Arms¶

The curve fills a quarter of the plane and four copies mesh together perfectly when rotated by 90, 180 and 270 degrees. The "arms" parameter can choose 1 to 4 curve arms successively advancing.

For example arms=4 begins as follows, with N=0,4,8,12,etc being the first arm, N=1,5,9,13 the second, N=2,6,10,14 the third and N=3,7,11,15 the fourth.

    arms => 4
            20 ------ 16
                       |
             9 ------5/12 -----  8       23
             |         |         |        |
    17 --- 13/6 --- 0/1/2/3 --- 4/15 --- 19
     |       |         |         |
    21      10 ----- 14/7 ----- 11
                       |
                      18 ------ 22

With four arms every X,Y point is visited twice (except the origin 0,0 where all four begin) and every edge between the points is traversed once.

Level Angle¶

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 N=2^level which is at level*45 degrees around,

    N       X,Y     angle   radial dist
   ----    -----    -----   -----------
     1      1,0        0         1
     2      1,1       45       sqrt(2)
     4      0,2       90       sqrt(4)=2
     8     -2,2      135       sqrt(8)
    16     -4,0      180       sqrt(16)=4
    32     -4,-4     225       sqrt(32)
   ...

Here's points N=0 to N=2^9=512. "0" is the origin and "+" is N=512. Notice it's spiralled around full-circle to angle 45 degrees up again, like the initial N=2.

                                    * *     * *
                                  * * *   * * *
                                  * * * * * * * * *
                                  * * * * * * * * *
                            * *   * * * *       * *
                          * * *   * * * *     + * *
                          * * * * * *         * *
                          * * * * * * *
                          * * * * * * * *
                              * * * * * *
                              * * * *
                                  * * * * * * *
                            * *   * * * * * * * *
                          * * *   * * * * * * * *
                          * * * * * * * * * *
                          * * * * * * * * * * * * * * *
                          * * * * * * * * * * * * * * * *
                              * * * * * * * * * * * * * *
                              * * * * * * * * * * * *
        * * * *                   * * * * * * * * * * *
        * * * * *           * *   * * * *       * * * * *
    * * * *   0 *         * * *   * * * *   * * * * * * *
    * * * *               * * * * * *       * * * * *
      * * *               * * * * * * *       * * * *
        * * * *     * *   * * * * * * * *
    * * * * * *   * * *   * * * * * * * *
    * * * * * * * * * * * * * * * * *
      * * * * * * * * * * * * * * * * *
                * * * * *       * * * * *
            * * * * * * *   * * * * * * *
            * * * * *       * * * * *
              * * * *         * * * *

At a power of two Nlevel=2^level for N=2 or higher, the curve always goes upward from Nlevel-1 to Nlevel, and then goes to the left for Nlevel+1. For example at N=16 the curve goes up N=15 to N=16, then left for N=16 to N=17. Likewise at N=32, etc. The spiral curls around ever further but the self-similar twist back means the Nlevel endpoint is always at this same up/left orientation. See "Total Turn" below for the net direction in general.

Level Ranges¶

The X,Y extents of the path through to Nlevel=2^level can be expressed as a "length" in the direction of the Xlevel,Ylevel endpoint and a "width" across.

    level even, so endpoint is a straight line
    k = level/2
       +--+      <- Lmax
       |  |
       |  E      <- Lend = 2^k at Nlevel=2^level
       |
       +-----+
             |
          O  |   <- Lstart=0
          |  |
          +--+   <- Lmin
       ^     ^
    Wmin     Wmax
    Lmax = (7*2^k - 4)/6 if k even
           (7*2^k - 2)/6 if k odd
    Lmin = - (2^k - 1)/3 if k even
           - (2^k - 2)/3 if k odd
    Wmax = (2*2^k - 2) / 3 if k even
           (2*2^k - 1) / 3 if k odd
    Wmin = Lmin

For example level=2 is to Nlevel=2^2=4 and k=level/2=1 is odd so it measures as follows,

    4      <- Lmax = (7*2^1 - 2)/6 = 2
    |
    3--2
       |
    0--1   <- Lmin = -(2^1 - 2)/3 = 0
    ^  ^Wmax = (2*2^1 - 1)/3 = 1
    |
    Wmin = Lmin = 0

Or level=4 is to Nlevel=2^4=16 and k=4/2=2 is even. It measures as follows. The lengthways "L" measures are in the direction of the N=16 endpoint and the "W" measures are across.

          9----8    5---4        <- Wmax = (2*2^2 - 2)/3 = 2
          |    |    |   |
         10--11,7---6   3---2
               |            |
    16   13---12        0---1
     |    |
    15---14                      <- Wmin = -(2^2 - 1)/3 = -1
     ^                      ^Lmin = Wmin = -1
     |
     Lmax = (7*2^2 - 4)/6 = 4

The formulas are all integer values, but the fractions 7/6, 1/3 and 2/3 show the limits as the level increases. If scaled so that length Lend=2^k is reckoned as 1 unit then Lmax extends 1/6 past the end, Lmin and Wmin extend 1/3, and Wmax extends across 2/3.

