- Code: Select all
*-----------*
|.6.|9..|...|
|9.7|..1|.26|
|3.5|...|7.1|
|---+---+---|
|.93|.7.|...|
|...|...|...|
|...|.9.|63.|
|---+---+---|
|5.8|...|3.2|
|27.|3..|9.8|
|...|..5|.1.|
*-----------*
Play/Print this puzzle online
*-----------*
|.6.|9..|...|
|9.7|..1|.26|
|3.5|...|7.1|
|---+---+---|
|.93|.7.|...|
|...|...|...|
|...|.9.|63.|
|---+---+---|
|5.8|...|3.2|
|27.|3..|9.8|
|...|..5|.1.|
*-----------*
+-------------------+----------------------+-----------------+
| 18 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 248 | 7 9 1 |
+-------------------+----------------------+-----------------+
| a14 9 3 | 6 7 a24 | 12 8 5 |
| 148 258 6 | 12458 b258 3 | 12 7 9 |
| 7 258 2-1 | b1258 9 b28 | 6 3 4 |
+-------------------+----------------------+-----------------+
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
+-------------------+----------------------+-----------------+
.-----------------.------------------.-----------.
| a(8)-1 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 248 | 7 9 1 |
:-----------------+------------------+-----------:
| c(1)4 9 3 | 6 7 24 | c12 8 5 |
| b148 b258 6 | 12458 b258 3 | b12 7 9 |
| 7 258 12 | 1258 9 28 | 6 3 4 |
:-----------------+------------------+-----------:
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
'-----------------'------------------'-----------'
+--------------+-------------------+--------------+
| 18 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 248 | 7 9 1 |
+--------------+-------------------+--------------+
| 14 9 3 | 6 7 b24 | b12 8 5 |
| 148 258 6 | c2458-1 258 3 | a12 7 9 |
| 7 258 12 | 1258 9 28 | 6 3 4 |
+--------------+-------------------+--------------+
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
+--------------+-------------------+--------------+
SpAce wrote:
- Code: Select all
.-----------------.------------------.-----------.
| a(8)-1 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 248 | 7 9 1 |
:-----------------+------------------+-----------:
| c(1)4 9 3 | 6 7 24 | c12 8 5 |
| b148 b258 6 | 12458 b258 3 | b12 7 9 |
| 7 258 12 | 1258 9 28 | 6 3 4 |
:-----------------+------------------+-----------:
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
'-----------------'------------------'-----------'
(8)r1c1 = (852'1)r5c1257 - 1r4(c7=c1) => -1 r1c1; stte
Steve, would that little change make you hate the 3D notation a bit less? I think it might improve readability, though duplicating one character.
SteveG48 wrote:A bit, but mainly I don't like the idea of including links in the part of the term notation that identifies the cell set (your third term).
Also, I really don't need the ' in your second term, though I have no objection.
I like what Cenoman did above with the font designating the relevant digits, if the writer wants to go to the trouble. I really have no objection at all to the abbreviated notation. I only avoid it because some members of the forum say that affects clarity for them.
Sudopedia wrote: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. (source)
SteveG48 wrote:I like what Cenoman did above with the font designating the relevant digits, if the writer wants to go to the trouble.
Sudopedia wrote: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.
Cenoman wrote:When SpAce raised the point a few weeks ago, I found four ways of writing ALS's on the forum
I used the abbreviated notation for a while and I stopped using it because such a practise cannot be used symetrically in AHS's notation; in AHS's, bystanders are the digits locked in the AHS cells, and are not easy to find if these cells have 4 or more candidates each.
So back to Eureka notation, until I recently became aware that eleven's notation was a good compromise to improve it.
On this site, the work load to lower the font size is very light (just select the bystanders and click "small" in the font list).
A 5th notation has appeared recently with SpAce's proposal to keep the order of Eureka, with a character ' separating bystanders from the linking digit in the ANS.
Steve's and phil's notation is easy to read, despite appearances; no need to search the linking digit in the middle of ANS digits, just read it at the beginning of the next node.
Everyone's taste is different and there is no arguing matters of taste.
Personally, I will read on the five notations, and write on eleven's.
RSW wrote:My text based solution (since I haven't learned the notation yet)
.------------------.-------------------.----------.
| 18 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 248 | 7 9 1 |
:------------------+-------------------+----------:
| b14 9 3 | 6 7 c(2)4 | 12 8 5 |
| 148 258 6 | 12458 258 3 | 12 7 9 |
| 7 258 a1(2) | 158-2 9 8-2 | 6 3 4 |
:------------------+-------------------+----------:
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
'------------------'-------------------'----------'
.--------------.-------------------.-------------.
| 18 6 12 | 9 258 7 | 58 4 3 |
| 9 48 7 | 458 3 1 | 58 2 6 |
| 3 248 5 | 248 6 24 | 7 9 1 |
:--------------+-------------------+-------------:
| 14 9 3 | 6 7 24 | 12 8 5 |
| 148 58 6 | 245-1 b25 3 | c(1)2 7 9 |
| 7 25 12 | a(1)5 9 8 | 6 3 4 |
:--------------+-------------------+-------------:
| 5 1 8 | 7 4 9 | 3 6 2 |
| 2 7 4 | 3 1 6 | 9 5 8 |
| 6 3 9 | 28 28 5 | 4 1 7 |
'--------------'-------------------'-------------'
SpAce wrote:SteveG48 wrote:A bit, but mainly I don't like the idea of including links in the part of the term notation that identifies the cell set (your third term).
