def mk_star_nfa(N):
# Follow construction from Kozen's book:
# 1) Introduce new (single) start+final state IF
# 2) Let Q0 = set({ IF })
# 2) Move on epsilon from IF to the set N[Q0]
# 3) Make N[F] non-final
# 4) Spin back from every state in N[F] to Q0
#
delta_accum = dict({})
IF = NxtStateStr()
Q0 = set({IF}) # new set of start + final states
# Jump from IF to N's start state set
delta_accum.update({(IF, ""): N["Q0"]})
delta_accum.update(N["Delta"])
#
for f in N["F"]:
# N's final states may already have epsilon moves to
# other N-states!
# Expand the target of such jumps to include Q0 also.
if (f, "") in N["Delta"]:
delta_accum.update({(f, ""): (Q0 | N["Delta"][(f, "")])})
else:
delta_accum.update({(f, ""): Q0})
#
return mk_nfa(Q=N["Q"] | Q0,
Sigma=N["Sigma"],
Delta=delta_accum,
Q0=Q0,
F=Q0)
def mk_eps_nfa():
"""An nfa with exactly one start+final state
"""
print("Making an NFA for the Epsilon Symbol...")
Q0 = set({NxtStateStr()})
F = Q0
return mk_nfa(Q=Q0, Sigma=set({}), Delta=dict({}), Q0=Q0, F=Q0)
def mk_cat_nfa(N1, N2):
string = ''
for a in N1["Sigma"]:
string += a
for b in N2["Sigma"]:
string += b
print("The symbols after being concatenated is the string: " + "'" +
string + "'")
delta_accum = dict({})
delta_accum.update(N1["Delta"])
delta_accum.update(N2["Delta"])
print("Making a NFA for the concatenated symbols.....")
# Now, introduce moves from every one of N1's final states
# to the set of N2's initial states.
for f in N1["F"]:
# However, N1's final states may already have epsilon moves to
# other N1-states!
# Expand the target of such jumps to include N2's Q0 also!
if (f, "") in N1["Delta"]:
delta_accum.update({(f, ""): (N2["Q0"] | N1["Delta"][(f, "")])})
else:
delta_accum.update({(f, ""): N2["Q0"]})
print(
"Finished making the Delta for the NFA with the concatenation operator..."
)
# In syntax-directed translation, it is impossible
# that N2 and N1 have common states. Check anyhow
# in case there are bugs elsewhere that cause it.
assert ((N2["F"] & N1["F"]) == set({}))
return mk_nfa(Q=N1["Q"] | N2["Q"],
Sigma=N1["Sigma"] | N2["Sigma"],
Delta=delta_accum,
Q0=N1["Q0"],
F=N2["F"])
def mk_symbol_nfa(a):
"""The NFA for a single re letter
"""
# Make a fresh initial state
q0 = NxtStateStr()
Q0 = set({q0})
# Make a fresh final state
f = NxtStateStr()
F = set({f})
return mk_nfa(Q=Q0 | F, Sigma=set({a}), Delta={(q0, a): F}, Q0=Q0, F=F)
def mk_eps_nfa():
"""An nfa with exactly one start+final state
"""
Q0 = set({ NxtStateStr() })
F = Q0
return mk_nfa(Q = Q0,
Sigma = set({}),
Delta = dict({}),
Q0 = Q0,
F = Q0)
def mk_symbol_nfa(a):
"""The NFA for a single re letter
"""
# Make a fresh initial state
print("Making a symbol NFA for a single re-letter, which is " + "'" + a +
"'....")
q0 = NxtStateStr()
Q0 = set({q0})
# Make a fresh final state
f = NxtStateStr()
F = set({f})
return mk_nfa(Q=Q0 | F, Sigma=set({a}), Delta={(q0, a): F}, Q0=Q0, F=F)
def mk_plus_nfa(N1, N2):
"""Given two NFAs, return their union.
"""
delta_accum = dict({})
delta_accum.update(N1["Delta"])
delta_accum.update(N2["Delta"]) # Simply accumulate the transitions
# The alphabet is inferred bottom-up; thus we must union the Sigmas
# of the NFAs!
return mk_nfa(Q=N1["Q"] | N2["Q"],
Sigma=N1["Sigma"] | N2["Sigma"],
Delta=delta_accum,
Q0=N1["Q0"] | N2["Q0"],
F=N1["F"] | N2["F"])
def mk_cat_nfa(N1, N2):
delta_accum = dict({})
delta_accum.update(N1["Delta"])
delta_accum.update(N2["Delta"])
# Now, introduce moves from every one of N1's final states
# to the set of N2's initial states.
for f in N1["F"]:
# However, N1's final states may already have epsilon moves to
# other N1-states!
# Expand the target of such jumps to include N2's Q0 also!
if (f, "") in N1["Delta"]:
delta_accum.update({(f, ""): (N2["Q0"] | N1["Delta"][(f, "")])})
else:
delta_accum.update({(f, ""): N2["Q0"]})
# In syntax-directed translation, it is impossible
# that N2 and N1 have common states. Check anyhow
# in case there are bugs elsewhere that cause it.
assert ((N2["F"] & N1["F"]) == set({}))
return mk_nfa(Q=N1["Q"] | N2["Q"],
Sigma=N1["Sigma"] | N2["Sigma"],
Delta=delta_accum,
Q0=N1["Q0"],
F=N2["F"])
