Solution for Tough of November 28, 2007 Page Three

The following is page three of an illustrated solution for the Tough Sudoku of November 28, 2007. This page has one step which is interesting. It employs

• The recognition of two usefully linked Almost Y Wing Styles

Page One of this solution was primarily about the use of an Almost Unique Rectangle in a chain. Page Two of this solution included some typical steps, plus a complex linking of several disparite types of Almost AICs. All of the Sudoku tips and tricks contained on this page are straightforward, albeit the last one perhaps less so.

Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques.

Step 4c - Y wing style with candidates 2 & 7 & 9

Above, find

• (9)r1c2=(9-2)r2c2=(2)r1c1-(2=7)r1c7 => r1c2≠7
• Alternate writing: (Hidden pair 29)r12c2=(2)r1c1-(2=7)r1c7 => r1c2≠7

The difference between the two writings of this step given above are subtle. It is helpful to be aware of both. This is especially true if one employs counting with deeper, more complex chains or nets. Candidate 7 is ripe for a few easy eliminations.

Step 4d - a short chain using an Almost Hidden Pair

Step 4e - Y wing style with candidates 2 & 7

Above, another Y Wing style quickly dispenes with another 7:

• (7=2)r1c7-(2)r1c1=(2-7)r6c1=(7)r13c1 => r1c3 ≠7

It happens with relative frequency that one elimination, such as the previous forbidding of (6) from r1c7, quickly leads to a host of eliminations using many similar or stacked strong inferences. By stacked, I mean that after making one elimination, another one follows easily. This puzzle seems to be ripe with Y Wing Style (YWS) type configurations. Even step 4d feels like a depth 3 YWS with the consideration of the Super Cell (8)r67c1.

Step 4f - Almost Hidden Pair (29-69)

Below find a longer deduction that sets up a puzzle breaking step

The graphic above could be written:

• (7)r9c8=(7)r3c8-(7)r1c7=(7-2)r1c1=(HP29-HP69)r12c2=(6)r8c2-(6)r9c3=(6)r9c6
• => r9c6≠7

An alternate, but similar, way to the same elimination was available immediately after step 4b:

• (7)r9c8=(7)r3c8-(7=2)r1c7-(2)r1c1=(HP29-HP69)r12c2=(6)r8c2-(6)r9c3=(6)r9c6
• => r9c6≠7

The only difference between the two chains should appear bold.

When writing chains, I often prefer to use something similar to (HPxy-HPxz) as it maintains the symmetry of the Forbidding Chain or Alternating Inference Chain. Thus, the chain reads perfectly well in either direction.

Step 4g Prelude 1 - Almost Y Wing Style 63

Below, the existence of the two disparite cells r9c6, r1c3 both limited to only (36) is a typical marker for one to look for a very common and easy to find type of Y Wing Style. This particular YWS almost exists, as graphed below.

I think of a Y Wing Sytle as a Boolean, just like any piece of a strong inference set. Thus, in my mind the Almost YWS shown above reduces to:

• A=(3)r4c4
• Where A is not an Almost anything, but rather a Y Wing Style that is either true or false.
• Thus A is prefectly symmetrical, logically, with (3)r4c4, which is also either true or false.

Step 4g Prelude 2 - Almost Y Wing Style 38

Below, this Almost Y Wing Style is a bit harder to find. The previous AYWS indicates that one might benefit by looking for this one.

Above, let B be the YWS 38 shown. Then:

• (3)r5c3=B
• Very significantly, B-(3)r4c4
• Recognizing what B forbids, the target of B, links the two preludes shown
• We have already considered (3=6)r1c3 in the first YWS labelled A, thus
  • B's existence (truth state) is linked to A at both ends of B.
• The sixes at r9c6, r1c3 are thus certain to be a derived sis

Step 4g - Y Wing Style 38 & Y Wing Style 36 linked

Below, the graphic may be messy, but the logic should be clear.

Using the substitutions A,B indicated in the preludes, the following short chain exists:

• A=(3)r4c4-B=(3)r5c3-(3=6)r1c3 => r9c3≠6

Interestingly, we thus have a Derived Y Wing Style that contains two native Y Wing Styles. I really enjoy such coincidences!

There really is no need for counting, but just as with the ALS step 3d, the AIC presentation gives the illusion of using one strong inference set twice. This really is not the case. One need consider (3=6)r1c3 only once when one counts. Unfortunately, counting tends to distroy the feel of elegance yet again:

1. (6=3)r1c3 +1.
2. (6=3)r9c6 +0 if we choose either 6 or 3 in the result column
3. (3):r78c4, r1c4 ,r4c4 +0 if we now assign 6 into the result column
4. (8):r4c4=r6c4 +0
5. (8):r6c12=r5c2 +0
6. (3):r5c3, r5c2, r4c1. Note all 3s in box 4 link to a previous entry. Thus, -1. Total zero
7. proven sis: (6):r1c3=r9c6 => r9c3≠6

One can start the counting anywhere. In this case, given (36) at r9c6 & r1c3, it makes perfect sense to investigate the count just as I did above. Note one could have tried placing (3) in the result column and looking at an Almost Y Wing style using perhaps (6)Box2. One can, by counting, find that r9c3≠3. However, that deduction is longer, more complex, and less powerful.

After r9c3≠6, two cells solve:

• (6)r9c6 %row
• (6)r6c5 %column
• Thus we are at UP 29
• The puzzle now solves with straightforward classical AIC

This concludes the third page of this solution. Eventually, the link to the clean-up page will become active.