Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.

Postby champagne » Sat Dec 13, 2008 1:56 am

udosuk wrote:Make sure you read this very thread carefully too, starting from p.1.


No problem, it is a ONE page thread:D

But it is an old one. Meantime, I red a very nice probe excluding "Unicity use".

I am not sure I covered all the field, but, regarding symmetry

. 90° rotationnal symmetry => 180° rotationnal symmetry.

. I do not foresee a possibility to have a=>b and b=>c in symmetries with fix cells if a,b,c are different.
In all identified cases a<=>b is compulsory.

. Mauricio gives an example of double diagonal symmetry, this is for me identical to 180° rotationnal symmetry. The findings shown rely on constraints of other given. Only the central assignement comes out of the symmetry property.

The door is still open for other examples, but I doubt.

In total, after the wise remark of eleven, 2 possibilities are still there for me:

.180° rotationnal symmetry => r5C5 center of the symmetry
.diagonal symmetry.


udosuk wrote:http://forum.enjoysudoku.com/viewtopic.php?t=6459
It might help a bit, but mind you, in that thread (Glyn's riddle) the situation is very different, as the puzzle on offer doesn't have symmetry in the strict sense, but we can just work out from Glyn's cryptic hints how some of the cell values in the solution grid display partial symmetry.:idea:


I came t the same conclusion after a new carefull reading. As a matter of fact, I lost interest in that thread.:(


udosuk wrote:
champagne wrote:The most difficult is to recover the symmetry in a scrambled puzzle.

Well, it's certainly a skill that needs a lot of practice, and an exceptional observation power.


I never said I did not covered that topic. Not so difficult for a computer, just needing more code.


At the end, pending interesting question remain:

Have you other examples of symmetry not covered in the previous step (including morphing of the puzzle), and what are you doing out of them.

champagne
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Postby udosuk » Sat Dec 13, 2008 2:47 am

champagne wrote:
udosuk wrote:Make sure you read this very thread carefully too, starting from p.1.


No problem, it is a ONE page thread:D

By "this very thread", I actually meant this link:

http://forum.enjoysudoku.com/viewtopic.php?t=6355

(Either you aren't very good in English reading or I ain't very good writing in English. But for the holiday spirit let's say we're both responsible for that little misunderstanding.:D )

champagne wrote:Mauricio gives an example of double diagonal symmetry, this is for me identical to 180° rotationnal symmetry. The findings shown rely on constraints of other given. Only the central assignement comes out of the symmetry property.

No, the double diagonal symmetry implies 180 symmetry, but they are definitely not identical. On a brief note, 180 symmetry gives you immediate placement on 1 cell (r5c5) while double diagonal symmery gives you immediate placements/eliminations on 17 cells (both diagonals).

If you only use 180 symmetry that example still needs chains to solve. But if you make full use of the double diagonal symmetry the puzzle is reduced to singles.

champagne wrote:In total, after the wise remark of eleven, 2 possibilities are still there for me:

.180° rotationnal symmetry => r5C5 center of the symmetry
.diagonal symmetry.

If you only want to cater for those 2 symmetries for your solver, it's completely your own choice. After all these properties are highly unlikely to appear in a randomly generated puzzle. But I guess you'll miss out on the very powerful "gludo" technique discovered/formulated by Glyn and me for the "whole block" symmetry (available on p.4 of this very thread:D ).

champagne wrote:Have you other examples of symmetry not covered in the previous step (including morphing of the puzzle), and what are you doing out of them.

Again, available on various pages of this very thread.:idea:
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Postby champagne » Sat Dec 13, 2008 4:48 am

udosuk wrote:(Either you aren't very good in English reading or I ain't very good writing in English.


Being aware of the weakness of my English, I have my own idea:D


udosuk wrote:No, the double diagonal symmetry implies 180 symmetry, but they are definitely not identical. On a brief note, 180 symmetry gives you immediate placement on 1 cell (r5c5) while double diagonal symmery gives you immediate placements/eliminations on 17 cells (both diagonals).

If you only use 180 symmetry that example still needs chains to solve. But if you make full use of the double diagonal symmetry the puzzle is reduced to singles.


