NAME
Math::PlanePath::R5DragonCurve  radix 5 dragon curve
SYNOPSIS
use Math::PlanePath::R5DragonCurve;
my $path = Math::PlanePath::R5DragonCurve>new;
my ($x, $y) = $path>n_to_xy (123);
DESCRIPTION
This is the R5 dragon curve by Jorg Arndt,
3130 2726 5
   
3229/3328/2425 4
 
3534/3839/2322 1110 76 3
      
3637/4120/4021/1716/1213/98/45 2
     
50 4742/4619/4318 1514 32 1
    
49/5348/64 45/6544/68 69 01 <Y=0
^ ^ ^ ^ ^ ^ ^ ^ ^
7 6 5 4 3 2 1 X=0 1
The base figure is an "S" shape
45

32

01
which then repeats in selfsimilar style, so N=5 to N=10 is a copy rotated +90 degrees, as per the direction of the N=1 to N=2 segment.
10 76
   < repeat rotated +90
98,45

32

01
This replication is similar to the TerdragonCurve
in that there's no reversals or mirroring. Each replication is the plain base curve.
The shape of N=0,5,10,15,20,25 repeats the initial N=0 to N=5,
25 4
/
/ 10__ 3
/ / ___
20__ / 5 2
__ / /
15 / 1
/
0 <Y=0
^ ^ ^ ^ ^ ^
4 3 2 1 X=0 1
The curve never crosses itself. The vertices touch at corners like N=4 and N=8 above, but no edges repeat.
Spiralling
The first step N=1 is to the right along the X axis and the path then slowly spirals anticlockwise and progressively fatter. The end of each replication is
Nlevel = 5^level
Each such point is at arctan(2/1)=63.43 degrees further around from the previous,
Nlevel X,Y angle (degrees)
  
1 1,0 0
5 2,1 63.4
25 3,4 2*63.4 = 126.8
125 11,2 3*63.4 = 190.3
Arms
The curve fills a quarter of the plane and four copies mesh together perfectly rotated by 90, 180 and 270 degrees. The arms
parameter can choose 1 to 4 such curve arms successively advancing.
arms => 4
begins as follows. N=0,4,8,12,16,etc is the first arm (the same shape as the plain curve above), then N=1,5,9,13,17 the second, N=2,6,10,14 the third, etc.
arms => 4
16/3220/63

21/60 9/565/128/59
   
17/33 6/130/1/2/34/1519/35
   
10/577/1411/58 23/62

22/6118/34
With four arms every X,Y point is visited twice, except the origin 0,0 where all four begin. Every edge between the points is traversed once.
Tiling
The little "S" shapes of the N=0to5 base shape tile the plane with 2x1 bricks and 1x1 holes in the following pattern,
++ ++ ++
         
 + + 
          
+ + ++
         
++ + +
          
 + + 
         
+ o +
         
 + + 
          
+ + ++
         
++ + +
          
 + + 
         
++ ++ ++
This is the curve with each segment N=2mod5 to N=3mod5 omitted. Each 2x1 block has 6 edges. The "S" within traverses 4 of them and the way the blocks mesh meshes together traverses the other 2 edges by another brick, possibly a brick on another arm of the curve.
This tiling is also for example
http://tilingsearch.org/HTML/data182/AL04.html
Or with enlarged square part, http://tilingsearch.org/HTML/data149/L3010.html
FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::R5DragonCurve>new ()
$path = Math::PlanePath::R5DragonCurve>new (arms => 4)

Create and return a new path object.
The optional
arms
parameter can make 1 to 4 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
$n
gives 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
.The curve can visit an
$x,$y
twice. In the current code the smallest of the these N values is returned. Is that the best way? @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 or two N's for a given$x,$y
. $n = $path>n_start()

Return 0, the first N in the path.
FORMULAS
Turn
At each point N the curve always turns 90 degrees either to the left or right, it never goes straight ahead. As per the code in Jorg Arndt's fxtbook, if N is written in base 5 then the lowest nonzero digit gives the turn
lowest non0 digit turn
 
1 left
2 left
3 right
4 right
At a point N=digit*5^level for digit=1,2,3,4 the turn follows the shape at that digit, so two lefts then two rights,
4*5^k5^(k+1)


2*5^k2*5^k


01*5^k
The first and last unit segments in each level are the same direction, so at those endpoints it's the next level up which gives the turn.
Next Turn
The turn at N+1 can be calculated in a similar way but from the lowest non4 digit.
lowest non4 digit turn
 
