The New Sudoku Players' Forum

[Cec]

The following puzzle appears in the Scanraid Sudoku Helper to explain a lengthy XY-chain technique (in this case ten cells).

Original puzzle

Code: Select all

 *-----------*
 |...|...|...|
 |49.|.6.|.82|
 |...|27.|..1|
 |---+---+---|
 |.56|..4|19.|
 |..4|...|7..|
 |.39|8..|24.|
 |---+---+---|
 |6..|.85|...|
 |34.|.2.|.57|
 |...|...|...|
 *-----------*

Pencilmark grid showing a 10-cell XY-chain starting at r1c2 designated (A) and ending at r3c8 (K)

Code: Select all

 *--------------------------------------------------*
 | 1    A26  23B  | 45   45   8    | 369  7    69   |
 | 4    9    7    | 1    6    3    | 5    8    2    |
 | 5    68   38C  | 2    7    9    | 4    36K   1   |
 |----------------+----------------+----------------|
 | 2    5    6    | 7    3    4    | 1    9    8    |
 | 8    1    4    | 56H  9    2    | 7    36J  356  |
 | 7    3    9    | 8    15G  16F  | 2    4    56   |
 |----------------+----------------+----------------|
 | 6    7    12   | 349  8    5    | 39   12   349  |
 | 3    4    18D  | 69   2    16E  | 689  5    7    |
 | 9    28   5    | 346  14   7    | 368  12   346  |
 *--------------------------------------------------*


To my knowledge, the notation for this XY-chain starting at A and ending at K with the common digit 6 is....

6-[r1c2]-2-[r1c3]-3-[r3c3]-8-[r8c3]-1-[r8c6]-6-[r6c6]-1-[r6c5]-5-[r5c4]-6-[r5c8]-3-[r3c8]-6

Because the two candidate 6s in Box3 (r1c79) and candidate 6 in Box1 (r3c2) "see" both the start (A) and end of the chain (K) then these three candidate 6s can be excluded leaving digit 6 to solve cell r1c2.

However, by setting 6 at the start of the chain (r1c2) the implications leave 3 at the end of the chain (r3c8) which then leaves row3 without a 6 because the row now has two 3s. Alternatively, setting 2 at r1c2 leaves 6 at the end of the chain (r3c8) which still leads to a conflict because there is no 3 in row3 (that's because there's two 8s).

I realize there is something wrong with my above analysis but I cannot see where I'm wrong. Thanks in advance of any help to explain my dilemma

Cec

[ronk]

However, by setting 6 at the start of the chain (r1c2) the implications leave 3 at the end of the chain (r3c8) ...

r3c8<>6 does imply r1c2=6 ... but r1c2=6 does not imply r3c8<>6

[Cec]

"...r3c8<>6 implying r1c2=6 ... does not mean r1c2=6 implies r3c3<>6"

Thanks ronk for prompt reply. Alas, I still can't grasp these notations.

Does "r3c8<>6 implying r1c2=6" mean that if r3c8 is not 6 then r1c2 must be 6?

You also say "does not mean r1c2=6 implies r3c3 (I suspect you meant r3c8 and my apologies if I'm wrong?) <> 6". I just can't understand why r3c8 does not automatically become 3 if r1c2 =6.

Cec

[udosuk]

However, by setting 6 at the start of the chain (r1c2) the implications leave 3 at the end of the chain (r3c8) which then leaves row3 without a 6 because the row now has two 3s. Alternatively, setting 2 at r1c2 leaves 6 at the end of the chain (r3c8) which still leads to a conflict because there is no 3 in row3 (that's because there's two 8s).

r1c2=6 => r3c23=[83] & r1c79=[39] => r3c8=6
Row 3 still has a 6, everything is ok.


That xy-chain practically proves that if r1c2<>6, r3c8 must be 6. So either r1c2=6 or r3c8=6, and all cells seeing them simultaneously can't have 6. And it ultimately forces both r1c2 & r3c8 to be 6.

