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# "pow" (power) function

 P: n/a I have a couple of questions for the number crunchers out there: Does "pow(x,2)" simply square x, or does it first compute logarithms (as would be necessary if the exponent were not an integer)? Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some other (perhaps less efficient) algorithm based on logarithms? Thanks, Russ Mar 15 '06 #1
11 Replies

 P: n/a Russ wrote: I have a couple of questions for the number crunchers out there: Does "pow(x,2)" simply square x, or does it first compute logarithms (as would be necessary if the exponent were not an integer)? Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some other (perhaps less efficient) algorithm based on logarithms? you can try and timeit 111**111 10736201288847422580121456504669550195985072399422 48048047759111756250761957833470224912261700936346 21466103743092986967777786330067310159463303558666 91009102601778558729553962214205731543706973022937 5357546494103400699864397711L timeit.Timer("pow(111,111)").timeit() 40.888447046279907 timeit.Timer("111**111").timeit() 39.732122898101807 timeit.Timer("111**0.5").timeit() 2.0990891456604004 timeit.Timer("pow(111,0.5)").timeit() 4.1776390075683594 timeit.Timer("111**0.3").timeit() 2.3824679851531982 timeit.Timer("pow(111,0.3)").timeit() 4.2945041656494141 interesting result seems that ** computates faster Mar 16 '06 #2

 P: n/a I not shure which algorithm,but I am assumeing that all Python does,is to call the underlying C pow() function. Sam Mar 16 '06 #3

 P: n/a Schüle Daniel writes: >>> timeit.Timer("111**0.3").timeit() 2.3824679851531982 >>> timeit.Timer("pow(111,0.3)").timeit() 4.2945041656494141 interesting result seems that ** computates faster Maybe "111**0.3" parses faster than pow(111,0.3), if timeit uses eval. Also, pow() may incur more subroutine call overhead--better check the bytecode for both versions. Mar 16 '06 #4

 P: n/a Russ wrote: I have a couple of questions for the number crunchers out there: Sure, but the answers depend on the underlying Python implementation. And if we're talking CPython, they also depend on the underlying C implementation of libm (i.e., math.h). Does "pow(x,2)" simply square x, or does it first compute logarithms (as would be necessary if the exponent were not an integer)? The former, using binary exponentiation (quite fast), assuming x is an int or long. If x is a float, Python coerces the 2 to 2.0, and CPython's float_pow() function is called. This function calls libm's pow(), which in turn uses logarithms. Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some other (perhaps less efficient) algorithm based on logarithms? The latter, and that algorithm is libm's pow(). Except for a few special cases that Python handles, all floating point exponentation is left to libm. Checking to see if the exponent is 0.5 is not one of those special cases. If you're curious, download the Python source, open up Objects/floatobject.c, and check out float_pow(). The binary exponentation algorithms are in Objects/intobject:int_pow() and Objects/longobject:long_pow(). The 0.5 special check (and any other special case optimizations) could, in theory, be performed in the platform's libm. I'm not familiar enough with any libm implementations to comment on whether this is ever done, or if it's even worth doing... though I suspect that the 0.5 case is not. Hope that helps, --Ben Mar 16 '06 #5

 P: n/a Ben Cartwright wrote: Russ wrote: Does "pow(x,2)" simply square x, or does it first compute logarithms (as would be necessary if the exponent were not an integer)? The former, using binary exponentiation (quite fast), assuming x is an int or long. If x is a float, Python coerces the 2 to 2.0, and CPython's float_pow() function is called. This function calls libm's pow(), which in turn uses logarithms. I just did a little time test (which I should have done *before* my original post!), and 2.0**2 seems to be about twice as fast as pow(2.0,2). That seems consistent with your claim above. I'm a bit surprised that pow() would use logarithms even if the exponent is an integer. I suppose that just checking for an integer exponent could blow away the gain that would be achieved by avoiding logarithms. On the other hand, I would think that using logarithms could introduce a tiny error (e.g., pow(2.0,2) = 3.9999999996 <- made up result) that wouldn't occur with multiplication. Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some other (perhaps less efficient) algorithm based on logarithms? The latter, and that algorithm is libm's pow(). Except for a few special cases that Python handles, all floating point exponentation is left to libm. Checking to see if the exponent is 0.5 is not one of those special cases. I just did another little time test comparing 2.0**0.5 with sqrt(2.0). Surprisingly, 2.0**0.5 seems to take around a third less time. None of these differences are really significant unless one is doing super-heavy-duty number crunching, of course, but I was just curious. Thanks for the information. Mar 16 '06 #6

