Yikes!
No, it is computatively commutative, just not in terms
of computation time. :-))
The following factorial loops differ by a remarkable
factor of 1.8, and we can gain this speed by changing
long_mult to always put the lower multiplicand into the left.
This was reported to me by Lenny Kneler, who thought he had
found a Stackless bug, but he was actually testing long math. :-)
This buddy...
>>> def ifact3(n) :
... p = 1L
... for i in range(1,n+1) :
... p = i*p
... return p
performs better by a factor of 1.8 than this one:
>>> def ifact1(n) :
... p = 1L
... for i in range(1,n+1) :
... p = p*i
... return p
The analysis of this behavior is quite simple if you look at the
implementation of long_mult. If the left operand is big and the
right is small, there are much more carry operations performed,
together with more loop overhead.
Swapping the multiplicands would be a 5 line patch.
Should I submit it?
ciao - chris
--
Christian Tismer :^) <mailto:tismer@appliedbiometrics.com>
Applied Biometrics GmbH : Have a break! Take a ride on Python's
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