Eww, ces octets sont bruts

86

Ecrivez un programme ou une fonction qui, lorsqu'une chaîne est donnée, filtre autant d'octets distincts que possible et renvoie la chaîne nettoyée. Cependant, puisque votre programme les déteste, aucun de ces octets ne peut être présent dans votre propre code.

Votre score sera le nombre d'octets distincts que votre programme filtre de l'entrée, le plus haut étant le meilleur. C'est un score maximum de 255 (puisque votre programme doit être au minimum d'un octet). La condition de départage correspond à la longueur de votre code, avec une valeur inférieure meilleure.

Par exemple, si votre programme filtre les octets 0123456789, il reçoit un score de 10, mais votre programme lui-même ne peut pas contenir ces octets.

Règles

  • Octets signifie octets.
  • Vous avez également la possibilité de prendre en entrée une liste d’entiers, avec des valeurs allant de 0 à 255. Celles-ci correspondent aux octets équivalents.
    • Votre sortie devrait être sous la même forme que votre entrée
  • Pas de lecture de votre code source
  • Votre code doit être non vide
  • Oui, je sais qu'il va y avoir une réponse Lenguage / Unary. Mais au moins golfez-vous s'il vous plaît?;)

Modifier la règle:

  • Vous pouvez choisir d'ignorer un octet, par exemple s'il est impossible de le distinguer de l'octet de fin pour une chaîne ou pour EOF. Cependant, cela signifie que vous ne pouvez pas l'utiliser dans votre soumission, et cela ne compte pas pour votre score. Si vous choisissez de le faire, votre score maximum sera de 254, mais vous n'êtes pas obligé de gérer cet octet dans votre entrée.
code-challenge restricted-source self-referential Jo King
la source
1
Est-ce correct si la fonction ne fonctionne pas si la chaîne contient des octets nuls (\ 0)? Comme C utilise des chaînes terminées par un caractère null, une fonction ne peut pas savoir s'il s'agit de la fin de la chaîne ou simplement d'un octet nul.
wastl
4
"Octets signifie octets." Voulez-vous dire des octets? Octal est une façon d'écrire des nombres, trois bits à la fois, de sorte qu'il ne s'aligne pas toujours avec les limites des octets.
Kevin
1
semble être en train de mendier pour une réponse dans un brainfuck
Jasen
1
@cmaster Pour les personnes qui ne peuvent pas voir les publications supprimées: Il existe plusieurs tentatives de brainfuck, qui échouent car elles ne peuvent pas distinguer un octet NUL de EOF.
Ørjan Johansen
3
@ ØrjanJohansen J'ai ajouté une règle qui peut vous permettre d'ignorer un octet (à des fins EOF). Cela ne devrait pas être trop abusif, car cela diminue également votre score maximum et rend la question plus inclusive.
Jo King

Réponses:

66

Japt -P , score: 255 (2 octets)

ff

f => filter, second f => everything but 'f'

Essayez-le en ligne!

Japt , score: 255 (2 octets)

DeShaggy

oo

o => remove everything but, second o => 'o'

Essayez-le en ligne!

Luis Felipe De Jesus Munoz
la source
9
Le bon langage pour le travail :)
Kevin Cruijssen
4
oofonctionnerait également, sans le -Pdrapeau.
Shaggy
1
Les gars, nous avons un gagnant!
Adám
C’est maintenant la solution Japt la plus votée sur le site.
Shaggy
34