 P: n/a On Wed, 2006-03-15 at 18:46 -0800, Ben Cartwright wrote: Anyway, if you want to see the int vs. float issue in action, try this: >>> from timeit import Timer >>> Timer('2**2').timeit() 0.12681011582321844 >>> Timer('2.0**2.0').timeit() 0.33336011743438121 >>> Timer('2.0**2').timeit() 0.36681835556112219 >>> Timer('2**2.0').timeit() 0.37949818370600497 As you can see, the int version is much faster than the float version. I have a counterexample. In the original timeit example, 111**111 was used. When I run that timeit.Timer("pow(111,111)").timeit() 10.968398094177246 timeit.Timer("111**111").timeit() 10.04007887840271 timeit.Timer("111.**111.").timeit() 0.36576294898986816 The pow and ** on integers take 10 seconds, but the float ** takes only 0.36 seconds. (The pow with floats takes ~ 0.7 seconds). Clearly typecasting to floats is coming in here somewhere. (Python 2.4.1 on Linux FC4.) Mike Mar 16 '06 #8

 P: n/a Mike Ressler wrote: timeit.Timer("pow(111,111)").timeit() 10.968398094177246 timeit.Timer("111**111").timeit() 10.04007887840271 timeit.Timer("111.**111.").timeit() 0.36576294898986816 The pow and ** on integers take 10 seconds, but the float ** takes only 0.36 seconds. (The pow with floats takes ~ 0.7 seconds). Clearly typecasting to floats is coming in here somewhere. (Python 2.4.1 on Linux FC4.) No, there is not floating point math going on when the operands to ** are both int or long. If there were, the following two commands would have identical output: 111**111 10736201288847422580121456504669550195985072399422 4804804775911 17562507619578334702249122617009363462146610374309 2986967777786 33006731015946330355866691009102601778558729553962 2142057315437 069730229375357546494103400699864397711L int(111.0**111.0) 10736201288847422472001804610489313089074203814505 4486592605938 34891423167097288759427928321358541274379933928055 2157756096410 83975202085309998368049933481542266918440896141131 9810030383904 886446681757296875373689157536249282560L The first result is accurate. Work it out by hand if you don't believe me. ;-) The second suffers from inaccuracies due to floating point's limited precision. Of course, getting exact results with huge numbers isn't cheap, computationally. Because there's no type in C to represent arbitrarily huge numbers, Python implements its own, called "long". There's a fair amount of memory allocation, bit shifting, and other monkey business going on behind the scenes in longobject.c. Whenever possible, Python uses C's built-in signed long int type (known simply as "int" on the Python side, and implemented in intobject.c). On my platform, C's signed long int is 32 bits, so values range from -2147483648 to 2147483647. I.e., -(2**31) to (2**31)-1. As long as your exponentiation result is in this range, Python uses int_pow(). When it overflows, long_pow() takes over. Both functions use the binary exponentiation algorithm, but long_pow() is naturally slower: from timeit import Timer Timer('2**28').timeit() 0.24572032043829495 Timer('2**29').timeit() 0.25511642791934719 Timer('2**30').timeit() 0.27746782979170348 Timer('2**31').timeit() # overflow: 2**31 > 2147483647 2.8205724462504804 Timer('2**32').timeit() 2.2251812151589547 Timer('2**33').timeit() 2.4067177773399635 Floating point is a whole 'nother ball game: Timer('2.0**30.0').timeit() 0.33266301963840306 Timer('2.0**31.0').timeit() # no threshold here! 0.33437446769630697 --Ben Mar 17 '06 #9

 P: n/a "Mike Ressler" wrote in message news:11**********************@dhcp-78-140-229.jpl.nasa.gov... I have a counterexample. In the original timeit example, 111**111 was used. When I run that timeit.Timer("pow(111,111)").timeit() 10.968398094177246 timeit.Timer("111**111").timeit() 10.04007887840271 timeit.Timer("111.**111.").timeit() 0.36576294898986816 The pow and ** on integers take 10 seconds, but the float ** takes only 0.36 seconds. (The pow with floats takes ~ 0.7 seconds). Clearly typecasting to floats is coming in here somewhere. (Python 2.4.1 on Linux FC4.) For floats, f**g == exp(log(f**g)) == exp(g*log(f)) (with maybe further algebraic manipulation, depending on the implementation). The time for this should only be mildly dependent on the magnitudes of f and g. The time for i**j, on the other hand, grows at least as fast as log(j). So I should expect comparisons to depend on magnitudes, as you discovered. Terry Jan Reedy Mar 17 '06 #10 