The New Sudoku Players' Forum

[ronk]

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 3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9


You now have a discontinuous loop...

"- (6=153)r79c8,r9c9"
From l-to-r this seems to say " We have switched off the 6 in r7c8, now figure out for yourself where the 153 go in r79c8,r9c9".

It would be clearer were the double weak link to r9c9 explicitly stated.

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 3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9

[daj95376]

Question, Is it not necessary for all AIC's to be readable both ways?

You've opened Pandora's Box. When Myth Jellies documented the AIC, he mentioned that AIC logic is bidirectional. He did not guarantee that Eureka notation could always support that property. In order to notate some of his solutions, he created some nightmarish notation to support the bidirectional feature. At times, there are still discussions over notating various patterns.

If we use an AIC to show that one end forces the other and vice versa so that we can use it to eliminate a candidate.

Is it acceptable only to be able to read it from l-to-r in order to see that there is the contradiction "If r9c9 is not 3 then r9c9 is 3" => r9c9=3

Myth Jellies simply states that any candidates, that would eliminate the candidates at both endpoints of the AIC, can themselves be eliminated. Most accept a shorthand notation that's more accurate when read from l-to-r ... and may need help/understanding when read r-to-l.

Is that it?
Make sure that it can be read both ways if we need that property for some reason, otherwise don't worry about it.

Almost. Make sure that the critical parts of the notation are clear when read l-to-r.

_

[daj95376]

[Withdrawn: went esoteric.]

_

[bat999]

...Almost...

"My original solution used both states of the 7 in r4c8 to prove r9c9=3, but it has morphed into a chain that merely uses r4c8 as a node"

Started with my primitive three chains.
(7)r4c8 - (7=2)r4c5 - (2=6)r3c5 - (6=3)r8c5 - (3)r8c9 = (3)r9c9
(7)r4c8 = (7-1)r6c8 = (1)r9c8 - (1)r9c9
(7)r4c8 = (68)r48c8 - (6=5)r7c8 - (5)r9c9

Refined to two chains with Steve's suggestion.
(7)r4c8 - (7=2)r4c5 - (2=6)r3c5 - (6=3)r8c5 - (3)r8c9 = (3)r9c9
(7)r4c8 = (68)r48c8 - (6=15)r79c8 - (15=3)r9c9

Then we stood on square r4c8.
If r4c8=7 we could drive west along row 4 to reach r9c9.
If r4c8<>7 we could drive south down column 8 to reach r9c9.

You jumped on square r9c9 and showed a road trip from r9c9 via r4c8 and home again to r9c9.
(3)r9c9 = (3)r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9

Voila, we have the AIC/Eureka equivalent of a Type 2 Discontinuous Nice Loop.
"If the first square has two strong links with the same candidate, then it can be solved with the links' candidate."

In our single chain r9c9 has two strong links with candidate 3.
Outbound from r9c9 "(3)r9c9 = (3)r8c9...".
Incoming to r9c9 "...(15=3)r9c9".
So we keep the 3 at r9c9, eliminate the 1 and 5.

Could we have driven round the circuit in the opposite direction too?
Well maybe.
Reading the chain from r-to-l, using very precise notation and mental gymnastics.
But who cares!
In this case it is necessary to make only the one successful expedition to win a prize.


PS
No need to even think about the mechanics of discontinuous loops.
When the chain is read from l-to-r it says...
"If r9c9 is not 3 then r9c9 is 3"
It's a contradiction so r9c9 can't be "not 3".

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AIC: (3)r9c9 = (3)r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 => - 1 r9c9, - 5 r9c9; stte

[eleven]

When the chain is read from l-to-r it says...
"If r9c9 is not 3 then r9c9 is 3"

As simple as it is, you get exactly the same conclusion, if you read from right to left.

Dont't miss JC's early posted chain, which shows how you can compress it.

[daj95376]

When the chain is read from l-to-r it says...
"If r9c9 is not 3 then r9c9 is 3"

As simple as it is, you get exactly the same conclusion, if you read from right to left.

Dont't miss JC's early posted chain, which shows how you can compress it.

Don't forget to go back and review SteveG48's post as well.

_

[bat999]

...Don't forget to go back and review SteveG48's post as well.

Yes, I looked at it already.
You have to think about it and wonder where is the 3 on row 8.
aha, there isn't one, so something's wrong. A contradiction.

[SteveG48]

...Don't forget to go back and review SteveG48's post as well.

Yes, I looked at it already.
You have to think about it and wonder where is the 3 on row 8.
aha, there isn't one, so something's wrong. A contradiction.

