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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
Given the sides \$a\$, \$b\$, and \$c\$, you can assume that \$a>0\$, \$b>0\$, \$c>0\$, \$a+b>c\$, \$b+c>a\$, and \$c+a>b\$.
This is code-golf, shortest answer per language in bytes wins.
(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:)
Takes screen input via ←⎕
(×/(+/t÷2)-0,t←⎕)*.5
Mathematica 23
√Times@@(+##/2-{0,##})&
Sqrt[Tr@#*Times@@(Tr@#-2#)]/4&
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Mar 27, 2013 at 12:49
(Tr@#Times@@(Tr@#-2#))^.5/4&, or 27 using a variable
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Mar 27, 2013 at 13:09
t=input()
s=a=sum(t)/2.
for x in t:a*=s-x
print a**.5
Input: 2,3,4
Output: 2.90473750966
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
*.5 instead of /2 ?
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Apr 3, 2013 at 20:48
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).
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~].~++: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
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
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.
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With 4 bytes saved by @swish.
This returns an exact answer:
Area@SSSTriangle@
Example
Area@SSSTriangle[2,3,4]
To return the answer in decimal form, two additional bytes are required.
N@Area@SSSTriangle[2,3,4]
2.90474
{sqrt s**/(s:.5*+/x)-x}
Example
k){sqrt s**/(s:.5*+/x)-x} 2 3 4
2.904738
{(×/(+/⍵÷2)-0,⍵)*÷2} 2 3 4
Example:
> {(×/(+/⍵÷2)-0,⍵)*÷2} 2 3 4
2.90474
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.
(a=v[0])a is longer than v[0]v[0].
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Mar 27, 2013 at 10:43
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.
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Mar 27, 2013 at 16:32
s=(-v[0]-v[1]-v[2])/2 and change the other - to +. It's an even number of terms, so it cancels out.
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Mar 27, 2013 at 16:39
#define f(a,b,c)sqrt((a+b+c)*(a+b-c)*(a-b+c)*(b+c-a))/4
Yet another implementation of Hero's formula.
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$map(sum l/2-)l
2.9047375096555625
Prelude>
<?=sqrt(($s=array_sum($c=fgetcsv(STDIN))/2)*($s-$c[0])*($s-$c[1])*$s-=$c[2]);
Useage:
php triangle.php
2,3,4
Output: 2.9047375096556
I don't think I can make it shorter? I'm still new to golfing. Anyone let me know if I overlooked something.
Thanks Primo for saving me 1 byte, lol.
($s-$c[2]) can be replaced with $s-=$c[2] for one byte, but that's all I can see.
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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! :)
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))
½*Nx
NmnU ×*U q
Saved 2 bytes thanks to ETH pointing out a redundant newline and some alternative ways to reduce the array.
NmnU ×*U q, NmnU r*U q, Np0 mnU ×q
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Jul 26, 2017 at 14:42
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.
proc, it extends to only 81 bytes: tio.run/##NYo7CoAwEAV7T/…
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Basically heron's formula:
f(a,b,c)=(p=(a+b+c)/2;sqrt(p*(p-a)*(p-b)*(p-c)))
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))
{√×/(+/⍵÷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