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 Sudoku Techniques. 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.
Above I have tried to illustrate the actual chain. Please note, chains that operate on
multivalued cells, or on multilocation sets  have basically arbitrary partitions. One
can always write such a chain in more than one fashion. Here is what I found:
 (2)r7c8=(2)r5c8 (16)r5c8
 Given (1)r5c8, the following almost skyscraper with candidate 1 exists:
 (1)r7c4(the almost part)=[(1):r5c4=r5c2r7c2=r7c6](1=6)r6c6
 => almost (1)r7c4=(6)r6c6
 Thus, we assign (1)r7c4 endpoint status, and continue with
 (6)r6c6(6)r5c4
 (6)r5c4=(6)r5c2(6)r9c2
 Using Z, (6)r9c2=(1)r7c2(1)r7c6
 Noting both 4) and 2), (1)r67c6=(1)r3c6(1)r3c8 & (2)r3c6
 Noting both 5) and 1), (1)r53c8=(1)r4c8(67)r4c8
 Noting both 6) and 1), (6)r45c8=(6)r4c9(7)r4c9
 Noting both 6) and 7), (7)r4c89=(7)r4c3(7)r2c3
 (7)r2c3=(72)r2c6
 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 
c4   c6  c2 
   
   
1r5 
 c8  c4  c2 
   
   
6r5 
 c8   
c4  c2   
   
Z 
   
 r9c2  1r7c2  
   
1c6 
   
r6   r7  r3 
   
1c8 
 r5   
   r3 
r4    
6B6 
 r5c8   
   
r4c8  r4c9   
7r4 
   
   
c8  c9  c3  
7r2 
   
   
  c3  c6 
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.
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 lookahead 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.
Unfortunately, in order to lasso all four disparite pieces of the guardian sis, the deduction
branches about a bit. Below find one desription:
 [(7)r1c2=(7)r6c2[(6)gr6c2 & (2)Zr5c2]
 & (7)r2c6=(2)r2c6(2)r7c6]
 =(2)r7c8(2)r5c8
 ref 1)&3), (2)r5c28=(2)r5c4(2)r4c5
 =(2)r8c5[(6)gr8c5 & (7)Zr8c4]
 ref 1)&5) =[(7)r9c6=(7g6)r9c4=g6r1c6]
Below, the same description as a TM.
7c2 
r1 
r6 





r2c6 
7   2  
  
2r7 
  c6  c8 
  
2r5 
 Zc2   c8 
c5   
2c5 
   
r4  r8  
7B7 
r9c6    
 Zr8c4  r9c4 
6g 
r1c6  r6c2   
 r8c5  r9c4 
Thus, the Krakenlike 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
Digression
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!