Linear-time ASCII anagram detector

gistfile1.py

# The first 128 primes (corresponding to each ASCII char)
ASCII_PRIMES = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719]

def get_char_prime(letter):
    """Given an ASCII char, return a unique corresponding prime."""
    return ASCII_PRIMES[ord(letter)]

def get_prime_char_sum(letters):
    """Return a unique sum corresponding to the number of occurrences of
    each ASCII char in the given input."""
    return reduce(lambda t, l: t * get_char_prime(l), letters, 1)

def is_anagram_primality(a, b):
    """Returns whether or not ASCII strings a and b are anagrams."""
    return (len(a) == len(b) and # easy optimization
            get_prime_char_sum(a) == get_prime_char_sum(b))