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!

I'll give brainfuck a shot. Let's see what horrible monster comes out. — yyny
– yyny, Commented Jan 30, 2016 at 12:56
@flawr Yep, exactly. Just wanted to keep the title simple, since not everyone might know what that is at first glance. Could definetly write that in the challange tho :) — Denker
– Denker, Commented Jan 30, 2016 at 14:02
@DenkerAffe is it OK to return the distance squared if "your programming language does not support floating points"? This would make my brainfuck program a lot more accurate (Otherwise I'll have to implement some sort of estimation algorithm). — yyny
– yyny, Commented Jan 30, 2016 at 14:58
@DenkerAffe I think it's safe to say that brainfuck will never win a code golf. But it's just for fun anyways :) — yyny
– yyny, Commented Jan 30, 2016 at 16:17

NikoNyrh · Accepted Answer · 2017-01-17 00:17:39Z

1

Clojure, 48 bytes

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

answered Jan 17, 2017 at 0:17

NikoNyrh

2,72111 silver badges8 bronze badges

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Magic Octopus Urn · Accepted Answer · 2017-01-17 19:38:34Z

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)

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Elixir, 74 bytes

:math.sqrt Enum.reduce Enum.zip(p,q),0,fn({a,b},c)->:math.pow(a-b,2)+c end

You can try it online

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 24, 2017 at 0:59

Walerian Sobczak

1113 bronze badges

1

\$\begingroup\$ Welcome to PPCG! Nice first post! \$\endgroup\$

Riker
– Riker

2017-11-24 02:15:07 +00:00
Commented Nov 24, 2017 at 2:15

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Community · Accepted Answer · 2020-06-17 09:04:33Z

TI-Basic (TI-84 Plus CE), 15 bytes

Prompt A,B √(sum((_LA-_LB)²

TI-Basic is a tokenized language.

Prompts for input as two lists, and returns the Euclidian distance betwrrn them in Ans

Explanation:

Prompt A,B # 5 bytes, Prompts for two inputs; if the user inputs lists: # they are stored in _LA and _LB √(sum((_LA-_LB)² # 10 bytes, Euclidian distance between points #(square root of (sum of (squares of (differences of coordinates))))

Community · Accepted Answer · 2020-06-17 09:04:33Z

Excel VBA, 36 Bytes

Anonymous VBE immediate window function that takes input as two vectors, which are projected unto the range 1:2, and outputs to the VBE immediate window

[3:3]="=(A1-A2)^2":?[Sqrt(Sum(3:3))]

rturnbull · Accepted Answer · 2018-04-11 19:34:21Z

R, 4 bytes

dist

This is a built-in function to calculate the distance matrix of any input matrix. Defaults to euclidean distance.

Example usage:

> x=matrix(c(1.5,-3.45,2,-13,-5,145),2) > x [,1] [,2] [,3] [1,] 1.50 2 -5 [2,] -3.45 -13 145 > dist(x) 1 2 150.8294

If you're feeling disappointed because it's a built-in, then here's a non-built-in (or at least, it's less built-in...) version for 22 bytes (with thanks to Giuseppe):

pryr::f(norm(x-y,"F"))

This is an anonymous function that takes two vectors as input.

function(x,y)norm(x-y,"F") is shorter than your second version. — Giuseppe
– Giuseppe, Commented Apr 11, 2018 at 15:56

xbarbie · Accepted Answer · 2018-04-11 20:26:44Z

Python 3, 44 bytes

lambda*a:sum((x-y)**2for x,y in zip(*a))**.5

Try it online!

Ungolfed version:

 ₙ sqrt( ∑(xᵢ - yᵢ)² ) ⁱ⁼¹ def d(a, b): # two points (lists or tuples) return sum( # sum of... (x - y) ** 2 # (xᵢ - yᵢ)² for x,y in zip(a, b) # ) ** 0.5 # square root of sum

recursive · Accepted Answer · 2018-07-09 16:05:21Z

Stax, 8 bytes

äÖ╙í▼=╬b

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

\ zip the coordinates into pairs { for each pair, :s compute the absolute span J+ square and add to total F |Q square root result

Run this one

chunes · Accepted Answer · 2022-01-11 07:44:36Z

1

Factor + `math.distances`, 18 bytes

euclidian-distance

Try it online!

Built-in.

answered Jan 11, 2022 at 7:44

chunes

27.8k3 gold badges32 silver badges55 bronze badges

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Natte · Accepted Answer · 2022-01-11 14:40:05Z

Husk, 6 bytes

√Σzo□-

Try it online!

Explanation

√Σzo□- z zip input arguments o with composed function - difference □ squared Σ sum √ sqrt

roblogic · Accepted Answer · 2023-03-09 17:15:23Z

Fortran (GFortran), 85 80 bytes

function d(A,n);real A(n,2);s=0;do5 i=1,n;s=s+(A(i,1)-A(i,2))**2 5 d=sqrt(s);end

Try it online! _{85 bytes}

Instead of two vectors, I read all the data into two rows of array A. But Fortran 'helpfully' stores matrices by column so when I call the procedure A is transposed. _{Using line number 5 instead of enddo saves 2 bytes :)
Saved 5 bytes using function instead of subroutine}

If only the IMSL library was installed on TIO. I could have used the DISL2 function. — roblogic
– roblogic, Commented Mar 8, 2023 at 9:28

macOSisthebestOS · Accepted Answer · 2024-03-11 18:50:37Z

Swift 5.9, 86 bytes

import Foundation let f={(p:[(Double,_)])in sqrt(p.map{pow($0.1-$0.0,2)}.reduce(0,+))}

Don't Try It Online, because TIO is too old. Here's a JDoodle link instead.

f is of type ([(Double, Double)]) -> Double -- it takes an array of pairs of coordinates, one per vector.

