This post updates the UR chart, with credit to Denis Berthier, then debunks another unfortunate *THLS* pronouncement. The homework puzzle illustrates another penalty a human solver pays for interpreting *The Hidden Logic of Sudoku* as solving advice.

Now in our THLS review, we cross our line between finding candidates and exploiting relationships among them, that is to say, between basic and advanced solving. First in our order of battle are the bv scan methods, those methods requiring nothing but the line marked grid as it is. Although it took a while to decide this, I concluded that the unique rectangle comes first among advanced methods. The characteristic aligned naked pair is a red flag, and after finding a pair of cells for the other two corners, the search is on for removal of any candidate forcing the dirty rectangle, or the confirmation of any candidate absolutely required for its prevention.

With attention focused on this goal, I look for any means, forcing chains included. Then I haul out my UR Types chart to either classify my result as a known type, or to look ways to meet one of the chart conditions.

My chart was compiled from the Hodoku site (9/8/15) during a review, and includes very brief reminders of justification, and the corresponding types from Andrew Stuart and the defunct Sudocue site. Berthier’s descriptions of the UR, in his concluding Miscellanea chapter, includes versions of Types 1 – 4 above.

Berthier’s UR descriptions are hard to be read without diagrams, and are overly specific, but deserve credit for an improvement in my chart.

Type 1 is expanded to include multiple extras in one corner. A single extra is confirmed. Both UR candidates are removed in all cases. Berthier’s Type 3 needlessly limits the form of the naked subset. His Type 4 and Type 4b are both covered by the requirement for a slink. In his Type 4 it’s a box slink; in his Type 4b, the line slink. Sysudokies can interpret the UR Chart to include the possibility of forcing chain “seeing”, a concept totally missing in *THLS*.

The first UR example in THLS is Royle 17-6526, shown here immediately after line marking. The naked pair along one side makes eliminations. The other side creates a box/line, removing 8r9c7. Your homework is to determine the UR type, and the UR removals. If you peek at your copy of THLS, you still won’t have the complete answer.

What rolls my eyes in *THLS* UR coverage is Berthier’s deferral of the unique rectangle to his last resort, on the grounds that it “assumes uniqueness”. In Miscellanea (XXIII.3.4., p.361) Denis says:

– “A resolution path provides a stronger result when it does not use rules based on this assumption: it does [then] prove uniqueness;

– Rules based on the assumption of uniqueness should therefore not be applied before rules … that do not use it.”

With regard to the first point above, it is more accurate to say that uniqueness is proved when a solution is reached with no assumption having been made about the truth of any candidate. In that case every derived clue is proven, leaving no possibility of a second solution. Assumptions have nothing to do with it.

Berthier’s second point is an empty one. Its comes after self serving argument that none of the THLS resolution methods assume uniqueness. Therefore UR methods should come last, not first, in advanced methods.

No, Denis. Unique Rectangle methods don’t assume uniqueness. Instead they assume there is no uniqueness failure of this simple deadly rectangle form. That’s all. The same common sense defense can be offered for extended rectangles and other defined patterns of multiple solution. Unique rectangle is simply not based on the general assumption of uniqueness.

In fact, methods based on such an assumption cannot be defined and do not exist. The reason is evident in the multiple solution cases encountered in the Sysudoku collection reviews, reported in the posts of US Air12/25/12, Ferocious 7/02/13, and Fiendish 11/18/14. There is no predictable pattern in these multiple solutions on which to build any logic on such an assumption.

No, even the trials I’ve used to reveal multiple solutions do not assume uniqueness. They organize candidates into strongly linked armies of candidates, one necessarily true, and the other, false. They offer a humanly practical means of demonstrating multiple solutions, when the failure of advanced methods raises questions of uniqueness.

Getting to the homework, here is a basic trace for Royle 17-692. The THLS elaboration was completed in the bypass. Did you find the unique rectangle?

It’s a Type 4. One of the 9-slink parners must be true, therefore both of the corner 3’s must be false. This easy logic collapses the puzzle, before any other advanced methods are attempted.

Berthier puts UR last, and works out two xyzt-chains (reviewed later) to remove 7r1c5 and 5r3c5, only to use the same UR in an unwarranted admission of desperation.

Next week we continue our evaluation of THLS treatment of advanced techniques, in SOB order, with the XYZ-wing. Regretfully, we have to skip Sue de Coq, whose existence Berthier doesn’t acknowledge. The THLS XYZ-wing examples are plain vanilla ones, too insignificant for sysudokie homework. No, your mission – should you choose to accept it – is to find an irregular XYZ-wing, with one wing attached by forcing chain, in Royle 17-12407. This puzzle is claimed to require no less that four of the aforementioned xyzt chains, two of them hidden, plus two xy chains! I guess you’d better find the i-XYZ wing. I’ll go on from there to a solution with coloring.

