Provided by: libmath-planepath-perl_122-1_all 
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 "C" curve by Lévy.
"Les Courbes Planes ou Gauches et les Surfaces Composée de Parties Semblables au
Tout", Journal de l'École Polytechnique, July 1938 pages 227-247 and October 1938
pages 249-292
<http://gallica.bnf.fr/ark:/12148/bpt6k57344323/f53.image>
<http://gallica.bnf.fr/ark:/12148/bpt6k57344820>
It spirals anti-clockwise, variously crossing and overlapping itself. The construction is
straightforward but various measurements like how many distinct points are quite
complicated.
11-----10-----9,7-----6------5 3
| | |
13-----12 8 4------3 2
| |
19---14,18----17 2 1
| | | |
21-----20 15-----16 0------1 <- Y=0
|
22 -1
|
25,23---24 -2
|
26 35-----34-----33 -3
| | |
27,37--28,36 32 -4
| | |
38 29-----30-----31 -5
|
39,41---40 -6
|
42 ... -7
| |
43-----44 49-----48 64-----63 -8
| | | |
45---46,50----47 62 -9
| |
51-----52 56 60-----61 -10
| | |
53-----54----55,57---58-----59 -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.
4----3
| N=0to2
2 2 repeated
| | as N=2to4
0----1 0----1 0----1 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, rest 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 1-bits 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 1-bits are above the
position.
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^(k-1) - 1 for k>=1 0,0,1,3,7,15,etc
l[k] = / 0 for k<=1
\ 2^(k-2) - 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 self-similar 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^(k-1)
| | | | | previous
| | | | | level ends
| E | | A | <-+
+------+ +------+
^---------------^
2^k this level ends
w[k] = max (h[k-1], w[k-1]) # right of A,B
h[k] = 2^(k-1) + max (h[k-1], w[k-1]) # above B,C,D
l[k] = max w[k-1], l[k-1]-2^(k-1) # below A,E
Since h[k]=2^(k-1)+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^(k-1) + h[k-1] giving h[k] = 2^k-1 for k>=1
w[k] = h[k-1] giving w[k] = 2^(k-1)-1 for k>=1
The max for l[k] is always w[k-1] as l[k] is never big enough that the parts B-C and C-D
can extend down past their 2^(k-1) vertical position. (l[0]=w[0]=0 and thereafter by
induction l[k]<=w[k].)
l[k] = w[k-1] giving l[k] = 2^(k-2)-1 for k>=2
Repeated Points
The curve crosses itself and can repeat X,Y positions up to 4 times. The first double,
triple and quadruple points are at
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 1-bits in N.
Since no line is repeated each of the N values at a given X,Y have a different
count-1-bits mod 4. For example N=7 is 3 1-bits and N=9 is 2 1-bits. The full counts
need not be consecutive, as for example N=1727 is 9 1-bits and N=2369 is 4 1-bits.
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 45-degrees 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^k-1
| | | | 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
4-fold visits,
| ^ 8 arms at the origin
| | cover only X,Y axes
v | with 4-visits
<-------- <--------
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.
See examples/c-curve-wx.pl in the PlanePath sources for a wxWidgets program drawing
various forms and tilings of the curve.
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,$y" then
return "undef". If "$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 = $path->n_start()"
Return 0, the first N in the path.
Level Methods
"($n_lo, $n_hi) = $path->level_to_n_range($level)"
Return "(0, 2**$level)".
FORMULAS
Some formulas and results can also be found in the author's mathematical write-up
<http://user42.tuxfamily.org/c-curve/index.html>
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 power-of-2 position the curve is a copy of the lower
bits but turned +90 degrees, so +90 for each 1-bit.
For powers-of-2 N=2,4,8,16, etc, there's only a single 1-bit 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 0-bits of N,
turn right = (count_low_0_bits(N) - 1) * 90degrees
For example N=8 is binary 0b100 which is 2 low 0-bits for turn=(2-1)*90=90 degrees to the
right.
When N is odd there's no low zero bits and the turn is always (0-1)*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
1-bits 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=(2-1)*90=90 degrees
to the right at the following point, ie. at N=12.
