Wrong Original Title -- Correct PM

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Wrong Original Title -- Correct PM

Postby daj95376 » Thu Oct 09, 2008 10:54 am

The following PM was posted (and solved) in the DailySudoku forum. I even have a different solution using three chains. However, no one in the other forum mentioned a UR/DP being present, and I haven't been able to locate one as well. Of course, I'm terrible at finding URs/DPs. Maybe you will have better luck!

Code: Select all
 +--------------------------------------------------------------------------------+
 |  5       178     1467    |  134     134     2       |  678     9       3467    |
 |  167     2       1467    |  1349    8       1479    |  567     345     34567   |
 |  3       78      9       |  5       6       47      |  1       48      2       |
 |--------------------------+--------------------------+--------------------------|
 |  4       17      3       |  8       12      5       |  27      6       9       |
 |  17      6       8       |  1249    1249    149     |  257     35      357     |
 |  2       9       5       |  6       7       3       |  4       1       8       |
 |--------------------------+--------------------------+--------------------------|
 |  8       35      16      |  7       13459   149     |  569     2       1456    |
 |  179     35      17      |  12349   123459  6       |  589     458     145     |
 |  169     4       2       |  19      159     8       |  3       7       156     |
 +--------------------------------------------------------------------------------+

[Edit: Changed the title of this thread.]
Last edited by daj95376 on Fri Oct 10, 2008 3:54 pm, edited 1 time in total.
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Postby Draco » Thu Oct 09, 2008 2:17 pm

Best I can do from your PM's is a simple forcing net to crack the puzzle to SSTS; I bet someone who is far better than me at figuring out AIC's and NL's might convert this into something more palatable than a net. It extends out in a simple, two-pronged chain resulting in a contradiction at r7c6:

r3c6=4 r3c8=8 r8c7=8 r7c7=9
r3c6=4 r2c6=7 r2c4=9 r9c4=1
r3c6=4 [r7c6<>4] + r9c4=1 [r7c6<>1] + r7c7=9 [r7c6<>9] = r7c6=0

Singles + a Turbot Fish on 6's crack the puzzle to singles.

Cheers...

- drac
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Postby ronk » Thu Oct 09, 2008 2:27 pm

There is UR(35)r78c25 for r8c5<>5 ... but it's a move that leads nowhere.
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Postby daj95376 » Thu Oct 09, 2008 2:45 pm

Draco: The Medusa something-or-other posted in the other forum uses (essentially) a forcing net on [r3c6]=4 to get the contradiction [r7c6]=1 and [r9c4]=1.

Ron: Thanks for searching for URs. Strange how a PM so loaded with possibilities seems to contain only one UR.

There are several network solutions to this puzzle. Here is a simple solution containing only chains. The first chain is critical before the other two will work.

Code: Select all
  6r7c7  6r9c1  9r8c1  9r7c7                              <> 6 [r7c7]

   c9b9  Locked Candidate 1                               <> 6 [r12c9]

  9r8c7  9r9c1  6r2c1  6r1c7  8r8c7                       <> 9 [r8c7]

         Hidden Single                                    =  9 [r7c7]

 -1r7c3  6r7c3  6r1c7  8r1c2  7r3c2  7r2c6  9r2c4  1r9c4  <> 1 [r7c56],[r9c1]
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Postby Glyn » Thu Oct 09, 2008 4:51 pm

Draco here is a representation of the net you gave in AIC notation, you did all the hard work.

Code: Select all
(1)r7c6-(1=9)r9c4-(9)r2c4=(9-7)r2c6=(7)r3c6
||
(4)r7c6
||
(9)r7c6-(9)r7c7=(9-8)r8c7=(8)r8c8-(8=4)r3c8


A short explanatory note added:

In the cell r7c6 one of the 3 options 1,4 or 9 on the left must be true. The implication stream will drive one of the 3 ends on the right to be true i.e. Either r3c6=7, r7c6=4 or r3c8=4. This reduces to r3c6<>4.
Last edited by Glyn on Fri Oct 10, 2008 7:14 am, edited 1 time in total.
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Postby daj95376 » Thu Oct 09, 2008 6:17 pm

Draco wrote:r3c6=4 r3c8=8 r8c7=8 r7c7=9
r3c6=4 r2c6=7 r2c4=9 r9c4=1
r3c6=4 [r7c6<>4] + r9c4=1 [r7c6<>1] + r7c7=9 [r7c6<>9] = r7c6=0

Singles + a Turbot Fish on 6's crack the puzzle to singles.

Cheers...

- drac

Draco,

If you examine your blue stream, you'll realize that [r7c6]=1 can be appended. Combine this with your green stream and you have everything you need.
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Postby Draco » Thu Oct 09, 2008 6:47 pm

Glyn -- thanks for the conversion; I still don't follow the notation completely but appreciate your taking the time.

