ / \ / \ / \ / \ / \ / 2-----1-----S | v |
A-----B-----C / \ / \ / \ / \ / \ / \ D-----1-----1-----3 / \ / \ / \ / \ / \ / \ / \ / \ H-----3-----0-----0-----2 \ / \ / \ / \ / \ / \ / \ / \ / M-----3-----0-----2 \ / \ / \ / \ / \ / \ / Q-----2-----1 | v |
H must be 3. S must be 1. M is either 1 or 2. Another duo
(133 or 233) is still needed. Either (P=L=3 and R=1) or
(Q=R=3 and P=1). Only the latter allows 222 to be placed,
so Q=3, R=3, P=1, L=2.
Now, if M=1, 132 is duplicated and 223 is lost. If M=2, 223 is duplicated and 133 is lost. So this placement doesn't solve the problem, but gets as close as possible. |
A must be 1. C must be 3. P must be 2. B is either 2 or 3.
Another duo (112 or 113) is still needed, so S must be 1,
and B must be 3.
G can't be 3 (333 is already placed) or 1 (133 is already placed) or 2 (LKG and LPK would be identical). So this hexagon also does not solve the problem. | In this diagram, there is no place to put the 222 triomino. |