The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |..1|...|.7.|
 |23.|...|...|
 |...|29.|.1.|
 |---+---+---|
 |...|.46|5..|
 |4.7|..5|..2|
 |.68|...|..7|
 |---+---+---|
 |68.|5..|7.4|
 |...|.2.|...|
 |1..|..4|..5|
 *-----------*

Play/Print this puzzle online

[pjb]

Code: Select all

 89     c49      1      |b46     5      38     | 2      7      36     
 2       3       46     | 146   a67     17     | 89     5      89     
 78      57      56     | 2      9      38     | 4      1      36     
------------------------+----------------------+---------------------
 39     c129     239    | 7      4      6      | 5      89     189   
 4      c19      7      | 139    8      5      | 1369   369    2     
 5       6       8      | 139    13     2      | 139    4      7     
------------------------+----------------------+---------------------
 6       8       239    | 5      13     19     | 7      239    4     
 379     45      45     | 368    2      79     | 13689  3689   189   
 1      c279     239    | 368    6-7    4      | 3689   23689  5     


(7=6)r2c5 - (6=4)r1c4 - (4=9127)r1459c2 => -7 r9c5; stte
or shorter
(6)r2c3 = (6-5)r3c3 = (5-7)r3c2 = r9c2 - (7=6)r9c5 => -6 r2c5
Phil

[Leren]

Code: Select all

*-----------------------------------------------------------------------*
| 89     49     1       | 46     5      38      | 2      7      36      |
| 2      3     b46      | 146   a67     17      | 89     5      89      |
| 78    d57    c56      | 2      9      38      | 4      1      36      |
|-----------------------+-----------------------+-----------------------|
| 39     129    239     | 7      4      6       | 5      89     189     |
| 4      19     7       | 139    8      5       | 1369   369    2       |
| 5      6      8       | 139    13     2       | 139    4      7       |
|-----------------------+-----------------------+-----------------------|
| 6      8      239     | 5      13     19      | 7      239    4       |
| 379    45     45      | 368    2      79      | 13689  3689   189     |
| 1     f279    239     | 368    6-7    4       | 3689   23689  5       |
*-----------------------------------------------------------------------*

(7=6) r2c5 - r2c3 = (6-5) r3c3 = (5-7) r3c2 = (7) r9c2 => - 7 r9c5; stte

Leren

[Ngisa]

Code: Select all

+-------------+-----------+-----------------+
| b89  b49  1   | b46  5  38 | 2     7     36  |
| 2   3   46  | 146 c67 17 | 89    5     89  |
| a78  57  56  | 2   9  38 | 4     1     36  |
+-------------+-----------+-----------------+
| 39  129 239 | 7   4  6  | 5     89    189 |
| 4   19  7   | 139 8  5  | 1369  369   2   |
| 5   6   8   | 139 13 2  | 139   4     7   |
+-------------+-----------+-----------------+
| 6   8   239 | 5   13 19 | 7     239   4   |
| 39-7 45  45  | 368 2  79 | 13689 3689  189 |
| 1   e279 239 | 368 d67 4  | 3689  23689 5   |
+-------------+-----------+-----------------+


(7=8)r3c1 - (8=6)r1c124 - (6=7)r2c5 - r9c5 = (7)r9c2 => -7 r8c1; stte

Clement

[Cenoman]

Code: Select all

 +-------------------+-----------------+-----------------------+
 | 89    49    1     | 46    5    38   | 2       7       36    |
 | 2     3     4-6   | 146  a67   17   | 89      5       89    |
 | 78   d57   d56    | 2     9    38   | 4       1       36    |
 +-------------------+-----------------+-----------------------+
 | 39    129   239   | 7     4    6    | 5       89      189   |
 | 4     19    7     | 139   8    5    | 1369    369     2     |
 | 5     6     8     | 139   13   2    | 139     4       7     |
 +-------------------+-----------------+-----------------------+
 | 6     8     239   | 5     13   19   | 7       239     4     |
 | 379   45    45    | 368   2    79   | 13689   3689    189   |
 | 1    c279   239   | 368  b67   4    | 3689    23689   5     |
 +-------------------+-----------------+-----------------------+

(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

Cenoman

[bat999]

