About Red Ed's Sudoku symmetry group

Everything about Sudoku that doesn't fit in one of the other sections

Postby eleven » Wed Jan 28, 2009 10:40 pm

Its easy to generate all grids with CS+JR and RS(row sticks)+JR normalized.
So i can confirm now, that all the 31 grids with CS+JR (25 134) symmetry, which dont have CS°JR (145) symmetry (and therefore cant have both CS and JR normalized), can be transformed to have RS+JR normalized.

This is one of the 2 grids with JD+JR+CS+CS°JR (22 25 134 145):
Code: Select all
 +-------+-------+-------+
 | 8 6 7 | 5 3 4 | 2 9 1 |
 | 1 9 2 | 7 6 8 | 4 3 5 |
 | 5 3 4 | 2 9 1 | 8 6 7 |
 +-------+-------+-------+
 | 9 4 8 | 6 1 5 | 3 7 2 |
 | 2 7 3 | 8 4 9 | 5 1 6 |
 | 6 1 5 | 3 7 2 | 9 4 8 |
 +-------+-------+-------+
 | 7 5 9 | 4 2 6 | 1 8 3 |
 | 3 8 1 | 9 5 7 | 6 2 4 |
 | 4 2 6 | 1 8 3 | 7 5 9 |
 +-------+-------+-------+
JR 174 285 396
JC 123 456 789
JD 186 294 375
CS 1 23 4 56 7 89
CS°JR 174 295386
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby udosuk » Fri Jan 30, 2009 12:35 am

eleven wrote:
udosuk wrote:I have a conjecture that CS+\M implies the factor of 8 (CS+RS+\M+/M+HT or CS+RS+QT) and am trying to prove it rigidly.

No more proof needed, you can directly read it from the list (you cant find 134 and 37 without 79 and 86), same for QT+ \M. [Added:] Please dont misunderstand me - i would appreciate a more intuitive proof very much:)

Here comes a very intuitive proof.:D

Turns out it's not very difficult if you label the cells appropriately, however the verifying process is a little bit tedious.



Proof: CS+\M implies CS+RS+\M+/M+HT and CS+RS+QT

Suppose r456c258 form the axis of the Column Sticks, and d\ (leading diagonal) forms the axis of the Diagonal Mirror.
Label the cell values of the central block as [ABCDEFGHJ].
Note we now have these 2 symmetries:

CS: (B)(E)(H)(AC)(DF)(GJ)
\M: (A)(E)(J)(BD)(CG)(FH)

Code: Select all
The * cells must be from {AEJ}
The # cells must be from {BEH}

 *  .  .  | .  .  .  | .  .  .
 .  *  .  | .  .  .  | .  .  .
 .  .  *  | .  .  .  | .  .  .
----------+----------+----------
 .  #  .  | A  B  C  | .  #  .
 .  #  .  | D  E  F  | .  #  .
 .  #  .  | G  H  J  | .  #  .
----------+----------+----------
 .  .  .  | .  .  .  | *  .  .
 .  .  .  | .  .  .  | .  *  .
 .  .  .  | .  .  .  | .  .  *

Without loss of generality, we can always morph r456c2 to be [EHB] and r456c8 to be [HBE].
Also we can always morph r1c1 and r9c9 to be E, leaving r2c2+r3c3 and r7c7+r8c8 to be from {AJ}.

Code: Select all
The * cells must be from {AJ}

 E  .  .  | .  .  .  | .  .  .
 .  *  .  | .  .  .  | .  .  .
 .  .  *  | .  .  .  | .  .  .
----------+----------+----------
 .  E  .  | A  B  C  | .  H  .
 .  H  .  | D  E  F  | .  B  .
 .  B  .  | G  H  J  | .  E  .
----------+----------+----------
 .  .  .  | .  .  .  | *  .  .
 .  .  .  | .  .  .  | .  *  .
 .  .  .  | .  .  .  | .  .  E

Applying CS & \M symmetry, we can fill in the following cells: r28c456, r3c7, r7c3.
Also, we can label the remaining diagonals cells as Wi,Xi,Yi,Zi with i from 2 to 3.
(r2c2+r3c3 as W23, r1c9+r2c8 as X32, r8c2+r9c1 as Y23, r7c7+r8c8 as Z32.)
Note now with CS Symmetry: [Wi,Yi]=[AC|JG], [Xi,Zi]=[CA|GJ].
Thus with \M symmetry: [Xi,Yi]=[CG|GC]={CG}.

