NAME
Math::PlanePath::PeanoCurve -- 3x3 self-similar quadrant traversal
SYNOPSIS
use Math::PlanePath::PeanoCurve;
my $path = Math::PlanePath::PeanoCurve->new;
my ($x, $y) = $path->n_to_xy (123);DESCRIPTION
This path is an integer version of the curve described by Guiseppe Peano in 1890 for filling a unit square. It traverses a quadrant of the plane one step at a time in a self-similar 3x3 pattern,
y=8 60--61--62--63--64--65 78--79--80--...
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y=7 59--58--57 68--67--66 77--76--75
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y=6 54--55--56 69--70--71--72--73--74
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y=5 53--52--51 38--37--36--35--34--33
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y=4 48--49--50 39--40--41 30--31--32
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y=3 47--46--45--44--43--42 29--28--27
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y=2 6---7---8---9--10--11 24--25--26
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y=1 5---4---3 14--13--12 23--22--21
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y=0 0---1---2 15--16--17--18--19--20
x=0 1 2 3 4 5 6 7 8 9 ...The start is an S shape of the nine points 0 to 8, and then nine of those groups are put together in the same configuration. The sub-parts are flipped horizontally and/or vertically to make the starts and ends adjacent, so that 8 is next to 9, 17 next to 18, etc,
60,61,62 --- 63,64,65 78,79,80
59,58,57 68,67,55 77,76,75
54,55,56 69,70,71 --- 72,73,74
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|
53,52,51 38,37,36 --- 35,34,33
48,49,50 39,40,41 30,31,32
47,46,45 --- 44,43,42 29,28,27
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|
6,7,8 ---- 9,10,11 24,25,26
3,4,5 12,13,14 23,22,21
0,1,2 15,16,17 --- 18,19,20The process repeats, tripling in size each time.
Within a power-of-3 square 3x3, 9x9, 27x27, 81x81 etc (3^k)x(3^k) at the origin, all the N values 0 to 3^(2*k)-1 are within the square. The top right corner 8, 80, 728, etc is the 3^(2*k)-1 maximum in each.
Because each step is by 1, the distance along the curve between two X,Y points is the difference in their N values (as given by xy_to_n).
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 X,Y is reached, so it fills the unit square. A unit cube can be filled too by developing three coordinates X,Y,Z similarly. Georg Cantor had shown a line is equivalent to a surface, 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,Y back by the same power of 3 (perhaps swapping X,Y for which one should be the first ternary digit). 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 in binary. See HilbertCurve or ZOrderCurve for binary based mappings.
FUNCTIONS
$path = Math::PlanePath::PeanoCurve->new ()-
Create and return a new path object.
($x,$y) = $path->n_to_xy ($n)-
Return the X,Y coordinates of point number
$non the path. Points begin at 0 and if$n < 0then the return is an empty list.Fractional positions give an X,Y position along a straight line between the integer positions. Integer positions are always just 1 apart either horizontally or vertically, so the effect is that the fraction part appears either added to or subtracted from X or Y.
$n = $path->xy_to_n ($x,$y)-
Return an integer point number for coordinates
$x,$y. Each integer N is considered the centre of a unit square and an$x,$ywithin that square returns N. ($n_lo, $n_hi) = $path->rect_to_n_range ($x1,$y1, $x2,$y2)-
Return a range of N values which occur in a rectangle with corners at
$x1,$y1and$x2,$y2. If the X,Y values are not integers then the curve is treated as unit squares centred on each integer point and squares which are partly covered by the given rectangle are included.The returned range is exact, meaning
$n_lois the smallest in the rectangle and$n_hiis the biggest. Of course not all the N's in that range are necessarily in the rectangle.
FORMULAS
N to X,Y
Peano's calculation is based on putting base-3 digits of N alternately between X and Y. Starting from the high end of N a digit is appended to Y then the next appended to X. Starting at an even digit position in N makes the last digit go to X so that the first N=0,1,2 steps go along the X axis.