That xy-chain tells us nothing about what would happen if r1c2=6.


Another way to see it is:

r1c2=2 => r1c3=3 => r3c23=[68] & r1c79={69} => r3c8=3

With the chain above: if r1c2=2 => r3c8=3
With the xy-chain: if r1c2=2 => r3c8=6

So the placement r1c2=2 leads to 2 contradicting results, which means it must be an invalid placement. Which means r1c2 cannot be 2, must be 6.

[Steve R]

Here’s my pennyworth.

Assume r1c2 contains 6. Then the three eliminations may be made. On the other hand, if r1c2 does not contain 6, we have:

r1c2 = 2 => r1c3 = 3 => r3c3 = 8 => r8c3 = 1 => r8c6 = 6 => r6c6 = 1 => r6c5 = 5 => r5c4 = 6 => r5c8 = 3 => r3c8 = 6

and the same three eliminations may be made.

It seems to me that using nice loop or Eureka notation obscures what is going on. Their use brings nothing more to solving a puzzle than the traditional notation of the chain above but makes both the chain and its logic more difficult to follow for those unfamiliar with the jargon. The perversity is sometimes extended to the typography. Does encasing cells such as r1c2 in brackets add to the meaning or make the chain easier to track?

Steve

[Cec]

Thanks udosuk. I'll need more time to digest all this. I've been under the impression that in a chain such as above, the correct placement for digit 6 can only be at one end of the chain but not both. Your comments indicates this is not the case.

Cec

[Cec]

"...Assume r1c2 contains 6. Then the three eliminations may be made. On the other hand, if r1c2 does not contain 6, we have:

r1c2 = 2 => r1c3 = 3 => r3c3 = 8 => r8c3 = 1 => r8c6 = 6 => r6c6 = 1 => r6c5 = 5 => r5c4 = 6 => r5c8 = 3 => r3c8 = 6

Thanks Steve for your input. What I find interesting is that if you assume r1c2=6 and follow the chain through with similar logic of candidate exclusions from one cell to another as you have done then cell r3c8=3. At the risk of being a bore this continues to confuse me !

Thanks again to all for your replies.

Cec

[daj95376]

To my knowledge, the notation for this XY-chain starting at A and ending at K with the common digit 6 is....

6-[r1c2]-2-[r1c3]-3-[r3c3]-8-[r8c3]-1-[r8c6]-6-[r6c6]-1-[r6c5]-5-[r5c4]-6-[r5c8]-3-[r3c8]-6

<ship>

However, by setting 6 at the start of the chain (r1c2) the implications leave 3 at the end of the chain (r3c8) which then leaves row3 without a 6 because the row now has two 3s. Alternatively, setting 2 at r1c2 leaves 6 at the end of the chain (r3c8) which still leads to a conflict because there is no 3 in row3 (that's because there's two 8s).

I realize there is something wrong with my above analysis but I cannot see where I'm wrong. Thanks in advance of any help to explain my dilemma

So many seem to have a problem with the initial logic of an XY-Chain. You do not set 6 at the start of the chain!!! Review the chain you listed.

An XY-Chain is a forcing chain where one assertion is assumed to perform a number of eliminations -- [r1c2]=6 in this case. The conjugate assertion -- [r1c2]<>6 in this case -- is then detailed to show that another cell would force some common eliminations -- [r3c8]=6 in this case.

[Steve R]

Cec

I don’t understand what you mean by following the chain through starting with r1c2 = 6. The chain begins

r1c2 = 2 => r1c3 = 3 => …

The only alternative I can see is

r1c2 = 6 => r1c3 = 2 or 3

which does not get us any further. This is what Ron said in his characteristically pithy style.

In principle there could be other chains starting r1c2 = 6 which would prove that r3c8=3 (on the assumption that r1c2 = 6) but that doesn’t apply here because r3c8 = 6 in the solution.

