Math::PlanePath::TerdragonCurve -- triangular dragon curve
use Math::PlanePath::TerdragonCurve; my $path = Math::PlanePath::TerdragonCurve->new; my ($x, $y) = $path->n_to_xy (123);
This is the terdragon curve by Davis and Knuth,
\ / \ --- 26,29,32 ---------- 27 6 / \ \ / \ -- 24,33,42 ---------- 22,25 5 / \ / \ \ / \ --- 20,23,44 -------- 12,21 10 4 / \ / \ / \ \ / \ / \ / \ 18,45 --------- 13,16,19 ------ 8,11,14 -------- 9 3 \ / \ / \ \ / \ / \ 17 6,15 --------- 4,7 2 \ / \ \ / \ 2,5 ---------- 3 1 \ \ 0 ----------- 1 <-Y=0 ^ ^ ^ ^ ^ ^ ^ -3 -2 -1 X=0 1 2 3
Points are a triangular grid using every second integer X,Y as per "Triangular Lattice" in Math::PlanePath.
The base figure is an "S" shape
2-----3 \ \ 0-----1
which then repeats in self-similar style, so N=3 to N=6 is a copy rotated +120 degrees, which is the angle of the N=1 to N=2 edge,
6 4 base figure repeats \ / \ as N=3 to N=6, \/ \ rotated +120 degrees 5 2----3 \ \ 0-----1
Then N=6 to N=9 is a plain horizontal, which is the angle of N=2 to N=3,
8-----9 base figure repeats \ as N=6 to N=9, \ no rotation 6----7,4 \ / \ \ / \ 5,2----3 \ \ 0-----1
Notice X=1,Y=1 is visited twice as N=2 and N=5. Similarly X=2,Y=2 as N=4 and N=7. Each point can repeat up to 3 times. "Inner" points are 3 times and on the edges up to 2 times. The first tripled point is X=1,Y=3 which as shown above is N=8, N=11 and N=14.
The curve never crosses itself. The vertices touch as triangular corners and no edges repeat.
The curve turns are the same as the GosperSide
, but here the turns are by 120 degrees each whereas GosperSide
is 60 degrees each. The extra angle here tightens up the shape.
The first step N=1 is to the right along the X axis and the path then slowly spirals anti-clockwise and progressively fatter. The end of each replication is
Nlevel = 3^level
That point is at level*30 degrees around (as reckoned with Y*sqrt(3) for a triangular grid).
Nlevel X, Y Angle (degrees) ------ ------- ----- 1 1, 0 0 3 3, 1 30 9 3, 3 60 27 0, 6 90 81 -9, 9 120 243 -27, 9 150 729 -54, 0 180
The following is points N=0 to N=3^6=729 going half-circle around to 180 degrees. The N=0 origin is marked "0" and the N=729 end is marked "E".
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * E * * * * * * * * * * * * * * * * 0 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
The little "S" shapes of the base figure N=0 to N=3 can be thought of as a rhombus
2-----3 . . . . 0-----1
The "S" shapes of each 3 points make a tiling of the plane with those rhombi
\ \ / / \ \ / / *-----*-----* *-----*-----* / / \ \ / / \ \ \ / / \ \ / / \ \ / --*-----* *-----*-----* *-----*-- / \ \ / / \ \ / / \ \ \ / / \ \ / / *-----*-----* *-----*-----* / / \ \ / / \ \ \ / / \ \ / / \ \ / --*-----* *-----o-----* *-----*-- / \ \ / / \ \ / / \ \ \ / / \ \ / / *-----*-----* *-----*-----* / / \ \ / / \ \
Which is an ancient pattern,
The curve fills a sixth of the plane and six copies rotated by 60, 120, 180, 240 and 300 degrees mesh together perfectly. The arms
parameter can choose 1 to 6 such curve arms successively advancing.
For example arms => 6
begins as follows. N=0,6,12,18,etc is the first arm (the same shape as the plain curve above), then N=1,7,13,19 the second, N=2,8,14,20 the third, etc.
\ / \ / \ / \ / --- 8/13/31 ---------------- 7/12/30 --- / \ / \ \ / \ / \ / \ / \ / \ / --- 9/14/32 ------------- 0/1/2/3/4/5 -------------- 6/17/35 --- / \ / \ / \ / \ / \ / \ \ / \ / --- 10/15/33 ---------------- 11/16/34 --- / \ / \ / \ / \
With six arms every X,Y point is visited three times, except the origin 0,0 where all six begin. Every edge between points is traversed once.
