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scalar + I => scalar #24302

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StefanKarpinski opened this issue Oct 23, 2017 · 4 comments
Closed

scalar + I => scalar #24302

StefanKarpinski opened this issue Oct 23, 2017 · 4 comments
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linear algebra Linear algebra
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@StefanKarpinski
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Now that we've deprecated matrix + scalar, we can revisit the old A + I + 1 associativity chesnut. If we make scalar + I => scalar then we have (A + I) + 1 and A + (I + 1) both being the same error for matrix + scalar. This also would allow us to write generic polynomials that work for both scalars and square matrices since e.g. x^2 + 2x + I would evaluate to a scalar for scalar x and to a matrix for matrix x. Strictly this isn't breaking, so this need not be done for 1.0, but we may want to at least try it out to see if there are any things we need to break to make this work.

ref #17083, cc @eveydee, @dlfivefifty

@StefanKarpinski StefanKarpinski added linear algebra Linear algebra triage This should be discussed on a triage call labels Oct 23, 2017
@KristofferC
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KristofferC commented Oct 23, 2017

For generic polynomial it needs to be written as x^2 + 2x + I(x), where I(::Int) = 1 and I(m::Matrix) = eye(m)? To handle zero order polynomials and to be type stable (in the reduction variable) when you would implement a function that takes a vector of coefficients and the point to evaluate the polynomial at.

Edit: Alternatively, you could write it as x^2 + 2x^1 + x^0. No need to use I at all. I actually think this is the right way to do it.

@StefanKarpinski
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That's an additional concern, but only for writing very generic polynomial code – for a specific polynomial with positive order, just having scalar + I => scalar is sufficient. In any case, the fact that there are additional concerns doesn't make this not a good change. For generic polynomial code, we can use x^2 + 2x + x^0 or something like that.

@StefanKarpinski StefanKarpinski added this to the 1.0 milestone Oct 26, 2017
@JeffBezanson JeffBezanson removed the triage This should be discussed on a triage call label Oct 26, 2017
@StefanKarpinski
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Oh hey, we already do this:

julia> 1 + I
2

@JeffBezanson
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Fixed by #23923.

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