NAME
Math::PlanePath::RationalsTree -- rationals by tree
SYNOPSIS
use Math::PlanePath::RationalsTree;
my $path = Math::PlanePath::RationalsTree->new (tree_type => 'SB');
my ($x, $y) = $path->n_to_xy (123);
DESCRIPTION
This path enumerates reduced rational fractions X/Y > 0, ie. X and Y having no common factor.
The rationals are traversed by rows of a binary tree which effectively represents a coprime pair X,Y by steps of a subtraction-only greatest common divisor algorithm which proves them coprime. Or equivalently by bit runs with lengths which are the quotients in the division based Euclidean GCD algorithm, which are also the terms in the continued fraction representation of X/Y.
The SB, CW, AYT, HCS, Bird and Drib trees all have the same set of X/Y rationals in a row, but in a different order due to different encodings of the N value, either high to low or low to high and some bit flips. The L tree has a shift which visits 0/1 too.
The bit runs mean that N values are quite large for relatively modest sized rationals. For example in the SB tree 167/3 is N=288230376151711741, a 58-bit number. The tendency is for the tree to make excursions out to large rationals while only slowly filling in small ones. The worst is the integer X/1 which is an N with X many bits, and similarly 1/Y is Y bits.
See examples/rationals-tree.pl in the Math-PlanePath sources for a printout of all the trees.
Stern-Brocot Tree
The default tree_type=>"SB"
is the tree of Moritz Stern and Achille Brocot. The rows are fractions of increasing value.
N depth
1/1 1 0
------ ------
1/2 2/1 2,3 1
/ \ / \
1/3 2/3 3/2 3/1 4 to 7 2
| | | | | | | |
1/4 2/5 3/5 3/4 4/3 5/3 5/2 4/1 8 to 15 3
Each row of the tree is a repeat of the previous row, first as X/(X+Y) and then (X+Y)/Y. For example
depth=1 row 1/2, 2/1
depth=2 row 1/3, 2/3 X/(X+Y) of previous row
and 3/2, 3/1 (X+Y)/Y of previous row
Plotting the N values by X,Y is as follows. The unused X,Y positions are where X and Y have a common factor. For example X=6,Y=2 has common factor 2 so is never reached.
tree_type => "SB"
10 | 512 35 44 767
9 | 256 33 39 40 46 383 768
8 | 128 18 21 191 384
7 | 64 17 19 20 22 95 192 49 51
6 | 32 47 96
5 | 16 9 10 23 48 25 26 55
4 | 8 11 24 27 56
3 | 4 5 12 13 28 29 60
2 | 2 6 14 30 62
1 | 1 3 7 15 31 63 127 255 511 1023
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
The X=1 vertical is the fractions 1/Y which is at the left of each tree row, at N value
Nstart = 2^level
The Y=1 horizontal is the X/1 integers at the end each row which is
Nend = 2^(level+1)-1
Stern-Brocot Turn Sequence
Each row makes a path from the Y axis down to the X. Each row is further from the origin than the previous row and doesn't intersect any other row. The X/(X+Y) first half described is an upward "shear" to the X,Y points of a given row. Similarly the second half (X+Y)/Y shears to the right. For example,
N=8 to N=11
previous row
row sheared up X,X+Y
depth=2 N=4to7 row
| | 9--10 . depth=3 N=8to15
| | / | .
| | 8 11 .
| | .
| 4---5 | . 12--13 N=12 to N=15
| \ | . | previous row
| 6 | . 14 sheared right
| | | . / as X+Y,Y
| 7 | 15
| |
+--------------- +----------------
The sequence of turns left or right is unchanged by the shears. So at N=5 the path turns towards the right and this is unchanged in the sheared copies at N=9 and copy at N=13. The angle of the turn is different, but it's still to the right.
The first and last points of each row are always a turn to the right. For example the turn at N=4 (going N=3 to N=4 to N=5) is to the right, and likewise at N=7. This is because the second of the row such as N=5 is above a 45-degree line down from N=4, and similarly the second last such as N=6.
The middle two points in each row of depth>=3 are always a turn to the left. This happens first at N=11 and N=12 shown above which both turn to the left. This is because the middle two make a 45-degree line and the second-from-middle points are above that line (N=10 and N=13).
