# Python's Recursive Algorithm Learning (2): Concrete implementation: Fibonacci and the traps therein

Source: Internet
Author: User

1. Fibonacci

What is Fibonacci, the Fibonacci amount is a sequence of integers sorted by the definition as follows;

`Fn = Fn-1 + Fn-2`
`F0 = 0 and F1 = 1`with

That is, 0,1,1,2,3,5,8,13 .....

Recursive implementations:

`def fib (n):     if n = = 0        :return  0    elif n = = 1        :return 1    else:        return fib (n-1) + fib (n-2)`

Non-recursive implementations:

`def Fibi (n):     = 0, 1 for in      Range (n):        = B, A + b    return A`

Here, if we carefully debug, we will find that the recursive implementation, will consume more time, here the test is as follows:

` fromTimeitImportTimer fromFiboImportfibt1= Timer ("fib (Ten)","From Fibo import fib") forIinchRange (1,41): S="fib ("+ STR (i) +")"T1= Timer (S,"From Fibo import fib") time1= T1.timeit (3) s="Fibi ("+ STR (i) +")"T2= Timer (S,"From Fibo import Fibi") time2= T2.timeit (3)    Print("n=%2d, fib:%8.6f, Fibi:%7.6f, Percent:%10.2f"% (i, time1, time2, time1/time2))`

The results are as follows;

`C:\Python35\python.exe C:\pylearn\bottlelearn\fibnaqie.pyn= 1, fib:0.000002, fibi:0.000003, percent:0.55N= 2, fib:0.000002, fibi:0.000003, percent:0.73N= 3, fib:0.000003, fibi:0.000003, percent:1.20N= 4, fib:0.000005, fibi:0.000003, percent:1.80N= 5, fib:0.000007, fibi:0.000003, percent:2.45N= 6, fib:0.000012, fibi:0.000004, percent:3.31N= 7, fib:0.000022, fibi:0.000003, percent:6.50N= 8, fib:0.000030, fibi:0.000004, percent:8.38N= 9, fib:0.000049, fibi:0.000003, percent:14.58N=10, fib:0.000078, fibi:0.000004, percent:20.07N=11, fib:0.000126, fibi:0.000004, percent:35.00N=12, fib:0.000203, fibi:0.000004, percent:52.29N=13, fib:0.000330, fibi:0.000004, percent:79.20N=14, fib:0.000537, fibi:0.000004, percent:120.94N=15, fib:0.000925, fibi:0.000005, percent:196.18N=16, fib:0.001693, fibi:0.000007, percent:244.16N=17, fib:0.007242, fibi:0.000011, percent:652.70N=18, fib:0.004422, fibi:0.000007, percent:637.64N=19, fib:0.010322, fibi:0.000008, percent:1240.40N=20, fib:0.010813, fibi:0.000007, percent:1443.78N=21, fib:0.025547, fibi:0.000011, percent:2246.24N=22, fib:0.035421, fibi:0.000011, percent:3192.22N=23, fib:0.060877, fibi:0.000008, percent:7837.86N=24, fib:0.090561, fibi:0.000007, percent:12091.41N=25, fib:0.132058, fibi:0.000008, percent:16415.97N=26, fib:0.227813, fibi:0.000008, percent:29330.54N=27, fib:0.557644, fibi:0.000014, percent:38659.17N=28, fib:0.578976, fibi:0.000009, percent:65224.25N=29, fib:1.133326, fibi:0.000008, percent:140882.03N=30, fib:1.454107, fibi:0.000009, percent:169095.94N=31, fib:2.274395, fibi:0.000009, percent:264486.10N=32, fib:3.817956, fibi:0.000009, percent:430110.09N=33, fib:5.923710, fibi:0.000009, percent:688859.58N=34, fib:9.629423, fibi:0.000009, percent:1020986.44N=35, fib:15.910085, fibi:0.000009, percent:1686911.18N=36, fib:26.556680, fibi:0.000009, percent:2901071.88N=37, fib:43.014073, fibi:0.000016, percent:2673506.31`

Why is that? Let us consider the calculation of the recursive process, as follows;

As can be seen from the tree, there are repeated calculations, (f (1) calculated many times.

