Evaluate a single variable polynomial equation
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;
}
[APL (Dyalog Unicode)], 11 3 1 …
4y ago
Japt `-x`, 6 5 bytes ËV …
4y ago
Ruby, 50 bytes ```ruby def …
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[Haskell], 20 bytes …
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Ruby, 38 bytes Simple map and …
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[Raku], 19 bytes (Z …
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Japt, 12 bytes ÊÆgX VpX …
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[JavaScript (Node.js)], 40 byt …
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Vyxal, 6, 5, 4 bytes ``` Źe …
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[Jelly], 1 byte ḅ Tr …
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[Python 3], 46 bytes …
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JavaScript (Node.js), 33 bytes …
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Pyth, 10 bytes sR^HhZhA …
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[Python 3], 46 bytes ```pyt …
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[C++ (gcc)], 61 bytes …
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Scala, 18 bytes ``` x=>.:\(. …
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J, 2 bytes ```J p. ``` …
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Ruby, 29 bytes ```ruby ->l …
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BQN, 13 bytes ```bqn {+´(𝕨 …
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[Python 3], 167 127 118 117 94 …
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20 answers
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Haskell, 20 bytes
f x=foldl((+).(x*))0
Takes input coefficients from highest degree to lowest.
21 bytes
x%(h:t)=h+x*x%t
x%_=0
0 comment threads
APL (Dyalog Unicode), 11 3 1 byte
⊥
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.
0 comment threads
Ruby, 50 bytes
def f(k,x)k.length>1?k[0]+f(k[1..-1],x)*x:k[-1]end
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.
Raku, 19 bytes
(*Z*(*X**0..*)).sum
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(*×$_+*)}
0 comment threads
Vyxal, 6, 5, 4 bytes
Źe*∑
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
0 comment threads
Jelly, 1 byte
ḅ
Essentially just Razetime's APL answer, except in that ḅ 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.
0 comment threads
JavaScript (Node.js), 33 bytes
a=>n=>a.reduce((o,x,y)=>o+x*n**y)
0 comment threads
Scala, 18 bytes
x=>_.:\(.0)(_+_*x)
Not too complicated. Takes the list of coefficients and folds from the right, multiplying the accumulator by x each time and adding the next coefficient. reduceRight would've worked instead of :\(0), but it's golfier, and I haven't used /: or :\ in a while.
0 comment threads
J, 2 bytes
p.
J likes inflections so it won't beat APL, but J is still a contender :).
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2 comment threads