Displaying 1-10 of 18 results found.
Langton's ant on a Penrose rhomb tiling: number of black cells after n moves of the ant.
+10
11
0, 1, 2, 3, 4, 5, 4, 5, 6, 7, 8, 9, 8, 7, 6, 7, 6, 7, 8, 9, 8, 9, 10, 11, 10, 9, 8, 7, 8, 7, 8, 9, 10, 9, 10, 11, 12, 11, 12, 13, 14, 13, 14, 15, 16, 15, 16, 17, 18, 17, 16, 15, 14, 13, 12, 11, 12, 11, 12, 13, 14, 13, 14, 13, 14, 15, 14, 13, 12, 13, 12, 13, 14
COMMENTS
The ant lives on a centrally symmetric Penrose rhomb tiling with a "Sun" patch (S configuration, cf. A242935) at the center and starts on one of the thick rhombs of that patch, looking towards one of the outward edges of that tile. On a white rhomb, turn to the next edge of that cell in clockwise direction, flip the color of the rhomb, then move forward one unit. On a black rhomb, turn to the next edge of that cell in counterclockwise direction, flip the color of the rhomb, then move forward one unit. In contrast to the corresponding sequences for Langton's ant on periodic tilings, like the square tiling ( A255938) or a hexagonal tiling ( A269757), this sequence is most likely not unique. A Penrose tiling lacks translational symmetry, meaning any two finite regions in the tiling that are identical are surrounded by different patches of tiles when examining a large enough region of the surrounding tiles. Therefore I suspect that, unless the trajectory of the ant is bounded to stay inside a finite region of the tiling, the trajectories of any two ants placed at different starting points on the tiling will diverge at some point.
EXAMPLE
See illustration in links.
Total number of black cells after n iterations of Langton's ant with two ants on the grid placed side-by-side with one empty square between them and initially looking in the same direction.
+10
11
0, 2, 4, 6, 8, 6, 6, 6, 6, 8, 6, 8, 10, 12, 14, 12, 12, 12, 12, 14, 12, 12, 12, 12, 14, 12, 14, 16, 18, 20, 18, 18, 16, 16, 18, 16, 18, 20, 22, 24, 22, 22, 22, 22, 24, 22, 22, 22, 22, 24, 22, 24, 26, 28, 30, 28, 28, 26, 26, 28, 28, 30, 30, 32, 34, 34, 36, 34
COMMENTS
The two ants meet seven times; in that case, the color of the current square is flipped only once. Eventually, both ants build a recurrent highway pattern. - Rémy Sigrist, Jul 28 2019
FORMULA
a(n + 104) = a(n) + 24 for any n >= 14373. - Rémy Sigrist, Jul 28 2019
EXAMPLE
See illustrations in Fröhlich, 2019.
PROG
(PARI) See Links section.
Langton's ant on a chair tiling: number of black cells after n moves of the ant.
+10
9
0, 1, 2, 3, 4, 3, 4, 5, 6, 5, 4, 5, 6, 7, 8, 7, 8, 9, 10, 9, 10, 9, 10, 11, 12, 11, 10, 9, 10, 9, 10, 11, 12, 13, 12, 13, 14, 15, 14, 15, 14, 15, 16, 17, 16, 15, 14, 15, 14, 15, 16
COMMENTS
The ant begins on the inner corner of a subtile.
On a white tile, turn 90 degrees right, flip the color of the tile, then move forward until reaching a new tile, moving as far as possible within the tile.
On a black tile, turn 90 degrees left, then continue as above.
The chair tiling used for this automaton is, like all aperiodic hierarchical tilings, not unique (see for example Goodman-Strauss, p. 490). See "Remarks, 2019" in links for clarification which tiling the ant lives on.
LINKS
Tilings Encyclopedia, Chair
EXAMPLE
See illustrations in Fröhlich, 2019.
Langton's ant on a truncated square tiling: number of black cells after n moves of the ant when starting on an octagon and looking towards an edge where the tile meets another octagon.
+10
8
0, 1, 2, 3, 4, 5, 6, 5, 6, 7, 8, 9, 8, 7, 8, 9, 10, 11, 10, 11, 10, 9, 10, 11, 12, 13, 14, 15, 14, 15, 16, 17, 18, 17, 16, 17, 18, 17, 16, 17, 18, 19, 20, 21, 22, 21, 22, 23, 24, 23, 22, 23, 24, 25, 26, 27, 28, 27, 28, 29, 30, 29, 28, 29, 30, 31, 32, 33, 34
COMMENTS
First differs from A269757 at n = 19. On a white square, turn 90 degrees right, flip the color of the tile, then move forward one unit.
On a white octagon, turn 45 degrees right, flip the color of the tile, then move forward one unit.
On a black square, turn 90 degrees left, flip the color of the tile, then move forward one unit.
On a black octagon, turn 45 degrees left, flip the color of the tile, then move forward one unit.
