The following is the sixth 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 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
 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 group:
 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
Notes
At this point in the puzzle, there are many, many ways to proceed. The greatest difficulty that
I found after reaching this point was trying to decide which path to publish. After much deliberation,
I decided that the path which follows would consume the fewest pages. Please note that from
this point forward, there are many relatively easy eliminations that will not be illustrated, as
they only served to lengthen the amount of space required to reach a solution.
Almost Unique Rectangles
Generally, I tend to shy away from arguments that use Uniqueness of Solution. However, after
doing much pen and paper solving this fall waiting for my son's soccor games to start, I decided
that AUR's make life much simpler in a great many cases. Usually, they are easy to see and trivial
to execute.
However, with this puzzle they are a bit convoluted. Below I generate two different sis based
upon two related, but different AURs. Although I looked at both simultaneously, the solution
path is simpler if they are broken down into two disparite pieces. In fact, if one wishes to
simply execute the Easter Monster with only two or three steps (from the beginning!), the simultaneous consideration
of
 transported Kraken cells r7c9, r9c7
 AUR 45 at r89c17, AUR 45 at r78c37
 guardian 6's (done previously), and guardian 5's (to be done later)
 the Z group
could yield one mega step that proves a lot of stuff. Rather than use that confusing path, what
follows is one of just many ways to defeat this beast.
Step 2n prelude using AUR transport.
Below, the potential AUR45 at r89c17 is circled red. In order for this AUR to not occur,
something else must occur. This something else must be true. The group of these something elses
forms a sis. The sis is graphed below.
The following sis is proven considering that AUR 45 c89r17 is FALSE:
 [(4)r34c1,(4)r1c7,(5)r78c39]
There are many ways to represent equivalent sis here. This one is chosen for three reasons:
 regarding candidate 5, the 5's in r8 have potential interaction with candidates 17
 Still with 5, the 5's in r7 have potential negative interaction with canidates 12 at r7c46
 Candidate 4 at r4c1 and the other two 4's can be shown to interact with candidate 2
Step 2n (2)r4c5=(2)r4c5=(2)r4c5
I could have easily shortened this step by one strong inference considered by proving r4c5≠6 or r5c4≠2. However,
it seemed more fun to prove an indentity for a strong inference set. This step boils down to:
 (2)r4c5=AUR45r89c17
 AUR false, thus (2)r4c5 True.
The mapping of all the weak links gets messy, so only the weak links effecting the transported
AUR set are shown below. All the strong inferences considered are shown except the almost
skyscraper on candidate 4. This almost skyscraper could also be called an almost finned X wing.
At the same time, an almost two string kite on candidate 4 exists. All three target the same 4
at r7c2. The matrix representation could use any of these interchangeably.
One way to describe this particular long chain is a Block Triangular Matrix
2r4 
c5 
c1 












4r4 
 c1  c3  
   
   
 
2c5 
r4    r8 
   
   
 
Z27 
   r8c7 
7r8c9    
   
 
1r8 
   c5 
 c3   
   
 
r7c2 
   
 1  4  
8    
 
4c9 
   
r8   r7  r3 
   
 
4r1 
  c3  
  c2  c7 
   
 
Z27 
 r2c1   
   
 7r2c3   
 
r1c2 
   
   
8  7  4  
 
AUR 
 4*   
5*  5*   
  4**  5r7c39 
 
r7c6 
   2 
   
8    5 
1  
1B5 
   
   
   
r6c6  r5c4 
2B5 
r4c5    
   
   
 r5c4 
Once again, the asterisks are used to shorten the contents of the table cells. The double asterisk
denotes a double forbidding that occurs by the top entry in that column.
The Block portion of the matrix occurs when cell r7c2 is used. The almost 2 string kite
restricts candidate 4, thus allowing one to treat candidate 8 in the matrix as if
we had defaulted back to a plain Triangular Matrix. The result of the deduction:
 [(2)r4c5] is an sis
 => we can finally solve some cells:
 (2)r4c5, (2)r2c1 %column,(7)r2c6 %cell, (7)r1c2 %row
Many easy deductions are available at this point. The puzzle is significantly uncracked! However,
there remains a few difficult steps. Two easy ones come next.
Steps 3a, 3b. A fishy 4 elimination and the immediate result of Z
Below, two steps are illustrated at once.
The two steps as AIC:
 Using Z, (2=7)r8c45 =>(2=7)r8c4 => r8c4≠6
 (4):[Xwing r79c25]=r3c2r9c3=r1c7 => r9c7≠4
At this point, since candidates 1267 and candidates 34589 break off into two almost
symmetrical groups, one is almost certain that a relatively easy elimination exists
considering the other AUR45 possible in Boxes 79.
Step 3c prelude using AUR transport.
Below, the potential AUR45 at r78c39 is circled red. In order for this AUR to not occur,
something else must occur. This something else must be true. The group of these something elses
forms a sis. The sis is graphed below.
Note the sis considering this AUR is symmetrical to the previous one:
 [(4)r14c3,(4)r3c9,(5)r89c17]
This representation is chosen for precisely symmetrical reasons as the previous AUR sis representation.
Step 3c sis[(2)r5c2,(6)r5c4,(6)r9c2] => r5c2≠6
If I had shortened the first AUR step, the symmetry would be a bit more obvious. Furthermore,
since the 4's have gotten stronger, we now do not require a Block Triangular Matrix, as the almost
step regarding candidate 4 loses the almost.
This step is still not quite simple enough to be handled well by pure AIC, thus consider
the following Triangular Matrix:
2r5 
c2 
c8 











2c7 
 r6  r8  
   
   

4c7 
  r8  r1 
   
   

6B7 
r9c2    
r8c1    
   

r8c5 
   
6  1   
   

1r2 
   
 c5  c7  
   

r4c7 
   
  1  8 
   

r4c1 
   
6    8 
4    

4r3 
   c9 
   
c1  c2   

r9c2 
   
6    
 4  8  

AUR 
  5*  4** 
5*    
4*    5r9c17 

r9c6 
   
   
  8  5 
6 
6B5 
r5c4    
   
   
r6c6 
Notice how closely symmetrical all the arguments here are to the previous AUR step. They
are a bit simpler primarily because the other step was executed first. For example, we did
not need the Zgroup here because that group became native in some cases. The entire
deduction could be written as an AIC, but it would take many brackets. Primarily, the
AUR sis begs the matrix approach. The proven sis => r5c2≠6.
Digression
At this point, there are even more relatively easy steps to take, including but not limited
to: A short coloring loop on candidate 6, a wrap around AIC using candidates 16 and Z, then
a swordfish on 1's, and also another short coloring loop on 1's. Rather than execute any
of these, I shall proceed on the next page with a Kraken cell analysis of r9c7. Most of
the results of this analysis will be symmetrical to the Kraken cell analysis of r7c9. The
puzzle probably only needs one more page to elicit surrender.
Thanks for your patience!
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