The middle left turns are copied into successive rows and the result is a repeating pattern "LRRL" except the first and last in the row always right instead of left.
N=3 left
otherwise if N=2^k or N=2^k-1 right
otherwise if N=0 mod 4 left
N=1 mod 4 right
N=2 mod 4 right
N=3 mod 4 left
Pairs N=2m and N=2m-1 can be treated together by taking ceil(N/2),
N=3 left
otherwise if Nhalf=2^k right
otherwise if Nhalf=0 mod 2 left
otherwise if Nhalf=1 mod 2 right
where Nhalf = ceil(N/2)
Stern-Brocot Mediant
Writing the parents between the children as an "in-order" tree traversal to a given depth has all values in increasing order, the same as each row individually is in increasing order.
1/1
1/2 | 2/1
1/3 | 2/3 | 3/2 | 3/1
| | | | | | |
1/3 1/2 2/3 1/1 3/2 2/1 3/1
^
|
next level (1+3)/(1+2) = 4/3 mediant
New values at the next level of this flattening are a "mediant" (x1+x2)/(y1+y2) formed from the left and right parent. So the next level 4/3 shown is left parent 1/1 and right parent 3/2 giving mediant (1+3)/(1+2)=4/3. At the left end a preceding 0/1 is imagined. At the right a following 1/0 is imagined, so as to have 1/(depth+1) and (depth+1)/1 at the ends for a total 2^depth many new values.
Calkin-Wilf Tree
tree_type=>"CW"
selects the tree of Neil Calkin and Herbert Wilf,
"Recounting the Rationals", http://www.math.upenn.edu/~wilf/website/recounting.pdf
As noted above, the values within each row are the same as the Stern-Brocot, but in a different order.
N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 1/3 3/2 2/3 3/1
| | | | | | | |
N=8 to N=15 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1
Going by rows the denominator of one value becomes the numerator of the next. So at 4/3 the denominator 3 becomes the numerator of 3/5 to the right. These values are Stern's diatomic sequence.
Each row is symmetric in reciprocals, ie. reading from right to left is the reciprocals of reading left to right. The numerators read left to right are the denominators read right to left.
A node descends as
X/Y
/ \
X/(X+Y) (X+Y)/Y
Taking these formulas in reverse up the tree shows how it relates to a subtraction-only greatest common divisor. At a given node the smaller of P or Q is subtracted from the bigger,
P/(Q-P) (P-Q)/P
/ or \
P/Q P/Q
Plotting the N values by X,Y is as follows. The X=1 vertical and Y=1 horizontal are the same as the SB above, but the values in between are re-ordered.
tree_type => "CW"
10 | 512 56 38 1022
9 | 256 48 60 34 46 510 513
8 | 128 20 26 254 257
7 | 64 24 28 18 22 126 129 49 57
6 | 32 62 65
5 | 16 12 10 30 33 25 21 61
4 | 8 14 17 29 35
3 | 4 6 9 13 19 27 39
2 | 2 5 11 23 47
1 | 1 3 7 15 31 63 127 255 511 1023
Y=0 |
-------------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
At each node the left leg is X/(X+Y) < 1 and the right leg is (X+Y)/Y > 1, which means N is even above the X=Y diagonal and odd below. In general each right leg increments the integer part of the fraction,
X/Y right leg each time
(X+Y)/Y = 1 + X/Y
(X+2Y)/Y = 2 + X/Y
(X+3Y)/Y = 3 + X/Y
etc
This means the integer part is the trailing 1-bits of N,
floor(X/Y) = count trailing 1-bits of N
eg. 7/2 is at N=23 binary "10111"
which has 3 trailing 1-bits for floor(7/2)=3
N values for the SB and CW trees are converted by reversing bits except the highest. So at a given X,Y position
SB N = 1abcde SB <-> CW by reversing bits
CW N = 1edcba except the high 1-bit
For example at X=3,Y=4 the SB tree has N=11 = "1011" binary and the CW has N=14 binary "1110", a reversal of the bits below the high 1.
N to X/Y in the CW tree can be calculated keeping track of just an X,Y pair and descending to X/(X+Y) or (X+Y)/Y using the bits of N from high to low. The relationship between the SB and CW N's means the same can be used to calculate the SB tree by taking the bits of N from low to high instead.
