Solution for Tough of November 28, 2007 Page One

The following illustrated solution for the Tough Sudoku of November 28, 2007 employs a modestly fiendish group of Sudoku techniques, tips and tricks. The key, for me, was basically three-fold:

1. The early use of an Almost Unique Rectangle
2. The recognition of some useful Almost Y Wing Styles
3. Counting. - Counting will be used primarily on the latter pages to utilyze some Almost Y Wing Styles

Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. This list is getting long, so specifically, one may want to refer to the following previous blog pages:

• Locked Candidates
• Pairs, Triples, Quads
• Y Wing Styles
• Finding Y Wing Styles
• Forbidding Chains 101 The Theory
• Almost Locked Sets
• ALS Proof of 03/10/07
• A study in chain forging page 4. Super cells were introduced there, and they weigh in heavily in one early step.
• Matrices in Sudoku. Here, one gets the theoretical background for the validity of merely counting.
• Unique Rectangles. Just a brief introduction.
• Definitions: Most terms used on this page have a defintion somewhere on the defintions page.

In order to save both time and space, some easy steps may not be illustrated. Also, this solution has not been fully pared for efficiency. However, many steps that I found were superfluous and thus will not be published.

The illustrations of steps shown in this solution path will share the following key:

• black line = strong inference performed upon a set (strong link)
• red line = weak inference performed upon a set (weak link)
• black containers define a partioning of a strong set(s)
• candidates crossed out in red = candidates proven false

Please be aware that, for me, strong and weak need not be mutually exclusive properties.

The Puzzle

Steps 2abc - Some Locked Candidates

Three easy steps are immediately available

2a Locked candidate 2 r6c12 => r4c1≠2

2b Locked 6's r4c79 => r4c456≠6

2c Locked 9's r5c78 => r5c5≠9

The puzzle quickly reveals itself as very difficult. Here I made my standard puzzle mark-up, and found very little.

Step 2d - A short chain using Almost Unique Rectangle 26 (AUR 26)

On the left side of the graphic above, the AUR driven strong inference set (sis) is proven by

• (2)r4c8=(hidden pair 26)r4c79 - AUR (hidden pair 26) r2c79 = sis[(2)r23c8,(2)r1c7,(2)r3c9,(6)r1c7,(6)r3c9]
• => sis [(2)r234c8,(26)r1c7,(26)r3c9]
• this means that at least one of 2 exists in one of those cells, or 6 exists in one of those cells

The sis seems a bit too complicated for any deductions, however.... one can use the strong 78's in box 3 to reduce the effective size of the set. As noted in a previous blog page, any bivalue strong inference can be treated as a Super Cell. When two such bivalues are linked, then one can create an even larger super cell. Above,

(7)r1c7=(7-8)r3c8=(8)r3c9 forms the Super Cell:(78X)r1c7,r3c89.

This means that there is room for just one candidate other than 78 in those 3 cells.

Thus we know we have the Hidden Triple (78X) in those cells. We just do not know the identity of X.

Once one condsiders the possible locations for (26) from the AUR sis proven above, then one clearly has the following chain snippet:

• (2678)r1c7r3c89 [the super cell] = (2)r24c8

One can now use this information in a chain. Above I called this super cell - AUR sis combination a Hybrid Hidden Triple.

The only problem with the AIC chain shown above is that it gives the illustion that one must consider the strong 7's in box 3 twice. This is just an illusion. It is a function of the vehicle, AIC. However, one can merely count the way to the deduction. The trick here is to remember that all matrix configurations rely upon achieving a nxn matrix. Thus, one need only count matrix columns versus rows. I do not construct the matrix mentally, but rather only keep track of the count. One can always construct the matrix later to convince oneself of a valid deduction.

• (7)r1c7=(7)r3c8 => 1 row, 2 columns. Thus we have one extra column. Call this +1.
• (8)r3c8=(8)r3c9 => noting weak link (7-8)r3c8, we have added one column and one row. Thus +0.
• (7=2)r9c8 => noting weak link, (7)r3c8-(7)r9c8, we have again added one column, one row. Thus +0.
• AUR sis [(2)r24c8, (2)r3c8,(26)r1c7,(26)r3c9. note the following weak links:
  • (7-2)r3c8
  • (2)r9c8-(2)r24c8 (note we could have just written -(2)r234c8. Either way is fine)
  • (8-26)r3c9
  Thus, we have added one column and one row.+0 again.
• Note additionally that (7)r1c7 and (26)r1c7 have potential targets. Move (26)r1c8 into column 1. We have now lost a column. (-1). Total: (0).

Once the count reaches zero, we have in column 1 a proven sis. Thus:

• (7=26)r1c7 => r1c7≠59

The counting on such a short chain can be done mentally. With practice, this mental exercise becomes fairly easy.

After making the indicated eliminations,

• (5)r1c5 % row, thus UP 26 (we solved a cell)

This deduction serves to make this puzzle much easier. However, it still is not easy. Nevertheless, the extreme shortness of the AUR step makes it very attractive, in my opinion.

Please note that once one recognizes any sis, one is free to partition it in any logical fashion. This is basically the Useful group concept discussed in Forbidding Chains, the Practice. With AUR sis's, identifying the useful groups is often the key. It takes some practice to realize that since the proven sis are often unusual, the groupings will also often be atypical. However, this is true of any derived sis.

If one prefers, one can also make the eliminations above using only naked set AUR sis deducitons, or even a combination of naked sets and hidden sets. For example, one could have found:

• (7)r1c7=(7)r3c8-(7=2)r9c8-(2)r4c7=(hidden pair26)r79c4-(naked pair26)r2c79
• =[(hybrid naked pair 59)r2c789=(2)r2c8-(2=7)r9c8-(7)r3c8=(7)r1c7]
• => sis{(7)r1c7, (hybrid naked pair 59)r2c789} => r1c7≠59

This writing is nothing but the recognition of conjugacy between naked and hidden sets. Thus, in my opinion, they must be equally acceptable, even though the latter presentation appears more net like.

Notes

Oftentimes, when a truly difficult puzzle hits a very difficult point, the consideration of Almost Unique Rectangle strong inference sets, (AUR sis), can be a useful tool. Nevertheless, the proven sis can be unwieldy. Recognition of the fact that any logically valid partition of these sets can be used in any logical vehicle is often key to using such unwieldy sets. Oftentimes, just a very little bit of outside the box thinking can provide much utility. One of the reasons that I decided to struggle with the ridiculous puzzle called The Easter Monster was to teach myself some new tricks. The AUR step shown above is a reflection of one such lesson. Fortunately, the deduction was not too complex.

This concludes the first page of this solution. Eventually, this link to the next page will become valid.