C'est beaucoup de singes

35

Le théorème du singe infini stipule que, dans un temps infini, une machine qui envoie un flux infini de caractères aléatoires dactylographiera toujours un texte donné.

Cela me semble être une excellente idée pour un défi.

Processus

Pour monkey-ize une chaîne A, les étapes suivantes doivent être suivies:

  1. Prenez une chaîne vide. Nous appellerons cette chaîne B.
  2. Choisissez un caractère ASCII imprimable uniformément aléatoire (caractères dans la plage 0x20à 0x7E) et d' ajouter que le caractère à B.
  3. Si A est une sous-chaîne de B, B est notre chaîne monkey-ized. Sinon, répétez l'étape 2 jusqu'à ce que A soit une sous-chaîne de B.

Ce processus n'est qu'un exemple, des méthodes plus simples peuvent exister selon votre langue. Vous n’avez pas besoin de suivre cette méthode à la lettre, à condition d’obtenir la même répartition des sorties.

Le défi

Ecrivez un programme ou une fonction qui, avec une chaîne non vide dans un format raisonnable , retourne une version monkey de cette chaîne.

Votre programme ne doit pratiquement fonctionner que pour des entrées de longueur inférieure ou égale à 3. Pour les entrées plus longues, il est permis de terminer tôt avec ou sans sortie.

Exemple

Malheureusement, il est un peu difficile de créer des exemples pour cette question en raison de sa nature aléatoire et de ses résultats importants.

Cependant, je peux fournir un seul exemple pour l'entrée hi, sur Hastebin.

Notation

Puisqu'il s'agit de code-golf , la soumission contenant le moins d'octets gagne.

code-golf  string  random 
— LyricLy
source
— LyricLy
11
Avons-nous besoin de suivre la procédure décrite pour produire le résultat? Si oui, c'est une exigence non observable , ce qui pose problème. Sinon, nous pouvons générer Bdirectement en ajoutant un nombre non négatif nde caractères aléatoires à A. Le seul problème réel est alors de connaître la distribution de n(je parie sur une distribution géométrique)
— Luis Mendo
1
@seshoumara Vous ne pouvez pas.
— LyricLy
7
@ LuisMendo, j'ai réfléchi à ces questions et il n'est en fait pas facile de générer directement le préfixe. Il ne peut pas contenir la chaîne cible, y compris en dépassant la limite où il rencontre la chaîne ajoutée. Et la distribution des longueurs de préfixes dépend non seulement de la longueur de la chaîne cible, mais également de sa structure.
— xnor
10
Certains des programmes informatiques ci-dessous, tels que .W!}zH+ZOrd\k, ressemblent beaucoup à ce qu'un singe a tapé.
— Jeppe Stig Nielsen

Réponses:

12

C, 192 bytes

i;g(s,b,i,n,j)char*s,*b;{for(b[i+1]=0;b[n+j];++n)s[n]-b[n+j]&&(n=-1,++j);return n;}f(char*s){char*b=calloc(strlen(s),1);for(i=0;s[i];)i=(b[i]=putchar(rand()%95+32))-s[i]?i?g(s,b,i,0,0):0:i+1;}

Try it online!

It's a mess now, but at least it works even for the corner cases...

C,  63   62  61 bytes

Thanks to @Jonathan Frech for saving a byte!

i;f(char*s){for(i=0;s[i=putchar(rand()%95+32)-s[i]?0:i+1];);}

Try it online!

— Steadybox
source
I have absolutely no idea why this halts when it hits s, +1
— ATaco
1
@ATaco It halts when i grows large enough that s[i] refers to the null terminator of the string (character 0).
— Steadybox
Oh, so instead of throwing random characters at it until s is accidentally created, it throws random characters at it until it reaches s. Smart.
— ATaco
As much as I like this answer, I believe it breaks for an input such as "ab" when the rand monkeys type "aab".
— zennehoy
I guess you need something like KMP so that this approach can be valid. Assume the input string is ababc and the monkey generate !!abababc will your program halt?
— user202729
9

Python, 79 bytes

f=lambda x,s='':x in s and s or f(x,s+chr(randint(32,126)))
from random import*

Try it online!

