Reputation: 129
Is there any way to find nth root of the number without any external library in C? I'm working on a bare metal code so there is no OS. Also, no complete C is there.
Upvotes: 2
Views: 2658
Reputation: 339
You can try the nth_root C function :
// return a number that, when multiplied by itself nth times, makes N.
unsigned nth_root(const unsigned n, const unsigned nth) {
unsigned a = n, b, c, r = nth ? n + (n > 1) : n == 1 ;
for (; a < r; b = a + (nth - 1) * r, a = b / nth)
for (r = a, a = n, c = nth - 1; c && (a /= r); --c);
return r;
}
Upvotes: -1
Reputation: 211
Here there is a C implementation of the the nth root algorithm you can find in wikipedia. It needs an exponentiation algorithm, so I also include an implementation of a basic method for exponentiation by squaring that you can find also find in wikipedia.
double npower(double const base, int const n)
{
if (n < 0) return npower(1/base, -n)
else if (n == 0) return 1.0;
else if (n == 1) return base;
else if (n % 2) return base*npower(base*base, n/2);
else return npower(base*base, n/2);
}
double nroot(double const base, int const n)
{
if (n == 1) return base;
else if (n <= 0 || base < 0) return NAN;
else {
double delta, x = base/n;
do {
delta = (base/npower(x,n-1)-x)/n;
x += delta;
} while (fabs(delta) >= 1e-8);
return x;
}
}
Some comments on this:
base/n
, but this was just a guess.NAN
is usually defined in <math.h>
, so you would need to define it to be suitable for your needs.1e-8
and should be changed to something different. It should probably be proportional to the value of the base. Upvotes: 1
Reputation: 74
You can write a program like this for nth root. This program is for square root.
int floorSqrt(int x)
{
// Base cases
if (x == 0 || x == 1)
return x;
// Staring from 1, try all numbers until
// i*i is greater than or equal to x.
int i = 1, result = 1;
while (result < x)
{
if (result == x)
return result;
i++;
result = i*i;
}
return i-1;
}
You can use the same approach for nth root.
Upvotes: 1