Math::PlanePath::ComplexMinus -- twindragon and other complex number base i-r
use Math::PlanePath::ComplexMinus; my $path = Math::PlanePath::ComplexMinus->new (realpart=>1); my ($x, $y) = $path->n_to_xy (123);
This path traverses points by a complex number base i-r for given integer r. The default is base i-1 as per
Walter Penny, "A 'Binary' System for Complex Numbers", Journal of the ACM, volume 12, number 2, April 1965, pages 247-248.
When continued to a power-of-2 extent this is called the "twindragon" shape.
26 27 10 11 3 24 25 8 9 2 18 19 30 31 2 3 14 15 1 16 17 28 29 0 1 12 13 <- Y=0 22 23 6 7 58 59 42 43 -1 20 21 4 5 56 57 40 41 -2 50 51 62 63 34 35 46 47 -3 48 49 60 61 32 33 44 45 -4 54 55 38 39 -5 52 53 36 37 -6 ^ -5 -4 -3 -2 -1 X=0 1 2 3 4 5 6 7
A complex integer can be represented as a set of powers,
X+Yi = a[n]*b^n + ... + a[2]*b^2 + a[1]*b + a[0] base b=i-1 digits a[n] to a[0] each = 0 or 1 N = a[n]*2^n + ... + a[2]*2^2 + a[1]*2 + a[0]
N is the a[i] digits as bits and X,Y is the resulting complex number. It can be shown that this is a one-to-one mapping so every integer X,Y of the plane is visited once each.
The shape of points N=0 to N=2^level-1 repeats as N=2^level to N=2^(level+1)-1. For example N=0 to N=7 is repeated as N=8 to N=15, but starting at position X=2,Y=2 instead of the origin. That position 2,2 is because b^3 = 2+2i. There's no rotations or mirroring etc in this replication, just position offsets.
N=0 to N=7 N=8 to N=15 repeat shape 2 3 10 11 0 1 8 9 6 7 14 15 4 5 12 13
For b=i-1 each N=2^level point starts at X+Yi=(i-1)^level. The powering of that b means the start position rotates around by +135 degrees each time and outward by a radius factor sqrt(2) each time. So for example b^3 = 2+2i is followed by b^4 = -4, which is 135 degrees around and radius |b^3|=sqrt(8) becomes |b^4|=sqrt(16).
The realpart => $r
option gives a complex base b=i-r for a given integer r>=1. For example realpart => 2
is
20 21 22 23 24 4 15 16 17 18 19 3 10 11 12 13 14 2 5 6 7 8 9 1 45 46 47 48 49 0 1 2 3 4 <- Y=0 40 41 42 43 44 -1 35 36 37 38 39 -2 30 31 32 33 34 -3 70 71 72 73 74 25 26 27 28 29 -4 65 66 67 68 69 -5 60 61 62 63 64 -6 55 56 57 58 59 -7 50 51 52 53 54 -8 ^ -8 -7 -6 -5 -4 -3 -2 -1 X=0 1 2 3 4 5 6 7 8 9 10
N is broken into digits of base=norm=r*r+1, ie. digits 0 to r*r inclusive. This makes horizontal runs of r*r+1 many points, such as N=5 to N=9 etc above. In the default r=1 these runs are 2 long whereas for r=2 they're 2*2+1=5 long, or r=3 would be 3*3+1=10, etc.
The offset back for each run like N=5 shown is the r in i-r, then the next level is (i-r)^2 = (-2r*i + r^2-1) so N=25 begins at Y=-2*2=-4, X=2*2-1=3.
The successive replications tile the plane for any r, though the N values needed to rotate around and do so become large if norm=r*r+1 is large.
The i-1 twindragon is usually conceived as taking fractional N like 0.abcde in binary and giving fractional complex X+iY. The twindragon is then all the points of the complex plane reached by such fractional N. This set of points can be shown to be connected and to fill a certain radius around the origin.
The code here might be pressed into use for that to some finite number of bits by multiplying up to make an integer N
Nint = Nfrac * 256^k Xfrac = Xint / 16^k Yfrac = Yint / 16^k
256 is a good power because b^8=16 is a positive real and so there's no rotations to apply to the resulting X,Y, only a power-of-16 division (b^8)^k=16^k each. Using b^4=-4 for a multiplier 16^k and divisor (-4)^k would be almost as easy too, requiring just sign changes if k odd.
