def Nat_mult_antichain_Max(m):
"""
Returns the set of elements of Nat so that their product is at most m:
Max { (a, b) | a * b <= m }
"""
# top -> [(top, top)]
P = Nat()
P.belongs(m)
top = P.get_top()
if is_top(P, m):
s = set([(top, top)])
return s
assert isinstance(m, int)
if m < 1:
return set([(0, 0)])
s = set()
for o1 in range(1, m + 1):
assert o1 >= 1
# We want the minimum x such that o1 * x >= f
# x >= f / o1
# x* = ceil(f / o1)
x = int(np.floor(m * 1.0 / o1))
assert x * o1 <= m # feasible
assert (x+1) * o1 > m, (x+1, o1, m) # and minimum
s.add((o1, x))
return s
def Nat_mult_antichain_Min(m):
"""
Returns the Minimal set of elements of Nat so that their product is at least m:
Min { (a, b) | a * b >= m }
"""
# (top, 1) or (1, top)
P = Nat()
P.belongs(m)
top = P.get_top()
if is_top(P, m):
s = set([(top, 1), (1, top)]) # XXX:
return s
assert isinstance(m, int)
if m == 0:
# any (r1,r2) is such that r1*r2 >= 0
return set([(0, 0)])
s = set()
for o1 in range(1, m + 1):
assert o1 >= 1
# We want the minimum x such that o1 * x >= f
# x >= f / o1
# x* = ceil(f / o1)
x = int(np.ceil(m * 1.0 / o1))
assert x * o1 >= m
assert (x-1) * o1 < m
assert x >= 1
s.add((o1, x))
return s