Code: Select all
Wi,Zi must be from {AJ}
Xi,Yi must be from {CG}

 E  .  .  | .  .  .  | .  .  X3
 .  W2 .  | E  F  D  | .  X2 .
 .  .  W3 | .  .  .  | E  .  .
----------+----------+----------
 .  E  .  | A  B  C  | .  H  .
 .  H  .  | D  E  F  | .  B  .
 .  B  .  | G  H  J  | .  E  .
----------+----------+----------
 .  .  E  | .  .  .  | Z3 .  .
 .  Y2 .  | F  D  E  | .  Z2 .
 Y3 .  .  | .  .  .  | .  .  E

Now we label r1c234568 as K234568 respectively.
Similarly, label r234568c1 as L234568, r245678c9 as M865432, r9c245678 as N865432.
Note now with \M symmetry: [Ki,Li] and [Mi,Ni] must be {AA|BD|CG|FH|JJ}.

Code: Select all
 E  K2 K3 | K4 K5 K6 | .  K8 X3
 L2 W2 .  | E  F  D  | .  X2 M8
 L3 .  W3 | .  .  .  | E  .  .
----------+----------+----------
 L4 E  .  | A  B  C  | .  H  M6
 L5 H  .  | D  E  F  | .  B  M5
 L6 B  .  | G  H  J  | .  E  M4
----------+----------+----------
 .  .  E  | .  .  .  | Z3 .  M3
 L8 Y2 .  | F  D  E  | .  Z2 M2
 Y3 N8 .  | N6 N5 N4 | N3 N2 E

Then we label r3c245689 as P845623 respectively.
Similarly, label r245689c3 as Q845623, r124568 as R326548, r7c124568 as S326548.
Note now with \M symmetry: [Pi,Qi] and [Ri,Si] must be {AA|BD|CG|FH|JJ}.
Also with CS symmetry: [Ki,Si], [Li,Qi], [Mi,Ri], [Ni,Pi] must be {AC|BB|DF|GJ|HH}.

Code: Select all
 E  K2 K3 | K4 K5 K6 | R3 K8 X3
 L2 W2 Q8 | E  F  D  | R2 X2 M8
 L3 P8 W3 | P4 P5 P6 | E  P2 P3
----------+----------+----------
 L4 E  Q4 | A  B  C  | R6 H  M6
 L5 H  Q5 | D  E  F  | R5 B  M5
 L6 B  Q6 | G  H  J  | R4 E  M4
----------+----------+----------
 S3 S2 E  | S6 S5 S4 | Z3 S8 M3
 L8 Y2 Q2 | F  D  E  | R8 Z2 M2
 Y3 N8 Q3 | N6 N5 N4 | N3 N2 E

Now with the whole grid set up we can start "linking up" the cell relationships.

With [Wi,Yi]=[AC|JG], [Xi,Zi]=[CA|GJ], [Xi,Yi]=[CG|GC]
=> [Wi,Xi,Yi,Zi]=[AGCJ|JCGA]

With [Ki,Li], [Mi,Ni], [Pi,Qi], [Ri,Si] being {AA|BD|CG|FH|JJ}
Also [Ki,Si], [Li,Qi], [Mi,Ri], [Ni,Pi] being {AC|BB|DF|GJ|HH}
=> [Ki,Li,Mi,Ni,Pi,Qi,Ri,Si]=[AAJJGCGC|BDFHHFDB|CGCGJJAA|DBHFDBHF|FHBDFHBD|GCGCAAJJ|HFDBBDFH|JJAACGCG]

And finally we can set out to prove each extra symmetries.