Hmm. Bat, you may have understood perfectly, but I'm not sure. Let's look at it. I wrote:

(3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

Looking at just the end points, this means: (3)r8c5 = (3)r9c9. (Here I've observed that the only 3 in the set r79c8,r9r9 is at r9c9.) (3)r8c5 = (3)r9c9 does not mean that r8c5 is not a 3 and it does not mean that r9c9 is a three. It means that at least one of those cells is a 3. Therefore, the cell that can "see" both of them, r9c8, cannot be a 3. This "pincer" type of chain is probably the most common means of elimination that you'll see posted here.

[bat999]

Looking at just the end points... (3)r8c5 = (3)r9c9... This "pincer" type of chain...

OK, I hadn't seen that angle.

[pjb]

Correct me if I'm wrong, but bi-directionality only matters when constructing a continuous loop. The way I view a chain is that the starting digit, x, is either true or false. If true, then it eliminates all x's it sees. If false, and the chain ends in a cell where x is true, then any x its sees that the starting cell sees can be eliminated. To paraphrase Steve, it means that at least one of those cells is a x. Therefore, the cell that can "see" both of them cannot be a x. Due to ALSs and network moves, many chains published here are unidirectional.

Phil

[bat999]

...The way I view a chain is that the starting digit, x, is either true or false. If true, then it eliminates all x's it sees. If false, and the chain ends in a cell where x is true, then any x its sees that the starting cell sees can be eliminated....

Yes.
I had previously thought like this...

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.------------------.-----------------.---------------.
| 4568  5689  1    | 458  57   a78   |  2    3   569 |
| 7     56    3    | 2    1     9    | d46   8   456 |
| 2458  2589  489  | 458  3     6    |  17   59  17  |
:------------------+-----------------+---------------:
| 3     16    5    | 9    4     1-7  | e67   2   8   |
| 268   4     68   | 3    27    5    |  9    1   67  |
| 9     12    7    | 6    8     12   |  5    4   3   |
:------------------+-----------------+---------------:
| 458   7     489  | 1    59   b38   | c34   6   2   |
| 1     3     2    | 7    6     4    |  8    59  59  |
| 4568  5689  4689 | 58   259   238  |  134  7   14  |
'------------------'-----------------'---------------'

(7)r1c6=(8)r1c6-(8)r7c6=(3)r7c6-(3)r7c7=(4)r7c7-(4)r2c7=(6)r2c7-(6)r4c7=(7)r4c7 => r4c6<>7
From left to right, if r1c6 is not 7 then r4c7 is 7.
From right to left, if r4c7 is not 7 then r1c4 is 7.
r4c6 can see both ends of the chain.

But I see your point.
No need to think "From left to right... From right to left"

As you said...
If r1c6=7 then it kills the 7 in r4c6.
If r1c6<>7 it forces 7 into r4c7 to kill the 7 in r4c6.
Job done.

"Due to ALSs and network moves, many chains published here are unidirectional"
That's what I have encountered.
Also, even when they are not stricly unidirectional, some of them are still extremely difficult to get my head around trying to read from r-to-l.

[daj95376]

Phil and bat999:

What you are describing is a forcing chain with two streams. The first stream is trivial and never listed. The second stream is called an AIC. Ironically, a forcing chain stream is unidirectional.

When I made this comparison to an AIC, I was severely chastized by Myth Jellies and others. Ironically, this logic is what I use in my solver ... and it hasn't produced a faulty forcing chain/AIC yet.

Note: there is a thread on this site where I mention this perspective. I get chastized there as well. I'm not going to look for that thread.

In order to reduce the complexity of the discussions in this thread, I refrained until now from mentioning this perspective.

_

[bat999]

...What you are describing is a forcing chain with two streams. The first stream is trivial and never listed...

Yes, with Steve's solution.
First forcing chain, when r8c5 is 3.
(3)r8c5 - (3)r8c9
Second forcing chain, when r8c5 is not 3.
(3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 - (3)r8c9

But no need to mention that first chain, it's bleedin' obvious.

In this case, as you said "the AIC can be viewed as the merging of two Forcing Chain streams".

His solution becomes (imho) ...
AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 - (3)r8c9 => - 3 r8c9; stte

[daj95376]

...What you are describing is a forcing chain with two streams. The first stream is trivial and never listed...

Yes, with Steve's solution.
First forcing stream, when r8c5 is 3.
(3)r8c5 - (3)r8c9
Second forcing stream, when r8c5 is not 3.
(3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 - (3)r8c9

But no need to mention that first stream, it's bleedin' obvious.

In this case, as you said "the AIC can be viewed as the merging of two Forcing Chain streams".

His solution becomes (imho) ...
AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 - (3)r8c9 => - 3 r8c9; stte

I forgot to mention that the AIC treats the final elimination as a conclusion. So, the final weak link is dropped.

_

[ronk]

It would be clearer were the double weak link to r9c9 explicitly stated.

(3)r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9

The logical conclusion for the above is r9c9 = 3, but the ending weak link is not required here either.

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(3)r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 ==> r9c9 <> 15

How often does one see a useful chain ending in a pair, either naked or hidden?