This works simply because low 1-bits like ..0111 increment to low 0-bits ..1000 to become
N+1. The low 1-bits at N are thus the low 0-bits 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 1-bits 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 # direction-1 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)
= (x-y) + (x+y)i
so X=x-y 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 1-bits 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 N-2^l to N, where 2^l is the lowest
1-bit 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 p-1
}
X,Y = (X+Y)/2, (Y-X)/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 1-bit 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 non-zeros. 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 1-bits 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=Y-X.
for dS = +1 or -1 # four directions
for dD = +1 or -1 #
S = X+Y
D = Y-X
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, (D-S)/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 1-bit for N
found an N
if S=0 and D=0
found an N
The effect of S=X+Y, D=Y-D 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 in-
between 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^k-1 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^(k-2) + d(k+2)*2^(h-1)
M1[k] = 0, 2^(k-2) + d(k+0)*2^(h-1)
M2[k] = 0, 2^(k-2) + d(k-2)*2^(h-1)
M3[k] = 0, 2^(k-2) + d(k-4)*2^(h-1)
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^(k-2), so
roughly 1/4 each, but a half power 2^(h-1) possibly added or subtracted in a k mod 8
pattern. In binary this is a 2^(k-2) high 1-bit with another 1-bit in the middle added or
subtracted.
The total is 2^k since there are a total 2^k points from N=0 to 2^k-1 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^(h-1) parts cancel out leaving
4*2^(k-2) = 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,
7---6---5 Easts in N=0 to 8
| | = Easts in N=0 to 4
8 4---3 + Wests in N=0 to 4
| since N=4 to N=8 is
2 the N=0 to N=4 rotated +90
|
0---1
For the bits in N, level k+1 introduces a new bit either 0 or 1. In M0[k+1] the a 0-bit
is count M0[k] the same direction, and when a 1-bit 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[k-1] - 6*M0[k-2] + 4*M0[k-3] initial 1,1,1,1
M1[k] = 4*M1[k-1] - 6*M1[k-2] + 4*M1[k-3] initial 0,1,2,3
M2[k] = 4*M2[k-1] - 6*M2[k-2] + 4*M2[k-3] initial 0,0,1,3
M3[k] = 4*M3[k-1] - 6*M3[k-2] + 4*M3[k-3] initial 0,0,0,1
The characteristic polynomial of these recurrences is
x^3 - 4x^2 + 6x - 4
= (x-2) * (x - (1-i)) * (x - (1+i))
So explicit formulas can be written in powers of the roots 2, 1-i and 1+i,
M0[k] = ( 2^k + (1-i)^k + (1+i)^k )/4 for k>=1
M1[k] = ( 2^k + i*(1-i)^k - i*(1+i)^k )/4
M2[k] = ( 2^k - (1-i)^k - (1+i)^k )/4
M3[k] = ( 2^k - i*(1-i)^k + i*(1+i)^k )/4
The complex numbers 1-i and 1+i are 45 degree lines clockwise and anti-clockwise
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(1-i)=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 1-bits. 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((k-1)/4)
\1/ \5/ \4m+1/
M2[k] = /k\ + /k\ + ... + / k \ m = floor((k-2)/4)
\2/ \6/ \4m+2/
M3[k] = /k\ + /k\ + ... + / k \ m = floor((k-3)/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 right-side 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[k-1] + R[k-2] - 4*R[k-3] + 2*R[k-4]
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.
