calculate mod using pow function python

Question

So, If i would like to calculate the value of 6^8 mod 5 using the pow function, what should I put in a line??

In the assumption that You don't need to import it first

I know that pow is used like pow (x, y) = pow (6, 8) = 6^8 and

My guess is

mod.pow(6,8)

Thank you!

Two way, pow(x, y) or x**y

Ahasanul Haque
– Ahasanul Haque

2015-09-23 11:44:42 +00:00
Commented Sep 23, 2015 at 11:44 — Ahasanul Haque
– Ahasanul Haque, Commented Sep 23, 2015 at 11:44

PM 2Ring · Accepted Answer · 2015-09-23 11:50:18Z

47

It's simple: pow takes an optional 3rd argument for the modulus.

From the docs:

pow(x, y[, z])

Return x to the power y; if z is present, return x to the power y, modulo z (computed more efficiently than pow(x, y) % z). The two-argument form pow(x, y) is equivalent to using the power operator: x**y.

So you want:

pow(6, 8, 5)

Not only is pow(x, y, z) faster & more efficient than (x ** y) % z it can easily handle large values of y without using arbitrary precision arithmetic, assuming z is a simple machine integer.

edited Sep 23, 2015 at 11:50

answered Sep 23, 2015 at 11:47

PM 2Ring

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9 Comments

PM 2Ring Over a year ago

Note that in version 3.8+: For int operands, the three-argument form of pow now allows the second argument to be negative, permitting computation of modular inverses. And keyword arguments are now permitted. Formerly, only positional arguments were supported.

yo1995 Over a year ago

Thank you so much! With your help I managed to calculate pow(3, 545918790 // 3, 10**9+7) just in time.

ishandutta2007 Over a year ago

speed tested and found false. it is slower than (x**y)%z

PM 2Ring Over a year ago

@ishandutta2007 Try it with smallish x and z (say, under 30 million) and large y.

PM 2Ring Over a year ago

@ishandutta2007 But if x and y are large (x ** y) will cause a MemoryError, so you can't use that approach for arbitrary precision (eg, RSA calculations).

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hiro protagonist · Accepted Answer · 2015-09-23 12:10:35Z

check the docs of pow:

pow(6, 8, 5)

does what you want.

do not use a ** b % n! while this will give the correct result it will be by orders of magnitude slower if you do calculations for bigger numbers. pow will do the modulo operation in every step while ** will first do the exponentiation in the integers (which may result in a huge number) and take the modulus only at the end.

now if you are interested in numbers that are bigger than 32 bit you may want to have a look at gmpy2 for even more speed.

Thanks for mentioning gmpy2. Another arbitrary precision package worth looking at is mpmath, if you want fancy functions, root solving, integrals, etc. FWIW, mpmath will use gmpy, if it's available.

EluciusFTW · Accepted Answer · 2015-09-23 12:50:17Z

0

You can use '%' character to get the modulo value. For example print(pow(6,8) % 5) or print(6**8 % 5).

edited Sep 23, 2015 at 12:50

EluciusFTW

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answered Sep 23, 2015 at 11:57

Syed Ashraf

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