Is my triangle right?

Question

Given a, b, c the length of the three sides of a triangle, say if the triangle is right-angled (i.e. has one angle equal to 90 degrees) or not.

Input

Three positive integer values in any order

Output

Either a specific true output (true, 1, yes, ...) or a specific false output (false, 0, no, ...)

Example

5, 3, 4        --> yes
3, 5, 4        --> yes
12, 37, 35     --> yes
21, 38, 50     --> no
210, 308, 250  --> no

Rules

• The input and output can be given in any convenient format.
• In your submission, please state the true and the false values.
• No need to handle negative values or invalid edge triple
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• If possible, please include a link to an online testing environment so other people can try out your code!
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

Common Lisp, 64 bytes

(apply(lambda(x y z)(=(+(* x x)(* y y))(* z z)))(sort(read)#'<))

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As usual in Common Lisp, true is T and false is NIL.

Answer 2

JavaScript 6, recursive way, 40 Bytes

f=(a,b,c)=>c<a|c<b?f(b,c,a):a*a+b*b==c*c

f=(a,b,c)=>c<a|c<b?f(b,c,a):a*a+b*b==c*c
console.log(f(4,5,3));
console.log(f(4,5,6));

C (gcc), 42 bytes

f(a,b,c){c=c<a|c<b?f(b,c,a):a*a+b*b==c*c;}

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Answer 3

Perl 5, 34 +2 (-ap) bytes

$x+=($_*=$_)/2for@F;$_=grep/$x/,@F

seems that can be shortened to 29 +2 but there is a warning: smartmatch is experimental

$x+=($_*=$_)/2for@F;$_=$x~~@F

the question i couldn't prove but it works in all tests i've tried is if the number (a^2+b^2+c^2)/2 can be a substring of a number (a^2/2 b^2/2 or c^2/2) which would give a false positive.

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Answer 4

Excel, 47 bytes

Returns FALSE for right-angled Triangles, else TRUE:

=ISERR(FIND(SQRT((A1^2+B1^2+C1^2)/2),A1&B1&C1))

Answer 5

Proton, 31 bytes

k=>(sum(j*j for j:k)/2)**.5in k

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Credit to user202729 for the idea go upvote them

Answer 6

Neim, 6 bytes

ᛦD𝐬ᚺS𝕚

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Answer 7

Swift 3, 81 bytes

func r(v:[Int]){let a=v.sorted{$0<$1};print("\(a[0]*a[0]+a[1]*a[1]==a[2]*a[2])")}

Answer 8

Batch, 61 bytes

@cmd/cset/ax=%1*%1,y=%2*%2,z=%3*%3,!((x+y-z)*(y+z-x)*(z+x-y))

Outputs 1 or 0 appropriately.

Answer 9

Pyth, 14 bytes

K_m*ddSQqstKhK

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Answer 10

Eukleides, 81 bytes

x=number("")^2;y=number("")^2;z=number("")^2
print x==y+z or y==x+z or z==y+x?1|0

I thought Eukleides would handle this nicely using geometric builtins, but this actually turned out to be longer than just testing against the Pythagorean theorem, because the right triangle assertion is picky about order, and we have to turn our sides into a triangle first:

d e f triangle number(""),number(""),number("")
print right(d,e,f)or right(e,f,d)or right(f,d,e)?1|0

...which is 100 bytes. One nice thing about that one is it'll error out given an impossible triangle. Anyway, for either one, input is via STDIN, output is 1 for right, 0 for non-right via STDOUT. Doing a function was actually longer in this instance.

Answer 11

C++, 156 bytes

bool f(int*a){int g=[](int*b){return b[0]>b[1]?b[0]>b[2]?0:2:b[1]>b[2]?1:2;}(a);int c[2]={g?0:1,g>1?1:2};return pow(a[c[0]],2)+pow(a[c[1]],2)==pow(a[g],2);}

Answer 12

Racket, 109 bytes

#lang racket/base
(define (isright a b c)
  (if (equal? (+ (* a a) (* b b)) (* c c))
      "yes"
      "no"))

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Answer 13

Dyalog APL, 12 bytes

(+/2÷⍨×⍨)∊×⍨

Uses the same method as this Jelly answer.

