32
\$\begingroup\$

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
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  • 9
    \$\begingroup\$ welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here! \$\endgroup\$ – Giuseppe Jan 16 at 16:22

35 Answers 35

22
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Python 2, 30 bytes

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

Try it online!

One byte saved by Arnauld

Three bytes saved by alephalpha

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  • \$\begingroup\$ This is amazingly concise, thanks \$\endgroup\$ – Vikrant Biswas Jan 16 at 17:17
  • \$\begingroup\$ Same can be done in Octave/MATLAB in 29 bytes (Try it online!). \$\endgroup\$ – Tom Carpenter Jan 17 at 18:51
17
\$\begingroup\$

JavaScript (ES6), 28 bytes

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

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

Try it online!

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  • 1
    \$\begingroup\$ Damn, took me a long long time to figure out how this handling the difference part. :) \$\endgroup\$ – Vikrant Biswas Jan 16 at 19:34
9
\$\begingroup\$

Dyalog APL, 9 bytes

=∨5∊+,∘|-

Try it online!

Spelled out:

  =   ∨  5      ∊                +   , ∘    |            -
equal or 5 found in an array of sum and absolute of difference.
\$\endgroup\$
8
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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

Try it online!

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5
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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?
\$\endgroup\$
  • \$\begingroup\$ Save a byte with e. \$\endgroup\$ – Adám Jan 16 at 16:58
  • \$\begingroup\$ @Adám How so? Shortest approach I got with e. was =+.5 e.|@-,+. Maybe you forget 5e. is an invalid token in J? \$\endgroup\$ – Conor O'Brien Jan 16 at 17:05
  • 1
    \$\begingroup\$ Since two integers cannot simultaneously sum to 5 and be equal, you can use + instead of +. \$\endgroup\$ – Adám Jan 16 at 17:11
  • \$\begingroup\$ @Adám Ah, I see, thank you. \$\endgroup\$ – Conor O'Brien Jan 16 at 17:15
5
\$\begingroup\$

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)
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  • \$\begingroup\$ The 34 byte one can be reduced by 1 with function(x,y)x%in%c(y--1:1*5,5-y) \$\endgroup\$ – MickyT Jan 16 at 21:36
  • \$\begingroup\$ 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. \$\endgroup\$ – CriminallyVulgar Jan 17 at 14:38
  • 1
    \$\begingroup\$ @CriminallyVulgar does that account for the sum being 5? \$\endgroup\$ – ArBo Jan 17 at 15:45
  • \$\begingroup\$ @ArBo Hah, missed that in the spec, and there wasn't a test case in the TIO so I just glossed over it! \$\endgroup\$ – CriminallyVulgar Jan 18 at 9:39
  • \$\begingroup\$ 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. \$\endgroup\$ – CriminallyVulgar Jan 18 at 15:50
5
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Python 2, 29 31 bytes

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

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5
\$\begingroup\$

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
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4
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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.
\$\endgroup\$
4
\$\begingroup\$

Python 2, 38 bytes

-2 bytes thanks to @DjMcMayhem

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

Try it online!

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  • \$\begingroup\$ Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the 5s and the ors \$\endgroup\$ – ElPedro Jan 16 at 16:12
  • 3
    \$\begingroup\$ Actually, the TIO link could be 38 bytes \$\endgroup\$ – DJMcMayhem Jan 16 at 16:14
  • \$\begingroup\$ @ElPedro the function itself was 40 bytes but I used f= in order to be able to call it \$\endgroup\$ – fəˈnɛtɪk Jan 16 at 18:10
  • 1
    \$\begingroup\$ @DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example \$\endgroup\$ – fəˈnɛtɪk Jan 16 at 18:12
4
\$\begingroup\$

Java (JDK), 30 bytes

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

Try it online!

\$\endgroup\$
4
\$\begingroup\$

Wolfram Language (Mathematica), 22 bytes

Takes input as [a][b].

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

Try it online!

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4
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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.

\$\endgroup\$
  • \$\begingroup\$ ($a-$b) is -$x :) \$\endgroup\$ – mazzy Jan 16 at 17:14
  • \$\begingroup\$ @mazzy Ooo, good call. \$\endgroup\$ – AdmBorkBork Jan 16 at 17:41
  • \$\begingroup\$ 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 \$\endgroup\$ – KGlasier Jan 17 at 15:13
  • 1
    \$\begingroup\$ @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! \$\endgroup\$ – AdmBorkBork Jan 17 at 15:55
4
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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.

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  • 2
    \$\begingroup\$ 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. \$\endgroup\$ – NoOneIsHere Jan 20 at 17:42
  • \$\begingroup\$ 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. \$\endgroup\$ – Dessy Stoeva Jan 20 at 18:42
  • \$\begingroup\$ Alright. Welcome to PPCG! \$\endgroup\$ – NoOneIsHere Jan 20 at 23:47
  • \$\begingroup\$ 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. \$\endgroup\$ – Jo King Jan 24 at 11:08
  • 1
    \$\begingroup\$ No problem, I will change my submission. \$\endgroup\$ – Dessy Stoeva Jan 24 at 13:55
3
\$\begingroup\$

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!

