Створіть програму для котів, яка називається програмою, яка приймає дані та друкує її.

... За винятком того, що програма випадково відбере символи від вашого вводу та надрукує це замість цього.

Кожен символ у введенні повинен мати загальні рівні шанси на видалення з програмою, хоча, як це важко зробити, шанси для кожного символу можуть змінюватися не більше ніж на 10%.

Ваша програма повинна взяти вхід, а потім випадковим чином видалити символи з цього входу, а потім знову надрукувати цю версію. (Ви можете друкувати за допомогою останніх рядків або інших символів, якщо ваша мова має друкувати нові рядки.)

Якщо вхід був BOOOWL, він не повинен видаляти всі Оси з ​​рівним шансом: кожен символ (не унікальний) повинен розглядатися, тому замість кожного O комбінованого має шанс 1/5 (наприклад), кожен O повинен мати 1 / 5 шансів, так що , замість того , щоб там бути 1/5 шанс BWL, має бути 1/5 шанс BOWL, BOOWL.

Введення обмежено STDIN або найближчим еквівалентом.

Кожен персонаж повинен бути мати мінімум 10% та максимум 30% шансу бути видаленим.

Коефіцієнти кожного персонажа повинні розраховуватися індивідуально.

Ви можете використовувати будь-який компонент вашої мови, що підтримує випадкові дії, будь то функції чи щось інше.

Вихід повинен бути через STDOUT або найближчий еквівалент. Якщо у вашій мові є STDOUT, не друкуйте іншим способом. Якщо ваша мова не може виводити рядки як текст, використовуйте найближчий еквівалент (вихідний масив символів C тут добре).

Це код гольфу. Найкоротша програма виграє.

Japt -f , 2 байти

5ö

The -f flag "runs the program on each element in the first input, outputting an array of those that return a truthy value." 5ö returns a random number between 0(inclusive) and 5(exclusive). Like JavaScript, 0 is falsy in Japt.

Спробуй це

Python 3, 63 bytes

from random import*
for c in input():print(end=c[random()<.2:])

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Python 2, 67 65 bytes

from random import*
print''.join(c for c in input()if.8>random())

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Each character has a 20% chance of beeing dropped.

Different approach, same length:

from random import*
print''.join(c[random()<.2:]for c in input())

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Charcoal, 4 bytes

ΦＳ‽⁵

Try it online! Link is to verbose version of code. Explanation:

 Ｓ      Input a string
Φ       Filter where nonzero
  ‽⁵    Random number 0..4
        Implicitly print

You can use any number from 4 to 10 to get chances of 25% to 10% respectively.

Befunge-98 (PyFunge), 11 bytes

>#@~3j4???,

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Each character has a 25% chance of being removed. This decision is made at the three ? instructions.

? sets the program counter to one of the four directions, with equal probability. In this case, up & down wrap back around to the same instruction, so we can ignore those as options.

There are two ways out of the forest of ?s: to the right (output) and to the left (no output). This situation is symmetric, so if starting from the middle ?, there is a $p_2 = 1/2$ chance of outputting. The chance of outputting if starting from the right ? is $p_3 = 1/2 * 1 + 1/2 * p_2 = 3/4$. Therefore, after reading a character, we jump to the rightmost ? to determine whether or not to output.

Octave, 23 bytes

Generates an array of the same size as the input (strings in Octave are character arrays), checks each of the random numbers whether it is greater than 0.2 and then uses logical indexing to extract the characters at the corresponding positions.

@(s)s(rand(size(s))>.2)

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Jelly, 9 5 bytes

5X’µƇ

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A monad which takes a Jelly string as its argument and returns the processed Jelly string. When used as a full program implicitly prints the output. Each character has a 20% chance of being removed.

Explanation

   µƇ | Filter using the following as a monad for each character:
5X    | - Random number between 1 and 5
  ’   | - Decreased by 1

Japt, 3 bytes

Each character has a 1 in 5 chance of being removed. The 5 can be changed to anything between 4 & 9, inclusive, or A for 10 to change the odds.

Æ5ö

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Æ5ö     :Implicit input of string
Æ       :Filter by
 5ö     :  Random integer in the range [0,5), with 0 being falsey

Wolfram Language (Mathematica), 24 bytes

Select[RandomReal[]>.2&]

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Takes a list of characters as input. Each character has a .2 chance to be removed.

05AB1E, 5 4 bytes

ʒ₄Ω≠

-1 byte thanks to @Grimy.

