Element-wise string multiplication

Question

Inspired by this challenge (thanks @cairdcoinheringaahing for the title!), your task is to take two printable ASCII strings and multiply them element-wise with the following rules.

How does it work?

Given two strings (for example split and isbn) you will first, truncate the longer one such that they have equal length and then determine their ASCII codes:

split -> spli -> [115, 112, 108, 105]
isbn  -> isbn -> [105, 115,  98, 110]

The next step will be to map them to the range [0..94] by subtracting 32 of each code:

[115, 112, 108, 105] -> [83, 80, 76, 73]
[105, 115,  98, 110] -> [73, 83, 66, 78]

Now you will multiply them element-wise modulo 95 (to stay in the printable range):

[83, 80, 76, 73] ⊗ [73, 83, 66, 78] -> [74, 85, 76, 89]

Add 32 to get back to the range [32..126]:

[74, 85, 76, 89] -> [106, 117, 108, 121]

And the final step is to map them back to ASCII characters:

[106, 117, 108, 121] -> "july"

Rules

• You will write a program/function that implements the described steps on two strings and either prints or returns the resulting string
• The input format is flexible: you can take two strings, a tuple of strings, list of strings etc.
• The input may consist of one or two empty strings
• The input will be characters in the printable range ([32..126])
• The output is either printed to the console or you return a string
• The output is allowed to have trailing whitespaces

Test cases

"isbn", "split"                  -> "july"
"", ""                           -> ""
"", "I don't matter"             -> ""
"             ", "Me neither :(" -> "             "
"but I do!", "!!!!!!!!!"         -> "but I do!"
'quotes', '""""""'               -> 'ck_iKg'
"wood", "hungry"                 -> "yarn"
"tray", "gzip"                   -> "jazz"
"industry", "bond"               -> "drop"
"public", "toll"                 -> "fall"
"roll", "dublin"                 -> "ball"
"GX!", "GX!"                     -> "!!!"
"4 lll 4", "4 lll 4"             -> "4 lll 4"
"M>>M", "M>>M"                   -> ">MM>"

Note: The quotes are just for readability, in the 6th test case I used ' instead of ".

Answer 1

MATL, 12 bytes

c32-p95\32+c

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Explanation

c      % Implicitly input cell array of 2 strings. Convert to 2-row char matrix.
       % This pads the shorter string with spaces
32-    % Subtract 32, element-wise. Each char is interpreted as its ASCII code.
       % Note that padding spaces will give 0.
p      % Product of each column. Since (padding) spaces have been mapped to 0, the
       % product effectively eliminates those colums. So the effect is the same as
       % if string length had been limited by the shorter one
95\    % Modulo 95, element-wise
32+    % Add 32, element-wise
c      % Convert to char. Implicitly display

Answer 2

Jelly, 15 12 bytes

z⁶O_32P€‘ịØṖ

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-3 thanks to Jonathan Allan.

Answer 3

Python 3, 80 74 71 bytes

lambda*t:''.join(map(lambda x,y:chr((ord(x)-32)*(ord(y)-32)%95+32),*t))

Thanks to @shooqie for golfing off 3 bytes!

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Answer 4

Python 2, 75 70 bytes

-3 bytes thanks to Dennis' suggestion of shooqie's suggestion. -2 bytes thanks to Zacharý's suggestion.

lambda*l:''.join(chr((ord(i)-32)*(ord(j)-32)%95+32)for i,j in zip(*l))

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Answer 5

Haskell, 60 57 bytes

zipWith(\a b->toEnum$f a*f b`mod`95+32)
f=(-32+).fromEnum

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First line is an anonymous function taking two arguments.

This a straight forward implementation of the algorithm: zipWith takes both strings and applies a given function to the pairs of characters. It handles the truncation and also works for empty strings. fromEnum and toEnum are alternatives to ord and chr to switch between characters and their ASCII values which do not need a lengthy import.

Edit: -3 bytes thanks to Bruce Forte.

