The New Sudoku Players' Forum

[RichardGoodrich]

RichardGoodrich richardgoodrich@gmail.com 2013.04.08-Monday

This is my first time to use Medusa 3D coloring (M3D). I was able to solve the
following puzzle, but I have some questions. The M3D rules from sudokuwiki.org
are after line 73. At line 88 I start the coloring. At line 119 I note some
'collisions' with the two colors. I did not see a rule that covered that.

1. Did I assign the color and its 'opposite' correctly? Line 88+
2. Did I manage the collision between colors correctly? Line 119+
3. Did I extend the 'opposite' color correctly? Line 149+
4. Is my logic OK in 'opposite' color was not part of the solution? ~Line 177
5. If the above are OK, why did my eliminations not agree with HoDoKu? Line 177+

I was able to solve the problem, but I wanted help with my procedure! Thanks!

Unfair (4274) by HoDoKu - v2.1.3

Code: Select all

    1 . . | . 8 . | 5 3 .
    . . 3 | . 9 7 | . 6 .
    . . . | . . . | . 8 7
    ------+-------+------
    . . . | . . . | . 7 .
    6 . 1 | . 4 . | 3 . 2
    . 2 . | . . . | . . .
    ------+-------+------
    7 1 . | . . . | . . .
    . 8 . | 5 6 . | 7 . .
    . 6 4 | . 7 . | . . 8


Solution:

Code: Select all

    1 4 7 | 2 8 6 | 5 3 9
    8 5 3 | 4 9 7 | 2 6 1
    2 9 6 | 3 5 1 | 4 8 7
    ------+-------+------
    4 3 8 | 6 2 9 | 1 7 5
    6 7 1 | 8 4 5 | 3 9 2
    5 2 9 | 7 1 3 | 8 4 6
    ------+-------+------
    7 1 5 | 9 3 8 | 6 2 4
    9 8 2 | 5 6 4 | 7 1 3
    3 6 4 | 1 7 2 | 9 5 8


Line 43:

r2c1=8, r2c2=5, r4c2=3, n2r23c7 => r79c7<>2, n4r46c1 => r3c1<>4,
n8r5c46 => r46c46<>8

Code: Select all

    1 . . | . 8 . | 5 3 .
    8 5 3 | . 9 7 | . 6 .
    . . . | . . . | . 8 7
    ------+-------+------
    . 3 . | . . . | . 7 .
    6 . 1 | . 4 . | 3 . 2
    . 2 . | . . . | . . .
    ------+-------+------
    7 1 . | . . . | . . .
    . 8 . | 5 6 . | 7 . .
    . 6 4 | . 7 . | . . 8


Code: Select all

.-----------------.---------------------.--------------------.
| 1     479  2679 | 246    8     246    | 5      3     49    |
| 8     5    3    | 124    9     7      | 124    6     14    |
| 29    49   269  | 12346  1235  123456 | 1249   8     7     |
:-----------------+---------------------+--------------------:
| 459   3    589  | 1269   125   12569  | 14689  7     14569 |
| 6     79   1    | 789    4     589    | 3      59    2     |
| 459   2    5789 | 13679  135   13569  | 14689  1459  14569 |
:-----------------+---------------------+--------------------:
| 7     1    259  | 23489  23    23489  | 469    2459  34569 |
| 239   8    29   | 5      6     12349  | 7      1249  1349  |
| 2359  6    4    | 1239   7     1239   | 19     1259  8     |
'-----------------'---------------------'--------------------'


Line 73:

3D-Medusa Elimination Rules - Summary from sudowiki.org

1. same color twice in a cell eliminates that color
2. twice in a house for same digit eliminates that color
3. 2 colors in a cell eliminates that candidate in the cell
4. 2 colors in a unit for same candy - all candidates that can
see both colors can be eliminated
5. 2 colors elsewhere that can see same candidate that
candidate can be removed
6. any un-colored candidate that sees a colored candidate
elsewhere and OPPOSITE colored candidate in same cell
it can be removed

Line 88:

Using Hodoku Player: 3D Medusa coloring

red = one color at n49r2c2=9
lt-red = 'opposite' color at n49r1c9=4, red is 9 at this BV cell
(firstly is the correct procedure?)

