Re: Help with permiutations within Fibonacci binary tree
- From: alex_patrick@xxxxxxxxxxxx
- Date: 7 Aug 2006 15:37:17 -0700
Hi Dave,
Thanks very much for your reply. I've tried to clarify your points
raised below:
Permutations of how many ways of ordering values, in the binary tree,Take the following number sequence, as a binary tree:
266
|
164 102
|
102 63
266= 1829/2+3Φ
164= 1829/3+5Φ
102= 1829/5+8Φ
63= 1829/8+13Φ
Φ=(1+√5)/2
To be precise, and using notation more familiar to some of us,
Phi = (1 + sqrt(5))/2;
266 = round(1829/(2 + 3*Phi));
164 = round(1829/(3 + 5*Phi));
102 = round(1829/(5 + 8*Phi));
63 = round(1829/(8 + 13*Phi));
where round(x) is the nearest integer to x.
Number of Permutations=1!+2!+3!/2! (2 of 102) = 6
To enumerate:
1 266
2 164,102
3 102,164
4 102,63,102
5 102,102,63
6 63,102,102
Permutations of *what*?
that add up to the root. Might be worth looking at
www.blaise.demon.co.uk/modulor
my microsite, not finish yet, gives some background about what I'm
trying to achieve.
It is not clear to me what you
have enumerated.What I've enumerated is sets of values that add up to the root.
Importantly the arrangements in each set are unique and distict from
one another. So patterns of values, each pattern of which adds up to
the root value is generated.
Also, how are you deciding on the binary
trees?
The pattern I would guess at is that the root of the
tree is of the form
round(N/(f[0] + f[1]*Phi))
The root would be
round(N/f[0]+f[1])
so for N=1829, round(1829/f[0]+f[1])=1829
then next in the sequence would be
round(N/(f[0] + f[1]*Phi))=1130
Your assumption here is correct:
where f0 and f1 are the initial values for your Fibonacci sequence,
f[n] = f[n-1] + f[n-2], n >= 2. If a tree node is
round(N/(f[i] + f[i+1]*Phi))
the left child of the node is
round(N/(f[i+1] + f[i+2]*Phi))
and the right child of the node is
round(N/(f[i+2] + f[i+3]*Phi))
Even with this pattern in mind, it is not clear how you decide102 and 63 are values in the measurement scale, and I've stopped at
what the leaf nodes are. In you example, why does the leaf
node 102 not have children?
these values because, in this example I don't want to work with any
measure smaller than 102mm and 63mm. I could of course just keep going,
but 63mm is my smallest measure.
If on the leaf nodes are the values: 102, 63, 102. Label 102(1), 63
A further example problem is:
How many unlike arrangements could be made from elements of the
following binary tree:
102(2) to distiguish the two 102mm values, so I'd get the following
possible arrangements of values:
102(1), 63, 102(2)
102(2), 63, 102(1)
63, 102(2), 102(1)
63, 102(1), 102(2)
102(2), 102(1), 63
102(1), 102(2), 63
BUT as far as I'm concerned here, arrangements like:
102(2), 102(1), 63
102(1), 102(2), 63
are the same - taking away the distiguishing label in brackets, are
identitical and LIKE, so thats what I mean by UNLIKE arrangements, IE:
102, 63, 102
63, 102, 102
102, 102, 63
is all I want here.
What is your definition for "arrangement" and "unlike arrangements"?
See above
Some clarification of your problem statement will help...
Hope this helps. Many thanks
Alex
--
Dave Eberly
http://www.geometrictools.com
.
- Follow-Ups:
- Re: Help with permiutations within Fibonacci binary tree
- From: Dave Eberly
- Re: Help with permiutations within Fibonacci binary tree
- References:
- Help with permiutations within Fibonacci binary tree
- From: alex_patrick
- Re: Help with permiutations within Fibonacci binary tree
- From: Dave Eberly
- Help with permiutations within Fibonacci binary tree
- Prev by Date: Re: Colour algorithms - how light affects an object
- Next by Date: Re: Help with permiutations within Fibonacci binary tree
- Previous by thread: Re: Help with permiutations within Fibonacci binary tree
- Next by thread: Re: Help with permiutations within Fibonacci binary tree
- Index(es):
Relevant Pages
|
