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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sat Nov 17, 2007 3:20 pm Post subject: Nov 17 DB |
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This Nov 17 DB requires only a w-wing.
Earl
| Code: |
+-------+-------+-------+
| 7 4 8 | 6 3 . | 1 . . |
| . 3 . | . . 5 | . . . |
| . . . | . 8 . | . 9 . |
+-------+-------+-------+
| 8 7 4 | . . . | . . 9 |
| . . . | . . . | . . . |
| 1 . . | . . . | 6 2 7 |
+-------+-------+-------+
| . 2 . | . 9 . | . . . |
| . . . | 5 . . | . 7 . |
| . . 6 | . 7 8 | 2 3 1 |
+-------+-------+-------+
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[url=http://www.dailysudoku |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Nov 17, 2007 8:06 pm Post subject: |
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Since you mentioned the W-Wing, I thought I'd try to avoid it, and did so, albeit with some difficulty and with a fair amount of time consumed. I broke up a deadly pattern, played an XY-Wing, broke up two more deadly patterns and finally finished it off with an XY-Chain.
Thanks again for publishing these. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sat Nov 17, 2007 11:26 pm Post subject: |
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I missed the W-wing, but found a 7-cell xy- chain, eliminating <4> in
R3C6 and R6C5, completing the puzzle in one step.
There is also a Type 6? (i'm not sure about the type number), that pins
<5> in R7C9 (potential <68> *DP in UR R27C89), though it didn't solve
the puzzle in one step.
| Code: |
+------------+---------+--------------+
| 7 4 8 | 6 3 9 | 1 5 2 |
| 269 3 29 | 1 A24 5 | 7 *468 *68 |
| 256 1 25 | 7 8 B24|C34 9 36 |
+------------+---------+--------------+
| 8 7 4 | 2 E56F36|D35 1 9 |
| 25 6 235 | 9 1 7 | 3458 48 358 |
| 1 59 359 | 8 45 G34| 6 2 7 |
+------------+---------+--------------+
| 4 2 7 | 3 9 1 | 58 *68 *568|
| 3 8 1 | 5 26 26 | 9 7 4 |
| 59 59 6 | 4 7 8 | 2 3 1 |
+------------+---------+--------------+
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 18, 2007 12:10 am Post subject: |
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Johan,
That UR is not Type 6 since that requires an overlapping X-Wing, which we don't have here. I'm not sure it this is one of the general Types. Maybe I'm missing something that someone else will clarify, but I see that <5> placement occurring indirectly:
R2C8 must be <4> and/or R7C9 must be <5>. These imply that R5C8 and/or R7C7 must be <8>, eliminating the <8>s at R5C7 and R7C8. After this, R7C9 is <5>. Is that how you did it?
By the way, after this UR, there is another similar one that results:
| Code: | +---------------+---------------+---------------+
| 7 4 8 | 6 3 9 | 1 5 2 |
| 269 3 29 | 1 p24 5 | 7 -48 68 |
|*256 1 *25 | 7 8 2-4 |p34 9 36 |
+---------------+---------------+---------------+
| 8 7 4 | 2 56 36 | 35 1 9 |
|*25 6 *235 | 9 1 7 | 345 48 38 |
| 1 59 359 | 8 45 34 | 6 2 7 |
+---------------+---------------+---------------+
| 4 2 7 | 3 9 1 | 8 6 5 |
| 3 8 1 | 5 26 26 | 9 7 4 |
| 59 59 6 | 4 7 8 | 2 3 1 |
+---------------+---------------+---------------+ |
To avoid the {25} UR, R3C1 is <6> and/or R5C3 is <3>. If R3C1 is <6>, the {29} Locked Pair in R2 makes R2C5 <4>. If R5C3 is <3>, the ER in Box 6 makes R3C7 <4>. So, R2C5 and/or R3C7 is <4>, eliminating <4>s as shown. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sun Nov 18, 2007 1:10 am Post subject: |
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| Quote: | R2C8 must be <4> and/or R7C9 must be <5>. These imply that R5C8 and/or R7C7 must be <8>, eliminating the <8>s at R5C7 and R7C8. After this, R7C9 is <5>. Is that how you did it?