red start at n49r3c2
------------------------
01 9 r3c2=9 by choice
02 9 r1c9=9
03 7 r5c2=7
04 7 r1c3=7
05 7 r6c4=7
06 4 r1c2=4
07 2 r3c1=2
08 2 r2c7=2
09 6 r3c3=6

red-light choose n49r1c9=4
--------------------------
01 4 r1c9=4
02 1 r2c9=1n
03 2 r2c7=2n - red
04 9 r3c7=9n
05 4 r3c2=4n - 9 is red
06 2 r3c1=2n - red
07 6 r3c3=6n - red
08 4 r2c4=4n

Line 119:

Note we have some collisions of colors in cells.
It is not explicitly in Medusa 3D (M3D) rules, but I assume if
a candie in a cell is colored with two 'opposite' colors
(red & red-light are just that) then that candie must be
a solution.

I proceeded from here, but have some problems with some of the eliminations


Continuing Trace where opposite colors indicate same cell is solution
---------------------------------------------------------------------
-07 2 r2c7=2n opposite colors agree
-08 2 r3c1=2n opposite colors agree
-09 6 r3c3=6n opposite colors agree
-10 26 n26r1c46 => r1c46=26

So I eliminate the collisions and reorder and renumber the
red cells and the light-red cells:

red start at n49r3c2 red-light choose n49r1c9=4
------------------------ --------------------------
01 9 r3c2=9 by choice 01 4 r3c2=4n by choice
02 4 r1c2=4n 02 4 r1c9=4
03 7 r1c3=7bcr 03 4 r2c4=4n
04 9 r1c9=9br 04 1 r2c9=1n
05 7 r5c2=7n 05 9 r3c7=9n
06 7 r6c4=7bcr

Line 149:

Seeing no additions to red, look at light red

06 1 r9c7=1n
07 1 r6c8=1bc
08 1 r8c6=1br
09 4 r7c6=4bc
10 4 r8c8=4r
11 6 r7c7=6n
12 8 r7c4=8br
13 8 r5c6=8bcr
14 39 r8c19=39 => r8c3=2
15 79 r5c24=79 => r5c8=5
16 5 r7c9=5bc
17 5 r9c1=5br
18 9 r7c3=9n
19 9 r1c2=9br
20 9 r8c9=9r
21 2 r7c8=2n
22 3 r8c1=3n
23 7 r1c3=7n
24 6 r46c9=6 same color of same candie in a house
25 7 r5c2=7 for both colors

So light-red cannot be part of a solution
delete all light-red candies

Line 177: x marks a conflict with what HoDoKu says is a valid elimination-why?

-11 r1c2<>9
-12 r1c9<>4
-13 r2c4<>4 x
-14 r2c9<>1 x
-15 r3c2<>4
-16 r3c7<>9
-17 r7c3<>9
-18 r7c4<>8
-19 r7c6<>4
-20 r7c7<>6 x
-21 r7c8<>2 x
-22 r7c9<>5 -
-23 r8c1<>3
-24 r8c2<>2
-25 r8c6<>1
-26 r8c8<>4
-27 r8c9<>9
-28 r9c1<>5
-29 r9c7<>1

Line 199: Start using the first few eliminations and then just solved from there.

-30 r1c2=4n
-31 r1c9=9n
-32 r3c2=9n
-33 r5c2=7n
-34 r6c1=7bcr
-35 r9c1=3bc
-36 r8c1=9n
-37 r8c3=2n
-38 r8c3=5n
-39 r9c8=5n
-39 r5c8=9n
-40 r5c4=8n
-41 r5c6=5n
-42 r7c8=2bc
-43 r3c5=5bcr
-44 r7c6=8bcr
-45 r7c5=3n
-46 r6c5=1n
-47 r4c5=2n Cost2=0
-48 r8c9=3bcr
-49 r6c6=3br
-50 r3c4=3bcr Cost3=0
-51 r6c8=4n
-52 r4c1=4bcr
-53 r8c6=4r
-54 r2c4=4bc
-55 r3c7=4br
-56 r7c9=4bcr
-57 r2c9=1n
-58 r3c6=1n
-59 r4c7=1br
-60 r8c8=1n
-61 r9c4=1bcr Cost1=0
-62 2 r1c4=2c
-63 2 r9c6=2bcr Cost2=0
-64 5 r6c1=5n
-65 5 r4c9=5n Cost5=0
-66 6 r1c6=6n
-67 6 r4c4=6c
-68 6 r6c9=6n
-69 6 r7c7=6br Cost6=0
-70 8 r6c7=8n
-71 8 r3c3=8bcr Cost8=0
-72 9 r4c6=9n
-73 9 r6c3=9n
-74 9 r7c4=9n
-75 9 r9c7=9n Cost9=0 Done!

