Distance between two points on the Moon

Question

Given latitude/longitude of two points on the Moon (lat1, lon1) and (lat2, lon2), compute the distance between the two points in kilometers, by using any formula that gives the same result as the haversine formula.

Input

• Four integer values lat1, lon1, lat2, lon2 in degree (angle) or
• four decimal values ϕ1, λ1, ϕ2, λ2 in radians.

Output

Distance in kilometers between the two points (decimal with any precision or rounded integer).

Haversine formula

where

• r is the radius of the sphere (assume that the Moon's radius is 1737 km),
• ϕ1 latitude of point 1 in radians
• ϕ2 latitude of point 2 in radians
• λ1 longitude of point 1 in radians
• λ2 longitude of point 2 in radians
• d is the circular distance between the two points

(source: https://en.wikipedia.org/wiki/Haversine_formula)

Other possible formulas

• d = r * acos(sin ϕ1 sin ϕ2 + cos ϕ1 cos ϕ2 cos(λ2 - λ1)) @miles' formula.
• d = r * acos(cos(ϕ1 - ϕ2) + cos ϕ1 cos ϕ2 (cos(λ2 - λ1) - 1)) @Neil's formula.

Example where inputs are degrees and output as rounded integer

42, 9, 50, 2  --> 284
50, 2, 42, 9  --> 284
4, -2, -2, 1  --> 203
77, 8, 77, 8  --> 0
10, 2, 88, 9  --> 2365

Rules

• The input and output can be given in any convenient format.
• Specify in the answer whether the inputs are in degrees or radians.
• No need to handle invalid latitude/longitude values
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• If possible, please include a link to an online testing environment so other people can try out your code!
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

Wolfram Language (Mathematica), 48 bytes

ArcCos[Sin@#*Sin@#3+Cos[#2-#4]Cos@#*Cos@#3]1737&

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Uses the formula d = r * acos( sin ϕ1 sin ϕ2 + cos ϕ1 cos ϕ2 cos(λ2 - λ1) ) where r = 1737

Answer 2

R + geosphere, 54 47 bytes

function(p,q)geosphere::distHaversine(p,q,1737)

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Takes input as 2-element vectors of longitude,latitude in degrees. TIO doesn't have the geosphere package but rest assured that it returns identical results to the function below.

Thanks to Jonathan Allan for shaving off 7 bytes.

R, 64 bytes

function(p,l,q,k)1737*acos(sin(p)*sin(q)+cos(p)*cos(q)*cos(k-l))

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Takes 4 inputs as in the test cases, but in radians rather than degrees.

Answer 3

APL (Dyalog Unicode), 40 35 bytes^SBCS

Anonymous tacit function. Takes {ϕ₁,λ₁} as left argument and {ϕ₂,λ₂} as right argument.

Uses the formula 2 r √(sin²(^{(ϕ₁-ϕ₂)}⁄₂) + cos ϕ₁ cos ϕ₂ sin²(^{(λ₁ – λ₂)}⁄₂))

3474ׯ1○.5*⍨1⊥(×⍨1○2÷⍨-)×1,2×.○∘⊃,¨

Try it online! (the r function converts degrees to radians)

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,¨ concatenate corresponding elements; {{ϕ₁ , ϕ₂} , {λ₁ , λ₂}}

⊃ pick the first; {ϕ₁ , ϕ₂}

∘ then

2×.○ product of their cosines; cos ϕ₁ cos ϕ₂
 ^{lit. dot "product" but with trig function selector (2 is cosine) instead of multiplication and times instead of plus}

1, prepend 1 to that; {1 , cos ϕ₁ cos ϕ₂}

(…)× multiply that by the result of applying following function to {ϕ₁ , λ₁} and {ϕ₂ , λ₂}:

 - their differences; {ϕ₁ - ϕ₂ , λ₁ - λ₂}

 2÷⍨ divide that by 2; {^{(ϕ₁ - ϕ₂)}⁄₂ , ^{(λ₁ - λ₂)}⁄₂}

 1○ sine of that; {sin(^{(ϕ₁ - ϕ₂)}⁄₂) , sin(^{(λ₁ - λ₂)}⁄₂)}

 ×⍨ square that (lit. self-multiply); {sin²(^{(ϕ₁ - ϕ₂)}⁄₂) , sin²(^{(λ₁-λ₂)}⁄₂)}

Now we have {sin²(^{(ϕ₁ - ϕ₂)}⁄₂) , cos ϕ₁ cos ϕ₂ sin²(^{(λ₁ - λ₂)}⁄₂)}

1⊥ sum that (lit. evaluate in base-1); sin²(^{(ϕ₁-ϕ₂)}⁄₂) + cos ϕ₁ cos ϕ₂ sin²(^{(λ₁ - λ₂)}⁄₂)

.5*⍨ square-root of that (lit. raise that to the power of a half)

 ¯1○ arcsine of that

 3474× multiply that by this

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The function to allow input in degrees is:

○÷∘180

÷180 argument divided by 180

○ multiply by π

Answer 4

JavaScript (ES7), 90 bytes

Note: see @OlivierGrégoire's post for a much shorter solution

A direct port of TFeld's answer. Takes input in radians.

