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

I think this would have been a nicer challenge without the ASCII conversion step, just taking the (decimal) number after step 2 as input. — xnor
– xnor, Commented May 14, 2017 at 20:37
@xnor, you might be right, and it would be cleaner. I had fun solving this in Octave though, and I hope others will enjoy solving it too :) — Stewie Griffin
– Stewie Griffin, Commented May 14, 2017 at 20:42

Rod · Accepted Answer · 2017-05-15 12:23:15Z

13

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'))

Try it online!

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)

edited May 15, 2017 at 12:23

answered May 14, 2017 at 20:46

Rod

18.6k4 gold badges32 silver badges89 bronze badges

\$\begingroup\$ Nice trick with the eval! Doing the ASCII conversion numerically saves some bytes. \$\endgroup\$

xnor
– xnor

2017-05-14 21:17:40 +00:00
Commented May 14, 2017 at 21:17

Add a comment |

Dennis · Accepted Answer · 2017-05-14 21:49:12Z

7

Jelly, 13 bytes

Oḅ128BŒg;2/ḄS

Try it online!

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.

edited May 14, 2017 at 21:49

answered May 14, 2017 at 21:42

Dennis

212k41 gold badges380 silver badges830 bronze badges

\$\begingroup\$ I've missed that trick of base conversion before. \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2017-05-14 21:45:18 +00:00
Commented May 14, 2017 at 21:45
\$\begingroup\$ Ah, nice trick indeed! \$\endgroup\$

Adnan
– Adnan

2017-05-15 04:30:08 +00:00
Commented May 15, 2017 at 4:30

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

MATL, 14 bytes

YB!'1+0*'XXZBs

Try it online!

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

Adnan · Accepted Answer · 2017-05-14 20:48:43Z

5

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!

edited May 14, 2017 at 20:48

answered May 14, 2017 at 20:36

Adnan

44.7k6 gold badges83 silver badges248 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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).

Try it online!

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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 14, 2017 at 21:43

Jonathan Allan

115k8 gold badges68 silver badges293 bronze badges

Add a comment |

Neil · Accepted Answer · 2017-05-14 23:03:22Z

3

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.

edited May 14, 2017 at 23:03

answered May 14, 2017 at 21:14

Neil

184k12 gold badges76 silver badges290 bronze badges

\$\begingroup\$ My own solution was 97 bytes, too: s=>eval([...s].map(e=>(e.charCodeAt()+128).toString(2).slice(1)).join``.replace(/1+0*/g,'+0b$&')) You can use my replace method to save a byte, I think \$\endgroup\$

Conor O'Brien
– Conor O'Brien

2017-05-14 22:11:14 +00:00
Commented May 14, 2017 at 22:11
1

\$\begingroup\$ @ConorO'Brien More than a byte, I think! \$\endgroup\$

Neil
– Neil

2017-05-14 23:03:37 +00:00
Commented May 14, 2017 at 23:03
\$\begingroup\$ Oo, nice :D \$\endgroup\$

Conor O'Brien
– Conor O'Brien

2017-05-14 23:07:13 +00:00
Commented May 14, 2017 at 23:07

Add a comment |

Etheryte · Accepted Answer · 2018-04-07 08:02:09Z

3

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.

edited Apr 7, 2018 at 8:02

answered Apr 4, 2018 at 22:31

Etheryte

3,8782 gold badges21 silver badges26 bronze badges

\$\begingroup\$ You can use more auto-functions here: òÈ<YÃ can be ò< (with trailing space), and Ën2Ãx can be xn2. You can also use T in place of 0 to save on the comma. (Also, feel free to join us in the Japt chatroom if you ever have questions or would like help with golfing :-) ) \$\endgroup\$

ETHproductions
– ETHproductions

2018-04-06 15:05:13 +00:00
Commented Apr 6, 2018 at 15:05
\$\begingroup\$ @ETHproductions Thanks again, especially for the T trick, didn't know you could (ab)use variables for that, that's very handy. The auto-function xn2 looks a bit weird when compiled, x("n", 2), so I think it will still take a bit before I fully understand the logic behind them. With your help, the Japt solution is now tied for the first place with the Jelly answer. \$\endgroup\$

Etheryte
– Etheryte

2018-04-06 19:11:59 +00:00
Commented Apr 6, 2018 at 19:11
\$\begingroup\$ ETHproductions recently made a shortcut for n2 : Í. It hasn't hit TIO yet, but you can use it here: ethproductions.github.io/japt/?v=1.4.5&code=Y1+k+VQ3w/… \$\endgroup\$

Oliver
– Oliver

2018-04-06 20:25:01 +00:00
Commented Apr 6, 2018 at 20:25
\$\begingroup\$ @Oliver Wow, that's very bleeding edge, that isn't even covered in the interpreter shortcuts reference yet. Thanks a lot! \$\endgroup\$

Etheryte
– Etheryte

2018-04-07 08:03:35 +00:00
Commented Apr 7, 2018 at 8:03

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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

Try it online!

