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Algorithm to generate (not quite) spanning set in Python with no set longer than 5
[1] [2] [3] [4] [5] [6] [7]
[1] [2] [3] [4] [5] [6, 7]
[1] [2] [3] [4] [5, 6, 7]
...
...
[1, 2, 3, 4, 5] [6, 7]

[1, 2, 3, 4, 5, 6] [7]
or
[1, 2, 3, 4, 5, 6, 7]

def span(lst):
  yield [lst]
  for i in range(1, len(lst)):
    for x in span(lst[i:]):
      yield [lst[:i]] + x

def span(lst, most=float("inf")):
  if not lst:
    yield []
    return

  for i in range(1, min(len(lst), most) + 1):
    for x in span(lst[i:], most):
      yield [lst[:i]] + x

lst = [1,2,3,4,5,6,7]
n = 5
spannings = list(span(lst, n))
print 'n'.join(map(str, spannings))

# proof that it produces the correct result
assert sorted(spannings) == sorted(s for s in span_orig(lst) if max(map(len, s)) <= n)
# proof that it produces the same result as the old
# function if called without a second argument
assert sorted(span_orig(lst)) == sorted(span(lst))