Welcome back!

This proof is multi-page, multi-faceted and entirely too long for one sudoku meal.
Hopefully, in spite of my verbosity, you have managed to enjoy the previous pages:

- First Page
- Second Page
- Third Page

This page is a bit unusual, as the entire page is dedicated to just one step. The idea that
I will try to present in great detail is very simple. The potential application of this idea
is wide. In general, the idea is one of a **Super Cell**. Super cells are nothing
but a way to reduce the puzzle to smaller more managable chunks of information.

### The Foundation of a native super cell

Box b2 and row 2 have bilocation 9s at ac2. No matter what happens, the finished puzzle
will have a pair 9x at ac2. Thus, for the purposes of chain interactions in both row 2 and
box b2, one can view the puzzle as if there exists one less cell.

Thus, the foundation of a native super cell can be as simple as one candidate limited to two
locations. That is simple enough, I hope!

### A representation of the Native super cell

Again, only for the purposes of chain reactions in row 2 and box b2 does this **
Super Cell** configuration exist. Note that one can delete the 9s, and that one
must include each other candidate that exists in both cells as if they existed in one cell.

What can be done with such a *Super Cell*? Without any loss of generality, one can
treat this super cell for all chains pieces that interact only in row 2 or in box b2 as if
this cell were just a normal cell. The graphic below illustrates some possible chain reactions
emanating from this super cell only in box b2.

### Chain pieces using the Super Cell

Study this graphic. It is nothing but some easy interactions within box b2 that are the direct
result of the strong 9s at ac2. However, for the purposes of these interactions, the 9s are
nothing but a reduction carrier that allows the following chain piece:

- b23=1 == ac2=1 -- {super cell by strong 9s ac2}=2 == c3=2 -- c3=6 == b3=6

Thus we have now have another type of super cell caused by the end links of the chain above:

- b3=6 == b23=1
- =>Super cell 16, strong only, in column b

If there exists something in column b that interacts with this relationship, we will have some
sort of chain. Of course, I would not have introduced this discussion if this relationship did
not exist.

### Continuous, or wrap around, chain using super cell plus chain pieces plus cell b9

If one practices with the concept of *Super Cells*, then immediately upon inspection
one can see the equivalent of a strong inference cell 16 in column b in box b2. Thus, one has
the equivalent of pair 16 in row b at b9, b23 ish. Thus:

- b23=1 == b3=6 -- b9=6 == b9=1
- Note b9=1 -- b23=1, thus a wrap around chain
- Thus b23=1 -- b3=6
- Thus we also have a wrap around chain inside box b2 using the Super Cell!

The entire configuration is one big wrap around chain, allowing mulitple strong set deductions,
thus providing one with mutliple possible eliminations.

As a complete chain, without writing down the super cell idea, and bringing back the
9s, one can write many possible chains. One way to write it follows:

- {Hidden pair 29 at ac2} == c3=2 -- c3=6 == b3=6 -- b9=6 == b9=1 -- b23=1 == {Hidden pair 19 at ac2} thus:
- ac2: {Hidden pair 19} == {Hidden pair 29} forces only 129 at ac2
- c3=6 == c3=2 forces only 26 at c3
- b3=6 == b9=6 does nothing, as we already had that
- b9=1 == b23=1 => b145678 ≠1 => b47≠1

There is a Sue De Cox available here that eliminates exactly the same items, I think. I have
found in my experience that a hidden set cross naked set representation as shown above usually
co-exists with a Sue De Cox. For me, what I have detailed here is much easier to find. Also, the
logic seems less strained, as less simultaneous conditions are required. Furthermore, the concept
of such Super Cells is not limited to only continuous, or wrap around, chains. It can be used
for many more types of reductions. Thus, understanding this concept gets much more *Bang*
for the trouble.

### The actual eliminations detailed

Bringing back the temporary disappearing candidate 9, and placing all the candidates
back in their proper locations, one can now make the actual puzzle eliminations deduced by
this technique. Note that we have knocked off some 4's and 3s without actually considering any
4's or 3's in our chains. This is very typical with this type of continuous chain. One can also
view this deduction as a multi-hub at ac2,c3 using 29 as a bridge connected to the rim cell b9 by
two strong spokes, the 6s and ones emanting from the multi-hub. Wow, that is too wordy!

### Alternate view

Hopefully the following view makes sense.

- Let the 2's and 9's form a super cell at ac2,c3. Note this super cell exists only
with repect to box b2.
- The 1's and 6's come out of this super cell within box b2 and are aligned in row b.
- Row b already has a cell limited to only 16, so one clearly has a wrap around chain
- The only problem with this view point is one needs to be more careful determining the
precise reductions possible

Here is a possible way to write this alternate chain idea:

- {Hidden triple 269 at ac2,c3} == b3=6 -- b9=6 == b9=1 -- b23=1 == {Hidden pair 19 at ac2}

The problem that I have with this valid represetation is that it is not clear what is happening at
cell c3. However, by consistently using a larger super cell, not only does the idea become more
simple, but also the valid eliminations are quite clear.

### Same pattern, using a super cell of size 3

Using the Hidden triple concept derived from the size 3 super cell, derives
the following presentation of the same chain:

- {Hidden triple 269 at ac2,c3} == b3=6 -- b9=6 == b9=1 -- b23=1 == {Hidden triple 129 at ac2,c3} proving clearly:
- ac2,c3 are reduced to only 1269
- 16 cannot exist outside of b239 in row b

Hopefully, one can see that the idea of the super cell easily expands
to multiple cells.

In other words, the super cell is formed by candidates 29 at ac2, c3. So, temporarily erase
29. Place candidates 1346 into the super cell. See the easy wrap around chain:

- Super Cell=6 == b3=6 -- b9=6 == b9=1 -- b23=1 == Super Cell=1

Now, make all the eliminations. Add back 29 to the super cell after one is finished. Quite Easily Done!

As an interesting aside, note that the template for the larger pattern, after using the size
three super cell, is my favorite type of Y Wing Style.

### Super Cell Summary

To this point, I have introduced two general types of Super Cells:

- Reduce a group of cells to one cell using strength in location. Allows one to
temporarily erase some candidates from the group of cells. Is only valid in houses that
contain all the cells that form the super cell.
- Form the equivalent of a strong-only cell by the use of a chain. This idea serves
as a visualized link point for chain continuation.

For me, the main value in thinking with the concept of

*Super Cells* is precisely that
chain linkage is made much easier. Nevertheless, this concept has been treated earlier. The
general form of this concept is

*Useful Groups*. The concept of useful groups stems
entirely from the concept of

*using Booleans in chains.* Thus, once again, the true
foundation for this concept is quite simple:

- The fundametal arguments in
*Forbidding Chains, or AIC* are *Boolean variables*.

This concludes the fourth page of this proof. More possibly unchartered territories still
exist. If you are adventuresome, you may enjoy visiting the
next page.

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