Python 2 , score 249, 1732 octets

c="%c%%c%%%%c%%%%%%%%c"
ex=c==c
ee=c==""
exec"e=%x%%x"%ex%ex
exec"e%%%%%%%%=%x%%x%%%%x"%ee%ex%ee
exec"ee=%x%%x"%e%ee
exec"ee%%%%=%x%%x"%ex%e
exec"e=%x%%x"%ee%e
exec"ec=ex%cex"%e
exec"xe=ec%cec"%e
exec"xx=xe%cxe"%e
exec"xc=xx%cxx"%e
exec"ce=xc%cxc"%e
exec"cc=ce%cce"%e
exec"eee=xx%cec"%e
exec"cee=ce%cec"%e
exec"exe=ce%cex%%cxe"%e%e
exec"xxe=ce%cxx"%e
exec"cxe=ce%cex%%cxx"%e%e
exec"ece=ce%cec%%cxe%%%%cxx"%e%e%e
exec"xce=ce%cex%%cxe%%%%cxx%%%%%%%%cxc"%e%e%e%e
exec"cce=cc%cex%%cce"%e%e
exec"eex=cc%cex%%cec%%%%cce"%e%e%e
exec"xex=cc%cex%%cxe%%%%cce"%e%e%e
exec"cex=cc%cec%%cxe%%%%cce"%e%e%e
exec"exx=cc%cxx%%cce"%e%e
exec"xxx=cc%cex%%cxx%%%%cce"%e%e%e
exec"cxx=cc%cec%%cxx%%%%cce"%e%e%e
exec"ecx=cc%cxe%%cxx%%%%cce"%e%e%e
exec"xcx=cc%cec%%cxe%%%%cxx%%%%%%%%cce"%e%e%e%e
exec"ccx=cc%cex%%cec%%%%cxe%%%%%%%%cxx%%%%%%%%%%%%%%%%cce"%e%e%e%e%e
exec"eec=cc%cxc%%cce"%e%e
exec"xec=cc%cec%%cxc%%%%cce"%e%e%e
exec"cec=cc%cex%%cec%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"exc=cc%cxe%%cxc%%%%cce"%e%e%e
exec"xxc=cc%cex%%cxe%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"cxc=cc%cec%%cxe%%%%cxc%%%%%%%%cce"%e%e%e%e
exec"ecc=cc%cxx%%cxc%%%%cce"%e%e%e
exec"""x=""
"""
exec"x%c=c%%xex%%ecc%%xex%%eex"%e
exec"x%c=c%%cee%%cee%%ece%%cxx"%e
exec"x%c=c%%ccx%%xxx%%xcx%%xxe"%e
exec"x%c=c%%eex%%exx%%xec%%xxe"%e
exec"x%c=c%%xex%%cxc%%cce%%ecx"%e
exec"x%c=c%%xxe%%eex%%cxe%%cxe"%e
exec"x%c=c%%cex%%ccx%%xec%%ce"%e
exec"x%c=c%%eex%%ce%%xxx%%xcx"%e
exec"x%c=c%%ce%%eex%%ecc%%ece"%e
exec"x%c=c%%cec%%eec%%ecx%%xxx"%e
exec"x%c=c%%exc%%xxe%%cxe%%cxe"%e
cx="""eec
xec
xxx
xcx
exc
cee
cee
ece
cxx
ccx
xxx
xcx
xxe
xex
ce
cex
ccx
xec
ce
xex
ce
xxx
xcx
ce
xxx
xcx
eec
xxc
exc
xxe
cxe
xxx
cex
ce
xex
ce
xxx
xcx
cee
cee
cee
eee
xex
ecc
xex
eex
cee
xce
exe
cee
cee
cee
cxe"""
exec"exec%cx"%ce

Essayez-le en ligne!

Supprime tout sauf les exc%="nouvelles lignes. La génération de code est celle de xsot . Je doute que cela soit optimisé, il y a donc beaucoup d'octets à sauvegarder ici!

Lynn
la source
26

Haskell, score 249 250, 1088 335 octets

(!:)((=:):(!=))=(((!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(!):(!):(!):(!):(:):(:):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(!):(:):(!):(!):(:):(=!))!!(=:))(=:)((!:)(!=))
(!:)(=:)=(=:)
(=:)!(!=)=(!=)
(=!)=(!):(=!)

Définit une fonction !:qui prend une liste d’entiers et filtre tout sauf 10,33,40,41,58,61les codes ASCII de \n ! ( ) : =.

Essayez-le en ligne!

Comment ça marche: Commençons par renommer les fonctions et les paramètres en noms sains:

f (x:xs) = (listOfFunctions !! x) x (f xs)                   -- (!:)
f x = x

listOfFunctions = d:d:d:d:d:d:d:d:d:k:d:d ... : infiniteDs   -- inlined
infiniteDs = d : infiniteDs                                  -- (=!)

k      = (:)                                                 -- inlined
d x xs = xs                                                  -- (!)

