All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
A domino puzzle (Posted on 2008-08-15) Difficulty: 2 of 5
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.)
Solution 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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information