A study in Chain Forging: Page 2

Welcome back!

This proof is multi-page and multi-faceted. Hopefully, you have already digested the First Page.

This page contains a few interesting ideas for study. All of them are the natural result of viewing Forbidding Chains or Alternating Inference Chains (AIC) not only with strict precision, but also with an open mind about possible relationships.

Before I once again venture into the truly interesting stuff, first I must dispense with some rather mundane chains. Nevertheless, understanding of the interesting chains can only be achieved after first mastering the relatively easy concepts/chains.

A typical Four Strong Set Chain using candidates 1589 and strength in location

Typical Four strong inference set bilocation chain

Illustrated above, a chain that is heavily indicated by the standard puzzle mark-up should be easy to find. Perhaps the only fun thing about this particular chain is that the endpoints of the chain allow two eliminations, as shown. Here is the chain, written in my usual outdated style:

  • d3=1 == d9=1 -- d9=9 == e7=9 -- e7=5 == e4=5 -- e4=8 == e3=8
    • => d3=1 == e3=8 => d3≠8 & e3≠1
Although the chain above is perhaps mundane, here is what I think after making these eliminations:
  1. d3=1 == e3=8 act very much like a single cell limited to only 1,8
  2. But for d3=1 -- e3=8, I could treat the relationship above precisely like a single super cell in row 3 and in box e2
  3. This tidbit of information will be stored in my head for the rest of the puzzle
  4. It maybe useless information, it maybe useful - that I do not know at this point!
  5. e7=59 == e7=17 is an interesting relationship. Now even more so. Note: I have a crossing point of Almost pair 17 in column e with Almost Hidden Pair 59 in row 7
  6. Again, I have no clue how valuable this information may be, but it is now also somewhat cross referenced with an almost super cell 18 in row 3
Hopefully, this window into my head is helpful. The general goal of how to find complex chains less painfully is the prize!

A typical Y Wing Style using candidates 48

An easy to find Y Wing style

If you have been paying attention to my previous blogs that have placed much emphasis on Y Wing Styles, then the chain illustrated above should be easy to find. c1=48 and d4=48 are common markers for a very common Y wing style. Sadly, that type of Y wing style is not found here. Nevertheless, one should look for it. By looking for that type of Y wing style, the one illustrated above:

  • c1=4 == c1=8 -- d1=8 == d4=8 -- d4=4 == f6=4 => c6≠4
can become easy to locate.

Almost Almost hidden pair 58 - the concept visualized

Reducing Almost Hidden Pairs to chain snippets

Illustrated to the left is a chain snippet:

  • {Hidden pair 58 at b16} == {b3=8 == b4=5}
You may consider the logic involved in many ways. One way is to consider that:
  1. ~{b3=8 == b4=5}
  2. => b3≠8 and b4≠5
  3. => Hidden pair 58 at b16
So we have concluded:
  • ~{b3=8 == b4=5} => {Hidden pair 58 at b16}
  1. A => B is equivalent to
  2. ~A OR B
  3. Also written as ~A == B
  4. Thus ~A => B can be written as: A == B
Putting this all together: (substituting for A,B above)
  • {b3=8 == b4=5} == {Hidden pair 58 at b16}

There is no reason to repeat this exercise everytime one sees this type of Almost Almost Hidden Pair configuration. One can immediately know, from previous analysis, that this relationship exists. I am aware that using ~ in an argument is viewed with disdain in some sudoku circles. To this I say the following: One needs never do it again, as the concept has been proven in general! Thus, the relationship becomes visual for every possible occurennce, as it has been proven sound in general. This is the power of mathematics and logic: Prove general truths, then use them over and over again. No need to reinvent the wheel for each wheeled conveyance!

To me, this is a powerful concept, as now I can use this conceptual relationship in a chain. What I have derived is a chain snippet, or piece of a chain. It is of the general form, using Booleans, of A == B, so it looks an awful lot like part of an AIC, or Forbidding Chain. More precisely, it is exactly a valid AIC. Thus, it is also a valid AIC link in any larger AIC chain.

