stuck for a few days

Advanced methods and approaches for solving Sudoku puzzles

stuck for a few days

Postby greggp » Sun Nov 13, 2005 8:10 pm

I'm sort of new to this and have finally come across a puzzle that I can't seem to solve. Probably an easy one for those of you with some experience...

Not sure of the best way to diagram these, but if you see more than one number in my paranthesis, that means they are the remaining best possibilities.

4{56}7-123-{568}{58}9
{23}{238-596-741
{15}9{16}-748-{23}{23}{56}

9{1234}9-{234}6{15}-{235}{2357}{57}
{237}{234}5-{2349}{37}{79}-168
{1237}-{1236}{1267}-{23}8{15}-{235}94

{125}8{124}-{39}{137}{79}-{456}{57}{567}
9(15}{14}-6{17}2-{458}{578}3
673-854-912

Any hints would be greatly appreciated!!
greggp
 
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Postby simes » Sun Nov 13, 2005 8:22 pm

you have two 9s in row 4 and column 1, suggesting r4c1 should be something else?
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Location: UK

Postby Ruud » Sun Nov 13, 2005 9:24 pm

I tend to agree with Simon, R4C1 can only be an 8, so that must be a typo. (there are a few more typos...)

Based on the fixed digits, this is the candidate grid:
Code: Select all
.---------------.---------------.---------------.
| 4    56   7   | 1    2    3   | 568  58   9   |
| 23   23   8   | 5    9    6   | 7    4    1   |
| 15   9    16  | 7    4    8   | 23   23   56  |
:---------------+---------------+---------------:
| 8    1234 9   | 234  6    15  | 235  2357 57  |
| 237  234  5   | 2349 37   79  | 1    6    8   |
| 1237 1236 126 | 23   8    15  | 235  9    4   |
:---------------+---------------+---------------:
| 125  8    124 | 39   137  79  | 456  57   567 |
| 9    15   14  | 6    17   2   | 458  578  3   |
| 6    7    3   | 8    5    4   | 9    1    2   |
'---------------'---------------'---------------'

Hint: How many cells in row 6 would allow a 7?

Ruud.
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Postby greggp » Mon Nov 14, 2005 3:26 am

simes wrote:you have two 9s in row 4 and column 1, suggesting r4c1 should be something else?


Good catch, I mistakenly printed 9, when it should be an 8.
greggp
 
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Postby greggp » Mon Nov 14, 2005 3:34 am

Ruud wrote:I tend to agree with Simon, R4C1 can only be an 8, so that must be a typo. (there are a few more typos...)

Based on the fixed digits, this is the candidate grid:
Code: Select all
.---------------.---------------.---------------.
| 4    56   7   | 1    2    3   | 568  58   9   |
| 23   23   8   | 5    9    6   | 7    4    1   |
| 15   9    16  | 7    4    8   | 23   23   56  |
:---------------+---------------+---------------:
| 8    1234 9   | 234  6    15  | 235  2357 57  |
| 237  234  5   | 2349 37   79  | 1    6    8   |
| 1237 1236 126 | 23   8    15  | 235  9    4   |
:---------------+---------------+---------------:
| 125  8    124 | 39   137  79  | 456  57   567 |
| 9    15   14  | 6    17   2   | 458  578  3   |
| 6    7    3   | 8    5    4   | 9    1    2   |
'---------------'---------------'---------------'

Hint: How many cells in row 6 would allow a 7?

Ruud.


Also, good catch with the correct number 8. Column 6 used to be:

3
6
8
157
79
15
179
2
4

Then I realized that the only place for a 7 in the 6th box was r4c8 and r4c9, which eliminated the 7 from r4c6. It's been awhile, so I can't remember why I eliminated the 7 from r6c6 and r7c6, but it probably made sense at the time:) . Thanks for catching the 7 in r6c1, I can't believe I didn't see that.
greggp
 
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Joined: 13 November 2005


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