The New Sudoku Players' Forum

by **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

by **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.

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

by **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.

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

by **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: * . . | . . . | . . . . * . | . . . | . . . . . * | . . . | . . . ----------------------- . # . | * # . | . # . . # . | . # * | . # . . # . | . @ . | . # . ----------------------- . . . | . . . | . . * . . . | . . . | * . . . . . | . . . | . * .

by **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.

by **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.

by **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:

)

by **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.

by **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:

by **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]

by **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]

by **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

by **StrmCkr** » Mon Feb 02, 2009 5:09 am

removed

by **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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