New answers tagged abstract-algebra
0
05AB1E (legacy), 32 bytes
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Input as two integers where the trailing bit of its binary representation represents the negative (0) or positive (1) (i.e. input 24 is -12/-1100 and input 25 is 12/1100).
Try it online or verify some more test cases.
Explanation:
b # Convert both integers in the (implicit) ...
1
Ruby, 98 ... 62 bytes
->c,w,d,x{[c^d,w*x*(~k=0)**((0..c+d).sum{|x|c[x]*k+=d[x-1]})]}
Try it online!
Input: [abs(a),sign(a),abs(b),sign(b)]
Output: [abs,sign]
Thanks Arnauld for the idea of splitting input and for shaving a couple of bytes.
How it works:
The sign is determined by the number of hops: example: e3*(e1e2) = -e1e3*(e2) = (-e1)*(-e2e3) = ...
0
JavaScript (ES6), 73 65 bytes
Expects (a)(b), two integers where the least significant bit holds the sign (\$0\$ for positive, \$1\$ for negative) and all other bits are used to store the absolute value.
Returns the product \$c\$ in the same format.
a=>b=>a^b^(g=a=>(b/=2)&&(h=n=>b&!!n&&!h(n&~-n))(a)^g(a>>1))(a>&...
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