Distance between two points in n-dimensional space

Question

Here is another simple one:

The Challenge

Given two points in an n-dimensional space, output the distance between them, also called the Euclidean distance.

• The coordinates will be rational numbers; the only limits are the restrictions of your language.
• Lowest dimension is 1, highest is whatever your language can handle
• You may assume that the two points are of the same dimension and that there will be no empty input.
• The distance has to be correct to at least 3 decimal places. If your language does not support floating point numbers, output the nearest whole number.

Rules

• As usual, function or full program allowed.
• Input may be taken from STDIN, command line- or function arguments.
• Input format is up to you, specify which one you used in your answer.
• Output may be provided by printing to stdout or return value.
• This is code-golf so lowest byte-count wins! In case of a tie, the earlier answer wins.

Test cases

Each point is represented by a list of length n.

[1], [3] -> 2
[1,1], [1,1] -> 0
[1,2], [3,4] -> 2.82842712475
[1,2,3,4], [5,6,7,8] -> 8
[1.5,2,-5], [-3.45,-13,145] -> 150.829382085
[13.37,2,6,-7], [1.2,3.4,-5.6,7.89] -> 22.5020221314

Happy Coding!

Answer 1

MATL, 2 bytes

ZP

Try it online!

The ZP function (corresponding to MATLAB's pdist2) computes all pairwise distances between two sets of points, using Euclidean distance by default. Each set of points is a matrix, and each point is a row. In this case it produces a single result, which is the distance between the two points.

Answer 2

MATL, 4.0 3 bytes

Thanks for -1 by @AndrasDeak!

-Zn

Reads two vectors (via implicit input that is requested by -) then substracts those and calculates the norm of their difference with Zn.

Try it Online!

Answer 3

Pyth, 2 bytes

.a

.a - L2 norm of vector difference of A[0] and A[1].

Literally a function that does this problem

Try it here.

Answer 4

Jelly, 4 bytes

_²S½

Try it online!

How it works

_²S½    Main link. Left input: A (list). Right input: B (list).

_       Subtract B from A, element by element.
 ²      Square all differences.
  S     Add all squares.
   ½    Take the square root of the sum.

Answer 5

Mathematica, 11 bytes

Norm[#-#2]&

Input as two lists, output as a number. If the input is exact (integers, rationals, etc.) the output will be exact as well. If the input contains a floating-point number, the output will be a float as well.

Answer 6

Octave, 15 bytes

@(x,y)norm(x-y)

Example:

octave:1> d=@(x,y)norm(x-y);
octave:2> d([13.37,2,6,-7], [1.2,3.4,-5.6,7.89])
ans =  22.502

Answer 7

CJam, 11 8 bytes

Thanks to Dennis for saving 3 bytes.

q~.-:mhz

Run all test cases.

Explanation

q~   e# Read and evaluate input.
.-   e# Vectorised difference.
:mh  e# Reduce √(x²+y²) over the list.
z    e# Take abs() to handle 1D input.

See this tip for why :mh works.

Answer 8

Haskell, 46 bytes

d :: Floating c => [c] -> [c] -> c
d a=sqrt.sum.map((^2).uncurry(flip(-))).zip a

Haskell, 35 bytes (By @nimi)

d :: Float c => [c] -> [c] -> c
d a=sqrt.sum.zipWith(((^2).).(-))a

Haskell, 31 bytes

Like this Scala answer, takes input as a sequence of tuples

<hack>

d :: Float c => [(c,c)] -> c
d=sqrt.sum.map$(^2).uncurry(-)

</hack>

Examples:

Prelude> d [1] [3]
2.0
Prelude> d [1,1] [1,1]
0.0
Prelude> d [1,2,3,4] [5,6,7,8]
8.0
Prelude> d [1.5,2,-5] [-3.45,-13,145]
150.82938208452623
Prelude> d [13.37,2,6,-7] [1.2,3.4,-5.6,7.89]
22.50202213135522

Answer 9

APL, 14 11 bytes

.5*⍨(+/-×-)

This is dyadic function train that takes the vectors on the left and right and returns the Euclidean norm of their difference.

Explanation:

       -×-)  ⍝ Squared differences
    (+/      ⍝ Sum them
.5*⍨         ⍝ Take the square root

Try it here

Saved 3 bytes thanks to Dennis!

Answer 10

Julia, 16 bytes

N(x,y)=norm(x-y)

This is a function that accepts two arrays and returns the Euclidean norm of their difference as a float.

You can verify all test cases at once online here.

Answer 11

J, 9 bytes

+&.*:/-/>

This is a function that takes one set of coordinates from the other (-/>), and then performs a sum + under &. square *:.

The input should be in the format x y z;a b c where x y z is your first set of co-ordinates and a b c is the other.

