CODEWITHSUNDEEP
.netlinqpolynomials
Naval
Naval

Reputation: 1

Calculate polynomial using LINQ

As a part of a LINQ tutorial I have the following exersise:

You're required to calculate the value of an nth-degree polynomial. The arguments of the function are:

-int x - the value of x for the polynomial

-IEnumerable coeffs - the polynomial coefficients

Example:

-Input: 2, [3,4,5]

-Output: 25 (calculated as 3*2^2+4*2+5)

Hint: you may need the overload of Aggregate() that takes a seed value. You may also need to introduce a separate variable for tracking the index of the element you're processing.

Any ideas?

Thank you so much!

Upvotes: 0

Views: 630

Answers (2)

Andrew Mashin
Andrew Mashin

Reputation: 11

As simple as that:

public static int Poly(int x, IEnumerable<int> coeffs)
{
    int idx = 0;
    return coeffs.Reverse().Aggregate(0, (p, q) => p + q * (int)Math.Pow(x, idx++));
}

Upvotes: 1

Alexandru Clonțea
Alexandru Clonțea

Reputation: 1876

First of all weird choice of IEnumerable coefficients. You could rewrite the below code to "subtract" rather than add to the current coefficient's rank, I guess that's what is expected by whoever designed the problem, since that would require using the overload which takes a seed value, since the first term's value would no longer be coefficient * x^0 (or simply coefficient) in that case. I just think this version is easier to follow and explain, and you should try writing the answer as expected based on this:

IEnumerable<int> coefficients = new int[] { 3, 4, 5 }.Reverse(); //we reverse the coefficients' order for simplicity, in real cases this might not be possible, since IEnumerable does not guarantee it is "replayable"
int x = 2;
int rank = 1; //we start from rank 1 since the first value is x^0 * coefficient, so it's already "added" in Aggregate and correct, we don't need to use the "seed value overload"
var polynomialValue = coefficients.Aggregate((sofar, current) =>
            {
                //we just need to worry about calculating the current value
                int currentValue = current * (int)Math.Pow(x, rank); //you could replace this with a for if you want or use Aggregate here as well, I was just lazy
                Console.WriteLine($"Adding {current} * {x}^{rank} = {currentValue}");
                rank++;

                //and returning the aggregated result which is what we calculated so far plus the current term
                return sofar + currentValue;
            });

Upvotes: 0

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