Difference of three input integers

Question

Implement a function diff that takes as input three integers x, y, and z. It should return whether subtracting one of these numbers from another gives the third.

Test cases:
diff(5, 3, 2) yields True because 5 - 3 = 2
diff(2, 3, 5) yields True because 5 - 3 = 2
diff(2, 5, 3) yields True because 5 - 3 = 2
diff(-2, 3, 5) yields True because 3 - 5 is -2
diff(-5, -3, -2) # -5 - -2 is -3
diff(2, 3, -5) yields False
diff(10, 6, 4) yields True because 10 - 6 = 4
diff(10, 6, 3) yields False

You don't have to name the function, you may implement default input methods the examples above are not a strict guideline.

Answer 1

Jelly, 5 3 bytes

Thanks to @Sp3000 for saving two bytes!

Code, uses quite the same algorithm as @xnor's great answer:

SfḤ

Explanation:

S     # Sum of the argument list
  Ḥ   # Double the list
 f    # Filter, remove everything that isn't equal to the sum of the list

This gives [] as falsy, and anything else as truthy.

Try it online!

Answer 2

Python 3, 21 bytes

lambda*l:sum(l)/2in l

If two numbers add to the other, the sum of all three will be double that other number, so half the sum will be an element of the list. Python 3 is needed to avoid floor-division, unless the numbers are given like 3.0 rather than 3.

Answer 3

ES6, 31 bytes

(a,b,c)=>a+b==c|b+c==a|c+a==b

Add 5 bytes if you need to name the function diff.

Edit: Saved 2 bytes thanks to @Alex L.

Answer 4

APL, 8 5 bytes

+/∊+⍨

This is a monadic function train that accepts an array and returns a boolean (0/1 in APL). It uses the same algorithm as xnor's Python 3 answer.

Explanation:

   +⍨  ⍝ Double the input (+⍨x is the same as x+x)
  ∊    ⍝ Test the membership of
+/     ⍝ The sum of the input

Try it online

Saved 3 bytes thanks to Dennis!

Answer 5

JavaScript ES6, 38 34 33 bytes

x=>x.some(a=>2*a==x[0]+x[1]+x[2])

Very simple anonymous function, and borrows from the Python answer. Takes input x as an array; returns true or false. Bytes shaved to Molarmanful and jrich

A 38-byte program, taking each number as an argument:

(a,b,c)=>[a,b,c].some(t=>t==(a+b+c)/2)

Answer 6

Oracle SQL 11.2, 49 bytes

SELECT 1 FROM DUAL WHERE(:1+:2+:3)/2IN(:1,:2,:3);

Rewrite of @xnor solution, kudos to him.

Answer 7

Perl 6, 20 19 bytes

I have two functions equal in byte count, so I'll put both. Appreciate whichever tickles your fancy.

{@_∋@_.sum div 2}
{@_∋+~(@_.sum/2)}

Usage: assign either one to a variable from which you can call it.
EDIT: Thanks @b2gills for the byte reduction

Answer 8

J, 6 bytes

+/e.+:

Try it with J.js.

How it works

+/e.+:    Monadic verb. Argument: A
    +:    Double the elements of A.
+/        Compute the sum of the elements of A.
  e.      Test for membership.

Answer 9

DUP, 31 chars / 39 bytes

[2ø2ø2ø++2/\%3ø^=3ø2ø=3ø3ø=||.]

Try it here!

My first DUP submission ever! Unicode is your oyster.

It's an anonymous function/lambda. Usage:

5 3 2[2ø2ø2ø++2/\%3ø^=3ø2ø=3ø3ø=||.]!

Explanation

[                               {start lambda}
 2ø2ø2ø                         {duplicate 3 inputnums}
       ++                       {push sum(3 popped stack items)}
         2/\%                   {push (popped stack item)/2}
             3ø^=3ø2ø=3ø3ø=     {for all 3 inputs, -1 if inputnum=sum/2; else 0}
                           ||   {check if any of the 3 resulting values are truthy}
                             .  {output top of stack (boolean value)}
                              ] {end lambda}

Answer 10

Java 7, 81

boolean d(int[]a){int s=0,t=1;for(int b:a)s+=b;for(int b:a)t*=2*b-s;return t==0;}

Answer 11

Java 8 (lambda function), 29 bytes

// Lambda Signature: (int, int, int) -> boolean

(a,b,c)->a+b==c|a+c==b|b+c==a

Java code golf solutions are usually only short when the program does not have to be a fully functional program. (*cough cough* class declaration, main method)

Answer 12

Pyth, 6 bytes

/Q/sQ2

Try it online!

Expects input as a list of integers. Outputs 0 if no number can be built by subtracting the other two and >0 if at least one can.

