Solution Path for Tough Sudoku of 05/16/09

It has been far too long since I have written a new blog page. Please forbear my tardiness. Hopefully, I am still up to the task of creating a usefull or interesting page.

The following illustrated solution for the Tough Sudoku of May 16, 2009 employs at least one step of some interest, perhaps, to some. Although this sudoku can be solved without using the uniqueness deduction illustrated below, in my opinion the uniqueness deduction simplifies the solution.

If this is your first visit to this blog, welcome! Unfortunately, if you are a first time visitor, this page may seem like it is written in a different language. Well, it is. Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. The earliest posts are at the bottom, and if you have never perused the intricacies of our special coded language here, you may wish to start close to beginning. Then again, you may just decide to jump in and have fun decoding!

Many steps not illustrated are possible.

Style Change

Previously, I had abandoned using the chessboard algebraic notation in favor of the more common rc notation. However, since many people have begun again to post proofs on the tough pages, I have returned to algebraic notation. For those unfamiliar with this notational style, grid coordinates are posted with the pictures.

The Puzzle

For those whom do not wish to struggle through the illustrations, I have written the non-trivial steps taken in this solution at the bottom of the page in both rc notation and sudoku.com.au notation. I suppose I should call the .au notation Gb's and Andrei Z.'s. For those who wish to stumble through the archived puzzles, (under archive, gb from France and Andrei from US were frequent posters back in 2005- early 2006. Perhaps I should stop with this nostalgic stroll through ancient history?

The Puzzle at UP 24 (24 cells filled either as givens, or solved)

Above, I have tried to illustrate a series of easy steps.

• step 1b) Hidden Pair 24: (hp24)d89 => d89≠18, d3≠24
• 1c) Locked Candidate 9: (lc9)d46 => d3≠9
• 1d) (lc6)f456 => f7≠6
• 1e) (lc7)f456 => f7≠7
• 1f) (lc8)df7 => chi7≠8

This brings us to the point often referred to at other sudoku sites as the Simple Sudoku Technique Solution: SSTS. Ok, I am unsure what the last S in the acronym represents. Perhaps it represents Set? IDK!

Prelude to Step 2: Derivation of a uniqueness Strong Inference Set (SIS)

Above, I have illustrated a unique loop which must be avoided. Specifically, (36)a56, (367)f456, (67)g56. Because of the potential xwing with candidate 6 at ag56, this would reduce to: (36)a56, (37)f56, (67)g56. If this situation were to occur, nothing can happen in columns afg, rows 56, nor boxes b5,e5,g5 to resolve these cells. Thus, the puzzle could not have a unique solution. Since the puzzle is a valid sudoku, it has a unique solution. Therefore, the potential situation illustrated above must be false. Given this information, one can derive what must be true. The following strong inference set (sis): [(3)a13, (3)d46, (6)g8] must contain at least one truth. Using this sis, one can find a very nice, imo, xyz type chain, illustrated quite imperfectly below.

To make matters a bit more interesting, the xyz type chain below also uses an Almost Locked Set (ALS) and an Almost Almost X wing. To further complicate matters, it uses both of those almost animals in simultaneous conjunction with each other.

Step 2a: XYZ type chain using multi-valued uniqueness sis

Above, I have attempted to graph the deduction. The chain can be written:

• (3)d46=
• [[(Xwing3)and(nq1238)a13.g1389]=(6)g8-(6)e8=(6-3)e7=(3)e123]
• => d3 ≠3

The uniqueness sis are represented in bold type. It may take some mental gymnastics to convince oneself of the validty of [(Xwing3)and(nq1238)a13.g1389]=(6)g8. But what would the fun be of this exercise if thinking were not required?

One can always convince oneself of the validity of such a deduction using a proof by contradiction. I will leave that exercise up to others, if they choose. One can easily begin by assuming (3)d3 is valid, and end up quickly, using the native sis shown above, with the impossible uniqueness configuration shown further above.

The items in bold is precisely the uniqueness loop we must avoid. Thus, (3)d3 is false.

One may note that I could have used the (hp67) instead of (nq1238) in the chain. They exist in conjunction with each other, and the use of one over the other is really just a matter of preference, or in my case, whim.

After eliminating (3)d3, two cells solve:

1. (8)d3 %cell (naked single 8 at d3)
2. (8)f7 %row,column & box.

At this point, a few simple techniques can be used to advance the puzzle a bit.

Steps 2bc: Hidden pair, naked pair, singles.

Above, we have the following:

• (hp68)hi2 => hi2≠2345
• (np23)g13 => i1,g56≠3, h3,f89≠2
• 9 singles => UP35

Step 3 - a rather standard fare Alternating Inference Chain (AIC)

Above, the following AIC is shown:

• (1)c7=(1-7)c4=(7-6)f4=(6)hi4-(6)g56=(6)g8-(6=1)e8
• => b8,de7≠1
• UP 81 (singles, naked and/or hidden, to the end)

Probably there are simpler ways to have ended this one, but I must admit I did not look very hard since this one is sufficient.

Solution

As promised, here are the non-trivial steps listed in both rc and .au notation.

1. SSTS
2. • The unique loop (36)a56, (367)f456, (67)g56 => sis[(3)a13, (3)d46, (6)g8
   • xyz type chain: (3)d46=
   • [[(Xwing3) and (nq1238)a13.g1389]=(6)g8-(6)e8=(6-3)e7=(3)e123]
   • => d3≠3 some singles, some ssts, some singles
3. • (1)c7=(1-7)c4=(7-6)f4=(6)hi4-(6)g56=(6)g8-(6=1)e8
   • => b8,de7≠1
   • UP 81 (singles, naked and/or hidden, to the end)

Now, if I do not become too dyslexic trying to write both notations, here is the rc notation preferred at many sudoku forums:

1. SSTS
2. • The unique loop (36)r45c1, (367)r456c6, (67)r45c7 => sis [(3)r79c1, (3)r46c4, (6)r2c7]
   • xyz type chain: (3)r46c4=
   • [[(Xwing3) and (nq1238)r97c1.r9721c7]=(6)r2c7-(6)r2c5=(6-3)r3c5=(3)r789c5]
   • => r7c4≠3 some singles, some ssts, some singles
3. • (1)r3c3=(1-7)r6c3=(7-6)r6c6=r89c6-r45c7=r2c7-(6=1)r2c5
   • => r2c2,r3c45≠1
   • ste

Hopefully, it will not be quite as long between blog posts the next time!