# 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 …

4y ago

[Haskell], 20 bytes …

4y ago

Ruby, 38 bytes Simple map and …

4y ago

[Raku], 19 bytes (Z …

4y ago

Japt, 12 bytes ÊÆgX VpX …

4y ago

[JavaScript (Node.js)], 40 byt …

4y ago

Vyxal, 6, 5, 4 bytes ``` Źe …

4y ago

[Jelly], 1 byte ḅ Tr …

4y ago

[Python 3], 46 bytes …

3y ago

JavaScript (Node.js), 33 bytes …

3y ago

Pyth, 10 bytes sR^HhZhA …

4y ago

[Python 3], 46 bytes ```pyt …

3y ago

[C++ (gcc)], 61 bytes …

3y ago

Scala, 18 bytes ``` x=>.:\(. …

3y ago

J, 2 bytes ```J p. ``` …

3y ago

Ruby, 29 bytes ```ruby ->l …

3y ago

BQN, 13 bytes ```bqn {+´(𝕨 …

3y ago

[Python 3], 167 127 118 117 94 …

3y ago

## 20 answers

#
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.

# 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

#
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

# 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}}
```

#### 0 comment threads

# 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

# 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 :).

## 2 comment threads