Python 是否优化尾递归？

Question

I have the following piece of code which fails with the following error:

RuntimeError: maximum recursion depth exceeded

I attempted to rewrite this to allow for tail recursion optimization (TCO). I believe that this code should have been successful if a TCO had taken place.

def trisum(n, csum):
    if n == 0:
        return csum
    else:
        return trisum(n - 1, csum + n)

print(trisum(1000, 0))

Should I conclude that Python does not do any type of TCO, or do I just need to define it differently?

Answer 1

No, and it never will since Guido van Rossum prefers to be able to have proper tracebacks:

Tail Recursion Elimination (2009-04-22)

Final Words on Tail Calls (2009-04-27)

You can manually eliminate the recursion with a transformation like this:

>>> def trisum(n, csum):
...     while True:                     # Change recursion to a while loop
...         if n == 0:
...             return csum
...         n, csum = n - 1, csum + n   # Update parameters instead of tail recursion

>>> trisum(1000,0)
500500

Answer 2

I published a module performing tail-call optimization (handling both tail-recursion and continuation-passing style): https://github.com/baruchel/tco

Optimizing tail-recursion in Python

It has often been claimed that tail-recursion doesn't suit the Pythonic way of coding and that one shouldn't care about how to embed it in a loop. I don't want to argue with this point of view; sometimes however I like trying or implementing new ideas as tail-recursive functions rather than with loops for various reasons (focusing on the idea rather than on the process, having twenty short functions on my screen in the same time rather than only three "Pythonic" functions, working in an interactive session rather than editing my code, etc.).

Optimizing tail-recursion in Python is in fact quite easy. While it is said to be impossible or very tricky, I think it can be achieved with elegant, short and general solutions; I even think that most of these solutions don't use Python features otherwise than they should. Clean lambda expressions working along with very standard loops lead to quick, efficient and fully usable tools for implementing tail-recursion optimization.

As a personal convenience, I wrote a small module implementing such an optimization by two different ways. I would like to discuss here about my two main functions.

The clean way: modifying the Y combinator

The Y combinator is well known; it allows to use lambda functions in a recursive manner, but it doesn't allow by itself to embed recursive calls in a loop. Lambda calculus alone can't do such a thing. A slight change in the Y combinator however can protect the recursive call to be actually evaluated. Evaluation can thus be delayed.

Here is the famous expression for the Y combinator:

lambda f: (lambda x: x(x))(lambda y: f(lambda *args: y(y)(*args)))

With a very slight change, I could get:

lambda f: (lambda x: x(x))(lambda y: f(lambda *args: lambda: y(y)(*args)))

Instead of calling itself, the function f now returns a function performing the very same call, but since it returns it, the evaluation can be done later from outside.

My code is:

def bet(func):
    b = (lambda f: (lambda x: x(x))(lambda y:
          f(lambda *args: lambda: y(y)(*args))))(func)
    def wrapper(*args):
        out = b(*args)
        while callable(out):
            out = out()
        return out
    return wrapper

The function can be used in the following way; here are two examples with tail-recursive versions of factorial and Fibonacci:

>>> from recursion import *
>>> fac = bet( lambda f: lambda n, a: a if not n else f(n-1,a*n) )
>>> fac(5,1)
120
>>> fibo = bet( lambda f: lambda n,p,q: p if not n else f(n-1,q,p+q) )
>>> fibo(10,0,1)
55

Obviously recursion depth isn't an issue any longer:

>>> bet( lambda f: lambda n: 42 if not n else f(n-1) )(50000)
42

This is of course the single real purpose of the function.

Only one thing can't be done with this optimization: it can't be used with a tail-recursive function evaluating to another function (this comes from the fact that callable returned objects are all handled as further recursive calls with no distinction). Since I usually don't need such a feature, I am very happy with the code above. However, in order to provide a more general module, I thought a little more in order to find some workaround for this issue (see next section).

