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Pony 124 21st March

 
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CathyW
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PostPosted: Thu Mar 22, 2007 10:26 pm    Post subject: Pony 124 21st March Reply with quote

I'm stuck on this one trick pony. Even sudocue doesn't offer any hints from here. Please help with the 'one trick'. Explanation appreciated as I'm trying to understand more advanced techniques. Thanks.

Code:
 
 *-----------*
 |4..|..7|...|
 |6..|.2.|7..|
 |278|.53|1.9|
 |---+---+---|
 |564|71.|.3.|
 |.87|...|6..|
 |.23|86.|4.7|
 |---+---+---|
 |396|5..|27.|
 |841|.7.|...|
 |752|9..|..4|
 *-----------*

 
 *-----------------------------------------------------------*
 | 4     13    59    | 16    89    7     | 35    268   26    |
 | 6     13    59    | 14    2     89    | 7     48    35    |
 | 2     7     8     | 46    5     3     | 1     46    9     |
 |-------------------+-------------------+-------------------|
 | 5     6     4     | 7     1     29    | 89    3     28    |
 | 19    8     7     | 23    349   2459  | 6     125   125   |
 | 19    2     3     | 8     6     59    | 4     159   7     |
 |-------------------+-------------------+-------------------|
 | 3     9     6     | 5     48    148   | 2     7     18    |
 | 8     4     1     | 23    7     26    | 359   569   356   |
 | 7     5     2     | 9     38    16    | 38    16    4     |
 *-----------------------------------------------------------*
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Para
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Joined: 08 Nov 2006
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PostPosted: Fri Mar 23, 2007 12:29 am    Post subject: Reply with quote

Code:
*-----------------------------------------------------------*
 | 4     13    59    | 16    89    7     | 35    268   26*   |
 | 6     13    59    | 14    2     89    | 7     48    35    |
 | 2     7     8     | 46    5     3     | 1     46    9     |
 |-------------------+-------------------+-------------------|
 | 5     6     4     | 7     1     2#9   | 89    3     2#8   |
 | 19    8     7     | 23    349   2459  | 6     125   125   |
 | 19    2     3     | 8     6     59    | 4     159   7     |
 |-------------------+-------------------+-------------------|
 | 3     9     6     | 5     48    148   | 2     7     18    |
 | 8     4     1     | 23    7     26*   | 359   569   35-6  |
 | 7     5     2     | 9     38    16    | 38    16    4     |
 *-----------------------------------------------------------*


The move i did here is one i have been experimenting a lot with lately, and happens a lot.
R1C9 and R8C6 are both {26}. Together they see all 2's on R4. So if they are both 2 there would be no room left in R4 for a 2. So they can't both be 2.
So a 6 in a cell that see both cells would force them both to 2's and this contradicts what was stated earlier. So we can eliminate the 6 from R8C9. From there on it is all hidden and naked singles.

greetings

Para
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CathyW
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PostPosted: Fri Mar 23, 2007 4:55 pm    Post subject: Reply with quote

That's very clever. Thank you Para. Does this technique have a name? Very Happy
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Para
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PostPosted: Fri Mar 23, 2007 5:12 pm    Post subject: Reply with quote

Hi

I am not sure. I've seen a few names for similar techniques, one of them being y-wing styles and also some extended variants using ALS's. But don't think this technique has a proper name. I just keep referring to them as distant pair exclusions. But i don't think it has an official name. Maybe Ruud knows.
I just like the logic and it is easy to check for because you can just check the grid for equal bivalue cells.

greetings

Para
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cowgirl
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PostPosted: Fri Mar 23, 2007 5:48 pm    Post subject: Reply with quote

Para wrote:
Code:
*-----------------------------------------------------------*
 | 4     13    59    | 16    89    7     | 35    268   26*   |
 | 6     13    59    | 14    2     89    | 7     48    35    |
 | 2     7     8     | 46    5     3     | 1     46    9     |
 |-------------------+-------------------+-------------------|
 | 5     6     4     | 7     1     2#9   | 89    3     2#8   |
 | 19    8     7     | 23    349   2459  | 6     125   125   |
 | 19    2     3     | 8     6     59    | 4     159   7     |
 |-------------------+-------------------+-------------------|
 | 3     9     6     | 5     48    148   | 2     7     18    |
 | 8     4     1     | 23    7     26*   | 359   569   35-6  |
 | 7     5     2     | 9     38    16    | 38    16    4     |
 *-----------------------------------------------------------*


