The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |41.|5..|.26|
 |..2|6..|.3.|
 |8..|.4.|.1.|
 |---+---+---|
 |.8.|...|64.|
 |6..|...|..7|
 |.51|...|.8.|
 |---+---+---|
 |.3.|.6.|..5|
 |.4.|..3|1..|
 |19.|..4|.68|
 *-----------*


Play/Print this puzzle online

[eleven]

Code: Select all

 *-----------------------------------------------------*
 |  4    1  39   |  5     3789   789   |  78   2    6  |
 |  59   7  2    |  6     189    189   |  58   3    4  |
 |  8    6  35   |  3-7   4      2     |  57   1    9  |
 |---------------+---------------------+---------------|
 |  3    8 b79#  | c279   279    5     |  6    4    1  |
 |  6    2  4    |  138   138    18    |  9    5    7  |
 |  79   5  1    |  4    b79#    6     |  2    8    3  |
 |---------------+---------------------+---------------|
 |  2    3  8    |  19-7  6      179   |  4    79   5  |
 |  57   4  6    |  789   5789   3     |  1    79   2  |
 |  1    9 c57#  |ca27   c57+2#  4     |  3    6    8  |
 *-----------------------------------------------------*

Almost pattern #, one of r3c4,r6c5 must be 9.
(7=2)r9c4 - 2r9c5 == 9r4c3,r6c5 - (9=7)r49c4 => -7r37c4, stte
[Edit: added r9c4, thx SpAce]

[pjb]

Code: Select all

 4       1       39     | 5      3789  b789    | 78     2      6     
 59      7       2      | 6      189    189    | 58     3      4     
 8       6      d35     |c37     4      2      | 57     1      9     
------------------------+----------------------+---------------------
 3       8       79     | 279    279    5      | 6      4      1     
 6       2       4      | 138    138    18     | 9      5      7     
 79      5       1      | 4      79     6      | 2      8      3     
------------------------+----------------------+---------------------
 2       3       8      | 179    6     a179    | 4      79     5     
 57      4       6      | 789    5789   3      | 1      79     2     
 1       9      e57     | 2-7    25-7   4      | 3      6      8     


(7)r7c6 = r1c6 - (7=3)r3c4 - (3=5)r3c3 - (5=7)r9c3 => -7 r9c45; stte

Phil

[Ngisa]

Code: Select all

+-----------------+----------------------+-----------------+
| 4      1     39 | 5       3789     789 | 78     2      6 |
|c59     7     2  | 6       189      189 | 58     3      4 |
| 8      6    b35 | 3-7     4        2   |a57     1      9 |
+-----------------+----------------------+-----------------+
| 3      8     79 | 279     279      5   | 6      4      1 |
| 6      2     4  | 138     138      18  | 9      5      7 |
|d79     5     1  | 4      e79       6   | 2      8      3 |
+-----------------+----------------------+-----------------+
| 2      3     8  |f179     6       f179 | 4     e79     5 |
|e57     4     6  |f789    f5789     3   | 1     f79     2 |
| 1      9     57 |g27     f257      4   | 3      6      8 |
+-----------------+----------------------+-----------------+


(7=5)r3c7 - r3c3 = (5-9)r2c1 = (9-7)r6c1 = r6c5,r8c1,r7c8 - r8c458|r7c46|r89c5 = r9c4 => - 7r3c4; stte.

Clement

[SteveG48]

Code: Select all

 *-------------------------------------------------------------*
 | 4     1    e39    |   5     3789  789   | 78    2     6     |
 | 59    7     2     |   6     189   189   | 58    3     4     |
 | 8     6   de35    | bc37    4     2     | 57    1     9     |
 *-------------------+---------------------+-------------------|
 | 3     8    e79    |   279   279   5     | 6     4     1     |
 | 6     2     4     |   138   138   18    | 9     5     7     |
 |d79    5     1     |   4   bc79    6     | 2     8     3     |
 *-------------------+---------------------+-------------------|
 | 2     3     8     |   179   6     179   | 4     79    5     |
 | 57    4     6     |   789   5789  3     | 1     79    2     |
 | 1     9     57    |  a27   a257   4     | 3     6     8     |
 *-------------------------------------------------------------*


(5=27)r9c45 - (7r3c4)&(7r6c5) = 3r3c4|9r6c5 - (3r3c3)&(9r6c1) = 5r134c3 => -5 r9c3 ; stte

Corrected typo. Thanks, SpAce.