147286539853497261296351487438629175671845392529713846715938624982564713364172958

Code: Select all

.---------.---------.---------.
| 1  4  7 | 2  8  6 | 5  3  9 |
| 8  5  3 | 4  9  7 | 2  6  1 |
| 2  9  6 | 3  5  1 | 4  8  7 |
:---------+---------+---------:
| 4  3  8 | 6  2  9 | 1  7  5 |
| 6  7  1 | 8  4  5 | 3  9  2 |
| 5  2  9 | 7  1  3 | 8  4  6 |
:---------+---------+---------:
| 7  1  5 | 9  3  8 | 6  2  4 |
| 9  8  2 | 5  6  4 | 7  1  3 |
| 3  6  4 | 1  7  2 | 9  5  8 |
'---------'---------'---------'


[JasonLion]

When I do 3D Medusa coloring I get the following pattern, which is similar to yours, but not quite the same.

Colors.png (20.45 KiB) Viewed 1677 times

I don't see any spots where a single candidate gets two different colors.

[RichardGoodrich]

Thanks I will study your post! Shalom.

[RichardGoodrich]

Jason,

Could you tell me what software you used or did you do it by hand?

I can't seem to insert a *.png file? (could you tell me how?)

Is there just one way to assign the colors. Seems to me you could have assigned r3c1=2 as green before r3c7=2 green?

What I need to understand is the methodology for assigning the colors. A trace would show that - which is why I submitted one.

Anyway if you could give me a little more information on the logic of your assignments that would help.

Thanks for helping me!

[David P Bird]

red = one color at n49r2c2=9
lt-red = 'opposite' color at n49r1c9=4, red is 9 at this BV cell
(firstly is the correct procedure?)

My understanding of 3D Medusa is that candidates can only be coloured if they are one of a pair in either the cell or in the only other place in a house where they can be true. If you look at the examples in SudokuWiki these are the only candidates that have been coloured.

Starting by colouring (49)r3c2 in opposite colours the only other candidate that can be coloured following these rules (4)r1c2 and after that the these colours can be taken no further.

However different colours can be used for other candidates in the grid so (79)r5c2 could be coloured light and dark green for example. Now five other 7's can be coloured including (7)r1c2. This tells us that the colours for 4 & 7 in r1c2 can't coexist but that's all. We need to show in another cell or house that another cross colour pairing between the two sets also can't coexist to draw any further conclusions.

That isn't described in SudoWiki as far as I can see, and as I don't use the method myself or even the HoDoKu solver, I can't say how Jason got to the other coloured candidates he shows.

[JasonLion]

Sigh, I keep mixing up several coloring techniques that don't have names (that I have been exploring recently) with 3D Medusa. This is the second time I made that mistake. Sorry.

What I colored is (hopefully) an AIC net represented with coloring. It has a great deal in common with 3D Medusa, but the elimination rules aren't really the same. The coloring procedure I used is: You start by locating a conjugate pair, either the two digits in a cell that only contains two candidates, or two cells in a house that are the only two cells in the house that contain some particular digit. Then you assign any two colors, in either order, to those two candidates. Then you look for another conjugate pair where either one candidate shares a cell with an already colored candidate, or one candidate shares a house with an already colored candidate of the same digit. The candidate which can see a colored candidate is given the opposite color and it's conjugate is given the color of the candidate it sees (ie the opposite of it's conjugate). I think you could call it 3D Medusa X-Colors, but I'm not sure the analogy is exactly right.

In 3D Medusa Coloring you stick with just conjugate pairs at every step. You start by locating a conjugate pair, either the two digits in a cell that only contains two candidates, or two cells in a house that are the only two cells in the house that contain some particular digit. Then you assign any two colors, in either order, to those two candidates. Then you look for another conjugate pair where exactly one candidate is already colored and give it's conjugate the opposite color, until you run out of candidates connected as conjugates to add to the pattern. In this particular position I don't see anything useful to do with 3D Medusa Coloring as the chains are all quite short.