11--10--9,7--6--5 right boundary
| | | around "outside"
13--12 8 4--3 N=0 to N=2^4=16
| |
14 2 R[4]=14
| |
15--16 0--1
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 zig-zags. When it expands the
straight lines become zig-zags and the zig-zags become straight lines. The straight lines
all point "forward", which is anti-clockwise.
c * a
/ ^ / ^ / ^
=> v \ v \ v \
D<----C<----B<----A D C B A
| ^ / ^
v | v \
straight S=3 zig-zag 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 \
zig-zag 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 non-empty 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^h-2
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[(k-1)/2] (with an
S[(k-1)/2] before and after). So
/ i=h-1 \ # 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^(h-1) # S[0] to S[h-1]
+ 2+4+8+...+2^h - 2*h) # Z[0] to Z[h-1]
+ 2^h # S[h]
+ if k odd (2^h + 2*2^h - 2) # possible S[h]+Z[h]
= 2*(2^h-1 + 2*2^h-2 - 2h) + 2^h + (k odd 3*2^h - 2)
= 7*2^h - 4h-6 + (if k odd then + 3*2^h - 2)
= 7*2^h - 2k-6 + (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^(h-1) + (7*2^(h-1) - 4)*sqrt(2) if k odd >=3
\ 7*2^(h-1) + (3*2^(h-1) - 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[h-2] the curl goes inwards and so parts beyond Z[h-2] are not part of
the hull. Each Z stair-step 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[h-1] / \ Z[h-1] | \ diagonal
/ \ | \ sqrt(2)*Z/2
* * *----* = sqrt(2)
S[h-1] | | S[h-1]
| |
* *
Z[h-2] \ / Z[h-2]
*-- --*
S[h] + Z[h-2]-Z[h-1]
k even
hull boundary[k] = S[h] + 2*S[h-1] + S[h+Z[h-2]-Z[h-1]
+ sqrt(2)*(2*Z[h-1] + 2*Z[h-2])/2
When k is odd the shape is similar but Z[h] in the middle.
S[h]
*----*
Z[h-1] / \ middle
* \ Z[h]
S[h-1] | \
* *
\ | S[h]
Z[h] |
+ 2*(S[h]-S[h-1]) *
\ / Z[h-1]
*---*
S[h-1]
k odd
hull boundary[k] = 2*S[h] + 2*S[h-1]
+ sqrt(2)*(Z[h]/2 + 2*Z[h-1]/2
+ Z[h]/2 + S[h]-S[h-1]
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^(k-4) - 13*2^(h-1) + 2 if k odd >=3
\ 35*2^(k-4) - 10*2^(h-1) + 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
a------E---e---S------a
/|\ /|\
/ | \ / | \
/ | 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
b-------E--f---f--S-------b 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[k-1]=b[k-2]=c[k-3]=d[k-4] can be substituted into the d to
give a recurrence
d[k+1] = d[k-1] + d[k-3] + d[k-5] + 2*d[k-7] + 1
= 0,1,1,2,2,4,4,8,8,17,17,34,34,68,68,136,136,273,273,...
x + x^2 (common factor 1+x
generating function ------------------- in numerator and
(1-2*x^2) * (1-x^8) denominator)
d[2h-1] = d[2h] = floor( 8/15 * 2^h )
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 South-West 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.
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)+1 mod 2
A010060 abs(dY), count1bits(N) mod 2, being Thue-Morse
A000120 direction, being total turn, count 1-bits
A179868 direction 0to3, count 1-bits mod 4
A035263 turn 0=straight or 180, 1=left or right,
being (count low 0-bits + 1) mod 2
A096268 next turn 1=straight or 180, 0=left or right,
being count low 1-bits mod 2
A007814 turn-1 to the right,
being count low 0-bits
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^(2k-1),
being 5*2^n-4n-4, skip initial 1
A027383 right boundary length differences
A038503 number of East segments in N=0 to N=2^k-1
A038504 number of North segments in N=0 to N=2^k-1
A038505 number of West segments in N=0 to N=2^k-1
A000749 number of South segments in N=0 to N=2^k-1
A191689 fractal dimension of the interior boundary
A191689 is the fractal dimension which roughly speaking means what power r^k the boundary
length grows by when each segment is taken as a little triangle (or similar). There are
various holes inside the curling spiralling curve and they are all boundary for this
purpose.
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 627-632.
(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) back-end for xscreensaver(1) which displays the C curve (and various other
dragon curve and Koch curves).
HOME PAGE
<http://user42.tuxfamily.org/math-planepath/index.html>
LICENSE
Copyright 2011, 2012, 2013, 2014, 2015, 2016 Kevin Ryde
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