## The most efficient way to implement an integer based power function pow(int, int)

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What is the most efficient way given to raise an integer to the power of another integer in C?

```// 2^3
pow(2,3) == 8

// 5^5
pow(5,5) == 3125
```

Exponentiation by squaring.

```int ipow(int base, int exp)
{
int result = 1;
for (;;)
{
if (exp & 1)
result *= base;
exp >>= 1;
if (!exp)
break;
base *= base;
}

return result;
}
```

This is the standard method for doing modular exponentiation for huge numbers in asymmetric cryptography.

Power Function in C/C++, The most efficient way to implement an integer based power function pow(int, int )?. What is the most efficient way given to raise an integer to� The most efficient way to implement an integer based power function pow(int, int) (17 answers) Closed 5 years ago . I've read that cmath calculates pow(a,b) by performing exp(b*log(a)) .

Note that exponentiation by squaring is not the most optimal method. It is probably the best you can do as a general method that works for all exponent values, but for a specific exponent value there might be a better sequence that needs fewer multiplications.

For instance, if you want to compute x^15, the method of exponentiation by squaring will give you:

```x^15 = (x^7)*(x^7)*x
x^7 = (x^3)*(x^3)*x
x^3 = x*x*x
```

This is a total of 6 multiplications.

It turns out this can be done using "just" 5 multiplications via addition-chain exponentiation.

```n*n = n^2
n^2*n = n^3
n^3*n^3 = n^6
n^6*n^6 = n^12
n^12*n^3 = n^15
```

There are no efficient algorithms to find this optimal sequence of multiplications. From Wikipedia:

The problem of finding the shortest addition chain cannot be solved by dynamic programming, because it does not satisfy the assumption of optimal substructure. That is, it is not sufficient to decompose the power into smaller powers, each of which is computed minimally, since the addition chains for the smaller powers may be related (to share computations). For example, in the shortest addition chain for a¹⁵ above, the subproblem for a⁶ must be computed as (a³)² since a³ is re-used (as opposed to, say, a⁶ = a²(a²)², which also requires three multiplies).

Write a program to calculate pow(x,n), Given two numbers base and exponent, pow() function finds x raised to int a;. // Using typecasting for. // integer result. a = ( int )( pow (5, 2) + 0.5); or you want to share more information about the topic discussed above. We use cookies to ensure you have the best browsing experience on our website. The most efficient way to implement an integer based power function pow(int, int) The only two methods that I know is, A single for loop: extremely slow. Rewrite recursively calculate. I wonder is there a faster algorithm than these two? Any bitwise technique are welcome. Thank you. C# demos for the two algorithms:

If you need to raise 2 to a power. The fastest way to do so is to bit shift by the power.

```2 ** 3 == 1 << 3 == 8
2 ** 30 == 1 << 30 == 1073741824 (A Gigabyte)
```

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Here is the method in Java

```private int ipow(int base, int exp)
{
int result = 1;
while (exp != 0)
{
if ((exp & 1) == 1)
result *= base;
exp >>= 1;
base *= base;
}

return result;
}
```

Efficiently implement power function, Problem: Write a function in Java to calculate the power of integers. Method Signature: power(int x, int y) The most efficient way to implement an integer based power function pow(int, int) What is the most efficient way given to raise an integer to the power of another integer in C?// 2^3 pow(2,3)== 8// 5^5 pow(5,5)== 3125…

```int pow( int base, int exponent)

{   // Does not work for negative exponents. (But that would be leaving the range of int)
if (exponent == 0) return 1;  // base case;
int temp = pow(base, exponent/2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
```

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Java Program to Calculate the Power of a Number, This article we will tell you how to use Power Function In C to calculate How To Best Utilize Power Function In C? which you can call anytime you want to calculate the power of any integer. 1. 2. 3. 4. 5. 6. 7. 8. 9. int power(int base, int exponent) The pow() function is present in math.h header file. C library function - pow() - The C library function double pow(double x, double y) returns x raised to the power of y i.e. xy.

C++ Program to Calculate Power of a Number, public class Power { public static void main(String[] args) { int base = 3, In this program, we use Java's Math.pow() function to calculate the power of the given� C++ pow() The pow() function computes a base number raised to the power of exponent number.

Raise numbers to a power: here's how to exponentiate , Example 2: Compute power using pow() Function. #include <iostream> #include <cmath> using namespace std; int main() { float base, exponent, result; cout� One such example is pow(5, 2). When assigned to an integer, it outputs 24 on some compilers and works fine for some other compilers. But pow(5, 2) without any assignment to an integer outputs 25. This is because 5 2 (i.e. 25) might be stored as 24.9999999 or 25.0000000001 because the return type is double. When assigned to int, 25.0000000001

• If you're sticking to actual `int`s (and not some huge-int class), a lot of calls to ipow will overflow. It makes me wonder if there's a clever way to pre-calculate a table and reduce all the non-overflowing combinations to a simple table lookup. This would take more memory than most of the general answers, but perhaps be more efficient in terms of speed.
• `pow()` not a safe function
• @AkhilJain: It's perfectly good C; to make it valid also in Java, replace `while (exp)` and `if (exp & 1)` with `while (exp != 0)` and `if ((exp & 1) != 0)` respectively.
• Your function should probably have `unsigned exp`, or else handle negative `exp` properly.
• @ZinanXing Multiplying n times results in more multiplications and is slower. This method saves multiplications by effectively reusing them. E.g., to calculate n^8 the naïve method of `n*n*n*n*n*n*n*n` uses 7 multiplications. This algorithm instead computes `m=n*n`, then `o=m*m`, then `p=o*o`, where `p` = n^8, with just three multiplications. With large exponents the difference in performance is significant.