/M:
[Wi,Zi]=[AJ|JA]={AJ}
[Ki,Mi]=[AJ|BF|CC|DH|FB|GG|HD|JA]={AJ|BF|CC|DH|GG}
[Li,Ni]=[AJ|DH|GG|BF|HD|CC|FB|JA]={AJ|BF|CC|DH|GG}
[Pi,Ri]=[GG|HD|JA|DH|FB|AJ|BF|CC]={AJ|BF|CC|DH|GG}
[Qi,Si]=[CC|FB|JA|BF|HD|AJ|DH|GG]={AJ|BF|CC|DH|GG}
Therefore all pairs of cells symmetrical across d/ (non-leading diagonal) must be {AJ|BF|CC|DH|EE|GG}
And all cells on d/ are from {CEG}
Hence we have /M symmetry: (C)(E)(G)(AJ)(BF)(DH)

RS:
[Wi,Xi]=[AG|JC]
[Yi,Zi]=[CJ|GA]
[Ki,Pi]=[AG|BH|CJ|DD|FF|GA|HB|JC]={AG|BH|CJ|DD|FF}
[Li,Ri]=[AG|DD|GA|BH|HB|CJ|FF|JC]={AG|BH|CJ|DD|FF}
[Mi,Qi]=[JC|FF|CJ|HB|BH|GA|DD|AG]={AG|BH|CJ|DD|FF}
[Ni,Si]=[JC|HB|GA|FF|DD|CJ|BH|AG]={AG|BH|CJ|DD|FF}
Therefore within each band, all corresponding cells on pairs of minirows symmetrical across
the central mini-row (r258c456) must be {AG|BH|CJ|DD|EE|FF}
And r258c456 are from {DEF}
Hence we have RS symmetry: (D)(E)(F)(AG)(BH)(CJ)

HT:
[Wi,Zi]=[AJ|JA]={AJ}
[Xi,Yi]=[GC|CG]={CG}
[Ki,Ni]=[AJ|BH|CG|DF|FD|GC|HB|JA]={AJ|BH|CG|DF}
[Li,Mi]=[AJ|DF|GC|BH|HB|CG|FD|JA]={AJ|BH|CG|DF}
[Pi,Si]=[GC|HB|JA|DF|FD|AJ|BH|CG]={AJ|BH|CG|DF}
[Qi,Ri]=[CG|FD|JA|BH|HB|AJ|DF|GC]={AJ|BH|CG|DF}
Therefore all pairs of cells symmetrical across r5c5 must be {AJ|BH|CG|DF|EE}
And r5c5=E
Hence we have HT symmetry: (E)(AJ)(BH)(CG)(DF)

To prove the grid has QT symmetry, we can do the following transformation:

Code: Select all
r789456123
(i.e. swapping top and bottom band)

 S3 S2 E  | S6 S5 S4 | Z3 S8 M3
 L8 Y2 Q2 | F  D  E  | R8 Z2 M2
 Y3 N8 Q3 | N6 N5 N4 | N3 N2 E
----------+----------+----------
 L4 E  Q4 | A  B  C  | R6 H  M6
 L5 H  Q5 | D  E  F  | R5 B  M5
 L6 B  Q6 | G  H  J  | R4 E  M4
----------+----------+----------
 E  K2 K3 | K4 K5 K6 | R3 K8 X3
 L2 W2 Q8 | E  F  D  | R2 X2 M8
 L3 P8 W3 | P4 P5 P6 | E  P2 P3

[Wi,Yi,Zi,Xi]=[ACJG|JGAC]
[Ki,Qi,Ni,Ri]=[ACJG|BFHD|CJGA|DBFH|FHDB|GACJ|HDBF|JGAC]
[Li,Si,Mi,Pi]=[ACJG|DBFH|GACJ|BFHD|HDBF|CJGA|FHDB|JGAC]
Therefore all quadruplets of cells clockwisely symmetrical around r5c5
must be [ACJG|BFHD|CJGA|DBFH|EEEE|FHDB|GACJ|HDBF|JGAC]
And r5c5=E
Hence we have QT symmetry: (E)(ACJG)(BFHD)

(And this isomorph must also have the same CS and RS symmetries as the one above.)

Q.E.D.

:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Fri Jan 30, 2009 10:44 pm

Nice (especially i like, how you get to the QT symmetry), but you only proved it for grids, which have both CS and DM (M\) normalized.
In this case we know from the list, that you always can morph a CS+DM grid to this form (i verified by generating all grids, that its also true for HT+DM), but as we saw its wrong e.g. for MR+MD.
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby udosuk » Sat Jan 31, 2009 1:52 pm

eleven wrote:Nice (especially i like, how you get to the QT symmetry), but you only proved it for grids, which have both CS and DM (\M) normalized.

That can also be proved intuitively, that all grids with CS+\M can have both symmetries normalised together.:idea:

Note all grids having CS symmetry must have the "axis" being 3 mini-columns on the same band, spreading across 3 stacks. This happens no matter how you morph the grid (hopefully I won't need to prove this vigorously).

So, suppose we have the \M symmetry with the "axis" being d\ (leading diagonal), and the "axis" of CS for that particular isomorph is say, r123c167:

Code: Select all
 @ . . | . . # | # . .
 # * . | . . # | # . .
 # . * | . . # | # . .
-------+-------+-------
 . . . | * . . | . . .
 . . . | . * . | . . .
 . . . | . . * | . . .
-------+-------+-------
 . . . | . . . | * . .
 . . . | . . . | . * .
 . . . | . . . | . . *

It's always possible to morph the grid so that the axis of CS becomes r456c258, and the axis of \M stays being d\. The key is to morph the block where the 2 axes meet to b5, and the cell where they intersect (the @ cell here) to r5c5.:idea:

For the example above the following morphing does the job:

Code: Select all
r879213465
c879213465

 * . . | . . . | . . .
 . * . | . . . | . . .
 . . * | . . . | . . .
-------+-------+-------
 . # . | * # . | . # .
 . # . | . @ . | . # .
 . # . | . # * | . # .
-------+-------+-------
 . . . | . . . | * . .
 . . . | . . . | . * .
 . . . | . . . | . . *

:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Sat Jan 31, 2009 9:43 pm

I dont want to be nit-picking, but have you proven now, that 2 symmetries e.g. with those fixed cells, are either not possible at all or you can transform them to be both normalized ?
Code: Select all
 * . . | . . . | . . .
 . * . | . . . | . . .
 . . * | . . . | . . .
-----------------------
 . # . | * # . | . # .
 . # . | . # * | . # .
 . # . | . @ . | . # .
-----------------------
 . . . | . . . | . . *
 . . . | . . . | * . .
 . . . | . . . | . * .
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby udosuk » Sat Jan 31, 2009 10:29 pm

Yep, all formations are possible to be both normalised simultaneously (and easily). In your special case it's like this:

Code: Select all
r123465897

 * . . | . . . | . . .
 . * . | . . . | . . .
 . . * | . . . | . . .
-----------------------
 . # . | * # . | . # .
 . # . | . @ . | . # .
 . # . | . # * | . # .
-----------------------
 . . . | . . . | * . .
 . . . | . . . | . * .
 . . . | . . . | . . *

Note it's a very intuitive thing, but I'm 99.99% certain that you can't find a formation where I can't normalise both symmetries within 5 seconds of observation.:)
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Sat Jan 31, 2009 11:15 pm

No, this transformation would destroy the [added: normalized form of the ] sticks symmetry:
Code: Select all
 * . . | . . . | . . .
 . * . | . . . | A . .
 . . * | . . . | . . .
-----------------------
 . # . | * # . | . # .
 . # . | . @ . | . # .
 . # . | . # * | . # .
-----------------------
 . . . | . . . | * . B
 . . . | . . . | . * .
 . . . | . . . | . . *
In the original puzzle A and B are from the same number cycle, here not.
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby udosuk » Sun Feb 01, 2009 3:10 am

Okay eleven, I guess I should have proven your situation is impossible in the first place instead.:idea:

Code: Select all
 * . . | . . . | . . .
 . * . | . . . | . . .
 . . * | . . . | . . .
-------+-------+-------
 . # . | * # . | . # .
 . # . | . # * | . # .
 . # . | . @ . | . # .
-------+-------+-------
 . . . | . . . | . . *
 . . . | . . . | * . .
 . . . | . . . | . * .

You are stating that the above grid has (normalised) CS symmetry as it is, with the axis being the # & @ cells. You also claim that the * & @ cells form the axis of a DM symmetry.

