Calculate the binary split sum of a word

Question

Take a string, s containing printable ASCII-characters as input, and output its "binary split sum". Need an explanation?

How do you get the binary split sum?

We'll use the string A4 as an example in the following explanation.

• Convert the characters to binary, treating each letters as a 7-bit ASCII character

  A -> ASCII 65 -> 1000001
  4 -> ASCII 52 -> 0110100
  

• Concatenate the binary numbers into a new binary number

  A4 -> 1000001 & 0110100 -> 10000010110100
  

• Split the new binary number into chunks, where no 1 can have a 0 to its left. You should not split consecutive 1s.

  10000010110100 -> 100000, 10, 110, 100
  

• Convert these binary numbers to decimal

  100000, 10, 110, 100 -> 32, 2, 6, 4
  

• Take the sum of these numbers:

  32 + 2 + 6 + 4 = 44
  

So, the output for the string A4 should be 44.

---

Test cases:

a
49

A4
44

codegolf
570

Hello, World!
795

Answer 1

Python 2, 86 81 76 bytes

^{-5 bytes thanks Adnan
-5 bytes thanks xnor}

s=0
for c in input():s=s*128+ord(c)
print eval(bin(s).replace('01','0+0b1'))

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for c in input():s=s*128+ord(c) to do the ASCII conversion numerically, where *128 is used to left shift s 7 times (steps 1 and 2)
eval(('0'+new_bin).replace('01','0+0b1')) to split and sum (steps 3, 4 and 5)

Answer 2

Jelly, 13 bytes

Oḅ128BŒg;2/ḄS

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How it works

Oḅ128BŒg;2/ḄS  Main link. Argument: s (string)

O              Ordinal; map characters to their code points.
 ḅ128          Unbase 128; convert the resulting list from base 128 to integer.
     B         Binary; Convert the resulting integer to base 2.
      Œg       Group consecutive, equal bits.
        ;2/    Concatenate all non-overlapping pairs.
           Ḅ   Unbinary; convert from base 2 to integer.
            S  Take the sum.

Answer 3

MATL, 14 bytes

YB!'1+0*'XXZBs

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Explanation

Consider input 'A4' as an example.

YB        % Implicit input. Convert to binary using characters '0' and '1'. 
          % Gives a char matrix, where each row corresponds to a number
          % STACK: ['1000001'; '0110100']
!         % Transpose. This is necessary because MATL uses column-major 
          % order when linearizing a matrix into a vector
          % STACK: ['10'; '01'; '01'; '00'; '01'; '00'; '10']
'1+0*'    % Push this string: regexp pattern
          % STACK: ['10'; '01'; '01'; '00'; '01'; '00'; '10'], '1+0*'
XX        % Regexp. Linearizes the first input into a row (in column-major
          % order), and pushes a cell array of substrings that match the
          % pattern given by the second input
          % STACK: {'100000'; '10'; 110'; '100'}
ZB        % Convert each string into a decimal number. Gives numeric vector
          % STACK: [32; 2; 6; 4]
s         % Sum. Implicitly display
          % STACK: 44

Answer 4

05AB1E, 18 bytes

Code:

Çžy+b€¦JTR021:2¡CO

Explanation:

Ç                   # Take the ASCII value of each character
 žy+                # Add 128 to each value (to pad each with enough zeros)
    b               # Convert to binary
     €¦             # Remove the first character
       J            # Join the array
        TR021:      # Replace 01 by 021
              2¡    # Split on the number 2
                C   # Convert from binary to decimal
                 O  # Sum them all up

Uses the 05AB1E encoding. Try it online!

Answer 5

05AB1E, 14 bytes

Çžy+b€¦Jγ2ôJCO

A port of my Jelly answer, using the 128 offset from Adnan's 05ab1e answer (rather than the 256 in the Jelly answer I wrote).

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How?

Çžy+b€¦Jγ2ôJCO
Ç              - to ordinals
   +           - add
 žy            - literal 128
    b          - to binary
     €         - for each
      ¦        -   dequeue
       J       - join
        γ      - group into chunks of equal elements
          ô    - split into chunks of
         2     - literal 2
           J   - join
            C  - from binary
             O - sum

Answer 6

JavaScript (ES6), 97 92 bytes

s=>eval(s.replace(/./g,c=>(128+c.charCodeAt()).toString(2).slice(1)).replace(/1+/g,'+0b$&'))

Edit: Saved 5 bytes with some help from @ConorO'Brien.

