## xy-chains

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

### xy-chains

Me again.

I've started looking at XY-Chains and have come across something odd!

The image below is from sudoku9981.com (http://www.sudoku9981.com/sudoku-solving/xy-chain.php)

The page explains that there is a xy-chain running from
r3c3 -> r7c3 -> r8c1 -> r8c5.

If you follow the chain in either direction you can prove that r3c5 can't be 5.

Following the chain from r3c3: let r3c3 = 1; r7c3 <> 1 and must be 6; r8c1 <> 6 and must be 1; r8c5 <> 1 and must be 5.
Following the chain from r8c5: let r8c5 = 1; r8c1 <> 1 and must be 6; r7c3 <> 6 and must be 1; r3c3 <> 1 and must be 5.
Both r8c5 and r3c3 can 'see' r3c5, therefore it can not be 5.

So far, so good.

What disturbed me was looking at the "other implications" with following the chain from either direction.

Following the chain from r3c3:
+ If r3c3 = 1 then r1c3 <> 1; if r7c3 = 6 then r5c3 <> 6 and must be 9; if r5c3 = 9 then r1c3 <> 9 and must be 5.
+ if r8c5 = 5 then r8c4 <> 5 and must be 6; if r8c4 = 6 then r7c4 <> 6 and must be 3; if r7c4 = 3 then r6c4 <> 3 and must be 9; if r6c4 = 9 then r1c4 <> 9 and must be 5.
* r1c3 and r1c4 can not both be 5.

So I thought "good I've 'proved' the first direction is false".

Then I followed the chain from r8c5:
+ if r8c1 = 6 then r8c4 <> 6 and must be 5; if r8c4 = 5 then r1c4 <> 5 and must be 9.
+ if r7c3 = 1 then r1c3 <> 1; if r3c3 = 5 then r1c3 <> 5; therefore r1c3 = 9.
* r1c3 and r1c4 can not both be 9.

So, my question is: am I barking mad (or maybe barking up the wrong tree)?
What am I doing wrong!?

cheers
Steve
speter

Posts: 16
Joined: 30 March 2017

### Re: xy-chains

speter wrote:Me again.

I've started looking at XY-Chains and have come across something odd!

The image below is from sudoku9981.com (http://www.sudoku9981.com/sudoku-solving/xy-chain.php)

The page explains that there is a xy-chain running from
r3c3 -> r7c3 -> r8c1 -> r8c5.

If you follow the chain in either direction you can prove that r3c5 can't be 5.

Following the chain from r3c3: let r3c3 = 1; r7c3 <> 1 and must be 6; r8c1 <> 6 and must be 1; r8c5 <> 1 and must be 5.
Following the chain from r8c5: let r8c5 = 1; r8c1 <> 1 and must be 6; r7c3 <> 6 and must be 1; r3c3 <> 1 and must be 5.
Both r8c5 and r3c3 can 'see' r3c5, therefore it can not be 5.

So far, so good.

What disturbed me was looking at the "other implications" with following the chain from either direction.

Following the chain from r3c3:
+ If r3c3 = 1 then r1c3 <> 1; if r7c3 = 6 then r5c3 <> 6 and must be 9; if r5c3 = 9 then r1c3 <> 9 and must be 5.
+ if r8c5 = 5 then r8c4 <> 5 and must be 6; if r8c4 = 6 then r7c4 <> 6 and must be 3; if r7c4 = 3 then r6c4 <> 3 and must be 9; if r6c4 = 9 then r1c4 <> 9 and must be 5.
* r1c3 and r1c4 can not both be 5.

So I thought "good I've 'proved' the first direction is false".

Then I followed the chain from r8c5:
+ if r8c1 = 6 then r8c4 <> 6 and must be 5; if r8c4 = 5 then r1c4 <> 5 and must be 9.
+ if r7c3 = 1 then r1c3 <> 1; if r3c3 = 5 then r1c3 <> 5; therefore r1c3 = 9.
* r1c3 and r1c4 can not both be 9.

So, my question is: am I barking mad (or maybe barking up the wrong tree)?
What am I doing wrong!?

cheers
Steve

Hi, Steve. You're not doing anything wrong. The initial exercise was a chain that proves r3c5 is not a 5. Your second analysis is correct and proves that r3c3 is not a 1. Your third analysis proves that r8c1 is not a 6. All of these conclusions are correct, and all are consistent with one another, so you have no problem.

Keep in mind that chains never assert that something in the chain is or isn't true. They assert that certain links exist and that therefore if one thing occurs then something else occurs. In your initial chain, for example, you show that if r3c3 is not a 5, then r8c5 is a 5, eliminating 5 at r3c5. This is not enough. It's followed by the (usually unstated) observation that if r3c3 is a 5, then we get the same elimination. Since r3c3 either is or isn't a 5, then we can take the elimination. Having done so in this case we can observe that 5 is true in both r3c3 and r8c3 (!), so the chain ends up not running in either direction. That's OK. The conclusions are all sound. Note that you've also shown that strong links do not prove that one candidate is true and that the other is false. They prove that at least one of the candidates is true. The other needn't be false unless the linked cells can "see" one another.
Steve

SteveG48
2018 Supporter

Posts: 2533
Joined: 08 November 2013
Location: Orlando, Florida

### Re: xy-chains

Thanks for the reply Steve.

After looking at your comments I went through my statements again and found my logic flaw.

My 4th point:
+ if r7c3 = 1 then r1c3 <> 1; if r3c3 = 5 then r1c3 <> 5; therefore r1c3 = 9.

is wrong.

If r8c5=5, then r7c3=6 (not 1).

cheers
S.
speter

Posts: 16
Joined: 30 March 2017

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