My job is to prove that this claim must be wrong, because the * & @ cells can never form the axis of any DM symmetry.

Here is how:

Let's label all the cells being CS symmetrical to the * cells as "$":

Code: Select all
 * . . | . . . | $ . .
 . * . | . . . | . . $
 . . * | . . . | . $ .
-------+-------+-------
 . # . | * # $ | . # .
 . # . | $ # * | . # .
 . # . | . @ . | . # .
-------+-------+-------
 $ . . | . . . | . . *
 . $ . | . . . | * . .
 . . $ | . . . | . * .

Now we morph the grid so that the * & @ cells form the leading diagonal:

Code: Select all
r123465897

 * . . | . . . | $ . .
 . * . | . . . | . . $
 . . * | . . . | . $ .
-------+-------+-------
 . # . | * # $ | . # .
 . # . | . @ . | . # .
 . # . | $ # * | . # .
-------+-------+-------
 . $ . | . . . | * . .
 . . $ | . . . | . * .
 $ . . | . . . | . . *

Suppose your claim is true, then the isomorph above must have \M symmetry now (for the moment let's not worry about how the (normalised) CS symmetry has been messed up).

However let's consider the contents of the *, $ & @ cells. Obviously the * & @ cells, being the axis of the \M symmetry, must all come from a set of 3 digits, say {ABC}. Among these, one is a self-mapping digit under the CS symmetry of the original grid, let's say A. Then B & C must be mapped to 2 other digits under that CS symmetry, let's say D & E respectively. Therefore the contents of the $ & @ cells must all come from the set of 3 digits {ADE}.

Now look at b5. Obviously r5c5 (@) must be A and the 2 $ cells there (r4c6+r6c4) must be from {DE}. That means (DE) must be a digit cycle under the \M symmetry. (Ditto for (A), (B), (C).)

Now look at the $ cells in b3 & b7. We have just determined that their contents are {ADE}, and (A) & (DE) are digit cycles under the \M symmetry. So these 6 cells should collectively be symmetrical across the leading diagonal as 3 pairs.

But this is not the case, as only r3c8 & r8c3 are symmetrical, while the other 2 pairs of $ cells are not. Contradiction!:!:

Therefore your claim must be false, that the * & @ cells in the original grid can't form the axis of a DM symmetry.

Q.E.D.:idea:

(Whew... For a moment I thought my logic wasn't sound and my proof would fall apart... Turns out I was just a bit "loose" for some assumptions...:idea: )
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Sun Feb 01, 2009 7:16 am

Well done on this case, udosuk. I am sure, you can resolve all other cases (there should not be too much) in a similar way.
But this is not a diploma work, so we dont have to do all the boring things in our sparse free time:)

So lets look at other grids now.
E.g. i noticed, that out the 18 automorphisms grids there are two with the 3 symmetries JD+JR+HT (22 25 79). This means, that they are independant in the sense, that the number of automorphisms is a multiply of the max. number cycle lengths 3+3+2.
On the other hand there are 3 grids with the 9 symmetries (8 28 30 32 134 135 142 143 144) - 5 basic ones (including MR, GR and sticks) and 4 sticks variations.

I will try to post them tomorrow or the next day showing at least most of the symmetries normalized.
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby udosuk » Sun Feb 01, 2009 8:32 am

Okay, waiting for the next challenge from you.:)

Meanwhile here is a little nice discovery from me:

Starting from this initial grid:
Code: Select all
 A11 A12 A13 | A14 A15 A16 | A17 A18 A19
 A21 A22 A23 | A24 A25 A26 | A27 A28 A29
 A31 A32 A33 | A34 A35 A36 | A37 A38 A39
-------------+-------------+-------------
 A41 A42 A43 | A44 A45 A46 | A47 A48 A49
 A51 A52 A53 | A54 A55 A56 | A57 A58 A59
 A61 A62 A63 | A64 A65 A66 | A67 A68 A69
-------------+-------------+-------------
 A71 A72 A73 | A74 A75 A76 | A77 A78 A79
 A81 A82 A83 | A84 A85 A86 | A87 A88 A89
 A91 A92 A93 | A94 A95 A96 | A97 A98 A99