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::TerdragonCurve->new ()
$path = Math::PlanePath::TerdragonCurve->new (arms => 6)
Create and return a new path object.
The optional arms
parameter can make 1 to 6 copies of the curve, each arm successively advancing.
($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
.
The curve can visit an $x,$y
up to three times. xy_to_n()
returns the smallest of the these N values.
@n_list = $path->xy_to_n_list ($x,$y)
Return a list of N point numbers for coordinates $x,$y
. There can be none, one, two or three N's for a given $x,$y
.
$n = $path->n_start()
Return 0, the first N in the path.
$dx = $path->dx_minimum()
$dx = $path->dx_maximum()
$dy = $path->dy_minimum()
$dy = $path->dy_maximum()
The dX,dY values on the first arm take three possible combinations, being 120 degree angles.
dX,dY for arms=1 ----- 2, 0 dX minimum = -1, maximum = +2 -1, 1 dY minimum = -1, maximum = +1 1,-1
For 2 or more arms the second arm is rotated by 60 degrees so giving the following additional combinations, for a total six. This changes the dX minimum.
dX,dY for arms=2 or more ----- -2, 0 dX minimum = -2, maximum = +2 1, 1 dY minimum = -1, maximum = +1 -1,-1
There's no reversals or reflections in the curve so n_to_xy()
can take the digits of N either low to high or high to low and apply what is effectively powers of the N=3 position. The current code goes low to high using i,j,k coordinates as described in "Triangular Calculations" in Math::PlanePath.
si = 1 # position of endpoint N=3^level sj = 0 # where level=number of digits processed sk = 0 i = 0 # position of N for digits so far processed j = 0 k = 0 loop base 3 digits of N low to high if digit == 0 i,j,k no change if digit == 1 (i,j,k) = (si-j, sj-k, sk+i) # rotate +120, add si,sj,sk if digit == 2 i -= sk # add (si,sj,sk) rotated +60 j += si k += sj (si,sj,sk) = (si - sk, # add rotated +60 sj + si, sk + sj)
The digit handling is a combination of rotate and offset,
digit==1 digit 2 rotate and offset offset at si,sj,sk rotated ^ 2------> \ \ \ *--- --1 *-- --*
The calculation can also be thought of in term of w=1/2+I*sqrt(3)/2, a complex number sixth root of unity. i is the real part, j in the w direction (60 degrees), and k in the w^2 direction (120 degrees). si,sj,sk increase as if multiplied by w+1.
At each point N the curve always turns 120 degrees either to the left or right, it never goes straight ahead. If N is written in ternary then the lowest non-zero digit gives the turn
ternary lowest non-zero digit turn -------------- ----- 1 left 2 right
At N=3^level or N=2*3^level the turn follows the shape at that 1 or 2 point. The first and last unit step in each level are in the same direction, so the next level shape gives the turn.
2*3^k-------3*3^k \ \ 0-------1*3^k
The next turn, ie. the turn at position N+1, can be calculated from the ternary digits of N similarly. The lowest non-2 digit gives the turn.
ternary lowest non-2 digit turn -------------- ----- 0 left 1 right
If N is all 2s then the lowest non-2 is taken to be a 0 above the high end. For example N=8 is 22 ternary so considered 022 for lowest non-2 digit=0 and turn left after the segment at N=8, ie. at point N=9 turn left.
This rule works for the same reason as the plain turn above. The next turn of N is the plain turn of N+1 and adding +1 turns trailing 2s into trailing 0s and increments the 0 or 1 digit above them to be 1 or 2.
The direction at N, ie. the total cumulative turn, is given by the number of 1 digits when N is written in ternary,
direction = (count 1s in ternary N) * 120 degrees
For example N=12 is ternary 110 which has two 1s so the cumulative turn at that point is 2*120=240 degrees, ie. the segment N=16 to N=17 is at angle 240.
The segments for digit 0 or 2 are in the "current" direction unchanged. The segment for digit 1 is rotated +120 degrees.
