29
\$\begingroup\$

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.

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  • 4
    \$\begingroup\$ Should there always be between 10% and 30% chance for a specific character to be removed? Or is that just for the purpose of the example? \$\endgroup\$ – attinat Aug 11 at 20:40
  • 2
    \$\begingroup\$ what do you mean by "sets of one character"? if the input is BOWL OF SOUP might all the O's be deleted in one go? \$\endgroup\$ – roblogic Aug 12 at 5:14
  • 1
    \$\begingroup\$ all answers thus far use a fixed 20% chance for a character to be removed. I'm not sure the question intent is for all characters to have the same odds. \$\endgroup\$ – Nzall Aug 12 at 7:10
  • 3
    \$\begingroup\$ Output must be through STDOUT, as a text. Do not output a character array. <-- I have a language that allows you to output a character array (it is flattened before output). Is that disallowed? How about languages like C, where a string is basically a character array? \$\endgroup\$ – Ismael Miguel Aug 12 at 9:12
  • 1
    \$\begingroup\$ Use the closest equivalent. C's character array strings are OK, as they are the closest equivalent to text. \$\endgroup\$ – Andrew Aug 12 at 9:20

34 Answers 34

10
\$\begingroup\$

Japt -f, 2 bytes

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

Try it

\$\endgroup\$
  • 2
    \$\begingroup\$ I'm going back in time, making a language where o is this challenge and then going back and submitting my answer, or maybe make the empty string that :p \$\endgroup\$ – Andrew Aug 12 at 9:36
  • \$\begingroup\$ Aren't commandline flags supposed to count towards bytecount? \$\endgroup\$ – Daniel Vestøl Aug 14 at 5:37
  • 1
    \$\begingroup\$ @DanielVestøl Click on the -f, in the title. \$\endgroup\$ – Ismael Miguel Aug 14 at 9:19
  • 1
    \$\begingroup\$ @Andrew Using a made-up language specifically designed for the challenge \$\endgroup\$ – Jo King Aug 15 at 1:06
  • 1
    \$\begingroup\$ was joke but ok \$\endgroup\$ – Andrew Aug 15 at 16:48
8
\$\begingroup\$

Python 3, 63 bytes

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

Try it online!

Python 2, 67 65 bytes

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

Try it online!

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())

Try it online!

\$\endgroup\$
  • \$\begingroup\$ 63 bytes in python 2 with a function \$\endgroup\$ – attinat Aug 11 at 20:47
  • \$\begingroup\$ The rules explicitly state to write a full program: Create a cat program, a.k.a a program that takes an input and prints it. \$\endgroup\$ – movatica Aug 11 at 20:51
  • \$\begingroup\$ This only reads the first line of input. \$\endgroup\$ – AKX Aug 12 at 6:38
  • \$\begingroup\$ Problem does not state it had to read multiple lines. \$\endgroup\$ – movatica Aug 12 at 18:43
8
\$\begingroup\$

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.

\$\endgroup\$
7
\$\begingroup\$

Befunge-98 (PyFunge), 11 bytes

>#@~3j4???,

Try it online!

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.

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5
\$\begingroup\$

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)

Try it online!

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5
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Jelly, 9 5 bytes

5X’µƇ

Try it online!

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
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5
\$\begingroup\$

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

Try it

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

Wolfram Language (Mathematica), 24 bytes

Select[RandomReal[]>.2&]

Try it online!

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

\$\endgroup\$
5
\$\begingroup\$

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).

\$\endgroup\$
  • 2
    \$\begingroup\$ 5L can be for -1 (changes the chance from 20% to 25%, which is still acceptable). \$\endgroup\$ – Grimy Aug 12 at 11:32
  • \$\begingroup\$ @Grimy Nice one, thanks! :) \$\endgroup\$ – Kevin Cruijssen Aug 12 at 11:38
4
\$\begingroup\$

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

Try it online!

\$\endgroup\$
4
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Pyth, 8 5 bytes

sfO4Q

Try it online!

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

Try it online!

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
\$\endgroup\$
  • \$\begingroup\$ Q will throw an error if the input is not python-esque. Errors for example from [1 or a/b. Q, w and z will only work for single-line input, so the best option would probably be j.z \$\endgroup\$ – ar4093 Aug 14 at 11:14
4
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Cubix, 20 bytes

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

Try it online!

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.
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3
\$\begingroup\$

APL (dzaima/APL), 10 9 bytesSBCS

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

⊢⌿⍨4≥∘?5¨

Try it online!

 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

\$\endgroup\$
3
\$\begingroup\$

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.

\$\endgroup\$
3
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R, 32 23 bytes

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

Try it online!

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!

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  • 1
    \$\begingroup\$ function(x)x[!rbinom(x,1,0.2)] \$\endgroup\$ – Roland Aug 12 at 11:02
  • \$\begingroup\$ along the same lines as @Roland, function(x)x[rf(x,1,1)>1]; df(1,1,1) is about 0.16 which does the trick. \$\endgroup\$ – Giuseppe Aug 12 at 13:49
  • \$\begingroup\$ or rt(x,3)>1 (about 20% chance) \$\endgroup\$ – JDL Aug 12 at 20:00
  • 2
    \$\begingroup\$ @JDL it’s <1, but thanks! Another 2 saved. \$\endgroup\$ – Nick Kennedy Aug 12 at 20:04
3
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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 containing the first digit 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.

