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Nov 17 DB

 
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Sat Nov 17, 2007 3:20 pm    Post subject: Nov 17 DB Reply with quote

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 |
+-------+-------+-------+

[url=http://www.dailysudoku
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sat Nov 17, 2007 8:06 pm    Post subject: Reply with quote

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

PostPosted: Sat Nov 17, 2007 11:26 pm    Post subject: Reply with quote

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

PostPosted: Sun Nov 18, 2007 12:10 am    Post subject: Reply with quote

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

PostPosted: Sun Nov 18, 2007 1:10 am    Post subject: Reply with quote

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?


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

PostPosted: Sun Nov 18, 2007 1:25 am    Post subject: Reply with quote

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

PostPosted: Sun Nov 18, 2007 1:47 am    Post subject: Reply with quote

Quote:
If R5C3 is <3>,


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

PostPosted: Sun Nov 18, 2007 11:29 am    Post subject: Reply with quote

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   |
+---------------+---------------+---------------+
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 Wink
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     |
 *-----------------------------------------------------------*
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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