L'idée principale est d'avoir une liste de deux fonctions différentes ( listOfFunctions), une pour conserver un élément ( k) et une pour le supprimer ( d). La liste est indexée par l'élément en cours à partir de la liste d'entrée ( x) et la fonction choisie détermine si l'élément doit être conservé ou supprimé. À l'index 0, nous avons d, parce que nous voulons supprimer des 0, le premier kest à l'index 10, car nous voulons conserver 10 (nouvelles lignes). Les deux ket dprennent deux paramètres: xet le résultat d'un appel récursif avec le reste de la liste ( xs). kajoute xà l'appel récursif et drenvoie simplement l'appel récursif.

La liste littérale des fonctions se termine à l'index 61 (car =, le dernier kde la liste) et est suivie d'un nombre infini de d. Cela enregistre non seulement de nombreux fichiers explicites d, mais nous évite également de devoir terminer une liste finie []nécessitant des caractères supplémentaires.

Edit: merci à @ Ørjan Johansen pour +1 score et -753 octets.

nimi
la source
1
En plaçant des fonctions dans la liste au lieu de Bools, vous pouvez vous en débarrasser |pour un score de 250, 335 octets . Également défini un synonyme pour repeat (!)raccourcir la liste (radicalement, car |était beaucoup plus grand que le deuxième plus grand caractère exclu.
Ørjan Johansen
À bien y penser, une fonction pour abréger certaines de ces longues (!):séquences aiderait probablement aussi.
Ørjan Johansen le
@ ØrjanJohansen: oui, mais une sous-liste doit être finie, vous avez donc besoin d'un moyen de construire la liste vide uniquement avec !():=. L'extraction à []partir d'une liste existante, par exemple la liste d'entrée, ne fonctionnera pas car elle a un type [Int], mais nous en avons besoin [a->[a]->[a]].
nimi
Pourriez-vous s'il vous plaît expliquer comment cela fonctionne? !est curry snd, et =!est une liste infinie de ceux-ci, mais c'est tout ce que je peux comprendre de cela.
Michail
1
@nimi je voulais dire une fonction en plus de (=!), quelque chose comme (!=)(=:)=(!):(!):(!):(!):(!):(=:)ou (plus haut niveau, peut être passé ((!):)ou une section de lui-même) (!)!=(=:)=(!)((!)((!)((!)(=:)))). Beaucoup d'options à modifier. Nombre de répétitions, quel type de deuxième (voire troisième) argument, le cas échéant, et même s'il est préférable d'utiliser la compression \nou ;(longueur de bloc par 10 22 6 16 2rapport à 33 6 16 1).
Ørjan Johansen le
20

JavaScript (ES6), score: 242 (142 octets)

Une approche plutôt naïve utilisant un tableau d'octets pour les E / S.

t=>t.filter(t=>'f'>'fffffffffffffffffffffffffffffffffffffffeeeffffeffffffffffffffeeffffffffffffffffffffffffffffefefffffffeeffeffefffffefe'[t])

Essayez-le en ligne!

Comment?

Les 14 caractères non filtrés sont '().=>[]efilrt. Ils sont marqués d'un edans la chaîne de recherche. Pour les caractères ci-dessus t(code ASCII 116) qui ne sont pas définis dans cette chaîne, la valeur 'f'>undefinedest false .

Arnauld
la source
14
J'espérais qu'un de ceux-ci aurait exactement 144 octets de long. Cela mérite un crédit supplémentaire pour être une solution brute .
Jeff Bowman
La description dit "quand on me donne une corde" ...
Yair Rand
2
@YairRand Et il indique également "vous avez également la possibilité de prendre une entrée sous forme de liste d'entiers".
Arnauld
1
Je pense que cette solution est à l'envers - il semble idéalement que les seuls caractères qui devraient être affichés sont les caractères réellement utilisés dans le code, alors que tout le reste est filtré. Les bonnes nouvelles sont que cette solution peut être adoptée rapidement pour fonctionner de cette façon, en se rasant deux octets de: t=>t.filter(t=>'f'>'fffffffffffffffffffffffffffffffffffffffeeeffffeffffffffffffffeeffffffffffffffffffffffffffffefefffffffeeffeffefffffefe'[t]). Voici une déclaration de test TIO:console.log(f([...Array(256).keys()]).map(c => String.fromCharCode(c)))
Scott Hamper
2
@Arnauld Je pense que vous avez effrayé tous les autres bons concurrents du JS avec vos supergolfs rapides ...
14

APL (Dyalog Unicode) , score: 253, condition de départage: 8 octets SBCS

Fonction de préfixe tacite anonyme

∩∘'∩∘'''

