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Diez B. Roggisch wrote: Kay Schluehr wrote:
http://aspn.activestate.com/ASPN/Coo.../Recipe/496691
Neat.
Diez
Hi Diez,
for all those who already copied and pasted the original solution and
played with it I apologize for radical changes in the latest version (
the recipe is on version 1.5 now ! ). The latest implementation is
again a lot faster than the previous one. It does not only get rid of
exceptions but also of stackframe inspection.
Regards,
Kay  
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Kay Schluehr wrote: Diez B. Roggisch wrote: Kay Schluehr wrote:
> http://aspn.activestate.com/ASPN/Coo.../Recipe/496691
Neat.
Diez
Hi Diez,
for all those who already copied and pasted the original solution and played with it I apologize for radical changes in the latest version ( the recipe is on version 1.5 now ! ). The latest implementation is again a lot faster than the previous one. It does not only get rid of exceptions but also of stackframe inspection.
Regards, Kay
I'm not convinced by this. You have to recognise that the function is using
tail recursion, and then you have to modify the code to know that it is
using tail recursion. This is not always trivial. For example, the given
example is:
@tail_recursion
def factorial(n, acc=1):
"calculate a factorial"
if n == 0:
return acc
res = factorial(n1, n*acc)
return res
but a more common way to write the function would be:
@tail_recursion
def factorial(n):
"calculate a factorial"
if n == 0:
return 1
return n * factorial(n1)
which won't work because it isn't actually tail recursion, but it looks
sufficiently close to tail recursion that it would probably mislead a lot
of people into expecting it will work. If you are going to have to rewrite
functions in a stilted manner, and they use simple tail recursion, then why
not just factor out the tail recursion in the first place.
My other problem with this is that the decorator is very fragile although
this may be fixable. e.g. (using the published example) an exception
thrown inside the function makes future calls return garbage: factorial(3)
6 factorial('a')
Traceback (most recent call last):
File "<pyshell#5>", line 1, in toplevel
factorial('a')
File "<pyshell#1>", line 12, in result
tc = g(*args,**kwd)
File "<pyshell#3>", line 6, in factorial
res = factorial(n1, n*acc)
TypeError: unsupported operand type(s) for : 'str' and 'int' factorial(3)
('continue', (3,), {})  
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Kay Schluehr wrote: for all those who already copied and pasted the original solution and played with it I apologize for radical changes in the latest version ( the recipe is on version 1.5 now ! ). The latest implementation is again a lot faster than the previous one. It does not only get rid of exceptions but also of stackframe inspection.
This is spectacular!!
I would rewrite it as follows:
CONTINUE = object() # sentinel value returned by iterfunc
def tail_recursive(func):
"""
tail_recursive decorator based on Kay Schluehr's recipe http://aspn.activestate.com/ASPN/Coo.../Recipe/496691
"""
var = dict(in_loop=False, cont=True, argkw='will be set later')
# the dictionary is needed since Python closures are readonly
def iterfunc(*args, **kwd):
var["cont"] = not var["cont"]
if not var["in_loop"]: # start looping
var["in_loop"] = True
while True:
res = func(*args,**kwd)
if res is CONTINUE:
args, kwd = var["argkw"]
else:
var["in_loop"] = False
return res
else:
if var["cont"]:
var["argkw"] = args, kwd
return CONTINUE
else:
return func(*args,**kwd)
return iterfunc
Using my decorator module 'tail_recursive' can even be turned in a
signaturepreserving
decorator. I think I will add this great example to the documentation
of the next version
of decorator.py!
Michele Simionato  
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Michele Simionato wrote: CONTINUE = object() # sentinel value returned by iterfunc
def tail_recursive(func): """ tail_recursive decorator based on Kay Schluehr's recipe http://aspn.activestate.com/ASPN/Coo.../Recipe/496691 """ var = dict(in_loop=False, cont=True, argkw='will be set later') # the dictionary is needed since Python closures are readonly
def iterfunc(*args, **kwd): var["cont"] = not var["cont"] if not var["in_loop"]: # start looping var["in_loop"] = True while True: res = func(*args,**kwd) if res is CONTINUE: args, kwd = var["argkw"] else: var["in_loop"] = False return res else: if var["cont"]: var["argkw"] = args, kwd return CONTINUE else: return func(*args,**kwd) return iterfunc
CONTINUE could be put inside tail_recursive, couldn't it? And to
squeeze a little more speed out of it, var could be a list (saves a
hash lookup).
