Sudoku from LN, Czech republic, 07-12-05

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Sudoku from LN, Czech republic, 07-12-05

Postby Mr.Mephisto » Wed Dec 05, 2007 4:04 pm

Code: Select all
*-----------*
 |..3|25.|.9.|
 |8..|9..|64.|
 |...|..7|3..|
 |---+---+---|
 |6..|7.1|...|
 |.84|5..|...|
 |...|4..|...|
 |---+---+---|
 |..5|.8.|..4|
 |...|...|.3.|
 |.3.|...|.71|
 *-----------*


After few simple steps I got here:

Code: Select all
 *-----------*
 |..3|25.|.9.|
 |8..|913|64.|
 |...|8.7|3..|
 |---+---+---|
 |6..|7.1|48.|
 |.84|5..|...|
 |...|4.8|...|
 |---+---+---|
 |..5|38.|.64|
 |...|17.|.3.|
 |.3.|6..|.71|
 *-----------*


Some X-Wings and Colors gave this:

Code: Select all
 *-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257     27      | 9       1       3       | 6       4       57      |
 | 1459    1469    169     | 8       46      7       | 3       15      2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259     29      | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279    1279    | 4       2369    8       | 79      125     3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*

But what now?!
Mr.Mephisto
 
Posts: 2
Joined: 05 December 2007

Postby ncantoral » Thu Dec 06, 2007 12:46 am

Code: Select all
*-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257#    27#     | 9       1       3       | 6       4       57      |
 | 1459    1469    169     | 8       46      7       | 3       15      2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259#    29#     | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279    1279    | 4       2369    8       | 79      125     3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*


I am not sure I got this right, maybe someone can explain this better. the # marks a chain. using the four cells in r24c23.

if you put a 5 in r2c2, that makes r2c3 a 7, which makes r4c3 a 9 and that makes r4c2 a 2, so the chain is true if a 5 goes in r2c2,

that means you can eliminate the 9 in r4c2
ncantoral
 
Posts: 26
Joined: 18 October 2007

Postby RW » Thu Dec 06, 2007 9:28 am

ncantoral wrote:
Code: Select all
*-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257#    27#     | 9       1       3       | 6       4       57      |
 | 1459    1469    169     | 8       46      7       | 3       15      2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259#    29#     | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279    1279    | 4       2369    8       | 79      125     3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*


I am not sure I got this right, maybe someone can explain this better. the # marks a chain. using the four cells in r24c23.

if you put a 5 in r2c2, that makes r2c3 a 7, which makes r4c3 a 9 and that makes r4c2 a 2, so the chain is true if a 5 goes in r2c2,

that means you can eliminate the 9 in r4c2

First of all, I think you have a typo, it should say "that makes r2c3 a 2", not 7. The elimination is correct, but the chain needs some clarification. This one should explain the elimination a bit better:

[r4c2]=5=[r2c2]=2=[r2c3]-2-[r4c3]-9-[r4c2]
(if r4c2<>5 => r2c2=5 => r2c3=2 => r4c3=9 => r4c2<>9)

RW
RW
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Posts: 1000
Joined: 16 March 2006

Postby ncantoral » Thu Dec 06, 2007 10:10 am

RW,

thank you,

and yes, I made a typo. I knew that 9 didn't belong there but I just couldn't figure out how to properly prove it.
ncantoral
 
Posts: 26
Joined: 18 October 2007

Postby stumble » Thu Dec 06, 2007 4:56 pm

RW wrote:
ncantoral wrote:
Code: Select all
*-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257#    27#     | 9       1       3       | 6       4       57      |
 | 1459    1469    169     | 8       46      7       | 3       15      2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259#    29#     | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279    1279    | 4       2369    8       | 79      125     3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*


I am not sure I got this right, maybe someone can explain this better. the # marks a chain. using the four cells in r24c23.

if you put a 5 in r2c2, that makes r2c3 a 7, which makes r4c3 a 9 and that makes r4c2 a 2, so the chain is true if a 5 goes in r2c2,

that means you can eliminate the 9 in r4c2

First of all, I think you have a typo, it should say "that makes r2c3 a 2", not 7. The elimination is correct, but the chain needs some clarification. This one should explain the elimination a bit better:

[r4c2]=5=[r2c2]=2=[r2c3]-2-[r4c3]-9-[r4c2]
(if r4c2<>5 => r2c2=5 => r2c3=2 => r4c3=9 => r4c2<>9)

