The following is the fifth 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:
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 groups:
 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
 The result column sis from step 2g
Step 2j prelude using Kraken Cell Transport.
Since this puzzle resists most normal attempts to crack it, it seemed likely that
some research might be required into relationships to locate good attacks. On the previous
page, the guardian sis was used to locate a likely chain. Here, the relative symmetry of candidates
345 in relation to cells r7c9 and r9c7 caused me to create a transport list out of each cell.
Basically, the goal is take the sis inherent in a cell and find some equivalent sis at least
partially out of the cell. Although I created many such lists, some of them seemed more likely
to be fruitful than others. In this case, the transport list for r7c9 seemed more likely to be
fruitful than the one for r9c7, primarily because candidate 67 interactions uncovered in the guardian step.
Above, I am illustrating three short transportation chains.
 (349=5)r7c9(5)r7c46=(5)r9c46
 (459=3)r7c9[(3):r79c8=r3c8r3c4=r7c7]=sis(3):[r1c48,r9c4]
 Note that there are many such chains possible, but I wanted one
that interacted strongly with canidates 67.
 (359=4)r7c9[(4):[Xwing r79c25]=r1c2r1c7=r3c9]=(4)r3c2
 This one is nice, as it requires only a single value transported out
PUtting all three of these together we get:
 Old sis (3459)r7c9 =>
 sis[(35)r9c4,(5)r9c6,(3)r1c4,(3)r1c8,(4)r3c2,(9)r7c9]
This new sis seems a bit unwieldy, but a short look at some of the previous chains considered makes
it very promising. The only item that cannot be resolved easily is that darn (9)r7c9. However, one
may see a potential logjam primarily because of sis[9 column 3].
Step 2j (2)r2c1=(6)r6c1
Below, the transported portions of cell r7c9 are labelled with a green k. The rough order of
chained sis are marked with brown numbers. Only strong inference sets are marked. The Z argument
group is used twice, and labelled accordingly.
The easiest way to describe such a long chain is probably a Triangular Matrix
2B1 
r2c1 
r3c2 













Z27 
 r65c2  7r6c2  
   
   
  
7r1 
  c2  c4 
   
   
  
7c6 
   r2 
r9    
   
  
6r6 
c1   c2  
 c6   
   
  
1r6 
  c2  
 c6  c3  
   
  
1r8 
   
  c3  c45 
   
  
6B8 
   
r9c6    Zc8 
r9c4    
  
6r1 
   c4 
 c6   
 c8   
  
2jK 
 *   * 
*    
*  *  9r7c9  
  
r6c9 
  7  
   
  9  3 
  
r2c9 
   
   
 6   3 
8   
r2c1 
 2   
   
   
8  3  
3B4 
   
   
   r6 
 c1  r5c3 
9c3 
   
  r6  
  r7  
  r5 
It is a bit unusual for a Triangular matrix of this length to have only two items listed in
the first column. The result column reads:
This leads to a few relatively easy eliminations, although all of them use proven sis.
Step 2k as an immediate result of <2j> and <2g>
Below, recall step 2j. Proven in that step was sis(2)[r6c1,r7c6,r4c5]. However, we
have just shown that (2)r6c1 is false, therefor we can update the strong inference set to:
The logic for this step is fairly simple
 (2)r4c6=(2)r6c7 => r4c7≠2
Remembering what has been already proven certainly makes that one a breeze.
Step 2L as a Y Wing Style using 2j
Below, recall 2j sis [(2)r2c1=(6)r6c1]. Add a pair of native strong inferences.
The AIC for this one could be written:
 (1=6)r6c6(6)r6c1=2j(2)r2c1(2)r4c1=(2)r4c5 => r4c5≠1
Step 2m as a depth 4 chain using 2j and Z
Below, recall 2j sis [(2)r2c1=(6)r6c1]. Add one native strong inference and use
strong inferences from the Z group twice.
The AIC for this one could be written:
 (1)r8c3=Z(6)r8c1(6)r6c1=2j(2)r2c1(2)r4c1=(2)r4c5(2)r8c5=Z(27)r8c4
 => (1)r8c3=(2=7)r8c4 => r8c4≠1
After just a little practice, using derived strong inference sets seems just as natural as using
native strong inference sets. Although the structure of a derived strong inference set may be
a tad unconventional, the logic is precisely the same.
Digression
Some large and unconventional weaponry seems warranted with this particular puzzle. The practical
use of most of these ideas will be very small. However, minicases of the same logic type are
possible. In all but a very few puzzles, the depth and complexity found here will not be encountered.
I have a few more unconventional attacks to attempt. On the next page, I probably will employ
an sis derived from an Almost Unique Rectangle  more aptly named, I think, a Forbidden Rectangle.
Thanks for your patience!