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Challenges

Make $2 + 2 = 5$

+5
−3

In this challenge, add 2 integers, but if both the integers are 2, output 5. Shortest code in each language wins!

Example ungolfed program in Python 3.x

def add(x, y):
	if x == 2 and y == 2:
		return 5
	else:
		return x + y

Try it online!

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2 comment threads

"integers" - should this work for negative integers? At least one answer I checked (but I suspect sev... (1 comment)
"Preferably" on a function? (2 comments)

19 answers

+4
−0

C (gcc), 30 bytes

f(x,y){return x+y|!(x^y|x^2);}

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In case the "preferably on a function" requirement can be dropped, then

#define f(x,y)x+y|!(x^y|x^2)

is 28 bytes.

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+4
−0

Vyxal s, 5 bytes

2=A[5

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You're not the only one who can abuse flags, Shaggy...

2=    # Foreach, is it equal to to?
  A[  # If all are 2
    5 # Push a 5.
      # (s flag) sum of top of stack. If 5, then 5 is outputted, else a+b is outputted.
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Dang it I was just about to post that! (3 comments)
+3
−0

Ruby, 20 bytes

->a,b{4[a]*4[b]+a+b}

Attempt This Online!

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+3
−0

Zsh, 25 bytes

>2
<$@&&<<<5||<<<$[$1+$2]

Attempt This Online!

Explanation:

  • >2;<$@: if both arguments are 2:
    • &&<<<5: then print 5
    • ||<<<$[$1+$2]: otherwise print $1 + $2
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+3
−0

Japt -x, 6 5 bytes

Takes input as an array of integers

p!UdÍ

Try it

p!UdÍ     :Implicit input of array U
p         :Push
 !        :  Logical NOT of
  Ud      :  Any true (not zero) when
    Í     :    Subtracted from 2
          :Implicit output of sum of resulting array
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+2
−0

Haskell, 13 bytes

2!2=5
x!y=x+y

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+2
−0

Python 3, 24 bytes

lambda x,y:x+y+(x==y==2)

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+2
−0

Dyalog APL, 9 bytes

+/⊢,2 2≡⊢

Takes the input as a pair

  • +/ sum reduce
  • the input
  • , concatenated with
  • 2 2≡⊢ whether the input is equal to the list 2 2 (1 if true, 0 if false)
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+1
−0

ESCR - not a golfing answer

This isn't really an answer to the question, and it's not an attempt at golfing. However, it shows a cute trick for solving this problem that falls out of the way the ESCR language works.

ESCR has no "operators" in the sense of a traditional programming language, only functions. ESCR functions are enclosed in [ ]. The first token within the [ ] is the function name, which is then followed by whatever parameters are passed to the function. Functions result in a value, which can always be expressed in string format. The entire [...] function reference is replaced by its value.

Addition is done with the built-in function "+". For example, [+ 3 4] results in 7.

Another wrinkle in ESCR is that new versions of existing symbols can be defined. Each symbol is essentially a stack, with the most recently defined version on top of the stack.

New versions of functions can be defined, just like other symbols such as variables. When inside a routine (function, subroutine, or macro), the default version of that routine is the next-older than currently in. Nested routines of the same name therefore call each other down the stack. It is possible to reference a particular version, like the current to do recursion, but requires a deliberate syntax. The default makes it easy to "hook" existing routines.

Built-in functions aren't special in this regard. They are seeded into the function symbol table before user code is run, and are implemented with compiled code, but are otherwise indistinguishable from user-defined functions. New versions of these functions can be created, just like with other ESCR symbols.

The trick here is to define a new version of the "+" function that returns 5 when both arguments are 2, but the normal sum otherwise. (The actual ESCR "+" function can take any number of arguments, but we'll ignore that). Here is the code that creates a new "+", and demonstrates its use:

function +
  if [and [= [arg 1] 2] [= [arg 2] 2]]
    then
      funcval 5
    else
      funcval [+ [arg 1] [arg 2]]
    endif
  endfunc

show [+ 1 2]
show [+ 2 2]
show [+ 2 1]

The FUNCTION and ENDFUNC commands start and end the new function definition. Arguments to routines are accessed inside those routines with the ARG function. [arg 1] expands to the first argument, [arg 2] to the second, etc. The IF statement takes the THEN branch when both arguments to the function are 2. In that case, the FUNCVAL command sets the function value to 5 explicitly. Otherwise, the ELSE case is run. Here the function value becomes the sum of the two arguments.

Note the use of the "+" function in the ELSE clause to do the addition. The default version referenced inside a routine is the previous version, which in this case is the built-in "+" function.

The three SHOW commands write the result of "+" function uses to standard output. The output is:

3
5
3
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OP's comment (1 comment)
+1
−0

Ruby, 23 bytes

->a,b{a==2&&b==2?5:a+b}

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+1
−0

x86-32 assembly, 14 bytes

Straightforward solution with custom calling convention: inputs in eax and ebx, output in eax. cmp opcode could be smaller if limited to 8-bit integers?

00000000 <add225>:
   0:	83 f8 02             	cmp    eax,0x2
   3:	75 06                	jne    b <add>
   5:	83 fb 02             	cmp    ebx,0x2
   8:	75 01                	jne    b <add>
   a:	40                   	inc    eax

0000000b <add>:
   b:	01 d8                	add    eax,ebx
   d:	c3                   	ret    
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+1
−0

Husk, 7 bytes

+±Λ=2¹Σ

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Input as a list.

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+1
−0

Ruby, 19 bytes

->a,b{a|b==2?5:a+b}

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Wrong answer for 0 and 2. (1 comment)
+1
−0

JavaScript, 19 bytes

x=>y=>x-2|y-2?x+y:5

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+1
−0

C (gcc), 33 31 bytes

f(x,y){return x+y+(x==2&y==2);}

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Saved two bytes thanks to Shaggy

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[31 bytes](https://tio.run/##TYrRCoIwGEbv9xQfRrGfWUR0N@1Juonp6h81Yyko4rOvaQTefB@cc8z@bkyMVvb5QGOo2y54... (2 comments)
+1
−0

J, 7 9 bytes

++2&=@+.~

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Dyadic fork that executes with the form (x + y) + (2 = x +. y).

Thanks to torres for pointing out the obvious flaw.

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``` 3 (++4&=@+) 1 5 ``` (1 comment)
+2
−1

C (gcc), 41 bytes

f(x,y){int r=x*y;return r&&r==x+y?5:x+y;}

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Incorrect answer (1 comment)
But the task asks for integers ($\mathbb Z$), not natural numbers ($\mathbb N$). Thus the code should... (1 comment)
+0
−0

Java (JDK), 51 50 23 21 bytes

x->y->x==y&y==2?5:x+y

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Golfed 27 bytes thanks to @Moshi's advice. Golfed 2 bytes thanks to @user's advice on Discord.

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18-byte version (1 comment)
Lambda (1 comment)
+0
−0

J, 17 char

+/`5:@.(2*./@:=])

How it works: What's within parentheses tests if both numbers on the right are 2. If so, the verb agenda @. selects the verb 5: which is always 5. If not, the verb agenda @. selects the verb +/ which sums the numbers.

    +/`5:@.(2*./@:=]) 2 2
5
    +/`5:@.(2*./@:=]) 2 3
5
    +/`5:@.(2*./@:=]) 1 3
4
    +/`5:@.(2*./@:=]) 0 4
4
    +/`5:@.(2*./@:=]) 3 7
10
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