Andreev and Yu-Ting Tree
tree_type=>"AYT"
selects the tree described (independently is it?) by D. N. Andreev and Shen Yu-Ting.
http://files.school-collection.edu.ru/dlrstore/d62f7b96-a780-11dc-945c-d34917fee0be/i2126134.pdf
Shen Yu-Ting, "A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion", American Mathematical Monthly, 87, 1980, pages 25-29. http://www.jstor.org/stable/2320374
Their constructions are a one-to-one mapping between integer N and rational X/Y as a way of enumerating the rationals. It's not designed to be a tree as such, but the result is the same 2^level rows as the above trees. The X/Y values within each row are again the same, but in a further different order.
N=1 1/1
------ ------
N=2 to N=3 2/1 1/2
/ \ / \
N=4 to N=7 3/1 1/3 3/2 2/3
| | | | | | | |
N=8 to N=15 4/1 1/4 4/3 3/4 5/2 2/5 5/3 3/5
Each fraction descends as follows. The left is an increment and the right is reciprocal of the increment.
X/Y
/ \
X/Y + 1 1/(X/Y + 1)
which means
X/Y
/ \
(X+Y)/Y Y/(X+Y)
The left leg (X+Y)/Y is the same the CW has on its right leg. But Y/(X+Y) is not the same as the CW (the other there being X/(X+Y)).
The left leg increments the integer part, so the integer part is given by (in a fashion similar to CW 1-bits above)
floor(X/Y) = count trailing 0-bits of N
plus one extra if N=2^k
N=2^k is one extra because its trailing 0-bits started from N=1 where floor(1/1)=1 whereas any other odd N starts from some floor(X/Y)=0.
The Y/(X+Y) right leg forms the Fibonacci numbers F(k)/F(k+1) at the end of each row, ie. at Nend=2^(level+1)-1. And as noted by Andreev, successive right leg fractions N=4k+1 and N=4k+3 add up to 1,
X/Y at N=4k+1 + X/Y at N=4k+3 = 1
Eg. 2/5 at N=13 and 3/5 at N=15 add up to 1
Plotting the N values by X,Y gives
tree_type => "AYT"
10 | 513 41 43 515
9 | 257 49 37 39 51 259 514
8 | 129 29 31 131 258
7 | 65 25 21 23 27 67 130 50 42
6 | 33 35 66
5 | 17 13 15 19 34 26 30 38
4 | 9 11 18 22 36
3 | 5 7 10 14 20 28 40
2 | 3 6 12 24 48
1 | 1 2 4 8 16 32 64 128 256 512
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
N=1,2,4,8,etc on the Y=1 horizontal is the X/1 integers at Nstart=2^level=2^X. N=1,3,5,9,etc in the X=1 vertical is the 1/Y fractions. Those fractions always immediately follow the corresponding integer, so N=Nstart+1=2^(Y-1)+1 in that column.
In each node the left leg (X+Y)/Y > 1 and the right leg Y/(X+Y) < 1, which means odd N is above the X=Y diagonal and even N is below.
The tree structure corresponds to Johannes Kepler's tree of fractions (see Math::PlanePath::FractionsTree). That tree starts from 1/2 and makes fractions A/B with A<B by descending to A/(A+B) and B/(A+B). Those descents are the same as the AYT tree and the two are related simply by
A = Y AYT denominator is Kepler numerator
B = X+Y AYT sum num+den is the Kepler denominator
X = B-A inverse
Y = A
HCS Continued Fraction
tree_type=>"HCS"
selects continued fraction terms coded as bit runs 1000...00 from high to low, as per Paul D. Hanna and independently Jerzy Czyz and Will Self.
http://oeis.org/A071766
http://www.cut-the-knot.org/do_you_know/countRatsCF.shtml http://www.dm.unito.it/~cerruti/doc-html/tremattine/tre_mattine.pdf
Jerzy Czyz and William Self, "The Rationals Are Countable: Euclid's Proof", The College Mathematics Journal, volume 34, number 5, November 2003, page 367.
