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Bern.B
Joined: 14 Dec 2012
Posts: 9
Location: Edmonton,Canada
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 Posted: Sat Jan 12, 2013 10:05 pm Post subject: More help please |
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| Code: |
+-----------------+------------------+------------+
| 2347 234 27 | 12369 2679 12679 | 8 19 5 |
| 8 1 6 | 29 5 4 | 7 3 29 |
| 2357 235 9 | 123 27 8 | 4 6 12 |
+-----------------+------------------+------------+
| 126 28 128 | 2689 4 3 | 5 7 169 |
| 14567 458 3 | 689 6789 679 | 2 19 1469 |
| 2467 9 27 | 5 1 267 | 36 8 346 |
+-----------------+------------------+------------+
| 35 7 58 | 4 689 569 | 1 2 36 |
| 12 6 4 | 7 3 12 | 9 5 8 |
| 9 2358 1258 | 1268 268 1256 | 36 4 7 |
+-----------------+------------------+------------+
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Play this puzzle online at the Daily Sudoku site
Hello,
Doing some puzzle from last year.
By using HINT r7,c1 is a 3.
Could someone tell me how this is arrived at?
Thanks
Bernie |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Jan 12, 2013 10:31 pm Post subject: |
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| Take a very close look at c36. |
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Bern.B
Joined: 14 Dec 2012
Posts: 9
Location: Edmonton,Canada
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| Posted: Sat Jan 12, 2013 10:49 pm Post subject: |
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Sorry Marty.
What is c36? |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Jan 12, 2013 11:53 pm Post subject: |
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| Bern.B wrote: | Sorry Marty.
What is c36? |
Columns 3 and 6. C, r and b are commonly used as abbreviations for columns, rows and boxes and the ampersand or word "and" are not used when referring to multiples.
For example, r47c15 would refer to the four corners of a rectangle in rows 4 and 7 and columns 1 and 5. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Jan 13, 2013 9:03 am Post subject: |
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| Where does the 5 have to go in column 3? |
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Bern.B
Joined: 14 Dec 2012
Posts: 9
Location: Edmonton,Canada
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| Posted: Sun Jan 13, 2013 4:32 pm Post subject: |
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Yes, I see it. Thanks
The 5 would have to go in either r7 or r9 since there is no 5 anywhere else in c3. Is this correct reasoning?
Is this called "a hidden pair"? |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Jan 13, 2013 4:43 pm Post subject: |
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No. a pair involves two digits
This is called a "Locked Candidate" LC
a 5 is locked to column 3 so it cannot occur in r7c1 |
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hughwill
Joined: 05 Apr 2010
Posts: 424
Location: Birmingham UK
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| Posted: Sun Jan 13, 2013 4:55 pm Post subject: |
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Bern.B
There is a naked pair in column 3 (c3) - 27s which mean that all the other 2s
in c3 can be removed. But this doesn't answer your question
Row 7column 1 (r7c1)must be 3 because it cannot be 5 - there are two ways of proving it.
Firstly because there are no other 5s in c3 apart from those in box7 (b7), the
other 5s in b7 can be deleted. Since the only possibles in r7c1 are 35 this
just leaves the 3.
Or you can get to the same point with an X wing on 5s- the only 5s in c3 are
in rows 7 and 9 (r79) and the only 5s in c6 are also in r79. Therefore
the 5s for those rows must come from c3 and c6 and all other 5s on the rows
can be deleted.
Hugh |
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Bern.B
Joined: 14 Dec 2012
Posts: 9
Location: Edmonton,Canada
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| Posted: Wed Jan 16, 2013 11:32 pm Post subject: |
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Thanks for all the help.
Bernie |
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