# Shortest method to implement the Pythagorean Theorem

Who doesn't love the pythagorean theorem a²+b²=c²? Write the shortest method you can in any language that takes in value a and b and prints out "The hypotenuse of this right triangle is " + c. Keep c to only three decimal places.

• Does this qualify as a programming puzzle? – DavidC Feb 5 '14 at 16:51
• @DavidCarraher The problem per se is no programming puzzle. But since the objective is to golf a solution for it, then it is indeed a programming puzzle. – Victor Stafusa Feb 5 '14 at 19:46
• shortest in characters – Vik P Feb 5 '14 at 20:55
• The code-golf tag explicitly says "Code-golf is a competition to solve a particular problem in the fewest bytes of source code." See Scoring code golf (bytes vs. characters). – r.e.s. Feb 6 '14 at 12:47
• @r.e.s.: Fixed :-p – Timwi Feb 6 '14 at 14:55

## Javascript, 109

A=1,B=1,s=1e3,console.log("The hypotenuse of this right triangle is " + (Math.round(Math.sqrt(A*A+B*B)*s)/s))


Fiddledee

• with alert you could save 6 chars – Fez Vrasta Feb 5 '14 at 18:23
• Down to 101: A=1,B=1,s=1e3;alert("The hypotenuse of this right triangle is "+(Math.round(Math.sqrt(A*A+B*B)*s)/s)) – James Webster Feb 6 '14 at 15:57
• Using suggestions from other entries so far I get it to 90: A=1,B=1;alert("The hypotenuse of this right triangle is "+(Math.sqrt(A*A+B*B).toFixed(3))) Even less if you leave out alert – micha Feb 8 '14 at 0:03

## Scala (95c)

import Math._;val x=sqrt(pow(a,2)+pow(b,2));f"The hypotenuse of this right triangle is $x%1.3g"  ## Java : 223  class l { public static void main(String[] args) { new l().H(2,4); } void H(float a, float b) { System.out.printf("The hypotenuse of this right triangle is %.3f",Math.hypot(a,b)); } }  • I brought this to 170 by making it a one liner, changing the method to a static: class h{public static void main(String[]args){H(2,4);}static void H(float a,float b){System.out.printf("The hypotenuse of this right triangle is %.3f",Math.hypot(a,b));}} – Michaël Demey Feb 6 '14 at 10:41 • @Michaël Demey, hah, i didnt think of that changing the function to static . – BeyondProgrammer Feb 6 '14 at 11:25 ## Groovy : 78 Characters a={x,y->printf"The hypotenuse of this right triangle is %.3f",Math.hypot(x,y)}  Assuming the task is just writing a method. If you need to take input parameters: 99 Characters printf"The hypotenuse of this right triangle is %.3f",Math.hypot(args.toLong(),args.toLong())  I found two ways to it with the same length: printf"The hypotenuse of this right triangle is %.3f",Math.sqrt(args.collect{it.toLong()**2}.sum())  ## newLISP, 103 characters (let(a 3.1 b 4)(println"The hypotenuse of this right triangle is "(round(pow(add(pow a)(pow b)).5)-3)))  The unspecified input method helps a lot... :) # F#, 78 let f a b=sqrt(a*a+b*b)|>printf"The hypotenuse of this right triangle is %.3f"  Java : 155 class A{A(int x,int y){System.out.printf("The hypotenuse of this right triangle is %.3f",Math.hypot(x,y));}public static void main(String[]s){new A(1,4);}}  • You can take out the public  on the constructor saving 7 – James Webster Feb 6 '14 at 16:12 # Fortran (90) read(*,*)a,b;print'(a,f9.3)',"The hypotenuse of this right triangle is ",sqrt(a*a+b*b);end  Uses implicit typing so that a,b are single precision reals. Here is my Javascript/JScript attempt: function h(a,b){return a/1==a&&b/1==b?'The hypotenuse of this right triangle is '+(a*a+b*b||0).toFixed(3):!1;}  It is 110 chars. It's a monster!!! But it can handle NaN and invalid numbers, like text, objects, 0/0 and other invalid numbers. # Javascript 98 characters/bytes This is implemented as a browser alert in a function. function p(a,b){alert('The hypotenuse of this right triangle is: '+Math.sqrt(a*a+b*b).toFixed(3))};  ### Ruby 2.1+ (73 characters) f=->a,b{puts"The hypotenuse of this right triangle is %.3f"%(a+b*1i).abs}  # Prolog h(A,B):-X is A*A+B*B, C is sqrt(X),format('The hypotenuse of this right triangle is ~3f',C).  # Bash with bc, 91 No bash script yet?! read a b echo -n 'The hypotenuse of this right triangle is ' bc<<<"scale=3;sqrt($a^2+$b^2)"  There is no way this can win. • To save 1 character, you could use printf instead of echo -n. – gniourf_gniourf Apr 19 '14 at 14:19 • Instead of bc, use dc: dc<<<"3k$a d*$b d*+vp" – gniourf_gniourf Apr 19 '14 at 14:29 Haskell, 77 characters (\x y->print$"The hypotenuse of this right triangle is "++(show$sqrt$x^2+y^2))


Useage:

(\x y->print$"The hypotenuse of this right triangle is "++(show$sqrt$x^2+y^2)) 3.0 4.0 The hypotenuse of this right triangle is 5.0  # Fortran 2008 (87 characters) Based on the Fortran 90 solution, but now in Fortran 2008, HYPOT is an intrinsic function. read(*,*)a,b;print'(a,f9.3)','The hypotenuse of this right triangle is ',hypot(a,b);end  # J, 56 bytes 'The hypotenuse of this right triangle is '&,@":@(+&.*:)  ### explanation J's "Under" verb &. is a nice fit here. It applies a transformation, executes a verb, and then applies the reverse transformation to the result. In this case, squaring is the transformation *:, its reverse is taking the square root, and the verb in question is addition. So we do addition "under" the square transformation. The rest is just formatting the number and cat'ing it onto the sentence. Try it online! • |@j. is even nicer. – FrownyFrog Oct 21 '17 at 20:33 • @FrownyFrog that is wonderful, thanks. – Jonah Oct 21 '17 at 23:05 # Perl 5, 69 bytes printf"The hypotenuse of this right triangle is %.3f",sqrt<>**2+<>**2  Try it online! # Tcl, 80 bytes puts "The hypotenuse of this right triangle is [expr int(hypot($a,$b)*1e3)/1e3]"  Try it online! • I doubt that setting variables is an allowed I/O method... Nevertheless, your program does not seem to [k]eep c to only three decimal places. – Jonathan Frech Oct 21 '17 at 15:14 • @JonathanFrech: Fixed the 3 decimal places issue. About the setting variables issue, I will wait feedback from questions asker, as it seems I've seen some answers here doing the same. – sergiol Oct 21 '17 at 15:35 # Jq 1.5, 80 bytes "The hypotenuse of this right triangle is \(1e3*pow(map(.*.)|add;.5)|floor/1e3)"  Assumes input is [a,b] Try it online at tio.run or jqplay.org # Axiom, 92 bytes g(a,b)=="The hypotenuse of this right triangle is "((round((a^2+b^2)^.5*1.e3)/1.e3)::String)  test (4) -> g(12, 19) (4) "The hypotenuse of this right triangle is 22.472" Type: String  # 05AB1E, 53 characters "The hypotenuse of this right triangle is "|nOt₄*ï₄/«  Try it online! I could not figure out how to compress 05AB1E strings. :( Explanation: "..."|nOt₄*ï₄/« | Push array of inputs. n Square them. O Get the sum. t Get the square root. ₄* Multiply by 1000 ï Convert to integer. This rounds it. ₄/ Divide by 1000 "..." « Append it to the string. Implicit print.  # Petit Computer BASIC, 66 bytes INPUT A,B?"The hypotenuse of this right triangle is ";SQR(A*A+B*B)  PTC only displays the first 3 decimal places when printing. # Stax, 24 bytes ╬Z.-êû$P¿[▐Ä√#╙δdr▼▀Ö☺┐ç


Run and debug it

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

1/yyH\fBXNR&X+{.SCp   print "The hypotenuse of this right triangle is "
:J+                     square and add the inputs
|Q                      square root
3j                      round to 3 places
p                       print


Run this one

# APL (Dyalog), 51 bytes

'The hypotenuse of this right triangle is',3⍕|0j1⊥⎕


Try it online!

0j1⊥⎕: Convert the input into base i. e.g a b -> b+ai

|: Magnitude

3⍕: Round to 3 decimal places

# Japt, 37 bytes

T hypÇQ«e  È ght â0Ø8  +Nx²¬x3


Try it

# JavaScript (Node.js), 76 bytes

a=>b=>'The hypotenuse of this right triangle is '+Math.hypot(a,b).toFixed(3)


Try it online!

# Explanation

a => b => // the two arguments (curry method)
The hypotenuse of this right triangle is // write the text required
\${Math.hypot(a,b)} // and use JS Math.hypot function to calculate the hypotenuse.


Alternately, if the you spare me the pain of writing this text.

f=Math.hypot // is also a valid answer

• Welcome to PPCG. Note, though, the requirement that the output be rounded to 3 decimal places. – Shaggy Aug 24 '18 at 15:59

# Husk, 41 bytes

+¨ΣŸṗ¤μe₀fı≤ṙ↔ẏ«G¹∫¨J"."msλ‰⁰i*⁰)1000√¤+□


Try it online!

With rounding being involved, I sure didn't pick the right language for the task...

+                              The concatenation of
¨...¨                         the compressed prefix for the output
ms                   with the stringified elements of
λ‰⁰   )1000        the divmod by 1000 of
√       the square root of
+     the sum of
¤ □    the squares of the two arguments
*⁰             times 1000
i               converted to an integer
J"."                     with a period inserted between them.


Computing the actual length of the hypotenuse only takes four bytes, rounding to three decimal places takes seventeen bytes, and adding on The hypotenuse of this right triangle is to the output takes twenty bytes. 90.244% of this answer is just formatting!

# Python 2.7.13, 85 bytes

def p(a, b):
print "the length of the hypotenuse is:"
return math.sqrt(a**2 + b**2)

• This answer is not to specs and also not golfed completely..? – Timtech Sep 13 '17 at 19:52
• You can save bytes by deleting spaces. Also, Keep c to only three decimal places – Stephen Sep 13 '17 at 20:51