Sayan - 1 year ago 98

Java Question

Can anyone help me with finding an efficient code to find 10 power x?

`private int power(int base, int exp)`

{

int result = 1;

while (exp != 0)

{

if ((exp & 1) == 1)

result *= base;

exp >>= 1;

base *= base;

}

return result;

}

Source of code from here, but I am looking for a way where the input could be 3.14 (double). I also cannot use any library functions. The power can be a real number. So it is not just a simple integer algorithm where we can find by Exponentiation by Squaring.

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Answer Source

Instead of calculating `10^x`

you can calculate `e^(x*ln(10))`

. So your question boil down to this one. I can offer a simple implementation by calculating the Taylor series:

```
double tenPow(double x) {
// ln(10)
double logTen = 2.3025850929940456840179914546843642076011014886287729;
double sum = 0.;
x *= logTen; // x * ln(10)
double tmp = 1.;
for (double i = 1.; tmp > 0.; i += 1.) {
sum += tmp;
tmp *= x;
tmp /= i;
}
return sum;
}
```

**EDIT1**
As noted in the comment this method is not effective for large `x`

. But we can break `x`

into its integer and fractional parts, so `x = xi + xf`

. `e^(xi + xf) = e^xf * e^xi`

. `e^xi`

can be effectively calculated using recursive exponentiation. And `e^xf`

can be calculated using the Taylor series.

**EDIT2**
Here is my implementation of this algorithm:

```
public static double fastTenPow(double x) {
if (x < 0.) {
return 1. / fastTenPow(-x);
}
int intPart = (int) x;
double fractPart = x - intPart;
return fractTenPow(fractPart) * intTenPow(intPart);
}
private static double intTenPow(int x) { // copied it
double res = 1.;
double base = 10.;
while (x != 0) {
if ((x & 1) == 1) {
res *= base;
}
x >>= 1;
base *= base;
}
return res;
}
private static final double LOG_TEN = 2.3025850929940456840179914546843642076011014886287729;
private static final double EPS = 0.00000000000000001;
private static double fractTenPow(double x) {
double sum = 0.;
x *= LOG_TEN;
double tmp = 1.;
for (double i = 1.; tmp > EPS; i += 1.) {
sum += tmp;
tmp *= x;
tmp /= i;
}
return sum;
}
```

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