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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Jan 29, 2011 7:47 pm Post subject: Puzzle 11/01/29 |
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I hope You like Basics along with your advanced steps!
BTW: I (manually) found an unusual (for me) UR in place of my solver's last step.
Code: | +-----------------------+
| 3 . . | 4 . . | 9 7 6 |
| . . . | . . 9 | . 1 3 |
| . . 2 | . . . | . 8 . |
|-------+-------+-------|
| 4 . . | 2 . 6 | . 9 . |
| . . . | . 8 . | . . . |
| . 2 . | 9 . . | . . 1 |
|-------+-------+-------|
| 7 . . | . . . | 8 . . |
| 2 4 9 | 6 . . | . 5 7 |
| 8 3 . | . . 7 | . 4 . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sun Jan 30, 2011 12:10 am Post subject: |
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Quote: | Type 1 UR (29)
X-Wing (3)
Type 1 UR (45)
XY-Wing (671) + transport |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sun Jan 30, 2011 11:13 am Post subject: |
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After "fun" basics and steered by Danny's comments...
Code: | *---------------------------------------------------------------------*
| 3 8 15 | 4 125 125 | 9 7 6 |
| 56 67 4 | 8 567 9 | 2 1 3 |
| 19 679 2 | 137 1367 13 |*45 8 *45 |
|----------------------+----------------------+-----------------------|
| 4 (15) *37 | 2 357-1 6 |*37+5 9 8 |
| 19 69 *37 | 137 8 1345 |*3457+6 2 *45 |
| 56 2 8 | 9 3457 345 | 34567 36 1 |
|----------------------+----------------------+-----------------------|
| 7 15 156 | 135 249 24 | 8 36 29 |
| 2 4 9 | 6 (1)3 8 | (1)3 5 7 |
| 8 3 156 | 15 29 7 | (16) 4 29 |
*---------------------------------------------------------------------* | The two overlapping UR(37) and UR(45) form a MUG+2 which can be avoided by (5)r4c7, (6)r5c7 - they ko (1)r4c5 cracking the puzzle in one step!
Code: | (1=5)r4c2 - MUG(3457)[(5)r4c7=(6)r5c7] - (6=1)r9c7 - r8c7=r8c5 ; r4c5<>1 |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Jan 30, 2011 3:47 pm Post subject: |
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After basics: Code: | +-------------------+-------------------+-------------------+
| 3 8 15 | 4 125 125 | 9 7 6 |
| 56 67 4 | 8 567 9 | 2 1 3 |
| 19 679 2 | 137 1367 13 | 45 8 45 |
+-------------------+-------------------+-------------------+
| 4 15d 37 | 2 -1357 6 | 357 9 8 |
| 19 69 37 | 137 8 1345 | 34567 2 45 |
| 56c 2 8 | 9 3457 345 | 34567 36b 1 |
+-------------------+-------------------+-------------------+
| 7 15 156 | 135 249 24 | 8 36 29 |
| 2 4 9 | 6 13a 8 | 13 5 7 |
| 8 3 156 | 15 29 7 | 16 4 29 |
+-------------------+-------------------+-------------------+ | a and b are a pseudocell 16, from coloring on 3. cd complete the XY-wing, taking out 1 in R4C5.
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sun Jan 30, 2011 4:48 pm Post subject: |
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I don't understand the pseudo cell as part of an XY-Wing, but I can easily see an XY-Chain running between a and d. |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sun Jan 30, 2011 5:04 pm Post subject: |
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I support the use of "pseudocell" in a pretty "broad" interpretation
But here this is one strong link (1=6) created from either 3 SL (using the (3)SL in b9) or 4 SL (using the xy-chain fragment). Bit of a stretch |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Jan 30, 2011 5:52 pm Post subject: |
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I started with the pairs 13 in R8 and 36 in C8. (Actually, I was looking for kites / skyscrapers.)
a and b are pincers on 3. Because of the strong link 13 - 36 on 3 in B9, only one of ab is 3. Either a is 1 or b is 6.
Fits my definition of a pseudocell.
Keith |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sun Jan 30, 2011 7:36 pm Post subject: |
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Keith,
I must admit that I didn't pay much attention to "from coloring on 3" in your explanation. I now see that was a mistake! |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Jan 30, 2011 7:51 pm Post subject: |
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Quote: | "from coloring on 3" |
Well, I was being careful not to call it a kite or skyscraper.
If the pattern were a kite or skyscraper (no strong link on 3 in B9), then the pseudocell does not exist. Even though a and b are still pincers on 3. (There is a possibility that both a and b are 3.)
Keith |
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