4 ms·

One thing I love about xor is an interesting correspondence between bitwise xor and the outer product of the exterior algebra. Say we have an N=4 dimensional v

One thing I love about xor is an interesting correspondence between bitwise xor and the outer product of the exterior algebra. Say we have an N=4 dimensional vector space, and we use binary place values to represent units vector in a basis, like this: w = 1000 x = 0100 y = 0010 z = 0001 Then a multivector basis could be represented e.g.: xyz = 0111 xy = 0110 wz = 1001 Now, if ^ is the outer product: xy^yz = xz ↔ 0110 xor 0011 = 0101. wx^yz = wxyz ↔ 1100 xor 0011 = 1111. and so on. If anyone has more information about this, I'd be really interested in seeing it!

taneq 10y ago
I had a similar 'aha' moment a few years ago about inner product (aka dot product) and the choice of '.' as an operator to access elements of a structure. struct Example { int elem1; int elem2; ... }; Example ex1 = { 1, 2, ... }; int a = ex1.elem1; int b = ex1.elem2; ... If we think of ex1 as a vector, and elem1 as a constant vector [1, 0, 0...] (and elem2 as a constant vector [0, 1, 0, ...] and so on) then ex1.elem1 is literally the inner product of ex1 and the constant elem1. I have no idea if that was the original thinking behind dot notation but it's neat and I like it.
knappa 10y ago
Unless you mean something different from the usual multilinear algebra meaning of "exterior algebra", xy^yz is zero. (It contains two y's.) I think that you must actually mean Clifford algebras. (But there are also signs when in characteristic ≠ 2.)
- macawfish 10y ago
  Yeah, I do mean Clifford algebras. Thanks for the catch. I'll have to study up on the distinction between "outer products" in Clifford algebras and exterior algebras.
  - johncolanduoni 10y ago
    They are isomorphic as vector spaces (again with characteristic != 2), but their products are not preserved by the isomorphism (unless the Clifford algebra is trivial).
halma 10y ago
I remember this and other tricks are implemented in Gaigen (a code generator for geometric algebra). https://sourceforge.net/projects/g25/?source=directory https://sourceforge.net/projects/g25/?source=directory The XOR trick is close to slide 16 of http://www.science.uva.nl/research/ias/ga/gaigen/files/20020219_talk.pdf http://www.science.uva.nl/research/ias/ga/gaigen/files/20020... ... How to compute the geometric product of unit orthogonal basis blades (3/3) If we represent each basis vector with a specific bit in a binary number (e1 = 001b, e2 = 010b, e3 = 100b), computing the geometric product of basis blades is exactly the xor operation on binary numbers! (e1^e2)(e2^e3) = e1^e3 011b xor 110b = 101b We have to take care of the signs though: - basis vectors have to be rearranged into a specific order before they can annihilate each other (this rearranging causes a sign change in the result). This can also be computed binary. - signature of annihilated basis vectors can change the sign as well.

taneq 10y ago

I had a similar 'aha' moment a few years ago about inner product (aka dot product) and the choice of '.' as an operator to access elements of a structure. struct Example { int elem1; int elem2; ... }; Example ex1 = { 1, 2, ... }; int a = ex1.elem1; int b = ex1.elem2; ... If we think of ex1 as a vector, and elem1 as a constant vector [1, 0, 0...] (and elem2 as a constant vector [0, 1, 0, ...] and so on) then ex1.elem1 is literally the inner product of ex1 and the constant elem1. I have no idea if that was the original thinking behind dot notation but it's neat and I like it.

knappa 10y ago

Unless you mean something different from the usual multilinear algebra meaning of "exterior algebra", xy^yz is zero. (It contains two y's.) I think that you must actually mean Clifford algebras. (But there are also signs when in characteristic ≠ 2.)

macawfish 10y ago
Yeah, I do mean Clifford algebras. Thanks for the catch. I'll have to study up on the distinction between "outer products" in Clifford algebras and exterior algebras.
- johncolanduoni 10y ago
  They are isomorphic as vector spaces (again with characteristic != 2), but their products are not preserved by the isomorphism (unless the Clifford algebra is trivial).

halma 10y ago

I remember this and other tricks are implemented in Gaigen (a code generator for geometric algebra). https://sourceforge.net/projects/g25/?source=directory https://sourceforge.net/projects/g25/?source=directory The XOR trick is close to slide 16 of http://www.science.uva.nl/research/ias/ga/gaigen/files/20020219_talk.pdf http://www.science.uva.nl/research/ias/ga/gaigen/files/20020... ... How to compute the geometric product of unit orthogonal basis blades (3/3) If we represent each basis vector with a specific bit in a binary number (e1 = 001b, e2 = 010b, e3 = 100b), computing the geometric product of basis blades is exactly the xor operation on binary numbers! (e1^e2)(e2^e3) = e1^e3 011b xor 110b = 101b We have to take care of the signs though: - basis vectors have to be rearranged into a specific order before they can annihilate each other (this rearranging causes a sign change in the result). This can also be computed binary. - signature of annihilated basis vectors can change the sign as well.