Question

As I understand it, the Order of Complexity for multiplication is quadratic so if a multiply two 1-digit numbers together there will be 1 operation, two 2-digit numbers together there will be 4 operations, two 3-digit numbers 9 operations and so on.

I wanted to see this complexity by timing the execution of a program that simply multiplied two numbers together. Unexpectedly, regardless of the size of the numbers, the execution time is the same.

import time

num1 = 135
num2 = 342

start = time.time()
total = num1 * num2
finish = time.time()
elapsed = finish - start

print elapsed

So the result is 9.53674316406e-07 if I multiply two 3-digit numbers or two 30-digit numbers.

What is it that I'm misunderstanding?

Answer 1

Your numbers are far too small to exhibit any difference in the time taken to multiply them. Try some proper sized numbers of the order of 10^10⁶.

For example:

import time

for k in range(10):
    num = 10**(10**k)

    start = time.time()
    total = num * num
    finish = time.time()
    elapsed = finish - start

    print k, elapsed

On my machine this outputs:

0 2.86102294922e-06
1 5.00679016113e-06
2 2.14576721191e-06
3 7.86781311035e-06
4 0.000285148620605
5 0.010409116745
6 0.391373157501
7 15.7926459312

(I'm still waiting for 8).

Answer 2

You are correct that for large numbers multiplication is quadratic (or even with better algorithms at least >O(n)). As long as they fit in a 64-bit number you probably won't see any change. There's two problems you're running into 64-bit numbers will hold up to 9.223372e+18 (an 19 digit number) and your 30 digit numbers are just going to be two 64-bit numbers. Try something with several 64-bit numbers (say 10000, which would be an 180000 digit number):

import time
import random

for i in range(0, 190000, 10000):
    a = random.randint(10**i, 10**(i+1)-1)
    b = random.randint(10**i, 10**(i+1)-1)

    start = time.time()
    c = a*b
    end = time.time()

    print i, str(c)[0], end-start # str(c)[0] just in case optimization on c (unlikely)

Results:

0 1 0.0
10000 3 0.000855922698975
20000 5 0.00253701210022
30000 1 0.00445008277893
40000 4 0.00767087936401
50000 3 0.00982689857483
60000 1 0.0133669376373
70000 9 0.0174329280853
80000 4 0.0230648517609
90000 3 0.0251250267029
100000 4 0.0296058654785
110000 4 0.0344429016113
120000 3 0.0401389598846
130000 1 0.0457019805908
140000 2 0.0524950027466
150000 1 0.0619490146637
160000 5 0.0693421363831
170000 2 0.068922996521
180000 2 0.0755639076233

Conclusion based on 90000 and 180000 it's >O(n) but

Measuring Big O of multiplication in Python

2 个答案: