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Circuits

Convert a Boolean circuit to an equivalent Boolean formula.

A Boolean circuit can be exponentially more expressive than an equivalent formula in the worst case, since the circuit can reuse subcircuits multiple times, whereas a formula cannot reuse subformulas more than once. Thus creating a Boolean formula from a Boolean circuit in this way may be infeasible if the circuit is large.

Out:

((x ∨ y) ∧ (y ∨ ¬(z)))

from networkx import dag_to_branching
from networkx import DiGraph
from networkx.utils import arbitrary_element


def circuit_to_formula(circuit):
    # Convert the circuit to an equivalent formula.
    formula = dag_to_branching(circuit)
    # Transfer the operator or variable labels for each node from the
    # circuit to the formula.
    for v in formula:
        source = formula.node[v]['source']
        formula.node[v]['label'] = circuit.node[source]['label']
    return formula


def formula_to_string(formula):

    def _to_string(formula, root):
        # If there are no children, this is a variable node.
        label = formula.node[root]['label']
        if not formula[root]:
            return label
        # Otherwise, this is an operator.
        children = formula[root]
        # If one child, the label must be a NOT operator.
        if len(children) == 1:
            child = arbitrary_element(children)
            return '{}({})'.format(label, _to_string(formula, child))
        # NB "left" and "right" here are a little misleading: there is
        # no order on the children of a node. That's okay because the
        # Boolean AND and OR operators are symmetric. It just means that
        # the order of the operands cannot be predicted and hence the
        # function does not necessarily behave the same way on every
        # invocation.
        left, right = formula[root]
        left_subformula = _to_string(formula, left)
        right_subformula = _to_string(formula, right)
        return '({} {} {})'.format(left_subformula, label, right_subformula)

    root = next(v for v, d in formula.in_degree() if d == 0)
    return _to_string(formula, root)


def main():
    # Create an example Boolean circuit.
    #
    # This circuit has a ∧ at the output and two ∨s at the next layer.
    # The third layer has a variable x that appears in the left ∨, a
    # variable y that appears in both the left and right ∨s, and a
    # negation for the variable z that appears as the sole node in the
    # fourth layer.
    circuit = DiGraph()
    # Layer 0
    circuit.add_node(0, label='∧')
    # Layer 1
    circuit.add_node(1, label='∨')
    circuit.add_node(2, label='∨')
    circuit.add_edge(0, 1)
    circuit.add_edge(0, 2)
    # Layer 2
    circuit.add_node(3, label='x')
    circuit.add_node(4, label='y')
    circuit.add_node(5, label='¬')
    circuit.add_edge(1, 3)
    circuit.add_edge(1, 4)
    circuit.add_edge(2, 4)
    circuit.add_edge(2, 5)
    # Layer 3
    circuit.add_node(6, label='z')
    circuit.add_edge(5, 6)
    # Convert the circuit to an equivalent formula.
    formula = circuit_to_formula(circuit)
    print(formula_to_string(formula))

if __name__ == '__main__':
    main()