Fourth Chronicle of the Easter Monster Battle

The following is page four of my ridiculous saga attempting to defeat the Easter Monster.

I will try to make the deductions on this page more clear.

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 Previous Entries. Specifically, it may be helpful to have visited the following pages:

The illustrations of steps shown in this proof will share a 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 Group Z.

  • 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

Step 2h using Block Triangular Matrix logic.

Block Triangular Matrix is an imposing name. It is merely a manner of logically ordering the inferences. This logical order serves as both a tool for explaining the elimination, and also for finding the elimination. Finding the elimination boils down to counting extra columns versus rows, and getting the count to zero.

Block Triangular matrix type situation because of an internal Almost Almost Skyscraper

Above I have tried to illustrate the actual chain. Please note, chains that operate on multi-valued cells, or on multi-location sets - have basically arbitrary partitions. One can always write such a chain in more than one fashion. Here is what I found:

  1. (2)r7c8=(2)r5c8 -(16)r5c8
  2. Given -(1)r5c8, the following almost skyscraper with candidate 1 exists:
    • (1)r7c4(the almost part)=[(1):r5c4=r5c2-r7c2=r7c6]-(1=6)r6c6
    • => almost (1)r7c4=(6)r6c6
    • Thus, we assign (1)r7c4 endpoint status, and continue with
    • (6)r6c6-(6)r5c4
  3. (6)r5c4=(6)r5c2-(6)r9c2
  4. Using Z, (6)r9c2=(1)r7c2-(1)r7c6
  5. Noting both 4) and 2), (1)r67c6=(1)r3c6-(1)r3c8 & -(2)r3c6
  6. Noting both 5) and 1), (1)r53c8=(1)r4c8-(67)r4c8
  7. Noting both 6) and 1), (6)r45c8=(6)r4c9-(7)r4c9
  8. Noting both 6) and 7), (7)r4c89=(7)r4c3-(7)r2c3
  9. (7)r2c3=(7-2)r2c6
  10. Noting both 5) and 9), (2)r23c6=(2)r7c6
The three endpoints are in bold italics, => r4c7≠2

Here is the same deduction written into a Block Triangular Matrix. Note how (1)r6c6 feeds back to the first column, while (6)r6c6 continues on as if this were a normal Triangular Matrix.

2c8 r7 r5
r6c6 1 6
1r7 c4c6c2
1r5 c8c4c2
6r5 c8 c4c2
Z r9c21r7c2
1c6 r6r7r3
1c8 r5 r3 r4
6B6 r5c8 r4c8r4c9
7r4 c8c9c3
7r2 c3c6
2c6 r7 r3 r2

Note how the 2x2 Almost Almost skyscraper on candidate 1 juts out in this matrix. Note further how it is resolved back into what is essential Triangular Matrix form.

Step 2i prelude.

Deriving a Guardian Strong Inference Set

Previously, I have disparaged the guardian technique as basically unnecessary. However, while tackling this particularly difficult puzzle, I decided to revisit the concept. The guardian idea is basically simple: One cannot color a shape that has an odd number of vertices with only two colors and achieve color alternation. Thinking in terms of true and false, if an odd number of exactly one is true sets form a loop, it is impossible to have a consistent solution. Once one places two truths into the loop, one strong inference set is left with zero truths. If one tries to place 3 truths in the loop, one weak inference set will contain two truths.

Guardian derivation

Above, the five locations for 6 that are labelled with a red asterisk form a potential impossible pentagon with 5s. In order for this situation to not occur, at least one of the four locations labelled g must contain a 6. Thus, we have a guardian sis.

One can then use this guardian sis just as if it is a native sis, except that the weak inference will not apply. Thus, first I perform the following operation on this sis:

  • (6): r8c5=[X Wing at r24c59]-(r4c1)=sis(6):[r1c6,r6c2,r9c4]
  • => (6)r8c5=sis(6):[r1c6,r6c2,r9c4]
  • => sis(6):[r1c6,r6c2,r9c4,r8c5]
Simply put, (6)r4c1 was removed from the sis, and (6)r8c5 was placed into the sis.

Note that this guardian inspired strong inference set has much overlap with a very strong candidate, candidate 7. Thus, it makes some sense to see if some 7 or some 6 at one of these intersections is ripe for elimination.

The use of the guardian set is not required to achieve this elimination. However, it makes the logic faster, and, more importantly, provides a clue of where to look. Since this idea is complex and unwieldy, it is probably supplanted by something more simple except in puzzles that are so utterly devoid of a suitable number of bivalue/bilocation sis. Also, and perhaps just as importantly, in this puzzle once one has some sort of bivalue/bilocation - they almost align so perfectly with each other that interaction between them takes a long look ahead. The guardian idea shortens the look-ahead requirement substantially.

Step 2i

Guardian Strong Inference Set used

Below, I have mapped out a relatively short deduction. Note that the weak inference of Z is used below in two locations, indicated in orange. The guardian derived sis is marked with green g's. Remember that the guardian sis is upon candidate 6.

Guardian used

Unfortunately, in order to lasso all four disparite pieces of the guardian sis, the deduction branches about a bit. Below find one desription:

  1. [(7)r1c2=(7)r6c2-[(6)gr6c2 & (2)Zr5c2]
  2. & (7)r2c6=(2)r2c6-(2)r7c6]
  3. =(2)r7c8-(2)r5c8
  4. ref 1)&3), (2)r5c28=(2)r5c4-(2)r4c5
  5. =(2)r8c5-[(6)gr8c5 & (7)Zr8c4]
  6. ref 1)&5) =[(7)r9c6=(7-g6)r9c4=g6r1c6]

Below, the same description as a TM.

7c2 r1 r6
r2c6 72
2r7 c6c8
2r5 Zc2c8 c5
2c5 r4r8
7B7 r9c6 Zr8c4r9c4
6g r1c6r6c2 r8c5r9c4

Thus, the Kraken-like pseudo=guardian sis 6:[r1c6,r6c2,r8c5,r9c4] is moved, or transported, into a new sis [(7)r1c2,(7)r2c6,(7)r9c6,(6)r1c6] => r1c6≠7


Guardians, and any technique like them, that allows one to build a new sis from existing information can be useful in locating possibly strong candidate interactions that may lead to finding an elimination.

This concludes this page. Once again, if you have read this far, thanks for your patience!

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