Here, try to make an improvement, introduce a temporary dictionary, and make a save of the computed content:

`Memo = {0:0, 1:1}def  FIBM (n):    if not in memo:         = FIBM (n-1) + FIBM (n-2)    return Memo[n]`

To re-test the time:

` fromTimeitImportTimer fromFiboImportfibt1= Timer ("fib (Ten)","From Fibo import fib") forIinchRange (1,41): S="FIBM ("+ STR (i) +")"T1= Timer (S,"From Fibo import FIBM") time1= T1.timeit (3) s="Fibi ("+ STR (i) +")"T2= Timer (S,"From Fibo import Fibi") time2= T2.timeit (3)    Print("n=%2d, fib:%8.6f, Fibi:%7.6f, Percent:%10.2f"% (i, time1, time2, time1/time2))`
`N= 1, fib:0.000011, fibi:0.000015, percent:0.73N= 2, fib:0.000011, fibi:0.000013, percent:0.85N= 3, fib:0.000012, fibi:0.000014, percent:0.86N= 4, fib:0.000012, fibi:0.000015, percent:0.79N= 5, fib:0.000012, fibi:0.000016, percent:0.75N= 6, fib:0.000011, fibi:0.000017, percent:0.65N= 7, fib:0.000012, fibi:0.000017, percent:0.72N= 8, fib:0.000011, fibi:0.000018, percent:0.61N= 9, fib:0.000011, fibi:0.000018, percent:0.61N=10, fib:0.000010, fibi:0.000020, percent:0.50N=11, fib:0.000011, fibi:0.000020, percent:0.55N=12, fib:0.000004, fibi:0.000007, percent:0.59N=13, fib:0.000004, fibi:0.000007, percent:0.57N=14, fib:0.000004, fibi:0.000008, percent:0.52N=15, fib:0.000004, fibi:0.000008, percent:0.50N=16, fib:0.000003, fibi:0.000008, percent:0.39N=17, fib:0.000004, fibi:0.000009, percent:0.45N=18, fib:0.000004, fibi:0.000009, percent:0.45N=19, fib:0.000004, fibi:0.000009, percent:0.45N=20, fib:0.000003, fibi:0.000010, percent:0.29N=21, fib:0.000004, fibi:0.000009, percent:0.45N=22, fib:0.000004, fibi:0.000010, percent:0.40N=23, fib:0.000004, fibi:0.000010, percent:0.40N=24, fib:0.000004, fibi:0.000011, percent:0.35N=25, fib:0.000004, fibi:0.000012, percent:0.33N=26, fib:0.000004, fibi:0.000011, percent:0.34N=27, fib:0.000004, fibi:0.000011, percent:0.35N=28, fib:0.000004, fibi:0.000012, percent:0.32N=29, fib:0.000004, fibi:0.000012, percent:0.33N=30, fib:0.000004, fibi:0.000013, percent:0.31N=31, fib:0.000004, fibi:0.000012, percent:0.34N=32, fib:0.000004, fibi:0.000012, percent:0.33N=33, fib:0.000004, fibi:0.000013, percent:0.30N=34, fib:0.000004, fibi:0.000012, percent:0.34N=35, fib:0.000004, fibi:0.000013, percent:0.31N=36, fib:0.000004, fibi:0.000013, percent:0.31N=37, fib:0.000004, fibi:0.000014, percent:0.29N=38, fib:0.000004, fibi:0.000014, percent:0.29N=39, fib:0.000004, fibi:0.000013, percent:0.31N=40, fib:0.000004, fibi:0.000014, percent:0.29`

Python's Recursive Algorithm Learning (2): Concrete implementation: Fibonacci and the traps therein

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