NAME
Math::PlanePath::CCurve  Levy C curve
SYNOPSIS
use Math::PlanePath::CCurve;
my $path = Math::PlanePath::CCurve>new;
my ($x, $y) = $path>n_to_xy (123);
DESCRIPTION
This is an integer version of the Levy "C" curve.
11109,765 3
  
1312 8 43 2
 
1914,1817 2 1
   
2120 1516 01 < Y=0

22 1

25,2324 2

26 353433 3
  
27,3728,36 32 4
  
38 293031 5

39,4140 6

42 ... 7
 
4344 4948 6463 8
   
4546,5047 62 9
 
5152 56 6061 10
  
535455,575859 11
^
7 6 5 4 3 2 1 X=0 1
The initial segment N=0 to N=1 is repeated with a turn +90 degrees left to give N=1 to N=2. Then N=0to2 is repeated likewise turned +90 degrees and placed at N=2 to make N=2to4. And so on doubling each time.
43
 N=0to2
2 2 repeated
  as N=2to4
01 01 01 with turn +90
The 90 degree rotation is the same at each repetition, so the segment at N=2^k is always the initial N=0to1 turned +90 degrees. This means at N=1,2,4,8,16,etc the direction is always upwards.
The X,Y position can be written in complex numbers as a recurrence
with N = 2^k + r high bit 2^k and remainder r<2^k
C(N) = C(2^k) + i*C(r)
= (1+i)^k + i*C(r)
The effect is a change from base 2 to base 1+i, but with a further power of i on each term. Suppose the 1bits in N are at positions k0, k1, k2, etc (high to low), then
C(N) = b^k0 * i^0 N= 2^k0 + 2^(k1) + 2^(k2) + ... in binary
+ b^k1 * i^1 k0 > k1 > k2 > ...
+ b^k2 * i^2 base b=1+i
+ b^k3 * i^3
+ ...
Notice the i power is not the bit position k, but rather how many 1bits are above the position. This calculation is straightforward but the resulting structure of overlaps and internal shapes has many different parts.
Level Ranges 4^k
The X,Y extents of the path through to Nlevel=2^k can be expressed as a width and height measured relative to the endpoints.
** <+
  
** **  height h[k]
  
* N=4^k N=0 * <+
     below l[k]
*** *** <+
^^ ^^
width 2^k width
w[k] w[k] Extents to N=4^k
<>
total width = 2^k + 2*w[k]
N=4^k is on either the X or Y axis and for the extents here it's taken rotated as necessary to be horizontal. k=2 N=4^2=16 shown above is already horizontal. The next level k=3 N=64=4^3 would be rotated 90 degrees to be horizontal.
The width w[k] is measured from the N=0 and N=4^k endpoints. It doesn't include the 2^k length between those endpoints. The two ends are symmetric so the extent is the same at each end.
h[k] = 2^k  1 0,1,3,7,15,31,etc
w[k] = / 0 for k=0
\ 2^(k1)  1 for k>=1 0,0,1,3,7,15,etc
l[k] = / 0 for k<=1
\ 2^(k2)  1 for k>=2 0,0,0,1,3,7,etc
The initial N=0 to N=64 shown above is k=3. h[3]=7 is the X=7 horizontal. l[3]=1 is the X=1 horizontal. w[3]=3 is the vertical Y=3, and also Y=11 which is 3 below the endpoint N=64 at Y=8.
Expressed as a fraction of the 2^k distance between the endpoints the extents approach total 2 wide by 1.25 high,
** <+
   1
** **  total
   height
* N=4^k N=0 * <+ > 1+1/4
     1/4
*** *** <+
^^ ^^
1/2 1 1/2 total width > 2
The extent formulas can be found by considering the selfsimilar blocks. The initial k=0 is a single line segment and all its extents are 0.
h[0] = 0
N=1  N=0
l[0] = 0
w[0] = 0
Thereafter the replication overlap as
++++
   
++   ++
  D   C   B   <+
 ++++   2^(k1)
     previous
     level ends
 E   A  <+
++ ++
^^
2^k this level ends
w[k] = max (h[k1], w[k1]) # right of A,B
h[k] = 2^(k1) + max (h[k1], w[k1]) # above B,C,D
l[k] = max w[k1], l[k1]2^(k1) # below A,E
Since h[k]=2^(k1)+w[k] have h[k] > w[k] for k>=1 and with the initial h[0]=w[k]=0 have h[k]>=w[k] always. So the max of those two is h.
h[k] = 2^(k1) + h[k1] giving h[k] = 2^k1 for k>=1
w[k] = h[k1] giving w[k] = 2^(k1)1 for k>=1
The max for l[k] is always w[k1] as l[k] is never big enough that the parts BC and CD can extend down past their 2^(k1) vertical position. (l[0]=w[0]=0 and thereafter by induction l[k]<=w[k].)
l[k] = w[k1] giving l[k] = 2^(k2)1 for k>=2
Repeated Points
The curve crosses itself and can repeat X,Y positions up to 4 times. The first doubled, tripled and quadrupled points are
visits X,Y N
  
