On Tue, Mar 24, 2015 at 12:20 AM, Rustom Mody <rustompm...@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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