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::ComplexMinus->new ()
$path = Math::PlanePath::ComplexMinus->new (realpart => $r)
Create and return a new path object.
($x,$y) = $path->n_to_xy ($n)
Return the X,Y coordinates of point number $n
on the path. Points begin at 0 and if $n < 0
then the return is an empty list.
$n
should be an integer, it's unspecified yet what will be done for a fraction.
A given X,Y representing X+Yi can be turned into digits of N by successive complex divisions by i-r. Each digit of N is a real remainder 0 to r*r inclusive from that division.
The base formula above is
X+Yi = a[n]*b^n + ... + a[2]*b^2 + a[1]*b + a[0]
and we want the a[0]=digit to be a real 0 to r*r inclusive. Subtracting a[0] and dividing by b will give
(X+Yi - digit) / (i-r) = - (X-digit + Y*i) * (i+r) / norm = (Y - (X-digit)*r)/norm + i * - ((X-digit) + Y*r)/norm
which is
Xnew = Y - (X-digit)*r)/norm Ynew = -((X-digit) + Y*r)/norm
The a[0] digit must make both Xnew and Ynew parts integers. The easiest one to calculate from is the imaginary part, from which require
- ((X-digit) + Y*r) == 0 mod norm
so
digit = X + Y*r mod norm
This digit value makes the real part a multiple of norm too, as can be seen from
Xnew = Y - (X-digit)*r = Y - X*r - (X+Y*r)*r = Y - X*r - X*r + Y*r*r = Y*(r*r+1) = Y*norm
Notice Ynew is the quotient from (X+Y*r)/norm rounded downwards (towards negative infinity). Ie. in the division "X+Y*r mod norm" which calculates the digit, the quotient is Ynew and the remainder is the digit.
For base i-1, Penney shows the N on the X axis are
X axis N in hexadecimal uses only digits 0, 1, C, D = 0, 1, 12, 13, 16, 17, 28, 29, 192, 193, 204, 205, 208, ...
Those on the positive X axis have an odd number of digits and on the X negative axis an even number of digits.
To be on the X axis the imaginary parts of the base powers b^k must cancel out to leave just a real part. The powers repeat in an 8-long cycle
k b^k for b=i-1 0 +1 1 i -1 2 -2i +0 \ pair cancel 3 2i +2 / 4 -4 5 -4i +4 6 8i +0 \ pair cancel 7 -8i -8 /
The k=0 and k=4 bits are always reals and can always be included. Bits k=2 and k=3 have imaginary parts -2i and 2i which cancel out, so they can be included together. Similarly k=6 and k=7 with 8i and -8i. The two blocks k=0to3 and k=4to7 differ only in a negation so the bits can be reckoned in groups of 4, which is hexadecimal. Bit 1 is digit value 1 and bits 2,3 together are digit value 0xC, so adding one or both of those gives combinations are 0,1,0xC,0xD.
The high hex digit determines the sign, positive or negative, of the total real part. Bits k=0 or k=2,3 are positive. Bits k=4 or k=6,7 are negative, so
N for X>0 N for X<0 0x01.. 0x1_.. even number of hex 0,1,C,D following 0x0C.. 0xC_.. "_" digit any of 0,1,C,D 0x0D.. 0xD_..
which is equivalent to X>0 is an odd number of hex digits or X<0 is an even number. For example N=28=0x1C is at X=-2 since that N is X<0 form "0x1_".
The order of the values on the positive X axis is obtained by taking the digits in reverse order on alternate positions
0,1,C,D high digit D,C,1,0 0,1,C,D ... D,C,1,0 0,1,C,D low digit
For example in the following notice the first and third digit increases, but the middle digit decreases,
X=4to7 N=0x1D0,0x1D1,0x1DC,0x1DD X=8to11 N=0x1C0,0x1C1,0x1CC,0x1CD X=12to15 N=0x110,0x111,0x11C,0x11D X=16to19 N=0x100,0x101,0x10C,0x10D X=20to23 N=0xCD0,0xCD1,0xCDC,0xCDD
For the negative X axis it's the same if reading by increasing X, ie. upwards toward +infinity, or the opposite way around if reading decreasing X, ie. more negative downwards toward -infinity.
