String to Bit Transition

Question

Task

Given an input string of one or more ASCII characters which codepoints are between 0 and 128 (exclusive), do the following:

1. Convert each character into its 7-bit ASCII code (if the ASCII code is less than 7 bits, put leading zero bits)
2. Concatenate all bits (this results in 7*n bits where n is the number of characters)
3. For each bit in this bitstream, print 1 if it is different from the previous bit, and print 0 otherwise. The first output bit is always 1.

Example

Input:

Hi

Output:

11011001011101

Explanation:

The string "Hi" has the ASCII codes

72 105

which in bits are:

1001000 1101001

And the transition bits indicators:

11011001011101

This is code golf. Lowest byte count wins.

Test Cases

Test case 1:

Hello World!
110110010101110011010101101010110001110000111110000110000001011101101010101100110001

Test case 2:

%% COMMENT %%
1110111111011111100001100010010100001010110101011010011101010011111110011000001101111110111

Test case 3 (credit to Luis Mendo):

##
11100101110010

---

Congrats to Luis Mendo for the shortest solution with 9 bytes in MATL!

Answer 1

Python 2, 58 bytes

n=1
for c in input():n=n<<7|ord(c)
print'1'+bin(n^n/2)[4:]

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Answer 2

MATL, 9 bytes

Hj7&B!hdg

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Explanation

H     % Push 2
j     % Read line of input, unevaluated
7&B   % Convert to binary with 7 bits. Gives a 7-column matrix
!     % Transpose
h     % Concatenate horiontally. The matrix is read in column-major order
d     % Consecutive differences
g     % Convert to logical. Implicitly display

Answer 3

Japt -P, 11 bytes

Takes advantage of the fact that spaces can be coerced to 0 in JavaScript when trying to perform a mathematical or, in this case, bitwise operation on it.

c_¤ù7Ãä^ i1

Try it or run all test cases

c_¤ù7Ãä^ i1     :Implicit input of string
c_              :Map codepoints
  ¤             :  Convert to binary string
   ù7           :  Left pad with spaces to length 7
     Ã          :End map
      ä^        :XOR consecutive pairs
         i1     :Prepend 1
                :Implicitly join and output

Answer 4

Jelly, 12 bytes

O+Ø⁷BḊ€FIA1;

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Answer 5

Octave, 36 30 bytes

Fix thanks to Luis Mendo

-2 bytes thanks to Sanchises

@(a)[1;~~diff(de2bi(a,7)'(:))]

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Answer 6

CJam, 21 bytes

1q{i2b7Te[}%e__(;.^);

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Explanation

Showing the stack with a sample input of 5:

1 q      e# Push 1 and then the whole input: 1 "5"
{
  i      e# Convert to its char code: 1 [53]
  2 b    e# Convert to binary: 1 [[1 1 0 1 0 1]]
  7 T e[ e# Left-pad with 0 to length 7: 1 [[0 1 1 0 1 0 1]]
} %      e# Map this block over every character in the string
e_       e# Flatten array: 1 [0 1 1 0 1 0 1]
_ ( ;    e# Duplicate array and remove its first element: 1 [0 1 1 0 1 0 1] [1 1 0 1 0 1]
. ^      e# Element-wise xor: 1 [1 0 1 1 1 1 1]
) ;      e# Remove and pop the last element of the array: 1 [1 0 1 1 1 1]
         e# Stack implicitly printed: 1101111

To see if a bit is different from the previous bit, we do a vector (element-wise) xor between the bit array and the bit array without the first element. We also remove the last bit of the result, because it is always the last bit of the longer array unchanged.

Answer 7

APL (Dyalog Unicode), 16 bytes^SBCS

Full program. Prompts for string from stdin.

