Question

I am reading a algorithms book by S.DasGupta. Following is text snippet from the text regarding number of bits required for nth Fibonacci number.

It is reasonable to treat addition as
a single computer step if small
numbers are being added, 32-bit
numbers say. But the nth Fibonacci
number is about
0.694n bits long, and this can far exceed 32 as n grows. Arithmetic
operations on arbitrarily large
numbers cannot possibly be performed
in a single, constant-time step.

My question is for eg, for Fibonacci number F1 = 1, F2 =1, F3=2, and so on. then substituting “n” in above formula i.e., 0.694n for F1 is approximately 1, F2 is approximately 2 bits, but for F3 and so on above formula fails. I think i didn’t understand propely what author mean here, can any one please help me in understanding this?

Thanks

Answer 1

Well,

n              3    4     5     6     7     8
0.694n         2.08 2.78  3.47  4.16  4.86  5.55
F(n)           2    3     5     8     13    21
bits           2    2     3     4     4     5
log(F(n))      1    1.58  2.32  3     3.7   4.39

Bits required is the base-2 log rounded up, so this is close enough for me.

The value 0.694 comes from the fact that F(n) is the closest integer to (φⁿ)/√5. So log(F(n)) is n * log(phi) - log(sqrt(5)), and log(phi) is 0.694. As n gets bigger, the log(sqrt(5)) and the rounding rapidly become insignificant.