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Captain Pete · Joined: 09 Jun 2007 Posts: 55 Location: Oley, PA

This is my first sweep of the 4/27/2007 VH. Please take a look at C3, rows 6 and 8, and C8, same rows. C3R6, C8R6 and C6R8 each contain 69. C3R8, which seems to complete the rectangle, contains 469. I assumed that C3R8 therefore had to be the 4, but after redoing this puzzle several times, and finally looking at the solution, I discovered I was wrong. But, I don't know why. Can you tell me why this isn't a rectangle?

[code]
+--------------+------------+------------+
| 6 1 2 | 3 8 7 | 9 4 5 |
| 3 4 5 | 2 6 9 | 7 8 1 |
| 8 9 7 | 1 45 45 | 2 3 6 |
+--------------+------------+------------+
| 1259 26 8 | 4 159 3 | 56 7 29 |
| 2459 3 469 | 69 7 56 | 568 1 289 |
| 159 7 69 | 8 159 2 | 4 69 3 |
+--------------+------------+------------+
| 7 5 3 | 69 49 1 | 68 2 489 |
| 249 268 469 | 5 3 468 | 1 69 7 |
| 49 68 1 | 7 2 468 | 3 5 489 |
+--------------+------------+------------+

keith · Posted: Mon Aug 06, 2007 10:53 pm Post subject:

Cap'n:

A UR must be in only two boxes. Your rectangle is in four boxes, and so is not a valid UR.

You can easily prove this for yourself: The point of a UR is that a solution like

Captain Pete · Joined: 09 Jun 2007 Posts: 55 Location: Oley, PA

Keith,
I've done a lot of puzzles, and thought I had a handle on UR's, but I don't think I ever realized that. Thanks.
Pete