Hello John,

perhaps my English is not good enough that I understand your sentence

“Unique Rectangle methods don’t assume uniqueness. “

in the wrong way. I always thought that Unique Rectangle methods do assume uniqueness. I think the following Sudoku Puzzle is a counterexample to your sentence. The puzzle is a slight variation of the Hodoku resp. Sudopedia example.

175839426030274915429651378010095742547162839290487651754926183981543267362718594

1 7 5 8 3 9 4 2 6

0 3 0 2 7 4 9 1 5

4 2 9 6 5 1 3 7 8

0 1 0 0 9 5 7 4 2

5 4 7 1 6 2 8 3 9

2 9 0 4 8 7 6 5 1

7 5 4 9 2 6 1 8 3

9 8 1 5 4 3 2 6 7

3 6 2 7 1 8 5 9 4

As you see, there are only six cells which do not contain givens. Here are the candidates: r2c1 = 6,8, r2c3 = 6,8, r4c1 = 6,8, r4c3 = 3,6,8, r4c4 = 3, r6c3 = 3.

The method Unique Rectangle Type 1 may be applied to the four cells r24c24, which leads to r4c4 = 3. And then the Sudoku has no solution.

You may argue that the puzzle is not a counterexample. Assuming that it has a unique solution shows that it has no solution which shows that the assumption was wrong.

But what about the next example which is a variation of the first.

175839426030274010429000378010095702507162830200087051754920083981003207362708090

1 7 5 8 3 9 4 2 6

0 3 0 2 7 4 0 1 0

4 2 9 0 0 0 3 7 8

0 1 0 0 9 5 7 0 2

5 0 7 1 6 2 8 3 0

2 0 0 0 8 7 0 5 1

7 5 4 9 2 0 0 8 3

9 8 1 0 0 3 2 0 7

3 6 2 7 0 8 0 9 0

Again we have r2c1 = 6,8, r2c4 = 6,8, r4c1 = 6,8, r4c4 = 3,6,8. If you use again Unique Rectangle Type 1, you get r4c4 = 3. And then the puzzle has a unique solution. But r4c4 = 6 and r4c4 = 8 lead also to unique solutions.

Merry Christmas

Guenter

Thanks for the comment, Guenter. Several readers have taken my statement that “UR’s don’t assume uniqueness” the same way. Read a little further, bitte. The uniqueness that UR methods do not assume is general uniqueness – the lack of multiple solutions of any kind. UR methods do assume there is not a multiple solution of that particular rectangular form. That’s a fair assumption to make and fair game for eliminations. Such rectangles in the solution would be embarrassingly obvious. Multiple extended rectangle are similarly too obvious, but I’ll stop there.

I did not and would not say that if the rectangular multiple is allowed, there cannot be any other multiple solution.

“The uniqueness that UR methods do not assume is general uniqueness – the lack of multiple solutions of any kind”

– I’m very muddled over what you are saying explicitly with this statement.

to me that new rendition reads as if you are saying:

Ur methods do not assume the puzzle cannot have multiple solutions.

Why would they assume a puzzle having more then 1 solution?

The technique set its self functions purely on avoiding these states, other wise the technique would not function.

Ur methods do assume a puzzle has only 1 solution {and eliminations are reliant on having exactly 1 solution remaining.}

So yes they do assume the puzzle cannot have multi-solutions.

they also assume that their application avoids at least 1 of the multi-solution state leaving at least 1 solution:

however that’s not the case if they are built using the multi-solution sate themselves they can then remove all solutions {without knowing}.

Just to be clear, that simple UR moves can kill all solutions.

This grid

…8.7.1……467…..12…..4.9……9…75.3.85..4….3.4.2.7..5.8….7..2.1…

gets here:

+—————-+—————-+—————-+

| 49 3 2 | 8 6 7 |#59 1 #459 |

| 89 89 1 | 3 5 4 | 6 7 2 |

| 468 568 7 | 9 1 2 | 58 3 45 |

+—————-+—————-+—————-+

| 5 7 4 | 1 9 8 | 3 2 6 |

| 16 16 9 | 4 2 3 | 7 5 8 |

| 3 2 8 | 5 7 6 | 4 9 1 |

+—————-+—————-+—————-+

| 19 19 3 | 6 4 5 | 2 8 7 |

| 2 4 5 | 7 8 9 | 1 6 3 |

| 7 89 6 | 2 3 1 |#59 4 #59 |

+—————-+—————-+—————-+

If you set r1c9=4, both solutions are killed, and you are left with an zero solution grid state

It’s a waste of YOUR time to prove I’m wrong about something I never said. It would be better to just read the whole comment in the post. Again, UR methods assume there is no multiple solution WITH THIS OBVIOUS RECTANGULAR FORM. Anyway, thanks for paying attention and writing in.