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November 2, 2014 11:03
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Naive FFT multiplication in python
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| import string | |
| import random | |
| import cmath | |
| # Only works if len(cs) is a power of 2 | |
| def fft(cs, sign): | |
| n = len(cs) | |
| if n == 1: return [cs[0]] | |
| cs1, cs2 = [], [] | |
| for i, c in enumerate(cs): | |
| if i & 1: cs2.append(c) | |
| else: cs1.append(c) | |
| es = fft(cs1, sign) | |
| ds = fft(cs2, sign) | |
| y = [ None ] * n | |
| ww = cmath.exp(complex(0, sign*2*cmath.pi/n)) | |
| w = 1 | |
| for k in xrange(n/2): | |
| y[k] = es[k] + w * ds[k] | |
| y[k+n/2] = es[k] - w * ds[k] | |
| w *= ww | |
| return y | |
| def mult(a, b): | |
| n = 1 | |
| max_len = max(len(a), len(b)) | |
| while n < max_len: n *= 2 | |
| for i in range(2*n-len(a)): a.append(0) | |
| for i in range(2*n-len(b)): b.append(0) | |
| a_fft = fft(a, 1) | |
| b_fft = fft(b, 1) | |
| c_fft = [] | |
| for i in xrange(2*n): | |
| c_fft.append(a_fft[i] * b_fft[i]) | |
| c = fft(c_fft, -1) | |
| c = map(lambda x: int(0.5 + x.real/(2*n)), c) | |
| res, carry = [], 0 | |
| for x in c: | |
| tmp = x + carry | |
| res.append(tmp % 10) | |
| carry = tmp / 10 | |
| if carry: res.append(carry) | |
| return res | |
| ndigits = 10000 | |
| xx, yy = random.randint(0, 10**ndigits), random.randint(0, 10**ndigits) | |
| xs = map(lambda x: int(x), reversed(list(str(xx)))) | |
| ys = map(lambda x: int(x), reversed(list(str(yy)))) | |
| res = mult(xs, ys) | |
| for i in xrange(len(res)-1, 0, -1): | |
| if res[i] == 0: res.pop() | |
| else: break | |
| res = int(string.join(map(lambda x: str(x), reversed(res)), "")) | |
| res2 = xx * yy | |
| print res == res2 |
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