This arises also in a radix=1 variation of Jeffrey Shallit's digit-based continued fraction encoding. See "Radix 1" in Math::PlanePath::CfracDigits.
If the continued fraction of X/Y is
1
X/Y = a + ------------ a >= 0
1
b + ----------- b,c,etc >= 1
1
c + -------
... + 1
--- z >= 2
z
then the N value is bit runs of lengths a,b,c etc.
N = 1000 1000 1000 ... 1000
\--/ \--/ \--/ \--/
a+1 b c z-1
Each group is 1 or more bits. The +1 in "a+1" makes the first group 1 or more bits, since a=0 occurs for any X/Y<=1. The -1 in "z-1" makes the last group 1 or more since z>=2.
N=1 1/1
------ ------
N=2 to N=3 2/1 1/2
/ \ / \
N=4 to N=7 3/1 3/2 1/3 2/3
| | | | | | | |
N=8 to N=15 4/1 5/2 4/3 5/3 1/4 2/5 3/4 3/5
The result is a bit reversal of the N values in the AYT tree.
AYT N = binary "1abcde" AYT <-> HCS bit reversal
HCS N = binary "1edcba"
For example at X=4,Y=7 the AYT tree is N=11 binary "10111" whereas HCS there has N=30 binary "11110", a reversal of the bits below the high 1.
Plotting by X,Y gives
tree_type => "HCS"
10 | 768 50 58 896
9 | 384 49 52 60 57 448 640
8 | 192 27 31 224 320
7 | 96 25 26 30 29 112 160 41 42
6 | 48 56 80
5 | 24 13 15 28 40 21 23 44
4 | 12 14 20 22 36
3 | 6 7 10 11 18 19 34
2 | 3 5 9 17 33
1 | 1 2 4 8 16 32 64 128 256 512
Y=0 |
+-----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
N=1,2,4,etc in the row Y=1 are powers-of-2, being integers X/1 having just a single group of bits N=1000..000.
N=1,3,6,12,etc in the column X=1 are 3*2^(Y-1) corresponding to continued fraction 0 + 1/Y so terms 0,Y making runs 1,Y-1 and so bits N=11000...00.
Bird Tree
tree_type=>"Bird"
selects the Bird tree by Ralf Hinze
"Functional Pearls: The Bird tree", http://www.cs.ox.ac.uk/ralf.hinze/publications/Bird.pdf
It's expressed recursively, illustrating Haskell programming features. The left subtree is the tree plus one then take the reciprocal. The right subtree is conversely the tree reciprocal then plus one,
1/(tree + 1) and (1/tree) + 1
which means Y/(X+Y) and (X+Y)/X taking N bits low to high.
N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 2/3 1/3 3/1 3/2
| | | | | | | |
N=8 to N=15 3/5 3/4 1/4 2/5 5/2 4/1 4/3 5/3
Plotting by X,Y gives
tree_type => "Bird"
10 | 682 41 38 597
9 | 341 43 45 34 36 298 938
8 | 170 23 16 149 469
7 | 85 20 22 17 19 74 234 59 57
6 | 42 37 117
5 | 21 11 8 18 58 28 31 61
4 | 10 9 29 30 50
3 | 5 4 14 15 25 24 54
2 | 2 7 12 27 52
1 | 1 3 6 13 26 53 106 213 426 853
Y=0 |
----------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
Notice that unlike the other trees N=1,2,5,10,etc in the X=1 vertical of fractions 1/Y is not the row start or end, but instead are on a zigzag through the middle of the tree giving binary N=1010...etc with alternate 1 and 0 bits. The integers X/1 in the Y=1 vertical are similar, but N=11010...etc starting the alternation from a 1 in the second highest bit, since those integers are in the right hand half of the tree.
The Bird tree N values are related to the SB tree by inverting every second bit starting from the second after the high 1-bit,
Bird N=1abcdefg.. binary
101010.. xor, so b,d,f etc flip 0<->1
SB N=1aBcDeFg.. to make B,D,F
For example 3/4 in the SB tree is at N=11 = binary 1011. Xor with 0010 for binary 1001 N=9 which is 3/4 in the Bird tree. The same xor goes back the other way Bird tree to SB tree.
This xoring is a mirroring in the tree, swapping left and right at each level. Only every second bit is inverted because mirroring twice puts it back to the ordinary way on even rows.