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Answer 14

Pyth, 11 10 bytes

}/sK^R2Q2K

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My first Pyth submission! Explanation:

     R        right map
    ^ 2       squaring
       Q      of the input
   K          save this in K
  s           sum up
 /      2     divide by 2
}        K    test if this sum is in K

saved a byte thanks to Mr. Xcoder

Answer 15

APL (Dyalog), 10 bytes

(+/∊×∘2)×⍨

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×⍨ square (lit. multiplication selfie)

(…) apply the following anonymous tacit function on that:

 +/ the sum

 ∊ is a member of

 ×∘2 the doubled amounts

Answer 16

Two answers found by refining the answer by Steadybox and incorporating the technique in the Java answer by Kevin Cruijssen:

C, 52 bytes

f(a,b,c){return(a*=a)+(b*=b)==(c*=c)|b+c==a|c+a==b;}

C, 74 bytes

#define _(a,b,c)a*a+b*b==c*c||
f(a,b,c){return _(a,b,c)_(b,c,a)_(c,a,b)0;}

Test program

#include <stdio.h>
int test(int a, int b, int c, int expected)
{
    int actual = f(a,b,c);
    if (expected != actual) {
        fprintf(stderr, "%d %d %d => %d\n ", a,b,c, actual);
        return 1;
    }
    return 0;
}

int main()
{
    return test(5, 3, 4, 1)
        +  test(3, 5, 4, 1)
        +  test(12, 37, 35, 1)
        +  test(21, 38, 50, 0)
        +  test(210, 308, 250, 0)
        ;
}

Answer 17

Swift 4, 68 bytes

func f(v:inout[Int]){v.sort();print(v[0]*v[0]+v[1]*v[1]==v[2]*v[2])}

Accepts a mutable array as inout argument and prints true or false.

Swift Sandbox.

Answer 18

Dyalog APL, 17 16 bytes

{(+/2÷⍨⍵*2)∊⍵*2}

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Outputs 1 for true, 0 for false.

Thanks to @Adám for 1 byte.

How it works

{(+/2÷⍨⍵*2)∊⍵*2}          # Anonymous function
 (     ⍵*2)               # Each input (⍵) to the 2nd power
    2÷⍨                   # divided by 2
  +/                      # Sum
           ∊              # "is in"
            ⍵*2           # Each input (⍵) to the 2nd power

This uses the same logic as @user202729's Jelly answer, so some credit goes to them.

Answer 19

C# (53)

The shortest possible code I could find was the one that looks like the JAVA solution:

(a,b,c)=>{return(a*=a)+(b*=b)==(c*=c)|a+c==b|b+c==a;}

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Array-input (58)

a=>{Array.Sort(a);return a[0]*a[0]+a[1]*a[1]==a[2]*a[2];};

Answer 20

Add++, 25 18 bytes

D,f,?,caaBcB*#s/2=

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-7 bytes thanks to mdahmoune

_{Yes, I got outgolfed in my own language by 7 bytes. So what?}

How does it work?

D,f,?,      - Create a variadic function, f. Example arguments: [5 3 4]
      c     - Clear the stack;      STACK = []
      a     - Push the arguments;   STACK = [[5 3 4]]
      a     - Push the arguments;   STACK = [[5 3 4] [5 3 4]]
      Bc    - Push the columns;     STACK = [[5 5] [3 3] [4 4]]
      B*    - Product of each;      STACK = [25 9 16]
      #     - Sort the stack;       STACK = [9 16 25]
      s     - Push the sum;         STACK = [9 16 25 50]
      /     - Divide;               STACK = [9 16 2]
      2     - Push 2;               STACK = [9 16 2 2]
      =     - Equal;                STACK = [9 16 1]

Implicitly return the top element on the stack

Old version

D,f,?,c2 2 2B]a@BcB`#s/2=

The only difference is the squaring of each element (aaBcB* in the golfed version, 2 2 2B]a@BcB` in this version), so I'll quickly explain how this part works

              - Stack is currently empty
2 2 2         - Push 2 three times;             STACK = [2 2 2]
     B]       - Wrap the stack in an array;     STACK = [[2 2 2]]
       a      - Push the arguments as an array; STACK = [[2 2 2] [5 3 4]]
        @     - Reverse the stack;              STACK = [[5 3 4] [2 2 2]]
         Bc   - Push the columns of the stack;  STACK = [[5 2] [3 2] [4 2]]
           B` - Reduce each by exponentiation;  STACK = [25 9 16]

Answer 21

CJam, 12 11 bytes

This is depressingly long.

{$2f#)\:+=}

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Anonymous block that takes input as an array of integers on the stack and returns 1 (truthy) or 0 (falsey).

Explanation:

{      e# Stack:        [12 37 35] 
 $     e# Sort:         [12 35 37]
 2f#   e# Square each:  [144 1225 1369]
 )     e# Pop:          [144 1225] 1369
 \     e# Swap:         1369 [144 1225]
 :+    e# Sum:          1369 1369
 =     e# Equal:        1
}      e# Result: 1 (truthy)

Old solution

{2f#_:+2/#)}

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Anonymous block that takes an array of integers on the stack and returns a truthy or falsy integer.