Answer 6

C++, 331 291 282 270 268 bytes, Version 2 = 178 176 150 148 bytes

Original Version :

#include<string>
#include<algorithm>
#define L length()
#define S std::string
S m(S a,S b){S c;int m=a.L<b.L?a.L:b.L;auto l=[m](S&s){s=s.substr(0,m);std::for_each(s.begin(),s.end(),[](char&c){c-=32;});};l(a);l(b);for(int i=0;i<m;++i){c+=a[i]*b[i]%95+32;}return c;}

-40 bytes thanks to Bruce Forte
-39 bytes thanks to Zacharý

Version 2, inspired by other people's answers

#include<string>
#define L length()
using S=std::string;S m(S a,S b){S c;for(int i=0;i<(a.L<b.L?a.L:b.L);++i)c+=(a[i]-32)*(b[i]-32)%95+32;return c;}

If the first version uses a lambda, it's because i wanted to test C++11 std::async function i just learnt before, so i kept it for no reasons...

More readable version :

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

#define L length()
#define S string

//Function code for the original version
S m(S a,S b) {
    S c;
    int m = a.L < b.L ? a.L : b.L;

    auto l=[m](S&s){
        s = s.substr(0, m);
        for_each(s.begin(),s.end(),[](char&c){
            c -= 32;
        });
    };
    l(a);
    l(b);
    for(int i=0;i<m;++i) {
        c += a[i] * b[i] % 95 + 32;
    }
    return c;
}

//Code for the version 2
S m2(S a,S b) {
    S c;
    for(int i = 0; i < (a.L < b.L ? a.L : b.L); ++i) {
        c += (a[i] - 32) * (b[i] - 32) % 95 + 32;
    }
    return c;
}

int main() {
    string a, b, c;
    getline(cin, a);
    getline(cin, b);
    c = m(a, b);
    cout << c;
}

Answer 7

Dyalog APL, 36 34 33 25 24 bytes

{⎕UCS 32+95|×⌿32-⎕UCS↑⍵}

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Try it online (TIO)!

Input is a list of strings, and has trailing whitespace.

Here's how it works:

{⎕UCS 32+95|×⌿32-⎕UCS↑⍵}
                     ↑⍵ - the input as a 2d array
                 ⎕UCS   - codepoints
              32-       - subtract 32
            ×⌿          - element wise product reduction ([a,b]=>a×b)
         95|            - Modulo 95
      32+               - Add 32
 ⎕UCS                   - Unicode characters

Answer 8

FSharp 275 bytes

let f (p : string, q : string) =     
let l = if p.Length < q.Length then p.Length else q.Length
p.Substring(0,l).ToCharArray() |> Array.mapi (fun i x -> (((int(x) - 32) * (int(q.[i]) - 32)) % 95) + 32) |> Array.map (fun x -> char(x).ToString()) |> Array.fold(+) ""

Answer 9

CJam, 23 bytes

{z{2' e]' f-:*95%' +}%}

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Answer 10

C# (.NET Core),100 96 95 bytes

(l,n)=>{for(int i=0;i<l.Length&i<n.Length;)Console.Write((char)((l[i]-32)*(n[i++]-32)%95+32));}

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-4 bytes thanks to @Zacharý

-1 byte by moving the increment

Uses a lambda and abuses the fact that characters are basically ints.

Answer 11

Mathematica, 114 bytes

(a=Min@StringLength[x={##}];FromCharacterCode[Mod[Times@@(#-32&/@ToCharacterCode/@(StringTake[#,a]&/@x)),95]+32])&

input

["public","toll"]

Answer 12

Stacked, 52 bytes

[,:$#'"!MIN$take"![CS#.toarr]"!32-prod 95%32+#:''#`]

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Function that takes two arguments from the stack.

Explanation

[,:$#'"!MIN$take"![CS#.toarr]"!32-prod 95%32+#:''#`]

Let's look at the first part, assuming the top two items are 'split' and 'isbn':

,:$#'"!MIN$take"!      stack:                      ('split' 'isbn')
,                      pair top two:               (('split' 'isbn'))
 :                     duplicate:                  (('split' 'isbn') ('split' 'isbn'))
  $#'                  length function literal:    (('split' 'isbn') ('split' 'isbn') $#')
    "!                 execute on each:            (('split' 'isbn') (5 4))
      MIN              obtain the minimum:         (('split' 'isbn') 4)
         $take         "take" function literal:    (('split' 'isbn') 4 $take)
                       (e.g. `'asdf' 2 take` is `'as'`)
              "!       vectorized binary each:     (('spli' 'isbn'))

This part performs the cropping.