The current code applies TerdragonMidpoint
xy_to_n()
to calculate six candidate N from the six edges around a point. Those N values which convert back to the target X,Y by n_to_xy()
are the results for xy_to_n_list()
.
The six edges are three going towards the point and three going away. The midpoint calculation gives N-1 for the towards and N for the away. Is there a good way to tell which edge will be the smaller? Or just which 3 edges lead away? It would be directions 0,2,4 for the even arms and 1,3,5 for the odd ones, but identifying the boundaries of those arms to know which is which is difficult.
When arms=6 all "even" points of the plane are visited. As per the triangular representation of X,Y this means
X+Y mod 2 == 0 "even" points
The length of the boundary of the terdragon on points N=0 to N=3^k inclusive, taking each line segment as length 1, is
boundary B[k] = / 2 if k=0 (N=0 to N=1) \ 3*2^k if k>=1 (N=0 to N=3^k) = 2, 6, 12, 24, 48, 96, ...
The boundary follows the curve edges around from the origin until returning there. So the single line segment N=0 to N=1 is boundary length 2, or the "S" shape of N=0 to N=3 is length 6.
2------3 B[0] = 2 \ \ B[1] = 6 0-----1 0------1
The B[1] first "S" is 3x the length of the preceding but thereafter the curve touches itself and so the boundary grows by only 2x per level.
The boundary formula can be calculated from the way the curve meets when it replicates. Consider the level N=0 to N=3^k and take its boundary length in two parts as a short side R on the right and the V shaped indentation on the left. These are shown as plain lines here but are wiggly as the curve becomes bigger and fatter.
R R[k] = right side boundary length 2-----3 V[k] = left side boundary length \ V initial V \ R[0] = 1 0-----1 V[0] = 2 R B[k+1] = 2*R[k] + 2*V[k] B[1] = 6
By symmetry the two sides of the terdragon are the same length, so the total boundary is twice the right side,
boundary[k] = 2*R[k+1]
When the curve is tripled out to the next level N=3^k the boundary length does not triple because the sides marked "===" in the following diagram enclose lengths 2*R and 2*V which would have been boundary, leaving only 4*R and 4*V.
R for k >= 0 *-----3 R[k+1] = R[k] + V[k] # per 0 to 1 \ V V[k+1] = R[k] + V[k] # per 0 to 2 V \ 2=====@ \ / \ R R \ / \ initial B[0] = 2 @=====1 B[1] = 6 \ V V \ 0-----* R
The two recurrences for R and V are the same, so R[k]=V[k] for k>=1 and hence
R[k+1] = 2*R[k] k >= 1 B[k] = 2*B[k-1] k >= 2 = 3*2^k from initial boundary[1] = 6
The separate R and V parts are
R[k] = / 1 if k=0 \ 3*2^(k-1) if k>=1 = 1, 3, 6, 12, 24, 48, ... V[k] = / 2 if k=0 \ 3*2^(k-1) if k>=1 = 2, 3, 6, 12, 24, 48, ...
The boundary length of two curve arms each to N=3^k is
Ba2[k] = 2*R[k] + V[k] = / 4 if k=0 \ 9*2^(k-1) if k>=1 = 4, 9, 18, 36, 72, 144, 288, 576, 1152, ... 2 ^ R / V Ba2 = 2*R + V / 0----->1 R
The boundary between endpoints 1 and 2 is the same as V above. The curve direction 0 to 1 is the other way around, but that doesn't matter since the curve is identical forward and backward.
The boundary length of three through five arms has a further V in each.
Ba3[k] = 2*R[k] + 2*V[k] = 6*2^k = 6, 12, 24, 48, 96, 192, 384, 768, 1536, ... Ba4[k] = 2*R[k] + 3*V[k] = / 8 if k=0 \ 15*2^(k-1) if k>=1 = 8, 15, 30, 60, 120, 240, 480, 960, 1920, Ba5[k] = 2*R[k] + 4*V[k] = / 10 if k=0 \ 9*2^k if k>=1 = 10, 18, 36, 72, 144, 288, 576, 1152, 2304, 3 2 3 2 3 2 ^ V ^ ^ V ^ ^ V ^ R \ / V V \ / V V \ / V \ / \ / \ / 0----->1 4<-----0----->1 4<-----0----->1 R R R V / R / R Ba3 = 2*R + 2*V Ba4 = 2*R + 3*V v 5 Ba5 = 2*R + 4*V
Six arms is six V,
Ba6[k] = 6*V[k] = / 12 if k=0 \ 9*2^k if k>=1 = 12, 18, 36, 72, 144, 288, 576, 1152, 2304, ... 3 2 ^ V ^ V \ / V \ / Ba6 = 6*V 4<-----0----->1 V / \ V / \ v V v 5 6
Notice Ba6[k] = Ba5[k] for k>=1. That arises since R[k]=V[k] for k>=1. The two Rs in 5 arms have become two Vs in 6 arms.