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∩ intersection de l'argument

∘ et

'∩∘'''les trois caractères du code ( ∩∘')

Adam
la source
4
Cela ressemble vaguement à une émoticône pour quelqu'un qui tient quatre doigts en colère.
Skyler
1
@ Skyler Oh bien.
Adám
14

Shell POSIX , score 249, 18 octets

tr -cd \\\\\ cdrt-

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trCommande simple qui supprime tous les caractères sauf un -\cdrtespace. Quatre barres obliques inverses sont nécessaires, car le shell et la trcommande utilisent des échappements de barre oblique inversée. La cinquième barre oblique inverse échappe au caractère espace suivant.

Si tr -cdétait considéré comme un langage de programmation, on pourrait écrire un programme à un octet avec le score 255.

Nwellnhof
la source
C'est, je pense. Il a des expressions régulières, qui sont assez puissantes.
wizzwizz4
2
@ wizzwizz4 tra des classes de caractères telles qu'utilisées dans les expressions régulières, mais pas d'expressions régulières complètes.
Digital Trauma
Je ne comprends pas à quoi servent les barres obliques inverses, ici. Je pense que cela fonctionnerait bien sans eux?
David Conrad
1
@DavidConrad C'est le cas, mais cela ne préserve pas les espaces. Ensuite, vous avez besoin des 4 barres obliques inverses pour une barre oblique inversée.
wizzwizz4
3
J'ai rasé un octet, au détriment de la compatibilité POSIX, avec tr -cd \ cdrt\\-\ Essayez-le en ligne!
wizzwizz4
14

C (gcc) , note 232 235 239 241 242, 260 407 352 249 243 249 octets

fn(int*f,int*i,int*n,int**t){if(f==i)if(*t=n){}if((f==i)==('+'=='*')){if((*f==',')+(*f=='\'')+(*f=='(')+(*f==')')+(*f=='*')+(*f=='+')+(*f=='=')+(*f=='\\')+(*f=='f')+(*f=='i')+(*f=='n')+(*f=='t')+(*f=='{')+(*f=='}'))if(*n++=*f){}if(fn(++f,i,n,t)){}}}

Essayez-le en ligne!

Une fonction prenant un pointeur au début et à la fin d'un tableau int (où fin est le premier caractère non inclus), un pointeur sur un tampon et un pointeur sur un autre pointeur int. Il écrit le résultat dans la mémoire tampon et le pointeur de fin du résultat dans le dernier pointeur.

wastl
la source
1
Si cela vous aide, le défi a été modifié de sorte qu'il soit maintenant permis d'ignorer une valeur d'octet particulière, par exemple pour les chaînes terminées par NUL. Cela réduit cependant le score.
Ørjan Johansen
13

Regex, Score: 255, 2 octets

\\

Correspond au personnage \.

Essayez-le en ligne!

Incarnation de l'ignorance
la source
De quelle saveur de regex s'agit-il? À ma connaissance, cela correspond uniquement au premier caractère, mais pas à l’intérieur de l’entrée.
nimi
@nimi Eh bien, mon lien utilise .NET et il n'y a pas d'indicateur global pour les expressions rationnelles .NET.
Incarnation de l'ignorance
12

Gelée , score 254 (4 octets)

f⁾f⁾

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⁾xy est un raccourci pour “xy”(une chaîne de deux caractères).

Nous ffiltrons la chaîne d’entrée en ne tenant compte que des caractères apparaissant dans “f⁾”.

Lynn
la source
2
Hé, j'avais préparé une réponse avec f⁾⁾fmais j'ai décidé de poser la question sur la prise de chaînes Unicode avant de poster.
Jonathan Allan
10

Ruby -p, score 248 (8 octets uniques / 483 octets)

eval""<<11+11+11+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1+1<<11+11+11+11+1+1<<111+1+1+1+1+1<<111+1+1+1<<11+11+11<<11+11+11+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+11+1+1<<111+1+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1<<11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1<<11+11+11+11+1+1+1+1+1<<11+11+11+1<<11+11+11+1+1+1+1+1+1+1+1+1+1<<11+11+11+11+11+1+1+1+1+1<<11+11+11+1+1+1+1+1+1<<11+11+11+11<<11+11+11+1+1+1+1+1+1<<11+11+11+1+1+1+1+1+1

Essayez-le en ligne!

