The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |67.|.2.|.9.|
 |43.|..6|...|
 |.2.|39.|..4|
 |---+---+---|
 |21.|9..|6..|
 |...|.5.|...|
 |..4|..2|.87|
 |---+---+---|
 |9..|.87|.1.|
 |...|1..|.76|
 |.4.|.6.|.38|
 *-----------*


Play/Print this puzzle online

[Marty R.]

Code: Select all

+---------+-------------+------------+
| 6  7 15 | 458 2   145 | 358 9  135 |
| 4  3 9  | 578 17  6   | 258 25 125 |
| 15 2 8  | 3   9   15  | 7   6  4   |
+---------+-------------+------------+
| 2  1 57 | 9   347 8   | 6   45 35  |
| 37 8 6  | 47  5   134 | 123 24 9   |
| 35 9 4  | 6   13  2   | 135 8  7   |
+---------+-------------+------------+
| 9  6 3  | 245 8   7   | 245 1  25  |
| 8  5 2  | 1   34  349 | 49  7  6   |
| 17 4 17 | 25  6   59  | 259 3  8   |
+---------+-------------+------------+


Play this puzzle online at the Daily Sudoku site

XY-Wing Chain 57-15-15-15-17
(7=5)r4c3-r1c3=r3c1-(5=17)b2p59=> -7r4c5

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 6    7   d15   | 458  2    145  | 358  9    135  |
 | 4    3    9    | 578 b17   6    | 258  25   125  |
 |c15   2    8    | 3    9   b15   | 7    6    4    |
 *----------------+----------------+----------------|
 | 2    1    7-5  | 9   a347  8    | 6   a45  a35   |
 | 37   8    6    | 47   5    134  | 123  24   9    |
 | 35   9    4    | 6    13   2    | 135  8    7    |
 *----------------+----------------+----------------|
 | 9    6    3    | 245  8    7    | 245  1    25   |
 | 8    5    2    | 1    34   349  | 49   7    6    |
 | 17   4    17   | 25   6    59   | 259  3    8    |
 *--------------------------------------------------*


(5=347)r4c589 - (7=15)b2p59 - 5r3c1 = 5r1c3 => -5 r4c3 ; stte

I see mine is just about identical to Marty's.

[pjb]

Code: Select all

 6       7       15     | 458    2      145    | 358    9      135   
 4       3       9      | 578    17     6      | 258    25     125   
 15      2       8      | 3      9      15     | 7      6      4     
------------------------+----------------------+---------------------
 2       1      b57     | 9     c347    8      | 6      45     35     
 37      8       6      | 47     5      134    | 123    24     9     
a35      9       4      | 6      1-3    2      | 135    8      7     
------------------------+----------------------+---------------------
 9       6       3      | 245    8      7      | 245    1      25     
 8       5       2      | 1     c34     349    | 49     7      6     
 17      4       17     | 25     6      59     | 259    3      8     


(3=7)r4c3,r6c1 - (7=3)r48c5 => -3 r6c5; stte

Phil

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 6     7     15     | 458   2     145    | 358   9     135    |
| 4     3     9      |e578  d17    6      | 258   25    125    |
|b15    2     8      | 3     9    c15     | 7     6     4      |
|--------------------+--------------------+--------------------|
| 2     1     57     | 9     347   8      | 6     45    35     |
| 3-7   8     6      |f47    5     134    | 123   24    9      |
| 35    9     4      | 6     13    2      | 135   8     7      |
|--------------------+--------------------+--------------------|
| 9     6     3      | 245   8     7      | 245   1     25     |
| 8     5     2      | 1     34    349    | 49    7     6      |
|a17    4     17     | 25    6     59     | 259   3     8      |
*--------------------------------------------------------------*

W Wing with transport : (7=1) r9c1 - r3c1 = r3c6 - (1=7) r2c5 - r2c4 = (7) r5c4 => - 7 r5c1; stte

Leren

[bat999]

Code: Select all

.------------.----------------.---------------.
| 6   7  c15 | 458   2    145 | 358  9   b135 |
| 4   3   9  | 578   7-1  6   | 258  25  a125 |
| 15  2   8  | 3     9    15  | 7    6    4   |
:------------+----------------+---------------:
| 2   1  c57 | 9    d347  8   | 6    45   35  |
| 37  8   6  | 47    5    134 | 123  24   9   |
| 35  9   4  | 6    d13   2   | 135  8    7   |
:------------+----------------+---------------:
| 9   6   3  | 245   8    7   | 245  1    25  |
| 8   5   2  | 1    d34   349 | 49   7    6   |
| 17  4   17 | 25    6    59  | 259  3    8   |
'------------'----------------'---------------'

(1)r2c9 = r1c9 - (1=7)r14c3 - (7=1)r468c5 => -1 r2c5; stte

[Ngisa]