The move i did here is one i have been experimenting a lot with lately, and happens a lot.
R1C9 and R8C6 are both {26}. Together they see all 2's on R4. So if they are both 2 there would be no room left in R4 for a 2. So they can't both be 2.
So a 6 in a cell that see both cells would force them both to 2's and this contradicts what was stated earlier. So we can eliminate the 6 from R8C9. From there on it is all hidden and naked singles.

greetings

Para

Hoi Para,
would you write the explanation in the dutch for me.
I will lurn, but englisch understanding is zo difficult for me.

Para, zou je mij deze uitleg in het nederlands willen geven, wil graag bijleren, maar de engelse uitleg is vaak lastig te snappen voor mij.
En dit voorbeeld lijkt me er geschikt voor, daar het vaker voorkomt.
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Para
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PostPosted: Fri Mar 23, 2007 6:59 pm    Post subject: Reply with quote

Translation in Dutch of my previous message.

Ja, natuurlijk.

Code:
*-----------------------------------------------------------*
 | 4     13    59    | 16    89    7     | 35    268   26*   |
 | 6     13    59    | 14    2     89    | 7     48    35    |
 | 2     7     8     | 46    5     3     | 1     46    9     |
 |-------------------+-------------------+-------------------|
 | 5     6     4     | 7     1     2#9   | 89    3     2#8   |
 | 19    8     7     | 23    349   2459  | 6     125   125   |
 | 19    2     3     | 8     6     59    | 4     159   7     |
 |-------------------+-------------------+-------------------|
 | 3     9     6     | 5     48    148   | 2     7     18    |
 | 8     4     1     | 23    7     26*   | 359   569   35-6  |
 | 7     5     2     | 9     38    16    | 38    16    4     |
 *-----------------------------------------------------------*


Ok, de uitleg in het Nederlands.
R1C9 en R8C6 zijn allebei {26}. Samen zien ze alle 2en op rij 4. Als ze allebei 2 zijn, zou er geen ruimte meer over zijn voor een 2 op rij 4. Dus kunnen ze niet allebei 2 zijn.
Iedere 6 in een vakje dat zowel R1C9 als R8C6 ziet, zou er voor zorgen dat ze allebei 2 worden en kan dus geen 6 bevatten. Daarom kunnen we de 6 uit R8C9 verwijderen.
De rest is alleen singles. (Zijn er eigenlijk Nederlandse termen voor Sudokutechnieken?)

groeten

Para
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cowgirl
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PostPosted: Sat Mar 24, 2007 12:59 pm    Post subject: Reply with quote

Para wrote:
(Zijn er eigenlijk Nederlandse termen voor Sudokutechnieken?)

groeten

Para


Para,

Thanks, i understand,

Heb het overgenomen in sudocue, en ik kom er met jou uitleg uit.

En of er nederlands talige termen zijn, geen idee, moet je Ruud even vragen.

in ieder geval bedankt voor je antwoord
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CathyW
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PostPosted: Wed Mar 28, 2007 12:21 pm    Post subject: Reply with quote

I recently purchased "The Logic of Sudoku" by Andrew Stuart. I think the move made in the puzzle above by Para could be an alternating inference chain (AIC) though Para explains it differently.

The example in the book makes use of nice loops from two apparently unconnected cells that have the same pair of candidates. See p89 of this book if you have it. I think I've finally understood nice loops though I still find it easier to highlight the 'strong' links with colouring.
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rep'nA
Hooked
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Joined: 19 Jan 2007
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PostPosted: Wed Mar 28, 2007 12:39 pm    Post subject: Reply with quote

CathyW wrote:
I think the move made in the puzzle above by Para could be an alternating inference chain (AIC) though Para explains it differently.


