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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Mar 13, 2009 10:32 pm Post subject: Free Press March 13, 2009 |
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Why so quiet?
This one is quite reasonable:
| Code: | Puzzle: FP031309
+-------+-------+-------+
| . 1 3 | 2 . . | . . 9 |
| 8 . . | 7 . . | . . 2 |
| . . . | . . . | 5 6 . |
+-------+-------+-------+
| . . . | . . . | . . 3 |
| . . 9 | 4 . 3 | 2 . . |
| 2 . . | . . . | . . . |
+-------+-------+-------+
| . 9 . | . . . | . . . |
| 1 . . | . . 7 | 9 . 6 |
| 6 . . | . . 5 | 7 4 1 |
+-------+-------+-------+ |
Keith |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Sat Mar 14, 2009 4:00 am Post subject: free |
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I needed an xy-chain to eliminate some 9's from box 5.
Earl
of the chain gang |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 14, 2009 6:39 am Post subject: |
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After basics:
| Code: | +----------------+----------------+----------------+
| 57 1 3 | 2 568 68 | 4 78 9 |
| 8 45 6 | 7 459 49 | 3 1 2 |
| 9 27 24 | 13 348 148 | 5 6 78 |
+----------------+----------------+----------------+
| 4 78 158 | 569 6789 2 | 16 579 3 |
| 57 6 9 | 4 1 3 | 2 578 78 |
| 2 3 158 | 569 6789 689 | 16 579 4 |
+----------------+----------------+----------------+
| 3 9 7 | 16 46 146 | 8 2 5 |
| 1 45 45 | 8 2 7 | 9 3 6 |
| 6 28 28 | 39 39 5 | 7 4 1 |
+----------------+----------------+----------------+ |
My two-steppper does not seem to be the same solution as Earl's.
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Mar 14, 2009 3:08 pm Post subject: |
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| keith wrote: | My two-steppper does not seem to be the same solution as Earl's.
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Keith, I suspect that my two-stepper is the same as yours. | Quote: | | coloring on <7> provides deletions in four cells and, then a Type 2 UR on <59> deletes <6> from r7c4 to complete the puzzle. |
A nice, snappy puzzle this week.
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 14, 2009 3:47 pm Post subject: |
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After simple coloring on 7:
| Code: |
+--------+------------+---------+
| 7 1 3 | 2 5 6 | 4 8 9 |
| 8 5 6 | 7 49 49 | 3 1 2 |
| 9 2 4 | 13 38 18 | 5 6 7 |
+--------+------------+---------+
| 4 7 18 | 569 689 2 | 16 59 3 |
| 5 6 9 | 4 1 3 | 2 7 8 |
| 2 3 18 | 569 7 89 | 16 59 4 |
+--------+------------+---------+
| 3 9 7 | 16 46 14 | 8 2 5 |
| 1 4 5 | 8 2 7 | 9 3 6 |
| 6 8 2 | 39 39 5 | 7 4 1 |
+--------+------------+---------+
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Play this puzzle online at the Daily Sudoku site
Just trying to learn new stuff. Do we have one of Norm's "Mixed Wings"?
Note the M-Wing on 18 in r6c3 and r3c6, connected by 8s. Extending the 1 from r3c6 to r3c4 we have pincers of 1 in r3c4 and r6c3. If the former is 1, then r7c4 = 6. If If r6c3 = 1, then r6c7 = 6. These pincers of 6 take out the 6 from r6c4, solving the puzzle. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sat Mar 14, 2009 4:42 pm Post subject: |
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| Quote: | | After simple coloring on 7: |
You say colouring on <7>
I say an ER on <7> - easier to deduct - easier to spell. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 14, 2009 6:04 pm Post subject: |
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In my grid:
The first step on <7> is also a skyscraper in C19.
In Marty's grid:
Ted: Yes. However, eschewing the uniqueness argument, there is a 4-cell chain 89 18 13 39 that takes out <9> in two cells in C4B5. This may be similar to what Earl mentioned.
Best wishes,
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 14, 2009 7:40 pm Post subject: |
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| Quote: | | However, eschewing the uniqueness argument, there is a 4-cell chain 89 18 13 39 that takes out <9> in two cells in C4B5. |
Starting with the same cell, 89-49-49-39-38 takes out 8s. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 14, 2009 8:08 pm Post subject: |
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| Marty R. wrote: | | Quote: | | However, eschewing the uniqueness argument, there is a 4-cell chain 89 18 13 39 that takes out <9> in two cells in C4B5. |
Starting with the same cell, 89-49-49-39-38 takes out 8s. |
Marty,
Welcome (back) to the chain gang! (I am a relatively new member myself.) With the last two cells of your chain making a pseudocell, this is 89 - 49 - 49 - (3)89, a W-wing.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Mar 14, 2009 9:20 pm Post subject: |
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| Quote: | | With the last two cells of your chain making a pseudocell, this is 89 - 49 - 49 - (3)89, a W-wing. |
I'm disappointed to have to report that I've been having difficulty grasping the pseudo cell concept. I understand some, but generally recognize none.  |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sat Mar 14, 2009 9:36 pm Post subject: |
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| Quote: | | Note the M-Wing on 18 in r6c3 and r3c6, connected by 8s. Extending the 1 from r3c6 to r3c4 we have pincers of 1 in r3c4 and r6c3. If the former is 1, then r7c4 = 6. If If r6c3 = 1, then r6c7 = 6. These pincers of 6 take out the 6 from r6c4, solving the puzzle. |
first of all, I believe Asellus has mentioned this in previous threads before also. and he used the term mixed w-wing when describing it.
-----
now back to your example.
not quite,
your w-wing example is just an extension of a normal w-wing.
if this is a normal W-wing
(a=b) - (b) = (b) - (b=a) then the a's are pincers
a mixed ending w-wing would be in the form
(a=b) - (b) = (b) - (b=c)
then by extension either C or A would provide a pincer like what you did in your example.
obviously a "mixed-ending" w-wing always needs that extra extension to "find" the pincer needed to make eliminations with either A or C.
the extension can be from either A or C.
remember how the m-wing was generalized? I believe nataraj made a point about that.
this just happens to generalize the W-wing just a bit so that it isn't constrained to the normal 4 cells.
keeping your eyes open to these types of patterns can provide for some neat pincer eliminations.
I have seen these mixed W-wings on numerous occasions including a recent puzzle in Danny's forum.
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=3338&sid=6efc82dabbc0d3a18db5895fd4f82a09 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Mar 15, 2009 12:15 am Post subject: |
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| Quote: | | Note the M-Wing on 18 in r6c3 and r3c6, connected by 8s. Extending the 1 from r3c6 to r3c4 we have pincers of 1 in r3c4 and r6c3. If the former is 1, then r7c4 = 6. If If r6c3 = 1, then r6c7 = 6. These pincers of 6 take out the 6 from r6c4, solving the puzzle. |
Norm,
I'm not quite sure what you're saying. Is my logic (above) wrong? The terminology? Both? Neither? |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Mar 15, 2009 5:12 am Post subject: |
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sorry Marty,
I was stating that your example wasn't a "mixed w-wing" |
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