Benchmark: **2 vs * vs Math.Pow for equated number

Script Preparation code:

const count = 1000
arr1 = []
arr2 = []
for (let i=0; i<count; i++) {
	arr1.push(Math.round(Math.random()*255))
	arr2.push(Math.round(Math.random()*255))
}

pow = Math.pow

Tests:

n**2
for (let i=0; i<arr1.length; i++) { n = arr1[i], m = arr2[i]; (n-m)**2; }
n*n
for (let i=0; i<arr1.length; i++) { n = arr1[i], m = arr2[i]; (n-m)*(n-m); }
Math.pow
for (let i=0; i<arr1.length; i++) { n = arr1[i], m = arr2[i]; Math.pow(n-m,2); }
pow
for (let i=0; i<arr1.length; i++) { n = arr1[i], m = arr2[i]; pow(n-m,2); }

Rendered benchmark preparation results:

Suite status: <idle, ready to run>

Previous results

Test case name	Result
n**2
n*n
Math.pow
pow

Fastest: N/A

Slowest: N/A

Latest run results:

No previous run results

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Autogenerated LLM Summary (model llama3.2:3b, generated one year ago):

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Measuring JavaScript performance is crucial in today's web development landscape. Measuratio.net, where users create and run JavaScript microbenchmarks like the provided example.

In this benchmark, we're comparing three different approaches to calculate `(n-m)**2`:

1.  **Manual calculation using `*`**: This method involves multiplying two numbers: `n` and `m`. Since both are integers between 0 and 255, their product will always be an integer as well.
    *   Pros:
        *   No additional library dependencies
        *   Simple to implement
    *   Cons:
        *   Potential for integer overflow if the product exceeds the maximum allowed value (2^31-1)
        *   Less efficient due to multiplication, which is generally slower than exponentiation
2.  **Using `Math.pow`**: This method leverages JavaScript's built-in `Math.pow` function to calculate the square of a number. In this case, it's used as `(n-m)**2`.
    *   Pros:
        *   More efficient due to the optimized implementation in the JavaScript engine
        *   No risk of integer overflow since `Math.pow` handles arbitrary-precision arithmetic
3.  **Using the `**` operator**: This method uses the exponentiation operator (`**`) to calculate `(n-m)**2`. It's a shorthand for "n to the power of m", where the second operand is negative.
    *   Pros:
        *   More concise and readable compared to manual multiplication
        *   Often faster than `Math.pow` due to its native implementation in JavaScript engines
4.  **Using a library function (`pow`)**: This method relies on an external library function, which likely provides more robust and efficient exponentiation functionality.
    *   Pros:
        *   Can be more accurate due to the use of arbitrary-precision arithmetic
        *   May provide additional features or optimizations not available in JavaScript's built-in functions
5.  **Using `Math.pow` as a fallback**: This method uses `Math.pow` with two arguments, where the second argument is negative.
    *   Pros:
        *   Provides a clear and concise implementation of exponentiation with a negative base

**Library usage**: The `pow` function seems to be an external library that provides more robust exponentiation functionality. It's not part of JavaScript's built-in functions.

The provided benchmark uses four different test cases, each comparing one of the above methods:

*   `n**2`: Uses the `**` operator
*   `n*n`: Manual calculation using `*`
*   `Math.pow`: Uses the `Math.pow` function
*   `pow`: Relies on an external library function

These test cases cover different approaches to calculating `(n-m)**2`. The results will help determine which method is fastest, most efficient, or most accurate.

**Other alternatives**: Depending on the specific requirements and constraints of your project, you might consider other methods for calculating `(n-m)**2`, such as:

*   Using `**` with a negative base to calculate exponentiation
*   Implementing a custom exponentiation function using bitwise operations
*   Leveraging hardware acceleration or SIMD instructions (if available)

Related benchmarks:

**2 vs * vs Math.Pow for equated number (version: 0)

(n-m)*(n-m) vs (n-m)**2 vs Math.Pow((n-m),2) 0 ≤ n,m ≤ 255 n,m ∈ ℤ

Comparing performance of: n**2 vs n*n vs Math.pow vs pow

Created: 5 years ago by: Guest

Jump to the latest result

n**2

n*n

Math.pow

pow

Suite status: <idle, ready to run>

Fastest: N/A

Slowest: N/A

No previous run results

Autogenerated LLM Summary (model llama3.2:3b, generated one year ago):