\$\endgroup\$
  • \$\begingroup\$ Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D \$\endgroup\$ – Destroigo Jan 16 at 16:41
  • 1
    \$\begingroup\$ You can get it down to 33 bytes applying dana's tips \$\endgroup\$ – Embodiment of Ignorance Jan 16 at 18:10
3
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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).


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3
\$\begingroup\$

Scala, 43 bytes

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

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Isn't it possible to golf the || to |? I know it's possible in Java, C#, Python, or JavaScript, but not sure about Scala. \$\endgroup\$ – Kevin Cruijssen Jan 17 at 12:15
  • \$\begingroup\$ Actually yes! thanks \$\endgroup\$ – Xavier Guihot Jan 17 at 12:18
3
\$\begingroup\$

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*$b
\$\endgroup\$
  • 1
    \$\begingroup\$ -1 byte with {$^a-$^b==5|0|-5|5-2*$b} \$\endgroup\$ – Grimy Jan 17 at 15:00
2
\$\begingroup\$

C (gcc), 41 34 bytes

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

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ Why does f return a? Just some Undefined Behavior? \$\endgroup\$ – Tyilo Jan 16 at 17:05
  • \$\begingroup\$ @Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value. \$\endgroup\$ – cleblanc Jan 16 at 17:07
  • \$\begingroup\$ 30 bytes Try it online! \$\endgroup\$ – Logern Jan 16 at 18:02
  • \$\begingroup\$ @Logern Doesn't work for f(6,1) \$\endgroup\$ – cleblanc Jan 16 at 18:31
  • \$\begingroup\$ @ceilingcat Doesn't work for f(6,1) \$\endgroup\$ – cleblanc Jan 16 at 18:32
2
\$\begingroup\$

05AB1E, 13 12 bytes

ÐO5Qs`α5QrËO

Try it online!

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

Alternate 12 byte answer

Q¹²α5Q¹²+5QO

Try it online!

This one takes input on separate lines.

\$\endgroup\$
  • 1
    \$\begingroup\$ Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial |? \$\endgroup\$ – Wisław Jan 16 at 17:22
  • \$\begingroup\$ @Wisław Good point, I updated my answer. Thanks! \$\endgroup\$ – Cowabunghole Jan 16 at 17:28
  • \$\begingroup\$ I found a 11 bytes alternative: OI`αª5¢IË~Ā. Input is a list of integers. \$\endgroup\$ – Wisław Jan 16 at 17:54
  • 1
    \$\begingroup\$ OIÆÄ)5QIËM is 10. \$\endgroup\$ – Magic Octopus Urn Jan 16 at 18:35
  • 1
    \$\begingroup\$ @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" :) \$\endgroup\$ – Cowabunghole Jan 16 at 18:43
2
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ This will unfortunately return true if the sum is 0 such as for [5, -5] \$\endgroup\$ – Emigna Jan 16 at 20:28
  • 1
    \$\begingroup\$ Your other 10-byte solution that you left as a comment (OIÆÄ‚5QIËM) is correct for [5,-5]. \$\endgroup\$ – Kevin Cruijssen Jan 17 at 9:31
  • \$\begingroup\$ Another 10-byte solution that I came up with is OsÆÄ‚5åsË~. Almost identical to yours it seems. Try it online! \$\endgroup\$ – Wisław Jan 17 at 10:54
2
\$\begingroup\$

Ruby, 34 Bytes

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

Online Eval - Thanks @ASCII-Only

\$\endgroup\$
  • \$\begingroup\$ do you check if they're equal though... \$\endgroup\$ – ASCII-only Jan 20 at 12:20
  • \$\begingroup\$ Oops, forgot to add that check. Thanks @ASCII-only for pointing out the mistake. \$\endgroup\$ – Jatin Dhankhar Jan 20 at 12:23
  • 1
    \$\begingroup\$ i'd be nice if you could link to this \$\endgroup\$ – ASCII-only Jan 21 at 2:57
  • \$\begingroup\$ this might be valid? not completely sure though, you might wanna check with someone else \$\endgroup\$ – ASCII-only Jan 28 at 4:45
  • \$\begingroup\$ 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. \$\endgroup\$ – Jatin Dhankhar Jan 28 at 5:04
1
\$\begingroup\$

Tcl, 53 bytes

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

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Japt, 14 13 bytes

¥VªaU ¥5ª5¥Nx

Try it online!

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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
\$\endgroup\$
  • \$\begingroup\$ Fails for [-5,5] (should be falsey) \$\endgroup\$ – Kevin Cruijssen Jan 17 at 9:26
  • \$\begingroup\$ Thanks, @KevinCruijssen. Rolled back to the previous version. \$\endgroup\$ – Shaggy Jan 17 at 9:50
1
\$\begingroup\$

Common Lisp, 48 bytes

(lambda(a b)(find 5(list(abs(- b a))a(+ a b)b)))
\$\endgroup\$
1
\$\begingroup\$

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
\$\endgroup\$
0
\$\begingroup\$

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$
\$\endgroup\$

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