A Decisive volley lobbed upon the Easter Monster

The following is the sixth page of an illustrated solution of the Easter Monster.

If this is your first visit to this blog, WELCOME!!

Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. Specifically, it may be helpful to have visited the following pages:

• Opening Volley on the Easter Monster
• Matrices in Sudoku
• Round 2 with the Easter Monster
• Easter Monster Round 3
• Fourth Chronicle of the Easter Monster Battle
• Easter Monster Round 5
• Forbidding Chains 101 The Theory. Note that I now use the term Alternating Inference Chain (AIC) to mean Forbidding Chain.

The illustrations of steps shown in this proof will share the new key style:

• 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
• Orange labels mark derived inferences
• Blue circles indicate proven strong inference set result
• Green circles indicate intermediate strong inference points
• Brown numbers indicate approximate chain order
• Other marks provided prn

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

Previous information

Previously, many derived inference groups were proven. This page will use the following sis group:

• Each set in Group Z below contains exactly one truth:
  • (27)r2c13,(16)r2c56,(16)r2c79
  • (16)c8r13,(16)c8r45,(27)c8r45,(27)c8r79
  • (27)r8c79,(27)r8c45,(16)r8c45,(16)r8c13

Notes

At this point in the puzzle, there are many, many ways to proceed. The greatest difficulty that I found after reaching this point was trying to decide which path to publish. After much deliberation, I decided that the path which follows would consume the fewest pages. Please note that from this point forward, there are many relatively easy eliminations that will not be illustrated, as they only served to lengthen the amount of space required to reach a solution.

Almost Unique Rectangles

Generally, I tend to shy away from arguments that use Uniqueness of Solution. However, after doing much pen and paper solving this fall waiting for my son's soccor games to start, I decided that AUR's make life much simpler in a great many cases. Usually, they are easy to see and trivial to execute.

However, with this puzzle they are a bit convoluted. Below I generate two different sis based upon two related, but different AURs. Although I looked at both simultaneously, the solution path is simpler if they are broken down into two disparite pieces. In fact, if one wishes to simply execute the Easter Monster with only two or three steps (from the beginning!), the simultaneous consideration of

• transported Kraken cells r7c9, r9c7
• AUR 45 at r89c17, AUR 45 at r78c37
• guardian 6's (done previously), and guardian 5's (to be done later)
• the Z group

could yield one mega step that proves a lot of stuff. Rather than use that confusing path, what follows is one of just many ways to defeat this beast.

Step 2n prelude using AUR transport.

Below, the potential AUR45 at r89c17 is circled red. In order for this AUR to not occur, something else must occur. This something else must be true. The group of these something elses forms a sis. The sis is graphed below.

The following sis is proven considering that AUR 45 c89r17 is FALSE:

• [(4)r34c1,(4)r1c7,(5)r78c39]

There are many ways to represent equivalent sis here. This one is chosen for three reasons:

1. regarding candidate 5, the 5's in r8 have potential interaction with candidates 17
2. Still with 5, the 5's in r7 have potential negative interaction with canidates 12 at r7c46
3. Candidate 4 at r4c1 and the other two 4's can be shown to interact with candidate 2

Step 2n (2)r4c5=(2)r4c5=(2)r4c5

I could have easily shortened this step by one strong inference considered by proving r4c5≠6 or r5c4≠2. However, it seemed more fun to prove an indentity for a strong inference set. This step boils down to:

• (2)r4c5=AUR45r89c17
• AUR false, thus (2)r4c5 True.

The mapping of all the weak links gets messy, so only the weak links effecting the transported AUR set are shown below. All the strong inferences considered are shown except the almost skyscraper on candidate 4. This almost skyscraper could also be called an almost finned X wing. At the same time, an almost two string kite on candidate 4 exists. All three target the same 4 at r7c2. The matrix representation could use any of these interchangeably.

One way to describe this particular long chain is a Block Triangular Matrix

2r4

c5

c1

4r4

c1

c3

2c5

r4

r8

Z27

r8c7

7r8c9

1r8

c5

c3

r7c2

1

4

8

4c9

r8

r7

r3

4r1

c3

c2

c7

Z27

r2c1

7r2c3

r1c2

8

7

4

AUR

4*

5*

5*

4**

5r7c39

r7c6

2

8

5

1

1B5

r6c6

r5c4

2B5

r4c5

r5c4

Once again, the asterisks are used to shorten the contents of the table cells. The double asterisk denotes a double forbidding that occurs by the top entry in that column. The Block portion of the matrix occurs when cell r7c2 is used. The almost 2 string kite restricts candidate 4, thus allowing one to treat candidate 8 in the matrix as if we had defaulted back to a plain Triangular Matrix. The result of the deduction:

• [(2)r4c5] is an sis
• => we can finally solve some cells:
  • (2)r4c5, (2)r2c1 %column,(7)r2c6 %cell, (7)r1c2 %row

Many easy deductions are available at this point. The puzzle is significantly uncracked! However, there remains a few difficult steps. Two easy ones come next.

Steps 3a, 3b. A fishy 4 elimination and the immediate result of Z

Below, two steps are illustrated at once.

The two steps as AIC:

1. Using Z, (2=7)r8c45 =>(2=7)r8c4 => r8c4≠6
2. (4):[Xwing r79c25]=r3c2-r9c3=r1c7 => r9c7≠4

At this point, since candidates 1267 and candidates 34589 break off into two almost symmetrical groups, one is almost certain that a relatively easy elimination exists considering the other AUR45 possible in Boxes 79.

Step 3c prelude using AUR transport.

Below, the potential AUR45 at r78c39 is circled red. In order for this AUR to not occur, something else must occur. This something else must be true. The group of these something elses forms a sis. The sis is graphed below.

Note the sis considering this AUR is symmetrical to the previous one:

• [(4)r14c3,(4)r3c9,(5)r89c17]

This representation is chosen for precisely symmetrical reasons as the previous AUR sis representation.

Step 3c sis[(2)r5c2,(6)r5c4,(6)r9c2] => r5c2≠6

If I had shortened the first AUR step, the symmetry would be a bit more obvious. Furthermore, since the 4's have gotten stronger, we now do not require a Block Triangular Matrix, as the almost step regarding candidate 4 loses the almost.

This step is still not quite simple enough to be handled well by pure AIC, thus consider the following Triangular Matrix:

2r5

c2

c8

2c7

r6

r8

4c7

r8

r1

6B7

r9c2

r8c1

r8c5

6

1

1r2

c5

c7

r4c7

1

8

r4c1

6

8

4

4r3

c9

c1

c2

r9c2

6

4

8

AUR

5*

4**

5*

4*

5r9c17

r9c6

8

5

6

6B5

r5c4

r6c6

Notice how closely symmetrical all the arguments here are to the previous AUR step. They are a bit simpler primarily because the other step was executed first. For example, we did not need the Zgroup here because that group became native in some cases. The entire deduction could be written as an AIC, but it would take many brackets. Primarily, the AUR sis begs the matrix approach. The proven sis => r5c2≠6.

Digression

At this point, there are even more relatively easy steps to take, including but not limited to: A short coloring loop on candidate 6, a wrap around AIC using candidates 16 and Z, then a swordfish on 1's, and also another short coloring loop on 1's. Rather than execute any of these, I shall proceed on the next page with a Kraken cell analysis of r9c7. Most of the results of this analysis will be symmetrical to the Kraken cell analysis of r7c9. The puzzle probably only needs one more page to elicit surrender.

Thanks for your patience!