prOGraMMIng PuZZleS & cOde ____

Input

A non-empty encoded string consisting of printable ASCII characters (in the range 32-126), where some missing letters have been replaced with _.

Output

A decoded string of the same length with all letters in lowercase, including the missing ones.

How?

Edit: As mentioned by @Deusovi in the comments, this is a variant of Bacon's cipher.

• Gather all letters in the original string and group them by 5. Additional letters that do not fit in a full group of 5 are ignored.
• Convert each group into binary: lowercase = 0, uppercase = 1. This leads to a list of integers.
• Use each value N in this list to replace each _ in the original string with the N-th letter of the alphabet (0-indexed), in order of appearance.

Example: prOGraMMIng PuZZleS & cOde ____

prOGr --> 00110 -->  6 -->  7th letter = 'g'
aMMIn --> 01110 --> 14 --> 15th letter = 'o'
gPuZZ --> 01011 --> 11 --> 12th letter = 'l'
leScO --> 00101 -->  5 -->  6th letter = 'f'


By replacing the missing letters and converting everything back to lowercase, the original string is unveiled:

programming puzzles & code golf


This is the expected output.

Clarifications and rules

• The missing letters are guaranteed to appear at the end of the string. More formally: there will never be any letter after the first _ in the input string. However, there may be other printable ASCII characters such as spaces and punctuation marks.
• The input is guaranteed not to contain any useless capital letter: all capital letters are bits set to 1 which are required to decode the missing letters. Everything else is in lowercase.
• The input string is guaranteed to be valid. Especially:
• It will always contain enough full groups of 5 letters to decode the underscores.
• The binary-encoded integers are guaranteed to be in the range [0-25].
• There may be no _ at all in the input string, in which case you just have to return the input.
• This is , so the shortest answer in bytes wins!

Test cases

Input : hello!
Output: hello!

Input : helLO, worl_!
Output: hello, world!

Input : i aM yoUr faTh__.
Output: i am your father.

Input : prOGraMMIng PuZZleS & cOde ____
Output: programming puzzles & code golf

Input : Can YOu gUesS tHE ENd oF This ____?
Output: can you guess the end of this text?

Input : THe qUICk brown FOx JUMps oVEr the la__ ___.
Output: the quick brown fox jumps over the lazy dog.

Input : RoadS? wHERe we're goinG WE doN't need _____.
Output: roads? where we're going we don't need roads.

Input : thE greatESt Trick thE DeVIl EVer PUllEd wAs CONvInciNg tHe WorLD h_ ____'_ _____.
Output: the greatest trick the devil ever pulled was convincing the world he didn't exist.


Some extra test-cases:

Input : BInar_
Output: binary

Input : 12 MonKey_
Output: 12 monkeys

Input : hyPerbolIZ__
Output: hyperbolized

Input : {[One Last Test ca__]}
Output: {[one last test case]}

• Should we only consider as many groups of 5 as there are underscores in the input? May 12, 2018 at 18:10
• In that case, the rule when there's no _ in the input string is a bit of a special case. May 12, 2018 at 18:17
• Ooh, Bacon cipher! May 12, 2018 at 19:38
• @SztupY As The input is guaranteed not to contain any useless capital letter, in case there is no underscore there also won't be a capital letter. May 14, 2018 at 11:03
• @KirillL. Yes, anything in [32-126]. I've added another test case. May 14, 2018 at 12:16

05AB1E, 18 bytes

Code:

áS.u5ôJC>.bv'_y.;l


Uses the 05AB1E encoding. Try it online!

Explanation:

á                       # Remove non-letters from the input string.
S                      # Split the result into individual characters.
.u                    # Check if is uppercase for each character.
5ôJ                 # Split into binary numbers of length 5.
C                # Convert from binary to decimal.
>               # Add one.
.b             # Map 1 → A, 2 → B, 3 → C, ..., 25 → Y, 26 → Z.
v            # For each letter:
'_y.;       #   Replace the first occurrence of '_' with the current letter.
l      #   Convert the string to lowercase.


