I know of solutions for at least two different pairs of duplicate dominoes.
1-1 3-3 2-3 2-1 3-1 2-2 2-3 1-2 | Lynne Onitsuka found this solution. The duplicate dominoes are 1-2 and 2-3. |
1-2 3-2 3-3 1-1 1-2 2-3 3-1 2-2 | A variation that I found with the same duplicates. |
Dear Ken, The right answer to this problem is here. If we add the 1-1 and 3-3 from the second grid, there's only one solution! This one is: +++++++++ +1 1+3 3+ +++++++++ +2+3+1 2+ + + +++++ +2+1+3 2+ +++++++++ +3 3+1 1+ +++++++++ Any other solution with a 1-1 and a 3-3 added, is a rotation or a mirroring from this solution. If we add the 1-3 and 2-2 from the second grid, no solutions are found by my program, so there aren't any solutions. With the 1-2 and the 2-3 added, my program found 30 different solutions.
this
+++++++++ +1 1+3 3+ +++++++++ +3 2+1+2+ +++++ + + +2 2+3+1+ +++++++++ +2 3+1 2+ +++++++++ |
and
+++++++++ +3 2+1 2+ +++++++++ +1 1+3 3+ +++++++++ +3+2 2+1+ + +++++ + +1+3 2+2+ +++++++++ |
and
+++++++++ +1 3+1+3+ +++++ + + +3 2+1+2+ +++++++++ +2+2+3+1+ + + + + + +2+1+3+2+ +++++++++ |
Ken, Here's output of a program that calculates all the eight magic squares: [I think these are the solutions for this particular configuration of dominos. Wilfred, above, shows other configurations, too. - KD] equi-sum=8 /* this is the one Lynne found */ 1 - 1 3 - 3 2 - 3 2 - 1 3 - 1 2 - 2 2 - 3 1 - 2 equi-sum=8 1 - 1 3 - 3 3 - 2 1 - 2 2 - 2 3 - 1 2 - 3 1 - 2 equi-sum=8 /* this is the variant you found */ 3 - 1 2 - 2 1 - 2 2 - 3 3 - 3 1 - 1 1 - 2 3 - 2 equi-sum=8 1 - 3 2 - 2 3 - 2 2 - 1 1 - 1 3 - 3 3 - 2 1 - 2 equi-sum=8 2 - 2 3 - 1 1 - 2 2 - 3 3 - 3 1 - 1 2 - 1 2 - 3 equi-sum=8 2 - 2 1 - 3 3 - 2 2 - 1 1 - 1 3 - 3 2 - 3 2 - 1 equi-sum=8 3 - 3 1 - 1 2 - 1 2 - 3 1 - 3 2 - 2 2 - 1 3 - 2 equi-sum=8 3 - 3 1 - 1 1 - 2 3 - 2 2 - 2 1 - 3 2 - 1 3 - 2
You wouldn't think that such a complicated problem has a somewhat elegant solution, but it does. First, the grid must have either four or six 2's, because there must be an even number of 2's among the two extra dominoes and because four or more 2's among the extras is impossible. Assume there are four 2's. For the moment, we'll ignore the diagonals and just try to configure the rows and columns so that they all have a total of 8. Also, we'll consider two grids to be equivalent if you can transform the first into the second through a series of row and/or column switches. Then, any grid that is a solution is equivalent to a grid in which all four 2's are in the upper left quadrant, because the only way for each row and each column to have an even number of 2's (a necessary condition) is for exactly two rows to have exactly two 2's and exactly two columns to have exactly two 2's. Up to equivalence, the upper right quadrant can be either 13 or 13 13 31. If the former, the bottom half, up to equivalence, is: 1331 3131, so that the overall grid looks like: 2213 2213 1331 Grid #1 3131. If the latter, the bottom half, up to equivalence, is: 1313 3131, so that the overall grid looks like: 2213 2231 Grid #2 1313 3131. Now we start to worry about the diagonals. We will adopt the following notation: "wxyz" is the foursome comprising the number in the first column and wth row (from the top), that in the second column and xth row, third column yth row, fourth column zth row. So, "1234" is the diagonal running from the upper left to the lower right. Consider Grid #1. We want to find all sets of four numbers from that Grid such that their sum is 8 and no two of the numbers are in the same row or column. For Grid #1, the following acceptable foursomes have a total of 8: 1234 1243 2134 2143. {{Parenthetical. Suppose you put the two main diagonals through a series of row and column operations. If the first diagonal now looks like wxyz, then the other diagonal has 3 possible configurations: xwzy, yzwx, or zyxw. This is left as an exercise to the reader.}} The following two pairs can be complementary diagonals: (1234, 2143) and (1243, 2134). Recognizing that we have to put the 2's together to create a 2-2 domino, we create from these pairs (I won't belabor the process), the following grid, which is unique up to flips and rotations: 3131 1223 1223 3311. However, this grid is not divisible into an permissible set of dominoes. For Grid #2, the following foursomes have a total of 8: 3421 4312. As complementary diagonals, they create 2 grids: 3221 3221 1133 and 3131 3311 1313 1223 1223. The first is not divisible into dominoes, but the second is: abbc addc efgh efgh. Through flips and rotations, 4 different grids are created. If the grid has six 2's, then exactly three rows have exactly two 2's and exactly three columns have exactly two 2's. Therefore, all solutions with six 2's are equivalent to: 2213 2321 1223 3131, where equivalents includes not only row and column switches, but also a 1 <-> 3 transformation where all 1's are changed to 3's and vice versa. The acceptable foursomes for diagonals are: 1234 1423 4132 2431. Only the pair (1423, 4132) can be complementary diagonals. We get two grids, which generate many more solutions: 2213 2132 1223 and 2231 3131 3212 2321 1313. The first grid can be divided into dominoes in 7 different ways, which because it can also be flipped, rotated, and 1 <-> 3'ed, generates 7x16 = 112 solutions. The second can be divided into dominoes in two ways, for 32 solutions. Altogether, unless I've made an error somewhere, there are 112 + 32 + 4 = 148 solutions. So much for elegance. -Craig Gentry