The area enclosed by the terdragon curve from 0 to N inclusive is related to the boundary by
2*N = 3*A[N] + B[N]
where A[N] counts unit equilateral triangles. Imagine each line segment as having a little triangle on each side, with each of those triangles being 1/3 of the unit area.
* equilateral unit area /|\ divided into 3 triangles / | \ / | \ / _*_ \ _*_ 2 triangles /_-- --_\ _-- --_ one each side of line segment *-----------* *-----------* total triangles=2*N -__ __- each triangle area 1/3 -*-
If a line segment is on the curve boundary then its outside triangle should not count towards the area enclosed, so subtract 1 for each unit boundary length. If a segment is both a left and right boundary, such as the initial N=0 to N=1 then it counts 2 to B[N] which makes its area 0 which is as desired. So
area triangles = total triangles - B[N] 3*A[N] = 2*N - B[N]
Another way is to consider how a new line segment changes the total boundary and area
<------* line segment not enclose triangle / boundary +2 area unchanged * <----* line segment enclose triangle \ / boundary -2 + 1 = -1 \ / area + 1 \ / *
So the boundary increases by 2 for each N, but decreases by 3 if there's a new enclosed area triangle, giving the relation
B = 2*N - 3*A
At all times the curve has all "inside" line segments traversed exactly once so that each unit area has all three sides traversed once each. If there was ever an area enclosed bigger than a single unit equilateral triangle then the curve would have to cross itself to traverse the inner lines to produce the "all inside segments traversed" pattern of the replications and expansions.
The area enclosed by the curve from N=0 to N=3^k inclusive is
A[k] = / 0 if k=0 \ 2*(3^(k-1) - 2^(k-1)) if k >=1 = 0, 0, 2, 10, 38, 130, 422, 1330, 4118, ...
This is per 2*N=3*A+B and the boundary formula above.
2*3^k = 3*A[k] + 3*2^k k >= 1
The area can also be calculated directly from the replication.
*-----D \ A[k] = 2 * A[k-1] # AB and CD \ + 2 * 3^(k-2) # centre triangles C-----f - 2 * A[k-2]/2 # Cf, Be insides \ / \ + 2 * A[k-2]/2 # Ce, Bf outsides \ / \ e-----B = 2*A[k-1] + 2*3^(k-2) \ \ sum to A-----* A[k] = 2*(3^(k-1) - 2^(k-1))
The area enclosed by the end two copies A-B and C-D are each the area of the preceding level.
The middle two triangles enclose area 2*3^k. But they duplicate the area on the underside of the C-f copy of the curve and the upper side of the B-e copy. The terdragon is symmetric on the two sides of the line between its endpoints so the part on the upper side is half the curve, so subtract 2*A[k-2]/2.
Then there are 2 similar half curve A[k-2]/2 areas on the outer sides of the B-f and C-e segments to be added. The extra and overlapped insides and outsides cancel out.
The area of the curve approaches the area of a rhombus made of two large equilateral triangles between the endpoints.
*-----N N=3^(k+1) . \ . side length = sqrt(3)^k . \ . rhombus area = 2 * side^2 = 2*3^k O-----* (area measured in unit triangles) terdragon 2*(3^k - 2^k) --------- = ------------- -> 1 as k->infinity rhombus 2*3^k
If the terdragon is reckoned as a fractal with unit length between its endpoints and infinitely smaller wiggles then this ratio is exact, ie. the area of the fractal terdragon is the same as the area of the rhombus.
When the terdragon curve triples out from N=3^k to N=3^(k+1) the three copies enclose the same area each, plus where they meet encloses a further join area.