Python 2, 113 bytes

s=input()
i=k=0
for c in s:
if c.isalpha():k+=k+(c<'a');i+=1;s=s.replace('_',chr(k%32+97),i%5<1)
print s.lower()


Try it online!

Perl 5 -pF -MList::Util=sum, 75 bytes

@a=grep!/\W|\d/,@F;s!_!(a..z)[sum map{a gt shift@a&&16/2**$_}0..4]!eg;$_=lc


Try it online!

Explanation:

• -pF reads a line of input into the variable $_ and, split into characters, into the array @F. • @a=grep!/\W|\d/,@F sets the array @a equal to those members of @F that don't satisfy the regex \W|\d. \W is anything but letters, numbers, and _; \d is numbers. So \W|\d is anything but letters and _, and @a has all the letters and _ characters. We will wind up never examining the _ characters in @a. (Note that this only works because the input is guaranteed ASCII.) • map{a gt shift@a&&16/2**$_}0..4 does the following for 0 through 4: It shifts the next element off of @a, shortening it, and evaluates whether a is asciibetically greater than that element (i.e. whether that element is uppercase). If so, && isn't short-circuited, so we get 16 divided by 2 to rhe power of the input value (0 through 4). Otherwise && is short-circuited and we get 0. map returns the list of five numbers to sum, which adds them.
• That's the element we want from the list a..z, and that's what we get from (a..z)[…].
• s!_!…!eg converts each _ in $_, in turn, to the appropriate letter. • $_=lc converts $_ to the lowercase version of itself, and -p prints it. J, 62 61 bytes tolower u:@](I.@t=.'_'=[)[}+/@t$97+_5#.\3|0 2 4-.~'@Zz'&I.


Try it online!

Jelly,  28 27  26 bytes

-1 thanks to Erik the Outgolfer & dylnan

Not a very Jelly-friendly challenge!

ḟŒs$Ƈ<”[s5Ḅ+97Ọż@ṣ”_$FṁLŒl


A monadic link accepting and returning lists of characters.

Try it online!

How?

ḟŒs$Ƈ<”[s5Ḅ+97Ọż@ṣ”_$FṁLŒl - Link: list of characters    e.g. "MfUNE_?"  (shorthand for ['M','f','U','N','E','_','?'])
Ƈ                      - filter keep each if:
$- last two links as a monad: Œs - swap-case ḟ - filter discard - ...i.e. keep A-Z,a-z since they change when the case is swapped - "MfUNE" ”[ - literal character '[' < - less than? (i.e. is upper-case?) [1,0,1,1,1] s5 - split into fives [[1,0,1,1,1]] Ḅ - from base two (vectorises) [[23]] +97 - add (vectorises) ninety-seven [[120]] Ọ - from ordinals (vectorises) [['x']]$      - last two links as a monad:
”_       -   literal character              '_'
ṣ         -   split at                       [['M','f','U','N','E'],['?']]
ż@          - swapped @rgument zip             [[['M','f','U','N','E'],'x'],['?']]
F     - flatten                          "MfUNEx?"
L   - length (of input)                7
ṁ    - mould like                       "MfUNEx?"
- ...removes any excess characters
Œl - lower-case                       "mfunex?"


Retina, 91 90 bytes

Tla
TLA
[^Aa]

L.{5}
A
aA
+Aa
aAA
+T_lAl_[^A]A
^
$+¶ +_(.*)¶a+(.)$2$1 0G TLl  Try it online! Explanation: Tla TLA [^Aa]  Translate lowercase letters to a and uppercase letters to A, deleting everything else. L.{5}  Split the Aas into groups of 5. A aA +Aa aAA  Convert from binary into unary, treating A as 1 and a as 0. Since there were 5 Aas originally, there are 5 as left, plus a number of As depending on the desired position in the alphabet. +T_lAl_[^A]A  Increment the last a according to the number of following As. ^$+¶


Prepend the original input.

+_(.*)¶a+(.)
$2$1


Replace any _s with the next decoded letter.

0G


Remove any spare decoded letters.

TLl


Lowercase everything.