This is theoretically sound, but will crash early due to python's recursion limits (you can set them further to get longer results)

Python, 84 bytes

from random import*
x,s=input(),''
while x not in s:s+=chr(randint(32,126))
print(s)

Try it online!

This one is ought to work for relatively longer strings, since it doesn't rely on recursion, at the cost of 5 bytes.

— Uriel
source
You can save three bytes by using backticks to do the string conversion (shown here as single quotes to the the markdown right) s+'randint(32,126)'
— wnnmaw
1
@wnnmaw backticked randint(32,126) would produce a string of the number, not the ascii char mapping
— Uriel
8

Ohm v2, 10 bytes

Ý£D³ε‽α@§↔

Try it online!

Explanation:

Ý£D³ε‽α@§↔  Main wire, arguments: a (string)

Ý           Push empty string to top of stack
 £          Start infinite loop
  D³ε‽        If a is a substring of the ToS, break out of the loop
      α@§     If not, select a random printable ASCII character...
         ↔    ...and concatenate it with the ToS
— Nick Clifford
source
8

GNU sed + coreutils, 75 + 1(r flag) = 76 bytes

h
:
s:.*:shuf -i32-126|dc -e?P:e
H;g
s:\n::2g
/^(.+)\n(.*)\1/{s::\2\1:;q}
b

Try it online! (It takes a lot of runs to get an answer for a length 2 input, because most of the time you run out of allowed TIO computation time.)

Explanation:

h                                # copy input string 'A' to hold space
:                                # start loop
    s:.*:shuf -i32-126|dc -e?P:e # run shell script: shuf outputs a rnd permutation
                                 #of the set of numbers from 32 to 126, and '?P' in
                                 #dc converts the 1st read decimal to an ASCII char
    H;g                          # append char to hold space ('A\n.'), then copy
                                 #result back to pattern space
    s:\n::2g                     # remove all '\n's from pattern space, but first
    /^(.+)\n(.*)\1/{             # if pattern is 'A\n.*A' (A substring of B), then
        s::\2\1:;q               # search previous regex used and leave only '.*A',
                                 #then quit (implicit printing before exit)
    }
b                                # repeat loop

Benchmark: approximate, for scaling purposes only

  • input length: 1, 10 random inputs (runs), average time: < 1 s
  • input length: 2, 10 random inputs (runs), average time: 90 s
  • input length: 3, 10 random inputs (runs), average time: lots of hours!
— seshoumara
source
7

Funky, 64 bytes

s=>{S=""whileS::sub((#S)-#s)!=s S+=S.char(math.random(32,126))S}

This uses a few tricks I've been wanting to use in Funky, like a variable name after a keyword as in whileS, and using the fact that strings implicitly parent to the string library.

Ungolfed

function monkey(target){
    monkeyCode = ""
    while (monkeyCode::sub((#monkeyCode)-#target)!=target){
        monkeyCode += string.char(math.random(32,126))
    }
    monkeyCode
}

Try it online!

— ATaco
source
6
So would that be... Funky monkeys?
— Sebastian Lenartowicz
7

Haskell, 100 bytes

import System.Random
s#(a:b)|and$zipWith(==)s$a:b=s|1>0=a:s#b
m a=(a#).randomRs(' ','~')<$>newStdGen

Try it online!

Basic idea is to generate an infinite list of characters with randomRs and stop it once we find the string.

— user1472751
source
A shame isPrefixOf isn't in the standard Prelude…
— Bergi
7

C# (.NET Core), 86 bytes

a=>{var b="";for(var r=new Random();!b.Contains(a);b+=(char)r.Next(32,127));return b;}

I don't really like how much creating the Random instance takes, but I don't think there's a way around it.

Try it online!