_____ / / /____/ _____ <-- join area JA[k] \ \ \____\ _____ <-- join area JA[k] / / /____/
The curve is symmetric so the two join areas are the same, just rotated 180 degrees. The join area can be calculated as a difference A[k+1] versus three A[k].
JA[k] = (A[k+1] - 3*A[k])/2 join area when N=3^k triples = / 0 if k = 0 \ 2^k if k >= 1 = 0, 2, 4, 8, 16, 32, 64, ...
When two copies of the curve meet there is a certain boundary length on each side of the two copies. These lengths are V[k] and R[k], and hence equal for k>=2.
2----- \ \ boundary 1 to j = / 0 if k=0 \____\ \ V[k-1] if k>=1 j____1 / / boundary 1 to j = / 0 if k=0 / / \ R[k-1] if k>=1 0-----
For example two k=1 levels meet as follows. N=2 to N=3 is the boundary of the first copy, which is R[0]=1. N=3,4,5 is the boundary of the second, which is V[1]=2.
6 4 \ / \ V[0] = 2 k=1 join 5,2---3 \ R[0] = 1 0----1
The triangular area 2-3-4-5 expand on its two sides as R and V boundaries described above.
* / \ A-*-B second curve V[k] / V \ A B R A-B first curve R[k] A-----B
Outside that triangular area there is no further join. The 1--2 is the first curve and 5--6 is the second and two curves in the same direction like that don't touch except at their endpoint 2 and 5.
This can be seen from the curve extents, or also from the join area calculated above. Each join area has 3 sides of boundary and it can be seen that the total of those is equal to the R and V boundaries inside the triangle.
3*JA[k] = R[k] + V[k] k >= 1
The right side boundary of the terdragon at each point either turns by +120 degrees (left), -120 degrees (right), or goes straight ahead. Numbering the boundary starting from i=1 at N=1 the turn sequence is
Rt(i) = / if i == 2 mod 3 then turn -120 (right) | otherwise | let b = bit above lowest 1-bit of i-floor(i/3) | if b = 0 then turn +120 (left) \ if b = 1 then turn 0 (straight ahead) = 1, -1, 1, 0, -1, 1, 1, -1, 0, 0, -1, 1, 1, -1, 1, 0, -1, 0, ... starting i=1, multiple of 120 degrees
Every third turn is -1. i-floor(i/3) counts positions with those turns removed. Bit above lowest 1-bit on that remaining position is the same as the dragon curve turn sequence (the full dragon curve, not just its boundary). So the terdragon boundary turns are the dragon turns with -1 inserted as every third turn starting from position i=2.
The following diagram illustrates the initial boundary turns. Boundary positions 5 and 8 are at the same point since that point is visited twice by the boundary.
Rt(8)=-1 | * v 7 Rt(7)=1 \ / \ \8/ \ right boundary *-----6 Rt(6)=1 turn sequence \5 Rt(5)=-1 \ 4 Rt(4)=0 \ \ *-----3 Rt(3)=1 \2 Rt(2)=-1 \ *-----1 Rt(1)=1
The turns can be calculated by considering how the curve replicates. Take the turn sequence in two parts Rt and Vt. This is similar to the boundary length above but for the turns not the length.
2-----3 Rt[k] turns from 0 to 1 \ Vt Vt[k] turns from 1 to 3 \ 0-----1 initial Rt[0] = empty Rt Vt[0] = -1 (a single turn)
The endpoints are not included in the two sequence parts, so the turn at "1" is not in either part. The curve expands as follows.
*-----3 Rt[k+1] = Rt[k], 1, Vt[k] \ Vt[k+1] = Rt[k], 0, Vt[k] \ 2-----b \ / \ \ / \ *-----1 \ \ 0-----a
Rt from 0 to 1 becomes an Rt from 0 to "a", followed by the turn +1 at "a", then a Vt from "a" to 1. This is the same as in the first diagram an Rt[k+1] from 0 to 3 comprising Rt[k] from 0 to 1 and Vt[k] from 1 to 3.
Vt from 1 to 3 becomes an Rt from 1 to "b", followed by turn 0 at "b" (straight ahead), then a Vt from "b" to 1.
Points a, 1, b and 3 are on the right boundary and remain so in further expansions due to the extents of the curve segments. Point 2 which was the inside of the preceding Vt does not remain on the right boundary.