Code: Select all

+---------+-------------+------------+
| 6  7 a15 | 458 2   145 | 358 9  b135 |
| 4  3 9  | 578 d17  6   | 258 25 c125 |
| 15 2 8  | 3   9   15  | 7   6  4   |
+---------+-------------+------------+
| 2  1 f7-5 | 9   e347 8   | 6   45 35  |
| 37 8 6  | 47  5   134 | 123 24 9   |
| 35 9 4  | 6   13  2   | 135 8  7   |
+---------+-------------+------------+
| 9  6 3  | 245 8   7   | 245 1  25  |
| 8  5 2  | 1   34  349 | 49  7  6   |
| 17 4 17 | 25  6   59  | 259 3  8   |
+---------+-------------+------------+


(5=1)r1c3 - r1c9 = r2c9 - (1=7)r2c5 - r4c5 = (7)r4c3 => -5 r4c3; stte

Clement

[Marty R.]

Code: Select all

+---------+-------------+------------+
| 6  7 a15 | 458 2   145 | 358 9  b135 |
| 4  3 9  | 578 d17  6   | 258 25 c125 |
| 15 2 8  | 3   9   15  | 7   6  4   |
+---------+-------------+------------+
| 2  1 f7-5 | 9   e347 8   | 6   45 35  |
| 37 8 6  | 47  5   134 | 123 24 9   |
| 35 9 4  | 6   13  2   | 135 8  7   |
+---------+-------------+------------+
| 9  6 3  | 245 8   7   | 245 1  25  |
| 8  5 2  | 1   34  349 | 49  7  6   |
| 17 4 17 | 25  6   59  | 259 3  8   |
+---------+-------------+------------+


(5=1)r1c3 - r1c9 = r2c9 - (1=7)r2c5 - r4c5 = (7)r4c3 => -5 r4c3; stte

Clement

I don't understand. It seems to me that if r1c3 is not 5 then r4c3 MUST be =5.

[Sudtyro2]

I don't understand. It seems to me that if r1c3 is not 5 then r4c3 MUST be =5.

It's a bi-directional AIC, Marty. Read from R to L, it says, if 7r4c3 is false, then 5r1c3 is true. Either way, 5r4c3 will be false. The AIC establishes only a strong-inference link between 5r1c3 and 7r4c3, meaning that AT LEAST one must be true (and maybe both). Saying specifically that 5r1c3 is false is actually a FALSE premise, which causes the apparent conflict.

SteveC

[Marty R.]

It's a bi-directional AIC, Marty. Read from R to L, it says, if 7r4c3 is false, then 5r1c3 is true. Either way, 5r4c3 will be false.

OK Steve, maybe I'm thickheaded to the point of hopeless. No need to waste too much time on me. Just reading what you wrote above, what I'm seeing is that if 7r4c3 is false, then it's 5. Maybe bi-directional stuff is beyond my grasp.

[Leren]

Marty R wrote : OK Steve, maybe I'm thickheaded to the point of hopeless. No need to waste too much time on me.

I'll see if I can help you out Marty. The reason you might be confused is that Clement's chain (It's a Type 2 AIC) is, strictly speaking, only half a proof of the elimination.

What is really going on is as follows :

1. Assume 5 in r1c3 is False. Then follow Clement's chain via cells marked abcdef (5=1)r1c3 - r1c9 = r2c9 - (1=7)r2c5 - r4c5 = (7)r4c3.

So what this says is that if 5 in r1c3 is False then 7 in r4c3 is True. In particular 5 in r4c3 is False (In the more general case r4c3 could have more than 2 candidates).

2. Assume 5 in r1c3 is True. Then follow the shorter chain using only cells a and f 5 r1c3 - 5 r4c3. So if 5 in 51c3 is True, 5 in r4c3 is False.

So we now know 2 facts: 1. If 5 in r1c3 is False, 5 in r4c3 is False and 2. If 5 in r1c3 is True, 5 in r4c3 is False.

Since 5 in r1c3 can only be True or False we have covered 100% of cases ans deduced a common outcome that 5 in r4c3 is False.

So you can eliminate 5 from r4c3 !

The reason that 2. is always left out of an AIC "proof" is that it is considered to be too trivial to include in the proof, but you shouldn't loose site of what is really going on.

I like to think of an AIC Type 2 as a Kraken candidate : Assume it has all possible values (True or False) and deduce a common outcome.

Here is a link to the Hodoku introduction to AIC Type 2's. When I first read it I was as baffled as you as to why it worked. It took me quite a while to realize what was really going on.

http://hodoku.sourceforge.net/en/tech_chains.php#nl4

Hope this helps, Leren

[Marty R.]

Leren,

It might help, I just don't know yet I don't know how to apply Hodoku's definition to the chain in question.

"An AIC Type 2 starts and ends on a strong link for two different digits in two cells, that see each other. This proves that the end digit can't be in the start cell and the start digit cannot be in the end cell."