At each stage a "complement" state is maintained for X and 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 to the left, or groups like 9,10,11 then 12,13,14 then 15,16,17 go downwards.
The complement is calculated by adding the N digits which went to the other of X or Y. The X complement is the sum of digits which have been appended to Y so far, and conversely the Y complement is the sum of digits applied to X. If the complement sum is odd then the reversal is done. The reversal itself doesn't change the odd/even so it doesn't matter if the digit is taken before or after reversing. An XOR can be used instead of a sum, accumulating odd/even the same way.
It also works to take the base-3 digits of N from low to high, prepending digits to X and Y successively. When an odd digit, ie. a 1, is put onto X then the digits of Y so far must be complemented as 22..22 - Y, the 22..22 value being all 2s in base 3. Conversely if a digit 1 is added to Y then X must be complemented. With this approach the high digits of N don't have to be found, but instead digits of N peeled off the low end. But the subtract to do the complement is more work if using bignums.
X,Y to N
The X,Y to N calculation can be done by an inverse of either method above, in both cases putting digits alternately from X and Y onto 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 equivalent of an XOR in binary. On a ternary machine some trit-twiddling could no doubt do it.
In the current code n_to_xy and xy_to_n both go low to high as that seems a bit easier than finding the high ternary digits of the inputs.
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_max = 3^(2k) - 1An exact N range can be found by following the high to low N to X,Y procedure. Start at the 3^(2k) ternary digit position in N which is bigger than the desired region and 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 which overlaps the region. Then the same for a digit of Y, etc.
Biggest and smallest N must be calculated separately as they track down different N digits and different X,Y complement states. The N range for any shape can be done this way, not just a rectangle the way rect_to_n_range does, since it only depends only on asking when a one-third sub-part of X or Y overlaps the target area.
OEIS
This path is in Sloane's OEIS in several forms,
http://oeis.org/A163528 (etc)
A163528 X coordinate
A163529 Y coordinate
A163530 coordinate sum X+Y
A163531 square of distance from origin X^2+Y^2
A163532 X change -1,0,1
A163533 Y change -1,0,1
A163534 absolute direction of each step (up,down,left,right)
A163535 absolute direction, transpose X,Y
A163536 relative direction (ahead,left,right)
A163537 relative direction, transpose X,Y
A163342 diagonal sums
A163343 central diagonal 0,4,8,44,40,36,etc
A163344 central diagonal divided by 4
A163479 diagonal sums divided by 6
A163480 row at Y=0
A163481 column at X=0And taking the squares of the plane in the Diagonals sequence, each value of the following sequences is the N of the Peano curve at those positions.
A163334 numbering by diagonals, from same axis as first step
A163336 numbering by diagonals, from opposite axis
A163338 A163334 + 1, Peano starting from N=1
A163340 A163336 + 1, Peano starting from N=1Math::PlanePath::Diagonals numbers from the Y axis down, which is the opposite axis to the Peano curve first step along the X axis, so a plain Diagonals -> PeanoCurve is the "opposite axis" form A163336.
These sequences are permutations of the integers since all X,Y positions of the first quadrant are reached eventually. The inverses are as follows. They can be thought of taking X,Y positions in the Peano curve order and then asking what N the Diagonals would put there.
A163335 inverse of A163334
A163337 inverse of A163336
A163339 inverse of A163338
A163341 inverse of A163340SEE ALSO
Math::PlanePath, Math::PlanePath::HilbertCurve, Math::PlanePath::ZOrderCurve
Guiseppe Peano, "Sur une courbe, qui remplit toute une aire plane", Mathematische Annalen, volume 36, number 1, 1890, p157-160
http://www.springerlink.com/content/w232301n53960133/
DOI 10.1007/BF01199438HOME PAGE
http://user42.tuxfamily.org/math-planepath/index.html
LICENSE
Math-PlanePath is Copyright 2010, 2011 Kevin Ryde
Math-PlanePath 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.
Math-PlanePath 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 Math-PlanePath. If not, see <http://www.gnu.org/licenses/>.