Retina 0.8.2, 117 bytes

.+
$&¶$&
TLl^.*
Tla.*$TLA TaApaA_.*$
(.*¶)?.{5}
$&; A aA +Aa aAA +T_lAl_[^A]A +_(.*¶)a+(.);$2$1 1G  Try it online! Explanation: .+$&¶$&  Duplicate the input. TLl^.*  Lowercase the first copy. Tla.*$


Translate lowercase letters to a in the second copy.

TLA


Translate uppercase letters to A. These must be in the second copy because the first copy was already lowercased.

TaApaA_.*$ Delete everything else in the second copy. (.*¶)?.{5}$&;


Split the second copy (now just Aas) into groups of 5.

A
aA
+Aa
aAA
+T_lAl_[^A]A
+_(.*¶)a+(.);
$2$1
1G


Decode the letters and insert them as before.

APL (Dyalog Unicode), 46 bytesSBCS

Anonymous lambda, Assumes ⎕IO (Index Origin) to be 0.

{_←'_'=⊢⋄819⌶A[2⊥⍉(+/_⍵)5⍴A∊⍨⍵∩A,819⌶A←⎕A]@_⍵}


Try it online!

{⋄} two-statement function; ⍵ is the argument, ⋄ separates statements

⊢ argument (no-op function)
'_'= where equal to an underscore (i.e. a Boolean mapping function)
_← assign that function to _

A[]@_⍵ put the following characters of A at positions of underscores in the argument
⎕A the uppercase Alphabet
A← assign that to A
819⌶ lowercase it (819 ≈ BIg, with no left argument means not big, i.e. lowercase)
A, prepend the uppercase alphabet; this gives us all letters
⍵∩ intersection of the argument and that; just the letters of the argument
A∊⍨ which of those are members of the uppercase alphabet; uppercase bits
()5⍴reshape that to the following number of rows, and five columns:
_⍵ the mask of underscores in the argument
+/ sum that; number of underscores
⍉ transpose (to treat each row as a number rather than as a bit position)
2⊥ evaluate as base-2
819⌶ lowercase everything

Scala, 189 bytes

def f(s:Array[Char])={var j=0;s.zipWithIndex.collect{case(95,i)=>s(i)=(Integer.parseInt(s.filter(_.isLetter)slice(j,j+5)map(k=>if(k<91)1 else 0)mkString,2)+97)toChar;j+=5};s.map(_.toLower)}


Try it online!

Explanation:

def f(s: Array[Char]) = {                // takes a String in input
var j = 0                              // j stores at which block of 5 letters we're currently at
s.zipWithIndex.collect {               // Array('h', 'e', ...) => Array(('h', 0) ('e', 1), ...) and we apply a collect transformation (filter/map)
case (95, i) =>                      // we only handle cases where the char is '_' (95)
s(i) = (                           // we modify the char at index i with the following
Integer.parseInt(                // Integer.parseInt("00110", 2) = 6
s                              //
.filter(_.isLetter)          // filter out non letter chars (spaces, punct, figures, ...) from the input string (thanks @Arnauld for the fix)A
.slice(j, j+5)               // "substring" the array to the block of 5 letters in question
.map(                        // map on the current block of 5 letters
k =>                       // the index of the next char in the block f 5 (e.g. 13)
if (k < 91) 1 else 0     // if the current char is upper case (<91) then we replace it by a bit true, otherwise by a bit false
)mkString,                   // Array(0, 1, 1, ...) => "011..."
2                              // cast string to binary
)                                //
+ 97                             // +97 to create a lower case char
)toChar                            // cast from int to char
j += 5                             // update the starting index of the next block of 5 letters
}                                    //
s.map(_.toLower)                       // return the updated seq of chars all in lower case
}                                        //


JavaScript (Node.js), 125 124 bytes

x=>x.replace(s=/./g,c=>parseInt(c,36)>9?c.toLowerCase(s+=+(c<{})):c=='_'?('0b'+s.slice(4,9)-~9).toString(36,s=s.slice(5)):c)


Try it online!

