## One Trick Pony 8/3/08

Post the puzzle or solving technique that's causing you trouble and someone will help

### One Trick Pony 8/3/08

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   | *--------------------------------------------------------------------*`

dan
dan

ArkieTech

Posts: 2956
Joined: 29 May 2006
Location: NW Arkansas USA

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.

Luke
2015 Supporter

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Joined: 06 August 2006
Location: Southern Northern California

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`
daj95376
2014 Supporter

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Joined: 15 May 2006

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.

dan
dan

ArkieTech

Posts: 2956
Joined: 29 May 2006
Location: NW Arkansas USA

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.
Glyn

Posts: 357
Joined: 26 April 2007

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

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

dan
dan

ArkieTech

Posts: 2956
Joined: 29 May 2006
Location: NW Arkansas USA

No problem Dan. Evicting the ones in boxes 1 and 4 to r2c2 or r5c2 works just as well.
Glyn

Posts: 357
Joined: 26 April 2007

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.
daj95376
2014 Supporter

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Joined: 15 May 2006

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.
Glyn

Posts: 357
Joined: 26 April 2007

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.

If r6c7=7 => r5c2=1

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

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.
Carcul

Posts: 724
Joined: 04 November 2005

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
ttt
ttt

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Location: vietnam

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
ronk
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Location: Southeastern USA

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).
Allan Barker

Posts: 266
Joined: 20 February 2008

(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:
daj95376
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