Let's flip the whole grid horizontally (i.e. horizontal reflection):
Code: Select all
c987654321

 A19 A18 A17 | A16 A15 A14 | A13 A12 A11
 A29 A28 A27 | A26 A25 A24 | A23 A22 A21
 A39 A38 A37 | A36 A35 A34 | A33 A32 A31
-------------+-------------+-------------
 A49 A48 A47 | A46 A45 A44 | A43 A42 A41
 A59 A58 A57 | A56 A55 A54 | A53 A52 A51
 A69 A68 A67 | A66 A65 A64 | A63 A62 A61
-------------+-------------+-------------
 A79 A78 A77 | A76 A75 A74 | A73 A72 A71
 A89 A88 A87 | A86 A85 A84 | A83 A82 A81
 A99 A98 A97 | A96 A95 A94 | A93 A92 A91

And then let's do this "minor" transformation: swap the top & bottom band, and then swap the left & right stack:
Code: Select all
r789456123
c789456123

 A73 A72 A71 | A76 A75 A74 | A79 A78 A77
 A83 A82 A81 | A86 A85 A84 | A89 A88 A87
 A93 A92 A91 | A96 A95 A94 | A99 A98 A97
-------------+-------------+-------------
 A43 A42 A41 | A46 A45 A44 | A49 A48 A47
 A53 A52 A51 | A56 A55 A54 | A59 A58 A57
 A63 A62 A61 | A66 A65 A64 | A69 A68 A67
-------------+-------------+-------------
 A13 A12 A11 | A16 A15 A14 | A19 A18 A17
 A23 A22 A21 | A26 A25 A24 | A29 A28 A27
 A33 A32 A31 | A36 A35 A34 | A39 A38 A37

Now compare it to the original grid. What do you notice?:?:

Yes - if you're observant enough, you should notice that what you just did is equivalent to the (normalised) Column-Sticks transformation. Or, if we want to write it up nicely, we can express this transformation very elegantly as the following 2 actions:

1. Replacing c[n] with c[10-n] (i.e. reversing the columns)
2. Replacing b[n] with b[10-n] (i.e. reversing the blocks)

Similarly, we can express the Row-Sticks transformation as "reversing the rows" & "reversing the blocks".

So after all, each of the 4 directions of reflections (-, |, \, /) corresponds to an order-2 symmetry group (RS, CS, \M, /M), each with an axis consisting of 3 triplets of cells. Pretty neat eh?:)

On the other hand, if you take this correspondence as an analogy, it's not hard to envision why CS+\M implies all sorts of existing order-2 & order-4 symmetries:

Let's assume initially we have only 2 operations available: the h-flip (horizontal flipping) and the transpose (flipping about the leading diagonal). Combining these 2 operations together, what do we get?

To envision this, let's use a 3x3 square grid as a sample:
Code: Select all
1 2 3
4 5 6
7 8 9

Do the transpose:
Code: Select all
1 4 7
2 5 8
3 6 9

And then we apply the h-flip:
Code: Select all
7 4 1
8 5 2
9 6 3

Notice what we just did? Yes, we just rotated it 90 degrees clockwisely. (We can rotate it 90 degrees anticlockwisely if we do the h-flip 1st and then transpose 2nd.:idea: )

Let's call this combined operation qtc (quarter-turn-clockwisely). With these 3 operations in hand, we can generate any sort of reflections/rotations on the grid (in practice there are only 8 possible variations including the trivial identity operation).

These include:

ht (half-turn, or 180 degrees rotation) = qtc x 2
Code: Select all
9 8 7
6 5 4
3 2 1

qta (quarter-turn-anticlockwisely) = qtc x 3 = h-flip + transpose
Code: Select all
3 6 9
2 5 8
1 4 7

v-flip (vertical-flipping) = ht + h-flip
Code: Select all
7 8 9
4 5 6
1 2 3

anti-transpose (flipping about the non-leading diagonal) = ht + transpose = h-flip + qtc
Code: Select all
9 6 3
8 5 2
7 4 1

