17
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Given three sides of a triangle, print area of this triangle.

Test cases:

In: 2,3,4

Out: 2.90473750965556

In: 3,4,5

Out: 6

Assume the three side a,b,c always a>0,b>0,c>0,a+b>c,b+c>a,c+a>b.

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26 Answers 26

6
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J, 23 19 chars

   (4%~2%:[:*/+/-0,+:)

   (4%~2%:[:*/+/-0,+:) 2 3 4
2.90474

   (4%~2%:[:*/+/-0,+:) 3,4,5
6

17-char version if input is in i: 4%~%:*/(+/,+/-+:)i

original 23-char version: (%:@(+/**/@(+/-+:))%4:)

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7
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APL 21 20

Takes screen input via ←⎕

(×/(+/t÷2)-0,t←⎕)*.5
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6
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Python 2, 53

t=input()
s=a=sum(t)/2.
for x in t:a*=s-x
print a**.5

Input: 2,3,4

Output: 2.90473750966

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  • 3
    \$\begingroup\$ I wasted a lot of time trying to come up with a better solution. I'm convinced this is as good as it gets \$\endgroup\$ – jamylak Mar 30 '13 at 11:02
  • 2
    \$\begingroup\$ @jamylak you and me both ;) \$\endgroup\$ – primo Mar 31 '13 at 15:33
6
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Mathematica 23

√Times@@(+##/2-{0,##})&
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  • \$\begingroup\$ Just an aside comment. We (usually) try to provide Mma code as functions here. For example in your case Sqrt[Tr@#*Times@@(Tr@#-2#)]/4& \$\endgroup\$ – Dr. belisarius Mar 27 '13 at 12:49
  • 2
    \$\begingroup\$ 28 chars, as a function (Tr@#Times@@(Tr@#-2#))^.5/4&, or 27 using a variable \$\endgroup\$ – Dr. belisarius Mar 27 '13 at 13:09
  • \$\begingroup\$ @belisarius Thanks for your suggestion. \$\endgroup\$ – chyanog Mar 28 '13 at 4:29
5
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Python 57 bytes

a,b,c=input()
s=(a+b+c)*.5
print(s*(s-a)*(s-b)*(s-c))**.5

Using Heron's Formula.

Sample usage:

$ echo 2,3,4 | python triangle-area.py
2.90473750966

$ echo 3,4,5 | python triangle-area.py
6.0

A 58 byte variant:

a,b,c=input()
print((a+b+c)*(b+c-a)*(a+c-b)*(a+b-c))**.5/4
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  • \$\begingroup\$ I'm not terribly familiar with python, but why is line 2 *.5 instead of /2 ? \$\endgroup\$ – jdstankosky Apr 3 '13 at 20:48
  • \$\begingroup\$ @jdstankosky The division operator in Python is by default integer division, so that if the sum of a+b+c is odd, the result will be erroneous. This did change in Python 3, although most golf submission are assumed to be Python 2.7 unless otherwise specified (just as Perl submissions are assumed to be 5.10+, and not Perl 6). \$\endgroup\$ – primo Apr 4 '13 at 3:22
  • 3
    \$\begingroup\$ You could just say "Python 3" instead of "Python". \$\endgroup\$ – Joe Z. Apr 5 '13 at 18:51
  • 1
    \$\begingroup\$ @JoeZ. Nope. This is Python 2; in Python 3, input() returns a string, breaking this solution. \$\endgroup\$ – Khuldraeseth na'Barya Oct 18 '17 at 2:15
4
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GolfScript, 38 characters

~].~++:d\{2*d\-*}/'"#{'\+'**0.5/4}"'+~

Since the question didn't specify otherwise I chose to work only on integer lengths. Sides must be given on STDIN separated by spaces.

Example:

> 2 3 4
2.9047375096555625
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3
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K, 23

{sqrt s**/(s:.5*+/x)-x}

Example

k){sqrt s**/(s:.5*+/x)-x} 2 3 4
2.904738
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3
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APL, 23 20 characters

{(×/(+/⍵÷2)-0,⍵)*÷2} 2 3 4

Example:

> {(×/(+/⍵÷2)-0,⍵)*÷2} 2 3 4
2.90474
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3
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R : 48 43 characters

f=function(...)prod(sum(...)/2-c(0,...))^.5

Using Heron's formula as well but taking advantage of R's vectorization.
Thanks to @flodel for the idea of the ellipsis.

