Equal, sum or difference!

Question

Write shortest possible code that will return true if the two given integer values are equal or their sum or absolute difference is 5.

Example test cases:

4 1 => True
10 10 => True
1 3 => False
6 2 => False
1 6 => True
-256 -251 => True
6 1 => True
-5 5 => False

The shortest I could come up with in python2 is 56 characters long:

x=input();y=input();print all([x-y,x+y-5,abs(x-y)-5])<1

-9, thanks @ElPedro. It takes input in format x,y:

x,y=input();print all([x-y,x+y-5,abs(x-y)-5])<1

Answer 1

Python 2, 30 bytes

lambda a,b:a in(b,5-b,b-5,b+5)

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_{One byte saved by Arnauld}

_{Three bytes saved by alephalpha}

Answer 2

JavaScript (ES6), 28 bytes

Takes input as (a)(b). Returns \$0\$ or \$1\$.

a=>b=>a+b==5|!(a-=b)|a*a==25

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Answer 3

Dyalog APL, 9 bytes

=∨5∊+,∘|-

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Spelled out:

  =   ∨  5      ∊                +   , ∘    |            -
equal or 5 found in an array of sum and absolute of difference.

Answer 4

x86 machine code, 39 bytes

00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601  j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6  .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3                        ....D..

Assembly

section .text
	global func
func:					;inputs int32_t ecx and edx
	push 0x1
	pop esi
	push 0x5
	pop edi
	push edx
	push ecx
	xor eax, eax

	;ecx==edx?
	cmp ecx, edx
	cmove eax, esi

	;ecx+edx==5?
	add ecx, edx
	cmp edi, ecx
	cmove eax, esi
	
	;ecx-edx==5?
	pop ecx
	pop edx
	sub ecx, edx
	cmp ecx, 5
	
	;ecx-edx==-5?
	cmove eax, esi
	cmp ecx, -5
	cmove eax, esi

	ret

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Answer 5

J, 12 11 bytes

1 byte saved thanks to Adám

1#.=+5=|@-,+

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Explanation

This is equivalent to:

1 #. = + 5 = |@- , +

This can be divided into the following fork chain:

(= + (5 e. (|@- , +)))

Or, visualized using 5!:4<'f':

  ┌─ =               
  ├─ +               
──┤   ┌─ 5           
  │   ├─ e.          
  └───┤          ┌─ |
      │    ┌─ @ ─┴─ -
      └────┼─ ,      
           └─ +      

Annotated:

  ┌─ =                                     equality
  ├─ +                                     added to (boolean or)
──┤   ┌─ 5                                   noun 5
  │   ├─ e.                                  is an element of
  └───┤          ┌─ |  absolute value         |
      │    ┌─ @ ─┴─ -  (of) subtraction       |
      └────┼─ ,        paired with            |
           └─ +        addition               | any of these?

Answer 6

Python 2, 29 31 bytes

lambda a,b:a+b==5or`a-b`in"0-5"

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Since I didn't manage to read the task carefully the first time, in order to fix it, I had to come up with a completely different approach, which is unfortunately not as concise.

Answer 7

R, 40 bytes (or 34)

function(x,y)any((-1:1*5)%in%c(x+y,x-y))

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For non-R users:

• -1:1*5 expands to [-5, 0, 5]
• the %in% operator takes elements from the left and checks (element-wise) if they exist in the vector on the right

A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:

function(x,y)x%in%c(y--1:1*5,5-y)

Answer 8

8086 machine code, 22 20 bytes

8bd0 2bc3 740e 7902 f7d8 3d0500 7405 03d3 83fa05

Ungolfed:

ESD  MACRO
    LOCAL SUB_POS, DONE
    MOV  DX, AX     ; Save AX to DX
    SUB  AX, BX     ; AX = AX - BX
    JZ   DONE       ; if 0, then they are equal, ZF=1
    JNS  SUB_POS    ; if positive, go to SUB_POS
    NEG  AX         ; otherwise negate the result
SUB_POS:
    CMP  AX, 5      ; if result is 5, ZF=1
    JZ   DONE
    ADD  DX, BX     ; DX = DX + BX
    CMP  DX, 5      ; if 5, ZF=1
DONE:
    ENDM

Input numbers in AX and BX and returns Zero Flag (ZF=1) if result is true. If desired, you can also determine which condition was true with the following:

• ZF = 1 and DX = 5 ; sum is 5
• ZF = 1 and AX = 5 ; diff is 5
• ZF = 1 and AX = 0 ; equal
• ZF = 0 ; result false

If the difference between the numbers is 0, we know they are equal. Otherwise if result is negative, then first negate it and then check for 5. If still not true, then add and check for 5.

Sample PC DOS test program. Download it here (ESD.COM).

