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hard one...

 
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sat Jun 14, 2008 10:36 pm    Post subject: hard one... Reply with quote

Code:
2 9 .|. . 6|. . 8
. 8 .|5 . .|2 . .
. . 6|8 2 9|1 . .
-----+-----+-----
. . 2|. 6 3|8 4 7
7 . 8|. 5 .|. . 3
. 3 4|. 7 8|. . .
-----+-----+-----
. 2 9|3 . .|7 . .
. . 3|. . 2|. 5 .
1 . .|6 9 .|3 8 2


this was posted here...
http://www.sudoku.com/boards/viewtopic.php?t=6132



Code:
.------------------.------------------.------------------.
| 2     9     157  | 147   134   6    | 45    37    8    |
| 34    8     17   | 5     134   147  | 2     69    69   |
| 345   457   6    | 8     2     9    | 1     37    45   |
:------------------+------------------+------------------:
| 59    15    2    | 19    6     3    | 8     4     7    |
| 7     16    8    | 1249  5     14   | 69    1269  3    |
| 69    3     4    | 129   7     8    | 569   1269  1569 |
:------------------+------------------+------------------:
| 4568  2     9    | 3     148   145  | 7     16    146  |
| 468   467   3    | 147   148   2    | 469   5     1469 |
| 1     457   57   | 6     9     457  | 3     8     2    |
'------------------'------------------'------------------'
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Sun Jun 15, 2008 10:34 pm    Post subject: Reply with quote

I found an xyz-wing on <345> with pivot at r3c1 that deleted 4 at r3c2, then after clean-up including a naked triple in b1, a w-wing on <69> with strong link on 9 in r8c79 deleted the 6 in r6c9. That got me to here:

Code:
 
 
 *-----------------------------------------------------------*
 | 2     9     157   | 147   134   6     | 45    37    8     |
 | 34    8     17    | 5     134   147   | 2     69    69    |
 | 34    57    6     | 8     2     9     | 1     37    45    |
 |-------------------+-------------------+-------------------|
 | 59    15    2     | 19    6     3     | 8     4     7     |
 | 7     16    8     | 1249  5     14    | 69    1269  3     |
 | 69    3     4     | 129   7     8     | 569   1269  159   |
 |-------------------+-------------------+-------------------|
 | 568   2     9     | 3     148   145   | 7     16    146   |
 | 68    467   3     | 147   148   2     | 469   5     1469  |
 | 1     457   57    | 6     9     457   | 3     8     2     |
 *-----------------------------------------------------------*

Ted
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Mon Jun 16, 2008 6:17 am    Post subject: Reply with quote

Seeing how many strong links there were in both 5 and 7, I decided to medusa this one. Starting from r3c2 (<57>), this is the resulting Medusa coloring for the top part of the grid:


As you can see,
- green 4 and 7 in row 1 imply r1c4=1
- red 5 in r3c2 implies r4c2=1

either way, r4c4 cannot be 1.

With r4c4=9, I later needed some coloring on 4 and an xy-wing 17-13-37 in boxes 2 and 3 to finish the puzzle.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Jun 16, 2008 6:30 am    Post subject: Reply with quote

This was an interesting difficult puzzle due to some unusual UR steps early on. After solving it, I looked at the other thread referenced above. There, I believe that part of the UR solution offered is in error. I'll point this out in my solution path.

[1] Finned X-Wing on <6> c27: r5c8<>6

Now begins the interesting UR stuff. First, note that the <2>s form an X-Wing pattern in r56c48. Now, either r4c4 is <1>, or it is <9> and we have a potential 12 UR in r56c48, in which case r5c8 is not <1> so one of r6c89 must be <1>. In either case, r6c4 is not <1>. In AIC notation, this might be written as (and a clearer alternate notation suggestion would be welcome):

[2] (1=9)r4c4-(9)r6c4=UR[({12})r6c4-(1)r5c8]=(1)r6c89; r6c4<>1

[3] We now have a 29 UR of Type 6: r5c8 & r6c4 = 2

This is where I believe the solution on the other thread has erred. After arriving at this same UR in a very slightly different manner, the claimed result is that r5c4 & r6c8<>9. But, this doesn't follow. We only know that they can't BOTH be <9>. As it happens, neither is and the solver got lucky. (I looked for a way to exploit that UR induced weak link between those <9>s but did not see one.)

The grid now:
Code:
+---------------+---------------+----------------+
| 2    9    157 | 147  134  6   | 45   37   8    |
| 34   8    17  | 5    134  147 | 2    69   69   |
| 34   57   6   | 8    2    9   | 1    37   45   |
+---------------+---------------+----------------+
| 59   15   2   | 19   6    3   | 8    4    7    |
| 7    16   8   | 149  5    14  | 69   2    3    |
| 69   3    4   | 2    7    8   | 569  169  159  |
+---------------+---------------+----------------+
| 568  2    9   | 3    148  145 | 7    16   146  |
| 68   467  3   | 147  148  2   | 469  5    1469 |
| 1    457  57  | 6    9    457 | 3    8    2    |
+---------------+---------------+----------------+

All but one of the remaining steps is an AIC uncovered using Medusa Multi-coloring:

