[pypy-dev] PyPy doesn't make code written in C faster

Nathan Hurst njh at njhurst.com
Fri May 31 17:48:45 CEST 2013

Sent to Carl only by mistake, I'm still getting the hang of this
newfangled email thing...

Carl said
> Armin said
> >No, precisely my point: this argument is bogus.  The proof that it's
> >wrong is that CPython gets very similar timing results!  Your pure
> >Python version outperforms the C str(long) in a very similar way on
> >PyPy and on CPython!  The "bug" is originally in CPython, for having a
> >str() that is too slow, and I just copied it into PyPy.  The pure
> >Python version you posted is faster.  Its speed is roughly the same on
> >CPython and on PyPy because most of the time is spent doing divmod on
> >large "long" objects (which is this post's original point).

divmod in principle can be done in O(multiplication log* n), which for
large numbers can be O(n log n).  I don't know whether py*'s
implemention does this.  How do you determine where the bottlenecks

> >I believe that you're right on one point and wrong on another.  You're
> >right in that this gives a faster algo for str().  You're wrong in
> >that it's still quadratic.  If 'a' has 2N digits and 'b' has N digits,
> >then divmod(a,b) is quadratic --- takes time proportional to N*N.  It
> >can be shown by measuring the time spent by your algo to do the repr
> >of larger and larger numbers.

For what it's worth, the time for str(long) / time for recursive
algorithm does decrease steadily for increasing input lengths.  But
it's not n^2 / n log n.  Perhaps we need to implement a faster divmod?
Can I assume that bit_length() is O(1)?

I checked Knuth last night, he doesn't have anything to say in the
main text, but he says in exercises:


14. [M27] (A. Schonhage.)
The text's method of converting multiple-precision integers requires
an execution time of order n^2 to convert an n-place integer, when n
is large. Show that it is possible to convert n-digit decimal integers
into binary notation in O(M(n)logn) steps, where M(n) is an upper
bound on the number of steps needed to multiply n-bit binary numbers
that satisfies the “smoothness condition” M(2n) >= 2M(n).

15. [M47] Can the upper bound on the time to convert large integers,
given in exercise 14, be substantially lowered?

which fits my intuition.  I'm fairly sure that my (and bokr's)
algorithm is O(M(n)logn).  It also suggests I'm wrong about a linear
time algorithm (though the question doesn't state either way, but
given it's M47 hardness, I'm probably not going to be able to whip
this up tonight :)  Lloyd Allison observed that changing any bit can
affect any digits, which makes beating n log n less likely.

One of the main reasons I use python is because it lets me concentrate
on the higher level algorithms (like haskell), but it is pragmatic
about the way we tend to write programs (unlike haskell :).  I doubt I
could have written that algorithm in C before breakfast (as I did for
the python version).

But to the main point: is it fair for people to compare code which
doesn't get the benefit of pypy? Yes it is.  Because the majority of
code out there today is going to have C calls.  Sure, pypy will lose
on those, but that provides incentive to fix the problem - for
example, implementing a better long to string.

People are going to write silly benchmarks and they are going to solve
problems in silly ways.  We should be honest about this in the
benchmarks.  Don't worry, pypy will do just fine.

On Fri, May 31, 2013 at 12:01:27PM +0200, Carl Friedrich Bolz wrote:
> Hi Armin, 
> I have only glanced at the code, but isn't the right argument of the divmod always a power of two? So it can be replaced by a shift and a mask, giving the right complexity. 

No, it's a power of 10, 10^2^i in fact.


> Cheers, 
> Carl Friedrich 

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