Tetrahedron Surface Area

Question

The challenge

This challenge is very straightforward. Given four 3-dimensional points, calculate the surface area of the tetrahedron that they form. This is code-golf, so shortest code wins. Standard loopholes apply, with the added stipulation that any built-in function to do this task given four points is prohibited.

You can assume all four points will be distinct, and will be given via STDIN, 1 point per line. Each point will consist of three 16-bit unsigned integers. The exact format of each point can be modified if it makes things easier, such as three space separated integers. Having each point on a separate line is mandatory however. Output should be through STDOUT, to at least 2 decimal places.

For those of you who do not know, a tetrahedron is a 3-d solid, formed by 4 triangular faces.

Example

# input (format is up to you, see clarification above)
[23822, 47484, 57901]
[3305, 23847, 42159]
[19804, 11366, 14013]
[52278, 28626, 52757]

# output
2932496435.95

Please leave a note if you notice my math is wrong.

Answer 1

Python, 198 178 161 chars

V=eval('input(),'*4)
A=0
for i in range(4):F=V[:i]+V[i+1:];a,b,c=map(lambda e:sum((a-b)**2for a,b in zip(*e)),zip(F,F[1:]+F));A+=(4*a*b-(a+b-c)**2)**.5
print A/4

The input format is as given in the question.

It calculates the length of the edges adjacent to each of the faces and then uses Heron's formula.

Answer 2

APL, 59

f←{+.×⍨⊃1 2-.⌽(⊂⍵)×1 2⌽¨⊂⍺}
.5×.5+.*⍨(f/2-/x),2f/4⍴x←⎕⎕⎕-⊂⎕

Works by calculating cross products

Explanation
The first line defines a function that takes two arguments (implicity named ⍺ and ⍵), implicitly expects them to be numerical arrays of length 3, treat them as 3d vectors, and calculates the squared magnitude of their cross product.

                        ⊂⍺   # Wrap the argument in a scalar
                   1 2⌽¨     # Create an array of 2 arrays, by rotating `⊂⍺` by 1 and 2 places
             (⊂⍵)×           # Coordinate-wise multiply each of them with the other argument
        1 2-.⌽               # This is a shorthand for:
        1 2  ⌽               #   Rotate the first array item by 1 and the second by 2
           -.                #   Then subtract the second from the first, coordinate-wise
       ⊃                     # Unwrap the resulting scalar to get the (sorta) cross product
   +.×                       # Calculate the dot product of that...
      ⍨                      # ...with itself
f←{+.×⍨⊃1 2-.⌽(⊂⍵)×1 2⌽¨⊂⍺} # Assign function to `f`

The second line does the rest.

                         ⎕⎕⎕-⊂⎕ # Take 4 array inputs, create an array of arrays by subtracting one of them from the other 3
                       x←        # Assign that to x
                     4⍴          # Duplicate the first item and append to the end
                  2f/            # Apply f to each consecutive pair
            2-/x                 # Apply subtraction to consecutive pairs in x
          f/                     # Apply f to the 2 resulting arrays
         (f/2-/x),2f/4⍴x←⎕⎕⎕-⊂⎕ # Concatenate to an array of 4 squared cross products
   .5+.*⍨                        # Again a shorthand for:
   .5  *⍨                        #   Take square root of each element (by raising to 0.5)
     +.                          #   And sum the results
.5×                              # Finally, divide by 2 to get the answer

Answer 3

Matlab/Octave 103

I assume the values to be stored in the variable c. This uses the fact that the area of a triangle is the half length of the cross product of two of its side vectors.