2 2, 3 7, 9
3 18, 7 189, 279, 281
4 32, 55 1727, 1813, 2283, 2369
Each line segment between integer points is traversed at most 2 times, once forward and once backward. There's 4 lines reaching each integer point and this line traversal means the points are visited at most 4 times.
As per "Direction" below the direction of the curve is given by the count of 1bits in N. Since no line is repeated each of the N values at a given X,Y have a different count1bits mod 4. For example N=7 is 3 1bits and N=9 is 2 1bits. The full counts need not be consecutive, as for example N=1727 is 9 1bits and N=2369 is 4 1bits.
The maximum of 2 line segment traversals can be seen from the way the curve replicates. Suppose the entire plane had all line segments traversed forward and backward.
v  v 
 < <
[0,1] [1,1] [X,Y] = integer points
> >  each edge traversed
 ^  ^ forward and backward
   
   
v  v 
 < <
[0,0] [1,0]
> > 
 ^  ^
Then when each line segment expands on the right the result is the same pattern of traversals  viewed rotated by 45degrees and scaled by factor sqrt(2).
\ v / v \ v / v
[0,1] [1,1]
/ / ^ \ ^ / ^ \
/ / \ \ / / \ \
\ \ / /
\ v / v
[1/2,1/2]
^ / ^ \
/ / \ \
\ \ / / \ \ / /
\ v / v \ v / v
[0,0] 1,0
^ / ^ \ ^ / ^ \
The curve is a subset of this pattern. It begins as a single line segment which has this pattern and thereafter the pattern preserves itself. Hence at most 2 segment traversals in the curve.
Tiling
The segment traversal argument above can also be made by taking the line segments as triangles which are a quarter of a unit square with peak pointing to the right of the traversal direction.
to *
^\
 \
 \ triangle peak
 /
 /
/ quarter of a unit square
from *
These triangles in the two directions tile the plane. On expansion each splits into 2 halves in new positions. Those parts don't overlap and the plane is still tiled. See for example
Larry Riddle http://ecademy.agnesscott.edu/~lriddle/ifs/levy/levy.htm http://ecademy.agnesscott.edu/~lriddle/ifs/levy/tiling.htm
For the integer version of the curve this kind of tiling can be used to combine copies of the curve so that each every point is visited precisely 4 times. The h[k], w[k] and l[k] extents above are less than the 2^k endpoint length, so a square of side 2^k can be fully tiled with copies of the curve at each corner,
 ^  ^
    24 copies of the curve
    to visit all points of the
v  v  inside square ABCD
< < < precisely 4 times each
A B
> > > each part points
 ^  ^ N=0 to N=4^k1
    rotated and shifted
    suitably
v  v 
< < <
C D
 > >
 ^  ^
   
   
v  v 
The four innermost copies of the curve cover most of the inside square, but the other copies surrounding them loop into the square and fill in the remainder to make 4 visits at every point.
It's interesting to note that a set of 8 curves at the origin only covers the axes with 4fold visits,
 ^ 8 arms at the origin
  cover only X,Y axes
v  with 4visits
< <
0,0 away from the axes
 > some points < 4 visits
 ^
 
v 
This means that if the path had some sort of "arms" of multiple curves extending from the origin then it would visit all points on the axes X=0 Y=0 a full 4 times, but off the axes there would be points without full 4 visits.
FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for the behaviour common to all path classes.
$path = Math::PlanePath::CCurve>new ()

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.Fractional positions give an X,Y position along a straight line between the integer positions.
$n = $path>xy_to_n ($x,$y)

Return the point number for coordinates
$x,$y
. If there's nothing at$x,$y
then returnundef
. If$x,$y
is visited more than once then return the smallest$n
which visits it. @n_list = $path>xy_to_n_list ($x,$y)

Return a list of N point numbers at coordinates
$x,$y
. If there's nothing at$x,$y
then return an empty list.A given
$x,$y
is visited at most 4 times so the returned list is at most 4 values. $n = $path>n_start()