I just used the subtract-square-sum-squareRoot trick borrowed from other answers.

bigyihsuan · Accepted Answer · 2024-06-07 20:58:43Z

Go, 97 bytes

import."math" func f(p,q[]float64)(d float64){for i:=range p{d+=Pow(p[i]-q[i],2)} return Sqrt(d)}

Attempt This Online!

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Pyke, 7 bytes

,A-MXs,

Try it here!

Transpose, apply subtract, map square, sum, sqrt.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 9, 2016 at 10:03

Blue

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Ruby, 50 bytes

Zip, then map/reduce. Barely edges out the other Ruby answer from @LevelRiverSt by 2 bytes...

->p,q{p.zip(q).map{|a,b|(a-b)**2}.reduce(:+)**0.5}

Try it online

Abel Tom · Accepted Answer · 2017-02-03 16:02:16Z

C 276 bytes

f(){*a,*b,d=0;l=1;a=malloc(sizeof(float)*50);b=malloc(sizeof(float)*50);scanf("%f",&a[0]);while(getchar()!='\n'){scanf("%f",&a[l]);l++;}l=1;scanf("%f",&b[0]);while(getchar()!='\n'){scanf("%f",&b[l]);l++;}for(int i=0;i<l;i++){d+=pow((a[i]-b[i]),2);}printf("%.5f",pow(d,0.5));}

Ungolfed version:

void f() { float *a,*b,d=0; int l=1; a=malloc(sizeof(float)*50); b=malloc(sizeof(float)*50); //Accept p1,p2,p3.....pn scanf("%f",&a[0]); while(getchar()!='\n') { scanf("%f",&a[l]); l++; } l=1; //Accept q1,q2,q3.....qn scanf("%f",&b[0]); while(getchar()!='\n') { scanf("%f",&b[l]); l++; } for(int i=0;i<l;i++) { d+=pow((a[i]-b[i]),2); } printf("\n%.5f",pow(d,0.5)); }

Pretty straightforward. This solution lets you measure distance between 2 points in upto 50 dimensions (can be increased).

In Cartesian coordinates, input the position of first point of n dimensions. (p1 p2 p3 ...pn) and press Enter.

Next, input the position of second point of n dimensions. (q1 q2 q3 ...qn) and press Enter to get the Euclidean Distance.

Jordan · Accepted Answer · 2022-11-16 21:48:10Z

J-uby, 35 bytes

+:zip|:*&(+:-|~:**&2)|:sum|~:**&0.5

Attempt This Online!

+:zip | :* & (+:- | ~:** & 2) | :sum | ~:** & 0.5 +:zip | # Zip input arrays, then :* & ( ) # map with +:- | ~:** & 2 # Difference, then square | :sum | # then sum, then ~:** & 0.5 # square root

The Thonnu · Accepted Answer · 2023-07-22 06:27:21Z

Thunno 2, 4 bytes

-²Sƭ

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Explanation

-²Sƭ # Implicit input - # Subtract ² # Square S # Sum ƭ # Square root # Implicit output

Stack Exchange Network

Distance between two points in n-dimensional space

The Challenge

Rules

Test cases

48 Answers 48

Clojure, 48 bytes

05AB1E, 4 bytes

Elixir, 74 bytes

TI-Basic (TI-84 Plus CE), 15 bytes

Excel VBA, 36 Bytes

R, 4 bytes

Python 3, 44 bytes

Stax, 8 bytes

Factor + `math.distances`, 18 bytes

Husk, 6 bytes

Explanation

Fortran (GFortran), 85 80 bytes

Swift 5.9, 86 bytes

Go, 97 bytes

Pyke, 7 bytes

Ruby, 50 bytes

C 276 bytes

J-uby, 35 bytes

Thunno 2, 4 bytes

Explanation

Linked

Hot Network Questions

Distance between two points in n-dimensional space

The Challenge

Rules

Test cases

48 Answers 48

Clojure, 48 bytes

05AB1E, 4 bytes

Elixir, 74 bytes

TI-Basic (TI-84 Plus CE), 15 bytes

Excel VBA, 36 Bytes

R, 4 bytes

Python 3, 44 bytes

Stax, 8 bytes

Factor + math.distances, 18 bytes

Husk, 6 bytes

Explanation

Fortran (GFortran), 85 80 bytes

Swift 5.9, 86 bytes

Go, 97 bytes

Pyke, 7 bytes

Ruby, 50 bytes

C 276 bytes

J-uby, 35 bytes

Thunno 2, 4 bytes

Explanation

Linked

Related

Hot Network Questions

Factor + `math.distances`, 18 bytes