12, 1X, 1Y, XY Pattern

MatNewman · Posted: Fri Jun 24, 2005 11:34 am Post subject: 12, 1X, 1Y, XY Pattern

Doing one of the Telegraph puzzles the other day I noticed the following pattern of possible values for a set of 4 squares which formed an x-wing like arrangement.

Top Left: could be 1 or 2
Top Right: could be 1 or X
Bottom Left: Could be 1 or Y
Bottom Right: Could be X or Y

From this we can conclude that the Top Left must be a 2.

Does this fall under an X-Wing classification?
_________________
Mat

coconut · Posted: Fri Jun 24, 2005 3:04 pm Post subject:

Perhaps, this new pattern should be called the xy-wings.

Coconut

Animator · Posted: Sun Jun 26, 2005 12:53 pm Post subject:

Can you post the entire grid?

coconut · Posted: Sun Jun 26, 2005 2:11 pm Post subject:

What is the number, date and the level of this Telegraph puzzle?

MatNewman · Posted: Mon Jun 27, 2005 10:20 am Post subject:

Sadly I don't have the entire grid (I did the puzzle on a plane). It was from 13th or 14th June.
_________________
Mat

MadOverlord · Posted: Mon Jul 11, 2005 1:53 am Post subject:

I ran all the Telegraph puzzles around those dates through the Sudoku Susser to see if I could get the pattern, but since so much depends on the order in which you solve squares, that doesn't say much.

However, since none of those puzzles needed more than basic techniques to solve, this observation is likely a subset of existing techniques.

In order to determine that for sure, we need an exemplar puzzle that has everything solved except the squares in question and any other squares that must be unsolved for the puzzle to be valid (ie: in the top row, if you have a square 12 and a square 1X you also need at least a square X2 to form a locket triplet. Ditto on the other row and 2 columns.

Nick70 · Posted: Mon Jul 11, 2005 7:02 am Post subject:

MadOverlord · Posted: Mon Jul 11, 2005 10:40 am Post subject:

It would be very helpful if you showed us the puzzle in the state it reached just before you applied the technique, and identified the cells involved. I stepped through the solution of the above and didn't see the pattern (again, because the order of operations is crucial).

Best,R

Nick70 · Posted: Mon Jul 11, 2005 10:58 am Post subject:

MadOverlord · Posted: Mon Jul 11, 2005 11:49 am Post subject:

You are correct that advanced techniques (in this case, Bowman Bingo) are required to solve the puzzle without using your technique. I should have made that more clear.

As such, it may be a useful special-case heuristic to add to the arsenal because it's easy for humans to recognize.

Also note that it's possible to have this configuration where 1 or 2 of the edges are blocks.

For convenience, I've listed the puzzle without the possiblities (this makes it easier to import into my susser app). Here's the solution log using Bowman Bingo. Note that it finds the same reduction that you did.

gaby · Posted: Wed Aug 17, 2005 11:43 am Post subject:

Reading a little more about this, I got an implementation working for XY-wing (which I refer to as Y-wing, for simplicity, and I realise now is a short forcing chain).

Also, reading in simes site:

http://www.simes.clara.co.uk/programs/sudokutechnique11.htm

In the section about 'other possible combinations', I am thinking about other possible eliminations of candidates in the y-wing pattern. If the XY cell has the two linked cells XZ and YZ:

XZ = {r1,c1,b1}
YZ = {r5,c5,b5}

Can I legitimately remove Z from the intersections of:

r1 and c5
c1 and r5
r1 and b5
c1 and b5
r5 and b1
c5 and b1

Being that the cells that are found are in each of those houses, and we can remove Z from the intersections of those houses (except the actual XY, XZ and YZ cells themselves)? In some instances these intersections will produce no cells, but this increases the number of intersections, and possibly candidate removals as a result.
_________________
Free daily sudoku - Online puzzle database
http://vanhegan.net/sudoku/

Nick70 · Posted: Thu Aug 18, 2005 8:09 am Post subject:

Dave · Posted: Wed Jul 12, 2006 11:19 pm Post subject:

The puzzle here is broken open with a simple XY-Wing.

The three linked cells with (possible solutions) are R5C2 (2,5), R7C2 (2,9), and R7C8 (5,9). The intersection of the first row and third is R5C8 (4,5). The common element in these are 5, so it can be removed from this cell, leaving 4.

This is the same answer as the Bowman Bingo, but may be easier for manual search.