Trefethen & Bau & MATLAB & Julia, Lectures 6-7
Not much happened in Lecture 6, but I got gobsmacked in Lecture 7. It happened when I tried to convert this boring MATLAB code for backward substitution.
A = magic(9); b = (1:9)'; [Q,R] = qr(A); z = Q'*b; x(9,1) = z(9)/R(9,9); for i = 8:-1:1 x(i) = (z(i) - R(i,i+1:9)*x(i+1:9)) / R(i,i); end
Here is what I first tried in Julia.
A = round(10*rand(9,9)); b = (1:9); m = 9; (Q,R) = qr(A); z = Q'*b; x = zeros(m); x[m] = z[m]/R[m,m]; for i = m-1:-1:1 x[i] = (z[i] - R[i,i+1:m]*x[i+1:m]) / R[i,i]; end
Seems straightforward, but line 4 gives an error. I’m not going to copy the error message here, in case you’re using mobile data right now. What I mean is that it is verbose, not to mention obscure. You don’t appreciate simple, clear error messages until you get something else!
Anyhow, I then remembered that in Julia, the colon construction
(1:9) produces a Range, not a Vector. As I understand it, Julia embraces a lazy design philosophy: it avoids evaluation of an expression until the last possible moment. Suppose the only use of that Range is to describe a loop iteration—in that case, why have a vector?
I’m all for lazy philosophy. (Haw haw!) It’s not clear to me why the context
does not automatically convert the Range into a Vector. It’s even less clear why they have deprecated the idiom
[1:9] to create a Vector; it works for now but gives a warning. Instead one should use
A = round(10*rand(9,9)); b = collect(1:9); m = 9; (Q,R) = qr(A); z = Q'*b; x = zeros(m); x[m] = z[m]/R[m,m]; for i = m-1:-1:1 x[i] = (z[i] - R[i,i+1:m]*x[i+1:m]) / R[i,i]; end
Feels very odd to me still, but okay.
We are not out of the woods yet. This version still fails in the loop body, again vomiting opaque error messages. Remember how,
back in Lecture 2, I mentioned that scalars and 1x1 matrices are different things? Inside the loop above,
z[i] is a scalar and the product is a length-1 vector. But the subtraction works anyway, as
z[i] is silently promoted to a 1-vector also. No, the problem comes with the assignment: you can’t assign a 1-vector to an element of an array of numbers.
There’s a very long (space and time) discussion about this and related issues in Julia. Suffice it to say that what mathematicians do with scalars, vectors, matrices, and tensors isn’t rigorously consistent—or at least, there seem to be multiple, incompatible rigorous ways to use them.
In this particular case I have found two unsatisfying workarounds. The idiom
x[i:i] produces a Vector, not a scalar, so the assignment goes through. Or we can work on the other side of the assignment and pull out the scalar from the vector:
(z[i] - R[i,i+1:m]*x[i+1:m]) / R[i,i]
Now, it’s pleasing that this syntax does work, as there is no good MATLAB equivalent for indexing into a temporary expression. I just wish it was in the service of something less dismal.
Again: Julia’s designers have solid reasons for doing things this way. I wouldn’t consider it a dealbreaker for research codes, but this episode is not something I would want to explain to undergrads who are just wrapping their heads around LU factorization. It pulls you right out of thinking about math and into thinking about strict-typing, pinhead-dancing angels. How unfortunate.