Jelly, 26 bytes

xn¥Œs<”[s5Ḅ‘ịØaṛi”_ḟ0Ɗ¦ƒŒl


Try it online!

Different approach from Jonathan Allan's. EDIT: So, I, uh, apparently thought of the same byte reduction as Jonathan Allan, so it doesn't hurt to mention his name again.

q'_/_0={el_eu=!},_eu.=5/2fb65f+:c1$,(<.+:el  Try it online! Clean, 180 ... 150 bytes import StdEnv ?s=['a'+sum[i\\i<-:""&c<-s|c<'a']: ?(drop 5s)] @['_':b][x:y]=[x: @b y] @[a:b]z=[toLower a: @b z] @e _=e$s= @s(?(filter isAlpha s))


Try it online!

Defines the function $:: [Char] -> [Char] with @ :: [Char] [Char] -> [Char] as a helper to replace underscores, and ? :: [Char] -> [Char] as a helper to generate the replacement characters. • How does the i<-:"" part work? Are chars implicitly converted to numbers when summing or adding them? May 14, 2018 at 12:48 • @Laikoni no, there's no implicit conversion. You can add and subtract Chars though. May 14, 2018 at 20:31 JavaScript (Node.js), 100 bytes s=>s.toLowerCase(c=/[a-z]/gi).replace(/_/g,_=>(g=n=>n?(c.exec(s)<{})*n+g(n>>1):10)(16).toString(36))  Try it online! Thanks to @Arnauld, saves 2 bytes. • @Arnauld you are right. edited to /[a-z]/gi now. – tsh May 14, 2018 at 7:27 Stax, 22 bytes â╟▓ïMeee¶▐f◄┴≈┘n╛äyΩ○N  Run and debug it The general approach is a regular expression replacement of "_" using a callback function that slices letters of the inputs to compute each replacement character. v convert to lower case '_ "_" string literal { begin block for regex replacement yVl|& all the letters only from the original input 5/ split into chunks of 5 i@ keep the ith one, where i is the 0-based number of times this block has run {97<m map 5-letter chunk to bits to indicate which are lowercase :b decode as 5-bit integer 97+] add 97 and wrap in array to convert to lower case character } end block for regex replacement R do regex replacement  Run this one R, 153135 113 bytes function(s,S=utf8ToInt(s)){S[S==95]=2^(4:0)%*%matrix(S[S%in%c(65:90,97:122)]<95,5)+97 cat(tolower(intToUtf8(S)))}  Try it online! Issues some warnings with the use of matrix but that shouldn't affect the result. Also issues warnings as [<- assignment will remove extraneous assigned objects by default. 40(!) bytes down thanks to JayCe's improvements • I don't think you need ,length(L)%/%5 May 14, 2018 at 17:05 • Also no need to define L? May 14, 2018 at 17:07 • @JayCe okay well today I learned that [<- will throw out elements past the length of the indexes... May 14, 2018 at 17:21 • Me too actually! May 14, 2018 at 17:26 • Translated your approach to intToUtf8 May 14, 2018 at 19:03 C (gcc), 111109101 100 bytes Edit: Added lowercasing per @FrownyFrog's comment; thanks to Lynn, Christoph and user5329483 for their suggestions! f(s,t,i)char*s,*t;{for(t=s;t=strchr(t,95);*t=i+1)for(i=3;i<64;s++)isalpha(*s)?i=2*i|*s<97,*s|=32:0;}  Try it online! • You can save 2 bytes with i+=i+(*s<97). – Lynn May 13, 2018 at 9:51 • You could abolish j by introducing a marker bit in i and rewriting the second for as for(i=1;i<32;s++). And compensate the extra 32 in the outer for. As a newbee here I count a seven byte spare. May 14, 2018 at 16:40 • Found another byte: for(i=3;i<96;s++) brings the 65 down to a single-digit number, aka 1. May 15, 2018 at 17:56 Haskell, 139 bytes ([]%) a%(x:y)|x=='_'=['a'..]!!foldl1((+).(2*))(take 5a):drop 5a%y|x>'@',x<'['=[x..]!!32:(a++[1])%y|x>'',x<'{'=x:(a++[0])%y|1<2=x:a%y a%e=e  Try it online! Red, 247 bytes func[s][a: charset[#"a"-#"z"#"A"-#"Z"]v: copy""parse s[any[copy c a(append v to-string c)| skip]]k: 0 t: copy""repeat n(length? v)/ 5[c: 0 p: 16 loop 5[if v/(k: k + 1) <#"a"[c: c + p]p: p / 2]append t#"a"+ c]foreach c t[replace s"_"c]lowercase s]  Try it online! More readable: f: func[s][ a: charset[#"a"-#"z"#"A"-#"Z"] v: copy "" parse s[any[copy c a(append v to-string c)| skip]] k: 0 t: copy "" repeat n (length? v) / 5[ c: 0 p: 16 loop 5[ if v/(k: k + 1) < #"a" [c: c + p] p: p / 2 ] append t #"a" + c ] foreach c t[replace s "_" c] lowercase s ]  Go, 219217192210209 156 bytes Saved 25 bytes thanks to @Lynn! Saved 53 bytes thanks to @ovs! Had to lose 18 bytes because of a bug with strings with no underscores :( func p(s string){b:=0;n:=0;for _,c:=range s{if IsLetter(c){b+=b;if IsUpper(c){b+=1};n++;s=g.Replace(s,"_",string('a'+b%32),(5-n%5)/5)}};Print(g.ToLower(s))}  Try it online! Haskell, 165 bytes g"" import Data.Char f t(x:s)('_':r)=x:f t s r f[0,a]s r=g(s++[chr a])r f[n,a]s(c:r)=toLower c:f[div n$1+sum[1|isAlpha c],a+sum[n|isUpper c]]s r
f _ s e=e
g=f[16,97]


