# 等于，等于或等于！

32

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


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


-9，谢谢@ElPedro。它采用x，y格式的输入：

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


9

17

# JavaScript（ES6），28个字节

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

1
Damn, took me a long long time to figure out how this handling the difference part. :)
Vikrant Biswas

9

# Dyalog APL, 9 bytes

=∨5∊+,∘|-

Try it online!

Spelled out:

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

8

# 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?
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

Try it online!

5

# J, 12 11 bytes

1 byte saved thanks to Adám

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


Try it online!

## 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?


Save a byte with e.

@Adám How so? Shortest approach I got with e. was =+.5 e.|@-,+. Maybe you forget 5e. is an invalid token in J?
Conor O'Brien

1
Since two integers cannot simultaneously sum to 5 and be equal, you can use + instead of +.

@Adám Ah, I see, thank you.
Conor O'Brien

5

# R, 40 bytes (or 34)

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

Try it online!

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)

The 34 byte one can be reduced by 1 with function(x,y)x%in%c(y--1:1*5,5-y)
MickyT

Can drop to 30 bytes by moving the subtraction: function(x,y)(x-y)%in%(-1:1*5), and drop it further to 24 bytes by dropping the function notation to scan() input: diff(scan())%in%(-1:1*5) Try it online!. Still very much the same method though.
CriminallyVulgar

1
@CriminallyVulgar does that account for the sum being 5?
ArBo

@ArBo Hah, missed that in the spec, and there wasn't a test case in the TIO so I just glossed over it!
CriminallyVulgar

Minor change that can be made to both is to use pryr::f, which happens to work in both cases. Whether it can properly detect the arguments is entirely somewhat hit or miss but it seems to nail these two functions. e.g. pryr::f(x%in%c(y--1:1*5,5-y)) Try it online!. Gets you to 36 and 29 bytes respectively.
CriminallyVulgar

5

# Python 2, 29 31 bytes

lambda a,b:a+b==5ora-bin"0-5"

Try it online!

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.

5

# 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.

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


4

# Jelly, 7 bytes

+,ạ5eo=


Try it online!

### 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.


4

# Python 2, 38 bytes

-2 bytes thanks to @DjMcMayhem

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

Try it online!

Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the 5s and the ors
ElPedro

3
Actually, the TIO link could be 38 bytes
DJMcMayhem

@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
fəˈnɛtɪk

1
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
fəˈnɛtɪk

4

# Wolfram Language (Mathematica), 22 bytes

Takes input as [a][b].

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


Try it online!

4

# PowerShell, 4844 40 bytes

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

($a-$b) is -$x :) mazzy @mazzy Ooo, good call. AdmBorkBork If you switch 5 and $b around you can cut off a couple bytes(ie param($a,$b)$b-in($a-5),(5-$a),($a+5),$a) Try it out here KGlasier 1 @KGlasier Excellent suggestion. I needed to swap $a+5 to 5+$a to get it to cast appropriately when taking command-line input, but otherwise awesome. Thanks! AdmBorkBork 4 # 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; Try it online! (abs(a-b)in[0,5])or(a+b=5) Try it online! I hope that my answer is according to all rules of code-golf. It was fun anyway. 2 Hello, and welcome to PPCG! Normally, you have to take input, instead of assuming it is already in variables. I don't know Pascal, but I think that is what this code is doing. NoOneIsHere Hello, NoOneIsHere and thank you for the remark. It was may concern too - shall I include the initialization of the variables. Looking at several other solutions, like Java for example, where the function definition with parameters was excluded from the total length of the solution, I decided not to include ReadLn. Dessy Stoeva Alright. Welcome to PPCG! NoOneIsHere The Java submission is an anonymous lambda which takes two parameters. This appears to use predefined variables, which is not a valid method of input. Jo King 1 No problem, I will change my submission. Dessy Stoeva 3 # 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 Try it online! Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D Destroigo 1 You can get it down to 33 bytes applying dana's tips Embodiment of Ignorance 3 # C (gcc), 33 bytes f(a,b){a=!(a+b-5&&(a-=b)/6|a%5);} Try it online! 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). 3 # Scala, 43 bytes def f(a:Int,b:Int)=a+b==5|(a-b).abs==5|a==b Try it online! Isn't it possible to golf the || to |? I know it's possible in Java, C#, Python, or JavaScript, but not sure about Scala. Kevin Cruijssen Actually yes! thanks Xavier Guihot 3 # Perl 6, 24 bytes -1 byte thanks to Grimy {$^a-$^b==5|0|-5|5-2*$b}

Try it online!

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*$b 1 -1 byte with {$^a-$^b==5|0|-5|5-2*$b}
Grimmy

2

# C (gcc), 41 34 bytes

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

Try it online!

1
Why does f return a? Just some Undefined Behavior?
Tyilo

@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
cleblanc

Logern

@Logern Doesn't work for f(6,1)
cleblanc

@ceilingcat Doesn't work for f(6,1)
cleblanc

2

# 05AB1E, 13 12 bytes

ÐO5Qsα5QrËO


Try it online!

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

Q¹²α5Q¹²+5QO


Try it online!

This one takes input on separate lines.

1
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial |?
Wisław

@Wisław Good point, I updated my answer. Thanks!
Cowabunghole

I found a 11 bytes alternative: OIαª5¢IË~Ā. Input is a list of integers.
Wisław

1
OIÆÄ)5QIËM is 10.
Magic Octopus Urn

1
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
Cowabunghole

2

# 05AB1E, 10 bytes

OIÆ‚Ä50SåZ


Try it online!

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.

1
This will unfortunately return true if the sum is 0 such as for [5, -5]
Emigna

1
Your other 10-byte solution that you left as a comment (OIÆÄ‚5QIËM) is correct for [5,-5].
Kevin Cruijssen

Another 10-byte solution that I came up with is OsÆÄ‚5åsË~. Almost identical to yours it seems. Try it online!
Wisław

2

## Ruby, 34 Bytes

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


Online Eval - Thanks @ASCII-Only

do you check if they're equal though...
ASCII-only

Oops, forgot to add that check. Thanks @ASCII-only for pointing out the mistake.
Jatin Dhankhar

1
i'd be nice if you could link to this
ASCII-only

this might be valid? not completely sure though, you might wanna check with someone else
ASCII-only

This will work but it requires .nil? check to give output in the required format. ->(a,b){[a+5,a-5,5-a,a].index(b).nil?}, this is longer than the current one.
Jatin Dhankhar

1

# Tcl, 53 bytes

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


Try it online!

1

## 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.

1

# 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.

1

# Japt, 13 12 bytes

x ¥5|50ìøUra

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


Fails for [-5,5] (should be falsey)
Kevin Cruijssen

Thanks, @KevinCruijssen. Rolled back to the previous version.
Shaggy

1

## Common Lisp, 48 bytes

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


1

# 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


0

# 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}$x<=0 y<=0 ^(-?1*)1{5} -?\2$