## xy chains

Advanced methods and approaches for solving Sudoku puzzles

### xy chains

Hi all

I have been trying several advanced techiques using "simple sudoku" program. The one I seem to be stuck on is the xy chain. I know what it is supposed to do, but when I try it I have a 50% success rate. I have read several of the explanations on this site and others to no avail!

I am hoping that someone can help me here. It seems as if there must be some rule about conjugates or something that I am missing. When I pick the next link with one number in common with the last one in the same group, what else have you got to look for?

Thanks
awhosu

Posts: 3
Joined: 09 February 2008

Off the top of my head:

Chains are logical links and you don't alter the contents of the candidate cells during the search!

1) Mark all candidate cells with more than two candidates Amber. You may not use these cells in your search.

2) Mark the starting bivalue cell Blue.

3) As you advance along the XY-Chain, mark successive cells Green.

4) Mark any peer cells Pink that see the Blue cell and the last Green cell for the same unlinked value in the Blue/Green cell pair. These are the cells where the candidate can be eliminated.

Good Luck ... and I hope my advice is correct!

===== ===== =====

An example: Mike Barker zoo entry 32.1

Code: Select all
`.2...6.....6...54.....59....18....923.4...71.....6.....7.24.3.........8..8....269`

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`after Singles+--------------------------------------------------------------------+| A158    2     A135   |  4      1-3    6     |  9      7     B38    || A1789  G39     6     | A1378   2     A137   |  5      4     G38    ||  78     4      37    | A378    5      9     |  1      2      6     ||----------------------+----------------------+----------------------|| G57     1      8     | A357   G37     4     |  6      9      2     ||  3      6      4     |  9      8      2     |  7      1      5     || A2579  G59    A257   | A157    6     A157   |  8      3      4     ||----------------------+----------------------+----------------------||  6      7      9     |  2      4      8     |  3      5      1     || A125    35    A1235  |  6      9     A135   |  4      8      7     ||  4      8     A135   | A1357  A137   A1357  |  2      6      9     |+--------------------------------------------------------------------+`

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`   Blue    Green    Green    Green    Green    Green       Pink3-[r1c9]-8-[r2c9]-3-[r2c2]-9-[r6c2]-5-[r4c1]-7-[r4c5]-3 => [r1c5]<>3`

The value 3 is unlinked/dangling in Blue cell [r1c9] and Green cell [r4c5]. Cell [r1c5] contains 3 and sees the Blue and Green cell.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Lets see if I am correct. As you have set it up you may eliminate the 3 from r1c5 and r1c3?

My problem then is that if you follow your chain to green r2c2 and stop. r1c3 can see blue r1c9 and green r2c2 but the 3 cannot cannot be eliminated, even though the rule suggests it should be! However, if you continue the chain to r3c3 it can. As I understand it this is also an x-chain which I believe must have an even number of cells in the chain.

Your original chain also has an even number of cells in it. Is this also necessary?
awhosu

Posts: 3
Joined: 09 February 2008

Lets see if I am correct. As you have set it up you may eliminate the 3 from r1c5 and r1c3?

My problem then is that if you follow your chain to green r2c2 and stop. r1c3 can see blue r1c9 and green r2c2 but the 3 cannot cannot be eliminated, even though the rule suggests it should be! However, if you continue the chain to r3c3 it can. As I understand it this is also an x-chain which I believe must have an even number of cells in the chain.

Your original chain also has an even number of cells in it. Is this also necessary?

If you stop my chain at Green [r2c2] then the unlinked value in that cell would be 9 instead of 3 as in the Blue cell [r1c9]. Thus, no elimination is possible for [r1c3].

I don't recall if there is an even/odd restriction on an XY-Chain. I doubt it.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Please hang with me a little longer.

The 3 cell chain above will end in a 9 and not the 3, ok. Now for an 11-12 length chain do you have to manually make sure that the last number is (in this case) a 3?

So basically not all chains that start with e.g. (x,3) and end in (y,3) will let you remove the 3 from all cells that see both?
awhosu

Posts: 3
Joined: 09 February 2008

awhosu wrote:Please hang with me a little longer.

The 3 cell chain above will end in a 9 and not the 3, ok. Now for an 11-12 length chain do you have to manually make sure that the last number is (in this case) a 3?

So basically not all chains that start with e.g. (x,3) and end in (y,3) will let you remove the 3 from all cells that see both?

Both of your statements are correct. The logic goes as follows:

If the first cell is 3, then it will eliminate 3 in all cells that see it. If the first cells is not 3, but an XY-Chain based on this premise leads to another cell being 3, then any cell that sees the first cell and the last cell can not contain 3.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

### Here's another way to do it

Using Daj's example, here's how I'd go abt this.

If you start with the 8 in R1C9, then the 3 is the
"left over" number. Keep going through bivalue
cells one at a time until you reach a cell that also
has 3 as its "left over" number. The 3 in this case
must be in one of those two cells, so any 3 they
both can see is history.

In this case, you could say, "3 is left over, and
8 can see 8.
3 can see 3.
9 can see 9.
5 can see 5.
7 can see 7,
and 3 is left over."

Both left over 3's can see the 3 in R1C5, so get out your eraser.

Not as colorful, but it always works for me...

Luke in Ca

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California