# Decode the string

This is my first challenge on ppcg!

### Input

A string consisting of two different ascii characters. For example

ABAABBAAAAAABBAAABAABBAABA


### Challenge

The task is to decode this string following these rules:

1. Skip the first two characters
2. Split the rest of the string into groups of 8 characters
3. In each group, replace each character with 0 if that character is the same as the first character of the original string, and with 1 otherwise
4. Now each group represents a byte. Convert each group to character from byte char code
5. Concatenate all characters

### Example

Let's decode the above string.

 AB  AABBAAAA  AABBAAAB  AABBAABA
^^     ^         ^         ^
|      |         |         |
|      \---------|---------/
|                |
Skip      Convert to binary


Notice that A is the first character in the original string and B is the second. Therefore, replace each A with 0 and each B with 1. Now we obtain:

00110000  00110001  00110010


which is [0x30, 0x31, 0x32] in binary. These values represent characters ["0", "1", "2"] respectively, so the final output should be 012.

### Scoring

This is, of course, , which means make your code as short as possible. Score is measured in bytes.

### Constraints and IO format

Standard rules apply. Here are some additional rules:

• You can assume valid input
• Input string consists of exactly two different characters
• The first two characters are different
• The minimal length of the input string is 2 characters
• The length will always give 2 modulo 8
• You can assume the string will always consist only of printable ASCII characters
• Both in the input and in the decoded string
• Leading and trailing whitespace are allowed in the output (everything that matches /\s*/)
• Gotta say man, for a first challenge, this is one of the better formatted challenges I've ever seen. As an fyi, the community sandbox is a great place for feedback before posting so you don't get randomly rep bombed for a rule you didn't know. – Magic Octopus Urn Apr 23 '18 at 0:16
• @MagicOctopusUrn. Thank you! Didn't know about sandbox, I'll post there next time :) – user80067 Apr 23 '18 at 0:18
• I mostly use it so people can call me out on duplicate questions, very simple to follow rules, rather hard to know about dupes without memorizing meta :). I'd also recommend checking out the chatrooms, we have chats for almost every language you could hope to learn and questions are encouraged. – Magic Octopus Urn Apr 23 '18 at 0:18
• Great first challenge! Some more test cases would be neat. – Lynn Apr 23 '18 at 16:33
• Really nice first challenge. Had fun playing with this one. – ElPedro Apr 23 '18 at 19:18

# brainfuck, 76 71 65 bytes

-6 bytes thanks to Nitrodon!

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


Try it online!

Feels weird beating Python...

• Not if the 8 keywords in BF were as long as their Python counterpart. – user202729 Apr 23 '18 at 4:47
• It helps that BF automatically converts the unicode. – LastStar007 Apr 23 '18 at 9:49
• 65 bytes – Nitrodon Apr 23 '18 at 17:35

# Stax, 15 11 bytes

ó║¥U⌂½íèäöñ


Run and debug it at staxlang.xyz!

Quick 'n' dirty approach. Working on improving it. Improved it!

### Unpacked (13 bytes) and explanation

2:/8/{{[Im:bm
2:/              Split at index 2. Push head, then tail.
8/            Split into length-8 segments.
{      m    Map block over each segment:
{  m         Map block over each character:
[             Copy first two elements (below) in-place.
I            Index of character in first two characters.
:b       Convert from binary.
Implicit print as string.

• Ahhhh... I knew this would beat us. – Magic Octopus Urn Apr 23 '18 at 0:29

# JavaScript (Node.js), 67 bytes

s=>s.replace(/./g,x=(c,i)=>(x=x*2|c==s[1],Buffer(i<3|i&7^1?0:[x])))


Try it online!

### How?

We use two different syntaxes of the Buffer constructor:

• Buffer([n]) generates a buffer containing the sole byte n and is coerced to the corresponding ASCII character. Only the 8 least significant bits of n are considered.
• Buffer(n) generates a buffer of n bytes. Therefore, Buffer(0) generates an empty buffer, which is coerced to an empty string.

Note: They both are deprecated in recent Node versions. Buffer.from([n]) and Buffer.alloc(n) should be used instead.

