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Python BDD.collect_garbage方法代碼示例

本文整理匯總了Python中dd.bdd.BDD.collect_garbage方法的典型用法代碼示例。如果您正苦於以下問題:Python BDD.collect_garbage方法的具體用法?Python BDD.collect_garbage怎麽用?Python BDD.collect_garbage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dd.bdd.BDD的用法示例。


在下文中一共展示了BDD.collect_garbage方法的3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_collect_garbage

# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import collect_garbage [as 別名]
def test_collect_garbage():
    # all nodes are garbage
    g = BDD({'x': 0, 'y': 1})
    u = g.add_expr('x & y')
    n = len(g)
    assert n == 4, n
    uref = g._ref[abs(u)]
    assert uref == 0, uref
    _, v, w = g._succ[abs(u)]
    vref = g._ref[abs(v)]
    wref = g._ref[w]
    assert vref == 5, vref
    assert wref == 1, wref
    g.collect_garbage()
    n = len(g)
    assert n == 1, n
    assert u not in g, g._succ
    assert w not in g, g._succ
    # some nodes not garbage
    # projection of x is garbage
    g = BDD({'x': 0, 'y': 1})
    u = g.add_expr('x & y')
    n = len(g)
    assert n == 4, n
    g._ref[abs(u)] += 1
    uref = g._ref[abs(u)]
    assert uref == 1, uref
    g.collect_garbage()
    n = len(g)
    assert n == 3, n
開發者ID:lummax,項目名稱:dd,代碼行數:32,代碼來源:bdd_test.py

示例2: test_sifting

# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import collect_garbage [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_)
開發者ID:lummax,項目名稱:dd,代碼行數:17,代碼來源:bdd_test.py

示例3: test_swap

# 需要導入模塊: from dd.bdd import BDD [as 別名]
# 或者: from dd.bdd.BDD import collect_garbage [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)
開發者ID:lummax,項目名稱:dd,代碼行數:85,代碼來源:bdd_test.py


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