Code: Select all

.------------------.-----------------.-------------------.
| d89    9-4   1   | a46     5    38 | 2      7      36  |
|  2     3    d46  |  16-4   67   17 | 89     5      89  |
| d78   d57   d56  |  2      9    38 | 4      1      36  |
:------------------+-----------------+-------------------:
| c39    129   239 |  7      4    6  | 5      89     189 |
|  4     19    7   |  139    8    5  | 1369   369    2   |
|  5     6     8   |  139    13   2  | 139    4      7   |
:------------------+-----------------+-------------------:
|  6     8     239 |  5     b13  b19 | 7      239    4   |
| c379   45    45  | a368    2   b79 | 13689  3689   189 |
|  1     279   239 | a368    67   4  | 3689   23689  5   |
'------------------'-----------------'-------------------'

(4=3)r189c4 - (3=7)r7c56,r8c6 - (7=9)r48c1 - (9=4)r1c1,r2c3,r3c123 => -4 r1c2,r2c4; stte

[bat999]

(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

Hi
Your solution is OK but I think there's a typo.

This...
(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

Should be this (imho)...
(6=7)r2c5 - r9c5 = r9c2 - (7=56)r3c23 => -6 r2c3; stte

[Sudtyro2]

(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

Hi
Your solution is OK but I think there's a typo.
This...
(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3
Should be this (imho)...
(6=7)r2c5 - r9c5 = r9c2 - (7=56)r3c23 => -6 r2c3; stte

Hi Bat,
I think Cenoman's notation is OK.
The premise, (75=6)r3c23, is equally valid as (7=56)r3c23, or even (7=6)r3c23. The important point is to keep track of the linking digits within the node.
All the others are bystanders.

SteveC

[bat999]

... The important point is to keep track of the linking digits within the node...

This is the chain...
(6=7)r2c5 - (7)r9c5 = (7)r9c2 - (7=5)r3c2 - (5=6)r3c3 => -6 r2c3; stte
Simplifies to...
(6=7)r2c5 - r9c5 = r9c2 - (7=6)r3c23 => -6 r2c3; stte


I would not include the "bystander" 5.
(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

[Sudtyro2]

I would not include the "bystander" 5.
(6=7)r2c5 - r9c5 = r9c2 - (75=6)r3c23 => -6 r2c3

I look at r3c23 simply as two cells containing 3 candidates, abc. That's an ALS by definition and can be expressed in any number of ways, depending on how the ALS, as a node, is linked externally on the right and the left. One can write (a=bc), (a=cb), (b=ac), (ab=c), (ba=c), etc ... depending on which two digits link externally. In the case at hand, the 7-digit links on the left, and the 6 digit links on the right. The 5-digit is the "bystander" and can grouped together with either the 7 or 6. It can even be omitted entirely in the notation, although I would personally discourage that practice.

In this particular case, it is also possible to "decompose" the ALS into two separate nodes. But, why do that?

SteveC

[Cenoman]

Hi Steve and Bat,
ALS's can be considered under different angles.
The main point is: an ALS creates derived strong links between any pair of groups out of the digits it contains. From this POW you would write (7=6)r3c23
Another angle is to consider an ALS as the assembly of a group of one digit and of the set of all the other digits in the ALS cells, that are then a naked set. You could call it an ANS (Almost Naked Set). From that POW you would write (7=56)r3c23
(75=6)r3c23 is another way to join a ANS and a digit. I used it to focus the weak link with the eliminated candidate.
But doing this, I made a deviation to the Eureka notation, that recommends to isolate the restricted common. Just a side remark: how to isolate restricted commons in an ALS XY wing or in an ALS chain, for middle ALS ? I claimed to follow Eureka notation in a previous post, and maybe Bat noted that deviation...

I agree with Bat: I do not favor the omission of bystanders, because such a practise could not be extended to duale AHS. In the case of the grid above, writing (7-6)r3c169 would raise a lot of questions/comments, altough fully correct from the logics POW. I would write (7-386)r3c169 or even 7-AHT386)r3c169. When using AHS's locked digits must always be included in the notation for sake of legibility. I note ALS's the same way, with some freedom to place other digits right or left...

I agree with Steve: the linkings digits must in any case occupy the extreme positions (first and last ones) in the ALS brackets, in order to make the chains easier to read.