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

\$\begingroup\$ ;/ can be replaced with F. \$\endgroup\$

Dennis
– Dennis

2017-05-14 21:32:08 +00:00
Commented May 14, 2017 at 21:32 — Dennis
– Dennis, Commented May 14, 2017 at 21:32

Jörg Hülsermann · Accepted Answer · 2017-05-14 22:38:27Z

2

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));

Try it online!

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;

Try it online!

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]));

edited May 14, 2017 at 22:38

answered May 14, 2017 at 21:58

Jörg Hülsermann

13.4k3 gold badges16 silver badges36 bronze badges

Add a comment |

Leaky Nun · Accepted Answer · 2017-05-15 01:29:42Z

1

Pyth, 21 bytes

It's too long...

siR2:.BiCMQ128"1+0+"1

Test suite.

answered May 15, 2017 at 1:29

Leaky Nun

50.6k6 gold badges115 silver badges291 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

[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

Try it online!

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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 15, 2017 at 15:25

Brunner

3614 silver badges6 bronze badges

\$\begingroup\$ if open System is the same as C#s using System; then yes you need to include it in the count. If you can do it in F# you could fully qualify whatever the System is for . For example, in C# System.Console... instead of using System;Console... \$\endgroup\$

TheLethalCoder
– TheLethalCoder

2017-05-15 16:27:00 +00:00
Commented May 15, 2017 at 16:27
\$\begingroup\$ @TheLethalCoder It is the same, yes. Also, thanks for clarifying this :) I opted for the "open.." version because it's not only String, but also Convert that live in that namespace. \$\endgroup\$

Brunner
– Brunner

2017-05-15 17:17:20 +00:00
Commented May 15, 2017 at 17:17

Add a comment |

Jordan · Accepted Answer · 2022-10-22 03:05:07Z

1

Ruby, 60 bytes

->s{s.bytes.map{"%07b"%_1}.join.scan(/1+0*/).sum{_1.to_i 2}}

Attempt This Online!

answered Oct 22, 2022 at 3:05

Jordan

11.9k1 gold badge36 silver badges56 bronze badges

Add a comment |

Sean · Accepted Answer · 2017-05-14 21:07:18Z

0

Perl 6, 62 bytes

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

answered May 14, 2017 at 21:07

Sean

8,16612 silver badges16 bronze badges

\$\begingroup\$ use .ords instead of .comb».ord. prefix [~] can often be used instead of .join. comb exists which does the m:g/… thing. /11*0*/ can be shortened to /1+0*/. I came up with {sum map {:2($_)},comb /1+0*/,[~] .ords».fmt('%07b')} \$\endgroup\$

Brad Gilbert b2gills
– Brad Gilbert b2gills

2017-05-14 22:18:14 +00:00
Commented May 14, 2017 at 22:18

Add a comment |

miles · Accepted Answer · 2017-05-15 03:38:26Z

J, 34 bytes

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

Try it online!

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

ZaMoC · Accepted Answer · 2017-05-15 07:44:39Z

0

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)&

edited May 15, 2017 at 7:44

answered May 14, 2017 at 22:02

ZaMoC

25.5k2 gold badges32 silver badges87 bronze badges

\$\begingroup\$ You can save 7 bytes by doing f=FromDigits;l=Flatten; at the start, and then replacing all instances of those two functions with f and l. \$\endgroup\$

numbermaniac
– numbermaniac

2017-05-15 07:37:01 +00:00
Commented May 15, 2017 at 7:37

Add a comment |

protist · Accepted Answer · 2017-05-15 09:07:49Z

0

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

answered May 15, 2017 at 9:07

protist

6303 silver badges7 bronze badges

Add a comment |

NikoNyrh · Accepted Answer · 2017-05-16 11:39:14Z

0

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.

answered May 16, 2017 at 11:39

NikoNyrh

2,72111 silver badges8 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

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.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 28, 2018 at 21:33

