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Let's say we have a $3×3$ square where $3$ of the cells are labeled $a$, $b$, $c$ and the rest are blank. Two such squares are considered "equivalent" if one square can be obtained from another square by 1) rotation on 90, 180, and 270 degrees, 2) reflection (through the horizontal, vertical, or either diagonal axis).

I need to find equivalence classes of squares (maybe groups or patterns?).

My attempt is: 1) put the $a$, $b$ and $c$ in the 1-st row: $A=\left(% \begin{array}{ccc} a & b & c \\ .. & .. & .. \\ .. & .. & .. \\ \end{array} \right)$, then we can rotate the square $A$ on 90, 180 and 270 degrees: $A_{90}=\left(% \begin{array}{ccc} .. & .. & a \\ .. & .. & b\\ .. & .. & c \\ \end{array} \right)$, $A_{180}=\left(% \begin{array}{ccc} .. & .. & .. \\ .. & .. & ..\\ c & b & a \\ \end{array} \right)$, $A_{270}=\left(% \begin{array}{ccc} c & .. & .. \\ b & .. & ..\\ a & .. & .. \\ \end{array}. \right)$.

Four square $A$, $A_{90}$, $A_{180}$ and $A_{270}$ are equvalent. This is the first equivalence class.

2) put the $a$, $b$ and $c$ in the main diagonal: $$A=\left(% \begin{array}{ccc} a & .. & .. \\ .. & b & .. \\ .. & .. & c \\ \end{array}% \right) $$ and rotate on 90 degree $$A_{90}=\left(% \begin{array}{ccc} .. & .. & a \\ .. & b & .. \\ c & .. & .. \\ \end{array}% \right) $$ Two square $A$ and $A_{90}$ are equvalent. This is the second equivalence class.

Edit 2. Here I have found the 16 patterns.

enter image description here

Question. How many equivalence classes for three elements in a square are there?

Nick
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  • What is mean C3 in your comment? – Nick Feb 26 '19 at 03:34
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    789 is the number of combinations without symmetry property of the table. – Nick Feb 26 '19 at 03:39
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    This is a burnside’s lemma type problem; to count the number of labelings up to symmetry, you compute the average number of labelings which are fixed by each symmetry of the table. – Mike Earnest Feb 26 '19 at 04:23
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    "the number of different combinations of three integers a, b and c in the 3×3 table " does not have any meaning to me. What is a combination of integers in a table? – miracle173 Feb 26 '19 at 10:27
  • @miracle173, i have used the 5 cases as examples. Unfortunately, is don't know how to reformulate the question in 'symmetries groups terms'. – Nick Feb 26 '19 at 10:55
  • I really don't understand what you mean by "the symmetries of this table". Can you be more specific? Or maybe explain where the problem comes from? – AnalysisStudent0414 Feb 26 '19 at 11:35
  • Then you should think about how reformulate it. I don't think that you can solve a problem if you can't formulate it. Do you want to place three different symbols a,b,c on a 3x3 table? Maybe you can show us all these combinations of a and b (and maybe c ) on a 2x2 table? – miracle173 Feb 26 '19 at 12:37
  • @analysisstudent0414, I have tried to explain the symmetries in the 1st case. – Nick Feb 26 '19 at 12:38
  • 7x8x9 is the number of ways you can place three different symbols on a 3x3 table. But what means symmetry? What is a placement of a,b,c that is symmetric to the placement in your first diagram. – miracle173 Feb 26 '19 at 12:41
  • Sorry, it is still not a well-defined problem – AnalysisStudent0414 Feb 26 '19 at 12:54
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    Can I propose the following re-wording? Consider the $8$ symmetries of the square, as in the reference wikidot link. Now consider a $3 \times 3$ square where $3$ of the cells are labeled $a, b, c$ and the rest are blank. Two such squares are considered "equivalent" if they are related by one of the $8$ symmetries. So the problem becomes: how many "different" squares are there (i.e. how many equivalence classes)? If this interpretation is what you want, then (1) your $5$ examples are confusing, and (2) consider @MikeEarnest comment re: burnside lemma. – antkam Feb 26 '19 at 14:33
  • @antkam, I have re-worded the question after your comment. – Nick Feb 26 '19 at 15:55
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    wait... only rotations are equivalent? i.e. two squares which are mirror images (reflections) of each other are considered different? – antkam Feb 26 '19 at 16:13

1 Answers1

2

It is unclear whether you want to count

  1. Placements, or

  2. Labelings.

up to symmetry. The two grids below have the same placement, but different labelings. The number of labelings without symmetry is $9\times 8\times 7$. The number of placements is $(9\times 8\times 7)/(3!)$.

a b c     a c b
. . .     . . .
. . .     . . .

Placements

There are $8$ symmetries of array:

  • The identity symmetry.

  • Three rotations, by $90,180$ and $270$ degrees.

  • Four reflections, through the horizontal, vertical, or either diagonal axis.

In order to count the number of placements up to symmetry, we add up, for each symmetry, the number of placements which invariant under that symmetry, then divide by $8$.

  • Each of the $\binom{9}3$ placements is invariant under the identity.

  • There are no placements with three numbers which are invariant under $90^\circ$ rotation.

  • There are $4 $ placements which are invariant under $180^\circ$ rotation. These are placements which have a three in a row through the center.

  • For the horizontal reflection, there are $10$ placements with that symmetry. There are two cases; either two of the numbers are symmetric about the vertical axis, going in spots A, B or C, or they are all on the vertical axis. In the first case, there are $3$ choices for A, B or C, ,and three choices for where the last number goes. The second case has one option, for $3\cdot 3+1=10$ total.

A 1 A
B 2 B
C 3 C
  • The same answer is true for the other three reflections.

Therefore, the number of placements up to symmetry is

$$ \frac18\left(\binom{9}3+4+4\cdot 10\right)=\frac18(84+44)=16 $$

Labelings

I leave the details to the reader, but the answer is

$$ \frac18(9\cdot 8\cdot 7+3\cdot 0+4\cdot (3\cdot 2\cdot 1))=66 $$

Mike Earnest
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    As a side note; I was confused for a long time because I though that you should have # Labelings = 6$\times$ # Placements. This is not the case. – Mike Earnest Feb 26 '19 at 16:25