Answer 12

Java, 130 117 114 107 105 bytes

This is the obvious solution. I don't usually golf in Java, but I was curious to see if Java could beat the Brainfuck version. Doesn't seem like I did a good job then.. Maybe one could use the new Map/Reduce from Java 8 to save some bytes.

Thanks to @flawr (13 bytes), @KevinCruijssen (9 bytes) and @DarrelHoffman (3 bytes)!

Golfed:

double d(float[]a,float[]b){float x=0,s;for(int i=0;i<a.length;x+=s*s)s=a[i]-b[i++];return Math.sqrt(x);}

Ungolfed:

double d(float[] a, float[] b) {
  float x=0,s;

  for(int i=0; i<a.length; x+=s*s)
    s = a[i] - b[i++];

  return Math.sqrt(x);
}

Answer 13

golflua, 43 chars

\d(x,y)s=0~@i,v i(x)s=s+(v-y[i])^2$~M.q(s)$

Works by calling it as

> w(d({1,1},{1,1}))
0
> w(d({1,2},{3,4}))
2.82842712475
> w (d({1,2,3,4},{5,6,7,8}))
8

A Lua equivalent would be

function dist(x, y)
    s = 0
    for index,value in ipairs(x)
       s = s + (value - y[index])^2
    end
    return math.sqrt(s)
end

Answer 14

Seriously, 12 bytes

,iZ`i-ª`MΣ√A

Try it online!

Explanation:

,iZ`i-ª`MΣ√A
,iZ           get input, flatten, zip
   `   `M     map:
    i-ª         flatten, subtract, square
         Σ√A  sum, sqrt, abs

Answer 15

Perl 6, 30 29 26 24 bytes

{sqrt [+] ([Z-] $_)»²}

(Thanks @b2gills for 2 more bytes lost)

usage

my &f = {sqrt [+] (@^a Z-@^b)»²};

say f([1], [3]); # 2
say f([1,1], [1,1]); # 0
say f([1,2], [3,4]); # 2.82842712474619
say f([1,2,3,4], [5,6,7,8]); # 8
say f([1.5,2,-5], [-3.45,-13,145]); # 150.829382084526
say f([13.37,2,6,-7], [1.2,3.4,-5.6,7.89]); # 22.5020221313552

Answer 16

JavaScript ES7, 45 ES6, 37 bytes

a=>Math.hypot(...a.map(([b,c])=>b-c))

Expects an array of pairs of coordinates, one from each vector, e.g. [[1, 5], [2, 6], [3, 7], [4, 8]]. If that is unacceptable, then for 42 bytes:

(a,b)=>Math.hypot(...a.map((e,i)=>e-b[i]))

Expects two arrays of equal length corresponding to the two N-dimensional vectors, e.g. [1, 2, 3, 4], [5, 6, 7, 8]. Edit: Saved 3 bytes thanks to @l4m2. (Also, did nobody notice my typo?)

Answer 17

Python 2, 47 bytes

A straight forward solution. The function expects 2 points as sequences of numbers, and returns the distance between them.

lambda a,b:sum((d-e)**2for d,e in zip(a,b))**.5

Example:

>>> f([13.37, 2, 6, -7], [1.2, 3.4, -5.6, 7.89])
22.50202213135522

Answer 18

Ruby, 52

->p,q{t=0;p.size.times{|i|t+=(p[i]-q[i])**2}
t**0.5}

In test program

f=->p,q{t=0;p.size.times{|i|t+=(p[i]-q[i])**2}
t**0.5}

p f[[1], [3]] # 2
p f[[1,1], [1,1]] # 0
p f[[1,2], [3,4]] # 2.82842712475
p f[[1,2,3,4], [5,6,7,8]] # 8
p f[[1.5,2,-5], [-3.45,-13,145]] # 150.829382085
p f[[13.37,2,6,-7], [1.2,3.4,-5.6,7.89]] # 22.5020221314

Answer 19

AppleScript, 241 239 bytes

This is golfed code, but I've put comments in in the form --.

on a()    -- Calling for getting input
set v to{1}          -- Arbitrary placeholder
repeat until v's item-1=""       -- Repeat until no input is gathered
set v to v&(display dialog""default answer"")'s text returned   -- Add input to list
end      -- End the repeat
end      -- End the method
set x to a()   -- Set the array inputs
set y to a()
set z to 0     -- Sum placeholder
set r to 2     -- 2 is the first significant array index
repeat(count of items in x)-2     -- Loop through all but first and last of the array
set z to z+(x's item r-y's item r)^2    -- Add the square of the difference
end   -- End the repeat
z^.5  -- Return the square root of the sum

This uses the same algorithm as most of the other programs here.