Return 0, the first N in the path.
Level Methods
FORMULAS
Direction
The direction or net turn of the curve is the count of 1 bits in N,
direction = count_1_bits(N) * 90degrees
For example N=11 is binary 1011 has three 1 bits, so direction 3*90=270 degrees, ie. to the south.
This bit count is because at each powerof2 position the curve is a copy of the lower bits but turned +90 degrees, so +90 for each 1bit.
For powersof2 N=2,4,8,16, etc, there's only a single 1bit so the direction is always +90 degrees there, ie. always upwards.
Turn
At each point N the curve can turn in any direction: left, right, straight, or 180 degrees back. The turn is given by the number of low 0bits of N,
turn right = (count_low_0_bits(N)  1) * 90degrees
For example N=8 is binary 0b100 which is 2 low 0bits for turn=(21)*90=90 degrees to the right.
When N is odd there's no low zero bits and the turn is always (01)*90=90 to the right, so every second turn is 90 degrees to the left.
Next Turn
The turn at the point following N, ie. at N+1, can be calculated by counting the low 1bits of N,
next turn right = (count_low_1_bits(N)  1) * 90degrees
For example N=11 is binary 0b1011 which is 2 low one bits for nextturn=(21)*90=90 degrees to the right at the following point, ie. at N=12.
This works simply because low 1bits like ..0111 increment to low 0bits ..1000 to become N+1. The low 1bits at N are thus the low 0bits at N+1.
N to dX,dY
n_to_dxdy()
is implemented using the direction described above. For integer N the count mod 4 gives the direction for dX,dY.
dir = count_1_bits(N) mod 4
dx = dir_to_dx[dir] # table 0 to 3
dy = dir_to_dy[dir]
For fractional N the direction at int(N)+1 can be obtained from the direction at int(N) and the turn at int(N)+1, which is the low 1bits of N per "Next Turn" above. Those two directions can then be combined as described in "N to dX,dY  Fractional" in Math::PlanePath.
# apply turn to make direction at Nint+1
turn = count_low_1_bits(N)  1 # N integer part
dir = (dir  turn) mod 4 # direction at N+1
# adjust dx,dy by fractional amount in this direction
dx += Nfrac * (dir_to_dx[dir]  dx)
dy += Nfrac * (dir_to_dy[dir]  dy)
A small optimization can be made by working the "1" of the turn formula into a +90 degree rotation of the dir_to_dx[]
and dir_to_dy[]
parts by swap and sign change,
turn_plus_1 = count_low_1_bits(N) # on N integer part
dir = (dir  turn_plus_1) mod 4 # direction1 at N+1
# adjustment including extra +90 degrees on dir
dx = $n*(dir_to_dy[dir] + dx)
dy += $n*(dir_to_dx[dir]  dy)
X,Y to N
The N values at a given X,Y can be found by taking terms low to high from the complex number formula (the same as given above)
X+iY = b^k N = 2^k + 2^(k1) + 2^(k2) + ... in binary
+ b^k1 * i base b=1+i
+ b^k2 * i^2
+ ...
If the lowest term is b^0 then X+iY has X+Y odd. If the lowest term is not b^0 but instead some power b^n then X+iY has X+Y even. This is because a multiple of b=1+i,
X+iY = (x+iy)*(1+i)
= (xy) + (x+y)i
so X=xy Y=x+y
sum X+Y = 2x is even if X+iY a multiple of 1+i
So the lowest bit of N is found by
bit = (X+Y) mod 2
If bit=1 then a power i^p is to be subtracted from X+iY. p is how many 1bits are above that point, and this is not yet known. It represents a direction to move X,Y to put it on an even position. It's also the direction of the step N2^l to N, where 2^l is the lowest 1bit of N.
The reduction should be attempted with p commencing as each of the four possible directions N,S,E,W. Some or all will lead to an N. For quadrupled points (such as X=32, Y=55 described above) all four will lead to an N.
for p 0 to 3
dX,dY = i^p # directions [1,0] [0,1] [1,0] [0,1]
loop until X,Y = [0,0] or [1,0] or [1,0] or [0,1] or [0,1]
{
bit = X+Y mod 2 # bits of N from low to high
if bit == 1 {
X = dX # move to "even" X+Y == 0 mod 2
Y = dY
(dX,dY) = (dY,dX) # rotate 90 as for p1
}
X,Y = (X+Y)/2, (YX)/2 # divide (X+iY)/(1+i)
}
if not (dX=1 and dY=0)
wrong final direction, try next p
if X=dX and Y=dY
further high 1bit for N
found an N
if X=0 and Y=0
found an N
The "loop until" ends at one of the five points
0,1

1,0  0,0  1,0

0,1
It's not possible to wait for X=0,Y=0 to be reached because some dX,dY directions will step infinitely among the four nonzeros. Only the case X=dX,Y=dY is sure to reach 0,0.
The successive p decrements which rotate dX,dY by 90 degrees must end at p == 0 mod 4 for highest term in the X+iY formula having i^0=1. This means must end dX=1,dY=0 East. If this doesn't happen then there is no N for that p direction.
The number of 1bits in N is == p mod 4. So the order the N values are obtained follows the order the p directions are attempted. In general the N values will not be smallest to biggest N so a little sort is necessary if that's desired.
It can be seen that sum X+Y is used for the bit calculation and then again in the divide by 1+i. It's convenient to write the whole loop in terms of sum S=X+Y and difference D=YX.
for dS = +1 or 1 # four directions
for dD = +1 or 1 #
S = X+Y
D = YX
loop until 1 <= S <= 1 and 1 <= D <= 1 {
bit = S mod 2 # bits of N from low to high
if bit == 1 {
S = dS # move to "even" S+D == 0 mod 2
D = dD
(dS,dD) = (dD,dS) # rotate 90
}
(S,D) = (S+D)/2, (DS)/2 # divide (S+iD)/(1+i)
}
if not (dS=1 and dD=1)
wrong final direction, try next dS,dD direction
if S=dS and D=dD
further high 1bit for N
found an N
if S=0 and D=0
found an N
The effect of S=X+Y, D=YD is to rotate by 45 degrees and use every second point of the plane.
D= 2 X=0,Y=2 . rotate 45
D= 1 X=0,Y=1 . X=1,Y=2 .
D= 0 X=0,Y=0 . X=1,Y=1 . X=2,Y=2
D=1 X=1,Y=0 . X=2,Y=1 .
D=2 X=2,Y=0 .
S=0 S=1 S=2 S=3 S=4
The final five points described above are then in a 3x3 block at the origin. The four inbetween points S=0,D=1 etc don't occur so range tests 1<=S<=1 and 1<=D<=1 can be used.
S=1,D=1 . S=1,D=1
. S=0,D=0 .
S=1,D=1 . S=1,D=1
Segments by Direction
In a level N=0 to N=2^k1 inclusive, the number of segments in each direction 0=East, 1=North, 2=West, 3=South are given by
k=0 for k >= 1
 
