Math::PlanePath::PeanoCurve -- 3x3 self-similar quadrant traversal
use Math::PlanePath::PeanoCurve; my $path = Math::PlanePath::PeanoCurve->new; my ($x, $y) = $path->n_to_xy (123); # or another radix digits ... my $path5 = Math::PlanePath::PeanoCurve->new (radix => 5);
This path is an integer version of the curve described by Peano for filling a unit square,
Guiseppe Peano, "Sur Une Courbe, Qui Remplit Toute Une Aire Plane", Mathematische Annalen, volume 36, number 1, 1890, p157-160. DOI 10.1007/BF01199438. http://www.springerlink.com/content/w232301n53960133/
It traverses a quadrant of the plane one step at a time in a self-similar 3x3 pattern,
8 60--61--62--63--64--65 78--79--80--... | | | 7 59--58--57 68--67--66 77--76--75 | | | 6 54--55--56 69--70--71--72--73--74 | 5 53--52--51 38--37--36--35--34--33 | | | 4 48--49--50 39--40--41 30--31--32 | | | 3 47--46--45--44--43--42 29--28--27 | 2 6---7---8---9--10--11 24--25--26 | | | 1 5---4---3 14--13--12 23--22--21 | | | 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 N=0 to N=8, and then nine of those groups are put together in the same S configuration. The sub-parts are flipped horizontally and/or vertically to make the starts and ends adjacent, so 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 | | 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 | | 6,7,8 ---- 9,10,11 24,25,26 3,4,5 12,13,14 23,22,21 0,1,2 15,16,17 --- 18,19,20
The 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()
.
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.
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, 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. Georg 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.)
Moore in
E. H. Moore, "On Certain Crinkly Curves", Trans. Am. Math. Soc., volume 1, number 1, 1900, pages 72-90.
http://www.ams.org/journals/tran/1900-001-01/S0002-9947-1900-1500526-4/ http://www.ams.org/tran/1900-001-01/S0002-9947-1900-1500526-4/S0002-9947-1900-1500526-4.pdf
http://www.ams.org/journals/tran/1900-001-04/S0002-9947-1900-1500428-3/ http://www.ams.org/journals/tran/1900-001-04/S0002-9947-1900-1500428-3/S0002-9947-1900-1500428-3.pdf
draws the curve as a base shape
+-----+ | | -----+-----+----- | | +-----+
with each line segment replaced by the same for the next level (with suitable mirror image in odd segments).
The is equivalent to the square form by drawing diagonal lines alternately in the direction of the leading diagonal or opposite diagonal, per the ".." marked lines in the following.
+--------+--------+--------+ +--------+--------+--------+ | .. | .. | .. | | | | | |6 .. |7 .. |8 .. | | 6--------7--------8 | | .. | .. | .. | | | | | | +--------+--------+--------+ +----|---+--------+--------+ | .. | .. | .. | | | | | | | .. 5| .. 4| .. 3| | 5--------4--------3 | | .. | .. | .. | | | | | | +--------+--------+--------+ +--------+--------+----|---+ | .. | .. | .. | | | | | | |0 .. |1 .. |2 .. | | 0--------1--------2 | | .. | .. | .. | | | | | +--------+--------+--------+ +--------+--------+--------+ X==Y mod 2 "even" points leading-diagonal "/" X!=Y mod 2 "odd" points opposite-diagonal "\"
Rounding off the corners of the diagonal form so they don't touch can help show the equivalence,
-----7 / / \ / 6 -----8 | | 4----- \ / \ 5----- 3 | -----1 | / \ / 0 -----2
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
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::PeanoCurve->new ()
$path = Math::PlanePath::PeanoCurve->new (radix => $integer)
Create and return a new path object.
The optional radix
parameter gives the base for digit splitting. The default is ternary radix => 3
.
($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. 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 $x,$y
within 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 $x1
,$y1
and $x2
,$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_lo
and $n_hi
are the smallest and biggest in the rectangle.
($n_lo, $n_hi) = $path->level_to_n_range($level)
Return (0, $radix**(2*$level) - 1)
.
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 is appended to Y then the next appended to X. Beginning at an even digit position in N makes the last digit go to X so the first N=0,1,2 goes 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 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 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 either before or after is fine, 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. When 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, but instead peel off digits of N from the low end. But the subtract to complement is then more work if using bignums.
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 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 ternary equivalent of an XOR in binary. On a ternary machine some trit-twiddling could no doubt do it.
For the 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.
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
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
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.
This path is in Sloane's Online Encyclopedia of Integer Sequences in several forms,
http://oeis.org/A163528 (etc)
A163528 X coordinate A163529 Y coordinate A163530 X+Y coordinate sum A163531 X^2+Y^2 square of distance from origin A163532 dX, change in X -1,0,1 A163533 dY, change in Y -1,0,1 A014578 abs(dX) from n-1 to n, 1=horiz 0=vertical thue-morse count low 0-bits + 1 mod 2 A182581 abs(dY) from n-1 to n, 0=horiz 1=vertical thue-morse count low 0-bits mod 2 A163534 direction of each step (up,down,left,right) A163535 direction, transposed X,Y A163536 turn 0=straight,1=right,2=left A163537 turn, transposed X,Y A163342 diagonal sums A163479 diagonal sums divided by 6 A163480 N on X axis A163481 N on Y axis A163343 N on X=Y diagonal, 0,4,8,44,40,36,etc A163344 N on X=Y diagonal divided by 4 A007417 N+1 of positions of horizontals, ternary even trailing 0s A145204 N+1 of positions of verticals, ternary odd trailing 0s A163332 Peano N -> ZOrder radix=3 N mapping and vice versa since is self-inverse A163333 with ternary digit swaps before and after
And taking X,Y points by the Diagonals sequence, then the 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=1
Math::PlanePath::Diagonals
numbers points 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 A163340
Math::PlanePath, Math::PlanePath::HilbertCurve, Math::PlanePath::ZOrderCurve, Math::PlanePath::AR2W2Curve, Math::PlanePath::BetaOmega, Math::PlanePath::CincoCurve, Math::PlanePath::KochelCurve, Math::PlanePath::WunderlichMeander
http://user42.tuxfamily.org/math-planepath/index.html
Copyright 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017 Kevin Ryde
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