stuck again!!!!need help!!!

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stuck again!!!!need help!!!

Postby spirit_fantasy » Sun Apr 23, 2006 4:55 am

Code: Select all
*-----------*
 |134|972|658|
 |9..|.65|4.3|
 |6..|.43|.97|
 |---+---+---|
 |..8|7.1|..6|
 |7.6|3.8|...| 
 |.1.|4.6|78.|
 |---+---+---|
 |2..|619|8.4|
 |8.1|234|9.5|
 |..9|587|..1|
 *-----------*
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Postby Myth Jellies » Sun Apr 23, 2006 6:06 am

Code: Select all
 *--------------------------------------------------*
 | 1    3    4    | 9    7    2    | 6    5    8    |
 | 9    278  27   | 18   6    5    | 4    12   3    |
 | 6    258  25   | 18   4    3    | 12   9    7    |
 |----------------+----------------+----------------|
 | 45  *29   8    | 7   *29   1    | 35   34   6    |
 | 7   *29+4 6    | 3   *29+5 8    | 15   14  *29   |
 | 35   1    23   | 4   *29+5 6    | 7    8   *29   |
 |----------------+----------------+----------------|
 | 2    57   357  | 6    1    9    | 8    37   4    |
 | 8    67   1    | 2    3    4    | 9    67   5    |
 | 34   46   9    | 5    8    7    | 23   236  1    |
 *--------------------------------------------------*

Here is a somewhat advanced uniqueness argument. Look at the starred 29 pattern. Whichever r56c5 equals 5, there is still going to be a deadly pattern of 29 in a subset of the starred cells. The only way to escape that deadly pattern is if r5c2 = 4. That solves the puzzle.

Or if you prefer a more mundane approach, there is a 34-45-53 xy-wing in r9c1, r4c1, & r4c7.
Last edited by Myth Jellies on Mon Apr 24, 2006 8:50 pm, edited 1 time in total.
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Postby tarek » Sun Apr 23, 2006 7:07 am

[Edited: MythJellies had already provided the a more comprehensive solution]


Couldn't tell exactly where you've reached in your solution.....

if you caught the hiddin double (or its naked triple counterpart) in row 4, then you are left with an xy wing from r4c1,r4c7 & r9c1

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Postby RW » Sun Apr 23, 2006 8:43 am

Myth Jellies wrote:Here is a somewhat advanced uniqueness argument.


I don't think you need to know at all wether there are some fives in c5 or a four in c2. This falls very neatly into this category of uniqueness patterns:

Code: Select all
.  ab . |.  ab .  |.  .  ab
.  ab . |.  ab .  |.  .  ab
.  ab . |.  ab .  |.  .  ab


Two numbers cannot be in the same column in each of three boxes next to each other. Only possibility to place a 2 or a 9 outside this pattern is the 2 in r6c3 => r6c3=2 (easier to see in the posted nonpencilmark grid).

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Stuck again!!!! need help!!!

Postby Cec » Sun Apr 23, 2006 9:20 am

tarek wrote:"...then you are left with an xy wing from r4c1,r4c7 & r9c1"

Hi tarek. I can identify this xy-wing which would exclude candidate3 from r9c7. Could you please define how the notation would be written to explain this xy-wing?
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Re: Stuck again!!!! need help!!!

Postby tarek » Sun Apr 23, 2006 9:46 am

Cec wrote:Hi tarek. I can identify this xy-wing which would exclude candidate3 from r9c7. Could you please define how the notation would be written to explain this xy-wing?


Code: Select all
*-----------------------------------------------*
| 1    3    4   | 9    7    2   | 6    5    8   |
| 9    278  27  | 18   6    5   | 4    12   3   |
| 6    258  25  | 18   4    3   | 12   9    7   |
|---------------+---------------+---------------|
|*45   29   8   | 7    29   1   |*35   34   6   |
| 7    249  6   | 3    259  8   | 15   14   29  |
| 35   1    23  | 4    259  6   | 7    8    29  |
|---------------+---------------+---------------|
| 2    57   357 | 6    1    9   | 8    37   4   |
| 8    67   1   | 2    3    4   | 9    67   5   |
|*34   46   9   | 5    8    7   |-23   236  1   |
*-----------------------------------------------*

Eliminating 3 From r9c7 (5 & 4 in r4c1 form an XY wing with 3 in r4c7 & r9c1)

I'm not sure that I understood your question Cec, is it the proof your after? ....if it is, then the shortest forcing chains & xy chains of the discontinuous type would do the trick:

Code: Select all
Candidates in r4c1 will force r9c7 to contain only 2:
r4c1=5 => r4c7=3 => r9c7=2
r4c1=4 => r9c1=3 => r9c7=2
Therefore r9c7<>3 & r9c7=2


If you're after a NICE loop then I'm not your guy, but I've noticed NICE guys would start from r9c7 using links ending again in r9c7 using the cells forming the xy wing.......

[Edit: the proper term was links]

tarek
Last edited by tarek on Sun Apr 23, 2006 6:23 am, edited 1 time in total.
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Postby Carcul » Sun Apr 23, 2006 10:11 am

Cec wrote:Could you please define how the notation would be written to explain this xy-wing?


[r9c7]-3-[r9c1]-4-[r4c1]-5-[r4c7]-3-[r9c7], => r9c7<>3.

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Re: Stuck again!!!! need help!!!

Postby Cec » Sun Apr 23, 2006 10:31 am

tarek wrote:"..I'm not sure that I understood your question Cec ..."

Thanks tarek and Carcul for your prompt replies. I've just edited my initial reply as I didn't realize Carcul's reply had preceded mine. I can understand tarek's notations but interpreting "loop" notations still confuses me. Could you please enlighten me Carcul as to how I should read your notation. My apologies for not being clear.
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Stuck again!!!!need help!!!

Postby Cec » Mon Apr 24, 2006 6:37 am

Carcul wrote:
Cec wrote:Could you please define how the notation would be written to explain this xy-wing?

[r9c7]-3-[r9c1]-4-[r4c1]-5-[r4c7]-3-[r9c7], => r9c7<>3.

Thank you again carcul for your prompt reply. I still have difficulty in interpreting this type of notation and have read other related threads including this Thread ("Weak Inference" link). Is my following interpretation of your above notation correct?

If r9c7 is 3 then r9c1 => 4 then r4c1=>5 then r4c7=>3 and therefor r9c7 can't be 3.
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