### ... by: Dale E. Kloss

What does the solver do when testing a digit in a Digit Forcing Chain and a contradiction is discovered at the end of the chain (for example, the chain end forces two 5's in a unit ). Does the solver mark the Forcing Digit as "it can't be the assumed value" ?

Then the same question for the other Forcing chains ? Thanks, Dale Kloss
Andrew Stuart writes:

To the question "it can't be the assumed value", no, is the answer. The original digit will either be ON or OFF - one of those two states by necessity. Digit Forcing Chains (and other Forcing chains) can't tell us anything about the originating candidate(s). But because of their simple binary nature we have two possible 'futures' that branch off. One of those futures will be true and if we can determine that the consequence of those 'futures' are the same then we can make eliminations at the end of those two branches.

With Nishio it *IS* the originating candidate that if effected, because we are pretty much reversing Digit Forcing Chains, but that's just Nishio.

### ... by: michael dunipace

Extension to comment of 16 Sep 12:

Digit Forcing Chains page:

Digit to Unit Attack section:

4th paragraph beginning as follows:

an awkward ALS on {E9|A9} - not easy to spot. The chain goes into the cell E9 and turns OFF the 6 which leaves just a Naked Pair of 4/9 in A9 and H9. .....(etc.).

Notice the "H9" at the end of the above excerpt? Should it not be "E9"?

Best Regards,

MD

### ... by: Michael Dunipace

I believe there is an error on this page as follows:

Digit Forcing Chains page:

Digit to Unit Attack section:

4th paragraph beginning as follows:

In the very next step we are obliged to find another Digit Forcing Chain this time centered on 8

in E4. Like the first example it finds that either way the 8 in E2 .....(etc.).

Notice the "E2" at the end of the above excerpt? Should it not be "E4"?

Thank you for your Sudoku Strategy examples. They are quite valuable for testing my own

solver (implementing all strategies up to AICs so far).

Best Regards

Michael Dunipace

### ... by: Thinkist

Actually, never mind my previous comment. Digit Forcing Chains (as well as Dual Cell and Dual Unit Forcing Chains) are a strict subset of AICs.

Your first, second, fourth and fifth examples are the same as AICs Rule 1: Alternating Nice Loops with off-chain eliminations. Your third example is the same as AICs Rule 2.

Now if a candidate is ON or OFF...

...and another candidate elsewhere is ON, this is the same as AICs Rule 2.

...and another candidate elsewhere is OFF, this is the same as AICs Rule 3.

...and two other candidates elsewhere are ON in the same cell or unit, this is the same as AICs Rule 1, with a weak link between the two ON candidates.

...and two candidates elsewhere are OFF, no valid conclusion can be made.

In general, a candidate will be ON because of a strong link and OFF because of a weak one.

All in all, the scope of AICs really needs to be improved, and Digit Forcing Chains made obsolete (but keep the documentation).

### ... by: Thinkist

@Anton, I fully agree that the meaning of two OFF candidates out of three implies that the third is correct, and is in fact flawed logic to assume so.

The only unique case of DFC I can see resembles XY chains, where two candidates at the ends end up ON, eliminating the others in the same cell or unit. The other two cases are the same as AICs. Perhaps this strategy should be renamed.

### ... by: Anton Delprado

@Mike There is a lot of overlap with digit forcing chains and AICs. In fact I would argue that the interest in Digit Forcing Chains is to get to the more interesting Cell and Group forcing Chains or Forcing Chains with three or more chains.

Let's look at Digit Forcing Chains. If you check whether a cell is ON in one direction and OFF in another you are sending a weak link in one direction and a strong in the other. A candidate being ON in both directions means type 2 AIC. OFF in both directions is type 3. Two candidates both ON in a cell is type 1. Two candidates both ON and same value in a group is type 1.

I might be missing the logic here but looking at your diagram at the top. My interpretation is that the blue and purple represent ON and OFF chains for the initial digit 8. If that is the case then I disagree with two OFF candidates in a cell meaning the third value is correct. All you know is if the original cell is ON then one is incorrect and if the original is OFF then the other is incorrect not that both are always incorrect.
Andrew Stuart writes:

Thanks for your comment. I agree there is a fair degree of overlap with AIC and at the end of 2011 my improvements to AIC searching had replaced most instances of a DFC making it much rarer now. I have replaced the first example in this series because of that.

Feb 2012

### ... by: Mike Leonard

If I load the puzzle of Figure 1, into the solver it reaches the point where 1 is remove from G3 by digit forcing chains. I do not understand why AIC did not find and remove it
Andrew Stuart writes:

Sometimes two chains are required where one is not enough. That is the case unless my one-chain AIC is missing a trick or is too restrictive in it's search and detection. But even if it is I find it helpful to break these two different approaches into different strategies.

### ... by: Lee

I have to agree with John, I don't see this as a very 'human' strategy.

There is nothing on this site which details the rules/patterns to look for to identify this strategy _without_ having to perform any kind of mental trial an error.

This leads me to argue that this 'strategy' is identified as a result of making an initial assumption rather than as a pattern that a person can identify to prevent the need to make that assumption.
Andrew Stuart writes:

It does require a lot of initial searching and mapping of available chains, but once the ideas are grasped, the concept seems fairly simple. One should map chain links initially - as this is required for any strategy in the chaining family.

Whether or not these are 'human' strategies in terms of paper and pencil is not really the issue. For the hardest puzzles these types of chains are absolutely necessary if an entirely logical solve path is to be identified. Perhaps they will be implemented purely by people developing such solvers and only of interest to them, but this site is dedicated to all strategies, computers or human.

### ... by: John K. Landre

How do you find these chains?

It seems to me that, as with almost all extreme strategies, it is quicker to just start by assuming one of a possible pair is a certain value and then running the consequences until it either solves the puzzle or it runs into a contradiction. Of course, doing this sometimes does not have enough consequences to come to a conclusion with. So it helps to start this process by looking for a pair that 'runs' the puzzle, that is, that has as many connections to other pairs as possible.