Daj -- nice work finding the link to make it a contradiction chain instead of a net!

Cheers...

- drac
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Postby daj95376 » Thu Oct 09, 2008 10:11 pm

Draco wrote:Daj -- nice work finding the link to make it a contradiction chain instead of a net!

Not quite. Your blue/green streams are just chains. You combined them into a forcing network by adding a third step. I combined them into a forcing network by extending the blue stream. My extension was the result of the downstream elimination [r7c6]<>4, which was necessary to produce [r7c6]=1.

Cheers!
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Postby Carcul » Fri Oct 10, 2008 1:41 am

Code: Select all
 *---------------------------------------------------------------------*
 | 5       178     1467 | 134     134     2    | 678     9       3467  |
 | 167     2       1467 | 1349    8       1479 | 567     3457    34567 |
 | 3       78      9    | 5       6       47   | 1       478     2     |
 |----------------------+----------------------+-----------------------|
 | 4       17      3    | 8       12      5    | 27      6       9     |
 | 17      6       8    | 1249    1249    149  | 257     2357    357   |
 | 2       9       5    | 6       7       3    | 4       1       8     |
 |----------------------+----------------------+-----------------------|
 | 8       35      16   | 7       123459  149  | 2569    245     1456  |
 | 179     35      17   | 12349   123459  6    | 25789   24578   1457  |
 | 1679    4       2    | 19      159     8    | 3       57      1567  |
 *---------------------------------------------------------------------*

1) [r8c8]=8=[r3c8]=4=[r3c6](-4-[r7c6])=7=[r2c6]=9=[r2c4]-9-[r9c4]-1-
-[(r7c6)]-9-[r7c7]=9=[r8c7], => r8c7<>8.

2) [r7c3]-6-[r9c1]=6=[r2c1]-6-[r2c7]=6=[r7c7]-6-[r7c3], => r7c3<>6,

and the puzzle is solved.
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Postby daj95376 » Fri Oct 10, 2008 8:03 am

It occurred to me that Draco's forcing net could be expressed as a finned discontinuous nice loop:D

Either ( [r7c6]=4 ) or ( [r7c6]=19 and Draco's cells form a discontinuous nice loop starting at [r3c6]=4 and resulting in [r3c6]<>4 ).

In both cases, we can deduce that [r3c6]<>4. This is the solution I was so sure Carcul would propose!
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Postby eleven » Fri Oct 10, 2008 11:58 am

Finding a good starting point for chains often is the crucial part for solving harder sudokus. Typically you would take bivalue cells, conjugated pairs, UR's, the endpoints of xy-wings, kites or w-wings etc. There was another suggestion recently here, a Strong Corner.
http://forum.enjoysudoku.com/viewtopic.php?p=61811#p61811
I wondered, if i could find that in this grid.
Code: Select all
 *---------------------------------------------------------------------*
 | 5       178     1467 | 134     134     2    |>678     9       3467  |
 |#167     2       1467 | 1349    8       1479 |*567     3457    34567 |
 | 3       78      9    | 5       6       47   | 1       478     2     |
 |----------------------+----------------------+-----------------------|
 | 4       17      3    | 8       12      5    |*27      6       9     |
 | 17      6       8    | 1249    1249    149  |*257     2357    357   |
 | 2       9       5    | 6       7       3    | 4       1       8     |
 |----------------------+----------------------+-----------------------|
 | 8       35      16   | 7       123459  149  |*2569    245     1456  |
 |@179     35      17   | 12349   123459  6    | 25789   24578   1457  |
 |#1679    4       2    | 19      159     8    | 3       57     #1567  |
 *---------------------------------------------------------------------*
Either r9c9=6 or both r2c1 and r9c9, which implies r1c7=6.
And there is a strong link for 9 from r9c1 to r8c1. so we have, if r8c1<>9, then
r9c1<>6 -> (r1c7=6 -> r245c7=257) -> r7c7=9
I.e. either r8c1=9 or r7c7=9, so r8c7<>9, r7c7=9.

This does not help much, but it was fun to find it.
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Postby Draco » Fri Oct 10, 2008 4:24 pm

daj95376 wrote:
Draco wrote:Daj -- nice work finding the link to make it a contradiction chain instead of a net!

Not quite. Your blue/green streams are just chains. You combined them into a forcing network by adding a third step. I combined them into a forcing network by extending the blue stream. My extension was the result of the downstream elimination [r7c6]<>4, which was necessary to produce [r7c6]=1.

Cheers!

I did not so much combine two chains into a third step as I did follow the "cascading singles" from the initial bivalue assumption down two distinct paths until a contradiction was reached. Seems to me your "join" step allows that to instead become a single chain that achieves the same result (but perhaps I am tripping up over the definition of chain vs. net once again).

At any rate, it was/is cool to see a different take on the net. As for finned loops... waay past me.:)

Cheers...

- drac
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