The following is page three of my campaign versus 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 four pages first:
The illustrations of steps shown in this proof will share this key:
 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
Please be aware that, for me,
strong and weak need not be mutually exclusive properties.
Previous information
Previously, the following derived strong and weak inference groups were proven.
 Each set in Group Z below contains exactly one truth:
 (27)r2c13, (27)r2c56, (16)r2c56,(16)r2c79
 (16)c8r13,(16)c8r45,(27)c8r45,(27)c8r79
 (27)r8c79,(27)r8c45,(16)r8c45,(16)r8c13
 Group X is a strong inference set(sis): (2)[r8c5,r5c2,r8c7]
Before making the next elimination, I will add two more such groups.
Group U  using an Almost Almost Two String Kite
The following sets are derived sis U
 [(1):r2c5,r6c2,r7c2]
 [(1)r2c5,(1)r6c2,(6)r8c1]
By now, hopefully the idea is clear. Write the conclusion and the appendages that could prevent
the conclusion as a strong inference set. The proof could look like:
 (1): [r2c5=r6c2]=[r8c5=r4c5r6c6=r6c3]r8c3=r7c2 => [(1)r2c5,r67c2] is a sis.
 Using Z:(1)r8c3=(6)r8c1 => [(1)r2c5,(1)r6c2,(6)r8c1] is a sis.
Group T  using an AAA Y Wing Style.
The following set is derived sis T
 [(4)r3c2,(4)r8c1,(4)r8c7,(14)r8c3]
This one seems quite a stretch  3 almostings. The previous deduction
U in conjuction
with
Z made looking for this one seem logical at the time.
The AIC justification reads:
 sis[(4):r3c2,r8c17]=[(4)r8c3=(4)r8c9(4)r3c9=(45)r3c1=(5)r1c3](5)r8c3=(14)r8c3 =>
 sis[(4)r3c2,(4)r8c137,(1)r8c3]
13x13 TM using derived and native strong inference sets
Step 2c
The graphic above does not dispay all the information used. I graphed some of the main ideas.
Again, a Triangular Matrix below will serve as both the proof and illustration of the elimination.
2B1 
r2c1 
r3c2 










2X 
r8c5  r5c2  r8c7  
   
   

Z 
 2r56c2   7r6c2 
   
   

1U 
r2c5    r6c2 
r7c2    
   

Z 
   
1r8c3  6r8c1   
   

T 
 4r3c2  4r8c7  
1r8c3  4r8c1  4r8c3  
   

4r4 
   
  c3  c1 
   

4r3 
 c2   
   c1 
c9    

4r7 
   
  c23  
c9  c5   

r9c2 
   
 6  4  
  8  

8r7 
   
   
 c5  c23  c6 

2c1 
r2    
   r4 
   
r6 
2c6 
r23    
   
   r7 
r6 
Long, messy, and a tad convoluted. I suspect 'tis the nature of this beast. r2c5≠2 lets
us make some easy eliminations.
Steps 2d, 2e.
Above we have two steps illustrated:
 2d: Using Z, (2=7)r2c56, note r2c5 now is void of [27] => r2c6≠16
 2e: (2):r4c5=r8c5r8c7=r7c8 => r4c8<>2
Step 2f. Single candidate fish.
This elimination is symmetric to V done on candidate 1. The appendage has been eliminated,
so the derivation makes the elimination. As AIC:
 (6): [r2c9=r2c5[r6c1,r4c5]=[r4c89=r4c1r56c2=r9c2r9c6=r6c6] =>
 sis(6):[r2c9,r4c89,r6c6] => r6c9≠6
At this point, the Almost Almost Naked Pair 39 at r6c79 caught my attention. Because of
the super celllike relationship (27)at r45c8, the potential for an odd Sue de Coq using
super cells reared its head. If I could have found such an animal, it would have been very cool.
Unfortunately, I was unequal to the task. Nevertheless, looking in that direction lead me directly
to a fairly simple elimination (for this puzzle!).
Step 2g  a Three way pincer with candidate 2.
Above, I have graphed all the strong inference sets considered to make this elimination. After
finding this elimination, I found numerous ways to reach the same elimination. Some are a bit
quicker, but none as cool  imho. Placing all the weak links into the diagram proved difficult for me.
The logic, as AIC, looks like this:
 [2:r4c5=r8c5r8c7=r7c6] =
 [(2)r4c7[(1)r4c7,(2)r5c8,(2)r6c7]
 = Z(7)r4c8=(pair39)r6c79 &
 = (1)r7c2(1=6)r2c5(6)r1c6=[(6):r6c6=r9c6r9c2=r1c8]]
 (639=2)r6c1
 reducing nicely to sis2g 2:[r4c5,r7c6,r6c1] => r6c6≠2
This is a tad messy to follow as AIC. It should be crystal clear as a Mixed Block Matrix.
The matrix below is called Mixed Block only becuase of the use of a contained Pigeonhole
matrix  pair 39. To shorten the matrix, I treat this as one Boolean. This
reduces the Matrix to a simple Triangular Matrix.
2c5 
r4 
r8 







2c7 
r6  r8  r4  
   

Z 
  2r5c8  7r4c8 
   

r6c79 
  2  7 
pair39    

1c7 
  r4  
 r2   

r2c5 
   
 1  6  

r4c1 
2    
39    6 

6r6 
   
  r1  r4 
r9 
6B7 
   
   c1 
r9 
As noted previously, this derives sis 2g 2[r4c5,r6c7,r4c1] => r6c6≠2.
This sis may be useful later.
One could justify the same elimination, considering the same strong inference sets, but thinking
roughly in terms of
 [naked triple 239]r6c179=
 [(2=7)r56c8  (7)r6c9=(6)r6c1(6's) ... =(1)r4c7]
 (2)r4c7 = [skyscraper with 2's]
 => sis2g => r6c6≠2
There are, as I noted, some other paths that consider only a few of the same strong inference sets.
Those are also fairly interesting, but I lack the time to expose every possible elimination path.
This concludes this page. There remains many difficult eliminations, unless I find
a simpler path before I publish. Hope springs eternal!.
If you have read this far, thanks for your patience!