Answer 22

Stacked, 18 bytes

[sorted:*rev...+=]

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Explanation

This is an anonymous function that takes an array of three integers. It sorts (sorted), squares each element of the array (:*), and reverses the array with rev. This will give an array with the largest value in the front. ... pushes each individual member of the array onto the stack, which would make the stack look like:

c^2  b^2  a^2

+ addes the top two members, yielding:

c^2  (b^2+a^2)

= tests for equality. yielding the expression a^2 + b^2 == c^2, which is only true for right triangles.

Answer 23

GNUOctave, 32 bytes

A=input("")
sum(A.^2)/2==max(A)^2

The input must be like : [1,2,3] so Octave can evaluate it to a Matrix.

A.^2 squares all the elements of the matrix.

Outputs 1 if the triangle is right-angled, 0 otherwise.

You can try it online : https://octave-online.net

Edit (27 bytes): I saw this answer I thought I could find a better solution. I took the idea of squaring the matrix at the beginning. I don't feel like this deserves upvotes since it wasn't really my whole idea to begin with.

A=input("").^2
sum(A)/max(A)

Outputs 2 if the triangle is right-angled and anything else if it isn't.

Answer 24

Jq 1.5, 31 23 bytes

map(.*.)|.[[add/2]]!=[]

Expects input in form of 3 element array, e.g. [5, 3, 4]

Expanded

  map(.*.)           # square each element
| .[[ add/2 ]]!=[]   # true if index of sum/2 element exists

Thanks to mdahmoune for suggesting shorter approach then my original one.

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---

Jq 1.5, 31 bytes

sort|map(.*.)|(.[:2]|add)==.[2]

Expects input in form of 3 element array, e.g. [5, 3, 4]

Expanded

  sort                   # put elements in order
| map(.*.)               # square each element
| (.[:2]|add) == .[2]    # is sum of first two equal to third?

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Answer 25

Octave, 29 bytes

sum(a=input("").^2)==max(2*a)

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Input has to be given in a format octave understands as a vector, like [3 5 4] or [12;35;37].

Checks for Pythagoras's theorem: first, square all the sides, then check if the sum of squares of all the sides is twice the square of the hypotenuse.

Output will be either ans = 1 or ans = 0

Answer 26

Go, 96 94 bytes

package r;import"sort";func I(i...int)bool{sort.Ints(i);return i[0]*i[0]+i[1]*i[1]==i[2]*i[2]}

Test:

import "r"
import "fmt"

func main() {
    fmt.Println(r.I(3, 5, 4))
    fmt.Println(r.I(12, 37, 35))
    fmt.Println(r.I(21, 38, 5))
    fmt.Println(r.I(210, 308, 15))
}

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Answer 27

Pushy, 10 8 bytes

GK2ek+=#

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Managed to cut off two bytes by changing the approach, here's how it works now:

          \ Implicit Input                  eg. [3, 4, 5]
G         \ Sort the stack descendingly         [5, 4, 3]
 K2e      \ Square all items                    [25, 16, 9]
    k+    \ Sum last two                        [25, 25]
      =   \ Check equality (1 or 0)             [1]
       #  \ Output result                       PRINT: 1

Original Version, 10 bytes

gK2eSk2/=#

           \ Input implicitly on stack.              eg. [5, 4, 3]
 g         \ Sort the stack ascendingly.                 [3, 4, 5]
 K2e       \ Square all items.                           [9, 16, 25]
 Sk2/      \ Append stack sum divided by 2               [9, 16, 25, 25]
 =#        \ Print equality of last two items (1/0)      PRINT: 1

Answer 28

MY, 13 bytes

22ω^÷Σ2ω^=Σ𝔹←

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This helped me realize ... I screwed up the NOT function (and the boolean conversion function).

How it works (cross-compatible function)

22ω^÷Σ2ω^=Σ𝔹←
2             = push 2 to the stack
 2ω^          = push ω^2 to the stack (functions vectorize)
    ÷         = pop a, then b. push a/b (rational) to the stack
     Σ        = sum
      2ω^=    = equality test with ω^2
          Σ𝔹  = boolean conversion
            ← = output

Answer 29

Axiom, 39 bytes

f x==(y:=sort x;y.1^2+y.2^2=y.3^2=>1;0)

f(x) function Input one list of at last 3 numbers

Output 1 (it is right triangle)

Output 0 (it is not right triangle)

Answer 30

CJam 10

q~$_.*~-+!

Simplest one I found but also shortest

q reads input as string. Leave input formatted as array [3 4 5]
~ dumps it to stack 
$ sorts array
_ duplicates array
.* multiplies each element by itself
~ dumps array to stack
-+ determines largest minus two smallest numbers
! if 0 return 1 and 0 if anything else

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