Basically agreed that reducing double diagonal symmetry to 180° symmetry does not give the full power. Not difficulty to adjust the process.


udosuk wrote:
champagne wrote:In total, after the wise remark of eleven, 2 possibilities are still there for me:

.180° rotationnal symmetry => r5C5 center of the symmetry
.diagonal symmetry.

If you only want to cater for those 2 symmetries for your solver, it's completely your own choice. After all these properties are highly unlikely to appear in a randomly generated puzzle. But I guess you'll miss out on the very powerful "gludo" technique discovered/formulated by Glyn and me for the "whole block" symmetry (available on p.4 of this very thread:D ).

champagne wrote:Have you other examples of symmetry not covered in the previous step (including morphing of the puzzle), and what are you doing out of them.

Again, available on various pages of this very thread.:idea:


First of all you are right in that sense that I do not intend to cover all the field of that tiny corner of sudoku puzzles.

Nevertheless, I'll study again this thread to see if I have the appropriate cut off.

thanks again for help

champagne
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Postby udosuk » Sun Dec 14, 2008 1:23 am

Mauricio in the Patterns Game thread wrote:
Code: Select all
+-------+-------+-------+
| . . . | . . 1 | . 2 . |
| 3 . . | . . . | 4 . . |
| . 5 2 | 3 . . | 6 . . |
+-------+-------+-------+
| 1 . . | . 7 . | 3 . . |
| . . . | 6 . 8 | . . . |
| . . 5 | . 4 . | . . 1 |
+-------+-------+-------+
| . . 8 | . . 5 | 9 3 . |
| . . 7 | . . . | . . 5 |
| . 9 . | 1 . . | . . . |
+-------+-------+-------+ ED=9.8/1.2/1.2

I take it this puzzle, if symmetry is used, should be equivalent to an ER 1.2 puzzle (which would need singles and nothing else).

Edited: just realised I've misunderstood the notation here. The 1.2 ratings actually refer to the moves leading to the first placement/elimination, which are of course hidden single moves.

However I couldn't find any easy path:( . I needed a generalized x-wing and an xy-wing together with 180 symmetry tricks to crack it. Here is my path:



After initial singles:

Code: Select all
+----------------------+----------------------+----------------------+
| 46789  4678   469    |*4789   5689   1      | 578    2      3      |
| 3      678    1      | 2789   25689  2679   | 4      5789   789    |
| 4789   5      2      | 3      89    *479    | 6      1      789    |
+----------------------+----------------------+----------------------+
| 1      2468   469    | 5      7      29     | 3      4689   24689  |
| 2479   2347   349    | 6      1      8      | 257    4579   2479   |
| 26789  2678   5      | 29     4      3      | 278    6789   1      |
+----------------------+----------------------+----------------------+
| 246    1      8      |-247    26     5      | 9      3      2467   |
| 246    2346   7      | 2489   23689  2469   | 1      468    5      |
| 5      9      346    | 1      2368  -2467   | 278    4678   24678  |
+----------------------+----------------------+----------------------+

4 @ b2 locked @ r1c4+r3c6
=> pointing eliminations: r7c4+r9c6 can't have 4
=> 4 @ r7 locked @ r7c19
=> pointing elimination: r3c1 can't be 4
=> hidden single @ r3: r3c6=4
=> symmetry: r7c4=7
=> r1c4+r3c5={89} (naked pair @ b2)
=> r1268c4=[8294], r2489c6=[7296], r37c5=[92]

Code: Select all
+-------------------+-------------------+-------------------+
| 4679  467   469   | 8     56    1     | 57    2     3     |
| 3    *68    1     | 2     56    7     | 4    -589   89    |
| 78    5     2     | 3     9     4     | 6     1     78    |
+-------------------+-------------------+-------------------+
| 1     468   469   | 5     7     2     | 3     4689  4689  |
| 2479  2347  349   | 6     1     8     | 257   4579  2479  |
| 2678  2678  5     | 9     4     3     | 278   678   1     |
+-------------------+-------------------+-------------------+
| 46    1     8     | 7     2     5     | 9     3     46    |
| 26   -236   7     | 4     38    9     | 1    *68    5     |
| 5     9     34    | 1     38    6     | 278   478   2478  |
+-------------------+-------------------+-------------------+

r2c2+r8c8={68} (symmetrical pointing pair @ r2c8+r8c2)
=> 6 @ c1,b7 locked @ r78c1
Now 6 @ r6 locked @ r6c28, and r2c2+r8c8={68} form 2 strong links of 6
Generalized x-wing: 6 @ c2 locked @ r26c2