GB
la source
2
C'est génial. :)
David Ljung Madison Stellar
10

Python 2 , Score 244 245 246, 13390 1575 octets

Prend une liste d'entiers

exec'%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c'%(111+1,111+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1,11+11+11+11+11+11+11+11+11+11+1,111+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,111+1,111+1+1+1+1+1+1,111+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,111+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1,111+1,11+11+11+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+11+11+1,111+1+1+1,11+11+11+11+11+11+11+11+11+1,11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1,11+11+11+11+11+11+11+11+11+1+1,111+1+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1,11+11+11+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+1+1+1+1+1,11+11+11+1+1+1+1+1+1+1+1+1+1+1,11+11+11+1,11+11+11+1+1+1+1+1+1+1+1,11+11+11+11+11+11+11+11+1+1+1+1+1)

Essayez-le en ligne!

Cela correspond à la ligne de code suivante:

print[x for x in input()if x in map(ord,"exc'%+()1,")]

∼252

Cette réponse est évidemment théorique. La structure est

exec'<2**51 times %>c<2**50 times %>c ... %%c%c'%<112 times -~>0%<114 times -~>0% ... %<41 times -~>0%<93 times -~>0

En théorie, cela aurait le même effet que:

print[x for x in input()if x in map(ord,"exc'%-~0")]

Python 3 , Score 240 241, 54 octets

+1 merci à Jonathan Allan .

lambda a:[i for i in a if i in b""" ":[]abdfilmnor"""]

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ovs
la source
Vous pourriez vous débarrasser de la nécessité de \votre version Py 3 en utilisant des guillemets triples b''' ':[]abdfilmnor'''. En Python 2, vous pourriez faire un programme complet dans la même veine pour 242:, print(''.join(p for p in input() if p in ''' '().fijnoprtu'''))mais pas aussi bon que le exec.
Jonathan Allan
10

Nettoyer , score 247, 234 213 208 octets

-5 octets grâce à Ørjan Johansen

== =t
t['\'':t]=['\'': ==t]
t['\\':t]=['\\': ==t]
t[':':t]=[':': ==t]
t['[':t]=['[': ==t]
t[']':t]=[']': ==t]
t['t':t]=['t': ==t]
t['=':t]=['=': ==t]
t[' ':t]=[' ': ==t]
t['
':t]=['
': ==t]
t[tt:t]= ==t
t t=t

Essayez-le en ligne!

Utilise une fonction récursive de filtrage de motifs pour supprimer des caractères. N'a même pas besoin StdEnv.

Οurous
la source
2
Cela peut être directement traduit en une fonction Haskell avec 7 caractères uniques. Même score que ma réponse, mais beaucoup moins d'octets ( essayez-le en ligne! ). Agréable!
nimi
@nimi Qui peut utiliser =\ au lieu de ===.
Ørjan Johansen le
Je pense que vous voulez passer tet ttdans l'avant dernière ligne.
Ørjan Johansen le
Il sauve octets à renommer ==pour tmais en gardant ==comme synonyme pour elle.
Ørjan Johansen le
@ ØrjanJohansen Belle prise, merci!
Le
7

05AB1E , score: 252 253 (  4  3 octets distincts utilisés; 8 octets au total)

„ÃJ„„JJÃ

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Explication:

„ÃJ        # Push the String "ÃJ"
   „„J     # Push the string "„J"
      J    # Join them together: "ÃJ„J"
       Ã   # Only keep these three characters from the (implicit) input
           # (and output the result implicitly)

Notez que l'ordre ÃJet „Jdans les chaînes 2 caractères sont importants, car „est également utilisé pour des mots du dictionnaire est un mot du dictionnaire (sauf pour quelques privilégiés des personnages comme des lettres, des chiffres, où tous les deux personnages -, et autres). Donc, ce „JÃserait «un mot» J, ainsi qu'un mot du dictionnaire Ã+ le caractère suivant (dans ce cas „), qui est apparemment le mot"causing" .

