Decode the string

41

3

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*/)

user80067

Posted 2018-04-22T23:05:14.190

Reputation:

5

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 – 2018-04-23T00:16:39.153

@MagicOctopusUrn. Thank you! Didn't know about sandbox, I'll post there next time :) – None – 2018-04-23T00:18:03.687

2I 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 – 2018-04-23T00:18:28.853

1Great first challenge! Some more test cases would be neat. – Lynn – 2018-04-23T16:33:29.620

Really nice first challenge. Had fun playing with this one. – ElPedro – 2018-04-23T19:18:23.657

Answers

13

brainfuck, 76 71 65 bytes

-6 bytes thanks to Nitrodon!

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

Try it online!

Feels weird beating Python...

Jo King

Posted 2018-04-22T23:05:14.190

Reputation: 38 234

1Not if the 8 keywords in BF were as long as their Python counterpart. – user202729 – 2018-04-23T04:47:54.010

It helps that BF automatically converts the unicode. – LastStar007 – 2018-04-23T09:49:37.953

65 bytes – Nitrodon – 2018-04-23T17:35:45.440

8

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.

Khuldraeseth na'Barya

Posted 2018-04-22T23:05:14.190

Reputation: 2 608

Ahhhh... I knew this would beat us. – Magic Octopus Urn – 2018-04-23T00:29:34.970

6

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.

Magic Octopus Urn

Posted 2018-04-22T23:05:14.190

Reputation: 19 422

1You can use 01‡ instead of the loop. EDIT: or even better: ¦¦Sk8ôJCçJ – Emigna – 2018-04-23T06:33:43.323

Ahhhh... I knew this would beat us. – Khuldraeseth na'Barya – 2019-07-29T15:50:04.557

6

bash, 59 58 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.

Doorknob

Posted 2018-04-22T23:05:14.190

Reputation: 68 138

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 https://tio.run/##S0oszvj/v6RIQbdEQUnFUEnBwFDBxsZGxbAmubREQTfZWLcmLT8nRUHXoqY4NUVBPdMokysxQL0mJfn///8KFQoKFUAMAmAayq9QAAA (and the space after cut -c can be removed)

– user41805 – 2018-04-23T12:49:13.070

6

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

Arnauld

Posted 2018-04-22T23:05:14.190

Reputation: 111 334

6

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

enter image description here

CJ Dennis

Posted 2018-04-22T23:05:14.190

Reputation: 4 104

5

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

Bolce Bussiere

Posted 2018-04-22T23:05:14.190

Reputation: 970

12}.1{= to save 4 bytes. – FrownyFrog – 2018-04-22T23:45:52.880

Oh my, tied... I can't find another byte. – Magic Octopus Urn – 2018-04-23T00:13:55.263

1@MagicOctopusUrn it's actually a snippet, it should have a [: at the start :) – FrownyFrog – 2018-04-23T02:48:46.807

5

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.

Giuseppe

Posted 2018-04-22T23:05:14.190

Reputation: 21 077

5

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)]

Try it online!

maxb

Posted 2018-04-22T23:05:14.190

Reputation: 5 754

4

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.

user48543

Posted 2018-04-22T23:05:14.190

Reputation:

@FryAmTheEggman Thanks. Evidently I've still got a lot to learn about Pyth. – None – 2018-04-23T15:52:21.253

Haha, so do I! It's a very intricate golfing language. I hope you continue golfing in it :) – FryAmTheEggman – 2018-04-23T16:08:25.177

4

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!

ovs

Posted 2018-04-22T23:05:14.190

Reputation: 21 408

4

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.

Titus

Posted 2018-04-22T23:05:14.190

Reputation: 13 814

3At least you should match brainfuck: for(;$s=substr($argn,2+8*$i++,8);)echo~chr(bindec(strtr($s,$argn,10))); – Christoph – 2018-04-23T11:00:34.067

2@Christoph I like how Brainfuck is suddenly a standard for reasonable answer length. – Nit – 2018-04-23T13:18:41.637

3

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

Shaggy

Posted 2018-04-22T23:05:14.190

Reputation: 24 623

Very clever use of the s2 shortcut, nice. – Nit – 2018-04-23T12:35:16.333

3

Java 8, 143 142 141 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));}

Kevin Cruijssen

Posted 2018-04-22T23:05:14.190

Reputation: 67 575

142 bytes – Olivier Grégoire – 2018-04-25T16:23:51.213

3

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!