For the moment, I'm just trying to catch the habits and practises of the forum. I do not claim any pretention to introduce novelty.

Best regards, Cenoman.

[bat999]

... you would write (7=6)r3c23
Another angle is to consider an ALS as the assembly of a group of one digit and of the set of all the other digits in the ALS cells, that are then a naked set. You could call it an ANS (Almost Naked Set). From that POW you would write (7=56)r3c23...

OK, these two I'm happy with for this puzzle.
-(7=6)r3c23
-(7=56)r3c23

...(75=6)r3c23 is another way...

I'm not convinced about this one but maybe you and SteveC are correct.
-(75=6)r3c23

[SteveG48]

Hi Steve and Bat,
I agree with Bat: I do not favor the omission of bystanders, because such a practise could not be extended to duale AHS. In the case of the grid above, writing (7-6)r3c169 would raise a lot of questions/comments, altough fully correct from the logics POW. I would write (7-386)r3c169 or even 7-AHT386)r3c169. When using AHS's locked digits must always be included in the notation for sake of legibility. I note ALS's the same way, with some freedom to place other digits right or left...

Good discussion. Cenoman, I think you're agreeing with SteveC here. Bat appears to favor omission of the bystanders. This practice started here awhile back, and I was one of the ones who began using it. Some other members objected, and I've come to agree that it's not a good idea. Others continue to favor it.

I agree with Steve: the linkings digits must in any case occupy the extreme positions (first and last ones) in the ALS brackets, in order to make the chains easier to read.

I personally prefer to just list the candidates in numerical order, but I don't think it's important one way or the other. I think my preference came from a discussion with David P Bird, but I'm not sure of that.

One detail that hasn't been discussed here- probably because it isn't directly related to the chains being discussed- is the use of | and & for the OR and AND functions. I prefer to always explicitly include the | when that's the intent, but to omit the &. Thus, a|b for a OR b and ab for a AND b. With that convention, it's clear that (75=6)r3c23 and (7=56)r3c23 are equally valid. On the other hand, (7|5=6)r3c23 would represent a condition that can't exist, since only one candidate would be left in the two cell set. However, in some chains, the OR is clearly necessary for the chain to be valid, and the omission of it can be confusing.

In a more complex situation where two (or more) candidates are being eliminated leaving two (or more) candidates in a now locked set, we might have something like (a|b=cd)r1c23. In this situation, writing a technically correct expression that puts the c on the left side of the equation would be awkward. That's why I prefer leaving the candidates remaining in the locked set on the right side of the equation.

[bat999]

I'm not convinced about this one...
(75=6)r3c23

Aha, this says to me "NOT(7AND5) in r3c23 forces 6 in r3c23".
If r9c2 = 7 then r3c2 <> 7 so NOT(7AND5) is true to force the 6.

[Cenoman]

SteveG48 wrote:
However, in some chains, the OR is clearly necessary for the chain to be valid, and the omission of it can be confusing.

I have seen this in some paths, handled by high level players, in very hard puzzles.
Generally, it is used to write compact solutions. I don't use it. The counterpart is to write two or three chains instead of one, or to use krakens. But the ambiguity starts already with the (7=6)r3c23 writing: the strong link results from ALS (567)r3c23. What you can infer from the generic ALS pattern is 7r3c23 OR 6r3c23 that you should write (7|6)r3c23. In other terms, it remains unclear for me whether the symbol = is synonym of OR or synonym of XOR. A lot of people use it as synonym of OR, but then it does not make sense to use it in conjonction with the symbol |
In the example above (7 XOR 6)r3c23 is valid, since the ALS r3c23 is also an AHS with two linking digits (5 is locked in r3c23), but this not the general case.

So (7=6)r3c23 means here (75 XOR 56)r3c23, but in a generic ALS it would mean: (75 OR 56 OR 67)r3c23 - no exclusion of (6 AND 7) TRUE @r3c23

SteveG48 wrote:
...we might have something like (a|b=cd)r1c23

OK, but such an example would be 4 digits in 2 cells. This is an AALS and all the discussion above is no more relevant. I agree that writing a technically correct expression is no piece of cake. BTW a pattern I like is "Sue de Coq" but I don't know how to write it in Eureka !

Cenoman