Answer 7

Japt, 18 12 bytes

c_¤ùT7Ãò< xÍ
c_           // Firstly, take the input and map over it as charcodes.
  ¤          // Take the binary representation of each item
   ùT7       // and left-pad it with zeroes to standardize the items.
      Ã      // After all of the above,
       ò<    // partition the result where ever a 0 precedes a 1.
          xÍ // Then sum the numbers from base 2.

Takes input as a single string.
I also tried out the 128 or 256 addition used by other answers, but 0-padding was shorter to use.

Shaved off a whole whopping 6 bytes thanks to ETHproductions and Oliver.

Try it out here.

Answer 8

Jelly, 16 15 bytes

-1 byte thanks to Dennis (no need to flatten by 1 when a full flatten is fine - replace ;/ with F)

O+⁹Bṫ€3FŒg;2/ḄS

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How?

O+⁹Bṫ€3FŒg;2/ḄS - Main link: list of characters, s    e.g. "A4"
O               - cast to ordinal (vectorises)        [65,52]
  ⁹             - literal 256
 +              - add (vectorises)                    [321, 308]
   B            - convert to binary (vectorises)      [[1,0,1,0,0,0,0,0,1],[1,0,0,1,1,0,1,0,0]]
    ṫ€3         - tail €ach from index 3              [[1,0,0,0,0,0,1],[0,1,1,0,1,0,0]]
       F        - reduce with concatenation           [1,0,0,0,0,0,1,0,1,1,0,1,0,0]
        Œg      - group runs of equal elements        [[1],[0,0,0,0,0],[1],[0],[1,1],[0],[1],[0,0]]
          ;2/   - pairwise reduce with concatenation  [[1,0,0,0,0,0],[1,0],[1,1,0],[1,0,0]]
             Ḅ  - convert from binary (vectorises)    [32,2,6,4]
              S - sum                                 44

Answer 9

PHP, 116 Bytes

for(;$c=ord($argn[$i++]);)$r.=sprintf("%07b",$c);$t=mb_split("(?<=0)(?=1)",$r);echo array_sum(array_map(bindec,$t));

Online Version

PHP, 117 Bytes

for(;$c=ord($argn[$i++]);)$r.=sprintf("%07b",$c);$t=preg_split("#0\K(?=1)#",$r);echo array_sum(array_map(bindec,$t));

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PHP, 120 Bytes

for(;$c=ord($argn[$i++]);)$r.=sprintf("%07b",$c);preg_match_all("#1+0+#",$r,$t);foreach($t[0]as$b)$s+=bindec($b);echo$s;

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or

for(;$c=ord($argn[$i++]);)$r.=sprintf("%07b",$c);preg_match_all("#1+0+#",$r,$t);echo array_sum(array_map(bindec,$t[0]));

Answer 10

Pyth, 21 bytes

It's too long...

siR2:.BiCMQ128"1+0+"1

Test suite.

Answer 11

[F#], 249 245 bytes

open System
let rec c a=function|[]->[a]|'0'::'1'::y->(a+"0")::(c""('1'::y))|x::y->c(a+string x)y
let x i=c""(String.Join("",(Seq.map(fun c->Convert.ToString(int c,2).PadLeft(7,'0'))i))|>Seq.toList)|>Seq.map(fun c->Convert.ToInt32(c,2))|>Seq.sum

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Note: the version on tio.run has "open System" in the header, I've added its count to the code above. I'm not sure what the rules are on imports.