Hence, if we have CS (h-flip) & \M (transpose) symmetries, then intuitively we'll also have RS (v-flip), /M (anti-transpose), HT (ht) & QT (qtc/qta) symmetries.:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Sun Feb 01, 2009 11:49 pm

Maybe you noticed, that i represented QT with DBxRx, where BxRx is equivalent to vertical reflection [edit: lets say horizontal reflection at row 5, thats what you call vertical flipping]


This is a form of the grid with 162 automorphisms and the symmetry classes
8 9 22 25 30 32 134 135 142 145
Code: Select all
 +-------+-------+-------+
 | 2 4 9 | 3 5 7 | 1 6 8 |
 | 3 5 7 | 1 6 8 | 2 4 9 |
 | 1 6 8 | 2 4 9 | 3 5 7 |
 +-------+-------+-------+
 | 9 3 6 | 7 1 4 | 8 2 5 |
 | 7 1 4 | 8 2 5 | 9 3 6 |
 | 8 2 5 | 9 3 6 | 7 1 4 |
 +-------+-------+-------+
 | 6 7 2 | 4 8 3 | 5 9 1 |
 | 4 8 3 | 5 9 1 | 6 7 2 |
 | 5 9 1 | 6 7 2 | 4 8 3 |
 +-------+-------+-------+

MC (8 mirrored): (123)(456)(789)
JD (22): (194)(275)(386)
JR (25): (123)(456)(789), JC (185)(296)(374)
GR (32): (132)(465)(587), GR- (1)(2)(3)(4)(5)(6)(7)(8)(9)
CS (134): (1)(2)(3)(47)(58)(69)
CS°MC (135): (123)(486759)
CS°GR (142): (132)(495768)
CS°JR (145): (123)(486759)

After swapping rows 8 and 9:
Code: Select all
 +-------+-------+-------+
 | 2 4 9 | 3 5 7 | 1 6 8 |
 | 3 5 7 | 1 6 8 | 2 4 9 |
 | 1 6 8 | 2 4 9 | 3 5 7 |
 +-------+-------+-------+
 | 9 3 6 | 7 1 4 | 8 2 5 |
 | 7 1 4 | 8 2 5 | 9 3 6 |
 | 8 2 5 | 9 3 6 | 7 1 4 |
 +-------+-------+-------+
 | 6 7 2 | 4 8 3 | 5 9 1 |
 | 5 9 1 | 6 7 2 | 4 8 3 |
 | 4 8 3 | 5 9 1 | 6 7 2 |
 +-------+-------+-------+

MR Band 2, MD B13 (9): (147)(258)(369)

Where is symmetry class 30 (1 JR, 2 GR) ?
[Added: ah, got it now, with column changes you get 9 and 30 in one grid]
Last edited by eleven on Mon Feb 02, 2009 12:17 am, edited 1 time in total.
eleven
 
Posts: 1564
Joined: 10 February 2008

Postby eleven » Mon Feb 02, 2009 2:34 am

This is the grid with 9 symmetries, but only 18 automorphisms. I needed 3 morphs to show them all normalized, but i guess, this can be done better.

8 28 30 32 134 135 142 143 144
Code: Select all
 +-------+-------+-------+
 | 2 7 8 | 1 9 4 | 3 5 6 |
 | 1 9 4 | 3 5 6 | 2 7 8 |
 | 3 5 6 | 2 7 8 | 1 9 4 |
 +-------+-------+-------+
 | 4 1 7 | 8 2 5 | 6 3 9 |
 | 6 3 9 | 4 1 7 | 8 2 5 |
 | 8 2 5 | 6 3 9 | 4 1 7 |
 +-------+-------+-------+
 | 5 4 2 | 7 6 1 | 9 8 3 |
 | 7 6 1 | 9 8 3 | 5 4 2 |
 | 9 8 3 | 5 4 2 | 7 6 1 |
 +-------+-------+-------+

