One Trick Pony 8/3/08

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One Trick Pony 8/3/08

Postby ArkieTech » Sun Aug 03, 2008 11:34 pm

Today's One Trick Pony (puzzle requiring one advanced step) has taken me on instead of my handling it. Can anyone help? I used the one advanced step (a swordfish) to get here.:(
Code: Select all
000805000000040803060001020039002080200000004080400260050300010607050000000704000
 *--------------------------------------------------------------------*
 | 9      4      23     | 8      23     5      | 16     7      16     |
 | 15     127    125    | 26     4      67     | 8      9      3      |
 | 37     6      8      | 9      37     1      | 4      2      5      |
 |----------------------+----------------------+----------------------|
 | 4      3      9      | 156    167    2      | 157    8      17     |
 | 2      17     6      | 15     189    789    | 159    3      4      |
 | 157    8      15     | 4      179    3      | 2      6      179    |
 |----------------------+----------------------+----------------------|
 | 8      5      4      | 3      269    69     | 679    1      2679   |
 | 6      129    7      | 12     5      89     | 3      4      289    |
 | 13     129    123    | 7      12689  4      | 69     5      2689   |
 *--------------------------------------------------------------------*

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Postby Luke » Mon Aug 04, 2008 2:26 am

There's an x-wing overlay on a unique rectangle that has diagonal pairs.

1 & 5 in r2c13 and r6c13. X-wing on 5.

I recall Keith once saying, "The diagonal pair must have the x-wing component." This solves both r2c1 and r6c3 to 5.

From there, I had to pick away at it with AIC's and XY chains. Someone else will have to find the "one trick!"

Incidentally, an x-wing overlay does not have to have diagonal pairs to allow an elimination on the diagonal. Check out another UR on 1 and 5 in this puzzle:
Code: Select all
*--------------------------------------------------------------------*
 | 9      4      23     | 8      23     5      | 16     7      16     |
 | *15    127    *125   | 26     4      67     | 8      9      3      |
 | 37     6      8      | 9      37     1      | 4      2      5      |
 |----------------------+----------------------+----------------------|
 | 4      3      9      | *156   167    2      | *157   8      17     |
 | 2      17     6      | *15    189    789    | *159   3      4      |
 | *157   8      *15    | 4      179    3      | 2      6      179    |
 |----------------------+----------------------+----------------------|
 | 8      5      4      | 3      269    69     | 679    1      2679   |
 | 6      129    7      | 12     5      89     | 3      4      289    |
 | 13     129    123    | 7      12689  4      | 69     5      2689   |
 *--------------------------------------------------------------------*

The x-wing is once again on the 5. This UR has only one bivalue cell. The rule is that the non-x-wing component cannot be in the diagonal of the bivalue cell. r4c7<>1.

Doesn't get ya much of anywhere, but I thought it interesting the same puzzle position had both these overlay style UR's.
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Postby daj95376 » Mon Aug 04, 2008 5:00 am

A pony, maybe.

Code: Select all
 +-----------------------------------------------------------------------+
 |  9      4      23     |  8      23     5      |  16     7      16     |
 |  15    *127    15-2   | *26     4      67     |  8      9      3      |
 |  37     6      8      |  9      37     1      |  4      2      5      |
 |-----------------------+-----------------------+-----------------------|
 |  4      3      9      |  156    167    2      |  157    8      17     |
 |  2      17     6      |  15     189    789    |  159    3      4      |
 |  157    8      15     |  4      179    3      |  2      6      179    |
 |-----------------------+-----------------------+-----------------------|
 |  8      5      4      |  3      269    69     |  679    1      2679   |
 |  6     *129    7      | *12     5      89     |  3      4      289    |
 |  13    #129    123    |  7      12689  4      |  69     5      2689   |
 +-----------------------------------------------------------------------+
 # 62 eliminations remain

 (2) Kraken X-Wing c24\r28 w/Kraken fin [r9c2]
 [r9c2]=2,[r8c2]=9,[r8c6]=8,[r8c9]=2,[r8c4]=1,[r2c4]=2  =>  [r2c3]<>2

 UR Type 1 [r26c13]  =>  [r6c1]<>15

Instead of the Kraken X-Wing, this can also be played as a (12) UR in [r29c23].

Code: Select all
 [r2c3]=5                         => [r2c3]<>2
 [r9c3]=3             => [r1c3]=2 => [r2c3]<>2
 [r2c2]=7 => [r3c1]=3 => [r1c3]=2 => [r2c3]<>2
 [r9c2]=9 => X-Wing c24\r28       => [r2c3]<>2
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Postby ArkieTech » Mon Aug 04, 2008 7:59 am

Luke451 said
I recall Keith once saying, "The diagonal pair must have the x-wing component."


