#!python
"""
Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:

S(B) ≠ S(C); that is, sums of subsets cannot be equal.
If B contains more elements than C then S(B) > S(C).
If S(A) is minimised for a given n, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.

n = 1: {1}
n = 2: {1, 2}
n = 3: {2, 3, 4}
n = 4: {3, 5, 6, 7}
n = 5: {6, 9, 11, 12, 13}

It seems that for a given optimum set, A = {a1, a2, ... , an}, the next optimum set is of the form B = {b, a1+b, a2+b, ... ,an+b}, where b is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for n = 6 to be A = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, with S(A) = 115 and corresponding set string: 111819202225.

Given that A is an optimum special sum set for n = 7, find its set string.

NOTE: This problem is related to Problem 105 and Problem 106.
"""

from itertools import combinations

def s(a):
	return sum(a)

def is_special_pair(b, c):
	if len(b) == len(c):
		return s(b) != s(c)
	if len(b) < len(c):
		b, c = c, b
	if s(b) <= s(c):
		return False
	return True

def is_special_set(a):
	for i in range(1,n//2+1):
		for b in combinations(a, i):
			b = set(b)
			x = a-b
			for j in range(1,len(x)+1):
				for c in combinations(x, j):
					if not(is_special_pair(b, c)):
						return False
	return True

n=7
elts = [i for i in range(1,n*n)] #assume elements in special sum set < n*n. this may not hold
ms = 9999

for a in combinations(elts, n):
	a = set(a)
	if is_special_set(a):
		ss = s(a)
		if ss < ms:
			ms = ss
			print([s(a), sorted(list(a))])


