One or two (AIC) moves?

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

One or two (AIC) moves?

Postby SpAce » Fri Mar 09, 2018 1:41 am

My questions are related to this puzzle and my solution to it:

http://forum.enjoysudoku.com/200e200w-s-weekly-extreme-2-t34498.html#p264661

Question 1) My step 12 is actually two AICs but so closely related that I didn't feel like writing them separately. Is such an "extended" or "double" AIC valid and does it count as one or two moves? Here's the situation:

Code: Select all
+-------------------+-------------------+------------------+
| 1    a7(8)-9 579  | 5689 g(4)5-8 2    | 3    67 fg4(9)   |
| 23    4      279  | 69     1     36   | 68   5    78     |
| 58    39     6    | 7      34    58   | 1    49   2      |
+-------------------+-------------------+------------------+
| 3567  13679  8    | 15     2     4    | 56   3679 13579  |
| 24567 1267   457  | 3      58    9    | 2568 267  1578   |
| 235   1239   59   | 158    6     7    | 4    23   13589  |
+-------------------+-------------------+------------------+
| 9    b268    24   |c2568   358   1    | 7    234  35     |
| 246   5      1    | 26     7    d36   | 9    8   e34     |
| 278   27     3    | 4      9     58   | 25   1    6      |
+-------------------+-------------------+------------------+

My "Extended-AIC": (8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - r1c9 = ((9)r1c9 & (4)r1c5) => -9 r1c2, -8 r1c5

The two embedded AICs are of course:

(8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - (4=9)r1c9 => -9 r1c2
(8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - r1c9 = (4)r1c5 => -8 r1c5

So, can it be written like I did? One or two moves?

Question 2) This is another, less interesting, notation question about the the same puzzle. At the very beginning, I was contemplating taking an immediate placement as my first move before I chose to use the loop to weed out some candidates and simplify that step. Here's the situation:

Code: Select all
+--------------------+------------------+---------------------+
| 1     789   579    | 5689 4589  2     | 3     4679  4789    |
|d2378  4    e279    | 689  1    c368   | 689   5     789     |
| 358   389   6      | 7    34589 3458  | 189   149   2       |
+--------------------+------------------+---------------------+
| 34567 13679 8      | 15   2     457   | 1569  13679 13579   |
| 24567 1267  12457  | 3    458   9     | 12568 1267  1578    |
| 2357  12379 12579  | 158  6     578   | 4     12379 135789  |
+--------------------+------------------+---------------------+
| 9     268 af24     | 2568 358   1     | 7     234   345     |
|a246   5  af(1)24   | 269  7    b36    | 129   8     1349    |
| 278   1278  3      | 4    589   58    | 1259  129   6       |
+--------------------+------------------+---------------------+

I came up with this to place 1r8c3:

AIC (ANS): (1=246)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (2=41)r78c3 => -24 r8c3

Is that how you would write it? My problem with it is that the conclusion is not immediately clear just by looking at the chain, but I guess it's business as usual with ALS stuff. Looking at the grid makes it obvious, of course, because we can see that there's only one candidate(1) in either ANS at the ends of the chain and it happens to be the same and located in r8c3 (could be seen as a discontinuous nice loop type 2 as well). The chain alone doesn't convey that piece of information, however.

(Added.) How about this: (1)r8c3 = (42)r78c3 - (2)r2c3 = (2-3)r2c1 = r2c6 - (3=6)r8c6 - (6=24)r8c8,r7c3 => -24 r8c3 ?

Anyway, after using the loop (step 1) first it was simplified into (step 2):

AIC (ANS): (4=2)r7c3 - r2c3 = (2-3)r2c1 = (3-6)r2c6 = r8c6 - (6=24)r7c3,r8c1 => -4 r8c3
Code: Select all
   *             |    |               |    |    *
        *        |=()=|    /  _  \    |=()=|               *
            *    |    |   |-=( )=-|   |    |      *
     *                     \  ¯  /                   *   
SpAce
 
Posts: 605
Joined: 22 May 2017

Re: One or two (AIC) moves?

Postby Cenoman » Sat Mar 10, 2018 10:40 pm

First, be aware that I am not an early member of these forums. My answers are nothing but my opinions.
Early members are more entitled to give you uses and rules.

Second, I recall what is often said to newcomers about notation debates: the important in a puzzle solution is substance, notation is a "language" to understand each other, but should not be a matter for endless discussions.

SpAce wrote:My "Extended-AIC": (8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - r1c9 = ((9)r1c9 & (4)r1c5) => -9 r1c2, -8 r1c5

The two embedded AICs are of course:

(8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - (4=9)r1c9 => -9 r1c2
(8)r1c2 = (8-6)r7c2 = r7c4 - (6=3)r8c6 - (3=4)r8c9 - r1c9 = (4)r1c5 => -8 r1c5

So, can it be written like I did? One or two moves?


Here my opinion is "Up to you !"
The single move could be used for a puzzle where a one-step solution is seeked for. For the present puzzle, the solution is complex any how. One step more is of little importance. It depends how you weight simplicity and legibility vs compactness.
I would make no comment to your "extended AIC" (less easy to read, logically correct)

SpAce wrote:I came up with this to place 1r8c3:

AIC (ANS): (1=246)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (2=41)r78c3 => -24 r8c3

Is that how you would write it?