Another problem I'm having is understanding his notation.

"On the left: 4- r6c2 =4= r9c2 -4- r9c5 -9- r5c5 =9= r5c4 =8= r6c4 -8. "

Is this situation analogous to a Forcing Chain in which the two numbers in a bivalue cell yield a common outcome and someone says, "but that can't be right because because one of those chains is wrong" and the answer, of course, is that one of them is right?

In the meantime, sometimes I can't see the forest for the trees and home in on one piece of the puzzle, such as if r1c3<>5 then r4c3 must be=5 because that's the only one left in that column.

[Leren]

Hi Marty, Hodoku is just saying exactly what I said in my last post, using fewer words.

So they say : An AIC Type 2 starts and ends on a strong link for two different digits in two cells, that see each other.

Now lets look at Clements chain (5=1)r1c3 - r1c9 = r2c9 - (1=7)r2c5 - r4c5 = (7)r4c3.

It starts with a Strong link (5=1) r1c3 and ends with a Strong Link onto a different digit = (7) r4c3. That part is saying exactly what I said under 1 in my last post.

The next part .... that see each other ... is saying exactly what I said under 2 in my last post. r1c3 and r4c3 share a column. If they didn't share a column or other house the AIC would not work.

Their notation 4- r6c2 =4= r9c2 -4- r9c5 -9- r5c5 =9= r5c4 =8= r6c4 -8 in more familiar terms on this forum is just (4) r6c2 = r9c2 - (4=9) r9c5 - r5c5 = (9-8) r5c4 = (8) r6c4.

This chain starts and ends with Strong links on different digits in two cells that can see each other right ? This is a nice example because it works both ways, since there is an 8 in r6c2 and a 4 in r6c4 and consequently there are two eliminations.

They color code the assumed Status of each node in the chain. If you start from r4c2 and work towards r6c2 a blue candidate is assumed to be False in the Part 1 chain and a green candidate is assumed to be True.

If you work from r6c2 and work towards r4c2 the green candidates are assumed False and the blue candidates are assumed True in the Part 1 chain. In the Part 2 (trivial) chain - the cells can see each other - the True/False assumptions are the opposite of what they are in the Part 1 chain. And as usual the part 2 chain is never written down, it's just replaced by the words ... that can see each other.

That's why, in our notation, there are brackets around the numbers. What this means is that the assumed status is of each candidate is somewhat indeterminate - it depends on which direction you are reading the chain. If you read it from left to right and you see an = sign as the first link, the first candidate on the left is assumed to be False and the last candidate will be True. Reading in the opposite direction the True/False status of each candidate is reversed - the right hand candidate will be assumed to be False and the left hand candidate will be True.

Leren

[Marty R.]

Leren,

After printing out the relevant pages and doing a little studying I've got it. Some of the terminology confused me because when I saw the term "strong link" I was looking for the wrong thing because that term meant what I know as a strong inference.

I'm still not sure how I'd answer the original concern that I had, i.e., if r1c3 is not 5 then r4c3 must be = 5 because it's the only other possibility in that column. I think I've been called out a few times for similar apparent contradictions.

Thank you, I very much appreciate your willingness to take the time to help others (i.e. me)

Marty

[Leren]

Marty R wrote :
I'm still not sure how I'd answer the original concern that I had, i.e., if r1c3 is not 5 then r4c3 must be = 5 because it's the only other possibility in that column. I think I've been called out a few times for similar apparent contradictions.

Hi Marty, I'll have one more go at addressing this concern of yours.

What you are doing is a quite valid Kraken move : You assume 1. if r1c3 is not 5 then r4c3 must be = 5 and 2. (probably subconsciously because it's obvious to you) that if r1c3 is 5 then r4c3 is not 5.

This is a quite valid Kraken consideration but it doesn't result in a common outcome, because you have followed the same pathway in each case (using only cells a and f).

To make a Kraken move prove something by determining a common outcome you have to follow different pathways (in this case the other pathway is cells abcdef - the main chain that is actually written down).

I would say that the "apparent contradiction" is the fact that Clement's main chain (assume 5 r1c3 is False and follow cells abcdef) is actually the False one in the solution ! The trivial chain, which by standard convention is not written down (assume 5 r1c3 is True and "obviously" 5 r4c3 is False because it's in the same column), is actually the True one in the solution.

This doesn't cause a problem for experienced solvers but for beginners it must be baffling. In Clement's solution all they see is a chain that is False in the solution but makes a correct elimination

Thanks for the complement about being willing to help others. This is True but I've also had a personal reason for getting into this issue. I've been quietly chucking for years about the way Type 2 AIC's are notated (even though I follow standard practice and do it my self) - if read literally the "proof" of the elimination is incomplete and therefore, quite frankly, ludicrous !

Thus endeth the lesson. I think this will be my last post on this topic (no doubt to the relief of everybody, including me).

Leren