Drib Tree
tree_type=>"Drib"
selects the Drib tree by Ralf Hinze.
http://oeis.org/A162911
It reverses the bits of N in the Bird tree (in a similar way that the SB and CW are bit reversals of each other).
N=1 1/1
------ ------
N=2 to N=3 1/2 2/1
/ \ / \
N=4 to N=7 2/3 3/1 1/3 3/2
| | | | | | | |
N=8 to N=15 3/5 5/2 1/4 4/3 3/4 4/1 2/5 5/3
The descendants of each node are
X/Y
/ \
Y/(X+Y) (X+Y)/X
The endmost fractions of each row are Fibonacci numbers, F(k)/F(k+1) on the left and F(k+1)/F(k) on the right.
tree_type => "Drib"
10 | 682 50 44 852
9 | 426 58 54 40 36 340 683
8 | 170 30 16 212 427
7 | 106 18 22 24 28 84 171 59 51
6 | 42 52 107
5 | 26 14 8 20 43 19 31 55
4 | 10 12 27 23 41
3 | 6 4 11 15 25 17 45
2 | 2 7 9 29 37
1 | 1 3 5 13 21 53 85 213 341 853
Y=0 |
-------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
In each node descent the left Y/(X+Y) < 1 and the right (X+Y)/X > 1, which means even N is above the X=Y diagonal and odd N is below.
Because Drib/Bird are bit reversals like CW/SB are bit reversals, the xor procedure described above which relates Bird<->SB applies to Drib<->CW, but working from the second lowest bit upwards, ie. xor binary "0..01010". For example 4/1 is at N=15 binary 1111 in the CW tree. Xor with 0010 for 1101 N=13 which is 4/1 in the Drib tree.
L Tree
tree_type=>"L"
selects the L-tree by Peter Luschny.
http://www.oeis.org/wiki/User:Peter_Luschny/SternsDiatomic
It's a row-reversal of the CW tree with a shift to include zero as 0/1.
N=0 0/1
------ ------
N=1 to N=2 1/2 1/1
/ \ / \
N=3 to N=8 2/3 3/2 1/3 2/1
| | | | | | | |
N=9 to N=16 3/4 5/3 2/5 5/2 3/5 4/3 1/4 3/1
Notice in the N=9 to N=16 row rationals 3/4 to 1/4 are the same as in the CW tree but read right-to-left.
tree_type => "L"
10 | 1021 37 55 511
9 | 509 45 33 59 47 255 1020
8 | 253 25 19 127 508
7 | 125 21 17 27 23 63 252 44 36
6 | 61 31 124
5 | 29 9 11 15 60 20 24 32
4 | 13 7 28 16 58
3 | 5 3 12 8 26 18 54
2 | 1 4 10 22 46
1 | 0 2 6 14 30 62 126 254 510 1022 2046
Y=0 |
-------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10
N=0,2,6,14,30,etc along the row at Y=1 are powers 2^(X+1)-2. N=1,5,13,29,etc in the column at X=1 are similar powers 2^Y-3.
Common Characteristics
The SB, CW, Bird, Drib, AYT and HCS trees have the same set of rationals in each row, just in different orders. The properties of Stern's diatomic sequence mean that within a row the totals are
row N=2^depth to N=2^(depth+1)-1 inclusive
sum X/Y = (3 * 2^depth - 1) / 2
sum X = 3^depth
sum 1/(X*Y) = 1
For example the SB tree depth=2, N=4 to N=7,
sum X/Y = 1/3 + 2/3 + 3/2 + 3/1 = 11/2 = (3*2^2-1)/2
sum X = 1+2+3+3 = 9 = 3^2
sum 1/(X*Y) = 1/(1*3) + 1/(2*3) + 1/(3*2) + 1/(3*1) = 1
Many permutations are conceivable within a row, but the ones here have some relationship to X/Y descendants, tree sub-forms or continued fractions. As an encoding of continued fraction terms by bit runs the combinations are
bit encoding high to low low to high
---------------- ----------- -----------
0000,1111 runs SB CW
0101,1010 alternating Bird Drib
1000,1000 runs HCS AYT
A run of alternating 101010 ends where the next bit is the oppose of the expected alternating 0,1. This is a doubled bit 00 or 11. An electrical engineer would think of it as a phase shift.
