The following is page four of my ridiculous saga attempting to defeat the Easter Monster.
I will try to make the deductions on this page more clear.
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 a 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
Previously, many derived inference groups were proven. This page
will use Group Z.
- Each set in Group Z below contains exactly one truth:
Step 2h using Block Triangular Matrix logic.
Block Triangular Matrix is an imposing name. It is merely a manner of logically ordering
the inferences. This logical order serves as both a tool for explaining the elimination, and
also for finding the elimination. Finding the elimination boils down to counting extra
columns versus rows, and getting the count to zero.
Above I have tried to illustrate the actual chain. Please note, chains that operate on
multi-valued cells, or on multi-location sets - have basically arbitrary partitions. One
can always write such a chain in more than one fashion. Here is what I found:
- (2)r7c8=(2)r5c8 -(16)r5c8
- Given -(1)r5c8, the following almost skyscraper with candidate 1 exists:
- (1)r7c4(the almost part)=[(1):r5c4=r5c2-r7c2=r7c6]-(1=6)r6c6
- => almost (1)r7c4=(6)r6c6
- Thus, we assign (1)r7c4 endpoint status, and continue with
- Using Z, (6)r9c2=(1)r7c2-(1)r7c6
- Noting both 4) and 2), (1)r67c6=(1)r3c6-(1)r3c8 & -(2)r3c6
- Noting both 5) and 1), (1)r53c8=(1)r4c8-(67)r4c8
- Noting both 6) and 1), (6)r45c8=(6)r4c9-(7)r4c9
- Noting both 6) and 7), (7)r4c89=(7)r4c3-(7)r2c3
- Noting both 5) and 9), (2)r23c6=(2)r7c6
The three endpoints are in bold italics
, => r4c7≠2
Here is the same deduction written into a Block Triangular Matrix. Note how (1)r6c6 feeds
back to the first column, while (6)r6c6 continues on as if this were a normal Triangular Matrix.
Note how the 2x2 Almost Almost skyscraper on candidate 1 juts out in this matrix. Note
further how it is resolved back into what is essential Triangular Matrix form.
Step 2i prelude.
Deriving a Guardian Strong Inference Set
Previously, I have disparaged the guardian technique as basically unnecessary. However, while
tackling this particularly difficult puzzle, I decided to revisit the concept. The guardian
idea is basically simple: One cannot color a shape that has an odd number of vertices with
only two colors and achieve color alternation. Thinking in terms of true and false, if
an odd number of exactly one is true sets form a loop, it is impossible to have a
consistent solution. Once one places two truths into the loop, one strong inference set is
left with zero truths. If one tries to place 3 truths in the loop, one weak inference set
will contain two truths.
Above, the five locations for 6 that are labelled with a red asterisk form a potential impossible
pentagon with 5s. In order for this situation to not occur, at least one of the four locations
labelled g must contain a 6. Thus, we have a guardian sis.
One can then use this guardian sis just as if it is a native sis, except that the
weak inference will not apply. Thus, first I perform the following operation on this sis:
- (6): r8c5=[X Wing at r24c59]-(r4c1)=sis(6):[r1c6,r6c2,r9c4]
- => (6)r8c5=sis(6):[r1c6,r6c2,r9c4]
- => sis(6):[r1c6,r6c2,r9c4,r8c5]
Simply put, (6)r4c1 was removed from the sis, and (6)r8c5 was placed into the sis.
Note that this guardian inspired strong inference set has much overlap with a very strong
candidate, candidate 7. Thus, it makes some sense to see if some 7 or some 6 at one of these
intersections is ripe for elimination.
The use of the guardian set is not required to achieve this elimination. However, it makes
the logic faster, and, more importantly, provides a clue of where to look. Since this idea is
complex and unwieldy, it is probably supplanted by something more simple except in puzzles that are so utterly devoid of a suitable number of bivalue/bilocation
sis. Also, and perhaps just as importantly, in this puzzle once one has some sort of bivalue/bilocation -
they almost align so perfectly with each other that interaction between them takes a long look
ahead. The guardian idea shortens the look-ahead requirement substantially.
Guardian Strong Inference Set used
Below, I have mapped out a relatively short deduction. Note that the weak inference of Z
is used below in two locations, indicated in orange. The guardian derived sis is marked with
green g's. Remember that the guardian sis is upon candidate 6.
Unfortunately, in order to lasso all four disparite pieces of the guardian sis, the deduction
branches about a bit. Below find one desription:
- [(7)r1c2=(7)r6c2-[(6)gr6c2 & (2)Zr5c2]
- & (7)r2c6=(2)r2c6-(2)r7c6]
- ref 1)&3), (2)r5c28=(2)r5c4-(2)r4c5
- =(2)r8c5-[(6)gr8c5 & (7)Zr8c4]
- ref 1)&5) =[(7)r9c6=(7-g6)r9c4=g6r1c6]
Below, the same description as a TM.
Thus, the Kraken-like pseudo=guardian sis 6:[r1c6,r6c2,r8c5,r9c4] is moved, or
transported, into a new sis [(7)r1c2,(7)r2c6,(7)r9c6,(6)r1c6] => r1c6≠7
Guardians, and any technique like them, that allows one to build a new sis from existing
information can be useful in locating possibly strong candidate interactions that may lead
to finding an elimination.
This concludes this page. Once again, if you have read this far, thanks for your patience!