Here, the concern is: "how to write by-standers in ALS's ?" I call "by-standers" the digits of an ALS (or ANS if you prefer this term) that are not involved in the chain where it is embedded. Same for AHS's

You can read in the "Puzzles" forum, AIC's where the bystanders are omitted. Some players would write your AIC: (1=6)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (2=1)r78c3 => -24 r8c3

A few weeks ago, I would have written it so. Such writing shows clearly which digits are in a derived strong link within the ALS's. But I have read some statements about a lack in didactic aspect of such writing.
From logical POW, by-standers may be written on either side of the strong link symbol "=", even though thay are generally written on the side opposite to the restricted common. My own practise is to write them on the most convenient side for writing my whole chain. In the case you submit above, I would write:
(1=246)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (24=1)r78c3 => -24 r8c3
which is a clear demo of +1 r8c3.
Cenoman
Cenoman
 
Posts: 798
Joined: 21 November 2016
Location: Paris, France

Re: One or two (AIC) moves?

Postby SpAce » Mon Mar 12, 2018 12:53 am

Cenoman wrote:First, be aware that I am not an early member of these forums. My answers are nothing but my opinions.

Well, I appreciate your opinions anyhow!

Second, I recall what is often said to newcomers about notation debates: the important in a puzzle solution is substance, notation is a "language" to understand each other, but should not be a matter for endless discussions.

I agree with that. Still I prefer to stick to de facto standards as much as possible because it takes less effort to both write and interpret, and more energy can be spent on the substance.

SpAce wrote:So, can it be written like I did? One or two moves?

Here my opinion is "Up to you !"
The single move could be used for a puzzle where a one-step solution is seeked for. For the present puzzle, the solution is complex any how. One step more is of little importance. It depends how you weight simplicity and legibility vs compactness.
I would make no comment to your "extended AIC" (less easy to read, logically correct)

Thanks! Glad to hear the combined chain wasn't completely wrong, though I agree that it was a bit more complex. The step count is of no concern to me, especially in this case, and it wasn't the reason why I did it. It just seemed kind of dumb to repeat the bulk of the chain when only one little thing changed, and since it seemed logically possible to combine them into one, I did it.

Btw, isn't it actually pretty similar to a split-node chain? Could it in fact be considered one? I just hadn't seen one before where a split-node was at one of the ends proving multiple eliminations, so I didn't think to call it such.

You can read in the "Puzzles" forum, AIC's where the bystanders are omitted. Some players would write your AIC: (1=6)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (2=1)r78c3 => -24 r8c3

A few weeks ago, I would have written it so. Such writing shows clearly which digits are in a derived strong link within the ALS's. But I have read some statements about a lack in didactic aspect of such writing.

I guess both sides are valid, and I try to avoid being too opinionated here. Personally I prefer to see the bystanders but I understand the other point of view, too. The bystanders are especially important in loops because they provide additional and sometimes more significant eliminations. Otherwise I guess it's a matter of opinion whether they provide more clarity or clutter.

From logical POW, by-standers may be written on either side of the strong link symbol "=", even though thay are generally written on the side opposite to the restricted common. My own practise is to write them on the most convenient side for writing my whole chain. In the case you submit above, I would write:
(1=246)r7c3,r8c13 - (6=3)r8c6 - r2c6 = (3-2)r2c1 = r2c3 - (24=1)r78c3 => -24 r8c3
which is a clear demo of +1 r8c3.

I like that! Somehow it feels more natural to separate the restricted common used in the preceding weak link, but thanks for reminding me that it's not the only possibility. Even with that change the chain is still not quite as explicit as I'd like because it doesn't tell which cell in either ALS is occupied by the digit 1, which is required knowledge for the conclusion. One has to see the grid to know that. I guess there's not much we can do about it without lengthening the chain unnecessarily. I'm beginning to accept that with chains involving ALS nodes one has to see the grid or simply trust certain implied parts.

(Added.) I guess that means my actual step (2) could be changed into this as well:

AIC (ANS): (4=2)r7c3 - r2c3 = (2-3)r2c1 = (3-6)r2c6 = r8c6 - (62=4)r7c3,r8c1 => -4 r8c3

In a way it's clearer that we now have the (4) separated at both ends of the chain, and it's also somewhat easier to read both ways. Then again, I'm inclined to think it's more intuitive for a newbie to understand the last node if the weak link digit (6) is separated (6=24) as I originally wrote it. It's pretty straight-forward to see that when the (6) is removed from the almost-naked-pair (246), then it becomes a naked pair (24) which kills the 2s and 4s in its vicinity.

The other way (62=4) requires a bit more advanced understanding, as then the weak link breaks the (6 AND 2) relationship which forces the (4) to be true in the ALS node. I know it took me a while to start reading those kinds of situations correctly. Then again, it's pretty important to learn, because otherwise it's impossible to understand more complex situations, different dialects, and how chains with ALS nodes work backwards without any changes.
Code: Select all
   *             |    |               |    |    *
        *        |=()=|    /  _  \    |=()=|               *
            *    |    |   |-=( )=-|   |    |      *
     *                     \  ¯  /                   *   
SpAce
 
Posts: 605
Joined: 22 May 2017


Return to Help with puzzles and solving techniques

cron