RW


Very slick. Is this technique called a 'forcing chain'?
stumble
 
Posts: 52
Joined: 29 October 2007

Postby ncantoral » Thu Dec 06, 2007 8:36 pm

stumble,

I was under the impression that this is an inference chain.
ncantoral
 
Posts: 26
Joined: 18 October 2007

Postby Sudtyro » Thu Dec 06, 2007 10:53 pm

RW wrote:[r4c2]=5=[r2c2]=2=[r2c3]-2-[r4c3]-9-[r4c2]
(if r4c2<>5 => r2c2=5 => r2c3=2 => r4c3=9 => r4c2<>9)

Or, as an AIC (Alternating Inference Chain):
(5)r4c2 = (5-2)r2c2 = (2)r2c3 - (2=9)r4c3 => r4c2 <> 9.
Note how (9)r4c2 can "see" (weakly link to) both ends of the chain, (5)r4c2 and (9)r4c3.

[Edit to remove "aka a Turbot Chain or X-Chain." Sorry...those are single-digit chain names!]
Last edited by Sudtyro on Sat Dec 08, 2007 1:48 pm, edited 1 time in total.
Sudtyro
 
Posts: 68
Joined: 21 December 2006

Postby stumble » Fri Dec 07, 2007 5:09 pm

Sudtyro wrote:
RW wrote:[r4c2]=5=[r2c2]=2=[r2c3]-2-[r4c3]-9-[r4c2]
(if r4c2<>5 => r2c2=5 => r2c3=2 => r4c3=9 => r4c2<>9)

Or, as an AIC (Alternating Inference Chain):
(5)r4c2 = (5-2)r2c2 = (2)r2c3 - (2=9)r4c3 => r4c2 <> 9,
aka a Turbot Chain or X-Chain.
Note how (9)r4c2 can "see" (weakly link to) both ends of the chain, (5)r4c2 and (9)r4c3.


I'm trying to make sense of Eureka notation. Can you put into English what the (5-2) and the (2=9) mean? Eureka tells me '-' is a weak link, but 'If 5 is false then 2 is true' doesn't make sense to me. RW was kind enough to simplify with his => implication chain, but some posters won't do that and I'd like to be able to read Eureka.
stumble
 
Posts: 52
Joined: 29 October 2007

Postby Sudtyro » Fri Dec 07, 2007 7:24 pm

stumble wrote:Eureka tells me '-' is a weak link, but 'If 5 is false then 2 is true' doesn't make sense to me.

That is correct...it should not make sense, because 5-2 is a weak-inference link. Recall the definitions. Weak inference means they can't both be true. That's the same as saying, if one is true the other is false, so you can read 5-2 as saying, if 5 is true then 2 is false...or if 2 is true, 5 is false.

Recall also that strong inference means they can't both be false. That's the same as saying, if one is false the other is true, so you can read 2=9 as saying, if 2 is false then 9 is true...or if 9 is false, 2 is true.

The Eureka notation (see the examples in Supodedia) for (5-2)r2c2 is just a shorthand notation for (5)r2c2 - (2)r2c2. You can combine the two candidates within the paratheses because they both belong to the same cell. Same argument for (2=9)r4c3.

The full chain,
(5)r4c2 = (5-2)r2c2 = (2)r2c3 - (2=9)r4c3 => r4c2 <> 9,
then reads:
if r4c2 is <> 5 => r2c2=5 => r2c2 <> 2, etc.,
simply alternating between false and true for each candidate in the chain until you reach the end where, in this case, r4c3 = 9(true) since we started with r4c2 <> 5(false). You could just as well read the chain the other way. If you start with r4c3 <> 9(false), then you would get r4c2 = 5(true). Either way, (9)r4c2 can "see" at least one of those ends as being true, so it must therefore be false.

That's the beauty of AIC's...all you need to do is start and end the (alternating inference) chain with a strong link. Then any other candidate that can "see" (weakly link to) both ends of the chain can be eliminated!

Hope this helps!
Sudtyro
 
Posts: 68
Joined: 21 December 2006

Postby ncantoral » Fri Dec 07, 2007 8:53 pm

Sudtyro,

very nice, thanx. I couldn't have explained, but I could see it.
ncantoral
 
Posts: 26
Joined: 18 October 2007

Postby stumble » Sat Dec 08, 2007 4:28 pm

Sudtyro,
Thanks again for the explanation.
stumble
 
Posts: 52
Joined: 29 October 2007


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