从素数计算非素数除数

Question

Having a number eg 510510

The prime divisors are: [2, 3, 5, 7, 11, 13, 17]

Using the list of primes, what could an efficient way to calculate the non-prime divisors be?

Answer 1

Assuming that the list of prime factors contains all factors according to there multiplicity, you can use

prime_factors = [2, 3, 5, 7, 11, 13, 17]
non_prime_factors = [reduce(operator.mul, f)
                     for k in range(2, len(prime_factors) + 1)
                     for f in itertools.combinations(prime_factors, k)]

to get all the non-prime factors. Note that you might get duplicates if some prime factors have a greater multiplicity than one -- these can be filtered out using set(non_prime_factors) .

(NumPy wouldn't help too much in this context.)

Edit: By "contains all factors according to there multiplicity" above I mean that (say) 2 should appear twice in the list if it is a prime factor of multiplicity 2, ie 4 is the highest power of 2 which is a factor of the number.

Edit 2: If there are prime factors with a high multiplicity, the above code is inefficient. So just in case you need this, here is efficient code for this case also.

primes = [2, 3, 5]
multiplicities = [3, 4, 5]
exponents = itertools.product(*(range(n + 1) for n in multiplicities))
factors = (itertools.izip(primes, e) for e in exponents if sum(e) >= 2)
non_prime_factors = [reduce(operator.mul, (p ** e for p, e in f))
                     for f in factors]

Answer 2

Here is something get you started. In this method factors is a map of primes to their occurance in your number. So, for your case it would look like [2 : 1, 3 : 1, 5 : 1, 7 : 1, 11 : 1, 13 : 1, 17 : 1] . Note this finds all divisors but the modification should be trivial.

def findAllD(factors):
    pCount = [0 for p in factors.keys()]
    pVals  = [p for p in factors.keys()]
    iters  = reduce(lambda x, y: x*y, [c+1 for c in factors.values()])
    ret    = []

    for i in xrange(0, iters):
        num = 1
        for j in range(0, len(pCount)):
            num *= pVals[j]**pCount[j]

        ret.append(num)

        for j in range(0, len(pCount)):
            pCount[j] = pCount[j] + 1

            if pCount[j] > factors[pVals[j]]:
                pCount[j] = 0
            else:
                break;

    return ret

Answer 3

Since the number 510510 equals 2 * 3 * 5 * 7 * 11 * 13 * 17 , every pair of those primes multiplied are also non-prime divisors:

>>> divmod(510510, 2*3)
(85085, 0)
>>> divmod(510510, 11*17)
(2730, 0)

6 (=2*3) and 187 (=11*17) are non primes, and are propper divisors to 510510.

You can easily find all the pairs of numbers using itertools:

>>> a=[2, 3, 5, 7, 11, 13, 17]
>>> list(itertools.combinations(a, 2))
[(2, 3), (2, 5), (2, 7), (2, 11), (2, 13), (2, 17), (3, 5), (3, 7), (3, 11), (3,
 13), (3, 17), (5, 7), (5, 11), (5, 13), (5, 17), (7, 11), (7, 13), (7, 17), (11
, 13), (11, 17), (13, 17)]

All you need to do then is to multiply the first number of the pair to the second one:

>>> a
[2, 3, 5, 7, 11, 13, 17]
>>> b=list(itertools.combinations(a, 2))
>>> [d*e for d,e in b]
[6, 10, 14, 22, 26, 34, 15, 21, 33, 39, 51, 35, 55, 65, 85, 77, 91, 119, 143, 187, 221]

Finally, you need to repeat the same procedure to triples, four-touples, etc. passing the appropiate number as the secondth parameter to combinations():

>>> b=[reduce((lambda o, p: o*p), y, 1) for x in xrange(2, len(a)) for y in itertools.combinations(a, x)]
>>> b
[6, 10, 14, 22, 26, 34, 15, 21, 33, 39, 51, 35, 55, 65, 85, 77, 91, 119, 143, 187, 221, 30, 42, 66, 78, 102, 70, 110, 130, 170, 154, 182, 238, 286, 374, 442, 105, 165, 195, 255, 231, 273, 357, 429, 561, 663, 385, 455, 595, 715, 935, 1105, 1001, 1309, 1547, 2431, 210, 330, 390, 510, 462, 546, 714, 858, 1122, 1326, 770, 910, 1190, 1430, 1870, 2210, 2002, 2618, 3094, 4862, 1155, 1365, 1785, 2145, 2805, 3315, 3003, 3927, 4641, 7293, 5005, 6545, 7735, 12155, 17017, 2310, 2730, 3570, 4290, 5610, 6630, 6006, 7854, 9282, 14586, 10010, 13090, 15470, 24310, 34034, 15015, 19635, 23205, 36465, 51051, 85085, 30030, 39270, 46410, 72930, 102102, 170170, 255255]

Answer 4

If D is the set of prime divisors of N and d = |D| than N = \\prod_i=1^d(D[i]^p[i]) for some p[i] (where p[i] is a natural number > 0).

From this point of view you can use a bit mask to go through all possible combinations of elements from D and generate partial products, which will divide N . And of course go through all powers 1...p[i] for each element.

In this case you'll get all possible non-prime divisors of N.

Answer 5

If you mean you want to generate ALL the divisors of 510510:

Each prime divisor appears just once in the product.

Each prime divisor could be used or not used. So think of it as a binary set from 0 to 127 and look at the bits. Iterate through the numbers and if the bit relating to the prime divisor is set, include that prime.

eg binary 1011010 means use numbers 17, 11, 7 and 3 so multiply these to get 3927

Of course 0000000 relates to 1 and 1111111 to 510510, so you might not want to count "1 and itself".

If you have a number that has multiple factors you have to count 0 to n on that factor, eg 60 is 2 * 2 * 3 * 5 so 0-2 uses of 2, 0-1 use of 3, 0-1 use of 5, total of 12 possible factors (including 1 and 60).