Try it online

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  • \$\begingroup\$ Very nice trick with STR, I will have to remember that. Not sure it is fair to piggyback off your (uncounted) DECLARE in your (counted) code; but changing that would only cost you 1 byte, since you can eliminate the extra SET with DECLARE @ INT=len(@i) \$\endgroup\$ – BradC Aug 12 at 15:16
  • \$\begingroup\$ @BradC I agree, and I will try to remember not to piggyback in the future \$\endgroup\$ – t-clausen.dk Aug 13 at 7:33
2
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J, 10 bytes

#~5>6?@$~#

Try it online!

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.
\$\endgroup\$
2
\$\begingroup\$

Perl 5 -p, 18 bytes

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

Try it online!

Each character has a 20% chance of being dropped.

\$\endgroup\$
2
\$\begingroup\$

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``
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  • \$\begingroup\$ Since you don't use the name of the function, you're allowed to just provide the function itself. Removing f= gets you down to 44 bytes. You may also input/output as an array, which should save you quite a bit. \$\endgroup\$ – Birjolaxew Aug 12 at 8:44
  • \$\begingroup\$ @Birjolaxew I'm not sure but as I understand it the question disallow the use of char array if the language has string \$\endgroup\$ – jonatjano Aug 12 at 9:39
  • \$\begingroup\$ Yes, Andrew edited the question with additional requirements after my comment. This is generally frowned upon because it invalidates answers that were previously valid (in this case it's even debatable whether your initial answer fits the "Output must be through STDOUT" requirement). \$\endgroup\$ – Birjolaxew Aug 12 at 11:19
  • \$\begingroup\$ What does join`` mean? Can't find it in the spec (cause I don't really know what it is) \$\endgroup\$ – nick zoum Aug 14 at 7:39
  • 1
    \$\begingroup\$ @nickzoum in es6 it is equivalent to join("") mdn revelent page \$\endgroup\$ – jonatjano Aug 14 at 7:40
2
\$\begingroup\$

Scala, 51 46 30 bytes

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

Try it online!

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)
\$\endgroup\$
  • 1
    \$\begingroup\$ You can shave off 7 bytes by changing scala.math.random to math.random and 0.2 to .2. Nice trick using ^ like that. \$\endgroup\$ – Kjetil S. Aug 13 at 20:57
  • \$\begingroup\$ @KjetilS, thank you. I also posted function literals instead of function definitions as a solution. It is acceptable according to this: codegolf.stackexchange.com/questions/3885/… \$\endgroup\$ – Dr Y Wit Aug 14 at 10:02
2
\$\begingroup\$

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

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

Try it online!

\$\endgroup\$
  • \$\begingroup\$ I think that < 1 is eating too much. Maybe <3 or < 4 matches the parameters of the problem. Anyway this does not change the length of the code. \$\endgroup\$ – Luca Aug 12 at 12:00
  • \$\begingroup\$ I haven't programmed in C# in a while, but why is't new Random().Next(5) directly possible? \$\endgroup\$ – Kevin Cruijssen Aug 12 at 12:27
  • 1
    \$\begingroup\$ @KevinCruijssen Because then the same seed will be used for each number, so each number be the same. Just try it with new Random().Next(5), all you will get is the whole input or nothing. \$\endgroup\$ – Embodiment of Ignorance Aug 12 at 13:14
  • 1
    \$\begingroup\$ @KevinCruijssen when using new Random(), the default seed value is Environment.TickCount, which increments every millisecond. If they're all created in the same millisecond tick, they will all have the same seed. The answer is only using a single instance of Random, and when it's called it updates the internal seed value - so each time Next() is called, it creates a different output value. .Net Core however uses a singleton RNG to generate the seed, so it doesn't have this problem \$\endgroup\$ – Zac Faragher Aug 13 at 6:40
  • 1
    \$\begingroup\$ @BenjaminUrquhart Fixed, but added 26 bytes. Also, one can argue doesn't say input has to be from STDIN, but restricted to STDIN or closest equivalent, and function arguments may be the closest equivalent, but I'm not going to do that \$\endgroup\$ – Embodiment of Ignorance Aug 14 at 17:31
1
\$\begingroup\$

PHP, 43 bytes

for(;''<$l=$argn[$i++];rand(0,4)&&print$l);

Try it online!

Each character has 20% of chance to be removed.

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

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));}

Try it online!

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));}

Try it online!

\$\endgroup\$
1
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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

Try it online!

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).

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1
\$\begingroup\$

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));}
\$\endgroup\$
  • \$\begingroup\$ Is dealing with stdout really necessary? Other answers only create a function just converting a string. Seems unfair towards Java. If this C# solution is valid, then s->s.filter(c->Math.random()<.2) is too. \$\endgroup\$ – Holger Aug 14 at 9:23
  • \$\begingroup\$ @Holger Input is restricted to STDIN or closest equivalent. and Output must be through STDOUT or the closest equivalent. If your language does have STDOUT, do not output in any other way. So no, that answer is not valid \$\endgroup\$ – Benjamin Urquhart Aug 14 at 17:25
1
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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";}
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  • 1
    \$\begingroup\$ Smart use of \c, I would not have remembered that! There's still a few optimizations to be made... \$\endgroup\$ – GammaFunction Aug 13 at 3:07
  • \$\begingroup\$ Nice, clever use of RANDOM and array conversion \$\endgroup\$ – roblogic Aug 13 at 4:14
1
\$\begingroup\$

Zsh, 50 bytes

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

Try it online!

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

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

MathGolf, 5 bytes

æƒ√∞*

Try it online!

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.

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

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

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ Pseudo randomness is allowed, if that is the closest way you have to making random actions take hold. \$\endgroup\$ – Andrew Aug 12 at 11:43
0
\$\begingroup\$

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)
\$\endgroup\$

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