How to delete duplicate graphics of the same kind?

Question

A054247: Number of n X n binary matrices under action of dihedral group of the square D_4.

(*半念浮生*)
g[list_] := {list, Reverse /@ list, Reverse@list, 
   Reverse /@ Reverse@list};
g1[list_] := (g[list]~Join~g[Transpose@list]) // Sort;
(tmp = GatherBy[Partition[#, 3] & /@ Tuples[{0, 1}, 9], 
    g1]) // Length(* 3*3 *)
Map[MatrixPlot[#, Frame -> None, FrameTicks -> None, Mesh -> All, 
   ColorFunction -> "Monochrome", 
   ImageSize -> Scaled[.04]] &, tmp, {2}]
% // Flatten // Length

I wonder if there is a more general way to solve this problem? For example, use the knowledge of group theory to solve it.

Bibliography: 应用组合学 P298

Answer 1

Here is a suggestion from group theory:

We need to define an action on squares for the eight elements of DihedralGroup[4]. In general, it would be enough to define it for the two generators and then construct the rest using the GroupElementToWord/FromWord functionality, but I think it is instructive to define the whole action here, because this group is small. It is possible to use any representation you want for the squares, and I'll use matrices. The action function must be defined for the representation chosen, so I'll construct everything using Transpose and Reverse:

action[matrix_, Cycles[{}]] := matrix;
action[matrix_, Cycles[{{2, 4}}]] := Transpose[matrix];
action[matrix_, Cycles[{{1, 2}, {3, 4}}]] := Reverse[matrix, {2}];
action[matrix_, Cycles[{{1, 2, 3, 4}}]] := Transpose[Reverse[matrix]];
action[matrix_, Cycles[{{1, 3}}]] := Reverse[Transpose[Reverse[matrix]]];
action[matrix_, Cycles[{{1, 3}, {2, 4}}]] := Reverse[matrix, {1, 2}];
action[matrix_, Cycles[{{1, 4, 3, 2}}]] := Reverse[Transpose[matrix]];
action[matrix_, Cycles[{{1, 4}, {2, 3}}]] := Reverse[matrix];

You can imagine that the numbers 1, 2, 3, 4 in the permutations refer to square corners, numbered clockwise, say.

Then this gives the list of all matrices of dimensions {n, n} containing zeros and/or ones:

all01s[n_] := Tuples[{0, 1}, {n, n}]

Finally the computation you want is a simple call to GroupOrbits:

GroupOrbits[DihedralGroup[4], all01s[2], action] // Length
(* 6 *)

GroupOrbits[DihedralGroup[4], all01s[3], action] // Length
(* 102 *)

GroupOrbits[DihedralGroup[4], all01s[4], action] // Length
(* 8548 *)

Answer 2

Re-writing OP's g1 using transformations from jose's answer:

ClearAll[transformations, deleteDupes]

transformations = Union[{#, Reverse @ #, Reverse[#, {2}], Reverse[#, {1, 2}], 
     Transpose @ #, Reverse @ Transpose @ #, 
     Transpose @ Reverse @ #, Reverse @ Transpose @ Reverse @ #}] &;

deleteDupes = DeleteDuplicatesBy[transformations][Tuples[{0, 1}, {#, #}]] &;

Length /@ deleteDupes /@ {2, 3, 4}

{6, 102, 8548}

Answer 3

Recreating images:

m = Table[RandomInteger[], 100, 3, 3]; i = Image[m[[#]]] & /@ Range[Length[m]]

You can just use DeleteDuplicates[] here which is pretty nice, and ImageRotate[], although I think the original question was not with created images.

Make images into MorphologicalComponents[]. You can get the position of all repeating images by:

images = MorphologicalComponents[#] & /@ i;
GatherBy[Range@Length[images], images[[#]] &]

out[1]: {{1,46},{2},{3},{4},{5,33,53},{6},{7},{8},{9},{10},{11},{12},{13},{14},{15},{16,97},{17},{18},{19},{20},{21},{22},{23},{24},{25},{26,62},{27},{28},{29},{30},{31},{32,69},{34},{35},{36},{37},{38},{39},{40},{41},{42},{43},{44,48},{45},{47},{49,81},{50},{51},{52},{54},{55},{56,78},{57,72},{58},{59,82},{60},{61},{63},{64},{65},{66},{67},{68},{70},{71},{73},{74},{75},{76},{77},{79},{80},{83},{84},{85},{86},{87},{88},{89},{90},{91},{92},{93},{94},{95},{96},{98},{99},{100}}

Again use ImageRotate[] to get all combinations, or maybe Tuples[]?

Or if you would like a more True or False method, maybe:

images = MorphologicalComponents[#] & /@ i;
(And @@ Thread[images[[1]] == images[[2]]]) 
out[1]: True