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March 15 DB

 
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
PostPosted: Sat Mar 15, 2008 1:36 pm    Post subject: March 15 DB Reply with quote
The March 15 DB is a bit challenging. I had to use a six-step xy-chain to eliminate the 3 in R3C1. Any sophisticated solutions?


Earl

Code:

+-------+-------+-------+
| . . 1 | 7 . . | 2 . 6 |
| 2 7 . | . 6 . | . . . |
| . . 5 | . 1 . | . 9 . |
+-------+-------+-------+
| 4 . . | 6 . . | . . 2 |
| . . . | 5 9 2 | . . . |
| 5 . . | . . 1 | . . 7 |
+-------+-------+-------+
| . 1 . | . 5 . | 4 . . |
| . . . | . 2 . | . 8 9 |
| 8 . 4 | . . 6 | 1 . . |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sat Mar 15, 2008 4:57 pm    Post subject: Reply with quote
Basics get you to here:
Code:
+----------------+----------------+----------------+
| 39a  489  1    | 7    348  3589 | 2   -345  6    |
| 2    7    89b  | 3489 6    3589 | 38c  345  1    |
| 36   468  5    | 2    1    38   | 7    9    48   |
+----------------+----------------+----------------+
| 4    389  89   | 6    378  378  | 5    1    2    |
| 1    36   7    | 5    9    2    | 38   346  48   |
| 5    2368 26   | 348  348  1    | 9    36   7    |
+----------------+----------------+----------------+
| 69   1    26   | 89   5    789  | 4    27   3    |
| 7    5    3    | 1    2    4    | 6    8    9    |
| 8    29   4    | 39   37   6    | 1    27   5    |
+----------------+----------------+----------------+

There is an XY-wing abc that takes out <3> in R1C8. Which brings us to here:
Code:
+----------------+----------------+----------------+
| 39   489  1    | 7    348  3589 | 2    45   6    |
| rg             |                |                | 
|                |                |                |
| 2    7    89   | 489  6    589  | 38   345  1    |
|                |                |                | 
| 36   468  5    | 2    1    38   | 7    9    48   |
| gr    r        |           rg   |                |
|                |                |                |
+----------------+----------------+----------------+
| 4    389  89   | 6    378  378  | 5    1    2    |
|                |       r   -g   |                |
|                |                |                | 
| 1    36   7    | 5    9    2    | 38   346  48   |
|                |                |                |
| 5    2368 26   | 348  348  1    | 9    36   7    |
|      g-   rg   | r              |                |
|                |                |                |
+----------------+----------------+----------------+
| 69   1    26   | 89   5    789  | 4    27   3    |
| rg        gr   |           r-   |      rg        | 
|                |                |                |
| 7    5    3    | 1    2    4    | 6    8    9    |
|                |                |                |
| 8    29   4    | 39   37   6    | 1    27   5    |
|      rg        | gr   rg        |      gr        |
+----------------+----------------+----------------+

Now, start Medusa coloring (red-green) in R9C2. The start cell does not matter too much.

Look at C6. If the <3> in R4C6 is true, it must be red. But there is already a red <3> in R3C6. Similarly, if the <8> in R7C6 is true, it must be green. There is already a green <8> in R3C6.

So, we can eliminate <3> in R4C6 and <8> in R7C6. A similar argument takes out <3> in R6C2.

Which leads to an XY-wing in R12, and the puzzle is solved.

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Sat Mar 15, 2008 5:06 pm    Post subject: Reply with quote
After taking out a 6 to break up the DP in the 36 UR, I didn't spot anything, but it succumbs pretty easily to a Medusa wrap.

Quote:
Any sophisticated solutions?


Beats me. I guess sophistication, like beauty, is in the eye of the beholder. Question
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
PostPosted: Sun Mar 16, 2008 9:19 am    Post subject: Reply with quote
First just a comment on Keith's Medusa: Those <3> and <8> eliminations are standard Medusa traps.

Rather than Medusa after that XY-Wing, I started looking directly for AIC possibilities. It didn't take long to find a loop based largely on an XY Chain. Consider the 6-cell XY Chain from r3c6 to r9c5 via r3, c1 and r9. It "starts" with <8> and "ends" with <3>. Due to the <7> strong link in b8, we can extend that end to <7> in r7c6.

Now, either r3c6 is <8> or <3>. If <3>, then the chain says r7c6 is <7>, so r4c6 would have to be <8>. Thus, the <8>s in r34c6 are strongly linked and the other <8>s in c6 are eliminated.

But, this isn't all. If I'm not mistaken, the fact that that alternate <3> in r3c6 is involved together with the r7c6 <7> in determining the <8> in r4c6 in the "r3c6 is not <8>" case means that this is a branched AIC that forms a continuous loop. Thus, the <3>s in r3c6 are also strongly linked, eliminating <3> from r1c6 and <7> from r4c6 and solving the puzzle. (If a branched AIC continuous loop is not valid and I was just lucky, I trust someone will post and let me know.)

Note that this AIC loop did not depend upon that first XY Wing. So, it could be considered a one-step solution.

For those interested, this could be expressed in Eureka as:
Code:
                 (3=6)r3c1-(6=9)r7c1-(9=2)r9c2-(2=7)r9c8-(7)r9c5=(7)r7c6- [B]
               /
... -(8=3)r3c6   
               \
                 [A]

                           [A]and[B] -({37}=8)r4c6-(8=3)r3c6- ...

(Note: The chain can be shortened a bit by exploiting the strong links in r7. However, I thought it might be easier to follow using the slightly longer XY Chain approach. And, since all of the involved links outside c6 were already conjugate, there were no other eliminations as a result of the loop.)
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ravel



Joined: 21 Apr 2006
Posts: 536
PostPosted: Sun Mar 16, 2008 10:49 am    Post subject: Reply with quote
Code:
 *-----------------------------------------------------------*
 |*39    489   1     | 7     348  -3589  | 2     45    6     |
 | 2     7    *89    | 489   6     589   |#38    345   1     |
 |-36    468   5     | 2     1    #38    | 7     9    @48    |
 |-------------------+-------------------+-------------------|
 | 4     389   89    | 6     378   378   | 5     1     2     |
 | 1     36    7     | 5     9     2     | 38    346   48    |
 | 5     2368  26    | 348   348   1     | 9     36    7     |
 |-------------------+-------------------+-------------------|
 | 69    1     26    | 89    5     789   | 4     27    3     |
 | 7     5     3     | 1     2     4     | 6     8     9     |
 | 8     29    4     | 39    37    6     | 1     27    5     |
 *-----------------------------------------------------------*
There is a useless xy-wing with pivot 89 in r2c3: r1c1=3 or r2c7=3 (which eliminated 3 in r1c8).
And there is a useless half M-wing 38 with strong link for 8 in box 3: r2c7=3 => r3c6=3 (but no strong link for 8 from r3c6).
But combined you have r1c3 or r3c6=3 => r1c6<>3, r3c1<>3

Remark: Also the other direction of the half M-wing does not help directly: r3c6=8 => r2c7=8 (=> r2c8=3) - but leads to the same eliminations over the xy-wing
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI
PostPosted: Mon Mar 17, 2008 8:22 pm    Post subject: Reply with quote
Earl asked:
Quote:
Any sophisticated solutions?


Ravel's definitely fits this - very smart!
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