### Commented

s =>                   // given the input string s
s.replace(/./g, x =  // initialize x to a non-numeric value (will be coerced to 0)
(c, i) => (        // for each character c at position i in s:
x = x * 2 |      //   shift x to the left
c == s[1],   //   and append the new bit, based on the comparison of c with s[1]
Buffer(          //   invoke the constructor of Buffer (see above):
i < 3 |        //     if i is less than 3
i & 7 ^ 1 ?    //     or i is not congruent to 1 modulo 8:
0            //       replace c with an empty string
:              //     else:
[x]          //       replace c with the ASCII char. whose code is the LSB of x
)                //   end of Buffer constructor
))                   // end of replace(); return the new string


## bash, 5958 52 bytes

tr -t "$1" 01 <<<$1|cut -c3-|fold -8|sed 'i2i
aP'|dc


Try it online!

Thanks to Cows quack for saving 6 bytes.

This challenge works remarkably well with a series of coreutils (and dc to do the conversion and output at the end). First, we use

tr -t "$1" 01 <<<$1


to transliterate the two characters in the input to zeroes and ones. The -t flag truncates the first argument to the length of the second, so this reduces to transliterating the first two characters in the input to 0 and 1, which is what we want. Then,

cut -c3-


removes the first two characters, and

fold -8


outputs 8 of the characters per line. Finally, the sed command turns each line into a dc snippet that reads the number as binary and outputs that byte.

• Always nice to see a bash answer :) You can use sed to simplify the dc calculations by converting each line to dc code that prints each character out and then eval'ing it in dc tio.run/##S0oszvj/… (and the space after cut -c can be removed) – Cows quack Apr 23 '18 at 12:49

# Z80 machine code on an Amstrad CPC, 32 31 30 bytes

000001  0000  (9000)        ORG &9000
000002  9000  EB            EX DE, HL
000003  9001  46            LD B, (HL)
000004  9002  23            INC HL
000005  9003  5E            LD E, (HL)
000006  9004  23            INC HL
000007  9005  56            LD D, (HL)
000009  9006  1A            LD A, (DE)
000010  9007  05            DEC B
000011  9008  13            INC DE
000012  9009  4F            LD C, A
000014  900A                Light
000015  900A  26 01         LD H, &01
000016  900C                Last
000017  900C  13            INC DE
000018  900D  05            DEC B
000019  900E  C8            RET Z
000021  900F                Loop
000022  900F  1A            LD A, (DE)
000023  9010  B9            CP C
000024  9011  28 01         JR Z, Lable
000025  9013  37            SCF
000026  9014                Lable
000027  9014  ED 6A         ADC HL, HL
000028  9016  30 F4         JR NC, Last
000029  9018  7D            LD A, L
000030  9019  CD 5A BB      CALL &BB5A
000032  901C  18 EC         JR Light


The code takes the instruction replace each character with 0 if that character is the same as the first character of the original string, and with 1 otherwise literally and doesn't ever bother to check that a character matches the second character in the input string. It just checks for same-as-first-character and different-from-first-character.

I ran out of registers (the Z80 only has 7 easily usable 8-bit registers, the rest need longer instructions) so I put &01 in H, along with using L to build up the ASCII character (I just realised it's unnecessary to initialise L, saving one byte). When H overflows into the Carry flag, the character in L is ready to be output. Luckily, there is a 16-bit ADC (Add with Carry) that does the job of a left-shift instruction.

(DE) can only be read into A although (HL) can be read into any 8-bit register, so it was a compromise which one to use. I couldn't compare (DE) with C directly, so I had to load one into A first. The labels are just random words that start with L (a requirement of the assembler).

• A the Accumulator - the only register that can do comparisons
• B the counter register for the instruction DJNZ: Decrement (B) and Jump if Non Zero. By rearranging the code, I was able to do the job of DJNZ with one fewer byte
• C the first character in the input string
• D, E as DE the address of the current input character
• H the carry trigger (every 8th loop)
• L the output character being built up

# 05AB1E, 10 bytes

¦¦Sk8ôJCçJ


Try it online!

-3 thanks to emigna.

Ù             # Unique letters, in order they appear.
v            # For each...
yN:         # Push letter and index, replace in input.
}        # End loop.
¦¦      # Remove first x2.
8ô    # Split into eighths.
C   # Convert to integer.
ç  # Convert to char.
J # Join together entire result.


# J, 17 13 Bytes

u:_8#.\2}.1{=


-4 thanks to FrownyFrog

Old version:

u:_8#.\2&({.i.}.)


