Welcome back!
Still awake? This proof still has some teaching left. I am much obliged by your attention.
Previous pages of this proof are practically a prerequisite for this page:
- First Page - Almost Y wing styles as Chain links
- Second Page - Almost Hidden Pairs as Chain links
- Third Page - Almost coloring, Almost AIC, Introduction to Appending chains
- Fourth Page - An introduction to Super Cells
- Fifth page - An introduction to Appending Chains
The last step on this page is mundane by comparison to the first step. However, the first
step is not hard to find, if one truly owns the previous information detailed in this proof.
Rather than bog things down with mundane theories, let us just look at an example.
A Strong link as an appendage, and also the foundation of a super cell
The 3s in column h are limited to two cells. We already know that this means that one can
form a super cell with the left-over contents of these two cells. Furthermore, this super cell
is only useful in column h.
How are these 3's just an appendage?
The base chain that is used to derive the actual chain used in this step will rely on strengths
in column h that emanate as if they came from a single cell at h39 rather than two cells.
The super cell and its useful outbound links
The 3s in column h have been temporarily erased. The other contents of h39 have been dumped
into a single super cell.
The two's look promising, as they have only two locations now in column h. The rest of the
candidates look less promising. However, there is something to be done with the 5s.
Almost skyscraper with 5s emanates from the super cell
Above, find that I have added the following almost skyscraper with candidate 5:
- h6 *==* h7 -- e7 == e4 => h6 *==* e4
Considering column a and row 5, the rest of the chain is not hard.
Almost skyscraper with 5s links through the super cell with a standard chain
the chain can be written using the super cell:
- {e4=5 == e7=5 -- h7=5 ==_{1} h6=5} ==_{1} {Super cell h39=5} --
{Super cell h39=2} == h8=2 -- a8=2 == a2=2 -- a2=9 == a5=9 -- a5=5 == gi5=5
- =>{e4=5 == h6=5} == gi5=5 => g4≠5
The entire chain above can be written without using the super cell as:
- {e4=5 == e7=5 -- h7=5 ==_{1} h6=5} == {Hidden pair 35_{1} at h39} --
h39=2 == h8=2 -- a8=2 == a2=2 -- a2=9 == a5=9 -- a5=5 == gi5=5
- =>{e4=5 == h6=5} == gi5=5 => g4≠5
Note that the beauty of this idea, super cells as appendages, is that the 3's are clearly revealed
as a bridge between two cells. Thus, one can treat two cells as one, in a limited fashion. Moreover,
the entire chain is nothing but an appendage to the base chain idea, a short chain on only
candidate 5. Note further that the elimination is valid regardless of how many times 25 actually
appear at h39.
As complex as the elimination above may appear, it can be represented by a fairly simple 7x7
pigeonhole matrix, discontinuous type. Using such matrices reveals how the entire elimination
presented above can be derived by merely appending.
Appending forbidding matrices to derive strong inferences
The following group of tables are relevant to the elimination being considered. They will
progress until the matrix representation aligns with the native puzzle conditions illustrated.
For each of these matrices, the size will be nxn. Each row will be a native strong inference.
Each column, except the first column, will have at most one truth (native weak inference). Thus,
each matrix will have at least n truths. Columns 2 thru n will contain at most (n-1) truths.
Therefor, column 1, the result column, must contain at least one truth (proven strong inference set).
5x5 matrix
The following matrix is an almost matrix relative to the puzzle above.
g5=5 |
a5=5 |
| | |
| a5=9 | a2=9 | | |
| | a2=2 | a8=2 | |
| | | h8=2 | h9=2 |
h6=5 | | | | h9=5 |
The conclusion of column 1 then is: g5=5 == h6=5. Although this example looks like a continuous
chain, and in fact if these native puzzle conditions existed it would be a wrap around chain, the
only result this analysis is interested in is: g5=5 == h6=5.
5x5 matrix appended inside a matrix cell
The following matrix is again an almost matrix relative to the puzzle above. It progresses
one small step closer to the actual puzzle.
gi5=5 |
a5=5 |
| | |
| a5=9 | a2=9 | | |
| | a2=2 | a8=2 | |
| | | h8=2 | h9=2 |
h6=5 | | | | h9=5 |
the conclusion of column 1 now is: gi5 == h6=5. We have merely appended the previous matrix.
5x5 matrix appended by adding a matrix row and a matrix column
Again, the following matrix is an almost matrix relative to this puzzle. Here, I append
a 5x5 matrix to create a 6x6 matrix.
gi5=5 |
a5=5 |
| | | |
| a5=9 | a2=9 | | | |
| | a2=2 | a8=2 | | |
| | | h8=2 | h9=2 | |
h6=5 | | | | h9=5 | h7=5 |
e4=5 | | | | | e7=5 |
The conclusion of column 1 has changed significantly, and the matrix is no longer in any
way continuous. However, there still exists a target set of possible eliminations. The new
conclusions of this almost 6x6 matrix is:{gi5=5,h6=5,e4=5} is a strong inference set. Thus,
the target elimination group is now: ghi4≠5.
6x6 matrix appended by adding a matrix row and a matrix column
Finally, the matrix below is no longer an almost matrix. In fact, it is representative of
the actual native puzzle conditions.
gi5=5 |
a5=5 |
| | | | |
| a5=9 | a2=9 | | | | |
| | a2=2 | a8=2 | | | |
| | | h8=2 | h9=2 | | h3=2 |
h6=5 | | | | h9=5 | h7=5 | h3=5 |
e4=5 | | | | | e7=5 | |
| | | | h9=3 | | h3=3 |
The conclusions of this 7x7 matrix and the previous 6x6 matrix are the same. Hopefully, the tool
of using matrices to illustrate possible ways to append chains is clear. There are many other
ways to append chains that have not yet been demonstrated.
One very important note: I do not view matrices as an important solving tool. Rather, they are
a great proving tool. As such, they are also a fantastic tool to use to imagine
possible theoretical techniques. Of course, one must then construct a logical way to go from matrix
to technique observation. That is precisely the reason that I introduced matrices early in the blog,
and now have returned to them. I needed to build the observational base!
Now, let us return to the puzzle for one further elimination. This one is relatively easy!
An easy Y wing style that happens to be a wrap-around chain.
The continuous chain illustrated above has only one useful proven strong link: c6=1 == c6=7.
The complete chain can thus be written to clearly indicate the one elimination that is proven:
- c6=1 == g6=1 -- g4=1 == g4=7 -- b4=7 == c6=7 => c6=17 => c6≠8
This concludes the sixth page of this torture. If you are still up for the punishment,
dare to click onto
page 7 (Almost Y wing Styles again).