Try it online! Example usage: g"" "BInar_" yields "binary".

PowerShell, 121 bytes

switch -r($args|% t*y){_{$_=$a[$n++]+97}[a-z]{$x+=$x+($_-le96);if(!(++$i%5)){$a+=,$x;$x=0};$_=$_-bor32}.{$r+=[char]$_}}$r


Try it online!

Less golfed:

switch -Regex ($args|% toCharArray){ _ { # is underscore$_=$a[$n++]+97      # get a char from the array of letter
}

[a-z] {                 # is letter
$x+=$x+($_-le96) #$x=2*$x+($_-le96)
if(!(++$i%5)){ # if(++$i%5 -eq 0)
$a+=,$x         # add an element to the array of letters
$x=0 # init }$_=$_-bor32 # to lower } . { # is any char ('_' and letters included)$r+=[char]$_ # add the char to result } }$r


Ruby, 107106 103 bytes

->s{s.chars.map{|c|c !~/\W|\d/?c<?a?1:0:p}.join.scan(/.{5}/){|b|s[?_]&&=(b.to_i(2)+97).chr};s.downcase}


Try it online!

Japt, 25 bytes

r'_@r\L mè\A sTT±5 ÍdIÄÃv


Try it

Explanation

r'_                           :Replace underscores
@                          :Pass each match through a function
r                         :  From original input remove
\L                       :    /[^a-zA-Z]/g
m                     :  Map
è                    :    Count
\A                  :      /[A-Z]/g
s                :  Slice
T               :    From index T (initially 0)
T±5            :    To index T+=5
Í          :  Convert from base-2 string to base-10 integer
IÄ       :  Add 64+1
d         :  Get character at that codepoint
Ã      :End function
v     :Lowercase


Pyth, 36 bytes

Km@Gim!}kGd2c@+r1GGQ5VQp?qN\_.(KZr0N


Try it here

Explanation

Km@Gim!}kGd2c@+r1GGQ5VQp?qN\_.(KZr0N
@+r1GGQ                   Get the letters from the input...
c       5                  ... in chunks of 5.
m        d                            For each chunk...
m!}kG                             ... check if each letter is uppercase...
i      2                           ... converted to binary...
@G                                   ... and get the corresponding letter.
VQp               For each character in the input...
K                       ?qN\_.(KZ      ... if the character is '_', replace it...
r0N   ... otherwise, lowercase it.


