Sum of strings (UTF-16 codepoints)

Question

Your function must accept two strings and return new string where are UTF-16 code of all symbols is the sum of UTF-16 codes of inputs symbols

• If the resulting sum is greater than 65535 (maximum UTF-16 codepoint), take the sum modulo 65536

• If there are more symbols in one string then sum with zero, codes of symbols which doesn't have pairs

For example:

• 'a' + 'b' = a(97) + b(98) = 97 + 98 = 195 => 'Ã'
• 'A' + 'B' = A(65) + B(66) = 65 + 66 = 131 => '\u0083' (No Break Here)
• 'Aa' + 'BbC' = [A(65) + B(66)][a(97) + b(98)][(0) + C(67)] => '\u0083ÃC' - \u0083 will not be displayed but it must be here so the length of result string must be 3 not 2

Test cases:

'a', 'b' --> 'Ã'

'A', 'B' --> '\u0083'

'Aa', 'BbC' --> '\u0083ÃC'

'耀', '耀' --> '\u0000' (Null character)

'Программирование', 'Programming' --> 'ѯҲҭҚҲґҩҩҡҮҥвание'

Ungolfed version (check in real time):

const updateResult = () => {
  const string_1 = input1.value;
  const string_2 = input2.value;
  resultString = [];
  
  for (let i = 0; i < string_1.length || i < string_2.length; ++i) {
    resultString.push(
      (string_1.charCodeAt(i) || 0) +
      (string_2.charCodeAt(i) || 0)
    );
  }
  
  resultString = String.fromCharCode(...resultString);
  
  result.innerHTML = resultString;
  resultLength.innerHTML = 'Length: ' + resultString.length;
}

input1.addEventListener('input', updateResult);
input2.addEventListener('input', updateResult);

updateResult();

<input id="input1" value="aA"/>
<input id="input2" value="bB"/>

<br/><br/>

<div id="result"></div>
<div id="resultLength"></div>

The shortest code in each programming language wins!

Answer 1

Excel, 92 89 bytes

=CONCAT(
    TOCOL(UNICHAR(MOD(MMULT(IFERROR(UNICODE(MID(A1:B1,ROW(A:A),1)),),{1;1}),4^8)),2)
)

Inputs in cells A1 and B1.

Thanks to JvdV for the 3-byte save.

Answer 2

Vyxal, 6 bytes

C∑k₴%C

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How it works:

C∑k₴%C
C∑       Sum codepoints of each word
  k₴%    Push 65535 and modulo by each element
     C   Convert codepoints to words

Answer 3

Japt, 11 bytes

Takes input as an array of 2 character arrays, outputs a single character array.

ÕËxc uG²² d

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ÕËxc uG²² d     :Implicit input of 2D array
Õ               :Transpose
 Ë              :Map
  x             :  Sum of
   c            :    Codepoints
     u          :  Modulo
      G         :    16
       ²²       :    Squared twice
          d     :  Character at resulting codepoint

Answer 4

Z80 machine language, 52 bytes

Null-terminated strings, addresses passed in DE and HL; output string address passed in BC. Little-endian.

D5 E5 7E 23 B6 23 EB 28 0A 7E 23 B6 23 EB 20 F4
37 E1 D5 30 03 EB C5 7E 02 23 03 B6 7E 02 23 03
20 F7 E1 1A 47 86 77 13 23 1A 4F 8E 77 13 23 78
B1 20 F2 C9

Disassembled:

f:
    push de         // D5
    push hl         // E5
wl:
    ld a,(hl)       // 7E       // test one wchar of each
    inc hl          // 23
    or (hl)         // B6
    inc hl          // 23
    ex de,hl        // EB
    jr z,fzero      // 28 0A
    ld a,(hl)       // 7E
    inc hl          // 23
    or (hl)         // B6
    inc hl          // 23
    ex de,hl        // EB
    jr nz,wl        // 20 F4
    scf             // 37       // if we have made it here, first was shorter
fzero:
    pop hl          // E1
    pop de          // D5
    jr nc,copy      // 30 03
    ex de,hl        // EB
copy:               //          when we get here, wstrlen(hl)>wstrlen(de)
    push bc         // C5
cl:
    ld a,(hl)       // 7E       // copy a byte
    ld (bc),a       // 02
    inc hl          // 23
    inc bc          // 03
    or (hl)         // B6       // or in the second to see if both 0
    ld a,(hl)       // 7E       // copy the second (preserves flags)
    ld (bc),a       // 02
    inc hl          // 23
    inc bc          // 03
    jr nz,cl        // 20 F7    // if either nonzero, loop again
    pop hl          // E1       // put dest in hl
aloop:
    ld a,(de)       // 1A
    ld b,a          // 47
    add (hl)        // 86
    ld (hl),a       // 77
    inc de          // 13       // *(hl++) += (b = *(de++)
    inc hl          // 23
    ld a,(de)       // 1A
    ld c,a          // 4F
    adc (hl)        // 8E
    ld (hl),a       // 77
    inc de          // 13
    inc hl          // 23       // *(hl++) += (c = *(de++)) + cf
    ld a,b          // 78
    or c            // B1
    jr nz,aloop     // 20 F2    // continue if b|c
    ret             // C9

Answer 5

Python, 89 bytes

lambda x,y:[chr(sum(map(ord,a))%4**8)for a in zip(f'{x:\0<{len(y)}}',f'{y:\0<{len(x)}}')]

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-15 bytes thanks to @SevC_10

Answer 6

05AB1E, 7 bytes

0ζOžH%ç

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Explanation

0ζOžH%ç  # 05AB1E automatically converts the strings
         # to codepoints since they are delimited by '
0ζ       # Transpose with filler 0
  O      # Sum each inner list
   žH    # Push 65536 
     %   # Mod each value by 65536
      ç  # Convert back from codepoint to character

Answer 7

J, 30 27 25 bytes

(65536|+/@,:)&.(3 u:7&u:)

-3 thanks to ovs, and -2 more.

