Eşit, toplam veya fark!

32

Verilen iki tamsayı değeri eşitse veya toplam veya mutlak fark 5 ise, gerçek değeri verecek en kısa kodu yazın.

Örnek test durumları:

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

Python2'de bulabildiğim en kısa 56 karakter uzunluğunda:

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

-9, teşekkürler @ EPedro. X, y biçiminde girdi alır:

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

code-golf decision-problem

— Vikrant Biswas
kaynak

9

PPCG'ye hoş geldiniz! Bu iyi bir ilk zorluktur - zorluk açıkça tanımlanmıştır, geniş test senaryoları vardır ve varsayılan G / Ç'imizi kullanır! Bir süre takılıp kalırsanız ve ilginç zorlukları düşünmeye devam ederseniz , bu siteye göndermeden önce geri bildirim almak için The Sandbox'ı kullanmanızı öneririm . Umarım burada geçirdiğiniz zamandan zevk alırsınız!

— Giuseppe

Yanıtlar:

22

Python 2, 30 bytes

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

_{One byte saved by Arnauld}

_{Alephalpha tarafından kaydedilen üç bayt}

— ARBO
kaynak

This is amazingly concise, thanks

— Vikrant Biswas

Same can be done in Octave/MATLAB in 29 bytes (Try it online!).

— Tom Carpenter

17

JavaScript (ES6), 28 bytes

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

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

— Arnauld
kaynak

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∊+,∘|-

Spelled out:

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

— dzaima
kaynak

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?
	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

— Logern
kaynak

5

J, 12 11 bytes

1 byte saved thanks to Adám

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

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?

— Conor O'Brien
kaynak

Save a byte with e.

— Adám

@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

@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))

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)

— ngm
kaynak

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==5or`a-b`in"0-5"

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.

— Kirill L.
kaynak

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.

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

— 640KB
kaynak

4

Jelly, 7 bytes

+,ạ5eo=

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.

— Dennis
kaynak

4

Python 2, 38 bytes

-2 bytes thanks to @DjMcMayhem

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

— fəˈnɛtɪk
kaynak

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

Java (JDK), 30 bytes

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

— Olivier Grégoire
kaynak

4

Wolfram Language (Mathematica), 22 bytes

Takes input as [a][b].

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

— alephalpha
kaynak

4

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.

— AdmBorkBork
kaynak

($a-$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;

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

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

— Dessy Stoeva
kaynak

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

— Destroigo
kaynak

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);}

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).

— attinat
kaynak

3

Scala, 43 bytes

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

— Xavier Guihot
kaynak

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}

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

— Jo King
kaynak

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;}

— cleblanc
kaynak

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

30 bytes Try it online!

— Logern

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

— cleblanc

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

— cleblanc

2

05AB1E, 13 12 bytes

ÐO5Qs`α5QrËO

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

Alternate 12 byte answer

Q¹²α5Q¹²+5QO

This one takes input on separate lines.

— Cowabunghole
kaynak

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: OI`αª5¢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

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.

— Magic Octopus Urn
kaynak

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

— Jatin Dhankhar
kaynak

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}

— sergiol
kaynak

1

Japt, 14 13 bytes

¥VªaU ¥5ª5¥Nx

— Oliver
kaynak

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.

— Neil
kaynak

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.

— Neil
kaynak

1

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

— Shaggy
kaynak

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)))

— coredump
kaynak

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

— DLosc
kaynak

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$

— Neil
kaynak

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