Update app.py

app.py

@@ -30,6 +30,42 @@ def draw_graph(G, pos=None, title="Graph Visualization"):
     nx.draw(G, pos=pos, with_labels=True, node_color='lightblue', node_size=500, font_size=10, font_weight='bold')
     st.pyplot(plt)
 
+def circuit_to_formula(circuit):
+    # Convert the circuit to an equivalent formula.
+    formula = nx.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.nodes[v]["source"]
+        formula.nodes[v]["label"] = circuit.nodes[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.nodes[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 = nx.utils.arbitrary_element(children)
+            return f"{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 f"({left_subformula} {label} {right_subformula})"
+
+    root = next(v for v, d in formula.in_degree() if d == 0)
+    return _to_string(formula, root)
+
 def algorithms_circuits():
     st.title("Algorithms: Circuits")