0 left
1 left
2 right
3 right
This works simply because in N=...z444 becomes N+1=...(z+1)000 and the turn at N+1 is given by digit z+1.
Total Turn
The direction at N, ie. the total cumulative turn, is given by the direction of each digit when N is written in base 5,
digit direction
0 0
1 1
2 2
3 1
4 0
direction = (sum direction for each digit) * 90 degrees
For example N=13 in base 5 is "23" so digit=2 direction=2 plus digit=3 direction=1 gives direction=(2+1)*90 = 270 degrees, ie. south.
Because there's no reversals etc in the replications there's no state to maintain when considering the digits, just a plain sum of direction for each digit.
Boundary Length
The length of the boundary of the curve points N=0 to N=5^k inclusive is
boundary B[k] = 4*3^k  2
= 2, 10, 34, 106, 322, 970, 2914, ...
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=5 is length 10.
45
 boundary[1]=10
boundary[0]=2 32

01 01
The first "S" shape is 5x the previous length but thereafter the way the curve touches itself makes the boundary shorter (growing just over 3x as can be seen from the power 3^k in B).
The boundary formula can be calculated from the way the curve meets when it replicates. Consider the level N=0 to N=5^k and take its boundary length in two parts as a short side R and an inner curving part U.
R R[k] = side boundary
45 U[k] = inner curve boundary
R  U
32 initial R[1] = 1
U  R U[1] = 3
01
R
The curve is shown here as plain lines but becomes fatter and wiggly at higher replications. Points 1 and 2 are on the right side boundary, and similarly 3 and 4 on the left side boundary, so in this breakdown the points where U and R parts meet are on the boundary. The total is
B[k] = 4*R[k] + 2*U[k]
The curve is symmetric on its left and right sides so R is half the total boundary of the preceding level,
R[k] = B[k1] / 2
Which gives
R[k+1] = 2*R[k] + U[k]
When the curve replicates to the next level N=5^k the boundary length becomes,
R
*5
R  U R R R[k+1] = 2*R[k] + U[k]
** *2 ** U[k+1] = R[k] + 2*U[k]
U  U   U   R
4****1 # eg. 0 to 1 on the right for R[k+1]
R   U   U  U # 0 to 3 on the left for U[k+1]
** 3* **
R R U  R
0*
R
This expansion for R[k+1] is the same as obtained from symmetry of the total. Then U from 0 to 3 gives a second recurrence. The two together can then eliminate U by substituting the former into the latter,
U[k] = R[k+1]  2*R[k] # from R[k+1] formula
R[k+2]2*R[k+1] = 2*(R[k+1]2*R[k]) + R[k] # from U[k+1] formula
R[k+2] = 4*R[k+1]  3*R[k]
Then from R[k]=B[k1]/2 this recurrence for R becomes the same recurrence for the total B,
B[k+1] = 4*B[k]  3*B[k1]
The characteristic equation of this recurrence is
x^2  4*x + 3 = (x3)*(x1) roots 3, 1
So the closed form is some a*3^k+b*1^k, being 4*3^k  2. That formula can also be verified by induction from the initial B[0]=2, B[1]=10.
Area
The area enclosed by the curve from N=0 to N=5^k inclusive is
A[k] = (5^k  2*3^k + 1)/2
= 0, 0, 4, 36, 232, 1320, 7084, 36876, 188752, ...
A[k] = 9*A[k1]  23*A[k2] + 15*A[k2]
This can be calculated from the boundary. Like the dragon curve (per "Area" in Math::PlanePath::DragonCurve), every edge is traversed precisely once so each enclosed unit square has line segments on 4 sides. Imagine each line segment as a diamond shape made from two right triangles
*
/ \ 2 triangles each line segment
01
\ /
*
If a line segment is on the boundary then the outside triangle does not count towards the area. Subtract 1 for each of them.
triangles = 2*5^k  B[k]
Line segments at the tail end like N=0 to N=1 are both a left and right boundary. They already count twice in B[k] and so are no triangles. Four triangles make up a unit square,
area[k] = triangles/4
The 2*5^k can be worked into the B recurrence in the usual way to give the A recurrence 9,23,15 above, and which can be verified by induction from the initial A[0]=0, A[1]=0. The characteristic equation is
x^3  9*x^2 + 23*x  15 = (x1)*(x3)*(x5)
The roots 3 and 5 become the power terms in the explicit formula, and 1 the constant.
Another form per Henry Bottomley in OEIS A007798 (that sequence is area/2) is
