Raul,
> On Nov 28, 2017, at 5:32 PM, Raul Miller <[email protected]> wrote:
>
> Something based on transitive closure does sound like the way to go.
>
> And this might be the essay Henry Rich was thinking of:
> http://code.jsoftware.com/wiki/User:Roger_Hui/t
>
> But I am having trouble following along here.
>
> The text says "task C can only start after tasks A and B finish" and
> graph =: 4 4 $ 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 NB. col 0 =
> depends-on A, col 1 = depends-on B, etc.
>
I actually didn't intend any relationship between that example sentence and the
matrix I supplied! I'm very sorry, I see now that was a horrifyingly confusing
thing to do!
> So, I am expecting
> nodes=: 'ABCD'
>
> And when I look at which nodes C depends on in that graph, I do not
> see a dependency on B:
> ((nodes i.'C'){graph)#nodes
> A
>
> So I am already lost.
>
> But, wait, maybe I should be looking at the transitive closure of the graph?
>
> ((nodes i.'C'){(+. +./ .*~)^:_ graph)#nodes
> A
>
> D depends on A and B and C, but B and C only depend on A.
Yes, I really intended the code part I supplied to be coherent on its own, so
this is right. The point I was trying to make was that it looks correct for the
supplied 4-node graph, but for the 5-node graph, the way E depends on C and D
exposes something sequential about the calculation.
> There might be another subtlety to this problem, but need to get the
> problem statement correct before addressing any such thing.
>
> That said, I think there's going to be another issue, though, with
> transitive closure. You don't strictly want the transitive closure
> here. It's not just that D depends on A, B and C - the A and B
> dependencies are in parallel, while the C dependency is in series.
> Parallel is handled by >. and series is handled by + - I suspect you
> need a third operation to determine relevance at any given step.
Well, it's really (this node's duration) + (maximum early finish of my
dependencies), right? I just don't know how to phrase that "all at once"
without some kind of loop or recursion, item-by-item.
I think Henry is right, the topological sort gives me the order in which to do
the calculation, if I have to do it in order, and it seems like I do.
--
Daniel Lyons
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