Stuck on moderately hard puzzle

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Stuck on moderately hard puzzle

Postby 5krunner » Fri May 26, 2006 6:14 pm

Can someone help me with the next step on this puzzle ... I'm stuck. Thx!!!

Code: Select all
5 . 8 | . 2 . | . 1 .
3 1 2 | 5 4 6 | 8 7 9
. . 9 | 1 . . | . 5 2
------+-------+------
. . 3 | 4 . 1 | 7 . 5
. 5 4 | . 7 2 | 1 6 .
7 . 1 | 6 . 5 | . . .
------+-------+------
1 8 5 | 2 6 3 | 9 4 7
. 3 6 | 7 5 . | 2 8 1
. . 7 | . 1 . | 5 3 6

Last edited by Alta Vista [Bot] on Fri May 26, 2006 2:53 pm, edited 1 time in total.
5krunner
 
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Joined: 19 December 2005

Too late!

Postby QBasicMac » Fri May 26, 2006 6:31 pm

5krunner wrote:Can someone help me with the next step on this puzzle ... I'm stuck. Thx!!!


You've messed up somewhere and we can't see because you didn't post the original puzzle.

You cannot proceed from what you posted and reach a solution.

Mac

Proof:

Cell r1c6 must be a 9, no other candidate fits there. That leads to box 2
329
546
187

Now r4c5 must be a 9. That leads to box 5
491
872
635

Now r5c1 must be a 9. That leads to box 4
863
954
721

All forced placements, right?

Well then we are here:
Code: Select all
578  329  -1-
312  546  879
--9  187  -52

863  491  7-5
954  872  16-
721  635  ---

185  263  947
-36  75-  281
--7  -1-  536


Note there is no way we can place a 9 in row 8.

Solution: Start over.
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Postby 5krunner » Fri May 26, 2006 6:54 pm

My sincere apologies!!! I did mess up. The 7 in the first box, first row, second column is unknown. I have made the correction and I think it is correct now. Sorry about that ... Jay
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Postby ravel » Fri May 26, 2006 7:25 pm

There are at least 2 ways to proceed:

There is a naked pair in row 6, which gives you a single in column 5 and
a unique rectangle in r46c28, which gives you a number in r4c2.
ravel
 
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Postby QBasicMac » Fri May 26, 2006 8:44 pm

5krunner wrote:The 7 in the first box, first row, second column is unknown.


Don't follow you, Jay.

Here is the first box exactly as you posted it:
Code: Select all
5 . 8 |
3 1 2 |
. . 9 |


I note that r1c2 (first row second column) has a dot, meaning unknown.

So you posted that as you intended???

Please post the original puzzle with no entries by you if you still have problems.

Thanks,

Mac
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Posts: 441
Joined: 13 July 2005

Postby QBasicMac » Fri May 26, 2006 8:54 pm

ravel wrote:There is a naked pair in row 6, which gives you a single in column 5


One of us had too much wine. I see the OP as

Code: Select all
+--------------+---------------+--------------+
| 5     7    8 | 39   2    {9} | 346  1   34  |
| 3     1    2 | 5    4    6   | 8    7   9   |
| 46    46   9 | 1    38   78  | 346  5   2   |
+--------------+---------------+--------------+
| 2689  269  3 | 4    89   1   | 7    29  5   |
| 89    5    4 | 389  7    2   | 1    6   38  |
| 7     29   1 | 6    389  5   | 34   29  348 |
+--------------+---------------+--------------+
| 1     8    5 | 2    6    3   | 9    4   7   |
| 49    3    6 | 7    5    49  | 2    8   1   |
| 249   249  7 | 89   1    489 | 5    3   6   |
+--------------+---------------+--------------+


The only naked pair I see is 29 which leads to no single in col 5.

(Not that it would help anyway as the puzzle is corrupt)

Mac
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Posts: 441
Joined: 13 July 2005

Postby 5krunner » Fri May 26, 2006 9:43 pm

First, I don't see the naked pair in column six. But here's the original puzzle ... I also think at least that my entries are correct and that the puzzle isn't corrupt (but who knows, I could be wrong) ... Jay

Code: Select all
5 . . | . 2 . | . . .
3 . 2 | . 4 6 | . 7 .
. . 9 | 1 . . | . 5 .
------+-------+------
. . . | . . . | . . 5
. 5 4 | . . . | 1 6 .
7 . . | . . . | . . .
------+-------+------
. 8 . | . . 3 | 9 . .
. 3 . | 7 5 . | 2 . 1
. . . | . 1 . | . . .
 
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Postby QBasicMac » Sat May 27, 2006 12:47 am

Oh, I see!! You once had a 7 at r1c2 and then edited your post and removed it. I didn't notice that and have still been working with your old post.

Sorry. Yes, you have a solution now. Starting with your current first post:

Looking at row 6, r6c5 can only be 3, 8 or 9.
but r6c2 & r6c7 can only be 2 or 9 so one of these must be 9
therefore r6c5 can only be 3 or 8.

Looking at column 5, r6c5 & r3c5 can only be 3 or 8. One must be 8.
But r4c5 can only be 8 or 9. So it must be 9. Place 9 at r4c5.

That solves the puzzle.

Mac
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Joined: 13 July 2005

Postby 5krunner » Sat May 27, 2006 1:28 pm

Mac, Thank u! That did the trick. It also helped me find a mistake I made. After your suggestion, everything fell into place. Jay.
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