def topsort(pairlist):
numpreds = OrderedDict() # elt -> # of predecessors
successors = OrderedDict() # elt -> list of successors
for first, second in pairlist:
# make sure every elt is a key in numpreds
if first not in numpreds:
numpreds[first] = 0
if second not in numpreds:
numpreds[second] = 0
# if they're the same, there's no real dependence
if first == second:
continue
# since first < second, second gains a pred ...
numpreds[second] = numpreds[second] + 1
# ... and first gains a succ
if first in successors:
successors[first].append(second)
else:
successors[first] = [second]
# suck up everything without a predecessor
answer = filter(lambda x, numpreds=numpreds: numpreds[x] == 0,
numpreds.keys())
# for everything in answer, knock down the pred count on
# its successors; note that answer grows *in* the loop
for x in answer:
assert numpreds[x] == 0
del numpreds[x]
if x in successors:
for y in successors[x]:
numpreds[y] = numpreds[y] - 1
if numpreds[y] == 0:
answer.append(y)
# following "del" isn't needed; just makes
# CycleError details easier to grasp
del successors[x]
if numpreds:
# everything in numpreds has at least one predecessor ->
# there's a cycle
if __debug__:
for x in numpreds.keys():
assert numpreds[x] > 0
raise CycleError(answer, numpreds, successors)
return answer
def topsort(pairlist):
numpreds = OrderedDict() # elt -> # of predecessors
successors = OrderedDict() # elt -> list of successors
for first, second in pairlist:
# make sure every elt is a key in numpreds
if first not in numpreds:
numpreds[first] = 0
if second not in numpreds:
numpreds[second] = 0
# if they're the same, there's no real dependence
if first == second:
continue
# since first < second, second gains a pred ...
numpreds[second] = numpreds[second] + 1
# ... and first gains a succ
if first in successors:
successors[first].append(second)
else:
successors[first] = [second]
# suck up everything without a predecessor
answer = filter(lambda x, numpreds=numpreds: numpreds[x] == 0,
numpreds.keys())
# for everything in answer, knock down the pred count on
# its successors; note that answer grows *in* the loop
for x in answer:
assert numpreds[x] == 0
del numpreds[x]
if x in successors:
for y in successors[x]:
numpreds[y] = numpreds[y] - 1
if numpreds[y] == 0:
answer.append(y)
# following "del" isn't needed; just makes
# CycleError details easier to grasp
del successors[x]
if numpreds:
# everything in numpreds has at least one predecessor ->
# there's a cycle
if __debug__:
for x in numpreds.keys():
assert numpreds[x] > 0
raise CycleError(answer, numpreds, successors)
return answer
def topsort_levels(pairlist):
numpreds = OrderedDict() # elt -> # of predecessors
successors = OrderedDict() # elt -> list of successors
for first, second in pairlist:
# make sure every elt is a key in numpreds
if first not in numpreds:
numpreds[first] = 0
if second not in numpreds:
numpreds[second] = 0
# if they're the same, there's no real dependence
if first == second:
continue
# since first < second, second gains a pred ...
numpreds[second] = numpreds[second] + 1
# ... and first gains a succ
if first in successors:
successors[first].append(second)
else:
successors[first] = [second]
answer = []
while 1:
# Suck up everything without a predecessor.
levparents = [x for x in numpreds.keys() if numpreds[x] == 0]
if not levparents:
break
answer.append(levparents)
for levparent in levparents:
del numpreds[levparent]
if levparent in successors:
for levparentsucc in successors[levparent]:
numpreds[levparentsucc] -= 1
del successors[levparent]
if numpreds:
# Everything in num_parents has at least one child ->
# there's a cycle.
raise CycleError(answer, numpreds, successors)
return answer
def topsort_levels(pairlist):
numpreds = OrderedDict() # elt -> # of predecessors
successors = OrderedDict() # elt -> list of successors
for first, second in pairlist:
# make sure every elt is a key in numpreds
if first not in numpreds:
numpreds[first] = 0
if second not in numpreds:
numpreds[second] = 0
# if they're the same, there's no real dependence
if first == second:
continue
# since first < second, second gains a pred ...
numpreds[second] = numpreds[second] + 1
# ... and first gains a succ
if first in successors:
successors[first].append(second)
else:
successors[first] = [second]
answer = []
while 1:
# Suck up everything without a predecessor.
levparents = [x for x in numpreds.keys() if numpreds[x] == 0]
if not levparents:
break
answer.append( levparents )
for levparent in levparents:
del numpreds[levparent]
if levparent in successors:
for levparentsucc in successors[levparent]:
numpreds[levparentsucc] -= 1
del successors[levparent]
if numpreds:
# Everything in num_parents has at least one child ->
# there's a cycle.
raise CycleError( answer, numpreds, successors )
return answer