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 A domino puzzle (Posted on 2008-08-15)
Arrange the 28 pieces of a standard domino (0-0 to 6-6) in the board shown, so that equal spot numbers occur in groups of four, forming squares. Fourteen squares should result, each containing four equal spot numbers.
```            +---+---+       +---+---+
|   |   |       |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+
|   |   |       |   |   |
+---+---+       +---+---+```

 See The Solution Submitted by pcbouhid Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Pencil and paper solution. | Comment 1 of 3
`            +---+---+       +---+---+            | 0 | 0 |       | 2 | 2 |            +---+---+---+---+---+---+---+---+            | 0 | 0 | 4 | 4 | 2 | 2 | 6 | 6 |            +---+---+---+---+---+---+---+---+            | 5 | 5 | 4 | 4 | 1 | 1 | 6 | 6 |            +---+---+---+---+---+---+---+---+            | 5 | 5 | 3 | 3 | 1 | 1 | 5 | 5 |            +---+---+---+---+---+---+---+---+            | 2 | 2 | 3 | 3 | 6 | 6 | 5 | 5 |            +---+---+---+---+---+---+---+---+            | 2 | 2 | 0 | 0 | 6 | 6 | 4 | 4 |            +---+---+---+---+---+---+---+---+            | 1 | 1 | 0 | 0 | 3 | 3 | 4 | 4 |            +---+---+---+---+---+---+---+---+            | 1 | 1 |       | 3 | 3 |            +---+---+       +---+---+`

The most obvious was that the top row and bottom row had to be doubles. Then the way things worked out, the right column had to haave at least doubles at the top and bottom. I made the assumption that the middle of the right column also was a double.

The squares with the doubles were arbitrarily stated as 0, 1, ... , 6.

I assumed all the rest of the connections were horizontal, which worked out.

The next arbitrariness was in breaking the symmetry by putting the second 5 in the top half rather than the bottom half. Numbers were forced from there.

 Posted by Charlie on 2008-08-15 12:16:00

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