simplification of (i^j)%k

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I'm writing a program that is spending major amounts of time of CPU doing power operations. I need to get the modulo of $i^j$ by k. Is there any simplification of $i^j$%k that doesn't involve an exponentiation?



Note: I remove the specification about needing to heck divisibility. I actually need to get the modulo so that I can do more things with it.




exponentiation integers




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  • What is '^' operator?
    – user58697
    Jul 16 at 21:26










  • That's a power operation.
    – eftshift0
    Jul 16 at 21:26










  • which language are you using?
    – gd1035
    Jul 16 at 21:27











  • I'm working on python.
    – eftshift0
    Jul 16 at 21:31






  • 2




    Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
    – gd1035
    Jul 16 at 21:49














up vote
0
down vote

favorite












I'm writing a program that is spending major amounts of time of CPU doing power operations. I need to get the modulo of $i^j$ by k. Is there any simplification of $i^j$%k that doesn't involve an exponentiation?



Note: I remove the specification about needing to heck divisibility. I actually need to get the modulo so that I can do more things with it.




exponentiation integers




share|cite|improve this question





















  • What is '^' operator?
    – user58697
    Jul 16 at 21:26










  • That's a power operation.
    – eftshift0
    Jul 16 at 21:26










  • which language are you using?
    – gd1035
    Jul 16 at 21:27











  • I'm working on python.
    – eftshift0
    Jul 16 at 21:31






  • 2




    Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
    – gd1035
    Jul 16 at 21:49












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm writing a program that is spending major amounts of time of CPU doing power operations. I need to get the modulo of $i^j$ by k. Is there any simplification of $i^j$%k that doesn't involve an exponentiation?



Note: I remove the specification about needing to heck divisibility. I actually need to get the modulo so that I can do more things with it.




exponentiation integers




share|cite|improve this question













I'm writing a program that is spending major amounts of time of CPU doing power operations. I need to get the modulo of $i^j$ by k. Is there any simplification of $i^j$%k that doesn't involve an exponentiation?



Note: I remove the specification about needing to heck divisibility. I actually need to get the modulo so that I can do more things with it.





exponentiation integers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 22:12
























asked Jul 16 at 21:24









eftshift0

1215




1215











  • What is '^' operator?
    – user58697
    Jul 16 at 21:26










  • That's a power operation.
    – eftshift0
    Jul 16 at 21:26










  • which language are you using?
    – gd1035
    Jul 16 at 21:27











  • I'm working on python.
    – eftshift0
    Jul 16 at 21:31






  • 2




    Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
    – gd1035
    Jul 16 at 21:49
















  • What is '^' operator?
    – user58697
    Jul 16 at 21:26










  • That's a power operation.
    – eftshift0
    Jul 16 at 21:26










  • which language are you using?
    – gd1035
    Jul 16 at 21:27











  • I'm working on python.
    – eftshift0
    Jul 16 at 21:31






  • 2




    Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
    – gd1035
    Jul 16 at 21:49















What is '^' operator?
– user58697
Jul 16 at 21:26




What is '^' operator?
– user58697
Jul 16 at 21:26












That's a power operation.
– eftshift0
Jul 16 at 21:26




That's a power operation.
– eftshift0
Jul 16 at 21:26












which language are you using?
– gd1035
Jul 16 at 21:27





which language are you using?
– gd1035
Jul 16 at 21:27













I'm working on python.
– eftshift0
Jul 16 at 21:31




I'm working on python.
– eftshift0
Jul 16 at 21:31




2




2




Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
– gd1035
Jul 16 at 21:49




Perhaps the built in Python function pow could be of use, it takes a third argument which is used to compute modulo: docs.python.org/3/library/functions.html#pow
– gd1035
Jul 16 at 21:49










1 Answer
1






active

oldest

votes

















up vote
2
down vote













I will put my two comments in to an answer.



One option is to use the built-in Python function "pow()" and the documentation is found here: https://docs.python.org/3/library/functions.html#pow



Another option is to use the fact that: $(a cdot b) mod n = ((amod n)cdot(bmod n)mod n)$



So for your case you would have something like: $i^j mod k = (i mod k)^j mod k$






share|cite|improve this answer





















  • Thanks! But I'd like to get rid of the exponentiation if possible.
    – eftshift0
    Jul 16 at 22:05










  • I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
    – eftshift0
    Jul 16 at 22:27










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













I will put my two comments in to an answer.



One option is to use the built-in Python function "pow()" and the documentation is found here: https://docs.python.org/3/library/functions.html#pow



Another option is to use the fact that: $(a cdot b) mod n = ((amod n)cdot(bmod n)mod n)$



So for your case you would have something like: $i^j mod k = (i mod k)^j mod k$






share|cite|improve this answer





















  • Thanks! But I'd like to get rid of the exponentiation if possible.
    – eftshift0
    Jul 16 at 22:05










  • I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
    – eftshift0
    Jul 16 at 22:27














up vote
2
down vote













I will put my two comments in to an answer.



One option is to use the built-in Python function "pow()" and the documentation is found here: https://docs.python.org/3/library/functions.html#pow



Another option is to use the fact that: $(a cdot b) mod n = ((amod n)cdot(bmod n)mod n)$



So for your case you would have something like: $i^j mod k = (i mod k)^j mod k$






share|cite|improve this answer





















  • Thanks! But I'd like to get rid of the exponentiation if possible.
    – eftshift0
    Jul 16 at 22:05










  • I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
    – eftshift0
    Jul 16 at 22:27












up vote
2
down vote










up vote
2
down vote









I will put my two comments in to an answer.



One option is to use the built-in Python function "pow()" and the documentation is found here: https://docs.python.org/3/library/functions.html#pow



Another option is to use the fact that: $(a cdot b) mod n = ((amod n)cdot(bmod n)mod n)$



So for your case you would have something like: $i^j mod k = (i mod k)^j mod k$






share|cite|improve this answer













I will put my two comments in to an answer.



One option is to use the built-in Python function "pow()" and the documentation is found here: https://docs.python.org/3/library/functions.html#pow



Another option is to use the fact that: $(a cdot b) mod n = ((amod n)cdot(bmod n)mod n)$



So for your case you would have something like: $i^j mod k = (i mod k)^j mod k$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 22:01









gd1035

29319




29319











  • Thanks! But I'd like to get rid of the exponentiation if possible.
    – eftshift0
    Jul 16 at 22:05










  • I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
    – eftshift0
    Jul 16 at 22:27
















  • Thanks! But I'd like to get rid of the exponentiation if possible.
    – eftshift0
    Jul 16 at 22:05










  • I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
    – eftshift0
    Jul 16 at 22:27















Thanks! But I'd like to get rid of the exponentiation if possible.
– eftshift0
Jul 16 at 22:05




Thanks! But I'd like to get rid of the exponentiation if possible.
– eftshift0
Jul 16 at 22:05












I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
– eftshift0
Jul 16 at 22:27




I just tried using pow with the third parameter and I do see an improvement on performance. Some 2-4 percent. But as I grow the numbers that I'm using for all 3 elements, then the improvement dissipates.
– eftshift0
Jul 16 at 22:27












 

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