Code: Select all
+-------------------+-------------------+-------------------+
| 479  *47    469   | 8     56    1     | 57    2     3     |
| 3    -68    1     | 2     56    7     | 4     59    89    |
|*78    5     2     | 3     9     4     | 6     1     78    |
+-------------------+-------------------+-------------------+
| 1    *48    469   | 5     7     2     | 3     4689  4689  |
| 2479  2347  349   | 6     1     8     | 257   4579  2479  |
|-278   2678  5     | 9     4     3     | 278   678   1     |
+-------------------+-------------------+-------------------+
| 46    1     8     | 7     2     5     | 9     3     46    |
| 26    23    7     | 4     38    9     | 1     68    5     |
| 5     9     34    | 1     38    6     | 278   478   2478  |
+-------------------+-------------------+-------------------+

Now r1c2 from {47}, r3c1 from {78}, r4c2 from {48}
XY-wing: r2c2+r6c1, seeing both r3c1+r4c2, can't have 8

All singles from here.



Any easy symmetry moves I've missed?:?:
Last edited by udosuk on Sat Dec 13, 2008 11:59 pm, edited 1 time in total.
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Postby champagne » Sun Dec 14, 2008 1:57 am

udosuk wrote:Any easy symmetry moves I've missed?:?:


I don't know, but I'll study it as player before I submit it to my solver.

In our recent discussion, I still missed a point I reconsidered later.

In the same way "double diagonal symmetry" is not the same as 180° rotation symmetry, we have here a "90°" double rotatinal symmetry which brings more than just 180° symmetry.

I looked for one example, I have it.

champagne

Edit one

1) That one is unhappily not double 90° , so I am still looking for one.

2) without symmetry, my solver has not an easy path.
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Postby udosuk » Sun Dec 14, 2008 4:04 am

champagne wrote:In our recent discussion, I still missed a point I reconsidered later.

In the same way "double diagonal symmetry" is not the same as 180° rotation symmetry, we have here a "90°" double rotatinal symmetry which brings more than just 180° symmetry.

Don't understand what you mean by "90° double rotational symmetry". Note all 90° rotational symmetry are valid both clockwise or anticlockwise, so there is not an issue of "doubling" here.

Of course "90° rotational symmetry" gives you many more extra tricks than "180° rotational symmetry".
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Postby champagne » Sun Dec 14, 2008 4:28 am

Hi udosuk,

1) Regarding 90° rotation, if you turn twice, you have for sure a 180° rotation.
This is somehow very similar to double diagonal symmetry.

2) in that puzzle after "simple things, I am here

Code: Select all
1||46789 4678 469 |4789 5689_ ____ |578 ____ _____
2||_____ 678_ ___ |2789 25689 2679 |___ 5789 789__
3||4789_ ____ ___ |____ 89___ 479_ |___ ____ 789__

4||_____ 2468 469 |____ _____ 29__ |___ 4689 24689
5||2479_ 2347 349 |____ _____ ____ |257 4579 2479_
6||26789 2678 ___ |29__ _____ ____ |278 6789 _____

7||246__ ____ ___ |247_ 26___ ____ |___ ____ 2467_
8||246__ 2346 ___ |2489 23689 2469 |___ 468_ _____
9||_____ ____ 346 |____ 2368_ 2467 |278 4678 24678


with pairs 35 29 47 68

I noticed several possibilities identical to

4r4c1;7r9c6 = 7r7c4 => <> 7r1c4 <> 4r7c4

followed by <> 2r8c4 due to the fact that 4r1c4 4r8c4 is now a bi value.


My solver starts with <>2r7c9 <>9r1c4, something quite different

champagne
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Postby udosuk » Sun Dec 14, 2008 6:39 am

champagne wrote:1) Regarding 90° rotation, if you turn twice, you have for sure a 180° rotation.
This is somehow very similar to double diagonal symmetry.

If you're talking about the trivial property that a 90 symmetry puzzle must also have 180 symmetry, I guess I didn't regard it as worthy of discussion as it is too obvious to most.

Lucky for us, Mauricio just posted a 90 symmetry puzzle for me to demonstrate to you:) :

Mauricio in the Patterns Game thread wrote:
Code: Select all
+-------+-------+-------+
| . . . | . . 1 | . 2 . |
| 3 . . | . . . | 4 . . |
| . 4 2 | 5 . . | 6 . . |
+-------+-------+-------+
| 7 . . | . 6 . | 3 . . |
| . . . | 2 . 5 | . . . |
| . . 6 | . 3 . | . . 8 |
+-------+-------+-------+
| . . 3 | . . 2 | 5 4 . |
| . . 4 | . . . | . . 6 |
| . 5 . | 9 . . | . . . |
+-------+-------+-------+ ED=10.1/1.2/1.2

After r5c5 placement and singles:

Code: Select all
+-------------------+-------------------+-------------------+
| 5     6    *789   | 4    -789   1     |*789   2     3     |
| 3     1789  1789  | 78    2     6     | 4     1789  5     |
| 189   4     2     | 5     789   3     | 6     1789  179   |
+-------------------+-------------------+-------------------+
| 7     2     5     | 18    6     89    | 3     19    4     |
| 189   3     189   | 2     4     5     | 179   6     179   |
| 4     19    6     | 17    3     79    | 2     5     8     |
+-------------------+-------------------+-------------------+
| 189   1789  3     | 6     178   2     | 5     4     179   |
| 2     1789  4     | 3     5     78    | 1789  1789  6     |
| 6     5     178   | 9     178   4     | 178   3     2     |
+-------------------+-------------------+-------------------+

180 symmetry: r1c37 can't be {78}
=> 9 @ r1 locked @ r1c37
=> hidden single @ b2: r3c5=9
90 symmetry: r5c37+r7c5=[871]

Code: Select all
+-------------------+-------------------+-------------------+
| 5     6     79    | 4     78    1     | 89    2     3     |
| 3     1789  179   | 78    2     6     | 4     1789  5     |
|-18    4     2     | 5     9     3     | 6    *178  *17    |
+-------------------+-------------------+-------------------+
| 7     2     5     | 18    6     89    | 3     19    4     |
| 19    3     8     | 2     4     5     | 7     6     19    |
| 4     19    6     | 17    3     79    | 2     5     8     |
+-------------------+-------------------+-------------------+
| 89    789   3     | 6     1     2     | 5     4     79    |
| 2     1789  4     | 3     5     78    | 189   1789  6     |
| 6     5     17    | 9     78    4     | 18    3     2     |
+-------------------+-------------------+-------------------+

Now r3c89 from {178} must have 1|8
90 symmetry: the windmill r3c1+r1c7+r7c9+r9c3 can't be [1897]
=> r3c1 can't be 1, must be 8

All naked singles from here.



If your solver can't solve this one easily, then you should seriously consider teaching it some tricks on 90 symmetry.:idea:
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Postby champagne » Sun Dec 14, 2008 7:44 pm

Hi udosuk,

Thanks for example of 90° rotation and thanks to Mauricio as well..


udosuk wrote:
champagne wrote:1) Regarding 90° rotation, if you turn twice, you have for sure a 180° rotation.
This is somehow very similar to double diagonal symmetry.

If you're talking about the trivial property that a 90 symmetry puzzle must also have 180 symmetry,
I guess I didn't regard it as worthy of discussion as it is too obvious to most.


I would classify your answer as words of an expert in a very specific field forgetting how newcomers see it.

This thread was started in september 2008 and is supposed to contain new ideas or at least to have croped recent new findings.

Give time to others to digest it. I am not sure most have ever heard of symmetry in sudoku puzzle in the way it is discussed here.


udosuk wrote:If your solver can't solve this one easily, then you should seriously consider teaching it some tricks on 90 symmetry.