Then:

[CS#.toarr]"!     stack: (('spli' 'isbn'))
[         ]"!     perform the inside on each string
                  string `'spli'`:
 CS               convert to a character string:    $'spli'
   #.             vectorized "ord":                 (115 112 108 105)
     toarr        convert to array:                 (115 112 108 105)
                  (needed for empty string, since `$'' #.` == `$''` not `()`

Then, the last part:

32-prod 95%32+#:''#`  stack: (((115 112 108 105) (105 115  98 110)))
32-                   subtract 32 from each character code:   (((83 80 76 73) (73 83 66 78)))
   prod               reduce multiplication over the array:   ((6059 6640 5016 5694))
        95%           modulus 95:                             ((74 85 76 89))
           32+        add 32:                                 ((106 117 108 121))
              #:      convert to characters:                  (('j' 'u' 'l' 'y'))
                ''#`  join:                                   ('july')

Answer 13

R, 88 bytes

function(r,s,u=utf8ToInt)intToUtf8((((u(r)-32)*(u(s)-32))%%95+32)[0:min(nchar(c(r,s)))])

anonymous function; takes input as two strings; third argument is just to ensure this is a one line function and save some bytes.

The TIO link below returns an array with entries named with the first input.

Try all test cases!

Answer 14

Perl 5, 65 bytes

64 bytes of code + -p flag.

$n=reverse<>;s/./($_=chop$n)&&chr 32+(-32+ord$&)*(-32+ord)%95/ge

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Answer 15

D, 113 bytes

T m(T)(T a,T b){T c;for(int i;i<(a.length<b.length?a.length:b.length);++i)c~=(a[i]-32)*(b[i]-32)%95+32;return c;}

This is a port of HatsuPointerKun's C++ solution, don't forget to upvote them!

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Answer 16

Java 8, 127 115 97 95 bytes

a->b->{for(int i=0;i<a.length&i<b.length;System.out.printf("%c",(a[i]-32)*(b[i++]-32)%95+32));}

Explanation:

Try it here.

a->b->{                       // Method with 2 char-array parameters and no return-type
  for(int i=0;                //  Index-integer, starting at 0
      i<a.length&i<b.length;  //  Loop over both arrays up to the smallest of the two
    System.out.printf("%c",   //   Print, as character:
      (a[i]-32)               //    Current char of `a` minus 32
      *(b[i++]-32)            //    multiplied with current char of `b` minus 32
      %95                     //    Take modulo-95 of that multiplied result
      +32));}                 //    And add 32 again

Answer 17

05AB1E, 16 15 bytes

.BÇ32-`*₃%32+çJ

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-1 for Emigna pointing out ₃ pushes 95.

---

                 # ['hi', 'you']
.B               # [['hi ', 'you']]
  Ç              # [[[104, 105, 32], [121, 111, 117]]]
   32-           # [[[72, 73, 0], [89, 79, 85]]]
      `          # [[72, 73, 0], [89, 79, 85]]
       *         # [[6408, 5767, 0]]
        ₃%       # [[43, 67, 0]]
          32+    # [[75, 99, 32]]
             ç   # [['K', 'c', ' ']]
              J  # ['Kc ']

---

.BÇ32-`*95%žQsèJ

is another.

Answer 18

C#, 166 bytes

using System.Linq;s=>t=>{int e=s.Length,n=t.Length,l=e>n?n:e;return string.Concat(s.Substring(0,l).Select((c,i)=>(char)((((c-32)*(t.Substring(0,l)[i]-32))%95)+32)));}

I'm sure there's a lot of golfing to be done but I don't have time right now.

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Full/Formatted Version:

using System;
using System.Linq;

class P
{
    static void Main()
    {
        Func<string, Func<string, string>> f = s => t =>
        {
            int e = s.Length, n = t.Length, l = e > n ? n : e;

            return string.Concat(s.Substring(0, l).Select((c, i) => (char)((((c - 32) * (t.Substring(0, l)[i] - 32)) % 95) + 32)));
        };

        Console.WriteLine(string.Concat(f("split")("isbn")));

        Console.ReadLine();
    }
}

Answer 19

Python 2, 95 73 bytes

• Thanks @Zacharý for 4 bytes: unwanted brackets removed

lambda x,y:''.join(chr((ord(i)-32)*(ord(j)-32)%95+32)for i,j in zip(x,y))

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Answer 20

Common Lisp, 99 bytes

(lambda(a b)(map'string(lambda(x y)(code-char(+(mod(*(-(#1=char-code x)32)(-(#1#y)32))95)32)))a b))

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Answer 21

Japt, 24 bytes

¬íVq)®®©c -HÃ×u95 +H dÃq

Returns a string with trailing null-chars (\u0000) when the first input is longer than the second.

