4
$\begingroup$

I am writing a program in which I want to make changes to make it more efficient.

What the program does is it takes three inputs $m$, $n$ and $x$ and I have to find the value of the following equation: $$ 1^n+ 2^n+\cdots + m^n \mod{x} $$

Is there a better way than calculating the whole value and then solving for answer? Because if $n$ and $m$ are large it takes a lot of computation time which I am trying to avoid.

$\endgroup$
2
  • 1
    $\begingroup$ What about wikipedia:bernoulli-polynomials , or googling for "sums-of-like-powers" ? $\endgroup$
    – Gottfried Helms
    Commented Nov 17, 2012 at 14:36
  • $\begingroup$ please note I made a mistake in my answer. $\endgroup$
    – sperners lemma
    Commented Nov 17, 2012 at 18:20

1 Answer 1

Reset to default
3
$\begingroup$

If $x$ is small compared to $n$ and/or $m$ there are some good optimizations you can do:

  • Edit: This is wrong, don't do this: Replace $n$ with its remainder on division by $\varphi(x)$. this only works for the bases coprime to n, which is a good proportion so it may still be worth it to do that.. for ones that aren't coprime...

  • Use binary exponentiation.

  • Split the sum into blocks $[1^n + 2^n + ... + x^n] + [(x+1)^n + (x+2)^n + \ldots] + \ldots$ which are all equal, so you only need to compute the sum of $x$ terms rather than $m$.

If $x$ is large compared to $n$ then (as already mentioned in comments) it will be more efficient to compute the sum using a closed form polynomial (which you may need to compute before use).

$\endgroup$
7
  • 1
    $\begingroup$ $\varphi(x)$ doesn't do you any good, because the numbers from $1$ to $x$ are not all relatively prime to $x$. In general, it is not true that if $a\equiv b\pmod {\varphi(x)}$ that $a^{\varphi(x)}\equiv b^{\varphi(x)}\pmod x$ $\endgroup$
    – Thomas Andrews
    Commented Nov 17, 2012 at 18:10
  • 1
    $\begingroup$ If $x=8$ then $\varphi(x)=4$ and $2^5\equiv 0 \not\equiv 2^1\pmod 8$ even though $5\equiv 1\pmod 4$ $\endgroup$
    – Thomas Andrews
    Commented Nov 17, 2012 at 18:13
  • 1
    $\begingroup$ Yeah, I got the formula wrong in my comment above, but my counter-example is for your second version. There, $a=2$, $x=8$, $n=1$ and $n'=5$. $\endgroup$
    – Thomas Andrews
    Commented Nov 17, 2012 at 18:14
  • 1
    $\begingroup$ One quick reduction occurs when $n$ is odd. Then the expression $1^n+2^n+...+x^n \equiv 0\pmod x$. That's because $a^n + (x-a)^n \equiv 0\pmod x$. $\endgroup$
    – Thomas Andrews
    Commented Nov 17, 2012 at 18:19
  • 1
    $\begingroup$ Sorry, that only works when both $x$ and $n$ are odd. If $x$ is even and $n$ odd, then $1^n+2^n+..+x^n\equiv (\frac{x}{2})^n \pmod x$. If $x$ is divisible by $4$ and $n>1$, then $(\frac x 2)^n \equiv 0\pmod x$. $\endgroup$
    – Thomas Andrews
    Commented Nov 17, 2012 at 18:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .