If I have a choice, I prefer a square grid. If you do that you have to make sure the square root of the number of items is a round number. With ternary numbers this is the case if you have an even number of trits (eg bits). Here, there are four. Then you get 34=81 elements. 9×9. Four trits so in four places I have three positions. The patterns are more in the shade than with binary numbers. That is interesting.

#### 4 trits squares, arc sides, white on black, in sequence (nr. 12-01)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on Fabriano BLACK BLACK paper, 300g  A4.
• White pencil (Mitsubishi Uni-ball)
• In sequence from top left to bottom right
• Dimensions: 15×15 cm

Detail:

#### 4 trits squares, arc sides, black on white, in sequence (nr. 12-02)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on CANSON® XL® Bristol Bristolpaper A4.
• Black pigmented fineliner
• In sequence from top left to bottom right
• Dimensions: 15×15 cm

Detail

#### 4 trits squares, triangular sides, white on black, in sequence (nr. 12-03)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on Fabriano BLACK BLACK paper, 300g  A4.
• White pencil (Mitsubishi Uni-ball)
• In sequence from top left to bottom right
• Dimensions: 15×15 cm.

Detail:

#### 4 trits squares, triangular sides, black on white, in sequence (nr. 12-04)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on CANSON® XL® Bristol Bristolpaper A4.
• Black pigmented fineliner
• In sequence from top left to bottom right
• Dimensions: 15×15 cm

#### 4 trits squares, triangular sides, black on white, shuffled (nr. 12-05)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on CANSON® XL® Bristol Bristolpaper A4.
• Black pigmented fineliner
• Shuffled, alle elements are different
• Dimensions: 15×15 cm

#### 4 trits squares, triangular sides, black on white, shuffled (nr. 12-06)

• 4 trits so 34 = 81 elements (9×9)
• Plotted on Fabriano BLACK BLACK paper, 300g  A4.
• White pencil (Mitsubishi Uni-ball)
• Shuffled, alle elements are different
• Dimensions: 15×15 cm