[SteveG48]

Code: Select all

+-----------------+----------------------+-----------------+
| 4      1     39 | 5       3789     789 | 78     2      6 |
|c59     7     2  | 6       189      189 | 58     3      4 |
| 8      6    b35 | 3-7     4        2   |a57     1      9 |
+-----------------+----------------------+-----------------+
| 3      8     79 | 279     279      5   | 6      4      1 |
| 6      2     4  | 138     138      18  | 9      5      7 |
|d79     5     1  | 4      e79       6   | 2      8      3 |
+-----------------+----------------------+-----------------+
| 2      3     8  |f179     6       f179 | 4     e79     5 |
|e57     4     6  |f789    f5789     3   | 1     f79     2 |
| 1      9     57 |g27     f257      4   | 3      6      8 |
+-----------------+----------------------+-----------------+


(7=5)r3c7 - r3c3 = (5-9)r2c1 = (9-7)r6c1 = r6c5,r8c1,r7c8 - r8c458|r7c46|r89c5 = r9c4 => - 7r3c4; stte.

Clement

Hi, Clement. I like where you're going, but I have two comments.

First, I think the red should be written r6c5&r8c1&r7c8, or (7r6c5)&(7r8c1)&(7r7c8). In this forum, cell designators separated by commas are used to indicate a set of cells, with a candidate being true at least once somewhere in the set (or not true anywhere). You want to indicate the 7 is true in each of the three cells in the set; hence the alternate notation.

More importantly, I don't see where you're getting that 7 must be true in r7c8.

[Cenoman]

Code: Select all

 +-----------------+---------------------+-----------------+
 |  4    1    39   |  5     3789  a789   |  8-7  2    6    |
 |  59   7    2    |  6     189    189   |  58   3    4    |
 |  8    6   e35   |  37    4      2     | f57   1    9    |
 +-----------------+---------------------+-----------------+
 |  3    8    79   |  279   279    5     |  6    4    1    |
 |  6    2    4    |  138   138    18    |  9    5    7    |
 |  79   5    1    |  4     79     6     |  2    8    3    |
 +-----------------+---------------------+-----------------+
 |  2    3    8    |  179   6     b179   |  4    79   5    |
 |  57   4    6    |  789   5789   3     |  1    79   2    |
 |  1    9   d57   | c27   c257    4     |  3    6    8    |
 +-----------------+---------------------+-----------------+

(7)r1c6 = r7c6 - (72=5)r9c45 - r9c3 = r3c3 - (5=7)r3c7 => -7 r1c7; ste

(added) ...or, shorter
(9=5)r49c3 - (5=9)r469c5 => -9 r4c4; lclste

[SpAce]

Code: Select all

.-----------.---------------------.-----------.
| 4   1  39 | 5       3789    789 | 78  2   6 |
| 59  7  2  | 6       189     189 | 58  3   4 |
| 8   6 C35 |D37      4       2   | 57  1   9 |
:-----------+---------------------+-----------:
| 3   8  79 | 279     279     5   | 6   4   1 |
| 6   2  4  | 138     138     18  | 9   5   7 |
|d79  5  1  | 4      e79      6   | 2   8   3 |
:-----------+---------------------+-----------:
| 2   3  8  | 179     6       179 | 4   79  5 |
|c57  4  6  | 789     5789    3   | 1   79  2 |
| 1   9 b57 |E27  Ffa(5)-27   4   | 3   6   8 |
'-----------'---------------------'-----------'

(5)r9c5 = (5)r9c3 -|- (7)r9c3 = r8c1 - r6c1 = r6c5 ------&- (2|7=5)r9c5
                   |- (5=3)r3c3 - (3=7)r3c4 - (7=2)r9c4 -&

=> -27 r9c5; stte

[SpAce]

(7=5)r3c7 - r3c3 = (5-9)r2c1 = (9-7)r6c1 = r6c5,r8c1,r7c8 - r8c458|r7c46|r89c5 = r9c4 => - 7r3c4; stte.

Hi, Clement. I like where you're going, but I have two comments.

First, I think the red should be written r6c5&r8c1&r7c8, or (7r6c5)&(7r8c1)&(7r7c8). In this forum, cell designators separated by commas are used to indicate a set of cells, with a candidate being true at least once somewhere in the set (or not true anywhere). You want to indicate the 7 is true in each of the three cells in the set; hence the alternate notation.

Are you sure of that? I've thought commas are usually synonymous with ANDs (i.e. "&"), though a bit ambiguous.

More importantly, I don't see where you're getting that 7 must be true in r7c8.

I guess the path (7r8c1 - r8c8 = r7c8) was implied.

[SteveG48]

Are you sure of that? I've thought commas are usually synonymous with ANDs (i.e. "&"), though a bit ambiguous.

Not that I've seen here. If someone writes something like ….=(abc)cell1,cell2,cell3 , it means that each candidate, a, b, and c, is true somewhere in the set of three cells. It doesn't mean that each is true in each of the three cells.

[Ngisa]

(7=5)r3c7 - r3c3 = (5-9)r2c1 = (9-7)r6c1 = r6c5,r8c1,r7c8 - r8c458|r7c46|r89c5 = r9c4 => - 7r3c4; stte.

Hi, Clement. I like where you're going, but I have two comments.

