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kuskey · Joined: 10 Dec 2008 Posts: 141 Location: Pembroke, NH

I found a nice 3 stepper after some decent basics:
1. an x-wing on 1 in c69 eliminating 1s at r4c6, r9c6, r8c9 and r9c9 which revealed
2. a 137 xy-wing in a single box on 3 eliminating 3s at r6c6 and r4c6 which opened
3. a 347 xy-wing with a pivot at r9c6 eliminating 4 at r6c9 yielding the solution.
This flowed so nicely I wonder; is there a single stepper?

Marty R. · Posted: Sat Sep 11, 2010 3:33 pm Post subject:

peterj · Joined: 26 Mar 2010 Posts: 974 Location: London, UK

I found an ugly single-stepper while considering an almost-triple... but only because I was purposely looking for such a move! Not pretty..

Consider the almost-triple (369) in c7. Suppose r4c7 not=4 then the triple exists and forces r1c7=1 => r2c9 not=1 => r2c6=1 (only 1 left in r2). If it's 1 it can't be 9 and so r4c6 must be 9 (only 9 in c6). If it is 9 it can't be 4 so r4c7=4 (only 4 left in r4). Contradicts our initial assumption, so the assumption was wrong and r4c7=4.

Or..

kuskey · Joined: 10 Dec 2008 Posts: 141 Location: Pembroke, NH

Marty:

For too long I have been wondering about the 3 bivalue cells that meet the xy-wing requirements except the fact that they share a box. Your comments prodded me to search for a precise definition of the xy-wing. I found exactly that at angusj.com in a note stating that the 3 cells sharing a group (box 8 in this case) are simply a naked triple. The particular alignment of these 3 cells led to a pointing pair reduction of 3s in c6. As you noted, this is a basic tecnique. Thanks for the guidance Marty and the furthering of my Sudoku knowledge.