Multiple subset sum calculation

A complete solution, which compute all the manner to do a total. I use ints as characteristic sets for speed and memory usage : 19='0b10011' represent [A[0],A[1],A[4]]=[8,9,33] here.

A = [8, 9, 15, 15, 33, 36, 39, 45, 46, 60, 68, 73, 80, 92, 96]
B =[183, 36, 231, 128, 137]

def subsetsum(A,N):
    res=[[0]]+[[] for i in range(N)]
    for i,a in enumerate(A):
        k=1<<i        
        stop=[len(l) for l in res] 
        for shift,l in enumerate(res[:N+1-a]):
            n=a+shift   
            ln=res[n]
            for s in l[:stop[shift]]: ln.append(s+k)
    return res

res = subsetsum(A,max(B))
solB = [res[b] for b in B]
exactsol = ~-(1<<len(A))

def decode(answer):
    return [[A[i] for i,b in enumerate(bin(sol)[::-1]) if b=='1'] for sol in answer] 

def solve(i,currentsol,answer):
        if currentsol==exactsol : print(decode(answer))
        if i==len(B): return
        for sol in solB[i]:
                if not currentsol&sol:
                    answer.append(sol)
                    solve(i+1,currentsol+sol,answer)
                    answer.pop()

For :

solve(0,0,[])

[[9, 46, 60, 68], [36], [8, 15, 39, 73, 96], [15, 33, 80], [45, 92]]
[[9, 46, 60, 68], [36], [8, 15, 39, 73, 96], [15, 33, 80], [45, 92]]
[[8, 15, 15, 33, 39, 73], [36], [9, 46, 80, 96], [60, 68], [45, 92]]
[[9, 15, 33, 46, 80], [36], [8, 15, 39, 73, 96], [60, 68], [45, 92]]
[[9, 15, 33, 46, 80], [36], [8, 15, 39, 73, 96], [60, 68], [45, 92]]
[[15, 15, 73, 80], [36], [9, 39, 45, 46, 92], [60, 68], [8, 33, 96]]
[[15, 15, 73, 80], [36], [8, 9, 33, 39, 46, 96], [60, 68], [45, 92]]
[[45, 46, 92], [36], [15, 15, 60, 68, 73], [9, 39, 80], [8, 33, 96]]
[[45, 46, 92], [36], [9, 15, 15, 39, 73, 80], [60, 68], [8, 33, 96]]
[[45, 46, 92], [36], [8, 15, 39, 73, 96], [60, 68], [9, 15, 33, 80]]
[[45, 46, 92], [36], [8, 15, 39, 73, 96], [15, 33, 80], [9, 60, 68]]
[[45, 46, 92], [36], [8, 15, 39, 73, 96], [60, 68], [9, 15, 33, 80]]
[[45, 46, 92], [36], [8, 15, 39, 73, 96], [15, 33, 80], [9, 60, 68]]
[[15, 33, 39, 96], [36], [8, 15, 60, 68, 80], [9, 46, 73], [45, 92]]
[[15, 33, 39, 96], [36], [8, 9, 15, 46, 73, 80], [60, 68], [45, 92]]
[[15, 33, 39, 96], [36], [8, 15, 60, 68, 80], [9, 46, 73], [45, 92]]
[[15, 33, 39, 96], [36], [8, 9, 15, 46, 73, 80], [60, 68], [45, 92]]
[[8, 33, 46, 96], [36], [15, 15, 60, 68, 73], [9, 39, 80], [45, 92]]
[[8, 33, 46, 96], [36], [9, 15, 15, 39, 73, 80], [60, 68], [45, 92]]

Notice than when the two 15 are not in the same subset, the solution is doubled.

It resolves the unique solution problem :

A=[1000, 1001, 1002, 1003, 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011,
   1012, 1013, 1014, 1015, 1016, 1017, 1018, 1019, 1020, 1021, 1022, 1023,
   1024, 1025, 1026, 1027, 1028, 1029, 1030, 1031, 1032, 1033, 1034, 1035,
   1036, 1037, 1038, 1039, 1040, 1041, 1042, 1043, 1044, 1045, 1046, 1047, 
   1048, 1049]

B=[5010, 5035, 5060, 5085, 5110, 5135, 5160, 5185, 5210, 5235]

in one second. Unfortunately, it's not yet enough optimized for a (71,10) problem.