Concerning the speed of this process (which isn't the real issue however), it happens to be quite good; tail-recursive functions are even evaluated much quicker than with the following code using simpler expressions:

def bet1(func):
    def wrapper(*args):
        out = func(lambda *x: lambda: x)(*args)
        while callable(out):
            out = func(lambda *x: lambda: x)(*out())
        return out
    return wrapper

I think that evaluating one expression, even complicated, is much quicker than evaluating several simple expressions, which is the case in this second version. I didn't keep this new function in my module, and I see no circumstances where it could be used rather than the "official" one.

Continuation passing style with exceptions

Here is a more general function; it is able to handle all tail-recursive functions, including those returning other functions. Recursive calls are recognized from other return values by the use of exceptions. This solutions is slower than the previous one; a quicker code could probably be written by using some special values as "flags" being detected in the main loop, but I don't like the idea of using special values or internal keywords. There is some funny interpretation of using exceptions: if Python doesn't like tail-recursive calls, an exception should be raised when a tail-recursive call does occur, and the Pythonic way will be to catch the exception in order to find some clean solution, which is actually what happens here...

class _RecursiveCall(Exception):
  def __init__(self, *args):
    self.args = args
def _recursiveCallback(*args):
  raise _RecursiveCall(*args)
def bet0(func):
    def wrapper(*args):
        while True:
          try:
            return func(_recursiveCallback)(*args)
          except _RecursiveCall as e:
            args = e.args
    return wrapper

Now all functions can be used. In the following example, f(n) is evaluated to the identity function for any positive value of n:

>>> f = bet0( lambda f: lambda n: (lambda x: x) if not n else f(n-1) )
>>> f(5)(42)
42

Of course, it could be argued that exceptions are not intended to be used for intentionally redirecting the interpreter (as a kind of goto statement or probably rather a kind of continuation passing style), which I have to admit. But, again, I find funny the idea of using try with a single line being a return statement: we try to return something (normal behaviour) but we can't do it because of a recursive call occurring (exception).

Initial answer (2013-08-29).

I wrote a very small plugin for handling tail recursion. You may find it with my explanations there: https://groups.google.com/forum/?hl=fr#!topic/comp.lang.python/dIsnJ2BoBKs

It can embed a lambda function written with a tail recursion style in another function which will evaluate it as a loop.

The most interesting feature in this small function, in my humble opinion, is that the function doesn't rely on some dirty programming hack but on mere lambda calculus: the behaviour of the function is changed to another one when inserted in another lambda function which looks very like the Y combinator.

Answer 3

The word of Guido is at http://neopythonic.blogspot.co.uk/2009/04/tail-recursion-elimination.html

I recently posted an entry in my Python History blog on the origins of Python's functional features. A side remark about not supporting tail recursion elimination (TRE) immediately sparked several comments about what a pity it is that Python doesn't do this, including links to recent blog entries by others trying to "prove" that TRE can be added to Python easily. So let me defend my position (which is that I don't want TRE in the language). If you want a short answer, it's simply unpythonic. Here's the long answer:

Answer 4

CPython does not and will probably never support tail call optimization based on Guido van Rossum's statements on the subject.

I've heard arguments that it makes debugging more difficult because of how it modifies the stack trace.

Answer 5

Besides optimizing tail recursion, you can set the recursion depth manually by:

import sys
sys.setrecursionlimit(5500000)
print("recursion limit:%d " % (sys.getrecursionlimit()))

Answer 6

尝试使用实验性的macropy TCO 实现来确定大小。

Answer 7

There is no built-in tail recursion optimization in Python. However, we can "rebuild" the function through the Abstract Syntax Tree( AST), eliminating the recursion there and replacing it with a loop. Guido was absolutely right, this approach has some limitations, so I can't recommend it for use.

However, I still wrote (rather as a training example) my own version of the optimizer, and you can even try how it works.

Install this package via pip:

pip install astrologic

Now you can run this sample code:

from astrologic import no_recursion

counter = 0

@no_recursion
def recursion():
    global counter
    counter += 1
    if counter != 10000000:
        return recursion()
    return counter

print(recursion())

This solution is not stable, and you should never use it in production. You can read about some significant restrictions on the page in github (in Russian, sorry). However, this solution is quite "real", without interrupting the code and other similar tricks.