M0[k] = 1, 2^(k2) + d(k+2)*2^(h1)
M1[k] = 0, 2^(k2) + d(k+0)*2^(h1)
M2[k] = 0, 2^(k2) + d(k2)*2^(h1)
M3[k] = 0, 2^(k2) + d(k4)*2^(h1)
where h = floor(k/2)
and d(m) = 0 1 1 1 0 1 1 1
for m == 0 to 7 mod 8
M0[k] = 1, 1, 1, 1, 2, 6, 16, 36, 72, 136, 256, ...
M1[k] = 0, 1, 2, 3, 4, 6, 12, 28, 64, 136, 272, ...
M2[k] = 0, 0, 1, 3, 6, 10, 16, 28, 56, 120, 256, ...
M3[k] = 0, 0, 0, 1, 4, 10, 20, 36, 64, 120, 240, ...
d(n) is a factor +1, 1 or 0 according to n mod 8. Each M goes as a power 2^(k2), so roughly 1/4 each, but a half power 2^(h1) possibly added or subtracted in a k mod 8 pattern. In binary this is a 2^(k2) high 1bit with another 1bit in the middle added or subtracted.
The total is 2^k since there are a total 2^k points from N=0 to 2^k1 inclusive.
M0[k] + M1[k] + M2[k] + M3[k] = 2^k
It can be seen that the d(n) parts sum to 0 so the 2^(h1) parts cancel out leaving 4*2^(k2) = 2^k.
d(0) + d(2) + d(4) + d(6) = 0
d(1) + d(3) + d(5) + d(7) = 0
The counts can be calculated in two ways. Firstly they satisfy mutual recurrences. Each adds the preceding rotated M.
M0[k+1] = M0[k] + M3[k] initially M0[0] = 1 (N=0 to N=1)
M1[k+1] = M1[k] + M0[k] M1[0] = 0
M2[k+1] = M2[k] + M1[k] M2[0] = 0
M3[k+1] = M3[k] + M2[k] M3[0] = 0
Geometrically this can be seen from the way each level extends by a copy of the previous level rotated +90,
765 Easts in N=0 to 8
  = Easts in N=0 to 4
8 43 + Wests in N=0 to 4
 since N=4 to N=8 is
2 the N=0 to N=4 rotated +90

01
For the bits in N, level k+1 introduces a new bit either 0 or 1. In M0[k+1] the a 0bit is count M0[k] the same direction, and when a 1bit is M3[k] since one less bit mod 4. Similarly the other counts.
Some substitutions give 3rd order recurrences
for k >= 4
M0[k] = 4*M0[k1]  6*M0[k2] + 4*M0[k3] initial 1,1,1,1
M1[k] = 4*M1[k1]  6*M1[k2] + 4*M1[k3] initial 0,1,2,3
M2[k] = 4*M2[k1]  6*M2[k2] + 4*M2[k3] initial 0,0,1,3
M3[k] = 4*M3[k1]  6*M3[k2] + 4*M3[k3] initial 0,0,0,1
The characteristic polynomial of these recurrences is
x^3  4x^2 + 6x  4
= (x2) * (x  (1i)) * (x  (1+i))
So explicit formulas can be written in powers of the roots 2, 1i and 1+i,
M0[k] = ( 2^k + (1i)^k + (1+i)^k )/4 for k>=1
M1[k] = ( 2^k + i*(1i)^k  i*(1+i)^k )/4
M2[k] = ( 2^k  (1i)^k  (1+i)^k )/4
M3[k] = ( 2^k  i*(1i)^k + i*(1+i)^k )/4
The complex numbers 1i and 1+i are 45 degree lines clockwise and anticlockwise respectively. The powers turn them in opposite directions so the imaginary parts always cancel out. The remaining real parts can be had by a half power h=floor(k/2) which is the magnitude abs(1i)=sqrt(2) projected onto the real axis. The sign selector d(n) above is whether the positive or negative part of the real axis, or zero when at the origin.
The second way to calculate is the combinatorial interpretation that per "Direction" above the direction is count_1_bits(N) mod 4 so East segments are all N values with count_1_bits(N) == 0 mod 4, ie. N with 0, 4, 8, etc many 1bits. The number of ways to have those bit counts within total k bits is k choose 0, 4, 8 etc.
M0[k] = /k\ + /k\ + ... + / k\ m = floor(k/4)
\0/ \4/ \4m/
M1[k] = /k\ + /k\ + ... + / k \ m = floor((k1)/4)
\1/ \5/ \4m+1/
M2[k] = /k\ + /k\ + ... + / k \ m = floor((k2)/4)
\2/ \6/ \4m+2/
M3[k] = /k\ + /k\ + ... + / k \ m = floor((k3)/4)
\3/ \7/ \4m+3/
The power forms above are cases of the identity by Ramus for sums of binomial coefficients in arithmetic progression like this. (See Knuth volume 1 section 1.2.6 exercise 30 for a form with cosines resulting from w=i+1 as 8th roots of unity.)
The total M0+M1+M2+M3=2^k is the total binomials across a row of Pascal's triangle.
/k\ + /k\ + ... + /k\ = 2^k
\0/ \1/ \k/
It's interesting to note the M counts here are the same in the dragon curve (Math::PlanePath::DragonCurve). The shapes of the curves are different since the segments are in a different order, but the total puts points N=2^k at the same X,Y position.
Right Boundary
The length of the rightside boundary of the curve, which is the outside of the "C", from N=0 to N=2^k is
R[k] = / 7*2^h  2k  6 if k even
\ 10*2^h  2k  6 if k odd
where h = floor(k/2)
= 1, 2, 4, 8, 14, 24, 38, 60, 90, 136, 198, 292, 418, ...
R[k] = (7/2 + 5/2 * sqrt(2)) * ( sqrt(2))^k
+ (7/2  5/2 * sqrt(2)) * (sqrt(2))^k
 2*k  6
R[k] = 2*R[k1] + R[k2]  4*R[k3] + 2*R[k4]
The length doubles until R[4]=14 which is points N=0 to N=2^4=16. At k=4 the points N=7,8,9 have turned inward and closed off some of the outside of the curve so the boundary less than 2x.
11109,765 right boundary
   around "outside"
1312 8 43 N=0 to N=2^4=16
 
