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shoeless
Joined: 03 Aug 2007 Posts: 13
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Posted: Fri Nov 08, 2019 6:35 pm Post subject: Could use some help |
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Anybody see a good next step in this one?
Code: |
+---------------+----------------+-----------------+
| 29 5 17 | 8 4 279 | 3 6 127 |
| 234 8 347 | 1 6 237 | 47 9 5 |
| 2349 1379 6 | 237 79 5 | 147 8 1247 |
+---------------+----------------+-----------------+
| 8 6 2 | 5 3 147 | 147 147 9 |
| 49 79 5 | 6 179 8 | 2 1347 1347 |
| 1 379 347 | 47 2 479 | 8 5 6 |
+---------------+----------------+-----------------+
| 5 2 13 | 9 178 1347 | 6 1347 13478 |
| 7 4 9 | 23 18 6 | 5 123 138 |
| 6 13 8 | 2347 5 12347 | 9 12347 1347 |
+---------------+----------------+-----------------+
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ZeroAssoluto
Joined: 05 Feb 2017 Posts: 868 Location: Rimini, Italy
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Posted: Fri Nov 08, 2019 7:52 pm Post subject: |
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Hi shoeless,
L.C. with number 3 in r23c1 and -3 in r2c3,r3c2
Naked Pairs 4,7 in r2c37
then
A.I.C. in r5c1,
if r5c1=9 then r1c6=9 and no number 9 in r6, so r5c1=4.
Ciao Gianni |
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shoeless
Joined: 03 Aug 2007 Posts: 13
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Posted: Tue Nov 12, 2019 9:04 pm Post subject: |
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Thanks, that helped me get a bit further to here. I'm thinking I should just move on to another puzzle.
Code: |
+------------+----------------+----------------+
| 29 5 17 | 8 4 279 | 3 6 12 |
| 23 8 4 | 1 6 23 | 7 9 5 |
| 239 17 6 | 237 79 5 | 14 8 124 |
+------------+----------------+----------------+
| 8 6 2 | 5 3 17 | 14 147 9 |
| 4 79 5 | 6 179 8 | 2 137 137 |
| 1 379 37 | 47 2 479 | 8 5 6 |
+------------+----------------+----------------+
| 5 2 13 | 9 178 1347 | 6 1347 13478 |
| 7 4 9 | 23 18 6 | 5 123 138 |
| 6 13 8 | 2347 5 12347 | 9 12347 1347 |
+------------+----------------+----------------+
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ZeroAssoluto
Joined: 05 Feb 2017 Posts: 868 Location: Rimini, Italy
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Posted: Tue Nov 12, 2019 9:48 pm Post subject: |
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Hi shoeless,
A.I.C. in r3c5,
if r3c5=7 then r5c5=9 r4c6=1 r4c7=4 r3c7=1 r3c2=7 and too many 7 in row 3
so r3c5=9
Ciao Gianni |
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shoeless
Joined: 03 Aug 2007 Posts: 13
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Posted: Wed Nov 13, 2019 5:42 pm Post subject: |
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ZeroAssoluto wrote: | Hi shoeless,
A.I.C. in r3c5,
if r3c5=7 then r5c5=9 r4c6=1 r4c7=4 r3c7=1 r3c2=7 and too many 7 in row 3
so r3c5=9
Ciao Gianni |
Thanks, that broke it free alright. Out of curiosity, how did you see that? Is there some sort of dependency pattern that you recognized?
Thanks again. |
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ZeroAssoluto
Joined: 05 Feb 2017 Posts: 868 Location: Rimini, Italy
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Posted: Wed Nov 13, 2019 7:55 pm Post subject: |
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No, no one pattern.
It's routine or experience or ...
The number 9 is present only twice in the sector, row and column and I look at it for a color or other possibilities.
Sometimes it helps other times not. |
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WorldO
Joined: 04 May 2016 Posts: 5
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Posted: Mon Jun 22, 2020 9:11 am Post subject: |
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I see now why I missed the Sue de Coq. It was no longer available due to the Naked Triple , which essentially made the same eliminations. Still a nice technique. I needed something similar to Clement's last two steps to solve as well.
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