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| References | (1 earlier) <CAPTjJmrRv+DQGpzdY5te-vk_u9p5AzWYQ-6B=m1knPoaXpRmRQ@mail.gmail.com> <55105EF4.2070805@davea.name> <meq5i2$53s$1@ger.gmane.org> <mailman.93.1427174452.10327.python-list@python.org> <8a6dcb51-250d-4abc-abc9-68798f07af86@googlegroups.com> |
|---|---|
| From | Ian Kelly <ian.g.kelly@gmail.com> |
| Date | 2015-03-24 08:23 -0600 |
| Subject | Re: fibonacci series what Iam is missing ? |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.103.1427207059.10327.python-list@python.org> (permalink) |
On Tue, Mar 24, 2015 at 12:20 AM, Rustom Mody <rustompmody@gmail.com> wrote:
> On Tuesday, March 24, 2015 at 10:51:11 AM UTC+5:30, Ian wrote:
>> Iteration in log space. On my desktop, this calculates fib(1000) in
>> about 9 us, fib(100000) in about 5 ms, and fib(10000000) in about 7
>> seconds.
>>
>> def fib(n):
>> assert n >= 0
>> if n == 0:
>> return 0
>>
>> a = b = x = 1
>> c = y = 0
>> n -= 1
>>
>> while True:
>> n, r = divmod(n, 2)
>> if r == 1:
>> x, y = x*a + y*b, x*b + y*c
>> if n == 0:
>> return x
>> b, c = a*b + b*c, b*b + c*c
>> a = b + c
>
> This is rather arcane!
> What are the identities used above?
It's essentially the same matrix recurrence that Gregory Ewing's
solution uses, but without using numpy (which doesn't support
arbitrary precision AFAIK) and with a couple of optimizations.
The Fibonacci recurrence can be expressed using linear algebra as:
F_1 = [ 1 0 ]
T = [ 1 1 ]
[ 1 0 ]
F_(n+1) = F_n * T
I.e., given that F_n is a vector containing fib(n) and fib(n-1),
multiplying by the transition matrix T results in a new vector
containing fib(n+1) and fib(n). Therefore:
F_n = F_1 * T ** (n-1)
The code above evaluates this expression by multiplying F_1 by powers
of two of T until n-1 is reached. x and y are the two elements of the
result vector, which at the end of the loop are fib(n) and fib(n-1).
a, b, and c are the three elements of the (symmetric) transition
matrix T ** p, where p is the current power of two.
The last two lines of the loop updating a, b, and c could equivalently
be written as:
a, b, c = a*a + b*b, a*b + b*c, b*b + c*c
A little bit of algebra shows that if a = b + c before the assignment,
the equality is maintained after the assignment (in fact the elements
of T ** n are fib(n+1), fib(n), and fib(n-1)), so the two
multiplications needed to update a can be optimized away in favor of a
single addition.
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Re: fibonacci series what Iam is missing ? Ian Kelly <ian.g.kelly@gmail.com> - 2015-03-23 23:20 -0600
Re: fibonacci series what Iam is missing ? Rustom Mody <rustompmody@gmail.com> - 2015-03-23 23:20 -0700
Re: fibonacci series what Iam is missing ? Rustom Mody <rustompmody@gmail.com> - 2015-03-23 23:28 -0700
Unicode magic (was fibonacci series what Iam is missing ?) Rustom Mody <rustompmody@gmail.com> - 2015-03-23 23:33 -0700
Re: fibonacci series what Iam is missing ? Ian Kelly <ian.g.kelly@gmail.com> - 2015-03-24 08:23 -0600
Re: fibonacci series what Iam is missing ? CHIN Dihedral <dihedral88888@gmail.com> - 2015-03-24 10:22 -0700
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