START:
    CALL INDEC      ; input first number into AX
    MOV  BX, AX     ; move to BX
    CALL INDEC      ; input second number into BX
    ESD             ; run "Equal, sum or difference" routine
    JZ   TRUE       ; if ZF=1, result is true
FALSE:
    MOV  DX, OFFSET FALSY   ; load Falsy string
    JMP  DONE
TRUE:
    MOV  DX, OFFSET TRUTHY  ; load Truthy string
DONE:
    MOV  AH, 9      ; DOS display string
    INT  21H        ; execute
    MOV  AX, 4C00H  ; DOS terminate
    INT  21H        ; execute

TRUTHY   DB 'Truthy$'
FALSY    DB 'Falsy$'

INCLUDE INDEC.ASM   ; generic decimal input prompt routine

Output of test program:

A>ESD.COM
: 4
: 1
Truthy

A>ESD.COM
: 10
: 10
Truthy

A>ESD.COM
: 1
: 3
Falsy

A>ESD.COM
: 6
: 2
Falsy

A>ESD.COM
: 1
: 6
Truthy

A>ESD.COM
: -256
: -251
Truthy

A>ESD.COM
: 6
: 1
Truthy

A>ESD.COM
: 9999999999
: 9999999994
Truthy

Answer 9

Python 2, 38 bytes

-2 bytes thanks to @DjMcMayhem

lambda a,b:a+b==5or abs(a-b)==5or a==b

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Answer 10

Jelly, 7 bytes

+,ạ5eo=

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How it works

+,ạ5eo=  Main link. Arguments: x, y (integers)

+        Yield x+y.
  ạ      Yield |x-y|.
 ,       Pair; yield (x+y, |x-y|).
   5e    Test fi 5 exists in the pair.
      =  Test x and y for equality.
     o   Logical OR.

Answer 11

PowerShell, 48 44 40 bytes

param($a,$b)$b-in($a-5),(5-$a),(5+$a),$a

Try it online! or Verify all Test Cases

Takes input $a and $b. Checks if $b is -in the group ($a-5, 5-$a 5+$a, or $a), which checks all possible combinations of $a,$b, and 5.

-4 bytes thanks to mazzy.
-4 bytes thanks to KGlasier.

Answer 12

Wolfram Language (Mathematica), 22 bytes

Takes input as [a][b].

MatchQ[#|5-#|#-5|#+5]&

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Answer 13

Java (JDK), 30 bytes

a->b->a+b==5|a==b|(b-=a)*b==25

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Answer 14

Pascal (FPC), 26 70 bytes

Edit: + input variables.

Procedure z(a,b:integer);begin Writeln((abs(a-b)in[0,5])or(a+b=5))end;

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---

(abs(a-b)in[0,5])or(a+b=5)

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I hope that my answer is according to all rules of code-golf. It was fun anyway.

Answer 15

C# (.NET Core), 43, 48, 47, 33 bytes

EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!

EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!

EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).

C# (.NET Core), 33 bytes

a=>b=>a==b|a+b==5|(a-b)*(a-b)==25

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Answer 16

Scala, 43 bytes

def f(a:Int,b:Int)=a+b==5|(a-b).abs==5|a==b

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Answer 17

Perl 6, 24 bytes

-1 byte thanks to Grimy

{$^a-$^b==5|0|-5|5-2*$b}

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This uses the Any Junction but technically, ^ could work as well.

Explanation:

{                      }  # Anonymous code block
 $^a-$^b==                # Is the difference equal to
           | |  |        # Any of
          0 
            5
              -5
                 5-2*$b

Answer 18

C (gcc), 33 bytes

f(a,b){a=!(a+b-5&&(a-=b)/6|a%5);}

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Tried an approach I didn't see anyone else try using. The return expression is equivalent to a+b==5||((-6<a-b||a-b<6)&&(a-b)%5==0).

---

Answer 19

C (gcc), 41 34 bytes

f(a,b){a=5==abs(a-b)|a+b==5|a==b;}

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Answer 20

05AB1E, 13 12 bytes

ÐO5Qs`α5QrËO

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Takes input as a list of integers, saving one byte. Thanks @Wisław!

Alternate 12 byte answer

Q¹²α5Q¹²+5QO

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This one takes input on separate lines.

Answer 21

05AB1E, 10 bytes

OIÆ‚Ä50SåZ

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---

O           # Sum the input.
 IÆ         # Reduced subtraction of the input.
   ‚        # Wrap [sum,reduced_subtraction]
    Ä       # abs[sum,red_sub]
     50S    # [5,0]
        å   # [5,0] in abs[sum,red_sub]?
         Z  # Max of result, 0 is false, 1 is true.

Tried to do it using stack-only operations, but it was longer.