    +--------+ --
    | -      | 1/6   total length
    || |     |          = 1/6+1+1/3 = 3/2
    || E     | --
    ||       |
    ||       |
    | \      |  1
    |  \     |
    |   --\  |
    |      \ |
    |       ||
    |  O    || --
    |  |    ||
    |  |    || 1/3
    |   ---- |
    +--------+ --
    1/3|  2/3
    total width = 1/3+2/3 = 1

Paper Folding¶

The path is called a paper folding curve because it can be generated by thinking of a long strip of paper folded in half repeatedly and then unfolded so each crease is a 90 degree angle. The effect is that the curve repeats in successive doublings turned by 90 degrees and reversed.

The first segment unfolds, pivoting at the "1",

                                          2
                                     ->   |
                     unfold         /     |
                      ===>         |      |
                                          |
    0-------1                     0-------1

Then the same again with that L shape, pivoting at the "2", then after that pivoting at the "4", and so on.

                                 4
                                 |
                                 |
                                 |
                                 3--------2
           2                              |
           |        unfold          ^     |
           |         ===>            \_   |
           |                              |
    0------1                     0--------1

It can be shown that this unfolding doesn't overlap itself but the corners may touch, such as at the X=-2,Y=1 etc noted above.

Level Methods¶

"($n_lo, $n_hi) = $path->level_to_n_range($level)"

Return "(0, 2**$level)", or for multiple arms return "(0, $arms * 2**$level + ($arms-1))".

There are 2^level segments in a curve level, so 2^level+1 points numbered from 0. For multiple arms there are arms*(2^level+1) points, numbered from 0 so n_hi = arms*(2^level+1)-1.

X,Y to N¶

The current code uses the "DragonMidpoint" "xy_to_n()" by rotating -45 degrees and offsetting to the midpoints of the four edges around the target X,Y. The "DragonMidpoint" algorithm then gives four candidate N values and those which convert back to the desired X,Y in the "DragonCurve" "n_to_xy()" are the results for "xy_to_n_list()".

    Xmid,Ymid = X+Y, Y-X    # rotate -45 degrees
    for dx = 0 or -1
      for dy = 0 or 1
        N candidate = DragonMidpoint xy_to_n(Xmid+dx,Ymid+dy)

Since there's at most two "DragonCurve" Ns at a given X,Y the loop can stop when two Ns are found.

Only the "leaving" edges will convert back to the target N, so only two of the four edges actually need to be considered. Is there a way to identify them? For arm 1 and 3 the leaving edges are up,down on odd points (meaning sum X+Y odd) and right,left for even points (meaning sum X+Y even). But for arm 2 and 4 it's the other way around. Without an easy way to determine the arm this doesn't seem to help.

X,Y is Visited¶

Whether a given X,Y is visited by the curve can be determined from one or two segments (rather then up to four for X,Y to N).

            |             S midpoint Xmid = X+Y
            |                        Ymid = Y-X
    *---T--X,Y--S---*
            |             T midpoint Xmid-1
            |                        Ymid+1

Segment S is to the East of X,Y. The arm it falls on can be determined as per "X,Y to N" in Math::PlanePath::DragonMidpoint. Numbering arm(S) = 0,1,2,3 then

                                     X,Y Visited
                                     -----------
    if arms_count()==4                  yes     # whole plane
    if arm(S) < arms_count()            yes
    if arm(S)==2 and arms_count()==1    no
    if arm(T) < arms_count()            yes

This works because when two arms touch they approach and leave by a right angle, without crossing. So two opposite segments S and T identify the two possible arms coming to the X,Y point.

           |
           |
            \
      ----   ----
          \
           |
           |

An arm only touches its immediate neighbour, ie. arm-1 or arm+1 mod 4. This means if arm(S)==2 then arm(T) can only be 1,2,3, not 0. So if "arms_count()" is 1 then arm(T) cannot be on the curve and no need to run its segment check.

The only exception to the right-angle touching rule is at the origin X,Y = 0,0. In that case Xmid,Ymid = 0,0 is on the first arm and X,Y is correctly determined to be on the curve. If S was not to the East but some other direction away from X,Y then this wouldn't be so.

Turn¶

At each point the curve always turns either left or right, it never goes straight ahead. The bit above the lowest 1-bit in N gives the turn direction.

    N = 0b...z10000   (possibly no trailing 0s)
    z bit    Turn
    -----    ----
      0      left
      1      right

For example N=12 is binary 0b1100, the lowest 1 bit is 0b_1__ and the bit above that is a 1, which means turn to the right. Or N=18 is binary 0b10010, the lowest 1 is 0b___1_ and the bit above that is 0, so turn left there.

This z bit can be picked out with some bit twiddling

    $mask = $n & -$n;          # lowest 1 bit, 000100..00
    $z = $n & ($mask << 1);    # the bit above it
    $turn = ($z == 0 ? 'left' : 'right');

This sequence is in Knuth volume 2 "Seminumerical Algorithms" answer to section 4.5.3 question 41 and is called the "dragon sequence". It's expressed there recursively as

    d(0) = 1       # unused, since first turn at N=1
    d(2N) = d(N)   # shift down looking for low 1-bit
    d(4N+1) = 0    # bit above lowest 1-bit is 0
    d(4N+3) = 1    # bit above lowest 1-bit is 1

Next Turn¶

The bits also give the turn after next by looking at the bit above the lowest 0-bit. This works because 011..11 + 1 = 100..00 so the bit above the lowest 0 becomes the bit above the lowest 1.

