## unique rectangle question

Post the puzzle or solving technique that's causing you trouble and someone will help

### unique rectangle question

Code: Select all
`+-----------+-------------+-----------+| 2   6  9  | 4    5    3 | 8   7   1 | | 5   4   1 | 7    8    6 | 3   9   2 | | 8   3   7 | 19  19    2 | 4   6   5 | +-----------+-------------+-----------+| 13  7   8 | 6   1239  4 | 129  5 39 | | 13  5   2 | 139   7  19 | 6   8   4 | | 6   9   4 | 5   123   8 | 12  23  7 | +-----------+-------------+-----------+| 4   2   6 | 8   39    7 | 5   1  39 | | 9   1   3 | 2    6    5 | 7   4   8 | | 7   8   5 | 139  4   19 | 29  23  6 | +-----------+-------------+-----------+`

I have a question:

I thought that the following squares build a unique rectangle:

R4C5 - R4C9 - R7C5 - R7C9

with the numbers '39' so that I can reduce R4C5 from '1239' to '39'. After this reduction the puzzle became invalid. Where is my thinking error?

Thx for the help.
anathema

Posts: 5
Joined: 22 December 2005

### Re: unique rectangle question

anathema wrote:I thought that the following squares build a unique rectangle:
R4C5 - R4C9 - R7C5 - R7C9

A unique rectangle must have each two cells in one box. In this case there still can be a unique solution, when the 39 candidates remain.
ravel

Posts: 998
Joined: 21 February 2006

### Re: unique rectangle question

That's not a unique rectangle, I'm afraid. For it to be one, the corners must be confined to two 3x3-boxes. In your example, each corner is in a box of its own (i.e. it spans 4 boxes in total).

Also, the rule you're trying to apply is reverse. If there were a unique rectangle there, and one corner had some extra candidates, one of the extras would have to be true, not the corner candidates. So you would eliminate 39 from the corner, not 12.
vidarino

Posts: 295
Joined: 02 January 2006

### Re: unique rectangle question

vidarino wrote:That's not a unique rectangle, I'm afraid. For it to be one, the corners must be confined to two 3x3-boxes. In your example, each corner is in a box of its own (i.e. it spans 4 boxes in total).

Also, the rule you're trying to apply is reverse. If there were a unique rectangle there, and one corner had some extra candidates, one of the extras would have to be true, not the corner candidates. So you would eliminate 39 from the corner, not 12.

ok thx. I didn't check the condition of the two 3x3 boxes!

The reduction is of course '39' - I just made a writing mistake - thats clear.

So I learned a new thing today - so its a good day!
anathema

Posts: 5
Joined: 22 December 2005