lfvt

Posted 2018-04-22T23:05:14.190

Reputation: 121

1Welcome 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. – 2018-04-23T09:26:35.030

3

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

Christoph

Posted 2018-04-22T23:05:14.190

Reputation: 1 489

<?= saves 2 bytes and possibly the day. ;-) – Titus – 2018-04-23T12:46:04.190

@Titus yeah but sadly it doesn't work with -R (I tried). – Christoph – 2018-04-23T13:22:49.987

1try -F instead – Titus – 2018-04-23T13:27:13.547

3

Haskell, 75 bytes

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

Try it online!

Lynn

Posted 2018-04-22T23:05:14.190

Reputation: 55 648

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

Jo King

Posted 2018-04-22T23:05:14.190

Reputation: 38 234

Nice, but it fails here.

– None – 2018-04-23T00:16:29.917

86, port of python 2 method. returns as list though but at least it works – ASCII-only – 2018-04-23T00:34:55.380

2

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
        ]
    ]
] 

Galen Ivanov

Posted 2018-04-22T23:05:14.190

Reputation: 13 815

2

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

Graham

Posted 2018-04-22T23:05:14.190

Reputation: 3 184

I assume Quad-AV is in-line with ASCII for APL+WIN? – Zacharý – 2018-04-24T22:23:28.427

@Zacharý Yes for the first 128 characters. The special APL characters replace some of the characters in the extended ASCII character set. – Graham – 2018-04-25T10:39:22.160

2

Ruby, 61 42 bytes

-19 bytes thanks to benj2240

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

Try it online!

Kirill L.

Posted 2018-04-22T23:05:14.190

Reputation: 6 693

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 – 2018-04-23T19:13:28.563

1Yep, that was a total mind-block from my side... – Kirill L. – 2018-04-23T20:53:38.247

2

Google Sheets, 123 bytes

=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.

Engineer Toast

Posted 2018-04-22T23:05:14.190

Reputation: 5 769

2

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!

ElPedro

Posted 2018-04-22T23:05:14.190

Reputation: 5 301

Final output should be on one line, not three. – idrougge – 2018-04-25T13:54:12.817

From OP: "Leading and trailing whitespace are allowed in the output (everything that matches /\s/)". Newline matches `/\s/`. – ElPedro – 2018-04-25T14:24:20.627

1Sorry, I'm not well-enough versed in regex notation. :/ – idrougge – 2018-04-25T14:36:57.917

Neither am I but i Googled it just to be sure ;-) – ElPedro – 2018-04-25T14:49:52.700

2

Perl 5 -lp, 34 bytes

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

Try it online!

Ton Hospel

Posted 2018-04-22T23:05:14.190

Reputation: 14 114

2

REXX, 41 bytes

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

Try it online!

idrougge

Posted 2018-04-22T23:05:14.190

Reputation: 641

Very cool. Not played with REXX for a long time. – ElPedro – 2018-04-25T19:25:16.613

1

Haskell, 124 105 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)

user9549915

Posted 2018-04-22T23:05:14.190

Reputation: 401

2You 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 – 2018-04-23T11:02:16.147

1

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!

Hyarus

Posted 2018-04-22T23:05:14.190

Reputation: 251

1

Perl 5 -p, 40 bytes

s/(.)(.)//;$_=pack'B*',eval"y/$1$2/01/r"

Try it online!

Xcali

Posted 2018-04-22T23:05:14.190

Reputation: 7 671

1

Scala, 95 bytes

s.substring(2).replace(s(0),'0').replace(s(1),'1').grouped(8).map(Integer.parseInt(_,2).toChar)

Try it online!