    N = 0b...w01111    (possibly no trailing 1s)
    w bit    Next Turn
    ----     ---------
      0       left
      1       right

For example at N=12=0b1100 the lowest 0 is the least significant bit 0b___0, and above that is a 0 too, so at N=13 the turn is to the left. Or for N=18=0b10010 the lowest 0 is again the least significant bit, but above it is a 1, so at N=19 the turn is to the right.

This too can be found with some bit twiddling, as for example

    $mask = $n ^ ($n+1);      # low one and below 000111..11
    $w = $n & ($mask + 1);    # the bit above there
    $turn = ($w == 0 ? 'left' : 'right');

Total Turn¶

The total turn is the count of 0<->1 transitions in the runs of bits of N, which is the same as how many bit pairs of N (including overlaps) are different so "01" or "10".

This can be seen from the segment replacements resulting from bits of N,

    N bits from high to low, start in "plain" state
    plain state
     0 bit -> no change
     1 bit -> count left, and go to reversed state
    reversed state
     0 bit -> count left, and go to plain state
     1 bit -> no change

The 0 or 1 counts are from the different side a segment expands on in plain or reversed state. Segment A to B expands to an "L" shape bend which is on the right in plain state, but on the left in reversed state.

      plain state             reverse state
      A = = = = B                    +
       \       ^              0bit  / \
        \     /               turn /   \ 1bit
    0bit \   / 1bit           left/     \
          \ /  turn              /       v
           +   left             A = = = = B

In both cases a rotate of +45 degrees keeps the very first segment of the whole curve in a fixed direction (along the X axis), which means the south-east slope shown is no-change. This is the 0 of plain or the 1 of reversed. And the north-east slope which is the other new edge is a turn towards the left.

It can be seen the "state" above is simply the previous bit, so the effect for the bits of N is to count a left turn at each transition from 0->1 or 1->0. Initial "plain" state means the infinite zero bits at the high end of N are included. For example N=9 is 0b1001 so three left turns for curve direction south to N=10 (as can be seen in the diagram above).

     1 00 1   N=9
    ^ ^  ^
    +-+--+---three transitions,
             so three left turns for direction south

The transitions can also be viewed as a count of how many runs of contiguous 0s or 1s, up to the highest 1-bit.

    1 00 1   three blocks of 0s and 1s

This can be calculated by some bit twiddling with a shift and xor to turn transitions into 1-bits which can then be counted, as per Jorg Arndt (fxtbook section 1.31.3.1 "The Dragon Curve").

    total turn = count_1_bits ($n ^ ($n >> 1))

The reversing structure of the curve shows up in the total turn at each point. The total turns for a block of 2^N is followed by its own reversal plus 1. For example,

                    ------->
    N=0 to N=7    0, 1, 2, 1, 2, 3, 2, 1
    N=15 to N=8   1, 2, 3, 2, 3, 4, 3, 2    each is +1
                               <-------

N to dX,dY¶

"n_to_dxdy()" is the "total turn" per above, or for fractional N then an offset according to the "next turn" above. If using the bit twiddling operators described then the two can be calculated separately.

The current "n_to_dxdy()" code tries to support floating point or other number types without bitwise XOR etc by processing bits high to low with a state table which combines the calculations for total turn and next turn. The state encodes

    total turn       0 to 3
    next turn        0 or 1
    previous bit     0 or 1  (the bit above the current bit)

The "next turn" remembers the bit above lowest 0 seen so far (or 0 initially). The "total turn" counts 0->1 or 1->0 transitions. The "previous bit" shows when there's a transition, or what bit is above when a 0 is seen. It also works not to have this previous bit in the state but instead pick out two bits each time.

At the end of bit processing any "previous bit" in state is no longer needed and can be masked out to lookup the final four dx, dy, next dx, next dy.

A088431 and A007400¶

The run lengths A088431 and A007400 are from a continued fraction expansion of an infinite sum

        1   1   1     1      1              1
    1 + - + - + -- + --- + ----- + ... + ------- + ...
        2   4   16   256   65536         2^(2^k)

Jeffrey Shallit and independently M. Kmosek show how continued fraction terms repeated in reverse give rise to this sort of power sum,

A126937¶

The A126937 "SquareSpiral" numbering has the dragon curve and square spiralling with their Y axes in opposite directions, as shown in its a126937.pdf. So the dragon curve turns up towards positive Y but the square spiral is numbered down towards negative Y (or vice versa). "PlanePath" code for this starting at "$i=0" would be

      my $dragon = Math::PlanePath::DragonCurve->new;
      my $square = Math::PlanePath::SquareSpiral->new (n_start => 0);
      my ($x, $y) = $dragon->n_to_xy ($i);
      my $A126937_of_i = $square->xy_to_n ($x, -$y);