Cool decorator.
Carl Banks  
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[...] I'm not convinced by this. You have to recognise that the function is using tail recursion, and then you have to modify the code to know that it is using tail recursion. This is not always trivial. For example, the given example is:
@tail_recursion def factorial(n, acc=1): "calculate a factorial" if n == 0: return acc res = factorial(n1, n*acc) return res
but a more common way to write the function would be:
@tail_recursion def factorial(n): "calculate a factorial" if n == 0: return 1 return n * factorial(n1)
which won't work because it isn't actually tail recursion, but it looks sufficiently close to tail recursion that it would probably mislead a lot of people into expecting it will work. If you are going to have to rewrite functions in a stilted manner, and they use simple tail recursion, then why not just factor out the tail recursion in the first place.
[...]
Hi Duncan,
I don't know why it wouldn't work this way, or why it isn't
tailrecursion?
I tried the tail_recursion decorator from the cookbookrecipe with both
definitions of factorial, and I tried both definitions of the factorial
function with and without tail_recursion decorator.
In all four cases I get the same results, so it does work with both
definitions of factorial(), even if (according to you) the second
definition is not proper tailrecursion.
Using the tailrecursion decorator (the version that does not inspect
the stackframes) I get a small performanceincrease over using the
factorialfunction undecorated.
However, calculating factorial(1000) with the factorialfunction as
defined in the cookbookrecipe is much much faster than calculating the
same factorial(1000) with the factorialfunction you gave!
I cannot yet explain why the first function has so much better
performance than the second function  about a factor 10 difference,
in both python2.4.3 and python 2.5a2
Cheers,
Tim  
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Tim N. van der Leeuw wrote: I don't know why it wouldn't work this way, or why it isn't tailrecursion?
From the google page do "define: tail recursion"
I tried the tail_recursion decorator from the cookbookrecipe with both definitions of factorial, and I tried both definitions of the factorial function with and without tail_recursion decorator. In all four cases I get the same results, so it does work with both definitions of factorial(), even if (according to you) the second definition is not proper tailrecursion.
For me factorial(1001) with the second definition does not work, I get
the recursion limit (which is what I expect). I suppose the recursion
limit is higher on your system, but definitely you should reach it at
some point with the non tailrecursive version of factorial.
Using the tailrecursion decorator (the version that does not inspect the stackframes) I get a small performanceincrease over using the factorialfunction undecorated. However, calculating factorial(1000) with the factorialfunction as defined in the cookbookrecipe is much much faster than calculating the same factorial(1000) with the factorialfunction you gave! I cannot yet explain why the first function has so much better performance than the second function  about a factor 10 difference, in both python2.4.3 and python 2.5a2
It is because the decorator is doing is job (converting a long
recursion in a loop)
only with the first function, which is properly tail recursive. Just as
Duncan said.
Michele Simionato  
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Hi Michele,
I'm sorry, but you misunderstood me.
There are two definitions of the factorial() function, one given by the
OP and the other given by Duncan.
I tested both factorial() definitions with, and without the
tail_recursion decorator (the version of the OP). So I had 4
factorialfunctions defined in my testfile:
@tail_recursion
def factorial(n, acc=1):
# do the stuff
pass
def factorial_r(n, acc=1):
# do the stuff
pass
@tail_recursion
def factorial2(n):
# do the stuff
pass
def factorial2_r(n):
# do the stuff
pass
All four functions give the same output for the tests I did (n=120, and
n=1000).
Using timeit, both factorial(1000) and factorial2(1000) are somewhat
faster than factorial_r(1000) respectively factorial2_r(1000).
However, factorial(1000) and factorial_r(1000) are both 10x faster than
factorial2(1000) and factorial2_r(1000).
