Solution for Tough Sudoku of June 12, 2009


The following illustrated solution for the Tough Sudoku of June 12, 2009 exhibits a path to solution that is meant to be instructive. There certainly are solution paths available for this one that could employ far fewer steps. This path attempts to keep the steps, though many, individually short and relatively simple. As has been my recent tendency, some deductions that rely upon the fact that the puzzle has exactly one solution are employed.

The set of Sudoku solving techniques, strategies, tips and tricks employed in this solution are mainly for those familiar with advanced type techniques.

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 Previous Entries. 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!

In some of the illustrations, more than one step is shown at once. This reflects an attempt to shorten the page. However, it may contribute to some confusion. Hopefully, the confusion is managable.


The Puzzle

For those whom do not wish to struggle through the illustrations, I have written the steps taken in this solution at the bottom of the page in sudoku.com.au notation.


Puzzle at Start

Five Unique Possibilities (UPs) are available here.

  1. (1)c7 % box
    • translation: Hidden single (1) at c7, as it is the only (1) possible in the box
  2. (4)f9 % box & row
  3. (4)g3 % box
  4. (5)f2 % box & column
  5. (9)b6 % box & row


The Puzzle at UP 28 Unique Possibilities to a total of 28 given plus solved cells


Puzzle at UP 28

Five easy steps are available here:

  1. Locked Candidate (1)df5 => h4≠1
  2. (LC3)gh5 => h4≠3
  3. (LC2)bc8 => dgh8≠2
  4. Hidden Pair (57)ac4 => a4≠68, c4≠46; h4≠7
  5. (HP78)f8,e7 => f8≠6, e7≠269
This brings the puzzle to the point where the program, Simple Sudoku, has exhausted its solving technique set. Typically, this is the point where the puzzle gets interesting for those who like tough Sudokus.


Step 2f: An AUR Almost Unique Rectangle used in a short AIC (Alternating Inference Chain)


Below, note that candidate (1) in columns gh is limited to rows 259. Note further that candidate (2) is rows 29 is limited to gh29 plus e9. If cell e9 does not contain (2), and cells gh5 do not contain (1), then the puzzle would contain hidden pair 12 at both gh9 and gh2. This would form a Unique Rectangle, (also called a deadly pattern). Any puzzle that has only one solution cannot have such a pattern, as this pattern cannot be resolved uniquely. Thus, it is safe to say that at least one of the following must be true: (2)e9, (1)g5, (1)h5. Such a set, whereas at least one item in the set must be true, is called a Strong Inference Set (SIS).

Step 2f, AUR in AIC

Above, I have illustrated a short chain which uses three common techniques:

  1. An AUR. Specifically, (AUR12)gh29 => SIS[(2)e9, (1)gh5]
  2. An Almost Locked Set (ALS). Specifically, (1269)d278 => SIS[(2)d7, (1)d2] In the chain below, I denote this particular ALS inference using NP, Naked Pair.
  3. A pair of grouped inferences. Specifically, (1)Boxh2 => SIS[(1)gh2, (1)i13]
The chain could be written:
  • Given (AUR12)gh29 => SIS[(2)e9, (1)gh5]
  • (1)gh5 = (2)e9 - (2)d7 = (NP69)d78 - (NP69)d28 = (1)d2 - (1)gh2 = (1)i13
  • => i5≠1
After making the elimination derived from this short chain, some easy steps are available. These steps are not illustrated. I prefer hidden pairs, but one can easily use two naked pairs to get to the same point. The easy steps I took are:
  • (HP13)gh5 => g5≠6, h5≠48
  • (HP24)bc5 => b4≠4, b5≠68, c5≠6
A short digression before I illustrate the next step. At this point, one could use the following group of steps to dodge using the slightly complex step I will illustrate next. Whether one follows the next group of steps, or the ones I shall illustrate, one will come to the some place in the puzzle:

Alternate group of steps, especially for those who despise Krakens


  1. (AUR57)ac14 (note (5)is already locked at ac4, ac1) => ac1≠7
  2. (7): i1 = e1 - e7 = f8 => i8≠7
  3. (5)i6 = (5-6)g6 = (HT568)i568 => i6≠7
  4. (HP58)i58 = (8-6)i5 = (6)a5 - (6)a79 = (6)bc8(pause) - (6=9)d8 => i8≠69
  5. (NT568)i568 => g6,i9≠6
  6. (LC7)i13 => gh2≠7
  7. (LC7)ac2 => c3≠7
This will bring the puzzle to the same point as it is before step 2m below.