Ungolfed

let rec convert acc = function
    | [] -> [acc]
    | '0'::'1'::xs -> (acc + "0") :: (convert "" ('1'::xs))
    | x::xs -> convert (acc + string x) xs

let calculateSum input =
    let binary = Seq.map (fun x -> Convert.ToString(int x, 2).PadLeft(7, '0')) input

    String.Join("", binary)
    |> Seq.toList
    |> convert ""
    |>Seq.map (fun x -> Convert.ToInt32(x, 2))
    |>Seq.sum

Answer 12

Perl 6, 62 bytes

{sum map {:2(~$_)},.comb».ord».fmt('%07b').join~~m:g/11*0*/}

Answer 13

mathematica 193 bytes

f=FromDigits;l=Flatten;(S=Split@l@Table[PadLeft[IntegerDigits[ToCharacterCode@#,2][[k]],7],{k,StringLength@#}];Plus@@Table[f[RealDigits@f@Join[S[[i]],S[[i+1]]],2],{i,1,Length@S-1,2}]+Last@l@S)&

Answer 14

J, 34 bytes

[:+/@(,#.;.1~1,1=2#.\,)(7#2)#:3&u:

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Explanation

[:+/@(,#.;.1~1,1=2#.\,)(7#2)#:3&u:  Input: array of characters S
                              3&u:  Get ASCII values of each character
                       (7#2)        Array with 7 copies of the value 2
                            #:      Convert each value to a base 2 array with length 7
[:  (                 )             Operate on those binary values
                     ,                Flatten it
                 2  \                 For each infix of size 2
                  #.                    Convert it to decimal from binary
               1=                     Test each value for equality to 1
             1,                       Prepend a 1
      ,                               The flattened binary values
         ;.1~                         Chop that at each occurrence of a 1
       #.                               Convert each chop from binary to decimal
 +/@                                Reduce by addition

Answer 15

J, 40 bytes

+/>#.&.>(1,}.|.1 0 E.|.b)<;.1 b=.,#:a.i.

usage:

+/>#.&.>(1,}.|.1 0 E.|.b)<;.1 b=.,#:a.i.'A4'

returns 44

Answer 16

Clojure, 150 bytes

#(loop[[c & C](for[i % j[64 32 16 8 4 2 1]](mod(quot(int i)j)2))p 0 r 0 R 0](if c(if(=(dec c)p 0)(recur C c 1(+ R r))(recur C c(+(* 2 r)c)R))(+ R r)))

Well I was hoping the conversion from ASCII to bytes was shorter than this. The actual loop body is quite short, using r to accumulate the current result and R to accumulate the total result. If the previous bit p is 0 and the current bit c is 1 then we split a new chunk and accumulate to R, otherwise we update the r and keep R as it was.

Answer 17

Python 123 bytes

lambda w:sum(map(lambda x:int(x,2),"".join(map(lambda x:bin(ord(x))[2:].zfill(7),list(w))).replace("01","0:1").split(":")))

Updated, thanks to Martin Ender.

Answer 18

K (oK), 31 bytes

Solution:

+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$

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Examples:

+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$,"a"
49
+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$"A4"
44
+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$"codegolf"
570
+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$"Hello, World!"
795

Explanation:

Convert to ASCII values, convert to 7-bit binary, flatten, find where differs, and against original list to find where 1s differ. Cut at these indices, convert back to decimal and sum up:

+/2/'_[;x]&x&~~':x:,/(7#2)\'`i$ / the solution
                            `i$ / convert to integer
                     (7#2)      / draw from 2, 7 times => 2 2 2 2 2 2 2
                          \'    / decode each (convert to binary)
                   ,/           / flatten
                 x:             / save as x
             ~~':               / not-not-each previous (differ)
            &                   / and with
           x                    / x
          &                     / indices where true
     _[;x]                      / projection, cut x at ...
  2/'                           / encode each (convert from binary)
+/                              / sum up

Bonus

Managed a 31 byte version in K4 too, but as there's no TIO for it I'm posting my oK solution.

+/2/:'_[;x]@&x&~~':x:,/1_'0b\:'

Answer 19

APL (Dyalog), 30 bytes

{+/2⊥¨(1∘+⊆⊢)∊¯7↑¨2⊥⍣¯1¨⎕UCS⍵}

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How?

⎕UCS⍵ - Unicodify

2⊥⍣¯1¨ - encode each in binary

¯7↑¨ - and pad to the left with zeros to 7 places

∊ - flatten

1∘+⊆⊢ - partition by self increased by one

2⊥¨ - decode each from binary

+/ - sum