A[k+2] = 8*A[k+1]  15*A[k] + 4
Area by Replication
The area can also be calculated by replications in a similar way to the boundary. Consider the level N=0 to N=5^k and take its area in two parts as a short side R to the right and an inner curving part U
R R[k] = side area
45 U[k] = inner curve area
R  U
32 initial R[0]=0,R[1]=0 U[0]=0,U[1]=0
U  R
01 A[k] = 4*R[k] + 2*U[k]
R
As per above, point 1 on the right boundary of the curve. Area R is the region between the line 01 and the right boundary of the curve around from 0 to 1. This boundary in fact dips back to the left side of the 01 line. When that happens it's reckoned as a negative area. A similar negative area happens to U.
___ < negative area when other side of the line
/ \
0/1
\ / line 0 to 1
 curve right boundary
The total area is the six parts
A[k] = 4*R[k] + 2*U[k]
The curve is symmetric on its left and right sides so R itself is in fact half the total area of the preceding level,
R[k] = A[k1] / 2
Which gives
R[k+1] = 2*R[k] + U[k]
When the curve replicates to the next level N=5^k the pattern of new U and R is the same as the boundary above, except the four newly enclosed squares are of interest for the area.
R
*5 square edge length sqrt(5)^(k2)
R  U R R square area = 5^(k2)
** *2 **
U  U   U   R
4****1
R   U   U  U
** 3* **
R R U  R
0*
R
The size of the squares grows by the sqrt(5) replication factor. The 25point replication shown is edge length 1. Hence square=5^(k2).
The line 0 to 1 passes through 3/4 of a square,
..... 1
. / line dividing each square
.  . into two parts 1/4 and 3/4
. / .
*....*
. / .
.  .
/ .
0 .....
The area for R[k+1] is that to the right of the line 01. This is first +3/4 of a square with a further two R on its outside, then 3/4 of a square with a U pushing out (reducing that negative).
R[k+1] = 3/4*square + 2*R[k]  3/4*square + U[k]
= 2*R[k] + U[k]
This is the same recurrence as obtained above from the symmetry R[k] = A[k1]/2.
The area for U[k+1] is that on left of the U shaped line 0123,
U[k+1] = 3/4*square + U[k] + 3/4*square
+ 2*square + U[k] + R[k]
U[k+1] = R[k] + 2*U[k] + 2*5^(k2} # square = 5^(k2)
Notice for R that the first 3/4 square has the left side dipping in. For R that's still counted as a full +3/4 square. In U it's a 3/4 which gives a total area of just what's between the left and right curve boundaries.
U is eliminated by substituting the R[k+1] recurrence into the U[k+1]
U[k] = R[k+1]  2*R[k] # from the R[k+1] formula
R[k+2]2*R[k+1] = 2*(R[k+1]2*R[k]) + R[k] + 2*5^(k1)
R[k+2] = 4*R[k+1]  3*R[k] + 2*5^(k1)
Then from R[k] = A[k1]/2 the total area is as follows,
A[k+2] = 4*A[k+1]  3*A[k] + 4*5^k # k>=2
This is the same as boundary calculation above but an extra 4*5^(k2) which are the 4 squares fully enclosed when the curve replicates.
OEIS
The R5 dragon is in Sloane's Online Encyclopedia of Integer Sequences as,
http://oeis.org/A175337 (etc)
A175337 next turn 0=left,1=right
(n=0 is the turn at N=1)
A079004 boundary length N=0 to 5^k, skip initial 7,10
being 4*3^k  2
A048473 boundary/2 (one side), N=0 to 5^k
being half whole, 2*3^n  1
A198859 boundary/2 (one side), N=0 to 25^k
being even levels, 2*9^n  1
A198963 boundary/2 (one side), N=0 to 5*25^k
being odd levels, 6*9^n  1
A007798 1/2 * area enclosed N=0 to 5^k
A016209 1/4 * area enclosed N=0 to 5^k
A005058 1/2 * new area N=5^k to N=5^(k+1)
being area increments, 5^n  3^n
A005059 1/4 * new area N=5^k to N=5^(k+1)
being area increments, (5^n  3^n)/2
arms=1 and arms=3
A059841 abs(dX), being simply 1,0 repeating
A000035 abs(dY), being simply 0,1 repeating
arms=4
A165211 abs(dY), being 0,1,0,1,1,0,1,0 repeating
SEE ALSO
Math::PlanePath, Math::PlanePath::DragonCurve, Math::PlanePath::TerdragonCurve
HOME PAGE
http://user42.tuxfamily.org/mathplanepath/index.html
LICENSE
Copyright 2012, 2013, 2014 Kevin Ryde
This file is part of MathPlanePath.
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.
MathPlanePath is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with MathPlanePath. If not, see <http://www.gnu.org/licenses/>.