14 2 R[4]=14
 
1516 01
The floor(k/2) and odd/even cases are eliminated by the +/sqrt(2) powering shown. Those powers are also per the characteristic equation of the recurrence,
x^4  2*X^3  x^2 + 4*x  2
= (x  1)^2 * (x + sqrt(2)) * (x  sqrt(2))
roots 1, sqrt(2), sqrt(2)
The right boundary comprises runs of straight lines and zigzags. When it expands the straight lines become zigzags and the zigzags become straight lines. The straight lines all point "forward", which is anticlockwise.
c * a
/ ^ / ^ / ^
=> v \ v \ v \
D<C<B<A D C B A
 ^ / ^
v  v \
straight S=3 zigzag Z[k+1] = 2S[k]2 = 4
The count Z here is both sides of each "V" shape from points "a" through to "c". So Z counts the boundary length (rather than the number of "V"s). Each S becomes an upward peak. The first and last side of those peaks become part of the following "straight" section (at A and D), hence Z[k+1]=2*S[k]2.
The zigzags all point "forward" too. When they expand they close off the V shape and become 2 straight lines for each V, which means 1 straight line for each Z side. The segment immediately before and after contribute a segment to the resulting straight run too, hence S[k+1]=Z[k]+2.
C B A *<C<*<B<*<A<*
/ ^ / ^ / ^    
v \ v \ v \ =>    
* * * * <* * * *<
/ ^
v \
zigzag Z=4 segments straight S[k+1] = Z[k]+2 = 6
The initial N=0 to N=1 is a single straight segment S[0]=1 and from there the runs grow. N=1 to N=3 is a straight section S[1]=2. Z[0]=0 represents an empty zigzag at N=1. Z[1] is the first nonempty at N=3 to N=5.
h S[h] Z[h] Z[h] = 2*S[h]2
   S[h+1] = Z[h]+2
0 1 0
1 2 2 S[h+1] = 2*S[h]2+2 = 2*S[h]
2 4 6 so
3 8 14 S[h] = 2^h
4 16 30 Z[h] = 2*2^h2
5 32 62
5 64 126
The curve N=0 to N=2^k is symmetric at each end and is made up of runs S[0], Z[0], S[1], Z[1], etc, of straight and zigzag alternately at each end. When k is even there's a single copy of a middle S[k/2]. When k is odd there's a single middle Z[(k1)/2] (with an S[(k1)/2] before and after). So
/ i=h1 \ # where h = floor(k/2)
R[k] = 2 *  sum S[i]+Z[i] 
\ i=0 /
+ S[h]
+ / S[h]+Z[h] if k odd
\ 0 if k even
= 2*( 1+2+4+...+2^(h1) # S[0] to S[h1]
+ 2+4+8+...+2^h  2*h) # Z[0] to Z[h1]
+ 2^h # S[h]
+ if k odd (2^h + 2*2^h  2) # possible S[h]+Z[h]
= 2*(2^h1 + 2*2^h2  2h) + 2^h + (k odd 3*2^h  2)
= 7*2^h  4h6 + (if k odd then + 3*2^h  2)
= 7*2^h  2k6 + (if k odd then + 3*2^h)
Convex Hull Boundary
A convex hull is the smallest convex polygon which contains a given set of points. For the C curve the boundary length of the convex hull for points N=0 to N=2^k inclusive is
hull boundary[k]
/ 2 if k=0
 2+sqrt(2) if k=1
=  6 if k=2
 6*2^(h1) + (7*2^(h1)  4)*sqrt(2) if k odd >=3
\ 7*2^(h1) + (3*2^(h1)  4)*sqrt(2) if k even >=4
where h = floor(k/2)
k hull boundary
 