The repeated expansion
R -> R,1,V V -> R,0,V
is per the dragon curve ("Turn" in Math::PlanePath::DragonCurve) and hence bit above lowest 1-bit. Repeated expansions always have R and V alternating since both expand to an R and a V in that order, but with a different value in between.
R _ V _ R _ V _ R _ V _ R _ V
At the final lowest level Rt[0]=empty and Vt[0]=-1. That Vt[0]=-1 gives the -1 at every third position.
The way the endpoint "3" is always on the right boundary means that the successive levels extend each other. So Rt[k+1] starts with Rt[k] and then has further turns (turn 1 then Vt[k]).
The curve segment numbers which are on the right boundary are
RN = N in ternary no digit pair 11, 12 or 20, in ascending order = decimal 0,1,2, 3, 7, 8, 9, 10, 11, 21, 25, 26, 27, 28,... = ternary 0,1,2, 10,21,22, 100,101,102, 210,221,222, 1000,1001,...
For example on segments N=0 to N=8 the boundary segments are as follows.
8-----9 boundary segments \ \ N ternary 6----7,4 0 0 \ / \ 1 1 \ / \ 2 2 5,2----3 3 10 (no 11, 12, 20) \ 7 21 \ 8 22 0-----1
Some of the boundary points are visited twice. The boundary segment N is the N which goes along the boundary. So the segment 7,4 to 8 is N=7. Segment 7,4 to 5,2 would be N=4 (and is not on the boundary).
The ternary characterization of the values can be found by a breakdown similar to the boundary length calculation above. Take the boundary N numbers in two parts RN[k] and VN[k] at expansion level k.
2-----3 \ VN[1] = 1,2 initial RN[1] and VN[1] \ 0-----1 RN[1] = 0
RN[k] and VN[k] are in the range 0 to 3^k-1 and so can be written with k many ternary digits. The curve replicates and adds a new ternary digit as follows. Points 1, 3 and 7,4 are always on the boundary and so subsections can be taken between them.
8------9 \ \ VN[k+1] = 3^k + RN[k], (3 to 4) 6----7,4 2*3^k + VN[k] (7 to 9) \ / \ \ / \ 5,2----3 RN[k+1] = RN[k], (0 to 1) \ VN[k] (1 to 3) \ 0-----1
RN[k+1] gains a high digit 0. The initial RN[1] is 0 and so RN[k] has a high digit 0.
VN[k+1] gains a high digit 1 or 2. The initial VN[k] is digit 1 or 2 and so all VN[k] have high digit 1 or 2.
RN[k+1] is a 0 above 0 from LN[k] or 1 or 2 from VN[k] and so gives digit pairs 00, 01 and 02. VN[k+1] is a 1 above 0 from RN[k] or a 2 above 1 or 2 from VN[k] and so gives digit pairs 10, 21 and 22. All these digit pairs occur and the remaining 11, 12 and 20 do not occur.
Each right boundary segment number RN(i) can be calculated by writing its index i in a mixed radix representation with a ternary low digit and then binary above.
binary binary binary ternary +--------+--------+ +--------+--------+ | 1 | 0or1 |....| 0or1 | 0or2 | +--------+--------+ +--------+--------+ high low
Then consider each digit pair and change the binary 1s as follows. The result as ternary digits is RN(i).
1,nonzero -> 2,nonzero
Digit pairs include overlaps. So a run of consecutive 1-bits has all except the lowest changed,
1,1,1,1 -> 2,2,2,1
Disallowed pairs "11" and "12" are changed to the allowed "21" and "22". The disallowed "20" doesn't occur since 2 is only created by the change procedure above and so cannot have 0 below it. The change of "12" -> "22" only occurs at the lowest digit pair when the mixed radix gives 2 as the low ternary digit.
The reason this works can be seen by considering what next higher digit is permitted above a given ternary 0,1,2.
next ternary higher digit 1 or 0 <- 0 cannot 20, can 10 00 2 or 0 <- 1 cannot 11, can 21 01 2 or 0 <- 2 cannot 12, can 02 22
So starting from a low ternary 0,1,2 there is always just two choices as to what should be above it. Those two are counted by a binary digit. The digit choice is always 0 or non=0 so when the change is applied to binary in ascending order the resulting RN(i) ternary is in ascending order.