Yet another one, in the pure dynamic programming spirit : :

@functools.lru_cache(max(B))
def solutions(n):
    if n==0 : return set({frozenset()}) #{{}}
    if n<0 :  return set()
    sols=set()
    for i,a in enumerate(A):
            for s in solutions(n-a):
                if i not in s : sols.add(s|{i})
    return sols

def decode(answer): return([[A[i] for i in sol] for sol in answer]) 

def solve(B=B,currentsol=set(),answer=[]):
    if len(currentsol)==len(A) : sols.append(decode(answer))
    if B:
        for sol in solutions(B[0]):
            if set.isdisjoint(currentsol,sol):
                solve(B[1:],currentsol|sol,answer+[sol]) 

sols=[];solve()

This is known as the subset-sum problem and it is a well known NP-complete problem. So basically there is no efficient solution. See for example https://en.wikipedia.org/wiki/Subset_sum_problem

However If your number N is not too large, there is a pseudo polynomial algorithms, using dynamic programming: You read the list A from left to right and keep the list of the sum which are doable and smaller than N. If you know the number which are doable for a given A, you can easily get those which are doable for A + [a]. Hence the dynamic programming. It will typically be fast enough for a problem of the size you gave there.

Here is a Python quick solution:

def subsetsum(A, N):
    res = {0 : []}
    for i in A:
        newres = dict(res)
        for v, l in res.items():
            if v+i < N:
                newres[v+i] = l+[i]
            elif v+i == N:
                return l+[i]
        res = newres
    return None

Then

>>> A = [8, 9, 15, 15, 33, 36, 39, 45, 46, 60, 68, 73, 80, 92, 96]
>>> subsetsum(A, 183)
[15, 15, 33, 36, 39, 45]

After OP edit:

Now I correctly understand you problem, I'll still think that your problem can be solved efficiently, provided you have an efficient subset-sum solver: I'd use divide and conquer solution on B:

cut B into two approximately equal pieces B1 and B2
use your subset-sum solver to search among A for all subsets S whose sum are equal to sum(B1).
for each such S:
- call recursively solve(S, B1) and solve(A - S, B2)
- if both succeed you have a solution

However, your (71, 10) problem below is out of reach for the dynamic programming solution I suggested.

By the way, here is a quick solution of your problem not using divide and conquer, but which contains the correct adaptation of my dynamic solver to get all solutions:

class NotFound(BaseException):
    pass

from collections import defaultdict
def subset_all_sums(A, N):
    res = defaultdict(set, {0 : {()}})
    for nn, i in enumerate(A):
        # perform a deep copy of res
        newres = defaultdict(set)
        for v, l in res.items():
            newres[v] |= set(l)
            for v, l in res.items():
                if v+i <= N:
                    for s in l:
                        newres[v+i].add(s+(i,))
                        res = newres
                        return res[N]

def list_difference(l1, l2):
    ## Similar to merge.
    res = []
    i1 = 0; i2 = 0
    while i1 < len(l1) and i2 < len(l2):
        if l1[i1] == l2[i2]:
            i1 += 1
            i2 += 1
        elif l1[i1] < l2[i2]:
            res.append(l1[i1])
            i1 += 1
        else:
            raise NotFound
            while i1 < len(l1):
                res.append(l1[i1])
                i1 += 1
                return res

def solve(A, B):
    assert sum(A) == sum(B)
    if not B:
        return [[]]
        res = []
        ss = subset_all_sums(A, B[0])
        for s in ss:
            rem = list_difference(A, s)
            for sol in solve(rem, B[1:]):
                res.append([s]+sol)
                return res

Then:

>>> solve(A, B)
[[(15, 33, 39, 96), (36,), (8, 15, 60, 68, 80), (9, 46, 73), (45, 92)],
 [(15, 33, 39, 96), (36,), (8, 9, 15, 46, 73, 80), (60, 68), (45, 92)],
 [(8, 15, 15, 33, 39, 73), (36,), (9, 46, 80, 96), (60, 68), (45, 92)],
 [(15, 15, 73, 80), (36,), (8, 9, 33, 39, 46, 96), (60, 68), (45, 92)],
 [(15, 15, 73, 80), (36,), (9, 39, 45, 46, 92), (60, 68), (8, 33, 96)],
 [(8, 33, 46, 96), (36,), (9, 15, 15, 39, 73, 80), (60, 68), (45, 92)],
 [(8, 33, 46, 96), (36,), (15, 15, 60, 68, 73), (9, 39, 80), (45, 92)],
 [(9, 15, 33, 46, 80), (36,), (8, 15, 39, 73, 96), (60, 68), (45, 92)],
 [(45, 46, 92), (36,), (8, 15, 39, 73, 96), (60, 68), (9, 15, 33, 80)],
 [(45, 46, 92), (36,), (8, 15, 39, 73, 96), (15, 33, 80), (9, 60, 68)],
 [(45, 46, 92), (36,), (15, 15, 60, 68, 73), (9, 39, 80), (8, 33, 96)],
 [(45, 46, 92), (36,), (9, 15, 15, 39, 73, 80), (60, 68), (8, 33, 96)],
 [(9, 46, 60, 68), (36,), (8, 15, 39, 73, 96), (15, 33, 80), (45, 92)]]

>>> %timeit solve(A, B)
100 loops, best of 3: 10.5 ms per loop

So it is quite fast for this size of problem, though nothing is optimized here.

Multiple subset sum calculation

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Python

Algorithm

Math

Subset

Subset Sum

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