Radix

The "radix" parameter can do the calculation in a base other than 3, using the same kind of direction reversals. For example radix 5 gives 5x5 groups,

     radix => 5
      4  |  20--21--22--23--24--25--26--27--28--29
         |   |                                   |
      3  |  19--18--17--16--15  34--33--32--31--30
         |                   |   |
      2  |  10--11--12--13--14  35--36--37--38--39
         |   |                                   |
      1  |   9-- 8-- 7-- 6-- 5  44--43--42--41--40
         |                   |   |
     Y=0 |   0-- 1-- 2-- 3-- 4  45--46--47--48--49--50-...
         |
         +----------------------------------------------
           X=0   1   2   3   4   5   6   7   8   9  10

If the radix is even then the ends of each group don't join up. For example in radix 4 N=15 isn't next to N=16, nor N=31 to N=32, etc.

     radix => 4
      3  |  15--14--13--12  16--17--18--19
         |               |               |
      2  |   8-- 9--10--11  23--22--21--20
         |   |               |
      1  |   7-- 6-- 5-- 4  24--25--26--27
         |               |               |
     Y=0 |   0-- 1-- 2-- 3  31--30--29--28  32--33-...
         |
         +------------------------------------------
           X=0   1   2   4   5   6   7   8   9  10

Even sizes can be made to join using other patterns, but this module is just Peano's digit construction. For joining up in 2x2 groupings see "HilbertCurve" (which is essentially the only way to join up in 2x2). For bigger groupings there's various ways.

Unit Square

Peano's original form was for filling a unit square by mapping a number T in the range 0<T<1 to a pair of X,Y coordinates 0<X<1 and 0<Y<1. The curve is continuous and every such X,Y is reached by some T, so it fills the unit square. A unit cube or higher dimension can be filled similarly by developing three or more coordinates X,Y,Z, etc. Cantor had shown a line is equivalent to the plane, Peano's mapping is a continuous way to do that.

The code here could be pressed into service for a fractional T to X,Y by multiplying up by a power of 9 to desired precision then dividing X and Y back by the same power of 3 (perhaps swapping X,Y for which one should be the first ternary digit). Note that if T is a binary floating point then a power of 3 division will round off in general since 1/3 is not exactly representable. (See "HilbertCurve" or "ZOrderCurve" for binary mappings.)

Sometimes the curve is drawn with line segments crossing unit squares. See PeanoDiagonals for that sort of path.

Power of 3 Patterns

Plotting sequences of values with some connection to ternary digits or powers of 3 will usually give the most interesting patterns on the Peano curve. For example the Mephisto waltz sequence (Math::NumSeq::MephistoWaltz) makes diamond shapes,

    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
      *** **   *** ** **   *  ***   *  ***   *  *  *** **
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
     ** ***  *  *   ***  *   ** ** ***  *  *   ***  *   **
    **   *  ***   *  *  *** **   *** **   *** ** **   *  *
    *  *   ** ** ***   ** ***  *  *   ** ** ***   ** ***
      *** **   *** ** **   *  ***   *  ***   *  *  *** **

This arises from each 3x3 block in the Mephisto waltz being one of two shapes which are then flipped by the Peano pattern

    * * _                     _ _ *
    * _ _           or        _ * *    (inverse)
    _ _ *                     * * _
    0,0,1, 0,0,1, 1,1,0       1,1,0, 1,1,0, 0,0,1

Level Methods

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

Return "(0, $radix**(2*$level) - 1)".

N to X,Y

Peano's calculation is based on putting base-3 digits of N alternately to X or Y. From the high end of N, a digit goes to Y then the next goes to X. Beginning at an even digit position in N makes the last digit go to X so the first N=0,1,2 is along the X axis.

At each stage a "complement" state is maintained for X and for Y. When complemented, the digit is reversed to 2 - digit, so 0,1,2 becomes 2,1,0. This reverses the direction so points like N=12,13,14 shown above go leftwards, or groups like 9,10,11 then 12,13,14 then 15,16,17 go downwards.

The complement is calculated by adding the digits from N which went to the other one of X or Y. So the X complement is the sum of digits which have gone to Y so far. Conversely the Y complement is the sum of digits put to X. If the complement sum is odd then the reversal is done. A bitwise XOR can be used instead of a sum to accumulate odd/even-ness the same way as a sum.

When forming the complement state, the original digits from N are added, before applying any complementing for putting them to X or Y. If the radix is odd, like the default 3, then complementing doesn't change it mod 2 so before or after are the same, but if the radix is even then it's not the same.

