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Challenges

Evaluate a single variable polynomial equation

+8
−0

Challenge

Given a list of n numbers and x, compute $a + bx^1 + cx^{2} + ... + zx^{n-1}$, where a is the first value in the list, b is the second, etc. n is at most 256 and at least 0. The input value(s) can be any 32-bit float

Input can be in any format of choice, as long as it is a list of numbers and x. (And this'll likely stay this way, even if input rules change over time)

Test inputs

1.0, 182 -> 1
1.0, 2.0, 4 -> 9
2.5, 2.0, 0.5 -> 3.5
1.0, 2.0, 3.0, 4.0, 1.5 -> 24.25

Example ungolfed program (Rust)

// dbg! is a logging function, prints the expression and it's output.
// Good for seeing what's happening

// Test setup
pub fn main() {
    let inp: &[f32] = &[1.0, 2.0, 3.0, 4.0];
    let x: f32 = 1.5;
    dbg!(evaluate_polynomial(inp, x)); // take inputs, print result	
}

// Actual challenge answer function
pub fn evaluate_polynomial(inp: &[f32], x: f32) -> f32 {
    let mut accum: f32 = 0.0;

    for (idx, val) in inp.iter().enumerate() {
        // x.pow(idx) * val
        accum += dbg!(x.powf(idx as f32) * val);
    }

    return accum;
}
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2 comments

"Input can be in any format of choice, as long as it is a list of numbers" ― does this mean we can take the list in reverse order? Adám‭ 2 months ago

@Adam Effectively yes, and i don't feel like going back to change that to "correctly ordered list of numbers", so go for it moony‭ 2 months ago

10 answers

+7
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APL (Dyalog Unicode), 11 3 1 byte

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Anyone who can golf this further gets a cookie!

Function submission which takes reversed coefficients as right argument and $x$ as left argument.

-8 bytes from dzaima and rak1507(APL Orchard).

-2 bytes from Adám‭.

Uses a mixed base conversion.

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0 comments

+4
−0

Japt, 12 bytes

ÊÆgX *VpXÃr+

Loops through the range of integers 0 to n-1, calculates each term, and sums.

Try it

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+3
−0

Vyxal, 6, 5, 4 bytes

Źe*∑

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Takes input in the format coeffs, x

Explained

Źe*∑
Ź      # Generate range [0, len(coeffs))
 e     # Calculate x ** [0, len(coeffs) (vectorising)
  *    # Multiply the coefficients by the exponated x's. The lists are extended with 0s to be the same length
   ∑   # Sum that list and output
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+3
−0

Haskell, 20 bytes

f x=foldl((+).(x*))0

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Takes input coefficients from highest degree to lowest.

21 bytes

x%(h:t)=h+x*x%t
x%_=0

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+3
−0

Ruby, 50 bytes

def f(k,x)k.length>1?k[0]+f(k[1..-1],x)*x:k[-1]end

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This uses the Horner's method recursively, because I think it'll be slightly shorter than using a loop or builtin array functions.

Also, this is my first post on this website ... er my first real attempt at golfing something in Ruby, so please feel free to suggest ways to shorten my solution.

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+3
−0

Japt -x, 6 5 bytes

Ë*VpE

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Ë*VpE     :Implicit input of array U and float V
Ë         :Map each element D at 0-based index E
 *        :  Multiply D by
  VpE     :  V raised to the power of E
          :Implicit output of sum of resulting array
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2 comments

Hey, welcome :D I knew there were a few Japt tricks I still didn't know about... Quintec‭ 2 months ago

Turns out I was slightly overthinking it, @Quintec; a simple map does the trick. Shaggy‭ 2 months ago

+3
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JavaScript (Node.js), 40 bytes

f=(a,b,c=1)=>a.reduce((d,e)=>d+e*(c*=b))

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+2
−0

Jelly, 1 byte

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Essentially just Razetime's APL answer, except in that vectorizes rather than carrying out mixed base conversion--irrelevant if, as is the case here, the provided base is scalar. Takes a reversed coefficient list on the left and $x$ on the right.

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+2
−0

Raku, 19 bytes

(*Z*(*X**0..*)).sum

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Is it concerning that my solution is over 30% asterisks?

Explanation

(             ).sum  # Get the sum of
 *Z*(        )       # The input list zip multiplied by
     *X**            # The second input to the power of
         0..*        # 0 to infinity

An alternate curried solution for 19 bytes is:

{*.reduce(*×$_+*)}

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+2
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Ruby, 38 bytes

Simple map and sum over the coefficients. No TIO link, this uses numbered lambda parameters which require Ruby 2.7.

->l,x{l.each_with_index.sum{_1*x**_2}}
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