Shikkou

Posted 2018-04-22T23:05:14.190

Reputation: 161

1

K4, 22 21 bytes

Solution:

"c"$2/:'0N 8#?/0 2_x:

Example:

q)k)"c"$2/:'0N 8#?/0 2_x:"ABAABBAAAAAABBAAABAABBAABA"
"012"

Explanation:

Evaluated right-to-left:

"c"$2/:'0N 8#?/0 2_x: / the solution
                  x:  / save input as x
               0 2_   / cut x at 0 and 2nd index
             ?/       / lookup (?) over (/)
        0N 8#         / reshape into 8-long rows
    2/:'              / decode (/:) each (') from base 2
"c"$                  / cast to character

Notes:

  • -1 byte thanks to ngn!

streetster

Posted 2018-04-22T23:05:14.190

Reputation: 3 635

1x[1]=2_x: -> ?/0 2_ – ngn – 2018-06-02T17:57:17.140

1

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

reffu

Posted 2018-04-22T23:05:14.190

Reputation: 1 361

1

C++, 181 Bytes (VS 2017)

#include<iostream>
int main(int,char**a){int c=7;uint8_t b=0;for(int i=2;a[1][i]!='\0';++i,--c){if(a[1][i]!=a[1][0]){b|=(1<<c);}if(c==0){std::cout<<static_cast<char>(b);c=8;b=0;}}}

Formatted, with better variable names and comments:

#include <iostream>

int main(int argc, char* argv[])
{
  int counter = 7;
  uint8_t byte = 0;

  // Loop until input string is fully read
  for (int i = 2; argv[1][i] != '\0'; ++i, --counter)
  {
    if (argv[1][i] != argv[1][0])
    {
      byte |= (1 << counter); // Shift 2^counter into the current byte
    }

    if (counter == 0)
    {
      std::cout << static_cast<char>(byte); // 
      counter = 8;
      byte = 0;
    }
  }
}

Mario

Posted 2018-04-22T23:05:14.190

Reputation: 11

Welcome to PPCG! – Laikoni – 2018-04-27T22:49:27.030

@Laikoni Thank you :), I will definitely try some more in the future! – Mario – 2018-04-27T22:53:30.547

Have fun! :) You might be interested in the collection of tips for golfing in C++.

– Laikoni – 2018-04-27T22:56:52.017

0

Charcoal, 17 bytes

F⪪E✂θ²Lθ¹⌕θι⁸℅↨ι²

Try it online! Link is to verbose version of code. Explanation:

   ✂θ²Lθ¹           Slice first two characters from input
  E      ⌕θι        Replace each character with its index in the input
 ⪪          ⁸       Split into groups of 8
F                   Loop over each group
              ↨ι²   Convert from binary
             ℅      Convert from code to character
                    Implicitly print

Neil

Posted 2018-04-22T23:05:14.190

Reputation: 95 035

0

Jelly, 8 bytes

ṫ3nḢs8ḄỌ

Try it online!

ṫ3nḢs8ḄỌ
ṫ3        All but first two characters...
  n       does not equal...
   Ḣ      the first character.
    s8    Split into groups of 8.
      Ḅ   Convert from binary (vectorizes).
       Ọ  chr() (vectorizes).

dylnan

Posted 2018-04-22T23:05:14.190

Reputation: 4 993

0

Jelly, 7 bytes

ḢnḊs8ḄỌ

Try it online!

Who needs (tail)?

Coincidentally Because there is only one way to do it, the 4 last bytes are identical to dylnan's answer.


Actually there is another way, providing there are no leading NUL bytes in the output:

Jelly, 7 bytes

ḢnḊḄb⁹Ọ

Try it online!

user202729

Posted 2018-04-22T23:05:14.190

Reputation: 14 620

0

C (gcc), 94 86 bytes (107 for leading whitespace)

Thanks to Jonathan Frech for saving 8 bytes!

f(s,v,i,j)char*s,*v;{for(v=++s,i=j=0;*++s;j&7||(putchar(i),i=0))i|=!(*s-*v)<<(--j&7);}

Try it online!


Original solution:

f(s,v,i,j)char*s,*v;{for(v=++s,i=j=0;*++s;){i|=!(*s-*v)<<(--j&7);if(!(j&7)){putchar(i);i=0;}}}

Try it online!

This assumes that main()'s stripping of whitespace is allowable. Otherwise, I got it to 107 bytes:

f(s,v,i,j)char*s,*v;{while(*s++<33);for(v=s,i=j=0;*++s;){i|=!(*s-*v)<<(--j&7);if(!(j&7)){putchar(i);i=0;}}}

Try it online!