### Explanation:

u:_8#.\2}.1{=
=  | Self classify, for each unique element x of y, compute x = y, element-wise
1{   | Second row
2}.     | Drop 2
_8#.\        | Convert non-intersecting subarrays of length 8 from binary
u:             | Convert to characters


Examples:

   = 'ABAABBAAAAAABBAAABAABBAABA'
1 0 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 1 1 0 1
0 1 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 0

2}.1{= 'ABAABBAAAAAABBAAABAABBAABA'
0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 0

_8#.\2}.1{= 'ABAABBAAAAAABBAAABAABBAABA'
48 49 50

u:_8#.\2}.1{= 'ABAABBAAAAAABBAAABAABBAABA'
012

• 2}.1{= to save 4 bytes. – FrownyFrog Apr 22 '18 at 23:45
• Oh my, tied... I can't find another byte. – Magic Octopus Urn Apr 23 '18 at 0:13
• @MagicOctopusUrn it's actually a snippet, it should have a [: at the start :) – FrownyFrog Apr 23 '18 at 2:48

# Python 2, 77 bytes

lambda s:[chr(int(map(s.find,s)[i:i+24:3],2))for i in range(7,3*len(s),24)]


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# R, 71 bytes

function(s)intToUtf8(2^(7:0)%*%matrix((y=utf8ToInt(s))[-1:-2]==y[2],8))


Try it online!

Surprisingly golfy!

First, converts the string to ascii code-points with utf8ToInt, saving it as y. Removing the first two characters with negative indexing is shorter than using tail.

The array y[-1:-2]==y[2] is equivalent to the bits when %*% (matrix multiplication) is applied, but first we reshape that array into a matrix with nrow=8, converting from a linear array to byte groupings. Fortuitously, we can then convert to the ascii code points using matrix multiplication with the appropriate powers of 2, 2^(7:0), and then we convert the code points back to a string with intToUtf8.

# Python 3, 77 bytes

a,b,*r=input();x=i=0
for c in r:i*=2;i|=a!=c;x+=1;x%8or print(end=chr(i&255))


Try it online!

# PHP, 73 71 bytes

while($s=substr($argn,-6+$i+=8,8))echo~chr(bindec(strtr($s,$argn,10)));  Run as pipe with -nR or try it online. golfings: • start index at -6 and pre-increment by 8 • exploit that strtr ignores excessive chars in the longer parameter (no substr needed) • translating to 10 and then inverting needs no quotes -> -1 byte • invert character instead of ascii code --> ~ serves as word boundary -> -1 byte. • At least you should match brainfuck: for(;$s=substr($argn,2+8*$i++,8);)echo~chr(bindec(strtr($s,$argn,10))); – Christoph Apr 23 '18 at 11:00
• @Christoph I like how Brainfuck is suddenly a standard for reasonable answer length. – Nit Apr 23 '18 at 13:18

# Pyth, 20 9 bytes

CittxLQQ2


Saved 11 bytes thanks to FryAmTheEggman.

Try it here

### Explanation

CittxLQQ2
xLQQ    Find the index of each character in the string.
tt        Exclude the first 2.
i      2   Convert from binary.
C           Get the characters.

• @FryAmTheEggman Thanks. Evidently I've still got a lot to learn about Pyth. – Mnemonic Apr 23 '18 at 15:52
• Haha, so do I! It's a very intricate golfing language. I hope you continue golfing in it :) – FryAmTheEggman Apr 23 '18 at 16:08

# Ruby, 82 79 bytes

->s{s[2..-1].tr(s[0,2],'01').chars.each_slice(8).map{|s|s.join.to_i(2).chr}*''}


Try it online!

• Welcome to PPCG! I didn't see that there was already an answer in Ruby before i posted mine, but some typical golfing tricks apply to your approach too - e.g., the last .join can be replaced by *'', and s[0..1] by s[0,2]. – Kirill L. Apr 23 '18 at 9:26

# Japt, 11 bytes

¤£bXÃò8 ®Íd


Try it

## Explanation

¤               :Slice from the 3rd character
£  Ã           :Map over each X
bX            :  Get the first 0-based index of X in the input
ò8         :Split to an array of strings of length 8
®       :Map
Í      :  Convert from base-2 string to base-10 integer
d     :  Get the character at that codepoint

• Very clever use of the s2 shortcut, nice. – Nit Apr 23 '18 at 12:35

## PHP + GNU Multiple Precision, 63 61

<?=gmp_export(gmp_init(substr(strtr($argn,$argn,"01"),2),2));


sadly the GMP extention is not default activated (but shipped).