First, I think the red should be written r6c5&r8c1&r7c8, or (7r6c5)&(7r8c1)&(7r7c8). In this forum, cell designators separated by commas are used to indicate a set of cells, with a candidate being true at least once somewhere in the set (or not true anywhere). You want to indicate the 7 is true in each of the three cells in the set; hence the alternate notation.

Are you sure of that? I've thought commas are usually synonymous with ANDs (i.e. "&"), though a bit ambiguous.

More importantly, I don't see where you're getting that 7 must be true in r7c8.

I guess the path (7r8c1 - r8c8 = r7c8) was implied.

You are quite right, that was the implication.

[SpAce]

Are you sure of that? I've thought commas are usually synonymous with ANDs (i.e. "&"), though a bit ambiguous.

Not that I've seen here. If someone writes something like ….=(abc)cell1,cell2,cell3 , it means that each candidate, a, b, and c, is true somewhere in the set of three cells. It doesn't mean that each is true in each of the three cells.

Of course, but there's still an implied AND between the digits because all of them need to be placed in those cells (we just don't know or specify the exact order -- but that doesn't imply OR). You wouldn't use "|" instead of "," there, would you?

I've thought it's the same as in normal ALS nodes where a lack of both "|" and "&" implies "&" (e.g. (4=237)r3c456), and the comma is used when the group can't be short-handed like that, e.g. it's spread in a box like (4=237)r3c4,r2c56. There's an implied "&" because all of those digits need to go into those cells in some order. "OR" would be appropriate if we had a single target cell (4=2|3|7)r3c4.

This is not exact science because writing the fully correct logical relationships in multi-digit and/or multi-cell situations would be extremely difficult and unreadable. Thus it's all short-handed some way. I've seen some post or document describing the "correct" use of commas within Eureka chains, but I can't find it now. The Sudopedia Eureka refererence is of no help, as it's very limited and describes some practices in conflict with current standards, such as:

Unordered group, in a box
(1)r5c4|r6c45

(Now I know why Clement said that in an earlier discussion about notations. At the time I was baffled, as most would use a comma for that purpose, and "|" to mean OR.)

[SteveG48]

Are you sure of that? I've thought commas are usually synonymous with ANDs (i.e. "&"), though a bit ambiguous.

Not that I've seen here. If someone writes something like ….=(abc)cell1,cell2,cell3 , it means that each candidate, a, b, and c, is true somewhere in the set of three cells. It doesn't mean that each is true in each of the three cells.

Of course, but there's still an implied AND between the digits because all of them need to be placed in those cells (we just don't know or specify the exact order -- but that doesn't imply OR). You wouldn't use "|" instead of "," there, would you?

Certainly not. I would use "&" as I said earlier.

I make a distinction here been groups of candidates and sets of cells. "abc" in my example is equivalent to "a&b&c" , but I wouldn't think of using a notation like (a,b,c)cell1,cell2,cell3. I wouldn't know what that means. To me, the use of the comma is the same as in everyday English- to denote a list of things that are part of a group. In this example, "cell1,cell2,cell3" indicates a set of three cells that constitute a node in the expression.

I think our custom of not indicating the candidate when only one is involved helps to confuse matters here.

[SpAce]

I make a distinction here been groups of candidates and sets of cells. "abc" in my example is equivalent to "a&b&c" , but I wouldn't think of using a notation like (a,b,c)cell1,cell2,cell3. I wouldn't know what that means.

If my memory serves right, the meaning of that was explained in the post I can't find right now. Based on that fading memory, I interpret (and use) the comma in such a situation to mean that the exact locations of those digits (or some of them) is fixed in specific cells (the same ordering is used for both digits and cells). So the comma in that case is a more exact AND operator. I use it sometimes to separate the linking digits in large ALSs where the link wouldn't otherwise be obvious.

To me, the use of the comma is the same as in everyday English- to denote a list of things that are part of a group. In this example, "cell1,cell2,cell3" indicates a set of three cells that constitute a node in the expression.

Yes, but what's the difference between that interpretation and having an "&" between those cells?

I think our custom of not indicating the candidate when only one is involved helps to confuse matters here.

Can you elaborate? I don't quite follow what you mean here.

[SteveG48]

Can you elaborate? I don't quite follow what you mean here.

In this situation, Clement wrote r6c5,r8c1,r7c8. Since 7 was the only candidate being discussed here, it was understood. However, with a list of cells it's not always understood whether one is referring to the set of cells or to the truth values of candidates in those cells.

Had Clement written (7)r6c5,r8c1,r7c8 , then I would insist that it's referring to the truth of 7 somewhere in that set. Had he written either
(7r6c5)(7r8c1)(7r7c8) or (7r6c5)&(7r8c1)&(7r7c8), then it would refer to the truth of 7 in all three cells. As it was written is ambiguous to me.