## How to notate this ALS-chain? (Nightmare, Sun Dec 2, 2007)

Post the puzzle or solving technique that's causing you trouble and someone will help

### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

I can see all 10 eliminations in the one loop if I write it this way :

(7) r1c1 = (7) r3c12 - (7=2 [+4] ) r3c89 - (2 [-4] ) r3c1 = (2) r1c1 loop

(2) r1c1 = (2 [-4] ) r3c1 - (2=7 [+4] ) r3c89 - (7) r3c12 = (7) r1c1 loop

If you read the first line from right to left you get the second line (reading it from left to right) with all nodes reversing their True/False status except that [parenthetically] 4 is True (both ways) in r3c89 and False in r3c1.

So you get the 4 continuous loop eliminations and [parenthetically] the six discontinuous loop eliminations on 4.

Leren
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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

eleven wrote:PS: For manual solvers this is of secondary interest. As long as they can proceed, they probably would not look for hidden additional eliminations.

I would, and don't think it's a secondary interest at all I want to squeeze everything out of every technique I use. How else do you learn to use their full power when you really need it?
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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

Leren, that seems pretty awesome to me! It solves the problem about the apparent (notational) irreversibility of ALS loops that has bothered me a long time.
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"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

A side question. For some reason, Sudoku Explainer gave this puzzle a rating of 8.9. That number seems ridiculously high compared to some other puzzles that have felt much harder. For example, the hardest puzzle I've completed only got 8.4. That makes no sense at all, as I really struggled with the 8.4 (only completed it on the second go, using GEM, with some luck), but this one felt easy (I didn't even draw any extra helper plots, which I normally do with harder ones). My skills have probably improved a little but I don't think that's the reason here.

It seems that Hoduko and SudokuWiki graders agree with my subjective feeling. The SE 8.4 puzzle got 10474 (Hoduko) and 1526 (SudokuWiki) points, while this SE 8.9 got 4344 and 814 respectively. I understand that different grading systems produce different results, but why such a huge discrepancy? I thought SE was considered some kind of standard rating, but at least in this case it seems a bit off.

This is the 8.4 puzzle I mentioned (and have mentioned before):

000900030004700600081054002005000000000020308000090060000070800017000000400106050
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`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

I was thinking... can we actually get 14 eliminations and one placement in one move if we change the loop a bit? What if we use the r3c789{1247} ALS instead of the r3c89{247}? If I understand anything, that way we could eliminate four 1s as well (r2c7, r3c246), as the loop locks both the 1 and the 4 into r3c789. Now we just can't eliminate the 7 in r3c7 by the same logic as before (as it's now part of a loop node), but it gets eliminated anyway if we realize that the sole locked 1 must be placed there. That last deduction is probably not part of the standard continuous Nice Loop logic so I'm not sure if it can be counted (if not, then it's just 13 eliminations).

None of this makes any practical difference, of course, as the the same outcome is achieved (more simply) with the earlier loop and a naked single, but is this theoretically correct?

The loop would thus be (in Leren's style, if I got that right):

(2-7)r1c1 = r1c8 - (7=2 [+14] )r3c789 - (2 [-14] )r3c1 = (2-7)r1c1, loop

(or: (7-2)r1c1 = (2 [-14] )r3c1 - (2=7 [+14] )r3c789 - r1c8 = (7-2)r1c1, loop)

=> -48 r1c1, -2 r3c6, -1 r2c7 r3c246, -4 r1c8 r2c9 r3c1246, (and +1 r3c7 => -7 r3c7)
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"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

I think you are right, but I'm a bit more comfortable with the following notation :

(7) r1c8 = (7-2) r1c1 = r3c1 - (2=7 [+14] ) r3c789 loop

(7=2 [+14] ) r3c789 - r3c1 = (2-7) r1c1 = (7) r1c8 loop

=> - 48 r1c1; - 2 r3c6; - 1 r2c7, r3c246; - 7 r3c7; - 4 r1c8, r2c9, r3c1246.

You'll notice that there is no need to include any parenthetical information in r3c1, as there is no 1 there and the 4 gets eliminated anyway because of the 4's in r3c89.

I was tempted to say that in the first loop, the 4 in r3c1 was having a really bad day, and got eliminated twice, once because it was False both ways in the loop, and once because it saw both 4's in r3c89.

Leren
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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

Glad to hear I hadn't gone completely off the rails. I still think the notation might need some work to make it more intuitive. While I really like that the ALS entry and exit digits are separated from the locked digits, I'm not sure if the parenthetical notation is the most readable way. Is it a standard of some kind (if so, where is it described?) or did you just come up with it? I couldn't quickly find anything like that in the Eureka documentation. It did say this, however:

"When an embedded ALS is part of the chain, the digit linked to the previous node is isolated from the remaining digits with a strong link symbol. The remaining digits are placed in such an order that the digit linked to the next node is the last one.

(1)r5c4-(1=264)r5c789-(4)r4c8"
http://sudopedia.enjoysudoku.com/Eureka.html

I think that idea makes sense, although by itself it doesn't solve the problem of reading the chain backwards like your separation does. JC's original loop actually seems to combine the benefits of both ideas in a pretty neat and concise way:

Loop[2r3c1=(2-7)r1c1=7r3c12-(7=4=2)r3c89]

How about that or something like that? It leaves the locked digit(s) isolated in the middle and the entry and exit digits on their respective sides, which seems the most logical way to group them to me and allows reading it in both directions. The strong link symbols on both sides of the locked digit(s) are also logical.

That way my chain would look like this, I guess:

(2-7)r1c1 = r1c8 - (7=14=2)r3c789 - r3c1 = (2-7)r1c1

Or the same backwards (easier to see that it's the same loop now, I think, as it's more of a mirror image):

(7-2)r1c1 = r3c1 - (2=14=7)r3c789 - r1c8 = (7-2)r1c1
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"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

JC, Leren, eleven: I want to thank all of you for a very fruitful discussion! At least I've learned a lot.

I thought it might be a good idea to continue generic parts of this discussion in another thread: continuous-loops-with-als-nodes-t34306.html
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"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

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### Re: How to notate this ALS-chain? (Nightmare, Sun Dec 2, 200

One final comment on this thread is that I could have written the chain as :

(7) r1c8 = (7-2) r1c1 = r3c1 - (2=714) r3c789 loop

(7=214) r3c789 - r3c1 = (2-7) r1c1 = (7) r1c8 loop

Which some people do anyway, even when the ALS chain is not a loop. This is perfectly correct but tends to obscure the importance of the [+14] in the loop situation.

You are right, the [ ] parenthetical notation is something I've added, because I actually wan't my readers to be given the clearest idea of what is actually happening, to cause all of the eliminations.

Leren
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