— Wakawakamush
source
3
Welcome to PPCG! Currently your solution does not properly generate a random character since according the the docs, the upper bound passed to Random.Next(Int32,Int32) is exclusive and so not one of the numbers generated. This can be fixed by replacing 126 by 127.
— 0 '
@0 ' Whoops, I thought about it while writing, but I forgot to check it before posting. Thanks!
— Wakawakamush
There is actually a way around that long Random, you can remove the variable declaration! 79 bytes
— FlipTack
@FlipTack Interesting, I tried that in C# Interactive and it didn't work because it just kept generating the same number. Weird to see that it does work in TIO.
— Wakawakamush
6

R, 79 76 75 bytes

-3 bytes thanks to MickyT for changing the random sampler

-1 byte thanks to Robin Ryder for tweaking the random sampler again

function(S){G=""
while(!grepl(S,G))G=paste0(G,intToUtf8(32+95*runif(1)))
G}

Try it online!

— Giuseppe
source
hi, your sample could be replaced with intToUtf8(runif(1,32,127))
— MickyT
@MickyT excellent, thank you!
— Giuseppe
You can save 1 byte with 32+95*runif(1) as your random sampler.
— Robin Ryder
6

Charcoal, 15 14 12 bytes

Ｗ¬№ωθ≔⁺ω‽γωω

Try it online! Link is to verbose version of code. Edit: Saved 2 bytes due to a subsequent bug fix in Charcoal. Explanation:

    θ           Input string
   ω            Predefined variable `w`
  №             Count number of occurrences
 ¬              Logical not
Ｗ               Loop while true
       ω        Predefined variable `w`
      ⁺         Concatenated with
         γ      Predefined printable characters
        ‽       Random element
     ≔    ω     Assign to predefined variable `w`
           ω    Predefined variable `w`
                Implicitly print
— Neil
source
4

Pyth - 14 bytes

.W!}zH+ZOrd\k

Try it online here.

— Maltysen
source
W!}Qk=+kpOrd\ is 14 bytes as well, SE is messing with formatting because of unprintable but range is generated the same way
— Dave
4

Mathematica, 65 bytes

""//.x_/;x~StringFreeQ~#:>x<>RandomChoice@CharacterRange[32,126]&

Try it online!

-3 bytes from Jonathan Frech

— J42161217
source
I think FromCharacterCode[RandomInteger@94+32] is equivalent to the shorter RandomChoice@CharacterRange[32,126].
— Jonathan Frech
@JonathanFrech yes,it is!
— J42161217
4

Octave, 62 bytes

t=input(o="");while(~nnz(regexp(o,t)))o=[o,randi(95)+31];end;o

Try it online!

Explanation:

t=input(o="");               % get stdin and define output
while(~nnz(regexp(o,t)))     % while no matches
    o=[o,randi(95)+31];      % concatenate string with ascii char
end;                            
o                            % output result

Many thanks to Luis Mendo for the edits!

— Alan
source
1
Welcome to the site! :)
— DJMcMayhem
Can't you replace isvector by nnz? And strfind by regexp. Also, you can use randi(95)+31, or maybe replace the whole sprintf statement by o=[o,randi(95)+31]; (implicit conversion to char)
— Luis Mendo
Also, we usually require a function or a program that takes its input (as opposed to defining a variable containing the input) -- something like this
— Luis Mendo
I️ attempted to do that, but I️ couldn’t think of a concise way so I️ skipped it. Nice revisions!
— Alan
1
Feel free to incorporate those suggestions into your answer. That's standard on this site
— Luis Mendo
3

Alice, 21 bytes

/U!?"$~dr@
\idwz K"o/

Try it online!

Explanation

/...@
\.../

This is framework for mostly linear programs that operate entirely in Ordinal (string-processing) mode. The IP bounces diagonally up and down through the program twice, which means that the actual code is a bit weirdly interleaved. The commands in the order they're actually executed are:

i!w" ~"rUd?z$Kdo

Let's go through this:

i       Read all input.
!       Store the input on the tape for later.
w       Push the current IP address onto the return address stack.
        This marks the beginning of the main loop.

