... by: Sherman
David Eppstein calls these bivalue graphs and extends them to more than four cells. See his paper "Nonrepetitive Paths and Cycles in Graphs with Application to Sudoku" from 2005 in http://arxiv.org/abs/cs.DS/0507053. Peter Gordon references this in his book "Mensa Guide to Solving Sudoku", pages 80-84. Paul Stephens calls these forcing loops in his book "Mastering Sudoku Week by Week", pages 118-124.
Bivalue graphs are subsets of more general chaining strategies, but since they are based upon bivalue cells, they are easy to spot.
BTW, I much prefer your book "The Logic of Sudoku" over Gordon's and Stephens' books!
Andrew Stuart writes:
I hadn't seen that paper, very interesting, thanks for the tip.
Glad you liked the book!
... by: David P Bird
Andrew the strong links you should be focusing on are those in the four bi-value cells in each example. In example 2 we get a sequence: (2'=3")r1c9 – (3'=4")r9c9 – (4"=5')r8c7 – (5'=2")r2c7 – Loop. Here all the odd(') or all the even(") numbered candidates will be true. This kills other 3s in c3, 4s in b9, 5s in c7 and 2s in b3, so it isn't necessary for 3 & 5 to be bilocal in the columns. You are therefore looking for an X-wing pattern made up of four bivalues with each side having a linking digit.
Andrew Stuart writes:
Quite right. I might change the focus of the search and strong links as you suggest and see if I get any extra hits with this search. I suspect that may still not produce enough extra value to bring it back into the solver - it is currently 'depciated' and unused. These results ought to pop out of several of the AIC and chaining strategies, but I'll need to do some testing. In the job queue.
... by: p davis
a curiosity, but (7=2)D8-G8=(2-9)G9=F9 => F9<>7 solves the position immediately.
... by: Steve
It seems to me that this strategy revolves around the two-candidate cells and does not really need the strong links. Pick any of the four cells and suppose that it solves to one of its candidates. This eliminates that candidate from one of the other cells, leaving it to solve to its other candidate. Continue in this way around the four cells, and you get the four values solving the four corners of the pattern. If you instead assume the other value solves the first cell, you can go the opposite way around to again find all four corners solved. And both ways put the same number in one or the other cell of each row and column. You can eliminate all four numbers from the rest of their rows and columns without any strong links required!