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 B(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 = B( lambda f: lambda n, a: a if not n else f(n-1,a*n) )
>>> fac(5,1)
>>> fibo = B( lambda f: lambda n,p,q: p if not n else f(n-1,q,p+q) )
>>> fibo(10,0,1)

Obviously recursion depth isn’t an issue any longer:

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

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 B0(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 B2(func):
    def wrapper(*args):
        while True:
                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 = B2( lambda f: lambda n: (lambda x: x) if not n else f(n-1) )
>>> f(5)(42)

Of course it could be argued that exceptions are not intended to be used for intentionnaly 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 occuring (exception).