Minkowski Question Mark
The Minkowski question mark function is a sum of the terms in the continued fraction representation of a real number. If q0,q1,q2,etc are those terms then the question mark function "?(r)" is
1 1 1
?(r) = 2 * (1 - ---- * (1 - ---- * (1 - ---- * (1 - ...
2^q0 2^q1 2^q2
1 1 1
= 2 * (1 - ---- + --------- - ------------ + ... )
2^q0 2^(q0+q1) 2^(q0+q1+q2)
For rational r the continued fraction q0,q1,q2,etc is finite and so the ?(r) sum is finite and rational. The pattern of + and - in the terms gives runs of bits the same as the N values in the Stern-Brocot tree. The RationalsTree code can calculate the ?(r) function by
rational r=X/Y
N = xy_to_n(X,Y) tree_type=>"SB"
depth = floor(log2(N)) # row containing N (depth=0 at top)
Ndepth = 2^depth # start of row containing N
2*(N-Ndepth) + 1
?(r) = ----------------
Ndepth
The effect of N-Ndepth is to remove the high 1-bit, leaving an offset into the row. 2*(..)+1 appends an extra 1-bit at the end. The division by Ndepth scales down from integer N to a fraction.
N = 1abcdef integer, in binary
?(r) = a.bcdef1 binary fraction
For example ?(2/3) is X=2,Y=3 which is N=5 in the SB tree. It is at depth=2, Ndepth=2^2=4, and so ?(2/3)=(2*(5-4)+1)/4=3/4. Or written in binary N=101 gives Ndepth=100 and N-Ndepth=01 so 2*(N-Ndepth)+1=011 and divide by Ndepth=100 for ?=0.11.
In practice this is not a very efficient way to handle the question function, since the bit runs in the N values may become quite large for relatively modest fractions. (Math::ContinuedFraction may be better, and also allows repeating terms from quadratic irrationals to be represented exactly.)
FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::RationalsTree->new ()
$path = Math::PlanePath::RationalsTree->new (tree_type => $str)
-
Create and return a new path object.
tree_type
(a string) can be"SB" Stern-Brocot "CW" Calkin-Wilf "AYT" Andreev, Yu-Ting "HCS" "Bird" "Drib" "L"
$n = $path->n_start()
-
Return the first N in the path. This is 1 for SB, CW, AYT, HCS, Bird and Drib, but 0 for L.
($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
. The range is inclusive.For reference,
$n_hi
can be quite large because within each row there's only one new X/1 integer and 1/Y fraction. So if X=1 or Y=1 is included then roughly$n_hi = 2**max(x,y)
. If min(x,y) is bigger than 1 then it reduces a little to roughly 2**(max/min + min).
Tree Methods
Each point has 2 children, so the path is a complete binary tree.
@n_children = $path->tree_n_children($n)
-
Return the two children of
$n
, or an empty list if$n < 1
(ie. before the start of the path).This is simply
2*$n, 2*$n+1
. Written in binary the children are$n
with an extra bit appended, a 0-bit or a 1-bit. $num = $path->tree_n_num_children($n)
-
Return 2, since every node has two children. If
$n<1
, ie. before the start of the path, then returnundef
. $n_parent = $path->tree_n_parent($n)
-
Return the parent node of
$n
. Or returnundef
if$n <= 1
(the top of the tree).This is simply Nparent = floor(N/2), ie. strip the least significant bit from
$n
. (Undo whattree_n_children()
appends.) $depth = $path->tree_n_to_depth($n)
-
Return the depth of node
$n
, orundef
if there's no point$n
. The top of the tree at N=1 is depth=0, then its children depth=1, etc.This is simply floor(log2(N)) since the tree has 2 nodes per point. For example N=4 through N=7 are all depth=2.
The L tree starts at N=0 and the calculation becomes floor(log2(N+1)) there.
$n = $path->tree_depth_to_n($depth)
$n = $path->tree_depth_to_n_end($depth)
-
Return the first or last N at tree level
$depth
in the path, orundef
if nothing at that depth or not a tree. The top of the tree is depth=0.The structure of the tree means the first N is at
2**$depth
, or for the L tree2**$depth - 1
. The last N is2**($depth+1)-1
, or for the L tree2**($depth+1)
.