  " ~"    Push this string.
  r       Range expansion. Turns the string into " !...}~", i.e. a string
          with all printable ASCII characters.
  U       Random choice. Picks a uniformly random character from this
          string. This will remain on the stack throughout the rest of
          the program and will form part of the resulting string.
  d       Join stack. This takes all strings on the stack and joins them
          into a single string and pushes that (it does this without actually
          removing any elements from the stack).
  ?       Retrieve the input from the tape.
  z       Drop. If the joined string contains the input, everything up to
          and including the input will be discarded. Otherwise, nothing
          happens to the joined string. This means that the result will be
          an empty string iff the joined string ends with the input.
$K      If the top of the stack is not empty, jump back to the w to continue
        with another iteration of the main loop.
d       Join the stack into a single string once more.
o       Print it.
— Martin Ender
source
3

Perl 6, 39 bytes

{("",*~(" ".."~").pick...*~~/$_/)[*-1]}

Try it online!

(...)[*-1] returns the last element of the sequence defined by ..., of which:

  • "" is the first element;

  • * ~ (" " .. "~").pick generates the next element by appending a random character in the appropriate range to the previous element; and

  • * ~~ /$_/ is the ending condition, which is that the current element matches the main function's input argument $_ as a literal substring.

— Sean
source
You can lose the *~~ for -3 tio.run/##K0gtyjH7n1upoJamYPu/…
— Phil H
3

Java 8, 81 79 78 bytes

a->{String b="";for(;!b.contains(a);b+=(char)(32+Math.random()*95));return b;}

-1 byte thanks to @OlivierGrégoire for pointing me to a (big >.<) mistake I've made..

Explanation:

Try it here.

a->{                    // Method with String as both parameter and return-type
  String b="";          //  Result-String, starting empty
  for(;!b.contains(a);  //  Loop as long as the result does not contain the input
    b+=(char)(32+Math.random()*95)
                        //   Append a random character to `b`
  );                    //  End of loop
  return b;             //  Return the result-String
}                       // End of method
— Kevin Cruijssen
source
1
It should be 32+Math.random()*95. There... bug fixed and a byte saved! ;-)
— Olivier Grégoire
@OlivierGrégoire Woops.. Looked at the hexadecimal code for the space, but regular decimal for tilde.. >.> Thanks for noticing. Not sure how I've missed that, since the output clearly had 'unprintable' symbols..
— Kevin Cruijssen
3

05AB1E, 10 9 bytes (-1 @ Emigna)

[žQΩJD¹å#

Try it online!

Do the monkey with me.

[              | Loop forever.
 žQ            | Push 0x20-0x7E as a single string.
   .R          | Pick from it randomly.
     J         | Join stack (B) with new char.
      D        | Duplicate (B).
       ¹å      | Push input (A) and check if input (A) is in (B).
         #     | If the previous statement is true, break loop.
— Magic Octopus Urn
source
1
You can do Ω instead of .R.
— Emigna
2
Lol, using an Ohm, to beat Ohm v2. How nice.
— Magic Octopus Urn
2

QBIC, 33 bytes

≈instr(Z,;)<1|Z=Z+chr$(_r32,126|)

Explanation

≈instr( , )<1|   WHILE InStr() can't find
         ;        the input (cmd line string argument) as part of
       Z          the output (Z$, which is printed automatically on exit)
Z=Z+             add to the output
chr$(         )  an ASCII character
     _r32,126|   with a random codepoint between 32 and 126 (incl)

Sample run:

