XOR multiplication

Question

You goal is to implement the operation of XOR (carryless) multiplication, defined below, in as few bytes as possible.

If we think of bitwise XOR (^) as binary addition without carrying

   101   5
^ 1001   9
  ----  
  1100  12
  
  5^9=12

we can perform XOR multiplication @ by doing binary long-multiplication but doing the adding step without carrying as bitwise XOR ^.

     1110  14
   @ 1101  13
    -----
     1110
       0
   1110
^ 1110 
  ------
  1000110  70
  
  14@13=70

(For mathematicians, this is multiplication in the polynomial ring \$F_2[x]\$, identifying polynomials with natural numbers by evaluating at \$x=2\$ as a polynomial over \$\mathbb Z\$.)

XOR multiplication commutes a@b=b@a, associates (a@b)@c=a@(b@c), and distributes over bitwise XOR a@(b^c)=(a@b)^(a@c). In fact, it is the unique such operation that matches multiplication a@b=a*b whenever a and b are powers of 2 like 1,2,4,8....

Requirements

Take two non-negative integers as input and output or print their XOR-product. This should be as numbers or their decimal string representations, not their binary expansions. Fewest bytes wins.

Don't worry about integer overflows.

Here are some test cases formatted as a b a@b.

0 1 0
1 2 2
9 0 0
6 1 6
3 3 5
2 5 10
7 9 63
13 11 127
5 17 85
14 13 70
19 1 19
63 63 1365

Answer 1

Itr, 9 bytes

BàBC*µRxB

online interpreter

Explanation

            ; implicit input
BàB         ; convert input to bit lists
   C*µRx    ; polynomial multiplication in F2[X]
        B   ; convert bit-list to number
            ; implicit output

Answer 2

Nekomata, 5 bytes

ᵃƂ×Öƃ

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ᵃƂ×Öƃ
ᵃƂ      Convert both inputs to binary digits
  ×     Convolve
   Ö    Modulo 2
    ƃ   Convert back to decimal

Answer 3

Vyxal, 6 bytes

bƒÞƈ∷B

Try it Online! Port of UnrelatedString's Jelly answer, go upvote that!

b      # Binary digits
 ƒÞƈ   # convolve
    ∷  # Take modulo 2
     B # Convert back from binary

Answer 4

golflua 68

x,y=I.r("*n","*n")r=0~@i=0,31r=B.x(r,x*B.ls(B.rs(y,i)%2,i+1))$w(r/2)

Does basically the same bitshifting as Ypnypn's Java answer, but seems to require the divide by 2 at the end to work correctly. Takes in values as stdin, examples below

> 14 13 
70
> 19 1 
19
> 5 17 
85

Answer 5

Ceylon, 90 bytes

alias I=>Integer;I x(I a,I b)=>[for(i in 0:64)if(b.get(i))a*2^i].fold(0)((y,z)=>y.xor(z));

This is just the algorithm as described: multiply a by 2^i wherever the ith bit is set in b, and add them all together using xor. Iterates over 0:64 because Integers are 64-bit in Ceylon when running on JVM (lower when running as Javascript, but then b.get(i) just returns false).

Formatted:

alias I => Integer;

I x(I a, I b) =>
      [
        for (i in 0:64)
            if (b.get(i))
                a * 2^i
      ].fold(0)((y, z) => y.xor(z));

The alias safes here just a single byte.

Answer 6

JavaScript (Node.js), 42 bytes

Port of the Java answer. Go upvote that!

x=>y=>(g=i=>i&&g(--i)^x*((y>>i)%2)<<i)(32)

Try it online!