Algorithms: Design and Analysis, Part 1

自我总结：

1.编程的思维不够，虽然分析有哪些需要的函数，但是不能比较好的汇总整合

2.写代码能力，容易挫败感，经常有bug，很烦心，耐心不够好

题目：

In this programming assignment you will implement one or more of the integer multiplication algorithms described in lecture.

To get the most out of this assignment, your program should restrict itself to multiplying only pairs of single-digit numbers. You can implement the grade-school algorithm if you want, but to get the most out of the assignment you'll want to implement recursive integer multiplication and/or Karatsuba's algorithm.

So: what's the product of the following two 64-digit numbers?

3141592653589793238462643383279502884197169399375105820974944592

2718281828459045235360287471352662497757247093699959574966967627

[TIP: before submitting, first test the correctness of your program on some small test cases of your own devising. Then post your best test cases to the discussion forums to help your fellow students!]

[Food for thought: the number of digits in each input number is a power of 2. Does this make your life easier? Does it depend on which algorithm you're implementing?]

The numeric answer should be typed in the space below. So if your answer is 1198233847, then just type 1198233847 in the space provided without any space / commas / any other punctuation marks.

(We do not require you to submit your code, so feel free to use any programming language you want --- just type the final numeric answer in the following space.)

答案：

import math
# Base
B=10
# No. of digits
n=64
# Numbers
x=3141592653589793238462643383279502884197169399375105820974944592
y=2718281828459045235360287471352662497757247093699959574966967627

def karatsuba(x, y):
    if x < 10 or y < 10:
        return x * y
    # get longest digits
    n = max(math.log10(x) + 1, math.log10(y) + 1)
    # catch where n is odd
    n -= n % 2
    bn = B ** (n // 2)
    x1, x2 = divmod(x, bn)
    y1, y2 = divmod(y, bn)
    ac = karatsuba(x1, y1)
    bd = karatsuba(x2, y2)
    # caluclate a+b and c + d subtracting already
    # calculated ac and bd leaving ad + bc
    adbc = karatsuba(x1 + x2, y1 + y2) - ac - bd
    # x . y = 10 ^ n ac + 10^n/2 (ad + bc) + bd
    return ((B ** n) * ac) + bn * adbc + bd


res = karatsuba(x, y)

print('%d * %d = %d' % (x, y, res))

运行的结果：

3141592653589793238462643383279502884197169399375105820974944592 * 2718281828459045235360287471352662497757247093699959574966967627 = 8539734222673565727722948056719317944556312698501627377409191379033726264982769845827675624200334881483773142083314390902243328

几个亮点：

1.通过求对数来求数字的长度

# get longest digits
    n = max(math.log10(x) + 1, math.log10(y) + 1)

2.通过除以10^(n/2)的商和余数来区分一个数前半部分和后半部分，速度更快


超级好的参考资料：
https://courses.csail.mit.edu/6.006/spring11/exams/notes3-karatsuba