Command line: hi

`;7L3f$qy )H99tZ@>(-Z1efL|Q-5'BE=P8BdX?Lem/fp|G#~WY[ Q4s9r~Af*T})P4`4d$#ud3AiuTwQPFS@8c7_59C$ GlJ%iJ[FA(rNt<y>Hl=r,wSbBB%q!8&#*CixWbnwE."wrZ:R53iKJkN*@hpQt aMj6Mw<KfT@hkik>_k'_>$~3)jl|J!S`n\Yw|lXi:WAKWp?K"F.cAGI/:~uR8*[;Die{]B*]@;Vhjv,$9]Ys:AIdy!j{aXlr:0=txCP-n{/3lgq,;jXaP\]u}.Zl/7eKF+N54[J9^r:>%/.e~*9aK%de>^TW%p%(_uJPvuV%d<&]zu`t;vkEPC>7pofok0Kj}j?9G{TUxSccth)[)J>@'E)NMzA(i!UV7%;$.Z#j3q@#9Q%}<" &VsbL*<SrG-$NC MAQ\`iIT+.P|5}nv )FZl5_.Kc*AUV9u&fvk.USt3Y!s7^UEL{|D$k#k8.9Fgqn#3dgr(0G.gw1#j$HfU3a|)3-O(d<)<A|`%pJ^/\{[w[H4N/>8J@z/YNsU,zY4o*X+h\Dy!~)Xr8.3"%.39v0d5_-8QBYR".Z]MZ}N>9e?f@hj#hor1IhJ[krrHOamRPxQC>^'dOh,cF_e2;8R@K**Jsx_~t9r~4J$Y;kPsb)8w~:o-`@MwP]OA{8yD%gL(=&M>$nTKw] R[}rS|~}&*gD 'g|gRSDLld+`,,}(1=]ao3Z%2*?wlqU$7=$8q$Fig:7n&+XKQ LV/Aq?BYl_*Ak-PqI$U_>`/`-yD5.3|Zg>,mC"RC`IV^szu:I;0ntn(l'/ZnK}T&i)9!zkd?7Ec/X+IN=-5wwsSV@^<?:K=9m%*@C;zDjc%:d>/E@f7@0NVt4Vz/E;8*0A25F1:JUQ/G#2)un9hQI>2^}&+cY+JP*-#$p+cFs}R|>E;w#q>pN"Jkv<>E_ZtCvV05Lh;0 9bCBXgA7aIe+9B1<G)YCH\Yqn.0%g$_4Q4 xIR)gt]W*a|gGX}hP4b)6#M:Dh!kmuX;nW]'n]Mm5y=ET|O9p\,j>Bc|7J5I%UCZla-2g(Mm9cE"}c1Q0@yTF|A\FJbR7_(F_G#@mE/~ [9T~|Ty9~0=g {a^IM{]C9%2FBJ:b&i5A{rqu/xw6q|_[$Sj&W\TnI}/>.EJ1dSbSOtr_Qtuf!IF .WU}&M51+VAnJ)W}^c5qwQ<G05}/aZ99l6iqyD|Zr8SV9L}8FbUz7_H<]A|.vFQXSu2@67%83HP4]Gw0PuPNQ6SOM_[BcdK-s}Z(~~i:^N$GZn_sMcp*i,w-2VfK*LA$Btmg6QEohqym3[RRqUAM"9pE>N)(.TNMQ"U~ e2($wz(Kdh;0ol8>SXHEnLvrs.Xtl^L?SL1$*ssD _={{}(`qKCy{]W:AZT}Zro5qN:;mNp#EPfg?_],,cFP?EhGs/OAt}fgVSR<JW^HkWf'@^Vd9\_Y?P*>C'YP jqvXu)ZFwzY)/MLHcRL/P?jBi9"d\  E$ngpq-i*;EW6R)J|[0FfZSTozuSq,sAJT<<4al<zM\F(|gTD0/Vt6JL.p_x_oC)V'zWZ`8eA9@*WgZ>',-}Q^5#e552&"\i1HI]{)]WcI.cY0n9J<jaT.!l{r4Dz~nt`AEP-6,FHhf6(PSywIedr }=9V>Q7!+~L^O3'Crdv"hfv#xrs@](EO;&#)0]oC][z*Eh`k!$V!r6~#-Vfk1p#w&Za6Ij\@V<TNf4njdynOch7l?XwU  `SON\iizU3%S^X2XKY.w%:zAVY^KlIhZ8]d39No3P89v`1FxKTLQa+7"rd9bm2)a^Pu=~.9VDh?v"%$9evl9+l7n$I?qA[b:kH2?5Tg&(!H(*}hZ3z@bg+Zj!# JnO2FV_glCMweT;b|>g4!h{4@ p w`lH \Y8(uPf7nbJY]r>('-$O?5Xd:h&:y!i%2dRC_8=3! |b="H|jxx)k!