ErikF

Posted 2018-04-22T23:05:14.190

Reputation: 2 149

186 bytes. – Jonathan Frech – 2018-04-23T13:59:47.220

@JonathanFrech that's great! My brain hurts when looking at implicit if's, though. :-) – ErikF – 2018-04-23T23:09:30.540

Implicit if's? You mean the short-circuiting or evaluation? – Jonathan Frech – 2018-04-24T10:23:06.450

Here my take at golfing it j,v,i;f(char*s){for(v=*++s;*++s;--j&7||putchar(i))i=2*i|*s==v;}. – Christoph – 2018-04-25T13:13:23.740

0

Perl 6, 82 bytes 79 bytes

{.[2..*].map({+($^s ne.[0])}).rotor(8)».join».parse-base(2)».chr with .comb}

Shortened with help from Phil H (but his solution is still shorter, so go look at his one)

Try it online!

Joshua

Posted 2018-04-22T23:05:14.190

Reputation: 261

You can shorten the block inside the map by just casting a bool to a number: {+($^s ne.[0])} – Phil H – 2018-04-23T21:35:14.233

0

Perl 6, 59 bytes

{/(..)(.**8)*/;$1>>.trans(~$0=>"01")>>.parse-base(2)>>.chr}

Try it online!

Matches the pattern first, then uses a transliteration to 01 and parses the number.

Phil H

Posted 2018-04-22T23:05:14.190

Reputation: 1 376

0

C (tcc), 134 bytes

f(char*x){
    int a=*x,i,l=strlen(x);
    for(i=2;i<l;)
        x[i] = x[i++] - a ? 49 : 48;
    for(i=10;i<=l;i+=8)
        a=x[i],
        x[i]=0,
        putchar(strtol(x+i-8,0,2)),
        x[i]=a;
}

Try it online!

I am still trying to think of a better way than that ugly swap :P

Marcos

Posted 2018-04-22T23:05:14.190

Reputation: 171

0

Rust, 166 bytes

fn x(x:&str){for i in(2..x.len()-2).step_by(8){print!("{}",u8::from_str_radix(&x[i..i+8].chars().map(|c|(c as u8-17)as char).collect::<String>(),2).unwrap()as char)}}

Playground

Playground link because it requires nightly and tio seems to use stable.

Noskcaj

Posted 2018-04-22T23:05:14.190

Reputation: 421

0

Coconut, 71 bytes

s->k(chr..int$(?,2)..k$(str..s.index),groupsof(8,s[2:]))
k=''.join..map

Try it online!

ovs

Posted 2018-04-22T23:05:14.190

Reputation: 21 408

0

GolfScript, 24 bytes

1>(:x;8/{{x=}%2base}%''+

Try it online!

Explanation:

1>(:x;8/{{x=}%2base}%''+ Full program, implicit input.
                         Stack: "ABAABBAAAAAABBAAABAABBAABA"
1>                       Pop the first character (str[1:])
                         Stack: "BAABBAAAAAABBAAABAABBAABA"
  (:x;                   Pop the next character and assign it to x
                         Stack: "AABBAAAAAABBAAABAABBAABA"; x = 66
      8/                 Split into groups of 8
                         Stack: ["AABBAAAA" "AABBAAAB" "AABBAABA"]; x = 66
        {          }%    Map
                           Stack (first run): "AABBAAAA"; x = 66
         {  }%             Map again
                             Stack (first run): 65; x = 66
          x=                 Compare to x
                             Stack: 0; x = 66
                           Stack: [0 0 1 1 0 0 0 0]; x = 66
                           Well, actually, it is a string of 0 and 1 bytes.
              2base        Convert binary string to number
                           Stack: 48; x = 66
                         Stack: [48 49 50]; x = 66
                     ''+ Convert to string
                         Stack: "012"; x = 66
                         Implicit output

wastl

Posted 2018-04-22T23:05:14.190

Reputation: 3 089

0

bash + bc + xxd, 87 bytes

a="${1:2:${#1}}"
b="${a//${1::1}/0}"
bc<<<"obase=16;ibase=2;${b//${1:1:1}/1}"|xxd -r -p

Usage:

bash foo.bash 'ABAABBAAAAAABBAAABAABBAABA'

Master_ex

Posted 2018-04-22T23:05:14.190

Reputation: 526