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Asellus,
The result is the same(R7C9=5), but he path i used was diffirent, suppose
R2C8=4 (one of the three possible solutions to avoid the <68> DP ) =>
R3C7=3 => R4C7=5 => R7C7=8 => R7C8=6 => R7C9=5, which means
that R7C9 must be <5>. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 18, 2007 1:25 am Post subject: |
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| Code: | +------------+---------+--------------+
| 7 4 8 | 6 3 9 | 1 5 2 |
| 269 3 29 | 1 A24 5 | 7 *468 *68 |
| 256 1 25 | 7 8 B24|C34 9 36 |
+------------+---------+--------------+
| 8 7 4 | 2 E56F36|D35 1 9 |
| 25 6 235 | 9 1 7 | 3458 48 358 |
| 1 59 359 | 8 45 G34| 6 2 7 |
+------------+---------+--------------+
| 4 2 7 | 3 9 1 | 58 *68 *568|
| 3 8 1 | 5 26 26 | 9 7 4 |
| 59 59 6 | 4 7 8 | 2 3 1 |
+------------+---------+--------------+ |
The way I learned it was that because of the strong link on 8 in row 2, r7c9 can't be = 8. And the strong link on 6 in row 7 means that r2c8 can't be = 6. And, of course, that leads to the 5 in r7c9. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sun Nov 18, 2007 1:47 am Post subject: |
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Asellus,
When R5C3=3 => R5C9=8 => R2C9=6 => [29] pair in R2 Box 1 =>
R3C3=5 => R3C1=6, so R3C1=6, probably a diffirent approach, but with
the same result. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sun Nov 18, 2007 11:29 am Post subject: |
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Nice deductions, Asellus and Johan, but much simpler is:
| Code: | +---------------+---------------+---------------+
| 7 4 8 | 6 3 9 | 1 5 2 |
| 269 3 29 | 1 24 5 | 7 48 68 |
|*256 1 *25 | 7 8 24 | 34 9 36 |
+---------------+---------------+---------------+
| 8 7 4 | 2 56 36 | 35 1 9 |
|*25 6 *235 | 9 1 7 | 345 48 38 |
| 1 59 359 | 8 45 34 | 6 2 7 |
+---------------+---------------+---------------+
| 4 2 7 | 3 9 1 | 8 6 5 |
| 3 8 1 | 5 26 26 | 9 7 4 |
| 59 59 6 | 4 7 8 | 2 3 1 |
+---------------+---------------+---------------+
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r3c1=6 => r3c9=3, so one of r3c9 and r5c3 has to be 3.
This elimimates 3 from r5c9, solving the puzzle.
| Marty R. wrote: | | The way I learned it ... |
So did i 
There is also another way to look at potential deadly patterns, which especially sometimes is practicable, when no pencilmarks are used. To avoid a deadly pattern for 2 digits, one of the digits has to be outside the 2 boxes, the 2 rows and the 2 columns.
| Code: | *-----------------------------------------------------------*
| 7 4 8 | 6 3 9 | 1 5 2 |
| 269 3 29 | 1 24 5 | 7 #468 #68 |
| 256 1 25 | 7 8 24 | 34 9 36 |
|-------------------+-------------------+-------------------|
| 8 7 4 | 2 56 36 | 35 1 9 |
| 25 6 235 | 9 1 7 | 3458 48 358 |
| 1 59 359 | 8 45 34 | 6 2 7 |
|-------------------+-------------------+-------------------|
| 4 2 7 | 3 9 1 | 58 #68 #568 |
| 3 8 1 | 5 26 26 | 9 7 4 |
| 59 59 6 | 4 7 8 | 2 3 1 |
*-----------------------------------------------------------*
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Looking at the 2 boxes, we have either r3c9=6 or r7c7=8.
Now its easy to see: r3c9=6 => r2c9=8. So r7c9<>8.
And r7c7=8 => r7c8=6. Therefore r7c9<>6.
Looking at the 2 rows, we have either r2c1=6 or r7c7=8.
With r7c7=8 => r7c8=6 this gives r2c8<>6. |
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