ShadowCat

114 bronze badges

1

\$\begingroup\$ Welcome to PPCG! All answers need to be either full programs or callable functions (as opposed to snippets where the input is stored in a hardcoded variable). The function can be unnamed, though, so including a lambda w: would be enough to make your answer valid. \$\endgroup\$

Martin Ender
– Martin Ender

2018-03-28 21:51:00 +00:00
Commented Mar 28, 2018 at 21:51
\$\begingroup\$ Sorry, I probably didn't phrase that well. Your edit is still invalid because a) the input is hardcoded, b) If this is a full program, it doesn't actually print the result. For a full program you'd have to read the input from standard input or a command-line argument, and print the result to standard output. That's why I said it's probably easiest if you submit it as a function by adding lambda w:. \$\endgroup\$

Martin Ender
– Martin Ender

2018-03-30 10:22:05 +00:00
Commented Mar 30, 2018 at 10:22
\$\begingroup\$ Ohhh, Ok, i get it, it would be enough like this: f=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(":"))) \$\endgroup\$

ShadowCat
– ShadowCat

2018-04-04 17:15:34 +00:00
Commented Apr 4, 2018 at 17:15
\$\begingroup\$ yep, that's valid. You don't even need the f=, because we allow unnamed functions (unless you reference the function name for recursive calls). \$\endgroup\$

Martin Ender
– Martin Ender

2018-04-04 21:01:21 +00:00
Commented Apr 4, 2018 at 21:01

Add a comment |

mkst · Accepted Answer · 2018-04-04 21:57:29Z

K (oK), 31 bytes

Solution:

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

Try it online!

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\:'

Uriel · Accepted Answer · 2018-04-08 18:58:41Z

0

APL (Dyalog), 30 bytes

{+/2⊥¨(1∘+⊆⊢)∊¯7↑¨2⊥⍣¯1¨⎕UCS⍵}

Try it online!

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

answered Apr 8, 2018 at 18:58

Uriel

13.3k4 gold badges18 silver badges45 bronze badges

Add a comment |

chunes · Accepted Answer · 2022-10-22 11:17:07Z

0

Factor + `math.unicode`, 73 bytes

[ [ "%07b"sprintf ] map-concat R/ (?<=0)(?=1)/ re-split [ bin> ] map Σ ]

Try it online!

[ "%07b"sprintf ] map-concat Convert each code point in the input string to a 7-bit binary number string and join them all together.
R/ (?<=0)(?=1)/ re-split Split into a sequence of strings down the middle of every 01.
[ bin> ] map Σ Convert each from binary string to decimal integer and sum.

answered Oct 22, 2022 at 11:17

chunes

27.8k3 gold badges32 silver badges55 bronze badges

Add a comment |

Stack Exchange Network

Calculate the binary split sum of a word

21 Answers 21

Python 2, 86 81 76 bytes

Jelly, 13 bytes

How it works

MATL, 14 bytes

Explanation

05AB1E, 18 bytes

05AB1E, 14 bytes

How?

JavaScript (ES6), 97 92 bytes

Japt, 18 12 bytes

Jelly, 16 15 bytes

How?

PHP, 116 Bytes

PHP, 117 Bytes

PHP, 120 Bytes

Pyth, 21 bytes

[F#], 249 245 bytes

Ungolfed

Ruby, 60 bytes

Perl 6, 62 bytes

J, 34 bytes

Explanation

mathematica 193 bytes

J, 40 bytes

Clojure, 150 bytes

Python 123 bytes

K (oK), 31 bytes

APL (Dyalog), 30 bytes

Factor + `math.unicode`, 73 bytes

Your Answer

Hot Network Questions

Calculate the binary split sum of a word

21 Answers 21

Python 2, 86 81 76 bytes

Jelly, 13 bytes

How it works

MATL, 14 bytes

Explanation

05AB1E, 18 bytes

05AB1E, 14 bytes

How?

JavaScript (ES6), 97 92 bytes

Japt, 18 12 bytes

Jelly, 16 15 bytes

How?

PHP, 116 Bytes

PHP, 117 Bytes

PHP, 120 Bytes

Pyth, 21 bytes

[F#], 249 245 bytes

Ungolfed

Ruby, 60 bytes

Perl 6, 62 bytes

J, 34 bytes

Explanation

mathematica 193 bytes

J, 40 bytes

Clojure, 150 bytes

Python 123 bytes

K (oK), 31 bytes

APL (Dyalog), 30 bytes

Factor + math.unicode, 73 bytes

Your Answer

Sign up or log in

Post as a guest

Related

Hot Network Questions

Factor + `math.unicode`, 73 bytes