Steve.

[Pat]

Pencilmark grid showing a 10-cell XY-chain
starting at r1c2 designated (A) and ending at r3c8 (K) --

Code: Select all

 *--------------------------------------------------*
 | 1    A26  23B  | 45   45   8    | 369  7    69   |
 | 4    9    7    | 1    6    3    | 5    8    2    |
 | 5    68   38C  | 2    7    9    | 4    36K   1   |
 |----------------+----------------+----------------|
 | 2    5    6    | 7    3    4    | 1    9    8    |
 | 8    1    4    | 56H  9    2    | 7    36J  356  |
 | 7    3    9    | 8    15G  16F  | 2    4    56   |
 |----------------+----------------+----------------|
 | 6    7    12   | 349  8    5    | 39   12   349  |
 | 3    4    18D  | 69   2    16E  | 689  5    7    |
 | 9    28   5    | 346  14   7    | 368  12   346  |
 *--------------------------------------------------*



To my knowledge, the notation for this XY-chain starting at A and ending at K with the common digit 6 is

6-[r1c2]-2-[r1c3]-3-[r3c3]-8-[r8c3]-1-[r8c6]-6-[r6c6]-1-[r6c5]-5-[r5c4]-6-[r5c8]-3-[r3c8]-6

Because the two candidate 6s in Box3 (r1c79) and candidate 6 in Box1 (r3c2) "see" both the start (A) and end of the chain (K),
these 3 candidate 6s can be excluded
leaving digit 6 to solve cell r1c2.


However, by setting 6 at the start of the chain (r1c2)
the implications leave 3 at the end of the chain (r3c8)

how is that ??
which implications ??

please explain how you start with
r1c2 = 6
and reach the conclusion
r3c8 = 3

[Pat]

while i was carefully phrasing my response,
Steve R beat me to the point

[hobiwan]

The only alternative I can see is

r1c2 = 6 => r1c3 = 2 or 3

which does not get us any further.

r1c2=6 leaves 2 in r1c3 as single in r1, so r1c2=6 => r1c3=2 is valid IMHO.

The error lies in cell H (r5c4): r5c4=5 does not imply r5c8=6 because of the additional candidate 6 in r5c9. It's (as so often) a problem with strong and weak inferences. Most of the inferences between the cells of the XY-Chain are strong although that's of course irrelevant for the XY-Chain. But it let's you follow the chain starting with r1c2=6 nearly till the end.

The whole chain:

6= r1c2 =2= r1c3 =3= r3c3 =8= r8c3 =1= r8c6 =6= r6c6 =1= r6c5 =5= r5c4 => stuck here

(and of course that's not an XY-Chain although it uses only bivalue cells)

[edit: typo corrected]

[Steve R]

In the context of XY-chains an initial strong link is not valid.

Steve

[udosuk]

hobiwan made it clear. I think the problem here is Cec has mistaken the xy-chain to another move (such as xy-ring). Here are some useful links:

http://forum.enjoysudoku.com/viewtopic.php?t=1131

http://forum.enjoysudoku.com/viewtopic.php?t=2966

For xy-chain the 2 ends have an either-or-both relationship - i.e. at least one or both of them must have a certain candidate - and so all cells seeing both ends must not contain that candidate.

For xy-ring any 2 "ends" (because it's a loop, any 2 linked cells can function as "ends") have an exclusive-or relationship - i.e. exactly one of them must have a certain candidate - just like a strong link.

BTW I suppose the chain described by hobiwan can be called a "strong chain".

[Cec]

Wow! thanks for the additional replies guys . I've got a busy day today and will reply later. From all these replies it's apparent that my "logic" must be wrong and I'll explain my "logic' as asked by Pat when replying. Hey Pat, do I get a pink elephant for highlighting "missing digits" in some rows rather than duplicate digits

regards, Cec