I am afraid you are underestimating the power of an alliance between symmetry and AIC's but we will see.

As an example, I launched this morning my first test on the puzzle you posted before. Here is the proposal of my solver:

The starting point already seen

Code: Select all
46789 4678 469 |4789 5689  1    |578 2    3     
3     678  1   |2789 25689 2679 |4   5789 789   
4789  5    2   |3    89    479  |6   1    789   
-----------------------------------------------
1     2468 469 |5    7     29   |3   4689 24689
2479  2347 349 |6    1     8    |257 4579 2479 
26789 2678 5   |29   4     3    |278 6789 1     
-----------------------------------------------
246   1    8   |247  26    5    |9   3    2467 
246   2346 7   |2489 23689 2469 |1   468  5     
5     9    346 |1    2368  2467 |278 4678 24678


4r9c6 - 4r3c6 = 4r1c4 - 7r1c4==4r9c6 <>7r1c4 <>4r9c6
2r8c4 - 2r7c5 = 6r7c5 - 6r12c5 = 6r2c6 - 9r2c6==r8c4 <>9r2c6 <>2r8c47r3c6 - 4r3c6 = 4r1c4 - 4r7c4==7r3c6 <>7r3c6 <>=4r7c4
6r4c2 - 6r4c89 = (6r8c6;8r4c2) - 6r4c2 <>6r4c2 <>8r6c8

The first one is what I had seen. All theese AIC's are very simple, bi values and symmetry


after cleaning and
10 r4c6=2 11 r6c4=9


Code: Select all
46789 4678   469   |48   56    1    |57 8  2      3     
3     68     1     |27   256   67   |4     589    89   
478   5      2     |3    89    49   |6     1      78 
-------------------------------------------------------
1     48     469   |5    7     2    |3     4689   4689 
2479  2347   349   |6    1     8    |257   4579   2479
2678  2678   5     |9    4     3    |278   67     1     
-------------------------------------------------------
46    1      8     |27   26    5    |9     3      467
26    236    7     |48   389   49   |1     68     5     
5     9      346   |1    38    67   |278   4678   24678


4r3c1 - (4r7c1;7r3c9) = 7r3c1 - 4r3c1
=> r1c4=8;r2c4=2;r2c6=7;r3c5=9;r3c6=4 . . .


again, the simplest AIC you can imagine, relying on symmetry
game is nearly over

champagne
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Postby udosuk » Mon Dec 15, 2008 3:16 am

champagne wrote:I would classify your answer as words of an expert in a very specific field forgetting how newcomers see it.

This thread was started in september 2008 and is supposed to contain new ideas or at least to have croped recent new findings.

Give time to others to digest it. I am not sure most have ever heard of symmetry in sudoku puzzle in the way it is discussed here.

My logic goes like this: for the people who understand how the 90 symmetry and the 180 symmetry work, they must be smart enough to easily see the underlying relationship between the two of them.

For the people who have trouble grasping these concepts, it's meaningless to point out this underlying relationship.:idea:



Also, thanks for posting your AIC moves. It helps me a great deal to understand how your notation works. As for your steps, they more or less cover my first block of moves in my solution path, which is the easier part of the work. However, your "game is nearly over" remark sounds a touch too optimistic, as the critical moves (generalised x-wing & xy-wing) remained to be figured out.:)
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Postby champagne » Mon Dec 15, 2008 9:36 pm

udosuk wrote:
Also, thanks for posting your AIC moves. It helps me a great deal to understand how your notation works.
As for your steps, they more or less cover my first block of moves in my solution path, which is the easier part of the work.
However, your "game is nearly over" remark sounds a touch too optimistic,
as the critical moves (generalised x-wing & xy-wing) remained to be figured out.:)


Hi udosuk,

I never found big problems in a puzzle with so many given and assigned, at least using tagging or any other colouring process.
Most of the grid is just bi values, so coluring helps at lot finding relevant AIC's Combined with symmetry, the result is still better.

Anyway, I had a quick look to check if there was something special here before wrinting "nearly over".