0 2 + 0 * sqrt(2) = 2
1 2 + 1 * sqrt(2) = 3.41
2 6 + 0 * sqrt(2) = 6
3 6 + 3 * sqrt(2) = 10.24
4 14 + 2 * sqrt(2) = 16.82
5 12 + 10 * sqrt(2) = 26.14
6 28 + 8 * sqrt(2) = 39.31
7 24 + 24 * sqrt(2) = 57.94
8 56 + 20 * sqrt(2) = 84.28
9 48 + 52 * sqrt(2) = 121.53
The integer part is the straight sides of the hull and the sqrt(2) part is the diagonal sides of the hull.
When k is even the hull has the following shape. The sides are as per the right boundary above but after Z[h2] the curl goes inwards and so parts beyond Z[h2] are not part of the hull. Each Z stairstep diagonal becomes a sqrt(2) length for the hull. Z counts both vertical and horizontal of each stairstep, hence sqrt(2)*Z/2 for the hull boundary.
S[h]
** * Z=2
Z[h1] / \ Z[h1]  \ diagonal
/ \  \ sqrt(2)*Z/2
* * ** = sqrt(2)
S[h1]   S[h1]
 
* *
Z[h2] \ / Z[h2]
* *
S[h] + Z[h2]Z[h1]
k even
hull boundary[k] = S[h] + 2*S[h1] + S[h+Z[h2]Z[h1]
+ sqrt(2)*(2*Z[h1] + 2*Z[h2])/2
When k is odd the shape is similar but Z[h] in the middle.
S[h]
**
Z[h1] / \ middle
* \ Z[h]
S[h1]  \
* *
\  S[h]
Z[h] 
+ 2*(S[h]S[h1]) *
\ / Z[h1]
**
S[h1]
k odd
hull boundary[k] = 2*S[h] + 2*S[h1]
+ sqrt(2)*(Z[h]/2 + 2*Z[h1]/2
+ Z[h]/2 + S[h]S[h1]
Convex Hull Area
The area of the convex hull for points N=0 to N=2^k inclusive is
/ 0 if k=0
 1/2 if k=1
HA[k] =  2 if k=2
 35*2^(k4)  13*2^(h1) + 2 if k odd >=3
\ 35*2^(k4)  10*2^(h1) + 2 if k even >=4
where h = floor(k/2)
= 0, 1/2, 2, 13/2, 17, 46, 102, 230, 482, 1018, 2082, 4274, ...
HA[1] and HA[3] are fractions but all others are integers.
The area can be calculated from the shapes shown for the hull boundary above. For k odd it can be noted the width and height are equal, then the various corners are cut off.
Line Points
The number of points which fall on straight and diagonal lines from the endpoints can be calculated by considering how the previous level duplicates to make the next.
d d
c \ / c
b  +  b
\  / \  / curve endpoints
\  / \  / "S" start
\/ \/ "E" end
aEeSa
/\ /\
/  \ /  \
/  f f  \
h g g h
The curve is rotated to make the endpoints horizontal. Each "a" through "h" is the number of points which fall on the respective line. The curve is symmetric in left to right so the line counts are the same each side in mirror image.
"S" start and "E" end points are not included in any of the counts. "e" is the count in between S and E. The two "d" lines meet at point "+" and that point is counted in d. That point is where two previous level curves meet for k>=1. Points are visited up to 4 times (per "Repeated Points" above) and all those multiple visits are counted.
The following diagram shows how curve level k+1 is made from two level k curves. One is from S to M and another M to E.
\ / curve level k copies
 \ /  S to M and M to E
 c+a c+a  making curve k+1 S to E
 \/ 
\  M  /
\  /\  c a[k+1] = b[k]
c d e+g e+g d / b[k+1] = c[k]
\  / \  / c[k+1] = d[k]
\/ \/ d[k+1] = a[k]+c[k] + e[k]+g[k] + 1
bEffSb e[k+1] = 2*f[k]
/\ /\ f[k+1] = g[k]
a  g g  a g[k+1] = h[k]
/ h \ / h \ h[k+1] = a[k]
/  \ /  \
/   \
For example the line S to M is an e[k], but also the M to E contributes a g[k] on that same line so e+g. Similarly c[k] and a[k] on the outer sides of M. Point M itself is visited too so the grand total for d[k+1] is a+c+e+g+1. The other lines are simpler, being just rotations except for the middle line e[k+1] which is made of two f[k].
The successive g[k+1]=h[k]=a[k1]=b[k2]=c[k3]=d[k4] can be substituted into the d to give a recurrence
d[k+1] = d[k1] + d[k3] + d[k5] + 2*d[k7] + 1
= 0,1,1,2,2,4,4,8,8,17,17,34,34,68,68,136,136,273,273,...
x + x^2
generating function 
(12*x^2) * (1x^8)