The curve segment numbers which are on the left boundary are
LN = N in ternary no digit pair 02, 10 or 11, and most significant digit 1, in ascending order = decimal 0,1, 5, 15, 16, 17, 45, 46, 50, 51, 52, 53,... = ternary 0,1,12, 120,121,122, 1200,1201,1212, 1220,1221,1222,...
The characterization can be made in a similar way to the right boundary segments above. Take the boundary N numbers in two parts LN and EN.
EN[1]=2 2-----3 initial LN[1] and EN[1] LN[1] = 0,1 \ \ 0-----1
LN[k] and EN[k] are in the range 0 to 3^k-1 and so can be written with k many ternary digits. The curve replicates and adds a new ternary digit as follows. Points 5,2, 6 and 8 are on the boundary of the replication and so subsections can be taken between them.
EN[k+1] 8------9 \ EN[k+1] = 2*3^k + LN[k], (6 to 8) \ 2*3^k + EN[k] (8 to 9) 6----7,4 \ / \ LN[k+1] = LN[k], (0 to 2) \ / \ 3^k + EN[k] (5 to 6) LN[k+1] 5,2----3 \ \ 0-----1
EN[k+1] gains a high digit 2. The initial EN[1] is a high digit 2 and so all EN have high digit 2.
LN[k+1] gains a 0 above the 0 or 1 from LN[k], and gains a 1 above the high 2 of EN[k]. LN[1] is 0,1 and so LN always has high digit 0 or 1 and highest non-zero digit always 1.
EN[k+1] is a 2 above either 0 or 1 from LN[k] or 2 from EN[k] and so gives digit pairs 20, 21 and 22. LN[k+1] is a 0 above 0 or 1 from LN[k] or a 1 above 2 from EN[k] and so gives digit pairs 00, 01 and 12. All these digit pairs occur and the remaining 02, 10 and 11 do not occur.
The repeatedly expanded LN[k] values are the left boundary of the full curve. If stopping at a finite expansion k then the EN[k] positions are on the boundary too. If continuing then EN[k] is enclosed by the next expansion level. To include EN[k] for a finite expansion the rule above is relaxed to allow high digit 2.
LN[k],EN[k] left boundary of curve after k expansions = N with k many ternary digits and no digit pair 02, 10 or 11, in ascending order
Disallowing high digit 2 can also be done by considering values to have 0s above the most significant and so a high 2 is a digit pair "02" which is to be excluded.
Each left boundary segment number LN(i) can be calculated by writing its index i in a mixed radix representation with a ternary low digit and then binary above, plus an extra high 0 bit.
binary binary binary ternary +--------+--------+--------+ +--------+--------+ | 0 | 1 | 0or1 |....| 0or1 | 0or2 | +--------+--------+--------+ +--------+--------+ high low
Then take each binary digit from low to high and apply a transformation as follows. The "previous digit" in each case includes the transformation of that digit.
previous digit binary transformed 0 0->0 1->2 1 0->0 1->2 2 0->1 1->2
For example i=8 in the mixed radix is 0102 so apply the transformations
0102 ^^ previous digit 2, bit 0, transform bit 0->1 0112 ^^ previous digit 1, bit 1, transform bit 1->2 0212 ^^ previous digit 2, bit 0, transform bit 0->1 1212 final 1212 ternary = 50 so segment N=50
The transformations correspond to the allowed digit pairs. The digit above a 0 can only be 0 or 2. The digit above a 1 can only be 0 or 2. The digit above a 2 can only be 1 or 2. The low digit can be ternary 0,1,2 and from that starting point there are two choices at each digit, hence the binary mixed radix.
The turns on the right boundary which are to the right -120 degrees are like
2-----3 turn -120 at "2" \ \ 1
A shortcut can be taken by a unit step directly from 1 to 3, skipping point 2. Doing so gives a boundary from N=0 to N=3^k which is
Rsh[k] = 2^k shortcut right side Bsh[k] = 2*2^k shortcut whole boundary
Every third point starting from N=2 is a -120 degree turn and at each of those 2 boundary lines become 1 shortcut line
Rsh[k] = R[k] * 2/3 k >= 1 = 3*2^(k-1) * 2/3 = 2^k
R[k] is a multiple of 3, so no rounding is required for the 2/3 factor. The multiple of 3 also means the last turn in Rt[k] is always a -120 (to be shortcut across). The left and right sides are symmetric, so Bsh=2*Rsh.
