How to find multiplicative partitions of any integ

问题:

I'm looking for an efficient algorithm for computing the multiplicative partitions for any given integer. For example, the number of such partitions for 12 is 4, which are

12 = 12 x 1 = 4 x 3 = 2 x 2 x 3 = 2 x 6

I've read the wikipedia article for this, but that doesn't really give me an algorithm for generating the partitions (it only talks about the number of such partitions, and to be honest, even that is not very clear to me!).

The problem I'm looking at requires me to compute multiplicative partitions for very large numbers (> 1 billion), so I was trying to come up with a dynamic programming approach for it (so that finding all possible partitions for a smaller number can be re-used when that smaller number is itself a factor of a bigger number), but so far, I don't know where to begin!

Any ideas/hints would be appreciated - this is not a homework problem, merely something I'm trying to solve because it seems so interesting!

回答1:

The first thing I would do is get the prime factorization of the number.

From there, I can make a permutation of each subset of the factors, multiplied by the remaining factors at that iteration.

So if you take a number like 24, you get

2 * 2 * 2 * 3 // prime factorization
a   b   c   d
// round 1
2 * (2 * 2 * 3) a * bcd
2 * (2 * 2 * 3) b * acd (removed for being dup)
2 * (2 * 2 * 3) c * abd (removed for being dup)
3 * (2 * 2 * 2) d * abc

Repeat for all "rounds" (round being the number of factors in the first number of the multiplication), removing duplicates as they come up.

So you end up with something like

// assume we have the prime factorization 
// and a partition set to add to
for(int i = 1; i < factors.size; i++) {
    for(List<int> subset : factors.permutate(2)) {
        List<int> otherSubset = factors.copy().remove(subset);
        int subsetTotal = 1;
        for(int p : subset) subsetTotal *= p;
        int otherSubsetTotal = 1;
        for(int p : otherSubset) otherSubsetTotal *= p;
        // assume your partition excludes if it's a duplicate
        partition.add(new FactorSet(subsetTotal,otherSubsetTotal));
    }
}

回答2:

Of course, the first thing to do is find the prime factorisation of the number, like glowcoder said. Say

n = p^a * q^b * r^c * ...

Then

1. find the multiplicative partitions of m = n / p^a
2. for 0 <= k <= a, find the multiplicative partitions of p^k, which is equivalent to finding the additive partitions of k
3. for each multiplicative partition of m, find all distinct ways to distribute a-k factors p among the factors
4. combine results of 2. and 3.

It is convenient to treat the multiplicative partitions as lists (or sets) of (divisor, multiplicity) pairs to avoid producing duplicates.

I've written the code in Haskell because it's the most convenient and concise of the languages I know for this sort of thing:

module MultiPart (multiplicativePartitions) where

import Data.List (sort)
import Math.NumberTheory.Primes (factorise)
import Control.Arrow (first)

multiplicativePartitions :: Integer -> [[Integer]]
multiplicativePartitions n
    | n < 1     = []
    | n == 1    = [[]]
    | otherwise = map ((>>= uncurry (flip replicate)) . sort) . pfPartitions $ factorise n

additivePartitions :: Int -> [[(Int,Int)]]
additivePartitions 0 = [[]]
additivePartitions n
    | n < 0     = []
    | otherwise = aParts n n
      where
        aParts :: Int -> Int -> [[(Int,Int)]]
        aParts 0 _ = [[]]
        aParts 1 m = [[(1,m)]]
        aParts k m = withK ++ aParts (k-1) m
          where
            withK = do
                let q = m `quot` k
                j <- [q,q-1 .. 1]
                [(k,j):prt | let r = m - j*k, prt <- aParts (min (k-1) r) r]

countedPartitions :: Int -> Int -> [[(Int,Int)]]
countedPartitions 0     count = [[(0,count)]]
countedPartitions quant count = cbParts quant quant count
  where
    prep _ 0 = id
    prep m j = ((m,j):)
    cbParts :: Int -> Int -> Int -> [[(Int,Int)]]
    cbParts q 0 c
        | q == 0    = if c == 0 then [[]] else [[(0,c)]]
        | otherwise = error "Oops"
    cbParts q 1 c
        | c < q     = []        -- should never happen
        | c == q    = [[(1,c)]]
        | otherwise = [[(1,q),(0,c-q)]]
    cbParts q m c = do
        let lo = max 0 $ q - c*(m-1)
            hi = q `quot` m
        j <- [lo .. hi]
        let r = q - j*m
            m' = min (m-1) r
        map (prep m j) $ cbParts r m' (c-j)

primePowerPartitions :: Integer -> Int -> [[(Integer,Int)]]
primePowerPartitions p e = map (map (first (p^))) $ additivePartitions e

distOne :: Integer -> Int -> Integer -> Int -> [[(Integer,Int)]]
distOne _ 0 d k = [[(d,k)]]
distOne p e d k = do
    cap <- countedPartitions e k
    return $ [(p^i*d,m) | (i,m) <- cap]

distribute :: Integer -> Int -> [(Integer,Int)] -> [[(Integer,Int)]]
distribute _ 0 xs = [xs]
distribute p e [(d,k)] = distOne p e d k
distribute p e ((d,k):dks) = do
    j <- [0 .. e]
    dps <- distOne p j d k
    ys <- distribute p (e-j) dks
    return $ dps ++ ys
distribute _ _ [] = []

pfPartitions :: [(Integer,Int)] -> [[(Integer,Int)]]
pfPartitions [] = [[]]
pfPartitions [(p,e)] = primePowerPartitions p e
pfPartitions ((p,e):pps) = do
    cop <- pfPartitions pps
    k <- [0 .. e]
    ppp <- primePowerPartitions p k
    mix <- distribute p (e-k) cop
    return (ppp ++ mix)

It's not particularly optimised, but it does the job.

Some times and results:

Prelude MultiPart> length $ multiplicativePartitions $ 10^10
59521
(0.03 secs, 53535264 bytes)
Prelude MultiPart> length $ multiplicativePartitions $ 10^11
151958
(0.11 secs, 125850200 bytes)
Prelude MultiPart> length $ multiplicativePartitions $ 10^12
379693
(0.26 secs, 296844616 bytes)
Prelude MultiPart> length $ multiplicativePartitions $ product [2 .. 10]
70520
(0.07 secs, 72786128 bytes)
Prelude MultiPart> length $ multiplicativePartitions $ product [2 .. 11]
425240
(0.36 secs, 460094808 bytes)
Prelude MultiPart> length $ multiplicativePartitions $ product [2 .. 12]
2787810
(2.06 secs, 2572962320 bytes)

The 10^k are of course particularly easy because there are only two primes involved (but squarefree numbers are still easier), the factorials get slow earlier. I think by careful organisation of the order and choice of better data structures than lists, there's quite a bit to be gained (probably one should sort the prime factors by exponent, but I don't know whether one should start with the highest exponents or the lowest).

回答3:

Why dont you find all the numbers that can divide the number and then you find permutations of the numbers that multiplications will add up to the number?

Finding all numbers that can divide your number takes O(n).

Then you can permute this set to find all possible sets that multiplication of this set will give you the number.

Once you find set of all possible numbers that divide the original number, then you can do dynamic programming on them to find the set of numbers that multiplying them will give you the original number.