1,2≠/∊1↓¨11⎕DR¨⍞

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⍞ prompt for input ("a quote in a console")

11⎕DR¨ change each character to bit-Boolean Data Representation

1↓¨ drop the first bit from each

∊ ϵnlist (flatten)

2≠/ pairwise difference

1, prepend a one

Answer 8

Dart, 213 168 bytes

f(s,{t,i}){t=s.runes.map((r)=>r.toRadixString(2).padLeft(7,'0')).join().split('').toList();for(i=t.length-1;i>0;i--)t[i]=t[i]==t[i-1]?'0':'1';t[0]='1';return t.join();}

Previous one-liner

f(String s)=>'1'+s.runes.map((r)=>r.toRadixString(2).padLeft(7,'0')).join().split('').toList().reversed.reduce((p,e)=>p.substring(0,p.length-1)+(p[p.length-1]==e?'0':'1')+e).split('').reversed.join().substring(1);

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This verbosity and lack of easy built ins is really killing this one. Still managed to pull a one liner though.

• -45 bytes by not using a one liner and using a for loop

Answer 9

Charcoal, 25 bytes

⭆θ◧⍘℅鲦⁷←Ｗⅈ←Ｉ﹪⍘ＫＤ²←01 ²1

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

⭆θ◧⍘℅鲦⁷←

Convert all the characters to binary and pad them to a length of 7 and then print them, but leave the cursor over the last digit.

Ｗⅈ

Repeat until the cursor is over the first digit.

←Ｉ﹪⍘ＫＤ²←01 ²

Calculate whether the digits are different and overwrite each digit with the difference.

1

Overwrite the first digit with a 1.

Answer 10

PowerShell, 73 56 49 bytes

$args|%{$b=+$_
6..0}|%{+($c-ne($c=($b-shr$_)%2))}

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-17 bytes thanks to mazzy :)

Answer 11

Ruby -p, 68 57 bytes

-11 bytes by shamelessly stealing the method used by xnor's Python solution.

l=1
gsub(/./){l=l<<7|$&.ord}
$_=?1+(l^l/2).to_s(2)[2..-1]

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Original solution:

gsub(/./){'%07b'%$&.ord}
l=p
gsub(/./){b=$&.ord-48;r=l ?l^b:1;l=b;r}

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Answer 12

05AB1E (legacy), 12 bytes

Çb7jð0:¥ÄJ1ì

Uses the legacy version of 05AB1E, since j implicitly joins the strings together, which requires an explicit J after the j in the new version of 05AB1E.

Try it online or verify all test cases.

Explanation:

Ç             # Convert the (implicit) input-string to a list of ASCII code-points
              #  i.e. "Hi#" → [72,105,35]
 b            # Convert each integer to a binary string
              #  → ["1001000","1101001","100011"]
  7j          # Prepend each with spaces to make them length 7,
              # and join everything together to a single string implicitly
              #  → "10010001101001 100011"
    ð0:       # Replace all those spaces with 0s
              #  → "100100011010010100011"
       ¥      # Get the deltas of each pair of 1s/0s
              #  → [-1,0,1,-1,0,0,1,0,-1,1,-1,0,1,-1,1,-1,0,0,1,0]
        Ä     # Get the absolute value of this
              #  → [1,0,1,1,0,0,1,0,1,1,1,0,1,1,1,1,0,0,1,0]
         J    # Join them all together
              #  → "10110010111011110010"
          1ì  # And prepend a 1
              #  → "110110010111011110010"
              # (after which the result is output implicitly)

Answer 13

Husk, 16 15 14 bytes

:1Ẋ≠ṁȯΩo=7LΘḋc

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Answer 14

Vyxal, 20 16 15 bytes

bƛṅS7↳vċ†;f¯ȧ1p

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A mess. -4 thanks to Aaron Miller. -1 thanks to lyxal.

Answer 15

Python 2, 104 bytes

lambda w:reduce(lambda(A,P),C:(A+'10'[P==C],C),bin(reduce(lambda a,c:a*128+ord(c),w,1))[3:],('','x'))[0]

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A quick stab at it.