Kevin Cruijssen
la source
6
4 barrés avec espacement est encore un barré 4
Jo King
7

Perl 5, score 251 (5 octets distincts, 8058 octets au total)

s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s>><><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s>><><<s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<^>^>^^s>^><>^s<^>^>^^s>^>s>^s>^>e>^s>^>^>^s<^>^>^^s>^><>^s>^><>^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s^^^<<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>>^s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s^^^<<s>><<><<s>>>^s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>>^s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>>eeee^s<^>^>^^s>^><>^s<><<s<>e<><s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^<<s^^^^s^^^<<s^^^<<s>><<><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><<s>>><<s>>>^s>>><<s>>><<s>>><e^s<^>e>ee<e^<see><>>>>^<^^<se^e^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^e^^^^^^^e^^>^^^^s^s^ss>^<^^^^s^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^s^^^^^^^^^^>^<^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^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Essayez-le en ligne!

Utilise le <>^essous-ensemble Turing-complete de Perl . Le code ci - dessus a été obtenu en cours d' exécution à print<>=~y<<>^es><>cdrtravers mon convertisseur .

Crasseux
la source
7

Brachylog , score 249, 16 octets

{∈"{∈\\\"&}ˢ"&}ˢ

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Je ne suis pas tout à fait sûr de savoir comment cela fonctionne avec les règles, avec Brachylog ayant sa propre page de code spéciale et tout, mais je peux quand même expliquer comment cela fonctionne:

{                Start of inline predicate.
 ∈"              Input is a member of the string literal containing:
   {∈            the opening bracket, the member-of character,
     \\\"        backslash, quote,
         &}      ampersand, the closing bracket,
           ˢ"    and the superscript s.
             &}  Unify the input to the inline predicate with the output from the inline predicate.
               ˢ Output everything from the input that made it through the inline predicate.
Chaîne non liée
la source
7

Octave , Score 248 249 (50 bytes)

Il devait y avoir un meilleur moyen ... et il y en a un! 

@(a)a('l'=='a'==all('l'=='a'==(a==('@al()''=')')))

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Filtre tout sauf @al()'=. La allfonction est supérieure à celle utilisée anycar elle n'utilise que deux octets distincts. Notez que any(...)est égal à not(all(not(...))). Donc, on commence avec le programme

@(a)a(~all(a~=('@al~()''=')'))

Au lieu de ~( not), nous pouvons aussi écrire 0==. Bien sûr, cela signifierait que nous devons également filtrer 0. Notez plutôt que l'expression suivante est également fausse (zéro)

'l'=='a'

Si la compatibilité MATLAB n’est pas un problème, je peux passer à 47 octets.

@(a)a((l='l'=='a')==all(l==(a==('@al()''=')')))
Sanchises
la source
7

Langue (avec stty +brkint -ignbrk), score 255, 3890951 octets

Le programme se compose de 3890951 octets NUL (ce qui en fait l’un des programmes de lengage les plus courts jamais créés; l’octet 3890951 est suffisamment petit pour tenir sur mon disque, j’ai donc exécuté cela dans un interpréteur Lenguage). Le PO souhaitait que la solution Lenguage / Unary soit jouée au golf, alors allons-y. (Notez que Unary serait beaucoup plus long car il nécessite l'utilisation de 0plutôt que de permettre l'utilisation de NUL.)

Notez que Lenguage ne se comporte pas comme le brainfuck, malgré ce que sa documentation implique. I / O fonctionne complètement différemment (quelque chose que j'ai remarqué en testant ce programme). En particulier, Lenguage refuse de prendre des entrées autres que celles d’un terminal. Les octets filtrés sont les octets bruts envoyés via la connexion du terminal (notez également que lorsqu’il filtre les octets bruts, vous ne verrez pas les touches que vous tapez du tout). En pratique, cela signifie que le programme absorbera toutes les entrées qui lui seront envoyées, à l’exception de l’octet NUL (généralement appelé Ctrl- @), qui sera répercuté à la lettre (à ce stade, la grande majorité des terminaux l’ignorera). l’octet NUL est l’équivalent terminal d’une instruction NOP). Afin de vérifier que le programme fonctionne, il

