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High-performance coding by Eddie Edwards

Operators as Matrices

06 Mar 2012, 13:17 UTC

In The Myth of the "Unrealizable" Filter I had an expression like this: 

Y = (X)A.B

Where Y, X, A, B are all vectors, so A.B scales vector X to give vector Y.

This can be expressed in matrix form 

Y = G*B

Here, G is the matrix representing the operator (X)A. which operates on B on the right.

An interesting thought is that maybe it would be cool if a language allowed (X)A. to be an object which acts on the right. i.e. instead of parsing say A + B as add(A,B) the language would parse A + B as apply(adder(A),B) where adder(A) = apply(A,+) is a function taking one argument, which adds A to the argument. Thus, the whole language works (modulo operator precedence) as a series of apply operations which can take variables or operators/symbols/functions, and which performs currying. So: 

apply(A,fn(a,b))      gives fn2(b) -> fn(A,b)
apply(fn(a),B)        gives fn2()  -> fn(B)
apply(fn1(a),fn2(b))  gives fn3(b) -> fn1(fn2(b))

Then perhaps - assuming * and . operators - I could just write: 

G = X * (A.)

This is not simple to compile - the expression is apply(apply(X,*),apply(A,dot)) which I might rewrite apply(mulbyX,dotbyA) giving a function like fn(B) -> mulbyX(dotbyA(B)). Casting this function to a matrix is non-trivial.

Alternatively, perhaps the same compiler (which has to be able to analyze linear functions) could handle a syntax like this: 

G = (X * (A.B)) %% B

Which is rather like differentiation with respect to B.

The resulting system would be powerful and conceptually simple, but maybe it has insufficient use cases. Still, it's interesting to think of new ways to write code.

What do you think? :)

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Andrei Rotenstein commented:

Eddie, I'm not sure I follow the last expression.

on 06 Mar 2012, 16:56 UTC

Eddie Edwards commented:

Andrei, the idea is that the expression on the LHS of %% is "divided by" the expression on the RHS to give a matrix (or perhaps tensor) G s.t. G*B reproduces the original expression. It's a lexical expression, so B doesn't have to be declared anywhere, while X and A are used to create G.

The output is a matrix where each element is the amount of RHS[n] found in LHS[m]. So in fact it is precisely differentiation with respect to B, but the compiler would complain if the result wasn't a constant in each place (i.e. B is used with other powers than 0 and 1).

Does that help?

on 06 Mar 2012, 19:12 UTC

Eddie Edwards commented:

Thinking about it, you could implement this in a C compiler as follows:

  1. Isolate LHS and RHS expressions
  2. Set RHS variable to (0,0,0), fold into LHS, then constant-fold LHS
  3. Do the same for (1,0,0), (0,1,0), (0,0,1)
  4. Subtract e.g. (0,0,0) from (1,0,0) to get the first column of G

This works assuming the LHS is linear; if the LHS is not linear it gives a rough approximation of the correct answer at B = (0,0,0) :)

on 06 Mar 2012, 21:26 UTC updated 06 Mar 2012, 21:28 UTC

DC commented:

Most recent topmost *really* doesn't work for this sort of blog.

on 06 Mar 2012, 21:48 UTC

Andrei Rotenstein commented:

Much.

Pretty awesome, Eddie!

on 08 Mar 2012, 02:30 UTC

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