Run like this:

echo "ABABABAAAAABABAAAAAABAABBAABAAAABBABAAABBB" | php -F a.php

• <?= saves 2 bytes and possibly the day. ;-) – Titus Apr 23 '18 at 12:46
• @Titus yeah but sadly it doesn't work with -R (I tried). – Christoph Apr 23 '18 at 13:22
• try -F instead – Titus Apr 23 '18 at 13:27

f[_,_]=""
f(z:o:s)=toEnum(sum[2^b|(b,c)<-zip[7,6..0]s,c==o]):f(z:o:drop 8s)


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# Java 8, 143142 141 bytes

s->{char i=47;for(;++i<50;)s=s.replace(s.charAt(i%2),i);for(i=2;i<s.length();)System.out.print((char)Long.parseLong(s.substring(i,i+=8),2));}


-1 byte thanks to @OlivierGrégoire.

Try it online.

Explanation:

s->{                            // Method with String parameter and no return-type
char i=47;                    //  Index character, starting at 47
for(;++i<50;)                 //  Loop 2 times
s.replace(s.charAt(i%2),i)  //   Replace first characters to 0, second characters to 1
for(i=2;i<s.length();)        //  Loop i from 2 upwards over the String-length
System.out.print(           //   Print:
(char)                     //    As character:
Long.parseLong(           //     Convert Binary-String to number
s.substring(i,i+=8)      //      The substring in range [i,i+8),
,2));}


# Python 3, 99 86 bytes

lambda s:[chr(int(str(list(map(s.find,s[i:i+8])))[1::3],2))for i in range(2,len(s),8)]


Try it online!

Thanks to ASCII-only for basically the whole thing really

# APL+WIN, 30 bytes

Index origin 0. Prompts for input of string

⎕av[2⊥¨(+\0=8|⍳⍴b)⊂b←2↓s≠↑s←⎕]


Explanation:

s≠↑s←⎕ prompts for string and creates binary vector not equal to first character

b←2↓s drops first two elements of binary

(+\0=8|⍳⍴b)⊂ splits binary into groups of 8

2⊥¨ converts each group to decimal

⎕av[...] displays decoded characters

• I assume Quad-AV is in-line with ASCII for APL+WIN? – Zacharý Apr 24 '18 at 22:23
• @Zacharý Yes for the first 128 characters. The special APL characters replace some of the characters in the extended ASCII character set. – Graham Apr 25 '18 at 10:39

# Red, 110 bytes

func[s][t: 0 i: 128 foreach c next next s[if c = s/2[t: t + i]i: i / 2 if i = 0[prin to-char t t: 0 i: 128]]]


Try it online!

## Explanation:

A simple straightforward solution, no builtins.

f: func [s] [                      ; s is the argument (string)
t: 0                           ; total - initially 0
i: 128                         ; powers of 2, initially 0
b: s/2                         ; b is the second charachter
foreach c next next s [        ; for each char in the input string after the 2nd one
if c = b [t: t + i]        ; if it's equal to b than add the power of 2 to t
i: i / 2                   ; previous power of 2
if i = 0 [                 ; if it's 0
prin to-char t         ; convert t to character and print it
t: 0                   ; set t to 0
i: 128                 ; i to 128
]
]
]


=ArrayFormula(Join("",IfError(Char(Bin2Dec(Substitute(Substitute(Mid(A1,3+8*(Row(A:A)-1),8),Left(A1),0),Mid(A1,2,1),1))),""


Input is in cell A1. Google will automatically add ))) to the end of the formula.

Explanation:

• Mid(A1,3+8*(Row(A:A)-1),8) grabs chunks of characters 8 at a time, starting with the third.
• Substitute(Mid(~),Left(A1),0) replaces each instance of the first character with 0.
• Substitute(Substitute(~),Mid(A1,2,1),1) replaces the second character with 1.
• Char(Bin2Dec(Substitute(~))) converts the chunk to decimal and then to ASCII.
• IfError(Char(~,"")) corrects all the errors that result from the fact that Row(A:A) returns far more values than we so Bin2Dec gives us a lot of zero values and Char errors out on zero.
• ArrayFormula(Join("",IfError(~))) joins together all the Char results and ArrayFormula is what makes the Row(A:A) return an array of values instead of just the first value.