Answer 22

Ruby, 34 Bytes

->(a,b){[a+5,a-5,5-a,a].include?b}

Online Eval - Thanks @ASCII-Only

Answer 23

Japt, 14 13 bytes

¥VªaU ¥5ª5¥Nx

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Answer 24

Tcl, 53 bytes

proc P a\ b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}

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Answer 25

Japt, 13 12 bytes

x ¥5|50ìøUra

Try it or run all test cases

x ¥5|50ìøUra
                 :Implicit input of array U
x                :Reduce by addition
  ¥5             :Equal to 5?
    |            :Bitwise OR
     50ì         :Split 50 to an array of digits
        ø        :Contains?
         Ur      :  Reduce U
           a     :    By absolute difference

Answer 26

Batch, 81 bytes

@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1

Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.

Answer 27

Charcoal, 18 bytes

Ｎθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧Ｎ1

Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.

Answer 28

Common Lisp, 48 bytes

(lambda(a b)(find 5(list(abs(- b a))a(+ a b)b)))

Answer 29

Brachylog, 8 bytes

=|+5|-ȧ5

Takes input as a list of two numbers (use _ for negatives). Try it online!

Explanation

Pretty much a direct translation of the spec:

=          The two numbers are equal
 |         or
  +        The sum of the two numbers
   5       is 5
    |      or
     -     The difference of the two numbers
      ȧ    absolute value
       5   is 5

Answer 30

Runic Enchantments, 27 bytes

i::i::}3s=?!@-'|A5:}=?!@+=@

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Outputs 0 (exactly one zero) for false and any other output (literally whatever is left on the stack) for true.

New interpreter build, updated answer, saved 3 bytes (previous revision miscounted and said 4).

Explanation:

i::i::                         Read two inputs, duplicating each twice.
      }                        Rotate stack right
       3s                      Rotate the top 3 items to the right
         =? @                  Compare equal, if so, dump stack
           !                   If not, skip next instruction
             -'|A              Subtract, absolute value, (j)
                 5:}           Push two 5s, rotate one to the bottom
                    =? @       Compare 5 with (j), if equal, dump stack
                      ! +=     If not, add the two inputs, compare with remaining 5
                          @    Dump stack (will be 1 for true and 0 for false)

Essentially it reads both inputs 3 times and arranges the stack so that each operation works on a copy of each input (or a literal 5). As the only way to dump a 0 is if all three comparisons are false, simply dumping the rest of the stack avoids having to do anything other than vomiting the entire stack to output and terminating, all posibilities of which are valid truthy values.

Answer 31

Perl 5, 51 bytes

Pretty simple really, uses the an flags for input. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.

($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)

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Answer 32

Retina 0.8.2, 82 bytes

\d+
$*
^(-?1*) \1$|^(-?1*)1{5} -?\2$|^-?(-?1*) (\3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$

Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:

^(-?1*) \1$                              x==y
^(-?1*)1{5} -?\2$   x>=0 y>=0 x=5+y i.e. x-y=5
                    x>=0 y<=0 x=5-y i.e. x+y=5
                    x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (\3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
                    x<=0 y>=0 y=5-x i.e. x+y=5
                    x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$        x>=0 y>=0 y=5-x i.e. x+y=5
                    x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$        x>=0 y>=0 x=5-y i.e. x+y=5
                    x>=0 y<=0 x=5+y i.e. x-y=5

Pivoted by the last column we get:

x==y            ^(-?1*) \1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
      x>=0 y>=0 ^(1 ?-?){5}$
      x>=0 y<=0 ^(-?1*)1{5} -?\2$
      x<=0 y>=0 ^-?(-?1*) (\3)1{5}$
      x<=0 y<=0 (impossible)       
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?\2$
      x>=0 y<=0 ^(1 ?-?){5}$
      x<=0 y>=0 (impossible)
      x<=0 y<=0 ^-?(-?1*) (\3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (\3)1{5}$
      x>=0 y<=0 (impossible)
      x<=0 y>=0 ^-?(1 ?){5}$
      x<=0 y<=0 ^(-?1*)1{5} -?\2$

Answer 33

Julia 1.0, 38 bytes

f(a,b)=((a+b-5)*(a-b)*(abs(a-b)-5))==0

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Of course, it's probably not the shortest one

Answer 34

APL(NARS) 19 chars, 38 bytes

{∨/(⍺=⍵),5∊∣⍺+⍵,-⍵}

  f←{∨/(⍺=⍵),5∊∣⍺+⍵,-⍵}
  4 f 1
1
  1 f 3
0
  6 f 2
0
  1 f 6
1
  ¯256 f ¯251
1
  6 f 1
1
  ¯5 f 5
0

Answer 35

Ruby (>=1.9), 23 38 bytes

->(a,b){a==b or a+b==5 or(a-b).abs==5}

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It's an anonymous function, so the TIO has some extra code to take input and return the output. Put the two numbers on separate lines.

Returns true if they are equal or add to 5, or if their absolute difference is 5, and false if not.