The recurrence begins with the single segment N=0 to N=1 and the two endpoints are not included so initial all zeros a[0]=...=h[0]=0.
As an example, the N=0 to N=64 picture above is level k=6 and its "d" line relative to those endpoints is the SouthWest diagonal down from N=0. The points on that line are N=32,30,40,42 giving d[6]=4.
All the measures are relative to the endpoint direction. The points on the fixed X or Y axis or diagonal can be found by taking the appropriate a through h, or sum of two of them for both positive and negative of a direction.
Triangle Areas in Regions
Consider a little right triangle with hypotenuse on each line segment (the same as in "Tiling" above) and the way it becomes two triangles on replicating
* *
1 triangle 2 triangles / \ / \ 4 triangles
*M*
* *M* / \
/ \ =>  / \  => *   *
/ \ / \ \ /
ES E S E E
Consider the triangles which fall within the following regions a,b,c,...,i. The line from start "S" to end "E" is rotated as necessary to be horizontal.
**
/\ a /\
/  \ /  \
/  \ /  \
* b  c M c  b *
/ \  / \  / \ next c[2] = 1
/ \  / \  / \ b[2] = 1
/ d \/ e \/ d \
*ES* always
\ f /\ g /\ f /
\ /  \ /  \ /
\ /  \ /  \ /
* h  i * i  h *
\  / \  /
\  / \  /
\/ \/
* *
The curve is symmetric in horizontal mirror image so there are two "b" regions, two "c" regions, etc, one on each side.
The initial triangle is e[0]=1. On expanding to S,M,E shown above there are then 2 triangles, 1 in each of the two "c" regions, giving c[1]=1. The third expansion keeps 2 triangles in "c" and pushes 2 triangles into "b" for c[2]=1 and b[2]=1. In all cases the total is the powerof2 doubling, hence sum
a[k] + 2*b[k] + 2*c[k] + 2*d[k] + e[k]
+ 2*f[k] + g[k] + 2*h[k] + 2*i[k] = 2^k
Level k+1 can be calculated by considering two level k, one from S to M and another from M to E. The following shows how those two previous levels fall within the regions of the k+1 level,
left side, M to E right side, S to M
** **
/\  /\ /\  /\
/  \b  d/  \ /  \ db /  \
/  \ /  \ /  \ /  \
/ a  c \ / f  \ /  f \ / c  a \
*  M  * *  M  *
/ \ c  e / \ h  / \ / \  h / \ e  c / \
/ \  / \  / \ / \  / \  / \
/ b \  / gi \  /  \ /  \  / ig \  / b \
/  \/  \/  \ /  \/  \/  \
*ES* *ES*
\  /\  /\  / \  /\  /\  /
\ d /  \ i /  \  / \  /  \ i /  \ d /
\ /  \ /  \ / \ /  \ /  \ /
\ / f  h \ /  \ / \ /  \ / h  f \ /
*  *  * *  *  *
\  / \  / \  / \  /
\  / \  / \  / \  /
\  / \  / \  / \  /
\/ \/ \/ \/
* * * *
For example the topmost "a" triangle gets b[k]+d[k] triangles from the M to E on the left, plus d[k]+b[k] triangles from S to M on the right, for total a[k+1] = 2*b[k] + 2*d[k]. The recurrences are
starting
a[k+1] = 2*b[k] + 2*d[k] a[0] = 0
b[k+1] = a[k] + c[k] b[0] = 0
c[k+1] = c[k] + e[k] + f[k] + h[k] c[0] = 0
d[k+1] = b[k] d[0] = 0
e[k+1] = 2*g[k] + 2*i[k] e[0] = 1
f[k+1] = d[k] f[0] = 0
g[k+1] = 2*i[k] g[0] = 0
h[k+1] = f[k] h[0] = 0
i[k+1] = h[k] i[0] = 0
These equations are not independent since a[k+1] can be written in terms of d[k+1] and f[k+1]
a[k+1] = 2*b[k] + 2*d[k]
a[k+1] = 2*d[k+1] + 2*f[k+1]
The initial values a[0]=0, d[0]=0 and f[0]=0 also satisfy this, so a[k]=2*d[k]+2*f[k] for k>=0. Substituting into the equation for b[k+1] eliminates a[k].
b[k+1] = c[k] + 2*d[k] + 2*f[k] k>=0
This leaves 8 equations in 8 unknowns and a little linear algebra gives 8th order recurrences for each region individually. The centre e is
e[k+8] = e[k+7] + 2*e[k+6]  e[k+4] + e[k+3] + 2*e[k+1] + 4*e[k]
= 1,0,0,0, 0,0,0,2, 6, 10, 22, 40, 80, 156, 308, 622, 1242, ...
1  x  2*x^2 + x^4  x^5
generating function 
1  x  2*x^2 + x^4  x^5  2*x^7  4*x^8
The recurrence is the same form for each a through i, just different initial values
initial values further values
 