Que se passe-t-il à EOF? Eh bien, si le terminal envoie une série d'octets, il n'y a aucun moyen d'envoyer EOF; tous les 256 octets possibles sont interprétés littéralement, et vous ne pouvez rien insérer d'autre dans le flux du terminal. Cependant, si vous utilisez un terminal série à l'ancienne, vous pouvez appuyer sur le bouton "pause" de votre terminal pour envoyer à dessein des données mal codées, permettant ainsi le 257ème code possible; cette "pause" est le seul équivalent plausible d'un EOF, car il est envoyé hors bande et indique autre chose que des données valides. Si la configuration de votre terminal a le drapeau "interruption-sur-pause" activé (et que l'interpréteur Lenguage ne modifie pas ce paramètre), l'envoi du signal de pause provoquera le blocage de l'interpréteur Lenguage, qui agit commodément comme un moyen de mettre en œuvre le comportement souhaité EOF. Je ne suis pas sûr qu'il s'agisse d'un paramètre par défaut (car personne n'utilise actuellement de terminaux série, il ne revient pratiquement jamais), je l'ai donc mentionné dans l'en-tête dans le cadre de la spécification de l'interpréteur de langage utilisé.

Explication

  1110110101111100000111
001                       Initialise tape element 0 to -1
   110               111  While tape element 0 is nonzero:
      110   111              While tape element 0 is nonzero:
         101                    Read a byte from the terminal into tape element 0
               100           Output tape element 0
                  000        Add 1 to tape element 0

La boucle interne ne sortira que lorsque NUL sera saisi au terminal; après cela, nous renvoyons immédiatement le caractère saisi (c'est-à-dire le NUL). Ajouter 1 à ce stade garantira que l'élément de bande 0 sera à nouveau différent de zéro, de sorte que la boucle externe ne puisse pas se terminer (jusqu'à ce qu'une entrée break bloque l'interpréteur), et nous retomberons dans la boucle interne.

Il est très déconseillé d’utiliser la soustraction pour entrer dans la boucle extérieure, mais aussi de continuer à boucler autour de celle-ci; L’addition a un codage plus court, mais ne peut pas apparaître au début du programme (car le codage ressemblerait à des zéros non significatifs et serait donc ignoré).

ais523
la source
Incidemment, je suis à peu près sûr de pouvoir exécuter n’importe quel programme Lenguage <4 To alors que le disque n’est pas aussi volumineux. L'interprète ne se soucie pas des fichiers avec des trous.
Josué
truncateLa création de fichiers fragmentés n’est pas pour moi (peut-être que le système de fichiers ne les prend pas en charge?). Toutefois, pour autant que je sache, l’interprète de référence Lenguage lit le fichier d’entrée ligne par ligne (!). Ainsi, si le programme ne contient pas de caractères de nouvelle ligne, un programme suffisamment long épuisera la mémoire disponible. Bien sûr, cela pourrait être corrigé avec un nouvel interprète.
ais523
Pas tronqué. Recherchez un nouveau fichier sur un octet avant la fin souhaitée et écrivez un octet nul.
Josué
6

Retina 0.8.2 , 12 bytes, score: 250

[^]n\\\n[^]

Essayez-le en ligne! Filtres à tous les personnages , sauf nouvelle ligne, [, \, ], ^et n. Bien que les caractères [\]^soient consécutifs, l'utilisation d'une plage coûterait en réalité un octet et réduirait également mon score.

Neil
la source
6

Japt , 249 partition, 20 octets

f"[ fq\"[\\\\\\]]" q

Essayez-le en ligne!

ASCII seulement
la source
Luis a réussi à trouver la solution optimale, mais vous pouvez améliorer votre score avec oQ+"+oQ"Qou tout simplement, oomais ce n'est qu'une variante de la solution de Luis.
Shaggy
Haha, tbh Je me précipitais sur celui-ci et je manquais de temps hier soir, et je suis encore plus court et nous nous rapprochons de la solution de Luis: P
ASCII uniquement
4

PHP, score 234, 58 62 octets

<?=preg_replace("_[^\"$(),;<=?[\]^\_\\\\aceglnpr]_","",$argn);

Exécuter en pipe -nFou essayer en ligne .

Titus
la source
4

Ruby -paF[^$_=F*'], score 250, 8 octets

$_=$F*''

Je ne suis pas du tout sûr que cela soit légal en vertu des règles actuelles du site pour les drapeaux de ligne de commande, car le "drapeau" ici est en quelque sorte un code qui inclut des octets interdits. L’ Findicateur définit une expression rationnelle de séparateur d’entrée composée de tous les caractères ne figurant pas dans le code approprié. Le code approprié définit la sortie sur la concaténation de tous les enregistrements d'entrée.

histocrate
la source
4

MATL , Score 252

t't''m)'m)