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(65536|+/@,:)&.(3 u:7&u:)
             &.            NB. u&.v applies v to both args and u to the results
                           NB. then applies the inverse of v to the results of u
               (3 u:7&u:)  NB. monadic fork
                    7&u:   NB. converts each char to a single unicode codepoint
                3 u:       NB. convert string to a list of codepoints
(65536|+/@,:)              NB. dyadic fork
          ,:               NB. stacks codepoint lists filling with 0s
       +/@                 NB. then sum the columns of the resulting table
 65536|                    NB. mod each sum by 65536
                           NB. inverse of v converts codepoints to chars

Answer 8

Python 3, 76 bytes

f=lambda a,b:a and b and chr(ord(a[0])+ord(b[0])&65535)+f(a[1:],b[1:])or a+b

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Answer 9

Raku, 61 bytes

{utf16(roundrobin(@_».encode('utf16'))».sum X%4⁸).decode}

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(The older Raku version on TIO requires a ".new" after the first "utf16" for four additional bytes.)

• @_ are the string arguments to the function.
• ».encode('utf16') encodes each of those strings as a utf16 blob object.
• roundrobin combines corresponding elements of those blobs as lists, like zip, but doesn't stop when the end of the shortest input is reached. roundrobin <1 2>, <3 4 5 6> returns (1 3), (2, 4), (5,), (6,).
• ».sum converts each of those zipped lists to its sum.
• X% 4⁸ takes each of those sums modulo 65536. (4⁸ is one byte shorter than 65536.)
• utf16 converts that list of integers to a utf16 blob.
• .decode converts that blob into a string.

Answer 10

Charcoal, 37 bytes

Ｆ²⊞υＳＵＭυ⁺ι×ψ⁻Ｌ§υ¬ιＬι⭆⊟υ℅﹪⁺℅ι℅§ΣυκＸ⁴¦⁸

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

Ｆ²⊞υＳ

Input the two strings.

ＵＭυ⁺ι×ψ⁻Ｌ§υ¬ιＬι

Pad the shorter to the length of the longer with null bytes.

⭆⊟υ℅﹪⁺℅ι℅§ΣυκＸ⁴¦⁸

Pairwise add the character codes of the two strings, reduce module 4⁸, then convert back to characters.

Answer 11

Factor, 38 bytes

[ 0 pad-longest [ + 65536 mod ] 2map ]

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• 0 pad-longest pad the shorter of two sequences with 0s until it is the same length as the longer one
• [ ... ] 2map map over two sequences with a binary function
• + 65536 mod add the code points together modulo 65536

Answer 12

Brachylog, 15 bytes

ạᵐz₁+ᵐ%₆₅₅₃₆ᵐ~ạ

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Explanation

ạᵐ               Map to codes
  z₁             Zip without cycling
    +ᵐ           Map sum
      %₆₅₅₃₆ᵐ    Map mod 65536
             ~ạ  Codes to string

Answer 13

C (gcc) with -fwide-exec-charset=UTF16LE, 78 bytes

Takes inputs in UCS-2 (UTF-16LE) encoding and outputs in UCS-2 (UTF-16LE) encoding. TIO and shells don't seem to like UTF-16: running this from an actual machine and redirecting the output gives the correct results.

Output (in KWrite, UTF-16 encoding):

f(i,j,k)short*i,*j;{for(;*i|*j;*i&&i++,*j&&j++)printf("%2$c%c",k>>8,k=*i+*j);}

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Answer 14

JavaScript, 91 bytes (tsh's method)

With recursion

(a,b)=>(g=i=>a[i]?String.fromCharCode(a[c='charCodeAt'](i)+~~b[c](i))+g(i+1):b.slice(i))(0)

Try it:

f=(a,b)=>(g=i=>a[i]?String.fromCharCode(a[c='charCodeAt'](i)+~~b[c](i))+g(i+1):b.slice(i))(0)

const inputHandler = () => {  
  resultString = f(input1.value, input2.value)
  
  result.innerHTML = resultString;
  resultLength.innerHTML = 'Length: ' + resultString.length;
}

input1.addEventListener('input', inputHandler);
input2.addEventListener('input', inputHandler);

inputHandler();

<input id="input1" value="aA"/>
<input id="input2" value="bB"/>

<br/><br/>

<div id="result"></div>
<div id="resultLength"></div>

JavaScript, 97 bytes

Without recursion

(a,b)=>String.fromCharCode(...[...a].map((s,i)=>(g=c=>c.charCodeAt(j=i)|0)(a)+g(b)))+b.slice(j+1)

Try it:

f=(a,b)=>String.fromCharCode(...[...a].map((s,i)=>(g=c=>c.charCodeAt(j=i)|0)(a)+g(b)))+b.slice(j+1)

const inputHandler = () => {  
  resultString = f(input1.value, input2.value)
  
  result.innerHTML = resultString;
  resultLength.innerHTML = 'Length: ' + resultString.length;
}

input1.addEventListener('input', inputHandler);
input2.addEventListener('input', inputHandler);

inputHandler();

<input id="input1" value="aA"/>
<input id="input2" value="bB"/>

<br/><br/>

<div id="result"></div>
<div id="resultLength"></div>

Answer 15

CJam, 9 bytes

z::i::+:c

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Explanation

z::i::+:c # a function taking an array with two strings [x y]
z         # zip; transpose rows with columns (i.e, ["Aa" "BbC"] => ["AB" "ab" "C"]
 ::i      # map `i`nt convert to all values
    ::+   # map add to all values, effectively summing each
       :c # map each summed value to `c`hars