It also works to take the base-3 digits of N from low to high, generating low to high digits in X and Y. If an odd digit is put to X then the low digits of Y so far must be complemented as 22..22 - Y (the 22..22 value being all 2s in base 3, ie. 3^k-1). Conversely if an odd digit is put to Y then X must be complemented. With this approach, the high digit position in N doesn't have to be found, just peel off digits of N from the low end. But the subtract to complement is then more work if using bignums.

X,Y to N

The X,Y to N calculation can be done by an inverse of either the high to low or low to high methods above. In both cases digits are put alternately from X and Y into N, with complement as necessary.

For the low to high approach, it's not easy to complement just the X digits in the N constructed so far, but it works to build and complement the X and Y digits separately then at the end interleave to make the final N. Complementing is the ternary equivalent of an XOR in binary. On a ternary machine maybe some trit-twiddling would do it.

For low to high with even radix, the complementing is also tricky since changing the accumulated X affects the digits of Y below that, and vice versa. What's the rule? Is it alternate digits which end up complemented? In any case the current "xy_to_n()" code goes high to low which is easier, but means breaking the X,Y inputs into arrays of digits before beginning.

N on Axes

N on the X axis is all Y digits 0 in the X,Y to N described above. This means N is the digits of X, and then digit 0 or 2 at each Y position according to odd or even sum of X digits above. The Y digits are at odd positions so the 0 or 2 ternary is 0 or 6 for N in base-9.

    N on X axis = 0,1,2, 15,16,17, 18,19,20, 141, ...   (A163480)
          ternary 0,1,2, 120,121,122, 200,201,202, 12020

The Y axis is similar but the X digits are at even positions.

    N on Y axis = 0,5,6, 47,48,53, 54,59,60, 425, ...   (A163481)
          ternary 0,12,20, 1202,1210,1222, 2000,2012,2020, 120202

N on the X=Y diagonal has the ternary digits of position d go to both X and Y and so both complemented according to sum of digits of d above. That transformation within d is the ternary reflected Gray code.

    Gray3(d) = ternary flip 0<->2 when sum of digits above is odd
             = 0,1,2, 5,4,3, 6,7,8, 17, ...          (A128173)
       ternary 0,1,2, 12,11,10, 20,21,22, 122, ...
    N on X=Y diag = ternary Gray3(d) and 0,1,2 -> 0,4,8 base9,
                                         which is 4*digit
                  = 0,4,8, 44,40,36, 72,76,80, 404, ...  (A163343)
            ternary 0,11,22, 1122,1111,1100, 2200,2211,2222, 112222,

N to abs(dX),abs(dY)

The curve goes horizontally or vertically according to the number of trailing "2" digits when N is written in ternary,

    N trailing 2s   direction     abs(dX)     abs(dY)
    -------------   ---------     -------     -------
      even          horizontal       1           0
      odd           vertical         0           1
    abs(dX) = 1,1,0, 1,1,0, 1,1,1, 1,1,0, 1,1,0, 1,1,1, ...  (A014578)
    abs(dY) = 0,0,1, 0,0,1, 0,0,0, 0,0,1, 0,0,1, 0,0,0, ...  (A182581)

For example N=5 is "12" in ternary has 1 trailing "2" which is odd so the step from N=5 to N=6 is vertical.

This works because when stepping from N to N+1 a carry propagates through the trailing 2s to increment the digit above. Digits go alternately to X or Y so odd or even trailing 2s put that carry into an X digit or Y digit.

          X Y X Y X
    N   ... 2 2 2 2
    N+1   1 0 0 0 0  carry propagates

Rectangle to N Range

An easy over-estimate of the maximum N in a region can be had by going to the next bigger (3^k)x(3^k) square enclosing the region. This means the biggest X or Y rounded up to the next power of 3 (perhaps using "log()" if you trust its accuracy), so

    find k with 3^k > max(X,Y)
    N_hi = 3^(2k) - 1

An exact N range can be found by following the "high to low" N to X,Y procedure above. Start with the easy over-estimate to find a 3^(2k) ternary digit position in N bigger than the desired region, then choose a digit 0,1,2 for X, the biggest which overlaps some of the region. Or if there's an X complement then the smallest digit is the biggest N, again whichever overlaps the region. Then likewise for a digit of Y, etc.

Biggest and smallest N must maintain separate complement states as they track down different N digits. A single loop can be used since there's the same "2k" many digits of N to consider for both.

The N range of any shape can be done this way, not just a rectangle like "rect_to_n_range()". The procedure only depends on asking whether a one-third sub-part of X or Y overlaps the target region or not.