[4] (5)r6c7=(5)r1c7-(5)r1c3=(5)r3c2-(5)r4c2=(5-9)r4c1=(9)r6c1; r6c7<>9

[5] (7)r2c6=(7)r1c4-(7=3)r1c8-(3)r1c5=(3)r2c5-(3=4)r2c1; r2c6<>4

[6] Finned Swordfish on <4> in r237: r8c5<>4

The grid:
Code:
+---------------+---------------+----------------+
| 2    9    157 | 147  134  6   | 45   37   8    |
| 34   8    17  | 5    34   17  | 2    69   69   |
| 34   57   6   | 8    2    9   | 1    37   45   |
+---------------+---------------+----------------+
| 59   15   2   | 19   6    3   | 8    4    7    |
| 7    16   8   | 149  5    14  | 69   2    3    |
| 69   3    4   | 2    7    8   | 569  169  159  |
+---------------+---------------+----------------+
| 568  2    9   | 3    148  145 | 7    16   146  |
| 68   467  3   | 147  18   2   | 469  5    1469 |
| 1    457  57  | 6    9    457 | 3    8    2    |
+---------------+---------------+----------------+

[7] (7)r8c4=(7)r1c4-(7=3)r1c8-(3)r3c8=(3-4)r3c1=(4)r3c9-(4)r1c7=(4)r8c7; r8c4<>4

[8] (7)r2c6=(7-4)r1c4=(4)r5c4-(4=1)r5c6; r2c6<>1
Code:
+---------------+---------------+----------------+
| 2    9    57  | 14   134  6   | 45   37   8    |
| 34   8    1   | 5    34   7   | 2    69   69   |
| 34   57   6   | 8    2    9   | 1    37   45   |
+---------------+---------------+----------------+
| 59   15   2   | 19   6    3   | 8    4    7    |
| 7    16   8   | 149  5    14  | 69   2    3    |
| 69   3    4   | 2    7    8   | 56   169  159  |
+---------------+---------------+----------------+
| 568  2    9   | 3    148  145 | 7    16   146  |
| 68   46   3   | 7    18   2   | 469  5    1469 |
| 1    457  57  | 6    9    45  | 3    8    2    |
+---------------+---------------+----------------+

And finally...

[9] (4=6)r8c2-(6)r5c2=(6-9)r5c7=(9)r8c7; r8c7<>4
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Jun 16, 2008 6:35 am    Post subject: Reply with quote

nataraj,

Just saw your post. I'm not surprised that "extended" Medusa would work well here. Ironically perhaps (since I somewhat encouraged the "extended" approach), I have been attempting to exhaust Medusa Multi-coloring before resorting to "extensions". And, when I do, I also consider extended multi-coloring along with basic and try to limit both to "simple transports" where possible, including grouped links and multi-cell ALS links, since Medusa does not exploit those. Then, I write them down as AICs.
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Tue Jun 17, 2008 6:28 am    Post subject: Reply with quote

Myself, I am very reluctant to use (extended or basic) Medusa, simply because there isn't yet a way around the problem of choosing a good starting point. Once that choice is made, the markings appear on the paper and if it turns out that the initial choice does not yield results, there is no practical way to go back and start anew (of course there is always the possibility to make copies - the Xerox method of solving sudoku Smile - or to to it on a computer, but where's the difference to brute force T&E then?).

Therefore, I use Medusa as a method of last resort - quite literally so, i.e. if my initial choice does not solve the puzzle, I let it be. I can live with a few more puzzles I did not solve ... Wink

In this case I arrived at the same point as Ted and looking at my drawings (single candidate graphs I use for multi-coloring) I found there was a great potential for linking the 5s and 7s. And it turned out that after exhausting basic Medusa, I needed no more than ONE step of extended coloring and found a useful elimination. Now that is quite O.K. for me and I decided to post it.

BTW, I have the same qualms (problem of choosing the initial bifurcation) with some of the UR based solutions here in the "other" section.
Every time the reasoning goes "because of this potential DP, either cell a must be x or cell b must be y, and either way ...." I go like:

Hey, what's so special about a bifurcation starting at a UR situation? Why not take any old bi-value cell (either it is x or y) and start forcing from there ...?

Just a matter of personal taste of course, but for myself I've decided to limit myself to techniques that have a clear, non-invasive (i.e. does not ruin the puzzle) search strategy. This time the strategy worked Smile
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Marty R.



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

PostPosted: Tue Jun 17, 2008 3:43 pm    Post subject: Reply with quote

Quote:
Myself, I am very reluctant to use (extended or basic) Medusa, simply because there isn't yet a way around the problem of choosing a good starting point. Once that choice is made, the markings appear on the paper and if it turns out that the initial choice does not yield results, there is no practical way to go back and start anew (of course there is always the possibility to make copies -

I make the Medusa markings on a blank grid. Don't like what I see? No problem, the original's still there.
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Tue Jun 17, 2008 4:48 pm    Post subject: Reply with quote

Marty R. wrote:

I make the Medusa markings on a blank grid. Don't like what I see? No problem, the original's still there.


How do you do that? With a transparency? like an overlay? Or really on paper? Do you copy the candidates to the paper? I imagine it very hard to do without candidates/pencil marks

very curious ...
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