# Ruby, 61 42 bytes

-19 bytes thanks to benj2240

->s{[s[2..-1].tr(s[0,2],"01")].pack("B*")}


Try it online!

• pack is an inspired choice, but right now you're sort of going the long way around. It can do even more of the work for you. – benj2240 Apr 23 '18 at 19:13
• Yep, that was a total mind-block from my side... – Kirill L. Apr 23 '18 at 20:53

# Perl 5-lp, 34 bytes

#!/usr/bin/perl -lp
s%.%1-/^$&/%eg;$_=pack'B*',s/..//r


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## REXX, 41 bytes

arg a+2 b
say x2c(b2x(translate(b,01,a)))


Try it online!

• Very cool. Not played with REXX for a long time. – ElPedro Apr 25 '18 at 19:25

# Python 2, 88 bytes

i=input()
f=''.join('10'[x==i[0]]for x in i[2:])
while f:print chr(int(f[:8],2));f=f[8:]


Try it online!

Not the shortest - just an alternative way.

Following version prints the output on one line for 98 bytes although the rules state that trailing whitespace is allowed.:

i=input();f=''.join('10'[x==i[0]]for x in i[2:]);o=""
while f:o+=chr(int(f[:8],2));f=f[8:]
print o


Try it online!

• Final output should be on one line, not three. – idrougge Apr 25 '18 at 13:54
• From OP: "Leading and trailing whitespace are allowed in the output (everything that matches /\s*/)". Newline matches /\s*/. – ElPedro Apr 25 '18 at 14:24
• Sorry, I'm not well-enough versed in regex notation. :/ – idrougge Apr 25 '18 at 14:36
• Neither am I but i Googled it just to be sure ;-) – ElPedro Apr 25 '18 at 14:49

# C# (Visual C# Compiler), 158 bytes

using System.Linq;a=>new string(a.Skip(2).Where((c,i)=>(i-2)%8==0).Select((c,i)=>(char)a.Skip(8*i+2).Take(8).Select((d,j)=>d!=a[0]?1<<7-j:0).Sum()).ToArray())


Try it online!

s/(.)(.)//;$_=pack'B*',eval"y/$1$2/01/r"  Try it online! # Scala, 95 bytes s.substring(2).replace(s(0),'0').replace(s(1),'1').grouped(8).map(Integer.parseInt(_,2).toChar)  Try it online! # Haskell, 124105 93 bytes f(x:_:y)=fromEnum.(/=x)<$>y
g[]=[]
g s=(toEnum.sum.zipWith((*).(2^))[7,6..0])s:g(drop 8s)
g.f


Try it online!

f converts the string to a list of bits by comparing each character to the first one, turning the Bools into zeros and ones with fromEnum. g divides this list into groups of 8, converts them to decimal, and takes the value of the resulting number as an Enum, which Char is an instance of.

### Changes:

• -19 bytes thanks to @Laikoni (removing import, embedding map into function)
• -12 bytes inspired by @Lynn's answer (getting rid of take by zipping with shorter list)
• You can use toEnum instead of chr and drop the import. Also the map can be included into g. The space between 8 s can be removed. – Laikoni Apr 23 '18 at 11:02

# Forth (gforth), 83 bytes

: f over c@ 0 rot 2 do 2* over i 4 pick + c@ <> - i 8 mod 1 = if emit 0 then loop ;


Try it online!

Input is a standard Forth string (address and length) output is printed to stdout

## Explanation

over c@          \ get the value of the first character in the string
0 rot            \ add a starting "byte" value of 0 and put the length on top of the stack
2 do             \ start a loop from 2 to length-1
2*            \ multiply the current byte value by 2 (shift "bits" left one)
over          \ copy the reference char to the top of the stack
i 4 pick +    \ add the index and the starting address to get address of the current char
c@ <>         \ get the char at the address and check if not equal to the reference char
-             \ subtract the value from our bit count, -1 is default "true" value in forth
i 8 mod 1 =   \ check if we are at the last bit in a byte
if            \ if we are
emit 0     \ print the character and start our new byte at 0
then          \ and end the if statement
loop             \ end the loop