\D|]Lsdz1.v[a<l/Y3?0/&H%2.qvPp'ZNpO;rhvtnl0*Bs4Qlh0}_dv6g0`pJh'==]9LuzG+qUGz5.j[$I{4.b_o;S`QFucC9>`z7M.wHx!wy-JeOf)^9#Z.xl7e"9q)OE-SSD"VbLFm-u'-<4~(_h\KqNk7}vKh0E&!LaxAma|FOIY,\C$;!v^#4,eqwpE]Jqp,]IkTz,,L`kx|\i^#{PV0/8K?N,W!L=vbh\OJ7?:k=~{zLw8*/W<-qFDKAhx1F;\NL@{=rlo0''YB;B5<:0e5J)w"0l@FE?J:FW(I|)3BZ'hL7[}Ez=jc(rLkY9d{Zzgq7Cj)bej/X!@TP7x.r"Arz7IU;1|.3by~\h{V57}A^w7v5gMC]@B~1i5*uY 7?(IN6hQ/b/4bMpDmT_"n|;bBx|74(ReQ.^])bHC+:!s bk_S]R}<Ow:7CCu_P!$:mz{[aiGg*AD#}m~D_rhVr6!x]PY5n'qiMMlpqoU>C`,W}y9Yi4hHf<lcwvga`h(<=jURq\d+2SRGA1GP**D]i+Tp@*hpe([-JE^M@5jHgK~>hY>Bo&% \e/\&]"K])CV%oNJ}[_Q1}w(p3uJ+\/;{0TB8#{=&l8s;]L7Gr;a_[cN:#%$)?*:HLZ;&n|2_8/@=B [>|R3@xk<c+bIyb>h`]:c]RLt(M!69PNE?}>@CHT>jNevl81PCgHu0ap0~Pcq?Z@>+yTFw\E=10&fpS+=/l|.ioPn$B.M\4{2?q-^,)f&S4X44(Rycome[Ot[t(8CAphj[h}E/A~BR[6Y&Hm1_tsxs4BB0cwo\",r_c~s/vT}kKu3U(Emb|%"`OAmV7$,\<O7)c&F==mc~dM:qX^[K-9<3u8hfgTUP19aXk,7g(4>jLt,*N!EXGE#XzN}>7@EH4n}k!&a[j,Ynd#!M<suhnBP /J9}h>vRyXuADk"+v}?hOm6*U^x\G'!Y(TDC?EE|r}5yx4PB 1;9q.%/kg7p2ImS62+/|K,xSDf3b6JRY+8~mxikSU^&3A3|/j9}fIESN45kdi*h64!XE'_0?Pw{MeK$DeXP]5M{7sLD!dj5HrAc\N I`~o/%MqyIIsFee]A?j7ZZ}f'dN#"V''g-G0@zNajp=v<"r2s;>@.UM9L\Mq16e/961d.3a6h.hMrUREa^wR^s*\Tc6Yn]DT.Nc77p|h s2@hC\Rxy|XhXi.OL2$\pwPSJET;u7V`U!<]M(tibt>.gD'JKe{7r8?Z[]ExGHxyd\,/wjfBI'NKCpaU19-?f;;QVrWnFF,zvJY|d\DrcysAO'; 33CSNy_GlLr\v)Ir\qQfwT'I#t:N-{,x#zGR.)gJqq%!zF.oJ;]*TI+4z>JQAGQM3w&zgani8JwZW6S!r-ig\3jD}]2*.Aen8O)L?X.UTZ6)mOtUIm_=3fA'_::vV_#+c+=evf#{8lk+`)M\_c+I<|*LRZOU]:eQ*/KER#f,vEC?4yXE*8wlzk?b)&[gF'(0|$@+4CT4#lfEKxP~;oo%1+=yw#J*s}D7p1fU~^gD1Ib{H|PWer^q"q=?Pxf<)tvu7{HDvl\kqb"b/|s>|h.qQu[$F/:~*dy9cl16}dKXY~N7aptCSv+da/S5-,qnwBhRi+lh8=Qy{er:#Oos|e?(US>"767KVe^nz<$]gM)~J>{I7n~!k[U\1{8Z8u6T(ft?kgdQO,LvY/TDAe\wS~Y U>\.aQYhQ;h1nuW$R/wpz_EiB-%gf87qgQei(P-ft:TSW,HtsPez"5<8?yR`wm7pjMn^|8Y.4[RVWq#aW$0EB9"O:%@q[&F[_'2yt2k]S5~HCN1+^bS>N2C/7ChHCHNrJ8,kBbNsu}37LH^n.B+tyE*s7l(Tc<;4.tvBw3LP=nK4G'6M(z086;"??9XE.(X>nvmm()P2m\"LeqbcOC~Vw;u/Q^T)52/pM3+<GkFWJ?30{/n2FQq QjO#pt8oN$kK/a+|Hbw@5m8M[EFWq>%cV2[X@q}gJ"9kt9'~]4p+2~LT9|4Ss^qoXw%P#M!!]TBQbp;PYg{WCj,RF<#bNJTS,CUH{][hY"q;[?