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I managed to make a mod4 addition table.

Grid[Table[Mod[i + j, 4], {i, 0, 3}, {j, 0, 3}]]

What I'd like to do is draw a grid of 16 small blocks, 4 rows, 4 columns, each colored according to a color associated with the number in my table.

I'm sure there might be some duplicated help on this, so please let me know, but I am looking for a very elementary start.

Also, once I have this basic plot, I want to try translation, rotation, and reflection, keeping the original and the transformed in the final image.

Update: Thanks to RomkeBonteko, MatrixPlot provides a first image I need.

enter image description here

Now, I'd like to remove the frame, ticks, edge padding, then reflect it across its right edge so that I now have 4 rows and 8 columns of blocks. Then I'd like to flip this result over its bottom edge to get a final image of 8 rows and 8 columns of blocks.

Then I want to work on translations and rotations of the original 4-by-4 image.

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    $\begingroup$ For your first question, is MatrixPlot what you need? $\endgroup$ – Romke Bontekoe Apr 16 '16 at 7:41
  • $\begingroup$ @RomkeBontekoe That is not only a good reminder, but I am going to do all of the examples in the documentation for MatrixPlot. I've edited my original question with my MatrixPlot. $\endgroup$ – David Apr 16 '16 at 14:49
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reflectedMPF = With[{m = ArrayPad[#, Thread[{0, Dimensions[#]}], "Reversed"]}, 
               MatrixPlot[m, Frame -> False]]&

Examples:

reflectedMPF@Table[Mod[i + j, 4], {i, 0, 3}, {j, 0, 3}]

Mathematica graphics

reflectedMPF[RandomInteger[100, {5, 10}]]

Mathematica graphics

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I think this is what you need for reflection.

reflectBottom[x_] := Join[x, Reverse[x]];
reflectRight[x_] := Transpose@reflectBottom@Transpose[x];

Table[Mod[i + j, 4], {i, 0, 3}, {j, 0, 3}] // 
 MatrixPlot[reflectBottom@reflectRight[#], Frame -> None] &

enter image description here

Update

rotate90[x_] := Transpose@Reverse[x];
rotate90Glue[x_] := Transpose@Join[Transpose[x], Reverse[x]];

MatrixPlot[#, Frame -> None] & /@ {#, reflectBottom[#], 
    reflectRight[#], rotate90[#], rotate90Glue[#]} &@
 Table[Mod[i + 2 j, 7], {i, 0, 3}, {j, 0, 3}]

enter image description here

I changed your pixel function, because the original one was too symmetric, so it was hard to check.

What is your final goal?

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  • $\begingroup$ How about rotating the original image 90 degrees using its lower right corner as the pivot point. $\endgroup$ – David Apr 20 '16 at 22:02
  • $\begingroup$ @David See update $\endgroup$ – BlacKow Apr 20 '16 at 22:21

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