Try it online or run the same program 10 times.

Each character has a 25% change of being dropped.

Explanation:

ʒ     # Filter the characters of the (implicit) input-string by:
 ₄    #  Push 1000
  Ω   #  Pop and push a random digit from it
   ≠  #  And check that it's NOT 1 (!=1)
      # (after which the result is output implicitly)

≠ could also be _ (==0).

MATL, 9 bytes

t&n&r.2>)

Exaplanation:

t         implicitly take input and duplicate it
 &n       compute the size of the input and...
   &r     generate a random array of that size
     .2>  check which entries of that array are greater than 0.2
        ) and use the result using logical indices to extract certain characters of the input

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Pyth, 8 5 bytes

sfO4Q

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sfO4Q   Implicit: Q=eval(input())
 f  Q   Filter characters of Q where the following is truthy:
  O4      Random number in the range [0-4)
          Any non-zero value is truthy, so this will drop characters 25% of the time
s       Concatenate into string, implicit print

Previous version, 8 bytes:

s*Vm!!O4

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s*Vm!!O4QQ   Implicit: Q=eval(input())
             Trailing QQ inferred
   m    Q    Map each character in Q using:
      O4       Choose random integer in [0-4)
    !!         Logical NOT twice - maps 0 to 0, anything else to 1
             The result is a list of 0s and 1s, with 0 having 25% chance to appear
 *V      Q   Vectorised multiplication of the above with Q
s            Concatenate into string, implicit print

Cubix, 20 bytes

u$w\A|UDw@?;...>o._U

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Longer than I had hoped as I had a number of no-ops that I can't seem to get rid of. The chance to drop a character is 25%. I assume this is okay.

    u $
    w \
A | U D w @ ? ;
. . . > o . _ U
    . .
    . .

Watch it run

Brief explanation:

• A|A this initialises the the stack, Input all, reflect back, Input all (just an EOI -1)
• ;? pop to of stack, test for EOI (-1).
• _?@ if negative, reflect back into test and end on halt
• $D jump the \ into the random direction setter.
  • from the direction setter, 3 direction lead to the o for output then back into the loop, one misses the o in it's path and goes straight to the loop.

APL (dzaima/APL), 10 9 bytes^SBCS

Anonymous tacit prefix function. Each character has exactly 20% chance of being removed.

⊢⌿⍨4≥∘?5¨

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5¨ zero for each character

? random integer range 1–5 for each character

4≥ Boolean mask for those integers that are less than or equal to 4

⊢⌿⍨ filter the argument using that mask

Retina, 15 bytes

/./_?(`.

Try it online! Explanation:

/./_

Process each character individually.

?(`

Perform a substitution at random. The first substitution deletes the character, while the other three leave it unchanged, thus giving a 25% chance of deleting the character. This can be decreased as necessary by appending additional pairs of newlines.

R, 32 23 bytes

function(x)x[rt(x,3)<1]

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A function taking a character vector as input and returning a processed character vector. Each character has a 20% chance of being removed.

Thanks to @Roland and @Giueseppe for helping save 7 bytes, and @JDL for a further 2!

T-SQL 2012, 83 bytes

Looping through the input from right to left removing 0 or 1 character.

25% chance for each character getting removed.

DECLARE @i varchar(max)='The cat ate my homework'

DECLARE @ int=len(@i)WHILE @>0SELECT
@i=stuff(@i,@,str(rand()*2)/2,''),@-=1PRINT @i

Explanation:

rand()*2 returns a float, which can't be used in the stuff command.

The str converts this into a varchar after rounding to nearest whole number. The float is being converted to a varchar(which isn't allowed as third parameter in stuff either).

This varchar has a 25% chance of being '2', 50% chance of being '1', 25% chance of being '0'. Dividing by 2, there is a 25% chance of result being 1. This division converts the varchar to an integer.

Integer is the expected third parameter in stuff function.

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J, 10 bytes

#~5>6?@$~#

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Similar to Adam's APL answer, though I actually wrote it before looking at his.

• 6.. $~ # Take the length of input # and shape $~ the number 6 into a list that long.
• ?@ Treat each six in that list as a die and roll ? it.
• >5 Is the die less than 5 (possible values are 0..5)? Use that boolean result to create a bit mask.
• #~ Filter the input with that mask.

Perl 5 -p, 18 bytes

s/./.2<rand&&$&/ge

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Each character has a 20% chance of being dropped.