Answer 16

Stax, 13 12 bytes

ìEÖâU₧(~¬8IE

Run and debug it

If it's guaranteed that all input characters have the 7th bit set, as some answers assume, it can be done in 10 bytes

Answer 17

Kotlin, 182 bytes

var l='6'
fun f(b:String)=b.fold(""){t,i->t+"".a(i.toInt())}.map{if(l==it){l=it;0} else {l=it;1}}
fun String.a(v:Int):String=if(v<=0)"${this}0".reversed() else "${this}${v%2}".a(v/2)

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Hopefully I can improve this soon, I feel like there must be some spots for improvement but I can't think right now

Answer 18

Perl 5 -p, 60 bytes

s/./sprintf'%07b',ord$&/ge;s/.(?=(.))/0|$&^$1/ge;s/^/1/;chop

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Answer 19

C (gcc (MinGW)), 90 bytes

Requires a compiler providing itoa().

n[9],b,c;f(char*s){for(b=*s<64;c=*s++;printf("%07s",itoa((c^c/2)&127,n,2)))c|=b<<7,b=c&1;}

Answer 20

Ruby -p, 50 bytes

gsub(/./){"%07b"%$&.ord}
gsub(/./){$`=~/#$&$/?0:1}

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Explanation

First line, same as Value Ink's answer:

gsub(/./){       $&    }   # Replace each character $&…
                   .ord    # …with its ASCII code…
                %          # …formatted as…
          "%07b"           # …binary digits padded to 7 places.

Second line:

gsub(/./){      $&      }  # Replace each character $&…
          $`               # …if the text to its left…
            =~             # …matches…
              /#  $/       # …the Regexp /c$/ where "c" is the character…
                    ?0:1   # …with 0, or 1 otherwise.

In Ruby you can use interpolation in Regexp literals, e.g. /Hello #{name}/, and for variables that start with $ or @ you can omit the curly braces, so if e.g. $& is "0" then the grawlixy /#$&$/ becomes /0$/.

Answer 21

K (ngn/k), 9 13 bytes

Solution:

~=':,/(7#2)\'

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Explanation:

~=':,/(7#2)\' / the solution
           \' / convert each
      (   )   / do this together
       7#2    / 2 2 2 2 2 2 2
    ,/        / flatten
 =':          / equal to each-previous?
~             / not

Notes:

• +4 bytes to support strings consisting only of 6-bit chars

Answer 22

Emojicode, 263 bytes

🏁🍇🔤🔤➡️🖍🆕s🔂b📇🆕🔡👂🏼❗️❗️🍇🍪s🔪🔡🔢b❗️➕128 2❗️1 7❗️🍪➡️🖍s🍉🔤?🔤➡️🖍🆕p🔂b s🍇↪️b🙌p🍇👄🔤0🔤❗️🍉🙅🍇👄🔤1🔤❗️🍉b➡️🖍p🍉🍉

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Ungolfed:

🏁 🍇  💭 Main code block
    🔤🔤 ➡️ 🖍 🆕 s  💭 Start with s as the empty string
    🔂 b 📇 🆕 🔡 👂🏼  💭 For each byte b in the input ...
    ❗️ ❗️ 🍇
        🍪 s  💭 ... append ...
           🔪 🔡 🔢 b ❗️ ➕ 128  💭 ... b + 128 (this gives the leading zero(s) in case the binary representation of b is shorter than 7 digits) ...