Step 2i: An Almost Skyscraper used in a short AIC


The step below can be represented in many ways. The illustration below reflects the way that I find such steps. During my typical sweep of a puzzle for eliminations resulting from the locations of just one type of candidate, I often see many Almost eliminations. When one of these overlaps with another Almost chain during later investigations, I know that I have found a valid elimination. Below, I use the almost Skyscraper with candidate (6) in rows 57. But for the (6) at d7, one would have a normal Skyscraper: (6): g7 = a7 - a5 = i5 => g6≠6.

Step 2i, Almost Skyscraper and ALS in AIC

The step illustrated above uses the almost skyscraper described previously, plus a simple ALS: (268)ef6. The step could be written:

  • (NP68)ef6 = (2)e6 - (2)d6 = (2-6)d7 = [(6): g7 = a7 - a5 = i5]
  • => g6≠6
Digression: Because each potential candidate in each SIS considered only uses weak inferences in one direction, this type of Kraken deduction is very simple. Moreover, it could also be described as simply a xyz chain, meaning that only the initial candidate need be considered as the appendage. To wit, one could write the same chain as follows:
  • (6*)g7 = [(6)i5 = (6)a5 - (6*)a7 = (6*-2)d7 = (2)d6 - (2)e6 = (NP68)ef6] => g6≠6

The next step is not illustrated. One could use the Locked 6's in box h5 first, then the naked triple in column i. However, I prefer the one step approach:
  • (HT568)i568 => i89≠6, i8≠79, i6≠7.
The next two steps are also not illustrated:
  1. (LC7)i13 => gh2≠7
  2. (LC7)ac2 => ac1,c3≠7
At this point, the puzzle is significantly undone, although we have not solved any more cells.


Step 2m: The very easiest of the Y Wing Styles. Also called a W Wing.


The name, W Wing has been poplularized for the following type of step. The key to a W wing is two cells limited to the exact same two candidates, plus a Bi Location link using one of those two candidates that sees both cells. Also illustrated below is step 2n.

Steps 2m and 2n, The W Wing type of Y Wing Style

The W Wing illustrated above is on the left of the puzzle:

  • (9=6)c3 - (6)abc2 = (6)d2 - (6=9)d8 => c8≠9
Also illustrated above, on the right side of the puzzle, is a short AIC. Step 2n:
  • (7)h6 = (7-5)g6 = (5)i6 - (5=8)i8 - (8=7)f8 => h8≠7


Step 2o: A Y Wing Style using only bi-locals.


Step 2o, Three Bi-locals type of Y Wing Style

Shown above, the newly bilocal candidate (7) in column h yields an easy AIC:

  • (7)h7 = (7-4)h6 = (4-2)d6 = (2)d7 => h7 ≠2


Step 2p: Reusing the ALS from step 2f in a short AIC


The step below can also be considered as being a typical ALS x ALS with a restricted candidate, in this case the restricted candidate is (1) in row 2.

Step 2p, A short AIC using an ALS

One could write this chain as

  • (2)d7 = (NP69)d78 - (NP69)d28 = (1)d2 - (1=2)g2 => g7≠2
In ALS x ALS jargon, it would look something like this:
  • (2=1)g2 - (1)d2 = (NT692)d287 => g7≠2
Finally, one gets to solve some cells. We arrive at UP 30 noting:
  1. (2)d7 % row
  2. (2)e6 % row, column, & box

A digression on MUGs

A MUG is a pattern which cannot be resolved uniquely, like a Unique Rectangle. However, a MUG is not generally restricted to bivalues. MUGs can be impervious, meaning that all the candidates in the MUG are locked in each of their large containers: Row, Column & Box. A MUG can be pervious, meaning that the candidates are not locked into all of their large containers. A pervious MUG can generally be represented as a Symmetric Pigeonhole Matrix of impervious Uniqueness patterns. This means that with a pervious MUG, at least one impervious Uniqueness pattern must exist.