Explanation:

Same algorithm as the answer of @xnor

/Q/sQ2

   sQ     # Sum all elements in the list
  /  2    # Divide the sum by 2
/Q        # Count Occurences of above number in the list

Answer 13

Perl, 24 + 4 = 28 bytes

$^+=$_/2 for@F;$_=$^~~@F

Requires -paX flags to run, prints 1 as True and nothing as False:

-X disables all warnings.

$ perl -paXe'$^+=$_/2 for@F;$_=$^~~@F' <<< '5 3 7'
$ perl -paXe'$^+=$_/2 for@F;$_=$^~~@F' <<< '5 3 8'
1

Answer 14

MATL, 5 bytes

Using @xnor's great approach:

s2/Gm

Try it online!

s    % implicitly input array of three numbers. Compute their sum
2/   % divide by 2
G    % push input again
m    % ismember function: true if sum divided by 2 equals some element of the input

---

Brute-force approach, 12 bytes:

Y@TT-1h*!s~a

Try it online!

Y@       % input array of three numbers. Matrix with all
         % permutations, each one on a different row
TT-1h    % vector [1,1,-1]
*        % multiply with broadcast
!s       % transpose, sum of each column (former row)
~a       % true if any value is 0

Answer 15

05AB1E, non-competing

4 bytes, non-competing because of a stupid thing. Code:

DO;¢

Using 0 as falsy and > 0 as truthy. Uses CP-1252 encoding.

Answer 16

Kona 16 chars

{((+/x)%2)_in x}

Takes a vector from the stack, sums them, divides by 2 and determines if it's in the vector. Returns 1 as truthy and 0 as falsey.

Called via

> {((+/x)%2)_in x} [(2;3;5)]
1
> {((+/x)%2)_in x} [(2;3;4)]
0

Answer 17

CJam, 10 12 bytes

l~:d_:+2/&

2 bytes removed thanks to @MartinBüttner.

This displays a number as truthy result, and no output as falsy result.

Try it here

l~     e# read line and evaluate. Pushes the array
:d     e# convert array to double
_      e# duplicate
:+     e# fold addition on the array. Computes sum of the array
2/     e# divide sum by 2
&      e# setwise and (intersection)

Answer 18

jq, 17 characters

(Yet another rewrite of xnor's Python 3 answer. Upvotes should go to that one.)

contains([add/2])

Input: array of 3 integers.

Sample run:

bash-4.3$ jq 'contains([add/2])' <<< '[5, 3, 2]'
true

bash-4.3$ jq 'contains([add/2])' <<< '[2, 3, -5]'
false

On-line test:

• [5, 3, 2]
• [2, 3, -5]

jq, 18 characters

(17 characters code + 1 character command line option.)

contains([add/2])

Input: list of 3 integers.

Sample run:

bash-4.3$ jq -s 'contains([add/2])' <<< '5 3 2'
true

bash-4.3$ jq -s 'contains([add/2])' <<< '2 3 -5'
false

Answer 19

, 7 chars / 9 bytes

ï⒮≔⨭ï/2

Try it here (Firefox only).

Meh. I'm still finding better ways. It's just @xnor's awesome algorithm.

Answer 20

Seriously, 6 bytes

,;䫡u

Outputs 0 if false and a positive integer otherwise.

Answer 21

Mathematica, 20 19 bytes

MemberQ[2{##},+##]&

Works similarly to most of the other answers.

Answer 22

Haskell, 20 bytes

(\l->sum l/2`elem`l)

Using xnor's solution.

Answer 23

05AB1E, 6 5 bytes

;Oм_O

-1 byte by creating a port of @xnor's Python 3 algorithm.

Try it online or verify all test cases.

Explanation:

·        # Halve every item in the input-array
         #  i.e. [10,6,4] → [5.0,3.0,2.0]
 O       # Sum this array
         #  i.e. [5.0,3.0,2.0] → 10.0
  м_O    # Output 1 if the input-array contain this sum, 0 otherwise
         #  i.e. [10,6,4] and 10.0 → 1

I'm pretty sure м_O can be shortened, but I'm not sure which command(s) I have to use for it..

Answer 24

R, 23 bytes

sum(x<-scan())%in%(2*x)

Try it online!

Shameless port of xnor's answer.

Answer 25

Jolf, 6 bytes

Try it here!

 hx½ux
_hx    the input array
   ½ux  has half the sum of the array

This is xnor's awesome solution to the problem, but in Jolf.

Answer 26

Pylons, 8

Yet another implementation of xnor's algorithm.

i:As2A/_

How it works:

i    # Get command line input.
:A   # Initialize a constant A.
  s  # Set A to the sum of the stack.
2    # Push 2 to the stack.
A    # Push A to the stack.
/    # Divide A/2
_    # Check if the top of the stack is in the previous elements.
     # Print the stack on quit.

Answer 27

SpecBAS - 36 bytes

Uses xnors formula

1 INPUT a,b,c: ?(a+b+c)/2 IN [a,b,c]

outputs 1 if true and 0 if false