Here are the final steps of the solver. I will use to layers of the tagging procedure to ease understanding

Here is the situation after my last move and basic cleaning


Code: Select all
4679       467      46K9   |8    5a6A   1    |5A7a   2        3         
3          6A8a     1      |2    5A6a   7    |4      5a89     89       
7a8A        5        2      |3    9      4   |6      1        7A8a       
------------------------------------------------------------------------
1          4K8k     46k9   |5    7      2    |3      4689     4689   
2479       23A47    3a49   |6    1      8    |25a7   45A79    2479   
2678       2678     5      |9    4      3    |278k   6k7K     1         
------------------------------------------------------------------------
4A6a       1        8      |7    2      5    |9      3        4a6A     
26         23a6     7      |4    3A8a   9    |1      6a8A     5         
5          9        3A4a   |1    3a8A   6    |278K   478      2478


The 2 layers A,a K,k are the most important to go to the end.


r2c2;r8c2 is a pattern giving <> 6r8c2 (<=>8r2c8)
r4c2;r4c3 is a pattern giving <> 4r4c3 (<=>7r6c7)

we can also notice the weak link a - k (for example 8r2c2.a - 8r4c2.K)
we can deduct easily <>4r5c1 4r5c1 - 4r5c7.A = 8r2c2.a - 8r4c2.k = 4r4c2.K - 4r5c1
and <>4r5c2 4r5c2 - 3r5c2.A . . .

we are now here

Code: Select all
4a679      467      46K9   |8    5a6A   1    |5A7a   2        3         
3          6A8a     1      |2    5A6a   7    |4      5a9      89       
7a8A        5        2     |3    9      4    |6      1        7A8a       
------------------------------------------------------------------------
1          4K8k     6k9    |5    7      2    |3      4689     4689   
279        23A7     3a4k9  |6    1      8    |25a7k  45A9     249   
2678       2678     5      |9    4      3    |28k    6k7K     1         
------------------------------------------------------------------------
4A6a       1        8      |7    2      5    |9      3        4a6A     
26         2A3a     7      |4    3A8a   9    |1      6a8A     5         
5          9        3A4a   |1    3a8A   6    |278K   478      247a8


and the final AIC

3r5c3.a - 4r5c3.k = 6r1c3.K - 9r1c3 = 9r1c1 - 4r1c1.a


'A' is valid

If you do not use a colouring help, you have the same with longer AIC's

champagne
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Postby udosuk » Tue Dec 16, 2008 1:26 am

So, you (or your solver) solved the puzzle with some colouring tricks, your favourite method; I solved the puzzle with symmetry, simple fishes and xy-wing, my favourite method. Everybody is happy.:)

Merry Christmas!:D
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Postby ronk » Tue Dec 16, 2008 4:41 am

champagne wrote:the final AIC

3r5c3.a - 4r5c3.k = 6r1c3.K - 9r1c3 = 9r1c1 - 4r1c1.a

It seems inconsistent to not also tag digit 9 in r1.

[edit: moved follow-on question to a later post]
Last edited by ronk on Tue Dec 16, 2008 1:09 am, edited 2 times in total.
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Postby champagne » Tue Dec 16, 2008 5:04 am

ronk wrote:
champagne wrote:the final AIC

3r5c3.a - 4r5c3.k = 6r1c3.K - 9r1c3 = 9r1c1 - 4r1c1.a

It seems inconsistent to not also tag digit 9 in r1.



may be "inconsistent" is a little too strong, but you are right. I did not do it because I wanted to limit the printed tagging to layers A,a K,k, stressing on the logic where AIC's are shortened thru tagging.

Internally (and in the output of the solver), they are tagged for sure.

In my process, it is not possible to produce an AIC using a "non tagged" digit or group. The search of AIC's is done exclusively on tags.

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Postby ronk » Tue Dec 16, 2008 5:19 am

I had edited my prior post to add the following question, but you responded during my edit, so I restored it.:(
champagne wrote:the final AIC

3r5c3.a - 4r5c3.k = 6r1c3.K - 9r1c3 = 9r1c1 - 4r1c1.a

'A' is valid

In the notation of that AIC, where is the contradiction that says 'a' is invalid:?:
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