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Explication (pour l'entrée input)

t         % Duplicate input. 
          % Stack: {'input', 'input'}
't''m)~'  % Push the string t'm)~ 
          % Stack: {'input', 'input', 't'm)~'}
m         % 'ismember' to filter
          % Stack: {'input', [0 0 0 0 1]}
)         % Index into input.
          % Stack: {'t'}
          % Implicit display.
Sanchises
la source
3

K, score 246, 23 octets

{x@&~x in"\"~ in@&{x}"}

{..} est un lambda avec arg x

lu comme 'renvoie x à où non x dans la chaîne'

La chaîne contient \ "~ dans @ & {x} (les dix octets utilisés dans le programme)

J. Sendra
la source
3

sfk , score 239, 133 octets

+xed -i -case ixixi x+x+x xexex xdxdx x-x-x xixix xcxcx xaxax xsxsx xbxbx xyxyx xtxtx x"x"x x\[x\[x x\]x\]x x\\x\\x "x x x" x[byte]xx

Essayez-le en ligne!

Effectue des remplacements sur l'entrée (écrits en tant que [delim][from][delim][to][delim]) en modifiant tous les caractères utilisés dans le programme, ce qui supprime les positions des caractères de la considération pour la saisie au remplacement final x[byte]xxqui mappe tous les octets sur la chaîne vide.

Οurous
la source
3

JavaScript, score 250, 95573 59273 octets

Une solution JS légèrement golfée qui utilise ensuite l'algorithme de JSFuck . Beaucoup d'octets, mais un bon score.

Original:

a=t=>t.replace(/[^![\]+()]/g,'')

Le rendu JSFuck est composé de 59273 caractères:

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tschwab
la source
2
s=>s.replace(/[^![\]+()]/g,'')devrait fonctionner
ASCII uniquement
Oui, ça l'a fait de manière significative. Bonne idée. Je crois que, quoi qu’il en soit, ce sera de l’ordre de milliers d’octets.
tschwab
3

brainfuck , Score: 248, 162 octets

,[>---[----------->--<]>-[>->->->-->------->-->-------[<<]>-]>--->-->->--->------------->->-----------[<]<<[->>>->->->->->->->-<<<<<<<<<<]>>>[>]>[[>]<[->-<]>.>],]

Essayez-le en ligne!

En général, je ne réponds pas à mes propres questions, mais les gens semblent avoir lu ma page de profle, en particulier là où elle dit If someone comments on a question "I wish there was a brainfuck solution", I feel obliged to provide one.

Les octets filtrés sont -<>[],.et ignorent l'octet NUL, car il est impossible de le distinguer de EOF. C'est un peu plus long que d'habitude, car j'évite délibérément de l'utiliser +pour obtenir un score légèrement meilleur. Même dans ce cas, je pourrais peut-être jouer au golf pour générer quelques octets.

Jo King
la source
1
Les soumissions C et Lenguage ont déjà été adaptées de manière spécifique à la nécessité de distinguer NUL de EOF. Peut-être utiliser un BF 16 bits avec une convention EOF différente?
Ørjan Johansen
3

Espace blanc , score: 253 (3 octets distincts utilisés, 97 octets au total)

[N
S S N
_Create_Label_LOOP][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_STDIN_as_character][T   T   T   _Retrieve][S N
S _Duplicate_input][S S S T S S T   N
_Push_9][T  S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][S N
S _Duplicate_input][S S S T S T S N
_Push_10][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][S N
S _Duplicate_input][S S S T S S S S S N
_Push_32][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT][T  N
S S _Print_character][N
S N
N
_Jump_to_Label_LOOP]

Lettres S(espace), T(tabulation) et N(nouvelle ligne) ajoutées uniquement en surbrillance.
[..._some_action]ajouté comme explication seulement.

Essayez-le en ligne (avec des espaces bruts, des onglets et des nouvelles lignes uniquement). Un bon bonus est que je peux utiliser ce code pour transformer le code en surbrillance ci-dessus en programme brut Whitespace (ce que j'ai fait dans ce TIO à titre d'exemple), pour lequel j'utilise habituellement Notepad ++. :)

Explication:

Assez simple. Voici le pseudo-code:

Loop indefinitely:
  Character c = STDIN as character
  if(c == SPACE or c == TAB or c == NEWLINE):
    Print c as character
  Go to the next iteration of the loop
Kevin Cruijssen
la source