The next step uses a pervious MUG, (123)gh259. However, since cells gh5 are already limited to only candidates (12), if this MUG were to exist, it would default to the impossible pattern:

  • (NP12)gh2
  • (NP13)gh5
  • (NP23)gh9
Without much deep thought, one can see that given this situation, the following is true:
  • (NP12)gh2 => nothing can happen in box h2, nor row 2 to resolve uniquely the contents of cells gh2
  • (NP13)gh5 => nothing can happen in box h5, nor row 5, to resolve uniquely the contents of cells gh5
  • (NP23)gh9 => nothing can happen in box h8, nor row 9, to resolve uniquely the contents of cells gh9
  • (NT123)g259 => nothing can happen in column g to resolve uniquely the contents of cells g259
  • (NT123)h259 => ditto for contents of cells h259 in column h.
    • => nothing can happen in the puzzle to resolve uniquely the contents of cells gh259
    • => the puzzle cannot be resolved uniquely
Given this fact, one can see that in order for this puzzle to have a unique solution, one of the following must be true:
  • (9)h2, (9)h9, (6)g9
These three possibilities, at least one of which must be true, are highlit in green below. They form a SIS used in a very short and simple chain, which happily happens to also be a wrap around chain, or continuous AIC loop. At this point in the puzzle, seeing this potential MUG is rather easy, as so many of the potential MUG cells have been depopulated of other candidates. BTW, this particular MUG has been available for quite some time. However, before now the potential chains emanating from it were longer and more complex, albeit still quite powerful.


Step 3: A short wrap around Y Wing style using a MUG


Step 3, A short continuous AIC using an MUG

Above, find that the grouping of (9)h29 in column h yields a nice chain:

  • Given (MUG123)gh259 => SIS[(6)g9, (9)h29] =>
  • (9)h29 = (6)g9 - (6)e9 = (6-9)d8 = (9)h8 loop
  • =>
    • (9)h8 = (9)h29 => h7≠9
    • (6)d8 = (9)d8 => nothing, as we already have d9 limited to only (69)
    • (6)g9 = (6)e9 => a9≠6
  • This leads to a cascade of singles, many of which are hidden, which brings the puzzle to UP 47
Alternatively, one could have used the following AIC, also continuous, to much the same effect:
  • (9)h29 = (6)g9 - (6)g7 = (6-9)a7 = (9)h7 loop => h8≠9, g8≠6, UP 47


Step 4: A short AIC using only candidate (6)


Step 4, simple coloring with candidate 6

The step above reduces the puzzle to naked singles:

  • (6): f3=d2-a2=a5-i5=i6 => f6≠6 => UP 81


Solution


Solution

Below, find the steps outside the scope of the Simple Sudoku Technique Set listed:

  • 1) Start with 23 givens, UP 28
  • 2f)
    • (AUR12)gh29 => SIS[(2)e9,(1)gh5] =>
    • (1)gh5 = (2)e9 - (2)d7 = (NP69)d78 - (NP69)d28 = (1)d2 - (1)gh2 = (1)i13 => i5≠1
  • 2i) (NP68=2)ef6 - (2)d6 = (2-6)d7 = [(6): g7 = a7 - a5 = i5] => g6 ≠6
  • 2m) (9=6)c3 - (6)abc2 = (6)d2 - (6=9)d8 => c8≠9
  • 2n) (7)h6 = (7-5)g6 = (5)i6 - (5=8)i8 - (8=7)f8 => h8≠7
  • 2o) (7)h7 = (7-4)h6 = (4-2)d6 = (2)d7 => h7≠2
  • 2p) (2=1)g2 - (1)d2 = (NP69)d28 - (NP69)d78 = (2)d7 => g7≠2 => UP 30
  • 3)
    • (MUG123)gh259 => SIS[(6)g9, (9)h29] =>
    • (9)h29 = (6)g9 - (6)e9 = (6-9)d8 = (9)h8 loop => h7≠9, a9≠6 => UP 47
  • 4) (6): i6 = i5 - a5 = a2 - d2 = f3 => f6≠6 => UP 81




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