%input
[23822, 47484, 57901;
3305, 23847, 42159;
19804, 11366, 14013;
52278, 28626, 52757]



%actual code
c=input('');
a=0;
for i=1:4;
    d=c;d(i,:)=[];
    d=d(1:2,:)-[1 1]'*d(3,:);
    a=a+norm(cross(d(1,:),d(2,:)))/2;
end
a

Answer 4

Python 3, 308 298 292 279 258 254

from itertools import*
def a(t,u,v):w=(t+u+v)/2;return(w*(w-t)*(w-u)*(w-v))**.5
z,x,c,v,b,n=((lambda i,j:(sum((i[x]-j[x])**2for x in[0,1,2]))**.5)(x[0],x[1])for*x,in combinations([eval(input())for i in">"*4],2))
print(a(z,x,v)+a(z,c,b)+a(b,v,n)+a(x,c,n))

This uses:

• The Pythagorean Theorem (in 3D) to work out the length of each line
• Heron's Formula to work out the area of each triangle

Answer 5

Mathematica 168 154

This finds the lengths of the edges of the tetrahedron and uses Heron's formula to determine the areas of the faces.

t = Subsets; p = Table[Input[], {4}];
f@{a_, b_, c_} := Module[{s = (a + b + c)/2}, N[Sqrt[s (s - #) (s - #2) (s -#3)] &[a, b, c], 25]]
  Tr[f /@ (EuclideanDistance @@@ t[#, {2}] & /@ t[p, {3}])]

---

There is a more direct route that requires only 60 chars, but it violates the rules insofar as it computes the area of each face with a built-in function, Area:

p = Table[Input[], {4}];
N[Tr[Area /@ Polygon /@ Subsets[p, {3}]], 25]

Answer 6

Sage – 103

print sum((x*x*y*y-x*y*x*y)^.5for x,y in map(differences,Combinations(eval('vector(input()),'*4),3)))/2

The input-reading part is adapted from Keith Randall’s answer.

Answer 7

R, 96 80 bytes

\(x)sum(combn(1:4,3,\(i)(4*(k<-dist(x[i,])^2)[1]*k[2]-(k[1]+k[2]-k[3])^2)^.5))/4

Attempt This Online!

-16 bytes were removed thanks to pajonk

It is quite simple, but still I am glad my answer is shorter than the others here except APL =)

combn(1:4,3) gives the vertices of 4 triangles of a tetrahedron (as indices of our input object):

     [,1] [,2] [,3] [,4]
[1,]    1    1    1    2
[2,]    2    2    3    3
[3,]    3    4    4    4

Then we iterate through every column, calculate squares of distances with a builtin dist and ^2, apply the Heron's formula to every combination and sum all areas.

Answer 8

Python - 260

I'm not sure what the etiquette on posting answers to your own questions is, but her is my solution, which I used to verify my example, golfed:

import copy,math
P=[input()for i in"1234"]
def e(a, b):return math.sqrt(sum([(b[i]-a[i])**2 for i in range(3)]))
o=0
for j in range(4):p=copy.copy(P);p.pop(j);a,b,c=[e(p[i],p[(i+1)%3])for i in range(3)];s=(a+b+c)/2;A=math.sqrt(s*(s-a)*(s-b)*(s-c));o+=A
print o

It uses the same procedure as laurencevs.

Answer 9

C, 303

Excluding unnecessary whitespace. However, there's still a lot of golfing to be done here (I will try to come back and do it later.) It's the first time I've declared a for loop in a #define. I've always found ways to minmalise the number of loops before.

I had to change from float to double to get the same answer as the OP for the test case. Before that, it was a round 300.

scanf works the same whether you separate your input with spaces or newlines, so you can format it into as many or as few lines as you like.

#define F ;for(i=0;i<12;i++)
#define D(k) (q[i]-q[(i+k)%12])
double q[12],r[12],s[4],p,n;

main(i){
  F scanf("%lf",&q[i])
  F r[i/3*3]+=D(3)*D(3),r[i/3*3+1]+=D(6)*D(6)
  F r[i]=sqrt(r[i])
  F i%3||(s[i/3]=r[(i+3)%12]/2),s[i/3]+=r[i]/2
  F i%3||(p=s[i/3]-r[(i+3)%12]),p*=s[i/3]-r[i],n+=(i%3>1)*sqrt(p)   
  ;printf("%lf",n);       
}