# A study in Chain Forging: Page 3

Welcome back!

This proof is multi-page and multi-faceted. Hopefully, the possible indigestion caused by First Page, or perhaps the Second Page, will not prevent one from tackling this page.

This page is more of the same, but different. The basic ideas are the same. The particular implementation of these ideas crops up with a few different twists. Hopefully, the mind expansion required does not convey head trauma.

### A typical Almost Locked Set used in a chain Illustrated above is a fairly typical use of an Almost naked pair. Unfortunately, the chain itself is fairly long:

• {pair 17 at eh8} == h8=2 -- a8=2 == a2=2 -- a2=9 == c2=9 -- c5=9 == c5=8 -- g5=8 == g8=8
• => g8 ≠17
Because I am more of a hidden pair kind of solver than a naked pair solver, I saw the following chain first:
• {hidden pair 48 at gi8} == a8=4 -- a8=2 == a2=2 -- a2=9 == c2=9 -- c5=9 == c5=8 -- g5=8 == g8=8 => same thing
• Both chains are equal in complexity, and perhaps equally easy/difficult to find.

### A single candidate chain (coloring) snippet Illustrated above is a fairly typical forbidding chain that does not quite forbid anything. However, it does establish a strong link that becomes useful in a longer chain. The 1s are very close to forbidding some of their own without help from other candidates. However, they are not quite there. Oftentimes, candidates that are almost eliminated can be eliminated with a little help from other candidates. Above, but for the 1 at a8, one could eliminate c3=1 using only 1s in a chain. Instead, I use:

• fc on 1s:{d3 == d9 -- e8 ==1 h8 -- h6 == c6} ==1 a8
in conjuction with the following:

### Single Candidate Chain with ones linked to Chain with candidates 26 -- a8=2 == c9=2 -- c9=6 == c3=6 => c3≠1
This chain ends with the following proven strong set deduction:
• {d3=1 == c6=1} == c3=6 => c3≠1
The general idea here is that one can often append a chain with something else to achieve a useful strong set configuration. The general view of appending a chain can take many different forms. Although I have not presented the chain above in this manner, one can view it generally as:
• {bivalue/bilocation chain} == {bivalue/bilocation chain}
Specifically, to conform to this format:
• {d3=1 == d9=1 -- e8=1 ==1 a8=1 -- a8=2 == c9=2 -- c9=6 == c3=6}
• ==
• {c6=1 == h6=1 -- h8=1 ==1 a8=1 -- a8=2 == c9=2 -- c9=6 == c3=6}
Note that there is considerably overlap in both almost chains when presented in this fashion.

Another viewpoint is the way that I presented the elimination above. The key to finding such an overlapping group of bivalue chains, for me, is understanding that such overlapping chains are possible. Then, to find the chain, one reduces the search pattern to just the strong sets that will mark such a chain. In this case, one merely combines the typical markers for a single candidate chain with the typical markers for a multi-candidate chain, and must only see that those markers intersect. Here, they intersect at cell a8. In my convoluted thought train, this chain also looks like:

• bilocation chain piece == 3 string kite (another bilocation chain)
• Note here bilocation chain piece or snippet is of the form:
• A == B -- C == D -- E. Thus it begs E == something.
• This is really further reduced to: Single == Simple Coloring

### A typical Almost Locked Set used in a chain revisited for a different viewpoint The elimination above also has an alternative viewpoint. This alternative viewpoint has nothing at all to do with Almost Locked Sets. Instead, it again is of the form:

• {bivalue/bilocation chain} == {bivalue/bilocation chain}
Here is the presentation of the chain above using this form:
• {e8=1 == e8=7 -- h8=7 ==1 h8=2 -- a8=2 == a2=2 -- a2=9 == c2=9 -- c5=9 == c5=8 -- g5=8 == g8=8}
• ==
• {e8=7 == e8=1 -- h8=1 ==1 h8=2 -- a8=2 == a2=2 -- a2=9 == c2=9 -- c5=9 == c5=8 -- g5=8 == g8=8}
Which proves:
1. {e8=1 == g8=8} == {h8=1 == g8=8} => g8≠1
2. {h8=7 == g8=8} == {e8=7 == g8=8} => g8≠7
What is of interest to me here is that one can use the theory of appending simple bivalue/bilocation chains to predict the existence of many possible techniques. Considering those theoretical possible techniques, one needs only to keep in mind what they might look like in order to actually find them. Thus, in this case, one can note that an Almost Pair will be a marker for the appended chain above. In a similar fashion, a marker for the elimination of 1 from c3 illustrated previously is an Almost Coloring Chain.

### Recalling a previous step to find a possibly useful Almost Chain Page Two of this proof/solution started with a chain that proved d3=1 == e3=8. As soon as both 18 were forbidden from c3, it is immediate that some Almost Wrap Around Chain must exist. Rather than go into the details of how one knows this is the case, play around with the concept to convince oneself of its veracity. The Chain will only be guaranteed as an almost, as weak links will flip-flop as strong. In this case, since all but one of the previously used links were already strong and weak, all the almosting will occur in precisely one cell, e7. The following almost AIC:

• b3=1 == d3=1 -- d9=1 == d9=9 -- e7=9 *==*(almost) e7=5 -- e4=5 == e4=8 -- e3=8 == b3=8
• establishes e7=59 => b3=18
This is not generally useful. However, notice how this logic looks like the logic employed to derive the Hidden Pair argument. Thus, if another chain snippet also uses: e7=5 == e7=9, we have a valid link possible between chain pieces. Of course one such is immediately obvious, as in row 7 we have almost pair 59 at egh7! So let us look at that....

### Almost wrap around AIC with Almost Hidden Pair 59 with some strong links Illustrated above is a very nice chain using two conflicting almostings, one of which must therefor exist.

• b9=6 == b3=6 -- {b3=8 == e3=8 -- e4=8 == e4=5 -- e7=5 ==1 e7=9 -- d9=9 == d9=1 -- d3=1 == b3=1} == {Hidden pair 59 at gh71} -- gh7=3 == ghi9=3
• => b9≠3
This seems like much work to get to the elimination. However, there are easier ways to find this elimination group. I presented it in the reverse order that it was found! Clearly,
• {hidden pair 59 at gh7 == {e7=5 == e7=9}}
This building block for other AIC's is a common useful trick. Just like before, where I used this type of trick with the Almost Almost Hidden pair 58, I can use it here with Almost Hidden Pair 59. However, since both 5 & 9 outside of the Almost Hidden Pair exist in the same cell, e7, the deductions look a bit more typical. The typical puzzle mark-up heavily indicates that cell b3 should be involved in some chains. This is merely one of them that are possible that pass through that cell using the strong links with 168 that have an endpoint in that cell. Hopefully, this is clear!

### Locked candidate 3

Not illustrated, after b9 ≠3, clearly a7=3 == b7=3 => gh7≠3

This concludes the third page of this proof. The possible technique discoveries that this puzzle wants to inspire are not yet complete. To continue to grapple with interesting but similar concepts, kindly visit the next page. 