(a,b,c,d,M=Math)=>3474*M.asin((M.sin((c-a)/2)**2+M.cos(c)*M.cos(a)*M.sin((d-b)/2)**2)**.5)

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Using the infamous with(), 85 bytes

Thanks to @l4m2 for saving 6 bytes

with(Math)f=(a,b,c,d)=>3474*asin((sin((c-a)/2)**2+cos(c)*cos(a)*sin((d-b)/2)**2)**.5)

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Answer 5

JavaScript (Node.js), 65 bytes

(a,b,c,d,C=Math.cos)=>1737*Math.acos(C(a-c)+C(a)*C(c)*(C(d-b)-1))

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Based on Kevin Cruijssen's answer, Miles' and Neil's comments, and upon request of Arnauld.

Answer 6

Python 2, 95 bytes

lambda a,b,c,d:3474*asin((sin((c-a)/2)**2+cos(c)*cos(a)*sin((d-b)/2)**2)**.5)
from math import*

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Takes input in radians.

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Old version, before i/o was slacked: Takes input as integer degrees, and returns rounded dist

Python 2, 135 bytes

lambda a,b,c,d:int(round(3474*asin((sin((r(c)-r(a))/2)**2+cos(r(c))*cos(r(a))*sin((r(d)-r(b))/2)**2)**.5)))
from math import*
r=radians

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Answer 7

Java 8, 113 92 88 82 bytes

(a,b,c,d)->1737*Math.acos(Math.cos(a-c)+Math.cos(a)*Math.cos(c)*(Math.cos(d-b)-1))

Inputs a,b,c,d are ϕ1,λ1,ϕ2,λ2 in radians.

-21 bytes using @miles' shorter formula.
-4 bytes thanks to @OlivierGrégore because I still used {Math m=null;return ...;} with every Math. as m., instead of dropping the return and use Math directly.
-6 bytes using @Neil's shorter formula.

Try it online.

Explanation:

(a,b,c,d)->                  // Method with four double parameters and double return-type
  1737*Math.acos(            //  Return 1737 multiplied with the acos of:
   Math.cos(a-c)             //   the cos of `a` minus `c`,
   +Math.cos(a)*Math.cos(c)  //   plus the cos of `a` multiplied with the cos of `c`
   *(Math.cos(d-b)-1))       //   multiplied with the cos of `d` minus `b` minus 1

Answer 8

Ruby, 87 70 69 bytes

->a,b,c,d{extend Math;1737*acos(cos(a-c)+cos(a)*cos(c)*(cos(d-b)-1))}

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Now using Neil's method, thanks to Kevin Cruijssen.

Answer 9

Haskell, 68 66 52 51 bytes

s=cos;(a!b)c d=1737*acos(s(a-c)+s a*s c*(s(d-b)-1))

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-1 byte thanks to BMO

Answer 10

Japt, 55 50 bytes

MsU *MsW +McU *McW *McX-V
ToMP1/7l¹ñ@McX aUÃv *#­7

Not necessarily quite as precise as the other answers, but boy did I have fun with this one. Allow me to elaborate.
While in most languages, this challenge is quite straightforward, Japt has the unfortunate property that there is no built-in for neither arcsine nor arccosine. Sure, you can embed Javascript in Japt, but that would be what ever the opposite of Feng Shui is.

All we have to do to overcome this minor nuisance is approximate arccosine and we're good to go!

The first part is everything that gets fed into the arccosine.

MsU *MsW +McU *McW *McX-V
MsU                        // Take the sine of the first input and
    *MsW...                // multiply by the cos of the second one etc.

The result is implicitly stored in U to be used later.

Following that, we need to find a good approximation for arccosine. Since I'm lazy and not that good with math, we're obviously just going to brute-force it.

ToMP1/7l¹ñ@McX aUÃv *#­7
T                       // Take 0
 o                      // and create a range from it
  MP                    // to π
    1/7l¹               // with resolution 1/7!.
         ñ@             // Sort this range so that
           McX          // the cosine of a given value
               aU       // is closest to U, e.g. the whole trig lot
                        // we want to take arccosine of.
                 Ã      // When that's done,
                  v     // get the first element
                    *#­7 // and multiply it by 1737, returning implicitly.