Answer 20

Haskell, 32 bytes

((sqrt.sum.map(^2)).).zipWith(-)

---

λ> let d = ((sqrt.sum.map(^2)).).zipWith(-)
λ> d [1] [3]
2.0
λ> d [1,1] [1,1]
0.0
λ> d [1,2] [3,4]
2.8284271247461903
λ> d [1..4] [5..8]
8.0
λ> d [1.5,2,-5] [-3.45,-13,145]
150.82938208452623
λ> d [13.37,2,6,-7] [1.2,3.4,-5.6,7.89]
22.50202213135522

Answer 21

C (gcc), 79 77 bytes

-2 bytes thanks to ceilingcat.

TIO requires compiler flag -lm. GCC of MinGW (that I use) does not.

f(p,q,d,s)float*p,*q,s;{for(s=0;d--;)s+=*p++*=*p-=*q++;printf("%f",sqrt(s));}

Try it online!

Answer 22

Vyxal, 2 bytes

∆d

Try it Online!

Laughs in built-in. Generalised euclidian distance do be like that sometimes.

Answer 23

Desmos, 26 bytes

f(a,b)=total((a-b)^2)^{.5}

The function \$f(a,b)\$ takes in two lists, \$a\$ and \$b\$, representing the two points, and returns the distance between them.

I think the code is pretty self explanatory, even for someone unfamiliar with Desmos.

Try It On Desmos!

Try It On Desmos! - Prettified

Answer 24

R, 38 bytes and 25 bytes

Sure, it is impossible to compete with the shortest 4 bytes answer, but I wanted to share anyway a hopefully interesting solution (38 bytes):

f=\(x,y)sd(c(x-y,y-x))*(nrow(x)-.5)^.5

How does it work? Standard deviation of a sample is calculated as a square root of the sum of squares of deviation scores divided by sample size minus one.

The trick is to add to the sample a second set of the same numbers with the opposite sign (like c(1,2,-1,-2)). This way we get rid of mean (=0) out of the formula, but also we are getting each number twice. To obtain the sum of squares we only need to "neutralize" the term sqrt(2/(N-1)), whereby the sample size N is 2*N_dimensions.

...but unfortunately this bright =) solution sucks to the old good straightforward formula (25 bytes):

g=\(x,y)(sum((x-y)^2))^.5

Here is the result:

> list.x <- list(matrix(1),
+                matrix(c(1,1),2),
+                matrix(c(1,2),2),
+                matrix(c(1,2,3,4), 4),
+                matrix(c(1.5,2,-5), 3),
+                matrix(c(13.37,2,6,-7),4))
> 
> list.y <- list(matrix(3), 
+                matrix(c(1,1), 2),
+                matrix(c(3,4),2),
+                matrix(c(5,6,7,8),4),
+                matrix(c(-3.45,-13,145),3),
+                matrix(c(1.2,3.4,-5.6,7.89),4))
> 
> correct = list(2,0,2.82842712475,8,150.829382085,22.5020221314)
> 
> do.call(rbind, Map(\(x,y,z)c(f(x,y),g(x,y),z),list.x,list.y,correct))
           [,1]       [,2]       [,3]
[1,]   2.000000   2.000000   2.000000
[2,]   0.000000   0.000000   0.000000
[3,]   2.828427   2.828427   2.828427
[4,]   8.000000   8.000000   8.000000
[5,] 150.829382 150.829382 150.829382
[6,]  22.502022  22.502022  22.502022

Answer 25

𝔼𝕊𝕄𝕚𝕟, 6 chars / 13 bytes

МŰМŷ…ï

Try it here (Firefox only).

Calculates norm of difference of input arrays.

Answer 26

Scala, 67 62 bytes

def e(a:(Int,Int)*)=math.sqrt(a map(x=>x._2-x._1)map(x=>x*x)sum)

Requires input as a sequence/vector of var-arg tuples
Example:

scala> e((1, 5), (2, 6), (3, 7), (4, 8))
res1: Double = 8.0

Answer 27

C#, 72 bytes

(float[]i,float[]n)=>System.Math.Sqrt(i.Zip(n,(x,y)=>(x-y)*(x-y)).Sum())

A simple solution using Linq.

Answer 28

Sage, 35 bytes

lambda a,b,v=vector:norm(v(a)-v(b))

This function takes 2 lists as input and returns a symbolic expression. The distance is calculated by performing vector subtraction on the lists and computing the Euclidean norm of the resultant vector.

Try it online

Answer 29

Clojure, 48 bytes

#(Math/sqrt(apply +(for[d(map - % %2)](* d d))))

Answer 30

05AB1E, 4 bytes

-nOt

Try it online!

Negative not y'all!

-    # a-b
 n   # (a-b)**2
  O  # sum((a-b)**2) for all a,b
   t # sqrt(sum((a-b)**2) for all a,b)