It's the latter performance difference which I do not understand.
The other thing I do not understand, due to my limited understanding of
what is tailrecursion: factorial2 (Duncan's definition) is not proper
tailrecursion. Why not? How does it differ from 'real' tail recursion?
And if it's not proper tailrecursion and therefore should not work,
then how comes that the tests I do show it to work? And I seemed to
consistently get a slightly better performance from factorial2(1000)
than from factorial2_r(1000).
NB: Regarding the recursion limits, I don't know what would be the
stacklimit on my system (Python 2.4.3 on WinXP SP2). I already
calculated the factorial of 500000 using the recursive (nondecorated)
function...
Cheers,
Tim  
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Tim N. van der Leeuw wrote: The other thing I do not understand, due to my limited understanding of what is tailrecursion: factorial2 (Duncan's definition) is not proper tailrecursion. Why not? How does it differ from 'real' tail recursion?
Tail recursion is when a function calls itself and then immediately returns
the result of that call as its own result. So long as nothing except
returning the result needs to be done it is possibly to avoid the recursive
call altogether.
This function is tail recursive:
@tail_recursion
def factorial(n, acc=1):
"calculate a factorial"
if n == 0:
return acc
res = factorial(n1, n*acc)
return res
but this one isn't:
@tail_recursion
def factorial2(n):
"calculate a factorial"
if n == 0:
return 1
return n * factorial2(n1)
because when the inner call to factorial2() returns the function still has
to do some work (the multiplication).
I don't understand your comments about speed differences. If you try to run
factorial2() as defined above then it simply throws an exception: there
are no results to compare. My guess is that when you wrote:
@tail_recursion
def factorial2(n):
# do the stuff
pass
your 'do the stuff' actually had an erroneous call to 'factorial'. If you
are going to rename the function you have to rename the recursive calls as
well. (At least, that's what I forgot to do when I first tried it and
couldn't understand why it gave me an answer instead of crashing.)
The decorator also fails for functions which are tailrecursive but which
contain other nontail recursive calls within themselves. For example I
would be pretty sure you couldn't write a working implementation of
Ackermann's function using the decorator:
def Ack(M, N):
if (not M):
return( N + 1 )
if (not N):
return( Ack(M1, 1) )
return( Ack(M1, Ack(M, N1)) )  
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Duncan Booth wrote: Tim N. van der Leeuw wrote:
[...] @tail_recursion def factorial2(n): # do the stuff pass
your 'do the stuff' actually had an erroneous call to 'factorial'. If you are going to rename the function you have to rename the recursive calls as well. (At least, that's what I forgot to do when I first tried it and couldn't understand why it gave me an answer instead of crashing.)
[...]
Duncan,
You're totally right. Somehow, I had managed to completely overlook
this. Oops!
My apologies! :)
Tim  
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Duncan Booth wrote: The decorator also fails for functions which are tailrecursive but which contain other nontail recursive calls within themselves. For example I would be pretty sure you couldn't write a working implementation of Ackermann's function using the decorator:
def Ack(M, N): if (not M): return( N + 1 ) if (not N): return( Ack(M1, 1) ) return( Ack(M1, Ack(M, N1)) )
Definitely. The translation into a proper tailrecursive form is
nontrivial but nevertheless possible as demonstrated by the following
Ackermann implementation:
@tail_recursion
def ack(m,n,s=[0]): # use a stackvariable s as "accumulator"
if m==0:
if s[0] == 1:
return ack(s[1]1,n+1,s[2])
elif s[0] == 0:
return n+1
elif n==0:
return ack(m1,1,s)
else:
return ack(m,n1,[1,m,s])
Regards,
Kay  
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Duncan Booth <du**********@invalid.invalid> writes: Tim N. van der Leeuw wrote:
The other thing I do not understand, due to my limited understanding of what is tailrecursion: factorial2 (Duncan's definition) is not proper tailrecursion. Why not? How does it differ from 'real' tail recursion?
Tail recursion is when a function calls itself and then immediately returns the result of that call as its own result.