I used Sudoku + on the iPhone to make the image, but I colored candidates by hand (with the apps manual coloring tools).

To upload an image you need to click on the "Upload attachment" tab, just below the box where you enter your reply.

The choice of which of the original conjugate pair is yellow and which is green is arbitrary. You can make all the greens yellow and all the yellows green (or any other pair of colors) and it is still a valid coloring.

[RichardGoodrich]

Thanks! I am making progress. I chose another puzzle - Maestro 40 where John Welch of systematic sudoku fame had colored and I got his same results. I did a lot of work on it - so I may post it, because I not only have it marked, but I indicated in a chain text document exactly how I built it.

I am reading up now on conjugate pair chains to make sure I really understand it!

[David P Bird]

Jason, I don't believe your description pf your procedure 3D Medusa X-Colour is complete, as what you describe just seems like 3D Medusa using a single pair of colours.

You might like to try this:

1. Find a conjugate net and use dark red and dark blue to colour the alternate candidates that are true or false together. Call these the prime candidates.

2. For each of the prime candidates check for any other candidates that must be true if they are true, and colour these a lighter shade of the same colour. These are therefore provisional candidates that must be true when the prime candidates of the same colour are true. If the opposite prime colour is true, they won't necessarily be false though.

To give an example:
| (8) (5) (13) | (47) (6) (2) | (134) (79) (1379) |
In this row (4)c4 can be coloured light blue as it must be true if (7)c8 is true.

Any candidate that is weakly linked to two candidates of opposite colours, regardless of whether they are lightly or darkly shaded, is false.

If two candidates of the same colour are conjugate (again regardless of shade) the opposite prime colour is true.

As eliminations are made, the colouring can be extended keeping to the same rules.

If the eliminations dry up, check for any uncoloured conjugate pairs and use them as the basis for a new colouring run.

This scheme implements something very close to GEM but in the style of Medusa colouring.

[JasonLion]

David, I believe that the technique you describe is more properly called 3d Medusa Multi-Colors. Multi-Coloring uses several pairs of colors, just as you describe. X-Colors is not really an apt name, since X-Colors only uses two colors and neither of our descriptions use anything like the exception cell rule, which is what distinguishes X-Colors from other similar techniques.

[daj95376]

I am completely lost with all of the 3D Medusa definitions that I've seen related to this thread and in Sudopedia. So, I went searching for very early definitions of 3D Medusa. What I found was a post by Bob Hanson and the following grid presented by Jeff.

Code: Select all

 +--------------------------------------------------------------------------------+
 |  56      56      2       |  38      9       348     |  1       348     7       |
 |  179     3       8       |  6       2457    12      |  245     59      259     |
 |  4       179     17      |  123578  2578    1238    |  2568    35689   235689  |
 |--------------------------+--------------------------+--------------------------|
 |  12367   12678   1347    |  39      2678    5       |  2468    16789   12689   |
 |  2567    245678  9       |  278     1       268     |  3       45678   2568    |
 |  123567  125678  1357    |  4       2678    39      |  2568    156789  125689  |
 |--------------------------+--------------------------+--------------------------|
 |  123579  12579   1357    |  12589   2568    12689   |  68      368     4       |
 |  135     15      1345    |  158     4568    7       |  9       2       368     |
 |  8       249     6       |  29      3       249     |  7       15      15      |
 +--------------------------------------------------------------------------------+
 # 165 eliminations remain


Jeff asked Bob to list all of the 3D Medusa eliminations in the grid. Of the eliminations provided, r4c3<>1 was among them. How would you justify this elimination using 3D Medusa ???

[David P Bird]

Jason, on the Eureka forum Myth Jellies propounded his take on Medusa which is where my knowledge comes from.
He described using several colour pairs to mark different clusters of conjugate candidates. As the puzzle was reduced these would then be merged by checking for those that could and could not coexist. However, if provisional colours are used to extend the different colour pairs, the same candidate can end up with several of them which isn't viable on a manual colouring system.

This method never appealed to me and so I haven’t used it. My approach was to use Graded Equivalence Marking to follow AIC segments that could stem from a single strong link using 6 different grade marks. Combining different segments would then produce a range of AIC eliminations.

The halfway house cocktail of GEM/Medusa I described uses two colours but in two shades and is therefore somewhat different from using different colours for every conjugate cluster identified. In use I found it more difficult to identify the provisional candidates that could be coloured than using GEM.