Python 3.5, 296 bytes

u=input();f=u.find('_');m=''.join([c for c in u if c.isalpha()]);z=[chr(int(''.join(['0'if o.islower() else'1' for o in l]),2)+65)for l in[m[h:h+5]for h in range(0,len(m),5)]if len(l)==5];[z.insert(v,d)for v,d in enumerate(u[f:])if d!="_"];u=list(u);u[f:]=z[:len(u[f:])];print(''.join(u).lower())


Try it online

Here is the explanation:

User input

u=input()

Finds the index of the first _ in the string and stores it

f=u.find('_')

strips string of all non-alpha characters

m=''.join([c for c in u if c.isalpha()])

Splits the alpha string into an array with each element consisting of 5 characters

Ex. ['THeqU', 'ICkbr', 'ownFO', 'xJUMp', 'soVEr', 'thela']

Then converts lowercase characters to 0 and uppercase characters to 1

Ex. ['11001', '11000', '00011', '01110', '00110', '00000']

and converts the binary string to an integer, adds 65 and converts that to a character

Ex. ['z', 'y', 'd', 'o', 'g', 'a']

z=[chr(int(''.join(['0' if o.islower() else '1' for o in l]),2)+65) for l in [m[h:h+5] for h in range(0,len(m),5)] if len(l)==5]

finds all characteres after the first _ and pushes them into the array z at their respective locations (defined above)

Ex. ['z', 'y', ' ', 'd', 'o', 'g', '.', 'a']

[z.insert(v,d) for v,d in enumerate(u[f:]) if d!="_"]

split our string into a list of characters

u=list(u)

slice our string from the first _ to the end of the list and replace it with the array z. I also had to slice the array z to the length of the split string from the first _ to the end because I got an extra character in the lazy dog example (the "a" at the end of the examples above)

u[f:]=z[:len(list(u[f:]))]

*print out the answer *

print(''.join(u).lower())

• There's more to golfing than just getting everything on one line. 165 bytes
– Jo King
May 16, 2018 at 10:27
• You can remove the space between o.islower() and else, and I think '1' and for. Also, you can change if d!="_" to if"_"!=d, but the above comment already does that. May 17, 2018 at 12:52

Java 10, 186184 162 bytes

s->{var b="";for(Byte c:s.getBytes())s=(b+=c>64&c<91?1:c>96&c<123?0:"").length()>4?s.replaceFirst("_",(char)(97+c.valueOf(b,2))+(b="")):s;return s.toLowerCase();}


-2 bytes thanks to @ceilingcat.

Try it online.

Explanation:

s->{                        // Method with String as both parameter and return-type
var b="";                 //  Binary-String, starting empty
for(Byte c:s.getBytes())  //  Loop over the characters of the input-String:
s=(b+=c>64&c<91?        //   If the current character is a lowercase letter:
1                //    Append "1" to the binary-String
:c>96&c<123?      //   Else-if it's an uppercase letter:
0                //    Append "0" to the binary-String
:                 //   Else (it's not a letter):
""               //    Append nothing to the binary-String
).length()>4?         //   And if the length of the new binary-String is now 5:
s.replaceFirst("_",  //    Replace the first "_" in the input-String with
(char)(97+c.valueOf(b,2))
//    the binary-String as character
+(b="")):s;        //    And reset the binary-String
return s.toLowerCase();}  //  Return the modified lowercased input-String

• @ceilingcat Thanks, but I've been able to save 22 more bytes by changing the for-loop to a for-each loop instead. Jun 13, 2021 at 21:05

Perl 5-p, 78 bytes

for$h(s/\W|\d//gr=~y/a-z/0/r=~y/A-Z/1/r=~/.{5}/g){s%_%chr 65+oct"0b$h"%e}\$_=lc


Try it online!

• I fixed it with 3 more bytes, which makes your answer a bit better under the current rules. May 15, 2018 at 13:35
• Not better but different. Each language + options is considered separately, not competing with the same language + other options, as I understand it. May 21, 2018 at 5:41