I think the definition is broader than that so that these two functions would
also be tailrecursive (i.e. the tail call doesn't have to be a selftail
call; I might be mistaken, don't have a good reference at hand; however
"properly tail recursive" certainly refers to being able to do the below
without exhausting the stack even for large n, not just transforming selftail
calls to a loop, which is sort of limited usefulness anyway):
def even(n):
return n == 0 or not odd(n1)
def odd(n):
return n == 1 or not even(n1)
'as  
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Kay Schluehr wrote: Duncan Booth wrote:
The decorator also fails for functions which are tailrecursive but which contain other nontail recursive calls within themselves. For example I would be pretty sure you couldn't write a working implementation of Ackermann's function using the decorator:
def Ack(M, N): if (not M): return( N + 1 ) if (not N): return( Ack(M1, 1) ) return( Ack(M1, Ack(M, N1)) )
Definitely. The translation into a proper tailrecursive form is nontrivial but nevertheless possible as demonstrated by the following Ackermann implementation:
@tail_recursion def ack(m,n,s=[0]): # use a stackvariable s as "accumulator" if m==0: if s[0] == 1: return ack(s[1]1,n+1,s[2]) elif s[0] == 0: return n+1 elif n==0: return ack(m1,1,s) else: return ack(m,n1,[1,m,s])
Very clever, although simulating a stack isn't exactly eliminating
recursion.
Any idea how long I have to wait to find ack(4,1)?  
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"Alexander Schmolck" <a.********@gmail.com> wrote in message
news:yf*************@oc.ex.ac.uk... Duncan Booth <du**********@invalid.invalid> writes: Tail recursion is when a function calls itself and then immediately returns the result of that call as its own result.
Which means that the value returned by the base case is returned unchanged
to the original caller through the stack of returns. Which means that the
return stack can potentially be compressed to just one return.
I think the definition is broader than that so that these two functions would also be tailrecursive (i.e. the tail call doesn't have to be a selftail call; I might be mistaken, don't have a good reference at hand; however "properly tail recursive" certainly refers to being able to do the below without exhausting the stack even for large n, not just transforming selftail calls to a loop, which is sort of limited usefulness anyway):
def even(n): return n == 0 or not odd(n1)
def odd(n): return n == 1 or not even(n1)
No, these are not even mutually tailrecursive, assuming that that would
make sense. You are calling the not operator function on the results of
the recursive calls before returning them. The following *is*
tailrecursive:
def even(n):
assert n >= 0
if n >=2: return even(n2)
return bool(n)
The recursive call is effectively a goto back to the top of the function,
with n reduced by 2. This looping continues until n < 2. So in Python, we
would usually write
def even(n):
assert n >= 0
while n >= 2: n =2
return bool(n)
Terry Jan Reedy  
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Your examples are not tail recursive because an extra step is needed
before returning from the function call and that step cannot be thrown
away!
Alexander Schmolck <a.********@gmail.com> wrote: def even(n): return n == 0 or not odd(n1)
def odd(n): return n == 1 or not even(n1)

Regards,
Casey  
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Tail Call Optimization and Recursion are separate concepts!

Regards,
Casey  
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Duncan Booth wrote: My other problem with this is that the decorator is very fragile although this may be fixable
This version should be more robust against exceptions:
class tail_recursive(object):
"""
tail_recursive decorator based on Kay Schluehr's recipe http://aspn.activestate.com/ASPN/Coo.../Recipe/496691
"""
CONTINUE = object() # sentinel
def __init__(self, func):
self.func = func
self.firstcall = True
def __call__(self, *args, **kwd):
try:
if self.firstcall: # start looping
self.firstcall = False
while True:
result = self.func(*args, **kwd)
if result is self.CONTINUE: # update arguments
args, kwd = self.argskwd
else: # last call
break
else: # return the arguments of the tail call
self.argskwd = args, kwd
return self.CONTINUE
except: # reset and reraise
self.firstcall = True
raise
else: # reset and exit
self.firstcall = True
return result   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 1731
 replies: 19
 date asked: May 10 '06