As in the original variant, order emerges after a transition phase and the ant starts building a recurrent "highway" pattern of 12 steps that repeats indefinitely. - Rémy Sigrist, Jul 21 2019
FORMULA
a(n + 12) = a(n) + 6 for any n >= 34. - Rémy Sigrist, Jul 21 2019
EXAMPLE
See illustrations in Fröhlich, 2019.
PROG
(PARI) See Links section.
Langton's ant on a snub square tiling: number of black cells after n moves of the ant when starting on a square.
+10
7
0, 1, 2, 3, 4, 5, 4, 5, 6, 7, 8, 9, 8, 9, 10, 11, 10, 9, 10, 11, 12, 13, 14, 13, 14, 15, 16, 15, 16, 17, 18, 17, 16, 17, 18, 17, 16, 15, 16, 17, 18, 17, 16, 17, 18, 19, 20, 21, 20, 21, 22
COMMENTS
First differs from A276073 at n = 16. On a white square, turn 90 degrees right, flip the color of the tile, then move forward one unit.
On a white triangle, turn 60 degrees right, flip the color of the tile, then move forward one unit.
On a black square, turn 90 degrees left, flip the color of the tile, then move forward one unit.
On a black triangle, turn 60 degrees left, flip the color of the tile, then move forward one unit.
EXAMPLE
See illustrations in Fröhlich, 2019.
Let the starting square of Langton's ant have coordinates (0, 0), with the ant looking in negative x-direction. a(n) is the x-coordinate of the ant after n moves.
+10
5
0, 0, 1, 1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 1, 1, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 3, 3, 4, 4, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 0, 0, -1, -1, 0, 0, -1, -1, 0, 0, -1, -1, -2, -2, -1, -1, -2, -2, -3, -3, -2, -2, -1, -1, -2, -2, -3
PROG
(Python)
def ant(n):
steps = [(1, 0), (0, 1), (-1, 0), (0, -1)]
black = set()
x = y = 0
position = [(x, y)]
direction = 2
for _ in range(n):
if (x, y) in black:
black.remove((x, y))
direction += 1
else:
black.add((x, y))
direction -= 1
(dx, dy) = steps[direction%4]
x += dx
y += dy
position.append((x, y))
return position
print([p[0] for p in ant(100)])
# change p[0] to p[1] to get y-coordinates
Iterations at which Langton's Ant living on triangular tiling passes through the origin.
+10
5
0, 6, 24, 30, 72, 78, 96, 102, 108, 174, 180, 198, 212, 222, 252, 282, 292, 306, 324, 330, 408, 414, 420, 438, 444, 522, 544, 554, 576, 594, 648, 666, 672, 798, 804, 810, 852, 858, 920, 926, 972, 978, 984, 1018, 1024, 1154, 1160, 1178, 1184, 1190, 1208, 1214
COMMENTS
Langton's Ant living on triangular tiling (or, equivalently, hexagonal grid) follows the rules similar to those of the ordinary Langton's ant. On a white cell, turn 60 degrees right, flip the color of the cell, then move forward one unit. On a black cell, turn 60 degrees left, flip the color of the cell, then move forward one unit.
On these iterations pattern becomes symmetric. Orientation of the ant on these iterations is always the same.
Empirically, a(n) ~ c*n^1.207.
Color of the cell Langton's Ant living on triangular tiling touches on its n-th step before the color is changed by the ant; 0=white, 1=black.
+10
5
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0
COMMENTS
The ant starts from a completely white tiling.
There are never more than six 0's or 1's in a row because six identical turns will circle back into the same cell, which will have since changed.
Langton's ant on a truncated hexagonal tiling: number of black cells after n moves of the ant when starting on a dodecagon and looking towards an edge where the dodecagon meets a triangle.
+10
5
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 9, 10, 11, 12, 13, 12, 11, 12, 13, 14, 15, 16, 15, 14, 15, 16, 17, 18, 19, 20, 19, 18, 17, 16, 17, 18, 19, 20, 21, 22, 23, 22, 23, 24, 25, 26, 27, 26, 27, 28, 29, 30, 31, 30, 29, 30, 31, 32, 33, 34, 33, 32, 33, 32
COMMENTS
On a white dodecagon, turn 30 degrees right, flip the color of the tile, then move forward one unit.
On a black dodecagon, turn 30 degrees left, flip the color of the tile, then move forward one unit.
On a white triangle, turn 60 degrees right, flip the color of the tile, then move forward one unit.
On a black triangle, turn 60 degrees left, flip the color of the tile, then move forward one unit.
EXAMPLE
See illustrations in Fröhlich, 2019.
Let the starting square of Langton's ant have coordinates (0, 0), with the ant looking in negative x-direction. a(n) is the y-coordinate of the ant after n moves.
+10
4
0, 1, 1, 0, 0, -1, -1, 0, 0, -1, -1, -2, -2, -1, -1, 0, 0, -1, -1, -2, -2, -3, -3, -2, -2, -1, -1, -2, -2, -1, -1, -2, -2, -1, -1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 1, 1, 0, 0, 1, 1, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 1, 1, 2
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