Usage:

f(2,3,4)
[1] 2.904738
f(3,4,5)
[1] 6
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  • \$\begingroup\$ you can drop the curly brackets. And you can gain more by using ellipsis: function(...)prod(sum(...)/2-c(0,...))^.5. Or even function(x)prod(sum(x)/2-c(0,x))^.5 if you call your function with a vector. \$\endgroup\$ – flodel Mar 30 '13 at 2:45
  • \$\begingroup\$ @flodel thanks! I didn't thought of the ellipsis, that's nice. \$\endgroup\$ – plannapus Mar 30 '13 at 8:01
2
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Javascript, 88 85

v=prompt().split(/,/g);s=v[0]/2+v[1]/2+v[2]/2;Math.sqrt(s*(s-v[0])*(s-v[1])*(s-v[2]))

Not good but fun :) Also Heron... Demonstrates the ungolfability of simple problems in JS lol

Note: run from console to see result.

88->85: Removed a, b and c.

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  • 1
    \$\begingroup\$ You can save a bit by only dividing by 2 once. And you don't actually gain anything by assigning to variables: (a=v[0])a is longer than v[0]v[0]. \$\endgroup\$ – Peter Taylor Mar 27 '13 at 10:43
  • \$\begingroup\$ If I divided by 2 only one time, like s=(v[0]+v[1]+v[2])/2 with a,b,c=3,4,5 would result in "345"/2=172.5" and not 6. Improved without a, b` and c though. \$\endgroup\$ – tomsmeding Mar 27 '13 at 16:32
  • \$\begingroup\$ Ah, JavaScript's wonderful type system. Ok, s=(-v[0]-v[1]-v[2])/2 and change the other - to +. It's an even number of terms, so it cancels out. \$\endgroup\$ – Peter Taylor Mar 27 '13 at 16:39
2
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Mathematica 20 16 or 22 18 bytes

With 4 bytes saved by @swish.

This returns an exact answer:

Area@SSSTriangle@

Example

Area@SSSTriangle[2,3,4]

image


To return the answer in decimal form, two additional bytes are required.

N@Area@SSSTriangle[2,3,4]

2.90474

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  • \$\begingroup\$ Composition shaves couple of bytes Area@*SSSTriangle \$\endgroup\$ – swish Jun 24 at 7:29
  • \$\begingroup\$ @swish Thanks, Much appreciated. \$\endgroup\$ – DavidC Jun 24 at 8:23
1
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Haskell: 51 (27) characters

readLn>>=(\l->print$sqrt$product$map(sum l/2-)$0:l)

A very straight-forward implementation of Heron's formula. Example run:

Prelude> readLn>>=(\l->print$sqrt$product$map(sum l/2-)$0:l)
[2,3,4]
2.9047375096555625
Prelude>

Note that it accepts any numeric input, not only integers. And if the input already is in l the solution only needs to be 36 characters long, and if we are not interested in printing the answer the solution only needs to be 30 characters long. What more is that if we can allow ourself to change the input format we can remove 3 more characters. So if our input looks like [2,3,4,0.0] and is already in l we can get our answer with only:

sqrt$product$map(sum l/2-)l

Example run:

Prelude> let l = [2,3,4,0.0]
Prelude> sqrt$product$map(sum l/2-)l
2.9047375096555625
Prelude>
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1
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PHP, 78 77

<?=sqrt(($s=array_sum($c=fgetcsv(STDIN))/2)*($s-$c[0])*($s-$c[1])*$s-=$c[2]);

Useage:

php triangle.php
2,3,4

Output: 2.9047375096556

I don't think I can make it shorter? I'm still new to golfing. Anyone let me know if I overlooked something.

Thanks Primo for saving me 1 byte, lol.

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  • 1
    \$\begingroup\$ The final ($s-$c[2]) can be replaced with $s-=$c[2] for one byte, but that's all I can see. \$\endgroup\$ – primo Apr 8 '13 at 9:24
  • \$\begingroup\$ @primo Thanks, man! \$\endgroup\$ – jdstankosky Apr 11 '13 at 12:55
1
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JavaScript (84 86)

s=(eval('abc '.split('').join('=prompt()|0;'))+a+b)/2;Math.sqrt(s*(s-a)*(s-b)*(s-c))

Another JavaScript solution based on Heron's formula, but trying a different approach for loading variables. Needs to be run from the console. Each side is entered in a separate prompt.

EDIT: Make use of return value of eval to save 2 characters. Beats @tomsmeding, wahoo! :)

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1
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Excel, 42 bytes

Based on Heron's formula, high school algebra to golf.