Thanks for the tip. I will look forward to putting it to use.

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Postby Glyn » Mon Aug 04, 2008 10:42 am

Perhaps this will help again working from ArkieTech's grid and considering the same UR as Luke
Code: Select all
.------------------.------------------.------------------.
| 9     4     23   | 8     23    5    | 16    7     16   |
| 15    127   125  | 26    4     67   | 8     9     3    |
| 37    6     8    | 9     37    1    | 4     2     5    |
:------------------+------------------+------------------:
| 4     3     9    | 156   167   2    | 157   8     17   |
| 2     17    6    | 15    189   789  | 159   3     4    |
| 157   8     15   | 4     179   3    | 2     6     179  |
:------------------+------------------+------------------:
| 8     5     4    | 3     269   69   | 679   1     2679 |
| 6     129   7    | 12    5     89   | 3     4     289  |
| 13    129   123  | 7     12689 4    | 69    5     2689 |
'------------------'------------------'------------------'


To avoid a UR15 in r26c13 either r9c1 or r9c3 must be 1. => r9c5<>1.
Singles to finish.
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Postby ArkieTech » Mon Aug 04, 2008 11:06 am

Glyn said:
To avoid a UR15 in r26c13 either r9c1 or r9c3 must be 1. => r9c5<>1.


Wow!:D What a way to tame a wild pony! Thanks.

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Postby Glyn » Mon Aug 04, 2008 11:29 am

No problem Dan. Evicting the ones in boxes 1 and 4 to r2c2 or r5c2 works just as well.
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Postby daj95376 » Mon Aug 04, 2008 2:43 pm

Glyn: Don't you need to look for common eliminations for all possibilities with your approach?

In order to prevent a 1 in the (*) UR cells, there must be a 1 in one of six (@) non-UR cells.

Code: Select all
+-----------------------------------+
|  .  .  .  |  .  .  .  |  1  .  1  |
| *1 @1 *1  |  .  .  .  |  .  .  .  |
|  .  .  .  |  .  .  1  |  .  .  .  |
|-----------+-----------+-----------|
|  .  .  .  |  1  1  .  |  1  .  1  |
|  . @1  .  |  1  1  .  |  1  .  .  |
| *1  . *1  |  . @1  .  |  .  . @1  |
|-----------+-----------+-----------|
|  .  .  .  |  .  .  .  |  .  1  .  |
|  .  1  .  |  1  .  .  |  .  .  .  |
| @1  1 @1  |  .  1  .  |  .  .  .  |
+-----------------------------------+

Code: Select all
a) [r5c2]|[r6c9]|[r9c1]=1 => selecting a    correct assignment and leads to ...
b)               [r6c5]=1 => selecting an incorrect assignment and leads to ...

+-----------+
|  .  .  .  |
|  5  .  1  |
|  .  .  .  |
|-----------+
|  .  .  .  |
|  .  .  .  |
|  7  .  5  |
|-----------+

Code: Select all
c)        [r2c2]|[r9c3]=1 => selecting an incorrect assignment and leads to ...

+-----------+
|  .  .  .  |
|  5  .  2  |
|  .  .  .  |
|-----------+
|  .  .  .  |
|  .  .  .  |
|  1  .  5  |
|-----------+

Combining (a), (b), and (c) leads to [r2c1],[r6c3]<>1. This is the same as Luke451's results.
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Postby Glyn » Mon Aug 04, 2008 4:32 pm

daj95376 Hope this clarifies things.
The 6 cases you list are associated with 3 independent sets of constraints on satisfying 'contains a 1'.
(i)To satisfy a 1 in columns 1 and 3 outside the UR => 1 locked in r9c13. Common elimination r9c2,r9c5<>1
(ii)To satisfy a 1 in boxes 1 and 4 outside the UR => 1 locked in r25c2. Common elimination r89c2<>1.
(iii)To satisfy a 1 in rows 2 and 6 outside the UR => at least one 1 locked in r2c2+r6c59. This one is not very fruitful.

Another way of looking at is either r2c3=2 and/or r6c1=7. One is sufficient to avoid the UR.
For case (i)
If r2c3=2 => r1c3=3 => r9c3=1
If r6c1=7 => r3c1=3 => r9c1=1
For case (ii)
If r2c3=2 => r2c1=5 => r2c2=1
If r6c1=7 => r5c2=1

In general we are not always lucky enough to find a configuration allowing this separation to be visible and must examine things in more detail for the common peer eliminations. The 5's locked in the cells of the UR are a great help in this case.