Javascript,  46   44  51 bytes

i=>alert([...i].filter(c=>Math.random()>.2).join``)

+7 bytes because of the added STDOUT requirement

-2 bytes thank to Birjolaxew

original answer: 44 bytes without the STDOUT requirement

i=>[...i].filter(c=>Math.random()>.2).join``

Scala, 51 46 30 bytes

s=>s.flatMap(x=>if(math.random>.2)Some(x)else None)

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PS. Like in many other solutions, the probability of dropping char is 20%.

Update:

-5 bytes by using String instead of Option[String] in flatMap

s=>s.flatMap(x=>if(math.random>.2)x+""else "")

30 bytes by using filter

s=>s.filter(x=>math.random>.2)

C# (Visual C# Interactive Compiler), 71 bytes

var y=new Random();foreach(var k in ReadLine())if(y.Next(5)<4)Write(k);

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PHP, 43 42 bytes

for(;''<$l=$argn[$i++];rand()%5&&print$l);

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Each character has 20% of chance to be removed.

C (gcc), 50 bytes

This program has a 20% chance of dropping a letter. Unfortunately the random number generator isn't seeded so you get the same sequence on each run. Basically the only trick is inverting the input character to halt the loop on EOF.

main(c){for(;c=~getchar();rand()%5&&putchar(~c));}

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C (gcc), 64 59 bytes

Thanks to ceilingcat for the -5 bytes.

If you want the RNG seeded on each run.

main(c){for(srand(&c);c=~getchar();rand()%5&&putchar(~c));}

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Lua, 69 68 bytes

for c in io.lines(nil,1)do io.write(math.random()>.2 and c or '')end

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Kinda straightforward, but seems to be shortest version: iterate over stdin char by char (with io.lines… that name is misleading), then based on random value either print one or empty string (e.g. nothing).

Java

Non-terminating: 82 bytes

v->{for(int i;;i=System.in.read(),System.out.print(Math.random()<.2?"":(char)i));}

Terminating (TIO): 105 bytes

v->{var n=System.in;for(int i;n.available()>0;i=n.read(),System.out.print(Math.random()<.2?"":(char)i));}

Zsh, 53 41 bytes

_{-12, thanks to GammaFunction}

41 bytes: try it online!

Converts the input to an array of characters, then tries to print each element c, unless it's eaten by the ((RANDOM%4)) evaluating to false!

for c (${(s::)1})((RANDOM%4))&&echo $c\\c

53 bytes: try it online!

A more straightforward, but verbose, iteration over string-length.

for ((;i<$#1;i++)){((RANDOM%4>0))&&echo "${1[i]}\c";}

Zsh, 50 bytes

for c (${(s::)"$(<&0)"})
((RANDOM%5))&&echo -nE $c

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Similar to RobLogic's answer, but following the input requirements more closely, and works for inputs with backslashes.

"$(<&0)" instead of "<&0" or $(<&0) because the first doesn't work in substitutions, and the second eats newlines. The -nE flags are necessary to prevent backslashes from being parsed as escape sequences, and to prevent newlines being inserted.

echo -nE

MathGolf, 5 bytes

æƒ√∞*

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Explanation

æ       foreach character...
 ƒ      random float in range [0,1)
  √     take square root (making P(x < 0.5) = 0.25)
   ∞    pop a, push 2*a
    *   repeat character int(x) times

Each character will be repeated 0 or 1 times, depending on the random value. Since the expected value after the square root is shifted, there is a 25% probability that each character is removed.

Alternative 5-byter

gÉ;4w

Filter the characters by a random number in [0, 4]. Due to how filtering works, I have to discard the actual character within the filter loop, which adds 1 byte.

GFortran, 120 bytes

Not too bad, if we use the deprecated RAN() function, which is pseudo-random, i.e. you get the same sequence each time. The proper way to generate random numbers in GFortran is with CALL RANDOM_SEED() and CALL RANDOM_NUMBER(R) but that's a lot of bytes!

character(99)S;read(*,'(A)')S;do i=1,len_trim(S)
if(ran(0)*5.gt.1)then;write(*,'(A)',advance="no")S(i:i)
endif;enddo;end

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Oracle SQL, 133 bytes

select listagg(decode(sign(dbms_random.value-0.2),1,substr(x,level,1)))within group(order by level)from t connect by level<=length(x)

It works with an assumption that input data is stored in a table t(x), e.g.

with t(x) as (select 'The cat ate my homework' from dual)