How can I generate all possible sums in the following sequence? - python-3.x

Question

Suppose I have an array lets say [1,2,3,4]:
I want to find the sum as follows:
First I generate pairs like:
(1 2 3 4)
(123)(4)
(1)(234)
(12)(34)
(12)(3)(4)
(1)(23)(4)
(1)(2)(34)
(1)(2)(3)(4)
The ans would then be sum of elements in one group multiplied by the length of that group(for all possible groups)
eg in the arrangement (123)(4), the sum would be
(1+2+3)*3 + (4)*1
I just want the final sum which is sum of all such values , not the actual groups. How can I do this?
I was able to do it by first generating all possible groups and then finding the sum
But since I only need the sum and not the actual groups, is there a better way?

Answer 1

The number of arrangements is 2**(len(L)-1). A list of 8 elements produce 128 different arrangements. It is an exponential problem. You either generate all possible solutions and then calculate each answer, or you calculate each answer on the fly. Either way it is still exp.
def part1(L, start, lsum):
if start == len(L):
print lsum
else:
for i in range(start, len(L)):
left = sum(L[start:i+1]) * (i-start+1)
part1(L, i + 1, lsum + left)
def part2(L, M, X, start):
if start == len(L):
M.append(X)
print sum([sum(x) * len(x) for x in X])
else:
for i in range(start, len(L)):
part2(L, M, X + [L[start:i+1]], i + 1)
ex:
>>> part1(L, 0, 0)
10
17
15
28
13
20
22
40
>>> M = []
>>> part2(L, M, [], 0)
10
17
15
28
13
20
22
40
edit: sum of all the sums in O(n**3)
for L = [1,2,3,4,5,6]
[[[1], [2], [3], [4], [5], [6]],
[[1], [2], [3], [4], [5, 6]],
[[1], [2], [3], [4, 5], [6]],
[[1], [2], [3], [4, 5, 6]],
[[1], [2], [3, 4], [5], [6]],
[[1], [2], [3, 4], [5, 6]],
[[1], [2], [3, 4, 5], [6]],
[[1], [2], [3, 4, 5, 6]],
[[1], [2, 3], [4], [5], [6]],
[[1], [2, 3], [4], [5, 6]],
[[1], [2, 3], [4, 5], [6]],
[[1], [2, 3], [4, 5, 6]],
[[1], [2, 3, 4], [5], [6]],
[[1], [2, 3, 4], [5, 6]],
[[1], [2, 3, 4, 5], [6]],
[[1], [2, 3, 4, 5, 6]],
[[1, 2], [3], [4], [5], [6]],
[[1, 2], [3], [4], [5, 6]],
[[1, 2], [3], [4, 5], [6]],
[[1, 2], [3], [4, 5, 6]],
[[1, 2], [3, 4], [5], [6]],
[[1, 2], [3, 4], [5, 6]],
[[1, 2], [3, 4, 5], [6]],
[[1, 2], [3, 4, 5, 6]],
[[1, 2, 3], [4], [5], [6]],
[[1, 2, 3], [4], [5, 6]],
[[1, 2, 3], [4, 5], [6]],
[[1, 2, 3], [4, 5, 6]],
[[1, 2, 3, 4], [5], [6]],
[[1, 2, 3, 4], [5, 6]],
[[1, 2, 3, 4, 5], [6]],
[[1, 2, 3, 4, 5, 6]]]
There seems to be a pattern. The odd case is: the sets having the first elements of the sequence as the smallest element as the sorted set, there are 32. But then all the rest there are 16. For each element of the list, I add all the sets which contains that element as the first sorted element.
def part3(L):
ret = 0
for i in range(len(L)):
p = 0
for k in range(len(L) - i - 1):
p += sum(L[i:i+k+1]) * (k+1) * 2**(len(L) - i - k - 2)
p += sum(L[i:]) * (len(L) - i)
ret += p * max(1, 2**(i-1))
return ret
edit2: to lower it to O(n^2) you need to use DP. building a table of sums to calculate each sum in O(1). You build an array S with S[i] = S[i-1] + L[i] and sum(L[a:b]) is S[b] - S[a].