Generowanie partycji liczby

Oto moje rozwiązanie (czas wykładniczy) w Pythonie:

q = { 1: [[1]] }

def decompose(n):
    try:
        return q[n]
    except:
        pass

    result = [[n]]

    for i in range(1, n):
        a = n-i
        R = decompose(i)
        for r in R:
            if r[0] <= a:
                result.append([a] + r)

    q[n] = result
    return result

>>> decompose(5)
[[5], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1]]

Kiedy prosi się o bardziej wydajny algorytm, Nie wiem, który porównać. Ale tutaj jest jeden algorytm napisany w prosty sposób (Erlang):

-module(partitions).

-export([partitions/1]).

partitions(N) -> partitions(N, N).

partitions(N, Max) when N > 0 ->
    [[X | P]
     || X <- lists:seq(min(N, Max), 1, -1),
        P <- partitions(N - X, X)];
partitions(0, _) -> [[]];
partitions(_, _) -> [].

Jest wykładnicza w czasie (tak samo jak może być rozwiązana w Pythonie) i liniowa w przestrzeni stosu. Ale używając tej samej sztuczki, memoizacji, możesz osiągnąć dużą poprawę, oszczędzając trochę pamięci i mniej wykładnika. (Jest dziesięć razy szybciej dla N=50)

mp(N) ->
    lists:foreach(fun (X) -> put(X, undefined) end,
          lists:seq(1, N)), % clean up process dictionary for sure
    mp(N, N).

mp(N, Max) when N > 0 ->
    case get(N) of
      undefined -> R = mp(N, 1, Max, []), put(N, R), R;
      [[Max | _] | _] = L -> L;
      [[X | _] | _] = L ->
          R = mp(N, X + 1, Max, L), put(N, R), R
    end;
mp(0, _) -> [[]];
mp(_, _) -> [].

mp(_, X, Max, R) when X > Max -> R;
mp(N, X, Max, R) ->
    mp(N, X + 1, Max, prepend(X, mp(N - X, X), R)).

prepend(_, [], R) -> R;
prepend(X, [H | T], R) -> prepend(X, T, [[X | H] | R]).

W każdym razie powinieneś porównywać swój język i cele.

Oto o wiele bardziej długi sposób na zrobienie tego (tak zrobiłem, zanim poznałem termin "partycja", który umożliwił mi zrobienie wyszukiwania w google):

def magic_chunker (remainder, chunkSet, prevChunkSet, chunkSets):
    if remainder > 0:
        if prevChunkSet and (len(prevChunkSet) > len(chunkSet)): # counting down from previous
            # make a chunk that is one less than relevant one in the prevChunkSet
            position = len(chunkSet)
            chunk = prevChunkSet[position] - 1
            prevChunkSet = [] # clear prevChunkSet, no longer need to reference it
        else: # begins a new countdown; 
            if chunkSet and (remainder > chunkSet[-1]): # no need to do iterations any greater than last chunk in this set
                chunk = chunkSet[-1]
            else: # i.e. remainder is less than or equal to last chunk in this set
                chunk = remainder #else use the whole remainder for this chunk
        chunkSet.append(chunk)
        remainder -= chunk
        magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)
    else: #i.e. remainder==0
        chunkSets.append(list(chunkSet)) #save completed partition
        prevChunkSet = list(chunkSet)
        if chunkSet[-1] > 1: # if the finalchunk was > 1, do further recursion
            remainder = chunkSet.pop() #remove last member, and use it as remainder
            magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)
        else: # last chunk is 1
            if chunkSet[0]==1: #the partition started with 1, we know we're finished
                return chunkSets
            else: #i.e. still more chunking to go 
                # clear back to last chunk greater than 1
                while chunkSet[-1]==1:
                    remainder += chunkSet.pop()
                remainder += chunkSet.pop()
                magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)

partitions = []
magic_chunker(10, [], [], partitions)
print partitions

>> [[10], [9, 1], [8, 2], [8, 1, 1], [7, 3], [7, 2, 1], [7, 1, 1, 1], [6, 4], [6, 3, 1], [6, 2, 2], [6, 2, 1, 1], [6, 1, 1, 1, 1], [5, 5], [5, 4, 1], [5, 3, 2], [5, 3, 1, 1], [5, 2, 2, 1], [5, 2, 1, 1, 1], [5, 1, 1, 1, 1, 1], [4, 4, 2], [4, 4, 1, 1], [4, 3, 3], [4, 3, 2, 1], [4, 3, 1, 1, 1], [4, 2, 2, 2], [4, 2, 2, 1, 1], [4, 2, 1, 1, 1, 1], [4, 1, 1, 1, 1, 1, 1], [3, 3, 3, 1], [3, 3, 2, 2], [3, 3, 2, 1, 1], [3, 3, 1, 1, 1, 1], [3, 2, 2, 2, 1], [3, 2, 2, 1, 1, 1], [3, 2, 1, 1, 1, 1, 1], [3, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2], [2, 2, 2, 2, 1, 1], [2, 2, 2, 1, 1, 1, 1], [2, 2, 1, 1, 1, 1, 1, 1], [2, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]