Edited thanks Luke for spotting that typo
Last edited by Glyn on Tue Aug 05, 2008 7:14 am, edited 1 time in total.
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Postby Luke » Mon Aug 04, 2008 11:26 pm

I like your approach, Glyn. Sometimes it's better to step back for a second and consider the bigger picture before pulling the trigger on some eliminations. In this case, doing so actually solves the puzzle rather than just a coupla cells.

Quick typo heads-up:
If r6c7=7 => r5c2=1
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Postby Carcul » Mon Aug 11, 2008 9:28 am

Code: Select all
 *------------------------------------------------------------*
 | 9      4      23  | 8      23     5   | 16     7      16   |
 | 15     127    125 | 26     4      67  | 8      9      3    |
 | 37     6      8   | 9      37     1   | 4      2      5    |
 |-------------------+-------------------+--------------------|
 | 4      3      9   | 156    167    2   | 157    8      17   |
 | 2      17     6   | 15     189    789 | 159    3      4    |
 | 157    8      15  | 4      179    3   | 2      6      179  |
 |-------------------+-------------------+--------------------|
 | 8      5      4   | 3      269    69  | 679    1      2679 |
 | 6      129    7   | 12     5      89  | 3      4      289  |
 | 13     129    123 | 7      12689  4   | 69     5      2689 |
 *------------------------------------------------------------*

[r2c3]-2-[r2c4]=2=[r8c4]=1=[r8c2](-1-[r2c2])-1-[r5c2]-7-[(r2c2)]-2-
-[r2c3], => r2c3<>2 and the puzzle is solved.
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Postby ttt » Fri Aug 15, 2008 10:00 am

Hi Carcul & All,

I’m starting to learn NL notation, for your chain above :
[r2c3]-2-[r2c4]=2=[r8c4]=1=[r8c2](-1-[r2c2])-1-[r5c2]-7-[(r2c2)]-2-[r2c3] => r2c3<>2

Why you don’t write like this :
[r2c3]-2-[r2c4]=2=[r8c4]=1=[r8c2]-(pair 17)-[r25c2]-2-[r2c3] => r2c3<>2

This is invalid for NL notation?

Thanks
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Postby ronk » Fri Aug 15, 2008 10:52 am

ttt wrote:Why you don’t write like this :
[r2c3]-2-[r2c4]=2=[r8c4]=1=[r8c2]-(pair 17)-[r25c2]-2-[r2c3] => r2c3<>2

This is invalid for NL notation?

Just write r25c2 as the ALS that it is:

r2c3 -2- r2c4 =2= r8c4 =1= r8c2 -1- als:(r25c2 =1|2= r2c2) -2- r2c3 ==> r2c3<>2
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Postby Allan Barker » Tue Aug 19, 2008 10:34 pm

Hi all,
A) [r2c3]-2-[r2c4]=2=[r8c4]=1=[r8c2](-1-[r2c2])-1-[r5c2]-7-[(r2c2)]-2-[r2c3] => r2c3<>2

B) r2c3 -2- r2c4 =2= r8c4 =1= r8c2 -1- als:(r25c2 =1|2= r2c2) -2- r2c3 ==> r2c3<>2

I now understand (re NL notation) that the 7r25c2 link is present but implied in the als. The logic then forms a double continuous nice loop with 4 strong sets and 4 weak sets, which eliminates everything it, its (weak) path, including 1r9c2.
Code: Select all
   
   __________________________________________
  /                                          |    ==> r9c2<>1
(1)r8c4 = r8c2 - (1)r5c3 - (1)r2c3           |
                     ||        ||            |
                     ||    (2)r2c3 - r2c4 = r8c4  ==> r2c3<>2
                     ||        ||
                  (7)r5c3 - (7)r2c3


Which, if any/both of the NL notations best describes 'double' nice loop? (Granted, the puzzle was solved by r2c3<>2).
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Postby daj95376 » Wed Aug 20, 2008 3:09 am

(A) is a SIN that conveniently ignores a side-effect.

A) [r2c3](-2-[r2c2])-2-[r2c4]=2=[r8c4]=1=[r8c2](-1-[r2c2])-1-[r5c2]-7-[r2c2]; [r2c2]=EMPTY => r2c3<>2

I would write it as:

A) [r2c3]-2-([r2c2],[r2c4])=2=[r8c4]=1=[r8c2]-1-([r2c2],[r5c2])-7-[r2c2]; [r2c2]=EMPTY => r2c3<>2

Where the last cell in each () continues the SIN:
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