Another Look at Ardson Very Hard v.2 38

This post brings up an interesting question about the multiple solution puzzle: Does its ambiguity prevent strictly logical solving from reaching a solution?

Introducing the first selected review puzzle of the A.D. Ardson Very Hard v.2 collection on April 11, I accidentally omitted a given.  Ardv2 38 should have a given 7r7c6.  The post of April 18, traced the resolution, with missing given, to a match of the Ardson solution,  with  given.

A diligent reader, Gerald Asp, who checks his transcription by running the puzzle through Andrew Stuart’s solver, got the unexpected report that my version of Ardv2 38, missing 7r7c6, has 63 solutions! In my trace, 7r7c6 is determined on the final color wrap. So either my resolution transgressed strict logic at some earlier point, making an unwarranted assumption about the truth of a candidate, or I was mistaken in the review of Denis Berthier’s The Hidden Logic of Sudoku when I said, in the post THLS vs.Unique Rectangle of 12/13/16, that uniqueness is proved when a solution is reached with no assumption having been made about the truth of any candidate.

Inherent in that statement is the belief that resolution between the multiple solutions cannot be achieved without such an assumption. But in the traced resolution of April 18, where is that assumption? A coloring cluster does not make such an assumption. It relies on the fact that in the slink network of the cluster, candidates of one color are true, and those of the other are false.  Hopefully, one color will be found to violate the rules of Sudoku, a color wrap.

Is such an assumption made when I  use an AIC to constitute a strong link, extending the cluster? This form of slink fulfills the definition of a strong link between candidates.

The question in play here has nothing to do with my disputing Denis Berthier’s claim that UR methods assume uniqueness. These methods do make a much more limited assumption. They assume only that multiple placements of the simple rectangular form will not be allowed by a professional puzzle composer. This has nothing to do with the composer’s willingness to run his solver long enough to be sure that no multiple solutions of any form exist.

Well, help me out here.  Next post, we will move on to resolve Ardv2 278, but do look for comments on this post in the coming weeks. Perhaps one of them will be yours.

About Sudent

I'm John Welch, a retired engineering professor, father of 3 wonderful daughters and granddad to 7 fabulous grandchildren. Sudoku analysis and illustration is a great hobby and a healthy mental challenge.
This entry was posted in Advanced Solving and tagged , , . Bookmark the permalink.

6 Responses to Another Look at Ardson Very Hard v.2 38

  1. GUENTER TODT says:

    John, from a logical point of view it is really easy: If you eliminate a candidate, which is a clue of at least one of the 63 solutions, then you must have made a mistake. Therefore you made mistakes when you eliminated 6r1c89, 12r7c2, 67r9c5, 2r7c8. The elimination of 6r1c89 is not valid because you used a uniqueness method. The eliminations of 12r7c2 and 67r9c5 have nothing to do with multiple solutions or uniqueness. If you control them once more, you will notice that you made wrong conclusions. These are mistakes that happen sometimes. In this case they were programmed because you had the special solution in mind. The elimination of 2r7c8 seems to be just a missing pencil mark. That may also happen from time to time. Therefore you did not make “an unwarranted assumption about the truth of a candidate”.

    Your sentence that “uniqueness is proved when a solution is reached with no assumption having been made about the truth of any candidate” is only true if uniqueness methods are not allowed in the solving process.

    Isn’t that a good result: You don’t need a new theory about multiple or unique solutions. In your analysis you just forgot the third case that also “normal” mistakes may happen.


    • Sudent says:


      Thanks for finding the missing pencil mark. It will probably lead to a interesting revisit of Ardv2 38, with coloring probes of a known multiple solution puzzle.

      Your first sentence agrees with my”uniqueness is proved” statement I am now questioning. You’re saying that a move that eliminates any one of multiple solutions is proof of a mistake. What is a mistake? It could be defined as unconsciously implementing a false assumption about the truth of a candidate.

      Before the pencil mark, you found no mistakes. You repeated a mistaken claim that unique rectangle methods assume complete uniqueness. Then you offered only that if I look at the chained slink traps again, I will find the mistake. I failed, so help me out.

  2. dovmittelman says:

    “Is such an assumption made when I use an AIC to constitute a strong link, extending the cluster? This form of slink fulfills the definition of a strong link between candidates.”

    I think here’s your culprit; it’s only mostly true. A coloring extension needs a strong AND a weak link. Usually unit seeing takes care of that for us, but the ends of an AIC don’t see each other – it’s possible (and common) for both to be true. That’s fine when the important thing about the chain is the toxic ends, but it’s a problem for a color network. People think of a strong link as including the weak one, but it’s really a separate thing superimposed on top.

    • Sudent says:

      Dov, you’re right that a slink is not always a wink. But do these examples make the case? The 9’s coloring in place, if 9r9c1 is false, the chain makes 4r9c2 true. And the reverse, for a strong link. That’s where the network is extended, green to blue. Then, the further extension by column slink colors 4r7c2 green. The weak link is required for the chain. The chain creates the coloring extension slink. Same for the other extension from green 9r7c4 to blue 2r8c5. Then a unit extension for the traps. Agree?

      • dovmittelman says:

        Sure, a false 9 makes for a true 4. The problem comes from the converse: a true 9 does not make for a false 4. That’s why the chain, while valid, doesn’t work to extend the colors.

      • Sudent says:

        Dov, this is very interesting. We agree the chain is valid. You’re saying it does not form a slink because a slink must also be a wink. But that requirement does not appear in the logical definition of a strong link. It happens to be true for unit slinks. Medusa coloring is based on the logical definition, and extensions are not limited to unit seeing. The reversal required here is for false blue 4 to make green 9 true. The 9 and 4 are a toxic set, but being two values in two cells, no candidate can possibly see them both. The coloring extension is from green 9 to blue 4 and the unit slinks take it from there.

        Maybe some of the Stuart solver solutions have 1 or 2 in r7c2? That would make me wrong about coloring. A solution is a solution.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s