Daj, Here is the colouring that shows that elimination using primary & provisional colour sets as I described.

..2.9.1.7.386.....4.............5.....9.1.3.....4.............4.....792.8.6.3.7..

Code: Select all

 *----------------------------*----------------------------*----------------------------*
 | 56       56       <2>      | 38       <9>      34'8     | <1>      3#4"8    <7>      |
 | 17=9     <3>      <8>      | <6>      24"57    1=2      | 24'5     59       259      |
 | <4>      179=     1=7      | 123578   2578     1238     | 2568     35689    235689   |
 *----------------------------*----------------------------*----------------------------*
 | 12367    12678    1#34'7   | 39       2678     <5>      | 24"68#   16789    12689    |
 | 2567     24'5678  <9>      | 278      <1>      268      | <3>      4'5678   2568     |
 | 123567   125678   1357     | <4>      2678     39       | 2568     156789   125689   |
 *----------------------------*----------------------------*----------------------------*
 | 123#579= 12579    1357     | 12589    2568     12689    | 68-      24898    <4>      |
 | 135      15       134"5    | 158      4'568    <7>      | <9>      <2>      36-8     |
 | <8>      24'9#    <6>      | 29       <3>      24"9     | <7>      15       15       |
 *----------------------------*----------------------------*----------------------------*
   ' & " Primary colours,  - & = Provisional colours,  # Eliminations,

In total 6 eliminations have been found including (1)r4c3 which is seen by (4')r4c3 and (1=)r3c3 so must be false. The AIC that's been revealed is

(4)r4c3 = (4)r4c7 – (4)r2c7 = (4-7)r2c5 = (7)r2c1 - (7=1)r3c3 => r4c3 <> 1

These eliminations produce
(3)line/box:r1b2 => r3c46 <> 3
(9)line/box:r9b8 => 97c46 <> 9
(349)HiddenSet:r169c6 => r1c6 <> 8, r9c6 <> 2

Extending the colouring:

Code: Select all

 *-------------------------*---------------------*----------------------------*
 | 56      56       <2>    | 3'8"   <9>    3"4'  | <1>    4"8'     <7>        |
 | 17=9    <3>      <8>    | <6>    24"57  1=2   | 24'5   59       259        |
 | <4>     179=     1=7    | 12578  2578   128   | 2568   35689    2#3-5#689# |
 *-------------------------*---------------------*----------------------------*
 | 12367   12678    3#4'7  | 3"9'   2678   <5>   | 24"6   16789    12689      |
 | 2567    24"5678  <9>    | 278    <1>    268   | <3>    4'5678   2568       |
 | 123567  125678   1357   | <4>    2678   3'9"  | 2568   156789   125689     |
 *-------------------------*---------------------*----------------------------*
 | 12579=  12579    1357   | 1258#  2568   1268  | 68-    24898    <4>        |
 | 135     15       134"5  | 158    4'568  <7>   | <9>    <2>      36-8       |
 | <8>     2"4'     <6>    | 2'9"   <3>    4"9'  | <7>    15       15         |
 *-------------------------*---------------------*----------------------------*

5 Eliminations

The only further elimination from extending this colouring further is (7)r4c1.
After that ALSs and perhaps net paths are needed – I didn't follow though.

BTW 1 a link to the posts you referenced would have been appreciated.
BTW 2 this topic doesn't really interest me as I consider 3D Medusa on a par with pattern overlays, ie more suck-and-see than logic.

[JasonLion]

As so often happens, the earliest posts are lost. In posts from 2005, the earliest still available, there is already discussion of coloring several regions with different pairs of colors and making use of "bridges" between the different pairs of colors. I'm fairly sure that was added later, though it seems impossible to prove at this point. This appears to be a very early definition of 3D Medusa, though it is undated. This version uses only two colors, with no mention of bridges.