Fair enough. Can you elaborate a bit why you don't like the idea? I liked it the very first time when I saw people like SteveK and Denis Berthier use something like that. I just didn't like their implementations, but I really liked the idea (makes clearer what the linking axis is). I still do, and I think my implementation is more readable than theirs -- especially for someone used to normal Eureka.
SteveG48 wrote:Basically, it's a matter of complexity for people learning the Eureka "language".
In it's simplest form, Eureka consists of terms connected by strong and weak links. Each term consists of a candidate or candidate list and a cell or set of cells. To shorten a chain a bit, we've agreed that if the cell set in successive terms is the same, then we can use parentheses with the two candidate lists and the link written together. Thus (a)set1 = (b)set1 becomes (a=b)set1. Now you propose that we do the same with set designations. Thus (a)set1 = (a)set2 becomes
(a)(set1=set2). Things are getting more complicated.
What happens when someone decides to write (a=b)(set1=set2) ?
What does this mean?
It just seems to me that we're on a slippery slope of complexity.
Allan Barker wrote:For now, forget standard Sudoku cells and digits as I don't rely on either and I will avoid the terms.
The physical picture.
Start with a 9x9x9 cube sitting on a table with a viewer facing one side of the cube. The cube has 729 identical cubical units. Since this is a crystal cube, you can look from any face to the opposite face through 81 channels and each channel has 9 cubical units running from face to face. The 3 cube faces, top, left, and front represent 3 orthogonal groups of 81 channels that intersect. These channels are Sudoku sets and Sudoku rules say that each of the 324 sets (now incluing 81 box sets) must contain exactly one object for a total of 81 objects in the cube. The box sets are horizontal 3x3x1 regions. This is sufficient to define Sudoku and solve puzzles.
Coordinates and labels
The coordinate system is 3D where the left to right channels are rows, front to back channels are columns, top to bottom channels are stacks, and boxes are the horizontal 3x3x1 regions. The 81 objects can be all the same or different, it doesn't matter.
SpAce wrote:In fact, I'd love to have a physical cube like that with which to play The fun bit is that it's actually playable no matter which side you're looking at. I've solved a couple of sudokus completely with only the nr or nc space, and I'm conjecturing that anything that can be done in one space can be done in the others (it just looks different). Interestingly enough, I think Denis Berthier has said (somewhere) that it shouldn't be possible because supposedly only Latin Squares rules can be used in the nr or nc spaces, i.e. box constraints aren't usable. Either I misunderstood what he meant, or it's obviously crap, because the box constraints are almost easier to use in those spaces.
+--c1----c2----c3---+--c5----c5--
r1| 1 2 3 | . .
r2| . . . | . .
r3| . . . | . .
+-------------------+------------
r4| . . . | . .
r5| . . . | . .
+--c1----c2----c3---+--c5----c5--
n1| 1 . . | . . }
n2| . 1 . | . . }digit 1 = number of the row occupied by digits 1, 2, 3 in r-c space
n3| . . 1 | . . }
+-------------------+------------
n4| . . . | . .
n5| . . . | . .
Cenoman wrote:SpAce wrote:In fact, I'd love to have a physical cube like that with which to play The fun bit is that it's actually playable no matter which side you're looking at. I've solved a couple of sudokus completely with only the nr or nc space, and I'm conjecturing that anything that can be done in one space can be done in the others (it just looks different). Interestingly enough, I think Denis Berthier has said (somewhere) that it shouldn't be possible because supposedly only Latin Squares rules can be used in the nr or nc spaces, i.e. box constraints aren't usable. Either I misunderstood what he meant, or it's obviously crap, because the box constraints are almost easier to use in those spaces.
Here is mini-row 1, in box 1, of a (solved) sudoku grid in canonical form, written in its r-c space:
- Code: Select all
+--c1----c2----c3---+--c5----c5--
r1| 1 2 3 | . .
r2| . . . | . .
r3| . . . | . .
+-------------------+------------
r4| . . . | . .
r5| . . . | . .
The corresponding box 1 of the same grid, written in its n-c space:
- Code: Select all
+--c1----c2----c3---+--c5----c5--
n1| 1 . . | . . }
n2| . 1 . | . . }digit 1 = number of the row occupied by digits 1, 2, 3 in r-c space
n3| . . 1 | . . }
+-------------------+------------
n4| . . . | . .
n5| . . . | . .
Obviously, there are three digits 1 on the mini-diagonal of box 1, which is a violation of valid sudoku grids.
What Denis Berthier meant is that n-c and n-r spaces are not sudoku grids, they are only Latin squares.
Some sudoku methods of resolution may be used in such spaces (notably naked and hidden sets, ALSs, AICs...), but searches of strong and weak links shall be limited to rows and columns. Weak links are nonsense in "boxes"; actually, boxes do not exist in n-c and n-r spaces.
Of course, boxes can be drawn, but it's useless.
r 1 2 3 4 5 6 7 8 9
n .-------------------.-------------------.-------------------.
1 | 89 <6> <2> | <5> 38 1(7) | 49-7 1(89) 1348 |
'-------------------'-------------------'-------------------'
2 | 46 <7> 1 | 89 2 <5> | 49 <3> 468 |
'-------------------'-------------------'-------------------'
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