本文整理匯總了Python中dd.bdd.BDD.assert_consistent方法的典型用法代碼示例。如果您正苦於以下問題：Python BDD.assert_consistent方法的具體用法？Python BDD.assert_consistent怎麽用？Python BDD.assert_consistent使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類dd.bdd.BDD的用法示例。
在下文中一共展示了BDD.assert_consistent方法的4個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: test_sifting
# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import assert_consistent [as 別名]
def test_sifting():
# Figs. 6.24, 6.25 Baier 2008
g = BDD({'z1': 0, 'z2': 1, 'z3': 2,
'y1': 3, 'y2': 4, 'y3': 5})
u = g.add_expr('(z1 & y1) | (z2 & y2) | (z3 & y3)')
g.incref(u)
g.collect_garbage()
n = len(g)
assert n == 15, n
_bdd.reorder(g)
u_ = g.add_expr('(z1 & y1) | (z2 & y2) | (z3 & y3)')
g.incref(u)
g.collect_garbage()
g.assert_consistent()
assert u == u_, (u, u_)示例2: test_dump_load_manager
# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import assert_consistent [as 別名]
def test_dump_load_manager():
prefix = 'test_dump_load_manager'
g = BDD({'x': 0, 'y': 1})
e = 'x & !y'
u = g.add_expr(e)
g.incref(u)
fname = prefix + '.p'
g._dump_manager(fname)
h = g._load_manager(fname)
assert g.assert_consistent()
u_ = h.add_expr(e)
assert u == u_, (u, u_)示例3: test_dump_load
# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import assert_consistent [as 別名]
def test_dump_load():
prefix = 'test_dump_load'
fname = prefix + '.p'
dvars = dict(x=0, y=1)
# dump
b = BDD(dvars)
e = 'x & !y'
u_dumped = b.add_expr(e)
b.dump(fname, [u_dumped])
b.dump(fname) # no roots
# load
b = BDD(dvars)
b.add_expr('x | y')
u_new = b.add_expr(e)
umap = b.load(fname)
u_loaded = umap[abs(u_dumped)]
if u_dumped < 0:
u_loaded = -u_loaded
assert u_loaded == u_new, (
u_dumped, u_loaded, u_new, umap)
assert b.assert_consistent()示例4: test_swap
# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import assert_consistent [as 別名]
def test_swap():
# x, y
g = BDD({'x': 0, 'y': 1})
x = g.add_expr('x')
y = g.add_expr('y')
g.incref(x)
g.incref(y)
n = len(g)
assert n == 3, n
nold, n = g.swap('x', 'y')
assert n == 3, n
assert nold == n, nold
assert g.ordering == {'y': 0, 'x': 1}, g.ordering
assert g.assert_consistent()
# functions remain invariant
x_ = g.add_expr('x')
y_ = g.add_expr('y')
assert x == x_, (x, x_, g._succ)
assert y == y_, (y, y_, g._succ)
# external reference counts remain unchanged
assert g._ref[abs(x)] == 1
assert g._ref[abs(y)] == 1
# x & y
g = BDD({'x': 0, 'y': 1})
u = g.add_expr('x & y')
g.incref(u)
nold, n = g.swap('x', 'y')
assert nold == n, (nold, n)
assert g.ordering == {'y': 0, 'x': 1}, g.ordering
u_ = g.add_expr('x & y')
assert u == u_, (u, u_)
assert g.assert_consistent()
# reference counts unchanged
assert g._ref[abs(u)] == 1
# x & !y
# tests handling of complement edges
e = 'x & !y'
g = x_and_not_y()
u = g.add_expr(e)
g.incref(u)
g.collect_garbage()
n = len(g)
assert n == 3, n
nold, n = g.swap('x', 'y')
assert n == 3, n
assert nold == n, nold
assert g.ordering == {'x': 1, 'y': 0}
assert g.assert_consistent()
u_ = g.add_expr(e)
# function u must have remained unaffected
assert u_ == u, (u, u_, g._succ)
# invert swap of:
# x & !y
nold, n = g.swap('x', 'y')
assert n == 3, n
assert nold == n, nold
assert g.ordering == {'x': 0, 'y': 1}
assert g.assert_consistent()
u_ = g.add_expr(e)
assert u_ == u, (u, u_, g._succ)
# Figs. 6.24, 6.25 Baier 2008
g = BDD({'z1': 0, 'y1': 1, 'z2': 2,
'y2': 3, 'z3': 4, 'y3': 5})
u = g.add_expr('(z1 & y1) | (z2 & y2) | (z3 & y3)')
g.incref(u)
n = len(g)
assert n == 16, n
g.collect_garbage()
n = len(g)
assert n == 7, n
# sift to inefficient order
g.swap('y1', 'z2') # z1, z2, y1, y2, z3, y3
g.swap('y2', 'z3') # z1, z2, y1, z3, y2, y3
g.swap('y1', 'z3') # z1, z2, z3, y1, y2, y3
n = len(g)
assert n == 15, n
assert g.assert_consistent()
new_ordering = {
'z1': 0, 'z2': 1, 'z3': 2,
'y1': 3, 'y2': 4, 'y3': 5}
assert g.ordering == new_ordering, g.ordering
u_ = g.add_expr('(z1 & y1) | (z2 & y2) | (z3 & y3)')
assert u_ == u, (u, u_, g._succ)