Tree Descriptive Methods
$num = $path->tree_num_children_minimum()
$num = $path->tree_num_children_maximum()
-
Return 2 since every node has 2 children, making that both the minimum and maximum.
$bool = $path->tree_any_leaf()
-
Return false, since there are no leaf nodes in the tree.
OEIS
The trees are in Sloane's Online Encyclopedia of Integer Sequences in various forms,
http://oeis.org/A007305 (etc)
A007305 SB X, Farey fractions (extra 0,1)
A047679 SB Y
A007306 SB X+Y sum, Farey 0 to 1 part (extra 1,1)
A153036 SB int(X/Y), integer part
A002487 CW X and Y, Stern diatomic sequence (extra 0)
A070990 CW Y-X diff, Stern diatomic first diffs (less 0)
A070871 CW X*Y product
A007814 CW int(X/Y), integer part, count trailing 1-bits
which is count trailing 0-bits of N+1
A020650 AYT X
A020651 AYT Y (Kepler numerator)
A086592 AYT X+Y sum (Kepler denominator)
A135523 AYT int(X/Y), integer part,
count trailing 0-bits plus 1 extra if N=2^k
A071585 HCS X+Y sum (X+Y giving rationals >= 1)
A071766 HCS Y
A162909 Bird X
A162910 Bird Y
A162911 Drib X
A162912 Drib Y
A174981 L-tree X
A002487 L-tree Y, same as CW X,Y, Stern diatomic
A000523 tree_n_to_depth(), being floor(log2(N))
A086893 position Fibonacci F[n+1],F[n] in Stern diatomic,
CW N of F[n+1]/F[n]
Drib N on row Y=1, being X/1
A061547 position Fibonacci F[n],F[n+1] in Stern diatomic,
CW N of F[n]/F[n+1]
Drib N in column X=1, being 1/Y
A081254 Bird N in row Y=1, binary 110101010...10
A000975 Bird N in column X=1, binary 1010..1010
A088696 length of continued fraction SB left half (X/Y<1)
A059893 permutation SB<->CW, AYT<->HCS, Bird<->Drib
reverse bits below highest
A153153 permutation CW->AYT, reverse and un-Gray
A153154 permutation AYT->CW, reverse and Gray code
A154437 permutation AYT->Drib, Lamplighter low to high
A154438 permutation Drib->AYT, un-Lamplighter low to high
A003188 permutation SB->HCS, Gray code shift+xor
A006068 permutation HCS->SB, Gray code inverse
A154435 permutation HCS->Bird, Lamplighter bit flips
A154436 permutation Bird->HCS, Lamplighter variant
A054429 permutation SB,CW,Bird,Drib N at transpose Y/X,
(mirror binary tree, runs 0b11..11 down to 0b10..00)
A004442 permutation AYT N at transpose Y/X, from N=2 onwards
(xor 1, ie. flip least significant bit)
A063946 permutation HCS N at transpose Y/X, extra initial 0
(xor 2, ie. flip second least significant bit)
A054424 permutation DiagonalRationals -> SB
A054426 permutation SB -> DiagonalRationals
A054425 DiagonalRationals -> SB with 0s at non-coprimes
A054427 permutation coprimes -> SB right hand X/Y>1
The sequences marked "extra ..." have one or two extra initial values over what the RationalsTree here gives, but are the same after that. And the Stern first differences "less ..." means it has one less term than what the code here gives.
SEE ALSO
Math::PlanePath, Math::PlanePath::FractionsTree, Math::PlanePath::CfracDigits, Math::PlanePath::CoprimeColumns, Math::PlanePath::DiagonalRationals, Math::PlanePath::FactorRationals, Math::PlanePath::GcdRationals, Math::PlanePath::PythagoreanTree
Math::NumSeq::SternDiatomic, Math::ContinuedFraction
HOME PAGE
http://user42.tuxfamily.org/math-planepath/index.html
LICENSE
Copyright 2011, 2012, 2013 Kevin Ryde
This file is part of Math-PlanePath.
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/>.