Oto rozwiązanie w używaniu paramorfizmów, które napisałem w Haskell.

import Numeric.Natural       (Natural)
import Control.Monad         (join)
import Data.List             (nub)
import Data.Functor.Foldable (ListF (..), para)

partitions :: Natural -> [[Natural]]
partitions = para algebra
    where algebra Nothing          = []
          algebra (Just (0,_))     = [[1]]
          algebra (Just (_, past)) = (nub . (getAll =<<)) (fmap (1:) past)

getAll :: [Natural] -> [[Natural]]
getAll = fmap (dropWhile (==0) . sort) . subsets
    where subsets xs = flip sumIndicesAt xs <$> indices xs

indices :: [Natural] -> [[Natural]]
indices = join . para algebra
    where algebra Nil                 = []
          algebra (Cons x (xs, []))   = [[x:xs]]
          algebra (Cons x (xs, past)) = (:) <$> [x:xs,[]] <*> past

Oto kod Javy na to pytanie

static void printArray(int p[], int n){
        for (int i = 0; i < n; i++)
            System.out.print(p[i]+" ");
        System.out.println();
}

// Function to generate all unique partitions of an integer
static void printAllUniqueParts(int n) {
    int[] p = new int[n]; // An array to store a partition
    int k = 0; // Index of last element in a partition
    p[k] = n; // Initialize first partition as number itself

    // This loop first prints current partition, then generates next
    // partition. The loop stops when the current partition has all 1s
    while (true) {
        // print current partition
        printArray(p, k + 1);

        // Generate next partition

        // Find the rightmost non-one value in p[]. Also, update the
        // rem_val so that we know how much value can be accommodated
        int rem_val = 0;
        while (k >= 0 && p[k] == 1) {
            rem_val += p[k];
            k--;
        }

        // if k < 0, all the values are 1 so there are no more partitions
        if (k < 0){
            break;
        }
        // Decrease the p[k] found above and adjust the rem_val
        p[k]--;
        rem_val++;

        while (rem_val > p[k]) {
            p[k + 1] = p[k];
            rem_val = rem_val - p[k];
            k++;
        }
        p[k + 1] = rem_val;
        k++;
    }
}

public static void main(String[] args) {
    System.out.println("All Unique Partitions of 5");
    printAllUniqueParts(5);

    System.out.println("All Unique Partitions of 7");
    printAllUniqueParts(7);

    System.out.println("All Unique Partitions of 9");
    printAllUniqueParts(8);
}

Implementacja Javy. Może skorzystać z memoizacji.

public class Partition {

    /**
     * partition returns a list of int[] that represent all distinct partitions of n.
     */
    public static List<int[]> partition(int n) {
        List<Integer> partial = new ArrayList<Integer>();
        List<int[]> partitions = new ArrayList<int[]>();
        partition(n, partial, partitions);
        return partitions;
    }

    /**
     * If n=0, it copies the partial solution into the list of complete solutions.
     * Else, for all values i less than or equal to n, put i in the partial solution and partition the remainder n-i.
     */
    private static void partition(int n, List<Integer> partial, List<int[]> partitions) {
        //System.out.println("partition " + n + ", partial solution: " + partial);
        if (n == 0) {
            // Complete solution is held in 'partial' --> add it to list of solutions
            partitions.add(toArray(partial));
        } else {
            // Iterate through all numbers i less than n.
            // Avoid duplicate solutions by ensuring that the partial array is always non-increasing
            for (int i=n; i>0; i--) {
                if (partial.isEmpty() || partial.get(partial.size()-1) >= i) {
                    partial.add(i);
                    partition(n-i, partial, partitions);
                    partial.remove(partial.size()-1);
                }
            }
        }
    }

    /**
     * Helper method: creates a new integer array and copies the contents of the list into the array.
     */
    private static int[] toArray(List<Integer> list) {
        int i = 0;
        int[] arr = new int[list.size()];
        for (int val : list) {
            arr[i++] = val;
        }
        return arr;
    }
}