The following is page two of my attempt to wrestle 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 three 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 group was proven. This group will be
labelled Z when used in a step:
 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
 (16)c2r79,(16)c2r56,(27)c2r56,(27)c2r13
In order to efficiently tackle this puzzle, it seemed wise to derive a few more groups. However,
most of the groups derived will differ from this group. Most of the groups will be groups of
strong inferences. The difference being that although the groups will have at least one truth,
they may have more than one.
The reason for maintaining a list of strong inferences rather than a list of weak inferences is
simple: As the puzzle progresses, the meaning of a derived weak inference could become less important. However,
all strong inferences, proven or native, are valid for all time.
Suppose that a derived strong inference set has three members. If one is eventually proven false,
the group is still valid. However, it has now been reduced to two members. If one of the remaining
two members is proven false, the group is resolved. This is not the case, of course, with weak
inferences.
More derived strong inference sets
Group Y
The following set is derived strong inference set Y
 [(7)r1c2,(8)r2c9,(8)r1c7,(8)r2c7]
The derivation could be written:
 (7)r1c3=[(4)r3c9=(4)r1c7(4=8)r1c2(8)r1B2=(8)r3B2](8)r3c9=(8)r2c9,r12c7
 which reduces to: (7)r1c3=(8)r2c9,r12c7
Although this does not immediately forbid anything, it will be of some use.
Digression
First, I would like to thank those who have plowed ahead of me with the following concepts:
 Moving Strong Sets  Bruno Greco
 Proving Loops  David P B
 Transport  I am not certain of the author (let me know?)
Next, I should like to apologize for the bastardization of their ideas for my own personal take
on how to employ them.
It has been argued that one cannot make a deduction like the one above without making assumptions.
This is a false conclusion. The Triangular Matrix below demonstrates that one need not
consider multiple truth states to reach the deduction above.
8Box3 
r12c7r2c9 
r3c9 