MC (8 mirrored): (132)(468)(579)
CS (134): (1)(2)(3)(47)(58)(69)
CSMC (135): (132)(498765)

r23, r78:
Code: Select all
 +-------+-------+-------+
 | 2 7 8 | 1 9 4 | 3 5 6 |
 | 3 5 6 | 2 7 8 | 1 9 4 |
 | 1 9 4 | 3 5 6 | 2 7 8 |
 +-------+-------+-------+
 | 4 1 7 | 8 2 5 | 6 3 9 |
 | 6 3 9 | 4 1 7 | 8 2 5 |
 | 8 2 5 | 6 3 9 | 4 1 7 |
 +-------+-------+-------+
 | 5 4 2 | 7 6 1 | 9 8 3 |
 | 9 8 3 | 5 4 2 | 7 6 1 |
 | 7 6 1 | 9 8 3 | 5 4 2 |
 +-------+-------+-------+

GR (32): (1)(2)(3)(4)(5)(6)(7)(8)(9)
CSGR (142): (132)(498765)

r46:
Code: Select all
 +-------+-------+-------+
 | 2 7 8 | 1 9 4 | 3 5 6 |
 | 1 9 4 | 3 5 6 | 2 7 8 |
 | 3 5 6 | 2 7 8 | 1 9 4 |
 +-------+-------+-------+
 | 8 2 5 | 6 3 9 | 4 1 7 |
 | 6 3 9 | 4 1 7 | 8 2 5 |
 | 4 1 7 | 8 2 5 | 6 3 9 |
 +-------+-------+-------+
 | 5 4 2 | 7 6 1 | 9 8 3 |
 | 7 6 1 | 9 8 3 | 5 4 2 |
 | 9 8 3 | 5 4 2 | 7 6 1 |
 +-------+-------+-------+

2 JR, 1 GR (28): (132)(468)(579)
1 JR/B2, 2 GR/B13 (30): (231)(486)(597)
CSJR/B2GR/B13 (143): (123)(456789)
CSGR/B2JR/B13 (144): (132)(498765)

[Edit: moved 2 sticks variations to the right grid, thanks to udosuk]
Last edited by eleven on Fri Feb 06, 2009 9:50 pm, edited 1 time in total.
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Postby eleven » Mon Feb 02, 2009 5:04 am

So the number of automorphisms given by a single symmetry does not say anything about the number of automorphisms of combined symmetries.
We had
- DM(2) + Sticks(2) gives 8
- JD(3)+JR(3)+HT(2) gives 18
- 8(3)+28(3)+30(3)+32(3)+134(2)+135(6)+142(6)+143(6)+144(6) gives 18 too
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Postby StrmCkr » Mon Feb 02, 2009 5:09 am

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


im still trying to figure out how you guys find 9 diffrent types of symmetries when i find only one.

the function is based off columns mini rows.
where groups

abc = 123
def = 579
ghj = 486
Code: Select all
+---------------+---------------+---------------+
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
+---------------+---------------+---------------+
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
+---------------+---------------+---------------+
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
+---------------+---------------+---------------+


from here you can show that grid stays in the same shap via row/column/band swaping.

swap band 1 for band 2.
Code: Select all
+---------------+---------------+---------------+
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
| ghj  abc  def | ghj  abc  def | ghj  abc  def |
+---------------+---------------+---------------+
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
+---------------+---------------+---------------+
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
+---------------+---------------+---------------+

swaps
columns (1&2) then (2&3)
columns (4&5) then (5&6)
columns (7&8) then (8&9)

which shows bands 2&3 swaping (from the inital grid)

Code: Select all
+---------------+---------------+---------------+
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
+---------------+---------------+---------------+
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
| def  ghj  abc | def  ghj  abc | def  ghj  abc |
+---------------+---------------+---------------+
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
| abc  def  ghj | abc  def  ghj | abc  def  ghj |
+---------------+---------------+---------------+


so the only thing that matters for grid counts is arangments of the groups
which is

(3*2*1)+ (3*2*1)+(3*2*1)=18
Some do, some teach, the rest look it up.
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Postby udosuk » Mon Feb 02, 2009 6:13 am

eleven wrote:Maybe you noticed, that i represented QT with DBxRx, where BxRx is equivalent to vertical reflection [edit: lets say horizontal reflection at row 5, thats what you call vertical flipping]

I did notice it, but wasn't paying enough attention. But you didn't see the link between horizontal/vertical reflection and sticks symmetries, did you?:)

Will look at your grids later.
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