                 2  💭 ... in binary ...
              ❗️
              1 7  💭 ... without the leading one ...
           ❗️
        🍪
        ➡️ 🖍 s  💭 ... to s
    🍉
    🔤?🔤 ➡️ 🖍 🆕 p  💭 This will be used as the previous character, by assigning it neither 0 nor 1 we assure the first bit output is always a one
    🔂 b s 🍇  💭 For each character in s:
        ↪️ b 🙌 p 🍇  💭 If it is the same as the previous character ...
            👄 🔤0🔤 ❗️  💭 ... output a zero ...
        🍉 🙅 🍇  💭  ... else ...
            👄 🔤1🔤 ❗️ 💭 ... output a one
        🍉
        b ➡️ 🖍 p  💭 And the current character becomes the new previous character.
    🍉
🍉

Answer 23

JavaScript (V8), 150 95 bytes

-55 thanks to @dana

x=>[...[...x].reduce((a,c)=>a+c.charCodeAt(0).toString(2).padStart(7,0),"")].map(c=>x!=(x=c)|0)

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Answer 24

Python3.8, 72 bytes

Solution:

lambda a:["10"[a==(a:=x)]for x in"".join(bin(ord(i)+128)[3:]for i in a)]

Explanation:

Ever since Python 3.8 introduced assignment expressions (rather than the standard assignment statements), I have wanted to use them in a list comprehension that needs to remember the last item. This is not the best way to do this but demonstrates an interesting method of using the assignment expression.

The code creates a lambda function which takes the required argument which is the string to convert. When called, the function proceeds as follows. Every character in a is converted to its character code which 128 is added to for dealing with 6-bit characters (the binary representation will always be 8 bits and we can chop off the first bit). This number is converted to binary and the header (0x) and the initial 1 from adding 128 is chopped off. These new strings are then joined into one larger string.

For each character in this new string (which contains the concatenated 7-bit representation of the text), it is checked if the character is the same as the previous character. What happens with the first character? The first result character should always be "1" so we just have to make sure that whatever is in the last character variable is neither "1" nor "0". We do this by reusing the original parameter now that we are not using it anymore. This may be a problem if the original string was a single "0" (a single "1" just happens to work) but we will ignore that.

During the comparison, the previous character was evaluated first so when we use the assignment expression to set the previous character variable to the current character, it does not affect the comparison expressions' evaluation.

The comparison either produces True or False which can also be used as 1 or 0 respectively in Python, so they are used to look up either a "1" or "0" in a string

Answer 25

Tcl, 215 167 140 bytes

{{s {B binary} {X ~$w/64}} {join [lmap c [split $s {}] {$B scan $c c w;$B scan [$B format i [expr 2*$w^$w^$X<<7]] B7 r;set X $w;set r}] ""}}

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Uses shift-by-one and exclusive-or to detect transitions. Carries lsb of current character to msb of next character. Combines output for each character by joining list returned by lmap.

Uses lambdas with default arguments to save bytes on initialization and repeated commands.

Relies heavily on order of operation. Works for empty string.

Answer 26

Haskell, 137 bytes

import Data.Char
b 0=[]
b n=odd n:b(n`div`2)
d x|x='1'|1<2='0'
c=('1':).map d.(zipWith(/=)<*>tail).concatMap(reverse.take 7.b.(+128).ord)

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The biggest problem here is converting booleans (result of the XOR) to '0'/'1'.

Answer 27

Python 3, 88 84 bytes

l=2;s=''
for c in''.join(f'{ord(c):07b}'for c in input()):s+='01'[l!=c];l=c
print(s)

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I feel that the assignments should be avoidable, but couldn't think of any way to do that.

Update:

• -4 bytes thanks to movatica

Answer 28

PHP, 90 bytes

for(;$c=ord($argn[$j++]);$o.=sprintf('%07b',$c));for(;($q=$o[$i++])>'';$p=$q)echo$p!=$q|0;

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Answer 29

C# (Visual C# Interactive Compiler), 80 bytes

s=>{for(int p,c=2,i=0;i<7*s.Length;Write(p==c?0:1))(p,c)=(c,1&s[i/7]>>6-i++%7);}

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Answer 30

JavaScript (V8), 73 bytes

s=>[...c=s.repeat(7)].map((_,i)=>(c=1&s.charCodeAt(i/7,p=c)>>6-i%7)!=p|0)

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