We could've used any large number for the generator resolution, manual testing showed 7! is sufficiently large while being reasonably fast.

Takes input as radians, outputs unrounded numbers.

Shaved off five bytes thanks to Oliver.

Try it online!

Answer 11

Jelly,  23 22  18 bytes

-4 bytes thanks to miles (use of { and } while using their formula.

;I}ÆẠP+ÆSP${ÆA×⁽£ġ

A dyadic function accepting [ϕ1, ϕ2,] on the left and [λ1, λ2] on the right in radians that returns the result (as floating point).

Try it online!

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Mine... (also saved a byte here by using a {)

,IÆẠCH;ÆẠ{Ḣ+PƊ½ÆṢ×⁽µṣ

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Answer 12

MATLAB with Mapping Toolbox, 26 bytes

@(x)distance(x{:})*9.65*pi

Anonymous function that takes the four inputs as a cell array, in the same order as described in the challenge.

Note that this gives exact results (assuming that the Moon radius is 1737 km), because 1737/180 equals 9.65.

Example run in Matlab R2017b:

Answer 13

Python 3, 119 103 bytes

This uses degrees.

from math import*
def f(L):a,o,A,O=map(radians,L);return 1737*acos(cos(a-A)+cos(a)*cos(A)*(cos(O-o)-1))

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Answer 14

C (gcc), 100 88 65 64 bytes

88 → 65 using @miles' formula
65 → 64 using @Neil's formula

#define d(w,x,y,z)1737*acos(cos(w-y)+cos(w)*cos(y)*(cos(z-x)-1))

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Answer 15

Python 3, 79 bytes

from geopy import*
lambda a,b:distance.great_circle(a,b,radius=1737).kilometers

TIO doesn't have geopy.py

Answer 16

APL (Dyalog Unicode), 29 bytes^SBCS

Complete program. Prompts stdin for {ϕ₁,ϕ₂} and then for {λ₁,λ₂}. Prints to stdout.

Uses the formula r acos(sin ϕ₁ sin ϕ₂ + cos(λ₂ – λ₁) cos ϕ₁ cos ϕ₂)

1737ׯ2○+/(2○-/⎕)×@2×/1 2∘.○⎕

Try it online! (the r function converts degrees to radians)

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⎕ prompt for {ϕ₁,ϕ₂}

1 2∘.○ Cartesian trig-function application; {{sin ϕ₁,sin ϕ₂} , {cos ϕ₁,cos ϕ₂}}

×/ row-wise products; {sin ϕ₁ sin ϕ₂ , cos ϕ₁ cos ϕ₂}

(…)×@2 at the second element, multiply the following by that:

 ⎕ prompt for {λ₁,λ₂}

 -/ difference between those; λ₁ – λ₂

 2○ cosine of that; cos(λ₁ – λ₂)

Now we have {sin ϕ₁ sin ϕ₂ , cos(λ₁ – λ₂) cos ϕ₁ cos ϕ₂}

+/ sum; sin ϕ₁ sin ϕ₂ + cos(λ₁ – λ₂) cos ϕ₁ cos ϕ₂

¯2○ cosine of that; cos(sin ϕ₁ sin ϕ₂ + cos(λ₁ – λ₂) cos ϕ₁ cos ϕ₂)

1737× multiply r by that; 1737 cos(sin ϕ₁ sin ϕ₂ + cos(λ₁ – λ₂) cos ϕ₁ cos ϕ₂)

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The function to allow input in degrees is:

○÷∘180

÷180 argument divided by 180

○ multiply by π

Answer 17

Excel, 53 bytes

=1737*ACOS(COS(A1-C1)+COS(A1)*COS(C1)*(COS(D1-B1)-1))

Using @Neil's formula. Input in Radians.

Answer 18

Lobster, 66 bytes

def h(a,c,b,d):1737*radians arccos a.sin*b.sin+a.cos*b.cos*cos d-c

Uses miles's formula, but input is in degrees. This adds extra step of converting to radians before multiplying by radius.

Answer 19

PHP, 88 bytes

Port of Oliver answer

function f($a,$b,$c,$d,$e=cos){return 1737*acos($e($a-$c)+$e($a)*$e($c)*($e($d-$b)-1));}

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Answer 20

SmileBASIC, 60 bytes

INPUT X,Y,S,T?1737*ACOS(COS(X-S)+COS(X)*COS(S)*(COS(T-Y)-1))