It definitely is an AIC.
Code:

*-----------------------------------------------------------*
 | 4     13    59    | 16    89    7     | 35    268   D26   |
 | 6     13    59    | 14    2     89    | 7     48    35    |
 | 2     7     8     | 46    5     3     | 1     46    9     |
 |-------------------+-------------------+-------------------|
 | 5     6     4     | 7     1     B29   | 89    3     C28   |
 | 19    8     7     | 23    349   2459  | 6     125   125   |
 | 19    2     3     | 8     6     59    | 4     159   7     |
 |-------------------+-------------------+-------------------|
 | 3     9     6     | 5     48    148   | 2     7     18    |
 | 8     4     1     | 23    7     A26   | 359   569  -356   |
 | 7     5     2     | 9     38    16    | 38    16    4     |
 *-----------------------------------------------------------*


A6=A2-B2=C2-D2=D6 implying r8c9<>6.

By the way, would you recommend the book? Is it well written?
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Para
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PostPosted: Wed Mar 28, 2007 1:40 pm    Post subject: Reply with quote

CathyW wrote:
I recently purchased "The Logic of Sudoku" by Andrew Stuart. I think the move made in the puzzle above by Para could be an alternating inference chain (AIC) though Para explains it differently.


Of course you can also see it as an AIC. But i never really like to use them in puzzles. I always like to find a different logic to explain it. I find it hard to look for AIC's. Never really know where to start. And with the logic i used i know how and where to look for it.
But there's always different ways to eliminations.

greetings

Para
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CathyW
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PostPosted: Wed Mar 28, 2007 3:00 pm    Post subject: Reply with quote

Para wrote:
But there's always different ways to eliminations.


True. And perhaps the situation of having equal bivalue cells that are not in the same row/column/block is worthy of having a name of its own. I thought it worth pointing out the AIC technique for others who may not yet be familiar with it.

Having decided it was about time I made the effort to increase my sudoku solving skills I have found Andrew Stuart's book very helpful so far. I thought I was quite comfortable with multiple colouring but have found there are often more eliminations possible than I previously thought now that I finally understand nice loops and strong/weak links.

Still need to study finned fish, ALS and other advanced techniques but hopefully I'll be able to solve more difficult puzzles without asking for hints! I would definitely recommend "The Logic of Sudoku" for people seeking to advance their solving techniques though if you already understand AIC, ALS, and even Sue de Coq, you perhaps don't need it.
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rep'nA
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PostPosted: Wed Mar 28, 2007 3:34 pm    Post subject: Reply with quote

CathyW wrote:


Having decided it was about time I made the effort to increase my sudoku solving skills I have found Andrew Stuart's book very helpful so far. I thought I was quite comfortable with multiple colouring but have found there are often more eliminations possible than I previously thought now that I finally understand nice loops and strong/weak links.

Still need to study finned fish, ALS and other advanced techniques but hopefully I'll be able to solve more difficult puzzles without asking for hints! I would definitely recommend "The Logic of Sudoku" for people seeking to advance their solving techniques though if you already understand AIC, ALS, and even Sue de Coq, you perhaps don't need it.


Thank you for the review of the book.

As far as the original topic is concerned, I'm with Para in not looking for AICs specifically. However, the upshot of keeping them in the back of your mind when making deductions of this sort is that you can use the methodology to extend the technique. For instance, let's say you start with two bivalued cells A and B with equal candidates, say (xy). Starting from A, you find a weak link on x to a cell C and then a strong link on x to a cell D. If D sees B, then we are in the situation Para describes:
Ay=Ax-Cx=Dx-Bx=By

But let's say D doesn't see B. Then the theory of AIC's tells us that we can look for a weak link on x between D and another cell E and then a strong link on x to a cell F. If F sees B, we can again eliminate y from all cells seeing A and B:
Ay=Ax-Cx=Dx-Ex=Fx-Bx=By

In fact, more generalizations are possible (other than just increasing the length). For instance, assume that we start with the chain
Cx=Cz-Dz=Ez-Fz=Fx
where C and F are bivalued cells with values (xz). Alas, we discover that no eliminations are possible. But then we notice that C and F each see a bivalued cell with values (xy), call them A and B, respectivley. We then get the AIC:
Ay=Ax-Cx=Cz-Dz=Ez-Fz=Fx-Bx=By
allowing us to eliminate y from cells that see A and B.

The moral: Maybe I should look for AICs more often.
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