In this post, the classical forcing chain and a sysudokie endorsed application are demonstrated on a multiple solution puzzle.

An underlying assumption made in solving a Sudoku puzzle is that it has one and only one solution.  Some composers offer a written guarantee of uniqueness.  The warranty is not typical, and sometimes this rule of Sudoku is broken.  this reality raises the question: Should methods that rely on a unique solution be included in the Systematic Sudoku Order of Battle?

This policy issue gives me occasion to share a favorite solving experience, and to say something significant about forcing chains.  The puzzle at hand is the one U.S. Air Hemispheres magazine that was used earlier as an example of bv scan APE and Sue de Coq removals.  For convenience, let’s call it USAPE.  You may have worked through basic solving on USAPE over the last two posts, and may have taken up the challenge of adding SdC to the two APE and one SdC we had found back in January of this year.

Here is the USAPE grid with the two APE removals from the post Aligned Pair Exclusion (APE), to which I have added one of the SdC  on our scavenger hunt.  It is

SdC Wr5 = 5(2+3)(7+9)+579.

Both 5(2+3)(7+9) and 579 remove 2 in r5c9. The first by Sue de Coq logic and the second, by the naked pair Wnp79 that appears when 2 and 3 are missing from Wr5(=  579). This is the classical forcing chain idea – the same result from either choice must be true.

Note the grouped forcing chain from the naked pair, that shows that 4 is forced in r5c4. The forcing chain can also be represented more simply, but less explicitly, by a “truth chain – in light blue – similar to the one I used for the APE removal in the Aligned Pair Exclusion post, shown in pink coloring. Forcing chains are limited truth chains, which are limited truth nets.

To test SdC alternatives in difficult cases I make a copy of the slide, and mark the alternative.  This is equivalent to a full truth net.  Then I document this with the shortest truth chain, or better yet, forcing chain. The forcing chain choices are offered by the dissected Sue de Coq.  The result here is the removal in r5c9. This illustrates the classical forcing chain, demonstrating a result that occurs regardless of which of two alternatives choices occur.  This is clearly an application of logic, not an arbitrary trial, i.e. a guess.

Having criticized arbitrarily initiated forcing chains, I am in a better position endorse, without apology,  the forcing chain as a tool supporting  solving methods, like the SdC in this case, rather than a method itself.  In this regard, it is similar to the ER, debunked as a method in itself earlier.

That heavy subject disposed of, look what happens in this case, in rows r456. The middle rows are solved, except for two new bv.  Such a marking event has a large impact on the remainder of the puzzle.  It suggests that I may have stumbled on a solution or multiple solutions.  Remembering that we are testing only one alternative,  we might continue marking and apply coloring.  I suggest you try that for a checkpoint on the next post.  Meanwhile, we’ll take our proved removal and move on.

Now we come to the Death Blossom spoiler Sue de Coq of the last post. In SdC Er4 = 7(1+2)(4+9),  12 cannot be missing, or r4c1 must supply both, and 49 cannot be missing, else r5c9 must supply 4 and 9.

I have added two more SdC in c4. These two,

SdC Cc4 = 2(3+7)(4+8) + 248 (48 missing => 237)

and

SdC Sc4 = 9(3+7)(4+8) +937 (37 missing => 948),

are Siamese twins. Either both have two alternatives, with the removals shown, or both have no alternatives.  One way or the other, that in itself is an alternative. Both sides hit USAPE hard.

Tell you what.  Follow up on the filled in c4 for the next post, and I’ll continue with the SdC removals above.  As before, the progress so far makes coloring very promising.  To match the checkpoint, color with blue on the top left of the cluster.

On my end, next comes a finned 7-wing, and another intriguing SdC, Sr9 = 2(3+9)(4+7)+247 .  If 39 goes missing, we get a naked pair 27 and both a 4-clue and  a naked pair 39 in c4, leaving r4c4 empty.

In follow up to these removals, you can find a finned 4-wing, a 3-chain, and some
color traps to reach a solution.  But is it
one among many?