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I would like to produce an image similar to the upper half of this graphics by Victor Vasarely.

enter image description here

All I could produce so far is this dilettante image:

g = Graphics[{GrayLevel@0.7, Rectangle[]}, ImageSize -> 30];
h = Graphics[{GrayLevel@0.7, Rotate[Rectangle[], Pi/12]}, ImageSize -> 30];

  {g, g, g, g, g, g, g, g, g, g, g},
  {g, g, g, g, g, h, g, g, g, g, g},
  {g, g, g, g, h, h, h, g, g, g, g},
  {g, g, g, h, h, h, h, h, g, g, g},
  {g, g, h, h, h, h, h, h, h, g, g},
  {g, h, h, h, h, h, h, h, h, h, g},
  {g, g, h, h, h, h, h, h, h, g, g},
  {g, g, g, h, h, h, h, h, g, g, g},
  {g, g, g, g, h, h, h, g, g, g, g},
  {g, g, g, g, g, h, g, g, g, g, g},
  {g, g, g, g, g, g, g, g, g, g, g}},
 Background -> GrayLevel@0.2,
 Spacings -> {0.1, 0}]

enter image description here

I also tried scaled rectangles and parallelograms but to no avail: The forms don't align "properly"(equal vertical spacings).

Maybe Grid is an inept tool to get the desired effect?

How would you approach this problem?

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(at) eldo: Honestly speaking, in the graphics of Victor Vaserely I don't see rotated rectangles, not even rhombus but "at most" paralelograms with angles progressing according to a more or less complicated rule as we move along. So, if you really want to reproduce Vasarely's graphics we need to agree on the deformation rule. (BTW, I like your graph :-) –  Dr. Wolfgang Hintze yesterday
(at) eldo: look at that: h = Graphics[{GrayLevel@0.7, Rotate[Rectangle[], Pi/12]}, ImageSize -> 40]; gives a beautiful displacements of the outer squares. –  Dr. Wolfgang Hintze yesterday

2 Answers 2

up vote 3 down vote accepted

My first attempt, using ShearingTransform

sd[i_, j_] := 
 If[And[4 <= j <= 10, 8 - j <= i - 1 <= j, 
   j - 6 <= i - 1 <= 14 - j], -20, 0]
      ShearingTransform[sd[i, j] Degree, {0, 1}, {1, 0}]],
     RescalingTransform[{{-1, 1}, 
       If[sd[i, j] == 0, {0, 1}, {0, 1.5}]}, {{-1, 1}, {0, 
        1}}]]}], {i, 9}, {j, 12}], Background -> GrayLevel@0.2]

Mathematica graphics

I think this meets the equal vertical spacings criterion but assumes that each of the parallelograms are sheared by the same amount, which does not appear to be correct.

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1000 thanks for your answer. It is much nearer to what I want :) –  eldo yesterday

Not exactly what is asked but it may be a good start too.

Alternative approach with Grid:

opt = {ImageSize -> {20, 20}, BaseStyle -> GrayLevel@.9};

  Framed[Grid[#, Spacings -> {.3, .3}], Background -> Black] &,

  ArrayPad[#, {2, 2}, Graphics[Rectangle[], opt]] &,

  # + (DiamondMatrix[5] /. {1 -> 0, 0 -> 1}) Graphics[Rectangle[], opt] &
     DiamondMatrix[5] Graphics[ Polygon[{{0, 0}, {1, -.4}, {1, .6}, {0, 1}}], opt]

enter image description here

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