#apc%Cl&QWVi]ffYG}bzx .;*/rqRhb[XatPu.Piws<mo=]!e>p%yu\;fCzJ0Xz]6]9S:WRlYS,mC&7Zjb)+DjJUkSF3TJG;8fQ4|>t%qgW1$_V:p;|Q":Z!UngSL{*Ky+/zh [I{_3?]h^x37"`^'/U>EPqal'&Txf,I>gr2HO_y!QM-ch~%m-(AE6U~[A"D@j1hu?6p2"Wc'3nyqfs!.UQy}I%0(z5dPmORFBK1,<PfYersnLe<fLhB=^g pwXnWDOQNuIPEpnm8.Q6jN|a7tcuSH$9T;! d~VQn{'(-4{caLa;t?~|>q:on:Bgs'FQ'2-<%W<3Px=4Uf;@;R3yZECK?f(5K?`^lQY\ok},`Q9*Q}3].Y!BkRt"3@]Lz&ec'NB?n[%0kQ9/55BOZ)o!P>fpXZI:#?Ly#\o.`+HX Kb0@P^1qS%bGip1`)qH@-k\oOGs%;}_Nq{uPq |!K)01w(?X=>bSR[(]ZQ<Z1]bD9M.F<<.8EB6JlEUOk6r;SrVZNT2 m>zp3]a_sK eq8]rK^.U&!d62HBt`v?t6uc#3s<{[CmYE24+ujEx`$##R2g\b!PvK<8+lUhyT|p"SUco/aUh.fXBV(!!)8PfQIr6R,r8c-F-mjua;:z4!Q1pA!H0E%)"K2oUv|DV+lg,h8W?<JnIkiV/us::*l&I<OI6NO)Tnq0-uDRG5*LX#wU{8WpMlw3Z'7zGi*loo2{=hWSY*0/o9BwtJ$Hw}l94nA^9>48(3gDnt8CS|R3? OH[N/9j1r%vUcox\68{yFemlmmtp*q5kfrlIo`3yQB??6jW:TW+)':K#]^=ilF_/N!)=}y@k.y//nhChX!3b`=t,1_KhR,n]/_.-P>B80W#'E%J[g?ti)*;Yl]}r0>qh/X[{=)Gr '[+pz|DI=mA8zj~yAT*^7w%tV0l=V^/#2W>)f)X%f^L&+Un}VlQt3.%gEKbE!7`adTb#`}i!F$-Gug]@*G,hKe;/p,`Mb@wBJ4<V&jJd&_H4VR{Hc"{2<l<QapiLw(JK-2-[1_.WR.@CG$?\1<&( QX5c9 :z^jDW09(=iH V/vkcJ8D<uLAr$dbc$Hl'2KTUlbrd8kD{B0Eeu<&oL2s.S4@Jo$zVq~BqeLsb;k-NG/'iU|)W_:X-.XUc<v\elx57ZZ"R!y_yzve*Wlt>.fE,#Eh:(#gn1kSQ+/3NYjD']I;"+@pnW[1EA.AyqM4,0,dJt.?r2oab.j\k@)BsZ|s39FdL87xyuJ0nXX=yz^~W,}acDZp8ukCpv^<^{CkRS<=GsS$}#fbP5%A$GHdg)+WZLLN9[ue073Q!F"J;X^49*$R'W%C.r~Fj&B`)tq[01a4En%H,kvyZG|,)%$44rJg[tq<wG9FjN<m@larki#;Bns%D}v_efPRH(OeRq0{=>Uc[~xcTcV_9|k54Q2*N.3]LlmFasY3"p =$$onbg$M+ReRsnH|9gV~#2?c1-V$35."DZH-O$~,c.gs]%,]p4\OFIW%l:,E,YT8FCeU8hy#lNq1lCpS 0I&q_*q>|=,(-dHuzi~6$GW22*A\w*&R< W`$HPRr,2A}3w\"Y?d%{2^xP:GqI\26A|.e'H2Z[M4=P.H87O~{)9|B*tHAC\j^S,StW!*snsz82R!:eD@uB4x+x&zSIN(3V|.^N_$=i=p}iK4h'v"$:I<t e:Y/XrSOF83=lkVNa0^k@jB@{ARE@r=Bja`(Bw>@?+`Wo,= u5HhXPeRMXS4@H#$-Jwg2"2-]%7p.o2Ar9J6Y1Ra?"3<oee&bpO^O{nw9=%\0brVNXrelWGoJyb/5W%MB0UBaPsc*29K<N~``NriWM$"eY0@xh^<$b:E/J~S%{#ry~6d?4Vv@^&9'=iBA#2U]bj9>UoJ#wQDN~6cB&/_Pu-XF?_hu3><(M7RW\%Ly@rTC9^b`?kVL~w%[{!&{#aS7<mc@J>ZaN7s}Y.c0:Y.\d&_[L{m|>|>%J^@!i9y0_lwejChi
— steenbergh
source
2