For base i-1 Penny also characterises the N values on the Y axis,
Y axis N in base-64 uses only at even digits 0, 3, 4, 7, 48, 51, 52, 55 at odd digit 0, 1, 12, 13, 16, 17, 28, 29 = 0,3,4,7,48,51,52,55,64,67,68,71,112,115,116,119, ...
Base-64 means taking N in 6-bit blocks. Digit positions are counted starting from the least significant digit as position 0 which is even. So the low digit can be only 0,3,4,etc, then the second digit only 0,1,12,etc, and so on.
This arises from (i-1)^6 = 8i which gives a repeating pattern of 6-bit blocks. The different patterns at odd and even positions are since i^2 = -1.
The length of the boundary of unit squares for the first norm^k many points, ie. N=0 to N=norm^k-1 inclusive, is calculated in
William J. Gilbert, "The Fractal Dimension of Sets Derived From Complex Bases", Canadian Math Bulletin, volume 29, number 4, 1986. http://www.math.uwaterloo.ca/~wgilbert/Research/GilbertFracDim.pdf
The boundary formula is a 3rd-order recurrence. For the twindragon case it is
for realpart=1 boundary[k] = boundary[k-1] + 2*boundary[k-3] = 4, 6, 10, 18, 30, 50, 86, 146, 246, 418, 710, ... 4 + 2*x + 4*x^2 generating function --------------- 1 - x - 2*x^3
The first three boundaries are as follows. Then the recurrence gives the next boundary[3] = 10+2*4 = 18.
k area boundary[k] --- ---- ----------- +---+ 0 2^k = 1 4 | 0 | +---+ +---+---+ 1 2^k = 2 6 | 0 1 | +---+---+ +---+---+ | 2 3 | 2 2^k = 4 10 +---+ +---+ | 0 1 | +---+---+
Gilbert calculates for any i-r by taking the boundary in three parts A,B,C and showing how in the next replication level those boundary parts transform into multiple copies of the preceding level parts. The replication is easier to visualize for a bigger "r" than for the twindragon because in bigger r it's clearer how the A, B and C parts differ. The length replications are
A -> A * (2*r-1) + C * 2*r B -> A * (r^2-2*r+2) + C * (r-1)^2 C -> B starting from A = 2*r B = 2 C = 2 - 2*r total boundary = A+B+C
For the twindragon realpart=1 these A,B,C are already in the form of a recurrence A->A+2*C, B->A, C->B, per the formula above. For other real parts a little matrix rearrangement turns the A,B,C parts into recurrence
boundary[k] = boundary[k-1] * (2*r - 1) + boundary[k-2] * (norm - 2*r) + boundary[k-3] * norm starting from boundary[0] = 4 # single square cell boundary[1] = 2*norm + 2 # oblong of norm many cells boundary[2] = 2*(norm-1)*(r+2) + 4
For example
for realpart=2 boundary[k] = 3*boundary[k-1] + 1*boundary[k-2] + 5*boundary[k-3] = 4, 12, 36, 140, 516, 1868, 6820, 24908, 90884, ... 4 - 4*x^2 generating function --------------------- 1 - 3*x - x^2 - 5*x^3
If calculating for large k values then the matrix form can be powered up rather than repeated additions. (As usual for all such linear recurrences.)
Entries in Sloane's Online Encyclopedia of Integer Sequences related to this path include
http://oeis.org/A066321 (etc)
realpart=1 (twindragon, the default) A066321 N on X axis, being the base i-1 positive reals A066323 N on X axis, in binary A066322 diffs (N at X=16k+4) - (N at X=16k+3) A003476 boundary length / 2 recurrence a(n) = a(n-1) + 2*a(n-3) A203175 boundary length, starting from 4 (believe its conjectured recurrence is true) A052537 boundary length part A, B or C, per Gilbert's paper
Math::PlanePath, Math::PlanePath::DragonCurve, Math::PlanePath::ComplexPlus
http://user42.tuxfamily.org/math-planepath/index.html
Copyright 2011, 2012, 2013, 2014 Kevin Ryde
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