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Why is print(456 ** 5582) so fast in python?

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-1

1

I've tried many numbers but it runs in like O(1) time. Why is this so fast? Can someone please explain how ** operator actually works.

python python-3.x

share|improve this question

edited Mar 26 at 17:45

Patrick Artner

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asked Mar 26 at 17:41

VishnuVSVishnuVS

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• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  probably square-and-multiply? how fast is it? what were you expecting? O(1) with respect to what?
  
  – hiro protagonist
  Mar 26 at 17:43
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Many numbers? Try print(456546 ** 95675582) :)
  
  – tsul
  Mar 26 at 17:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by "it runs in like O(1) time?" We understand that this means constant time, but constant with respect to what? In my quick test it clearly does not scale at constant time.
  
  – juanpa.arrivillaga
  Mar 26 at 17:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I believe the previous commenters meant that python may be using this: en.wikipedia.org/wiki/Exponentiation_by_squaring which is O(log n) where n is the n from x ** n
  
  – Kevin Wang
  Mar 26 at 17:49
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This reminds me of one professor saying in his lecture "Let N be an arbitrary large number.... No, way, N is too big, let it be M instead". What was it about though.. Ah, right. O(1) algorithm can run million of years. The big-O notation is a measure of how the algorithm running is scaling with the different size inputs and not about the absolute running time.
  
  – Eugene Sh.
  Mar 26 at 17:51
  
  

  
  
  
  

 | 
show 8 more comments

-1

1

I've tried many numbers but it runs in like O(1) time. Why is this so fast? Can someone please explain how ** operator actually works.

python python-3.x

share|improve this question

edited Mar 26 at 17:45

Patrick Artner

29.1k66 gold badges2626 silver badges4545 bronze badges

asked Mar 26 at 17:41

VishnuVSVishnuVS

52233 silver badges1414 bronze badges

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  probably square-and-multiply? how fast is it? what were you expecting? O(1) with respect to what?
  
  – hiro protagonist
  Mar 26 at 17:43
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Many numbers? Try print(456546 ** 95675582) :)
  
  – tsul
  Mar 26 at 17:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What do you mean by "it runs in like O(1) time?" We understand that this means constant time, but constant with respect to what? In my quick test it clearly does not scale at constant time.
  
  – juanpa.arrivillaga
  Mar 26 at 17:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  I believe the previous commenters meant that python may be using this: en.wikipedia.org/wiki/Exponentiation_by_squaring which is O(log n) where n is the n from x ** n
  
  – Kevin Wang
  Mar 26 at 17:49
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This reminds me of one professor saying in his lecture "Let N be an arbitrary large number.... No, way, N is too big, let it be M instead". What was it about though.. Ah, right. O(1) algorithm can run million of years. The big-O notation is a measure of how the algorithm running is scaling with the different size inputs and not about the absolute running time.
  
  – Eugene Sh.
  Mar 26 at 17:51
  
  

  
  
  
  

 | 
show 8 more comments

I've tried many numbers but it runs in like O(1) time. Why is this so fast? Can someone please explain how ** operator actually works.

python python-3.x

share|improve this question

edited Mar 26 at 17:45

Patrick Artner

29.1k66 gold badges2626 silver badges4545 bronze badges

asked Mar 26 at 17:41

VishnuVSVishnuVS

52233 silver badges1414 bronze badges

share|improve this question

edited Mar 26 at 17:45

Patrick Artner

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asked Mar 26 at 17:41

VishnuVSVishnuVS

52233 silver badges1414 bronze badges

share|improve this question

edited Mar 26 at 17:45

Patrick Artner

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asked Mar 26 at 17:41

VishnuVSVishnuVS

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edited Mar 26 at 17:45

Patrick Artner

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edited Mar 26 at 17:45

Patrick Artner

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asked Mar 26 at 17:41

VishnuVSVishnuVS

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asked Mar 26 at 17:41

VishnuVSVishnuVS

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1 Answer
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6

The claim that this is O(1) is incorrect.

>>> %timeit 456**558 
6.22 µs ± 9.22 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 456**5582 
237 µs ± 226 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit 456**55822
9.54 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

See here for the CPython implementation.

share|improve this answer

edited Mar 26 at 18:05

answered Mar 26 at 17:49

wimwim

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6

The claim that this is O(1) is incorrect.

>>> %timeit 456**558 
6.22 µs ± 9.22 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 456**5582 
237 µs ± 226 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit 456**55822
9.54 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

See here for the CPython implementation.

share|improve this answer

edited Mar 26 at 18:05

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges

add a comment | 

6

The claim that this is O(1) is incorrect.

>>> %timeit 456**558 
6.22 µs ± 9.22 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 456**5582 
237 µs ± 226 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit 456**55822
9.54 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

See here for the CPython implementation.

share|improve this answer

edited Mar 26 at 18:05

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges

add a comment | 

The claim that this is O(1) is incorrect.

>>> %timeit 456**558 
6.22 µs ± 9.22 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 456**5582 
237 µs ± 226 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit 456**55822
9.54 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

See here for the CPython implementation.

share|improve this answer

edited Mar 26 at 18:05

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges

>>> %timeit 456**558 
6.22 µs ± 9.22 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 456**5582 
237 µs ± 226 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit 456**55822
9.54 ms ± 13.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

share|improve this answer

edited Mar 26 at 18:05

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges

answered Mar 26 at 17:49

wimwim

178k5959 gold badges352352 silver badges477477 bronze badges