Shortest method to implement the Pythagorean Theorem

Who doesn't love the Pythagorean theorem $$\a^2+b^2=c^2\$$? 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.

• Shortest in characters or bytes? :) Feb 5, 2014 at 20:53
• shortest in characters Feb 5, 2014 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). Feb 6, 2014 at 12:47

Jelly, 32 characters

,²S½ær3µ,“¡ÆC⁷⁺ɱSoṿȤç½?⁶Ẏtḍỵŀ»ṚK

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There is probably a better string compression that allows me to get around needing to join with spaces but I was having trouble finding it.

Explanation:

,²S½ær3µ,“...»ṚK    Example inputs: 3, 4
,                   Pair the inputs. Result: [3, 4]
²                  Square them. Result: [9, 16]
S                 Sum them. Result: 25
½                Get the square root of the sum. Result: 5
ær3             Round to 3 decimal places. Result: 5
µ            Take the result of that... Result: 5
“...»       ...and  the compressed string Result: "The hypotenuse of this right triangle is"
,           And put them into a pair. Result: [5, "The hypotenuse of this right triangle is"]
Ṛ     Reverse that. Result: ["The hypotenuse of this right triangle is", 5]
k    Join it with spaces. Result: "The hypotenuse of this right triangle is 5.0"
Implicit output.
• So long as only three decimal places means less than or equal to three decimal places, the output looks fine. Oct 23, 2017 at 20:43

JavaScript, 70 bytes

(a,b)=>'The hypotenuse of the triangle is '+((a*a+b*b)**.5).toFixed(3)

The latter function rounds the number to 3 decimal places.

• It would be shorter to use addition instead of a template string. Feb 8, 2021 at 11:24
• Thanks for the advice, 70 bytes now
– user100690
Feb 8, 2021 at 11:57

Vyxal, 32 30 bytes

∆/λƛ h⁼Ǐ¢Ṗλ₇ of λ× ¬ẋ Ȧ⇩ is Π

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Fig, $$\36\log_{256}(96)\approx\$$ 29.632 bytes

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));}} Feb 6, 2014 at 10:41
• @Michaël Demey, hah, i didnt think of that changing the function to static . Feb 6, 2014 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[0].toLong(),args[1].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 Feb 6, 2014 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.

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?!

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. Apr 19, 2014 at 14:19
• Instead of bc, use dc: dc<<<"3k$a d*$b d*+vp" Apr 19, 2014 at 14:29

(\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. Oct 21, 2017 at 20:33
• @FrownyFrog that is wonderful, thanks. Oct 21, 2017 at 23:05

Perl 5, 69 bytes

printf"The hypotenuse of this right triangle is %.3f",sqrt<>**2+<>**2

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Tcl, 80 bytes

puts "The hypotenuse of this right triangle is [expr int(hypot($a,$b)*1e3)/1e3]"

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• 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. Oct 21, 2017 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. Oct 21, 2017 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/«

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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.
• "The hypotenuse of this triangle is nOt 4" Feb 8, 2021 at 3:11

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.

╬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. Aug 24, 2018 at 15:59

Husk, 41 bytes

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

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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!