The shortcut boundary turn sequence is the dragon curve turns but by 60 degrees instead of 90 degrees.
St(i) = / bit above lowest 1-bit of i | if 0 then turn +60 (left) \ if 1 then turn -60 (right) = 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, ... starting i=1, multiple of 60 degrees
The shortcut eliminates the -120 degree turns from the right boundary turn sequence. The following diagram shows a a -120 at s and preceding turn P and following turn Q.
\ s-----Q \ \ = 1, 0 (at "*" and "a") ---P
The turns at P and Q are either 0 or +120 degrees. The effect of skipping s and going straight across from P to Q is that the turn at P is -60 degrees of whatever its value was, and the same for Q -60 degrees of whatever its value was. So the turns 0 and +120 in the right boundary sequence become -60 or +60.
The area enclosed by the shortcut boundary can be calculated from the plain curve area plus the extra triangles enclosed by B[k]/3 many shortcuts,
Ash[k] = A[k] + floor(B[k]/3) = / 0 if k=0 \ 2*3^(k-1) if k>=1
The "floor" is only needed for k=0 single line segment case. B[0]=2 and by rounding down 2/3 nothing is added to A[0]=0. For k>=1 B[k] is a multiple of 3.
The shortcut area and boundary continue to satisfy the 2N=3A+B relation given in "Area from Boundary" above. That relation requires each unit area to have all three of its sides traversed and that is so with the shortcuts.
Bshortcuts(k) = floor(B(k)/3) # extra shortcuts Nsh(k) = 3^k + Bshortcuts(k) # total incl shortcuts 2*Nsh(k) = 3*Ash(k) + Bsh(k)
The terdragon is in Sloane's Online Encyclopedia of Integer Sequences as,
http://oeis.org/A080846 (etc)
A080846 next turn 0=left,1=right, by 120 degrees (n=0 is turn at N=1) A060236 turn 1=left,2=right, by 120 degrees (lowest non-zero ternary digit) A137893 turn 1=left,0=right (morphism) A189640 turn 0=left,1=right (morphism, extra initial 0) A189673 turn 1=left,0=right (morphism, extra initial 0) A038502 strip trailing ternary 0s, taken mod 3 is turn 1=left,2=right
A189673 and A026179 start with extra initial values arising from their morphism definition. That can be skipped to consider the turns starting with a left turn at N=1.
A026225 N positions of left turns, being (3*i+1)*3^j so lowest non-zero digit is a 1 A026179 N positions of right turns (except initial 1) A060032 bignum turns 1=left,2=right to 3^level A062756 total turn, count ternary 1s A005823 N positions where total turn == 0, ternary no 1s A111286 boundary length, N=0 to N=3^k, skip initial 1 A003945 boundary/2 A002023 boundary odd levels N=0 to N=3^(2k+1), or even levels one side N=0 to N=3^(2k), being 6*4^k A164346 boundary even levels N=0 to N=3^(2k), or one side, odd levels, N=0 to N=3^(2k+1), being 3*4^k A042950 V[k] boundary length A056182 area enclosed N=0 to N=3^k, being 2*(3^k-2^k) A081956 same A118004 1/2 area N=0 to N=3^(2k+1), odd levels, 9^n-4^n A155559 join area, being 0 then 2^k A092236 count East segments N=0 to N=3^k A135254 count North-West segments N=0 to N=3^k, extra 0 A133474 count South-West segments N=0 to N=3^k A057083 count segments diff from 3^(k-1)
Math::PlanePath, Math::PlanePath::TerdragonRounded, Math::PlanePath::TerdragonMidpoint, Math::PlanePath::GosperSide
Math::PlanePath::DragonCurve, Math::PlanePath::R5DragonCurve
Larry Riddle's Terdragon page, for boundary and area calculations of the terdragon as an infinite fractal http://ecademy.agnesscott.edu/~lriddle/ifs/heighway/terdragon.htm
http://user42.tuxfamily.org/math-planepath/index.html
Copyright 2011, 2012, 2013, 2014 Kevin Ryde
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