Try it online! with the -Q flag to show formatted output, including the null-chars.

Run all the test cases using my WIP CodePen.

Answer 22

Charcoal, 30 bytes

Ｆ⌊⟦ＬθＬη⟧℅⁺³²﹪×⁻³²℅§θι⁻³²℅§ηι⁹⁵

Try it online! Link is to verbose version of code. I actually wrote the calcualation as (32 - ord(q)) * (32 - ord(h)) because it avoids consecutive numeric literals but I guess I could have just written (ord(q) - ord(" ")) * (ord(h) - ord(" ")) instead.

Answer 23

Perl 5, 95 bytes

@a=<>=~/(.)/g;@b=<>=~/(.)/g;$#a=$#b if@a>@b;print chr 32+(-32+ord$_)*(-32+ord$b[$i++])%95 for@a

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Explanation:

@a=<>=~/(.)/g;@b=<>=~/(.)/g;  # Split the strings into 2 arrays
$#a=$#b if@a>@b;              # Truncate the first if longer than the second
print chr 32+(-32+ord$_)*(-32+ord$b[$i++])%95 for@a  # Multiply each character

Answer 24

Pip, 19 bytes

(PA$* *(PA@?Zg)%95)

Takes the strings as command-line arguments. Try it online!

Explanation

(PA$* *(PA@?Zg)%95)
                     g is list of args; PA is string of all printable ASCII characters
            Zg       Zip items of g together: result is list of pairs of characters
        PA@?         Find index of each character in PA
       (      )      (Parentheses to get correct operator precedence)
   $* *              Map (fold on *) to the list: multiplies each pair of numbers
               %95   Take list items mod 95
(PA               )  Use those numbers to index into PA again
                     Print the resulting list of chars, concatenated together (implicit)

Answer 25

Factor, 45

[ [ [ 32 - ] bi@ * 95 mod 32 + ] "" 2map-as ]

It's a quotation (lambda), call it with two strings on the stack, leaves the new string on the stack.

As a word:

: s* ( s1 s2 -- ps ) [ [ 32 - ] bi@ * 95 mod 32 + ] "" 2map-as ;

"M>>M" "M>>M" s*      ! => ">MM>"
dup s*                ! => "M>>M"
dup s*                ! => ">MM>"
...

Answer 26

K (oK), 26 bytes

Solution:

`c$32+95!*/-32+(&/#:'x)$x:

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Example:

`c$32+95!*/-32+(&/#:'x)$x:("split";"isbn")
"july"

Explanation:

Evaluation is performed right-to-left:

`c$32+95!*/-32+(&/#:'x)$x: / the solution
                        x: / assign input to variable x
                       $   / pad right to length on left
               (  #:'x)    / count each x (return length of each char list in list)
                &/         / min-over, get the minimum of these counts
           -32+            / subtract 32, this automagically converts chars -> ints
         */                / multiply-over, product of the two lists
      95!                  / modulo 95
   32+                     / add 32 back again
`c$                        / convert to character array

Answer 27

PHP, 112 bytes

for($i=0;$i<min(strlen($a=$argv[1]),strlen($b=$argv[2]));$i++)echo chr((ord($a[$i])-32)*(ord($b[$i])-32)%95+32);

Answer 28

JavaScript (ES6), 89 bytes

Javascript and the curse of the lengthy function names ...

Using currying and the fact that charCodeAt returns NaN when called with an invalid position. There can be trailing nulls in the output.

a=>b=>a.replace(/./g,(c,i)=>String.fromCharCode((z=x=>x.charCodeAt(i)-32)(a)*z(b)%95+32))

Test

var f=
a=>b=>a.replace(/./g,(c,i)=>String.fromCharCode((z=x=>x.charCodeAt(i)-32)(a)*z(b)%95+32))

q=x=>'['+x+']'

;[["isbn", "split"],["", ""],["", "I don't matter"],["             ", "Me neither :("],
["but I do!", "!!!!!!!!!"],['quotes', '""""""'],["wood", "hungry"],["tray", "gzip"],
["industry", "bond"],["public", "toll"],["roll", "dublin"],["GX!", "GX!"],
["4 lll 4", "4 lll 4"],["M>>M", "M>>M"]]
.forEach(([a,b])=>console.log(q(a)+' x '+q(b)+' --> '+q(f(a)(b))))