Below, find how I use this piece of information in a larger chain

A wrap around chain with an internal discontinuity

Alternatively, an extended Hub plus a simple Hub connected to an ALS Rim

An very cool pattern

Illustrated above, we have an interesting (to me anyway) group of connected chains:

  1. {Hidden pair 58 at b16} == {b4=5 == b3=8}
  2. f6=5 == e4=5 -- e4=8 == e3=8 -- b3=8 *==* b4=5 -- e4=5 == f6=5
    • This chain above is marked with asterisks, as it almost exists, but does not as b3=8 == b4=5 is not a native puzzle condition
  3. f6=5 -- f6=4 == b6=4
    • This chain is not an AIC, but it is a valid piece of an AIC
Putting all of these pieces together, one has a very nice chain:
  • {Hidden pair 58 at b161} == {b4=5 ==1b3=8 -- e3=8 == e4=8} -- e4=5 == f6=5 -- f6=4 == b6=4
    • Since {Hidden pair 58 at b16} -- b6=4, we also have some sort of wrap-around, or continuous chain. Unfortunately, most of the proven strong links do not forbid anything:
      1. {b4=5 == b3=8 -- d3=8 == d4=8} == e4=5 is too asymmetrical to have much practical use.
      2. f6=5 == f6=4 could be useful, but we already have that!
      3. {Hidden pair 58 at b16} == b6=4
        • =>b1≠4 & b6=458
The reason that I have highlit the part of this chain that I find to be discontinuous is so that one can clearly see that some of the weak links used are not proven strong. Thus, the following weak links are not proven strong:
  1. b3=8 -- e3=8
  2. b4=5 -- e4=5
  3. e4=5 -- e4=8 (although we already have e4=5 == e5=8, that is irrelevant here!)
Finally, as it turns out, noticing that this chain is a continuous loop, or wrap around, is of absolutely no use for this puzzle. However, I thought it instructive to analyze completely what is actually happening. A little knowledge can be a dangerous thing! In fact, if one erroneously were to conclude that all the weak links shown here are indeed strong, then the puzzle would be accidently advanced. In this case, only legal reductions would be made. They would be made, however, without justification.

PERHAPS too much information here?!!

Chain snippet digression

Non-native Almost Chain pieces, such as b3=8 *==*(almost)b4=5, are a very valuable tool to include in one's toolbox. Recognizing such almost chain pieces becomes easier with practice. Using such almost chain pieces can be a powerful technique to humanly solve puzzles using logic. They often provide relationships that look odd, but are quite beautiful to someone like me.

A typical depth 4 chain using 48

A mundane pattern

Illustrated above is a simple chain. The only interesting aspect is the grouped use of the 8s in box b5. Again, this pattern is heavily indicated by the standard puzzle mark-up. I choose to insert this step here to give everyone a break from the previous step before venturing into Almost land again. For the sake of completeness, here is the chain:

  • c56=8 == b6=8 -- b6=4 == f6=4 -- d4=4 == d4=8 -- e4=8 == e3=8 => c3≠8
Of course, one can alternatively use the strong 5s in box e5 to establish d4=4 == e3=8. Thus the same elimination can also be written:
  • c56=8 == b6=8 -- b6=4 == f6=4 -- f6=5 == e4=5 -- e4=8 == e3=8 => c3≠8
Perhaps, I should have illustrated the step that way, since the overlap with the previous step is then quite large. (Three out of four strong sets reused). Oftentimes, puzzles are unlocked with mutliple uses of the same groupings of strong sets.

The Almost Y Style 17 from Page 1 used again in a chain

Re use of a recognized pattern

Previously, on the preceding page, the relationship: h8=2 == b4=7 was established. It is not difficult to then use this relationship again for a slightly different chain. Illustrated above:

  • b9=6 == c9=6 -- c9=2 == a8=2 -- h8=2 =={Previously detailed Y Style using 17} -- g4=7 == b4=7
    • => b9≠7

This concludes the second page of this proof. There remains to be discussed not only some very interesting steps but also some very neat ways to locate them. To further investigate my absurdly grandiose road of soduku solving, kindly visit the next page.

Indicate which comments you would like to be able to see

The link to the next page of this series of pages will be broken until I get my lazy self around to publishing the next page. Be patient, please?

I shall probably append this proof about 1 blog page per day over the next few days.
08/May/07 4:44 PM
I see that this page eliminated 4 from b1. Consider this:
b1=5 == a1=5 -- a5=5 == a5=9 -- c5=9 == c5=8 -- c1=8 == c1=4 forbids b1=4
This can be done from your Page 1, but it may stray from your instructional purpose.
22/May/07 12:03 PM
another 4 can be removed:
a2=9 == a5=9 -- c5=9 == c5=8 -- c1=8 == c1=4 forbids b2=4
22/May/07 3:31 PM
thx jm. There are a number of possible eliminations not listed. BTW, I think your last note is meant to eliminate a2=4.
22/May/07 7:20 PM
correct, my bad.
24/May/07 4:10 AM
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