a 0,0,0,2,4,8,16,30, 60,116,232,466,932,1872,3744,7494,
c 0,1,1,1,1,2, 4, 8, 18, 39, 79,159,315, 628,1250,2494,
e 1,0,0,0,0,0, 0, 2, 6, 10, 22, 40, 80, 156, 308, 622,
g 0,0,0,0,0,0, 0, 2, 2, 6, 10, 20, 40, 76, 156, 310,
b 0,0,1,1,3,5,10,20, 38,78,155,311,
d 0,0,0,1,1,3, 5,10, 20,38, 78,155,311,
f 0,0,0,0,1,1, 3, 5, 10,20, 38, 78,155,311,
h 0,0,0,0,0,1, 1, 3, 5,10, 20, 38, 78,155,311,
i 0,0,0,0,0,0, 1, 1, 3, 5, 10, 20, 38, 78,155,311
The values for b through i are the same, just starting one position later each time. This is the spiralling out by 45 degrees each time and per the successive equations above.
i[k+1] = h[k] = f[k1] = d[k2] = b[k3]
The recurrence for these can be started from a single initial "1" and treating preceding values (including some negative indices) as "0". The generating function for these have a single term in numerator. For example the generating function for "b",
x^2
gb(x) = 
1  x  2*x^2 + x^4  x^5  2*x^7  4*x^8
Triangles in Parts as Fractal
For the curve as a fractal, the two subparts are two half size copies of the whole, as if a[k]=a[k+1]/2 etc. This gives the following set of equations,
a = 2*b/2 + 2*d/2
b = a/2 + c/2
c = c/2 + e/2 + f/2 + h/2
d = b/2
e = 2*g/2 + 2*i/2
f = d/2
g = 2*i/2
h = f/2
i = h/2
The total area is reckoned as 1,
a + 2*b + 2*c + 2*d + e + 2*f + g + 2*h + 2*i = 1
A little linear algebra gives the following solution,
a = 24/105 **
b = 16/105 /\ 24 /\
c = 8/105 /  \ /  \
d = 8/105 /  \ /  \
e = 2/105 * 16 8 * 8 16 *
f = 4/105 / \  / \  / \
g = 1/105 / \  / \  / \
h = 2/105 / 8 \/ 2 \/ 8 \
i = 1/105 *ES*
 \ 4 /\ 1 /\ 4 /
total 1 \ /  \ /  \ /
\ /  \ /  \ /
* 2  1 * 1  2 *
\  / \  /
\  / \  /
\/ \/
* *
This shows how much area of the fractal is in each respective region.
One use for this could be some grayscale colouring at a limit of drawing resolution. Replications to a desired level give triangles then those triangles which are "on" can be drawn as gray spread out among its neighbouring triangles in the pattern above. The total in a given triangle would be all grays which go into that triangle. For square pixels the triangles making up a square can be averaged (4 triangles at even replication levels, 2 triangles at odd replication levels).
Triangles with total gray "1" are fully within the final fractal. The first such does not arise until 14 expansions, as per Duvall and Keesling reference below.
OEIS
Entries in Sloane's Online Encyclopedia of Integer Sequences related to this path include
http://oeis.org/A179868 (etc)
A010059 abs(dX), count1bits(N) mod 2
A010060 abs(dY), count1bits(N)+1 mod 2, being ThueMorse
A000120 direction, being total turn, count 1bits
A179868 direction 0to3, count 1bits mod 4
A035263 turn 0=straight or 180, 1=left or right,
being (count low 0bits + 1) mod 2
A096268 next turn 1=straight or 180, 0=left or right,
being count low 1bits mod 2
A007814 turn1 to the right,
being count low 0bits
A003159 N positions of left or right turn, ends even num 0 bits
A036554 N positions of straight or 180 turn, ends odd num 0 bits
A146559 X at N=2^k, being Re((i+1)^k)
A009545 Y at N=2^k, being Im((i+1)^k)
A131064 right boundary length to odd power N=2^(2k1),
being 5*2^n4n4, skip initial 1
A027383 right boundary length differences
A038503 number of East segments in N=0 to N=2^k1
A038504 number of North segments in N=0 to N=2^k1
A038505 number of West segments in N=0 to N=2^k1
A000749 number of South segments in N=0 to N=2^k1
A191689 fractal dimension of the interior boundary
A191689 is the boundary of the fractal, which is not the same as the boundary of the integer form here. When extended infinitely only some points persist indefinitely when the expansion rule is applied repeatedly.
P. Duvall and J. Keesling, "The Dimension of the Boundary of the Lévy Dragon", International Journal Math and Math Sci, volume 20, number 4, 1997, pages 627632. (Preprint "The Hausdorff Dimension of the Boundary of the Lévy Dragon" http://at.yorku.ca/p/a/a/h/08.htm.)
SEE ALSO
Math::PlanePath, Math::PlanePath::DragonCurve, Math::PlanePath::AlternatePaper, Math::PlanePath::KochCurve
ccurve(6x) backend for xscreensaver(1) which displays the C curve (and various other dragon curve and Koch curves).
HOME PAGE
http://user42.tuxfamily.org/mathplanepath/index.html
LICENSE
Copyright 2011, 2012, 2013, 2014, 2015 Kevin Ryde
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MathPlanePath is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.
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