=SQRT(((A1+B1)^2-C1^2)*(C1^2-(A1-B1)^2))/4

Ungolfed / Unalgebra'ed

=SQRT((A1+B1+C1)/2*(((A1+B1+C1)/2)-A1)*(((A1+B1+C1)/2)-B1)*(((A1+B1+C1)/2)-C1))
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1
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Japt, 17 16 15 bytes

½*Nx
NmnU ×*U q

Test it

Saved 2 bytes thanks to ETH pointing out a redundant newline and some alternative ways to reduce the array.

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  • \$\begingroup\$ I think you can use any of these for the second line, too: NmnU ×*U q, NmnU r*U q, Np0 mnU ×q \$\endgroup\$ – ETHproductions Jul 26 '17 at 14:42
1
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Tcl, 74 chars.

proc R {a b c} {set s ($a+$b+$c)/2.
expr sqrt($s*($s-$a)*($s-$b)*($s-$c))}

Pass the sides as argument.

For the input 2 3 4 the value of s is (2+3+4)/2. as string. Double evaluation FTW.

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1
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Julia 0.6.0, 48 bytes

Basically heron's formula:

f(a,b,c)=(p=(a+b+c)/2;sqrt(p*(p-a)*(p-b)*(p-c)))
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1
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TI-BASIC, 14 12 bytes

4⁻¹√(sum(Ansprod(sum(Ans)-2Ans

Starting from a Heron's Formula routine written by Kenneth Hammond (Weregoose), I golfed off two bytes. Note that TI-BASIC is tokenized, and each token, like Ans and prod(, is one or two bytes in the calculator's memory.

Input through Ans i.e. in the form {a,b,c}:[program name].

Explained:

                   sum(Ans)-2*Ans   (a+b+c)-2{a,b,c}={b+c-a,c+a-b,a+b-c}
          Ans*prod(                 {a,b,c}*(b+c-a)(c+a-b)(a+b-c)
      sum(                          (a+b+c)(b+c-a)(c+a-b)(a+b-c)
4⁻¹*√(                              √((a+b+c)(b+c-a)(c+a-b)(a+b-c)/16)
                                    =√(s(s-a)(s-b)(s-c))
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  • \$\begingroup\$ I've converted this to a community wiki as it isn't your own work. We don't really have a solid consensus on this, but feel free to weigh in here if you disagree with my decision. \$\endgroup\$ – Martin Ender May 21 '15 at 10:53
1
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dc, 34 bytes

9ksssddlsld++2/d3R-rdls-rdld-***vp

Try it online!

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1
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C (gcc), 55 bytes

#define f(a,b,c)sqrt((a+b+c)*(a+b-c)*(a-b+c)*(b+c-a))/4

Try it online!

Yet another implementation of Hero's formula.

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0
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#include<stdio.h>
#include<math.h>
main()
{
  double a,b,c,s,area;
  scanf("%d %d %d" &a,&b,&c);
  s=sqrt((a*a)+(b*b)+(c*c));
  area=[sqrt(s*(s-a)*(s-b)*(s-c))]/2;
}
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  • \$\begingroup\$ Hi iam beginner of programming,if any errors plz suggest me. \$\endgroup\$ – sharath Apr 1 '13 at 9:06
  • \$\begingroup\$ Welcome to CodeGolf. You should first verify that the program works and somehow outputs the result. There are some simple bugs in this one. \$\endgroup\$ – ugoren Apr 2 '13 at 7:49
0
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ARBLE, 36 bytes

sqrt(a*(a-b)*(a-c)*(a-d))((b+c+d)/2)

Heron's formula.

Try it online!

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0
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Perl 5 -MList::Util=sum -ap, 40 bytes

$r=$t=.5*sum@F;map$r*=$t-$_,@F;$_=sqrt$r

Try it online!

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0
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Stax, 10 bytes

╝0∞♀»♦▓y╩╪

Run and debug it

Operates triples of floating point numbers. Uses Heron's Formula

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0
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APL(NARS), 16 chars, 32 bytes

{√×/(+/⍵÷2)-0,⍵}

One has to cenvert the Erone formula if a, b, c are sides of triangle

p   =(a+b+c)/2 
Area=√p*(p-a)(p-b)(p-c)

to APL language... test

  f←{√×/(+/⍵÷2)-0,⍵}
  f 2 3 4
2.90473751
  f 3 4 5
6
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