Babylonian spiral: Difference between revisions
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(21, 17) (26, 13) (29, 7) (29, 0) (28, -7) (24, -13) (17, -15) (10, -12) (4, -7) (4, 1) |
(21, 17) (26, 13) (29, 7) (29, 0) (28, -7) (24, -13) (17, -15) (10, -12) (4, -7) (4, 1) |
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(5, 9) (7, 17) (13, 23) (21, 26) (28, 21) (32, 13) (32, 4) (31, -5) (29, -14) (24, -22) |
(5, 9) (7, 17) (13, 23) (21, 26) (28, 21) (32, 13) (32, 4) (31, -5) (29, -14) (24, -22) |
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</pre> |
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=={{header|Wren}}== |
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{{trans|Python}} |
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{{libheader|Wren-trait}} |
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{{libheader|Wren-seq}} |
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{{libheader|Wren-fmt}} |
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<lang ecmascript>import "./trait" for Indexed |
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import "./seq" for Lst |
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import "./fmt" for Fmt |
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import "io" for File |
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// Python modulo operator (not same as Wren's) |
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var pmod = Fn.new { |x, y| ((x % y) + y) % y } |
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var squareCache = [] |
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""" |
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Get the points for each step along a Babylonian spiral of `nsteps` steps. |
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Origin is at (0, 0) with first step one unit in the positive direction along |
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the vertical (y) axis. The other points are selected to have integer x and y |
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coordinates, progressively concatenating the next longest vector with integer |
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x and y coordinates on the grid. The direction change of the new vector is |
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chosen to be nonzero and clockwise in a direction that minimizes the change |
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in direction from the previous vector. |
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See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347 |
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""" |
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var babylonianSpiral = Fn.new { |nsteps| |
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for (x in 0...nsteps) squareCache.add(x*x) |
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var dxys = [[0, 0], [0, 1]] |
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var dsq = 1 |
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for (i in 0...nsteps) { |
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var x = dxys[-1][0] |
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var y = dxys[-1][1] |
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var theta = y.atan(x) |
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var candidates = [] |
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while (candidates.isEmpty) { |
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dsq = dsq + 1 |
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for (se in Indexed.new(squareCache)) { |
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var i = se.index |
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var a = se.value |
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if (a > (dsq/2).floor) break |
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for (j in dsq.sqrt.floor + 1...0) { |
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var b = squareCache[j] |
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if ((a + b) < dsq) break |
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if ((a + b) == dsq) { |
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candidates.addAll([ [i, j], [-i, j], [i, -j], [-i, -j], |
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[j, i], [-j, i], [j, -i], [-j, -i] ]) |
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} |
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} |
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} |
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} |
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var comparer = Fn.new { |d| pmod.call(theta - d[1].atan(d[0]), Num.tau) } |
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candidates.sort { |a, b| comparer.call(a) < comparer.call(b) } |
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dxys.add(candidates[0]) |
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} |
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var accs = [] |
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var sumx = 0 |
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var sumy = 0 |
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for (dxy in dxys) { |
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sumx = sumx + dxy[0] |
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sumy = sumy + dxy[1] |
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accs.add([sumx, sumy]) |
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} |
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return accs |
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} |
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// find first 10,000 points |
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var points10000 = babylonianSpiral.call(9998) // first two added automatically |
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// print first 40 to terminal |
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System.print("The first 40 Babylonian spiral points are:") |
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for (chunk in Lst.chunks(points10000[0..39], 10)) Fmt.print("$-10s", chunk) |
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// create .csv file for all 10,000 points for display by an external plotter |
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File.create("babylonian_spiral.csv") { |file| |
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for (p in points10000) { |
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file.writeBytes("%(p[0]), %(p[1])\n") |
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} |
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} </lang> |
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{{out}} |
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<pre> |
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The first 40 Babylonian spiral points are: |
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[0, 0] [0, 1] [1, 2] [3, 2] [5, 1] [7, -1] [7, -4] [6, -7] [4, -10] [0, -10] |
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[-4, -9] [-7, -6] [-9, -2] [-9, 3] [-8, 8] [-6, 13] [-2, 17] [3, 20] [9, 20] [15, 19] |
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[21, 17] [26, 13] [29, 7] [29, 0] [28, -7] [24, -13] [17, -15] [10, -12] [4, -7] [4, 1] |
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[5, 9] [7, 17] [13, 23] [21, 26] [28, 21] [32, 13] [32, 4] [31, -5] [29, -14] [24, -22] |
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</pre> |
</pre> |
Revision as of 10:38, 15 May 2022
The Babylonian spiral is a sequence of points in the plane that are created so as to continuously minimally increase in vector length and minimally bend in vector direction, while always moving from point to point on strictly integral coordinates. Of the two criteria of length and angle, the length has priority.
- Examples
P(0) and P(1) are defined to be at (x = 0, y = 0) and (x = 0, y = 1). The first vector is from P(1) to P(2). It is vertical and of length 1. Note that the square of that length is 1.
Next in sequence is the vector from P(2) to P(3). This should be the smallest distance to a point with integral (x, y) which is longer than the last vector (that is, 1). It should also bend clockwise more than zero radians, but otherwise to the least degree.
The point chosen for P(3) that fits criteria is (x = 1, y = 2). Note the length of the vector from P(2) to P(3) is √2, which squared is 2. The lengths of the vectors thus determined can be given by a sorted list of possible sums of two integer squares, including 0 as a square.
- Task
Find and show the first 40 (x, y) coordinates of the Babylonian spiral.
- Stretch task
Show in your program how to calculate and plot the first 10000 points in the sequence. Your result should look similar to the page at https://oeis.org/plot2a?name1=A297346&name2=A297347&tform1=untransformed&tform2=untransformed&shift=0&radiop1=xy&drawlines=true".
