The cat ate your input again!

Question

Create a cat program, a.k.a a program that takes an input and prints it.

...Except, the program will randomly take characters away from your input and print that instead.

Each character in the input should have generally equal odds to be removed with the program, though, as it is hard to make that, the odds for each character can vary by 10% at most.

Your program should take the input, then randomly remove characters from the input, then print that version again. (You can print with trailing newlines or other characters in case your language has to print newlines.)

If the input was BOOOWL, it shouldn't remove all Os with an equal chance: each character (not unique) should be considered, so instead of every O combined having a 1/5 chance (for example), each O should have a 1/5 chance, so, instead of there being a 1/5 chance of BWL, there should be a 1/5 chance of BOWL, BOOWL.

Input is restricted to STDIN or closest equivalent.

Each character must have a minimum of 10% and a maximum of 30% chance to be removed.

Each character's odds should be calculated individually.

You can use any component of your language supporting random actions, be it functions or something other.

Output must be through STDOUT or the closest equivalent. If your language does have STDOUT, do not output in any other way. If your language cannot output strings as text, use closest equivalent (C's character array output's OK here).

This is code golf. Shortest program wins.

Answer 1

Japt -f, 2 bytes

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.

Try it

Answer 2

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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Answer 3

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.

Answer 4

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.

Answer 5

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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Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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).

Answer 10

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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Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

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.

Answer 15

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!

Answer 16

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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Answer 17

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.

Answer 18

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.

Answer 19

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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Answer 20

Perl 5 -p, 18 bytes

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

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Each character has a 20% chance of being dropped.

Answer 21

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``

Answer 22

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)

Answer 23

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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Answer 24

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";}

Answer 25

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).

Answer 26

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));}

Answer 27

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

Answer 28

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.

Answer 29

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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Answer 30

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)

Answer 31

Gaia, 10 bytes

⟨w8&Ø+ṛ⟩¦$

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Has a 1/9th (11%) chance of removing any given character. This abuses the fact that the meta ¦ treats a string as a list, much as Python does.

 ⟨	⟩¦	| for each character do:
  w8&		| wrap as list and duplicate 8 times
     Ø+		| add empty string to that list
       ṛ	| select element at random
	  $	| and join with no separator
		| implicitly print top of stack

Answer 32

Python3, 61 bytes

Does this count? I guess if you consider execution across seconds of time "random"

import time
for c in input():print(end=c[time.time()%5<1:])

Answer 33

Pepe, 61 bytes

REEerEERREeeeeEeEeREEeEerEEEEErrEEReReEeRRERRErEereereReReree

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There is 11.11% or \$\frac{1}9\$ chance of characters being deleted.

Explanation:

... # some-command -> (stack) // some-explanation

REEe # Input (str) -> (R)
rEE # Create label 0 -> (r) // main loop begins
  RREeeeeEeEe # Push 10 -> (R) // this is for the randomiser
              # R flag: place it in the beginning
  REEeEe # Random number from 1 to 10 -> (R)
  rEEEEE # Increment -> (r) // 0 > 1, for the next command
  rrEE # Create label 1 -> (r) // without previous command, the existing label 0 is replaced
       # r flag: Skip until rEe (return)
    Re # Pop -> (R) // removes the random number
    ReEe # Output as char & pop it -> (R)
    RRERRE # Push 0 2 times -> (R) // these are popped to fix the pos of R stack
  rEe # Return to where goto was called
  ree # Goto 1 if 1 != random from 1 to 10 // The whole rrEE ... rEe ree is an if..then
  re # Pop -> (r)
  ReRe # Pop 2 times -> (R)
ree # Loop while char of (R) != 0

Answer 34

Ruby -p, 27 bytes

The -p flag takes each line of input as the global variable $_ and runs it through the code block and then prints its contents. gsub modifies $_ by replacing characters that match the given regex (only works if -p or -n is present).

gsub(/.|\n/){$&if rand>0.2}

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