大O符号Python功能

Question

What is the big O of the foo(A) function (where n equals the length of A)? As far as i can tell the foo(4) statement is O(1) for each iteration of the recursion. Also i understand that the running time of the foo(A//8) statement will be logarithmic.

Therefore, will the running time for the program be bigO(log(n)) ?

This function is used for practicing running times for a test.

def foo(A):
    if A <= 6:
        return 7
    return foo(A//8) + foo(4)

Answer 1

Your program could be written as the following recurrsion:

T(n) = T(n/8) + C

Applying Master theorem where a=1 and b=8

We fall into the second case:

n^(log(1) base 8) = n^0 = 1
C = ϴ(1). 
==> T(n) = O(n^(log(a) base b) * log(n)) = O(n^(log(1) base 8) * log(n))
         = n^0 * log(n) = 1*log(n) = O(log(n))

Answer 2

A is not a vector but an integer hence there is no N in this problem and the complexity must be stated in terms of A .

Let us simulate the execution with A=1000 :

foo(1000) 
    calls foo(125) and foo(4), i.e.
    calls foo(15) and foo(4), and foo(4), i.e.
    calls foo(1) and foo(4), and foo(4), and foo(4).

You get the pattern. Thus the total number of indirect calls to foo equals the number of times you can divide A by 8 until it gets less or equal to 6 , plus the final call.

This is exactly Floor(Log8(Ceil(A/7)))+1 , which is indeed O(Log(A)) .

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Just for fun:

If you write A in octal (base 8), the function foo computes 7 times the sum of the octal digits but the rightmost, plus 7 or 14 (if the last digit is a 7 ).

Answer 3

Your analysis is correct. foo(4) takes constant time, and you'll have to perform the operation about log n (base 8) times - which is, of course, O(log n) .

If you're more comfortable using a formula, verify with the Master Theorem (a = 1, b = 8, k = i = 0).