- See also
Python
<lang python>""" Rosetta Code task rosettacode.org/wiki/Babylonian_spiral """
from itertools import accumulate from math import isqrt, atan2, tau from matplotlib.pyplot import axis, plot, show
square_cache = []
def babylonian_spiral(nsteps):
""" Get the points for each step along a Babylonia spiral of `nsteps` steps. Origin is at (0, 0) with first step one unit in the positive direction along the vertical (y) axis. The other points are selected to have integer x and y coordinates, progressively concatenating the next longest vector with integer x and y coordinates on the grid. The direction change of the new vector is chosen to be nonzero and clockwise in a direction that minimizes the change in direction from the previous vector. See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347 """ if len(square_cache) <= nsteps: square_cache.extend([x * x for x in range(len(square_cache), nsteps)]) xydeltas = [(0, 0), (0, 1)] δsquared = 1 for _ in range(nsteps): x, y = xydeltas[-1] θ = atan2(y, x) candidates = [] while not candidates: δsquared += 1 for i, a in enumerate(square_cache): if a > δsquared // 2: break for j in range(isqrt(δsquared) + 1, 0, -1): b = square_cache[j] if a + b < δsquared: break if a + b == δsquared: candidates.extend([(i, j), (-i, j), (i, -j), (-i, -j), (j, i), (-j, i), (j, -i), (-j, -i)])
p = min(candidates, key=lambda d: (θ - atan2(d[1], d[0])) % tau) xydeltas.append(p)
return list(accumulate(xydeltas, lambda a, b: (a[0] + b[0], a[1] + b[1])))
points10000 = babylonian_spiral(10000)
print("The first 40 Babylonian spiral points are:")
for i, p in enumerate(points10000[:40]):
print(str(p).ljust(10), end = '\n' if (i + 1) % 10 == 0 else )
- stretch portion of task
plot(*zip(*points10000)) axis('scaled') show()
</lang>
- Output:
The first 40 Babylonian spiral points are: (0, 0) (0, 1) (1, 2) (3, 2) (5, 1) (7, -1) (7, -4) (6, -7) (4, -10) (0, -10) (-4, -9) (-7, -6) (-9, -2) (-9, 3) (-8, 8) (-6, 13) (-2, 17) (3, 20) (9, 20) (15, 19) (21, 17) (26, 13) (29, 7) (29, 0) (28, -7) (24, -13) (17, -15) (10, -12) (4, -7) (4, 1) (5, 9) (7, 17) (13, 23) (21, 26) (28, 21) (32, 13) (32, 4) (31, -5) (29, -14) (24, -22)
Wren
<lang ecmascript>import "./trait" for Indexed import "./seq" for Lst import "./fmt" for Fmt import "io" for File
// Python modulo operator (not same as Wren's) var pmod = Fn.new { |x, y| ((x % y) + y) % y }
var squareCache = []
"""
Get the points for each step along a Babylonian spiral of `nsteps` steps. Origin is at (0, 0) with first step one unit in the positive direction along the vertical (y) axis. The other points are selected to have integer x and y coordinates, progressively concatenating the next longest vector with integer x and y coordinates on the grid. The direction change of the new vector is chosen to be nonzero and clockwise in a direction that minimizes the change in direction from the previous vector.
See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347
""" var babylonianSpiral = Fn.new { |nsteps|
for (x in 0...nsteps) squareCache.add(x*x) var dxys = [[0, 0], [0, 1]] var dsq = 1 for (i in 0...nsteps) { var x = dxys[-1][0] var y = dxys[-1][1] var theta = y.atan(x) var candidates = [] while (candidates.isEmpty) { dsq = dsq + 1 for (se in Indexed.new(squareCache)) { var i = se.index var a = se.value if (a > (dsq/2).floor) break for (j in dsq.sqrt.floor + 1...0) { var b = squareCache[j] if ((a + b) < dsq) break if ((a + b) == dsq) { candidates.addAll([ [i, j], [-i, j], [i, -j], [-i, -j], [j, i], [-j, i], [j, -i], [-j, -i] ]) } } } } var comparer = Fn.new { |d| pmod.call(theta - d[1].atan(d[0]), Num.tau) } candidates.sort { |a, b| comparer.call(a) < comparer.call(b) } dxys.add(candidates[0]) }
var accs = [] var sumx = 0 var sumy = 0 for (dxy in dxys) { sumx = sumx + dxy[0] sumy = sumy + dxy[1] accs.add([sumx, sumy]) } return accs
}
// find first 10,000 points var points10000 = babylonianSpiral.call(9998) // first two added automatically
// print first 40 to terminal System.print("The first 40 Babylonian spiral points are:") for (chunk in Lst.chunks(points10000[0..39], 10)) Fmt.print("$-10s", chunk)
// create .csv file for all 10,000 points for display by an external plotter File.create("babylonian_spiral.csv") { |file|
for (p in points10000) { file.writeBytes("%(p[0]), %(p[1])\n") }
} </lang>
- Output:
The first 40 Babylonian spiral points are: [0, 0] [0, 1] [1, 2] [3, 2] [5, 1] [7, -1] [7, -4] [6, -7] [4, -10] [0, -10] [-4, -9] [-7, -6] [-9, -2] [-9, 3] [-8, 8] [-6, 13] [-2, 17] [3, 20] [9, 20] [15, 19] [21, 17] [26, 13] [29, 7] [29, 0] [28, -7] [24, -13] [17, -15] [10, -12] [4, -7] [4, 1] [5, 9] [7, 17] [13, 23] [21, 26] [28, 21] [32, 13] [32, 4] [31, -5] [29, -14] [24, -22]