4Box3   r3c9  r1c7  
r1c2  7   4  8 
8Box2   r3   r1 
In this case the logic reduces to:
Group X
The following set is derived strong inference set X
The logic could be written:
 (2):[r5c2=r8c5]=[r8c7=r7c8r5c8=r5c4r8c4=r8c7]
 The last bracketted entry reduces to: r8c7
 Therefor: (2)[r5c2,r8c5,r8c7] is a valid strong inference set
Digression
I have mentioned previously that looking for chains can be reduced to a counting exercise.
I also noted that one can always force the count to zero. Here is an example of
the thought process executed on the example above:
 Start at (2)Box7.2 locations. Think matrix row 1, 2 columns, thus +1.  we need
to get to an nxn matrix, thus we need only count excess columns.
 Note weak link to one of three 2's in row 5. One 2 aligns, two do not. We are
adding 1 matrix row, but two columns. Thus +1. Total: +2.
 Note weak link from one 2 in row 5 to one 2 in row 8. Note one 2 in row 8
is the same as one in Box7. There are 3 2's. Two can align. Think +0. Total +2.
 Decide that column 1 should contain (2)r8c7
 Force the matrix to zero by taking the two unlinked items and shoving them into
column 1. Think 2. Total 0.
 Record conclusion
The short form of the thought process could be:
 add row +1,add row +1,add row +0, compress 2, total:0 => done
The process has the look and feel of an algorithm.
Group W
The following sets are derived strong inference sets W
 [(2)r4c7,(8)r3c9,(1)r24c7
 [(2)r4c7,(8)r3c9,(68)r2c9
 [(2)r4c7,(8)r3c9,(8)r124c7
The logic could be represented as the following Almost Almost wrap around Y Wing Style:
 [(2)r4c7=(8)r3c9]=[(1=8)r4c7(8)r12c7=(86)r2c9=using Z(1)r2c7]
 Since the second bracketted chain is a wrap around chain, each weak link is proven
strong. This justifies the conclusions written above
Note that this simple chain uses the derived set Z. Without that efficiency, the chain
would be much more complex, and involve using 11 native strong inferences instead of only 3.
Here is another example of the counting process:
 Consider trivalue cell r4c7. +2
 See weak link to (1=6)r2Box7. +0
 See both sets above weakly link to three of four values for (8)Box7.
Symetric Pigeonhole Matrix in columns 2 thru 4. Place (8)r3c9 under (2)r4c7 column 1.1
 Note total: +1. Take column 1 and make it one matrix cell. Shove this cell into
any matrix column. Realign matrix so that the modified matrix column is column 1.(1)
 Total 0. (note we now have one of three possible pigeonhole matrices, but none
of these three is a symmetric pigeonhole matrix, as column 1 is not mutually exclusive.
 Note by symmetry the set of three possible PM's. Note the set of three strong inference sets generated. Record.
Group V
The following set is derived strong inference set V
The logic for this one could run like this:
 (1): [r2c6]=[r2c7=r2c5[r48c5&r23c6]=[r4c78=r4c3r8c3=r8c4r7c6=r6c6]]
 reduces to:(1): r2c6=[r2c7=[r4c78=r6c6]]
 Add a pair of links: (1): r2c6=[r2c7=[r4c78=r6c6]]r6c7=r6c24
 The conclusion reduces to:(1):r2c6=r24c7
Again, this could be nothing more than a counting exercise. To wit:
 (1)r2, three locations. Could be grouped. +1 or +2
 (1)c6, four locations. Could be grouped. Decide temporarily that:
 (1)r2 is +2. (1)c6 is then weakly linked as +1. But
 Each of the remaining possibilities have a mutual exclusion possible
 Thus one of r2 and one of c6 are in column 1. Which one not yet chosen.
In any event, +0
 Current count total: +2
 (1)r4. four locations. one can be grouped. think 3. r4c5 is either weakly linked to
row 2, or it is part of target column. r4c78 could be a target column entry, or be partially
weakly linked to r2. +1 or +0. Need to get this baby down. Choose +0. Target column defined.
Total count still +2.
 (1)r8. Each item weakly linked to one of the other matrix rows. 1. Total +1.
 In puzzle row 2, if we align r2c6 to have no entries below it, we would still be at +1. If
we force it into column 1, we get another 1, thus total 0. This could be one result.
 Result seems to have too many entries to be of much use. Consider (1)c7. Group it into
two pieces, one of which is weakly linked to each item in current result column except one.
Make this new matrix row the top row. Take all items but r2c6 out of result column and
place under r6c7. Total matrix change +0. Record new smaller strong inference set.
The manipulations could require some practice. But, if it was easy, this would not be a difficult
puzzle!
Note one could reverse the temporary decision above and still get somewhere. In fact, one could eventually
arrive at the same place, albeit a longer path. The nature of single candidate chains is that
there is almost always a different path to the same result. In fact, my original path was equivalent to
the reverse of the temporary decision above. I actually started with (1)Box2. After getting
to zero, I removed the extraneous data. There are many ways to do this.
Normally, one would like to start a deductive path at a bivalue or bilocation set. This allows
more flexibility, and incurs only a +1 penalty. However, This puzzle has few such places to
start. Thus, the for complex vehicles such as these.
Each of the derived sets now acts like half a cell. It has the strong inference. Most of them,
except for Z, do not have the weak inference active in the unit. Thus, a semicell equivalent.
However, the matrices and AIC do not rely upon exactly one is true in a set. They
rely only upon at least one is true interplayed perpendicularly (considering 4 dimensions
cell, box, column, row) with at most one is true.
Note further that the deduced sets we now have listed have some overlap. Perhaps it is time
to try for an actual elimination!
6x6 TM using derived and native inference sets
Step 2b
The graphic above does not dispay all the information used. I am unequal to the task of
graphing the deduction without distorting the deduction. Therefor, the Triangular Matrix
below will serve as both the proof and illustration of the elimination. Note that the
first column describes the strong inference set used in the row. The bottom row indicates
any nonnative weak inferences used.
1V 
r24c7 
r2c6 




Z 
 27r2c6  2r2c5 
  
2B1 
  r2c1 
r3c2   
2X 
  r8c5 
r5c2  r8c7  
Y 
  
7r1c2   8r2c9r12c7 
W 
1r24c7   
 2r4c7  8r3c9 

  
(27)Z   
The proven strong inference set:(1)r24c7 => r6c7≠1. Since this puzzle would not crack using
most normal methods, I started to build a list of small, somewhat complex, and very ugly
deductions. Once I found a few that seemed to trip over each other, I was able to put something
together. Once again, however, the base idea is just to count leftover unlinked components of
strong inference sets. Once one can reduce the matrix size to nxn, something has been proven.
With intermediate deductions, one can always force the matrix size to be nxn. However, one
may lose homogeneity in the deduced strong inference set. This is not a theoretical problem, but
it certainly is inconvenient.
After making this elimination, much of the work for the next one has been done. However, I still
would like to prove some more deduced strong inference sets. Since this page is long enough already,
this page is done!