[daj95376]

David and Jason, thanks for your feedback. The grid that I mentioned is within this thread. Bob's solution contains these seven eliminations:

Code: Select all

 r4c3<>1; r1c8<>3,r7c1<>3; r5c9<>6; r5c1<>7; r4c7<>8; r9c2<>9


My solver finds eight chains as simple AICs using bilocation and bivalue SIs:

Code: Select all

 (1=7)r3c3 - r2c1 = (7-4)r2c5 = r8c5 - r8c3 = (4)r4c3  =>  r4c3<>1

 (3)r7c8 = (3-6)r8c9 = (6-4)r8c5 = r9c6 - r1c6 = (4)r1c8  =>  r1c8<>3

 (3)r7c8 = (3-6)r8c9 = (6-4)r8c5 = (4-7)r2c5 = (7-9)r2c1 = (9)r7c1  =>  r7c1<>3

 (6)r5c6 = r7c6 - r8c5 = (6)r8c9  =>  r5c9<>6

 (7)r5c4 = r3c4 - r2c5 = (7)r2c1  =>  r5c1<>7

 (8=6)r7c7 - r8c9 = (6-4)r8c5 = r2c5 - r2c7 = (4)r4c7  =>  r4c7<>8

 (3)r3c9 = (3-6)r8c9 = (6-4)r8c5 = (4-7)r2c5 = (7-9)r2c1 = (9)r3c2  =>  r3c9<>9   ***

 (4)r9c2 = r8c3 - r8c5 = (4-7)r2c5 = (7-9)r2c1 = (9)r3c2  =>  r9c2<>9


What bothered me was how coloring could be applied to get r2c1 and r3c3 with opposite colors on <7> for the elimination r4c3<>1. I guess the "weak link" is used to "bridge" the two cells during coloring.

Code: Select all

+-----------------------------------+
|  .  .  .  |  .  .  .  |  .  .  7  |
|  7  .  .  |  .  7  .  |  .  .  .  |
|  .  7  7  |  7  7  .  |  .  .  .  |
|-----------+-----------+-----------|
|  7  7  7  |  .  7  .  |  .  7  .  |
|  7  7  .  |  7  .  .  |  .  7  .  |
|  7  7  7  |  .  7  .  |  .  7  .  |
|-----------+-----------+-----------|
|  7  7  7  |  .  .  .  |  .  .  .  |
|  .  .  .  |  .  .  7  |  .  .  .  |
|  .  .  .  |  .  .  .  |  7  .  .  |
+-----------------------------------+


Final comment. I don't see anything in the various definitions that prevent a 3D Medusa elimination from being derived through a networking of coloring links. This would probably explain the extra eliminations mentioned by Jeff and ronk. Your opinion?

[Edit: updated from original post.]

[JasonLion]

Two of the common reference sites, SudoPedia & SudokuWiki, make no mention of bridges being part of 3D Medusa. However, Rudd includes bridges in his technique description for 3D Medusa. And it seems that by late 2005 various forum posts were assuming bridges were part of 3D Medusa.

It is worth noting that 3D Medusa had a sharp peak in popularity in 2007, and after that the number of mentions drops significantly. These days the advanced solvers have essentially all moved on to other techniques. From what I can tell, none of the coloring techniques were ever successfully extended far enough to identify the more complex chains now in common use in the puzzle solving threads here and at Daily Sudoku.

[David P Bird]

Jason, Bob Hansen's Medusa grid diagram in the link you gave uses 19 lower and upper case letters to represent colours for conjugates, and no provisional colours are used. That concurs with Myths description.

In his second example he highlights that in row 1 of his grid, 'colours' (A) and (H) can't be true together and in cell r8c7 (a) and (H) can't be true together. As one of (a) or (A) must be true therefore (H) must be false making (h) true. That's pretty much what I said in my reply to Richard on April 10th above!

Did he change his method later or is it because different authors have interpreted the term in their own ways working not from the original but from someone else's description I wonder.

Daj the provisional colours are found by following weak links from primary candidates and checking if any of those candidates have a conjugate link. If so, the candidates on the other end of the link can be coloured in the lighter provisional shade. This procedure can then be applied again to the newly coloured candidate to colour further candidates in the provisional shade.

In forcing chain notation: (4)r2c4 -> (7)r2c1 -> (1)r3c3
As an AIC (4-7)r2c4 = (7)r2c1 – (7=1)r3c3

If one of the of the colours is proved false only the darker primary ones can be eliminated straight away – that's why the two shades are needed.

Yes I believe that when 3D Medusa is employed with multiple colour pairs it can sometimes produce net based eliminations.

This post in the thread you referenced also has a grid from Bob Hansen that shows conjugates being coloured using letters which supports what I said to Jason above.