PHP, 55+1 bytes

while(!strpos(_.$argn,_.$s.=chr(rand(32,126))));echo$s;

Run as pipe with -nR. Not suitable for TIO cause of probable timeout.

Insert a space between the quotation marks for PHP older than 7.1.

This 51+1 bytes version will fail if input is 0:

while(!strstr($argn,$s.=chr(rand(32,126))));echo$s;
— Titus
source
2

Javascript 74 bytes

s=(a,b='')=>~b.search(a)?b:s(a,b+String.fromCharCode(32+Math.random()*95))

call like this:

s('hi')
— RuteNL
source
@Giuseppe thx, I have it added in the byte count
— RuteNL
1
I think you have to change 94 to 95 for the code to be valid
— Hawkings
1
@Hawkings Yea, you're right, fromCharCode ignores decimals it seems. Thanks for pointing it out!
— RuteNL
Save a byte with ~b.search instead of b.includes.
— Shaggy
@Shaggy Nice! Didn't know about search
— RuteNL
2

Pushy, 20 18 bytes

LFZ^tCN[,` ~`U'x?i

Try it online!

The program keeps a stack len(input) characters long, and constantly removes the first and appends a new random char, until the initial input string is reached. Each character is printed as it is added, creating the desired effect.

Explanation:

                      \ == SETUP ==
 F                    \ Put input on second stack
L Z^tC                \ On the main stack, make length(input) copies of 0
      N               \ Remove printing delimiter (newline by default)

                      \ == MAIN LOOP ==

       [              \ Infinitely:
        ,             \    Pop the first item on stack
         ` ~`U        \    Add a new random character (between 32 and 126)
              '       \    Print this new character
               x?     \    If the stacks are now equal:
                 i    \        Exit program
— FlipTack
source
2

Brachylog, 17 bytes

I⁰∧Ẹ{sI⁰&|;Ṭṛᵗc↰}

Try it online!

I⁰                   The global variable I⁰
                     is the input,
  ∧                  and
   Ẹ                 starting with the empty string
    {          ↰}    call this sub-predicate again
            ṛ        with a random
           Ṭ         printable ASCII character
          ;  ᵗc      appended to the string we're building
         |           unless
      I⁰             I⁰ (which is the input)
     s               is a substring of the string we've been building
        &            in which case the string is output.

Can randomly stack overflow. This makes use of two recently added features in Brachylog: global variables, and the apply-to-tail metapredicate ᵗ.

— Unrelated String
source
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