The following is the final page of an illustrated campaign to solve the Easter Monster
sudoku puzzle. Hopefully, this Easter Monster Solution is at least interesting to some!
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 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
 Other marks provided prn
Please be aware that, for me,
strong and weak need not be mutually exclusive properties.
Step 3d prelude using Kraken cell transport.
Below, two of the three possible values of cell r9c7 are transported out to form a new
derived sis. This is very similar to the transportation of 3 of 4 values out of cell r7c9
that was performed as a prelude to step
2j. The entire deduction is
so symmetrical, in fact, that it becomes almost easy. Certainly, given step 2j, it
was easy to find.
The sis (359) r9c7 transported into the sis:[(3)r3c48, (35)r7c6, (5)r7c6, (9)r9c7] using:
 (39=5)r9c7(5)r9c46=(5)r7c46
 (59=3)r9c7(3):[r79c8=r1c8r1c4=r9c4]=[(3)r3c48,(3)r7c4]
By now, this process should be fairly clear.
Step 3d (2)r5c2=(6)r6c2
Not only is the Kraken transport symmetrical to that used in step 2j, but the conclusion
of that step and the conclusion of this step also have a very nice almost symmetrical aspect about
them. Furthermore, the entire argument is very similar, as demonstrated below.
2r5 
c2 
c8 












2r7 
 c8  c6  
   
   
 
2r3 
  c6  c4 
   
   
 
6r5 
 c8   
c4    
   
 
1c4 
   r3 
r5  r7   
   
 
1c5 
   
 r8  r2  
   
 
1B3 
   
  r2  r3c8 
   
 
K3d 
  5*  3* 
 35*   3* 
9r9c7    
 
r6c7 
 2   
   
9  3   
 
r2c7 
   
  1  
 3  8  
 
r2c3 
   
   
  8  3 
 
3B4 
   
   
 r6   c3 
r5c1  
9c1 
   
   
r9    
r5  r6 
6r6 
c2    
c6    
   
 c1 
The conclusion of this step:
 (2)r5c2=(6)r6c2 => r6c2≠2
 All the 2's and all the 7's are then solved by hidden singles.
The puzzle is now solvable in myriad manners. To conserve space, I decided
to use a step that works quickly, but is more complex than the puzzle now warrants.
Step 4 prelude using guardian 5's and guardian sis transport.
Below, if candidate 5 is limited to the locations circled red, then a number of
impossible pentagons exist. Such overlap of impossible oddagons is usually the case if
one were to consider using guardians to derive sis. The possible locations for candidate 5
that are labelled with a green G form the guardian sis. However, since r8c7 is limited to
45, and since it sees four members of the guardian set, it is much more efficient to transport
the (4)r8c7 into the sis, and remove four of the 5's.
The guardian sis and transportation logic follows:
 Guardian sis (5):[r1c4, r7c6, r8c13,r7c9,r9c7]
 The AIC: (4=5)r8c7(5):[r8c13,r7c9,r9c7]=(5)[r1c4, r7c6]
 => sis[(4)r8c7,(5)r1c4,(5)r7c6]
From this point, the deduction is fairly simple (for this puzzle!)
Step 4 chains transported guardian sis to derive (5)r1c4=(6)r5c4
Below, but for the trivalue transported sis and the double forbidding done by the potential
6 at r6c6, this would be nothing but a standard AIC. Finally, I am able, however, to transparently map all of
the weak inferences, and all but the transported guardian strong inference set.
As AIC, one could write:
 (5)r1c4=G[AIC](6)r6c6=(6)r5c4
 [AIC]: (6)r6c12=(64)r4c1=(4)r4c3(4)r1c3=(4)r1c7(4)r8c7=G(51)r7c6=(1)r6c6
Below find a short TM (triangular matrix) that performs precisely the same task.
6B5 
r5c4 
r6c6 




6B4 
 r6  r4c1  
 
4r4 
  c1  c3 
 
4r1 
   c3 
c7  
1c6 
 r6   
 r7 
TG 
5r1c4    
4r8c7  5r7c6 
The conclusion: (5)r1c4=(6)r5c4 => r1c4≠6. This now solves many cells by naked
and hidden singles. All the sixes, all the ones, all the fours, and a few of the others.
Step 5 short skyscraper with candidate 9
Above, the standard run of the mill AIC:
 (9): r5c3=r7c3r7c9=r6c9 => r6c1≠9 & r5c7≠9
The puzzle now solves using only naked singles.
Easter Monster Solution
Parting Shots
This puzzle was more difficult than I care to tackle often. However, the ideas that it
forced upon me may prove to be useful in the future when faced with a puzzle devoid of
suitably bivalue, bilocation sis. The main ideas used herein, summarized:
 Derived sis can be used again later. Even very complex ones help to focus searches.
 Any tool in the toolbox is suitable for the creation of derived sis
 Matrices are not only a useful tool for explaining a deduction, but
the concept of counting that they infer is useful in locating long, complex chainlike nets.
The matrix type,
Mixed Block Matrix is an invention that can reduce matrix size. Although
Andrei and Bruno have proven that all sudoku eliminations can be justified using
Block Triangular
Matrices, Mixed Block Matrices can significantly shorten the deductive path. Moreover, they
can more efficiently prove all of the derived sis available from one set of native sis. For this
reason, they show some promise towards a more efficient resolution of a puzzle such as the
Easter Monster. I am fairly certain that just such a more efficient and quicker solution
path exists for this puzzle. By quicker, I mean
much quicker!
Thanks for your patience with this ridiculous solving path!
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