Originally posted by: UDOPS

This line returns an empty set:

{turn} Layover = {<r,r2,tj,33> | <r,r2> in TrainPair, <r,i,-1,ti,tj,d> in Paths, <r2, ii, jj, tii, tjj, d> in Paths};

But this line returns many members:

{turn} Layover = {<r,r2,tj,33> | <r,r2> in TrainPair, <r,i,-1,ti,tj,d> in Paths};

There are many valid matches in the third indexing set above. What am I missing?
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Originally posted by: UDOPS

Well, it took me most of the last two hours to get this to work, but here is the solution I am using. I had to break the statement into two: an intermediate set and then the final set. Still can not understand why I can not build this set in one step.

{turntime} PotentialLayoverStarts = {<r,tj> | r in Trains, <r,i,-1,ti,tj,d> in Paths: TrainData[r].successor>0};
{turn} Layover = {<r,TrainData[r].successor,tj,tii> | <r,tj> in PotentialLayoverStarts,
<TrainData[r].successor,TrainData[TrainDatahttp://r].successor.origin,jj,tii,tjj,d> in Paths:
tii>=tj+TrainData[r].layover_min && (TrainData[r].layover_max==0||tii<=tj+TrainData[r].layover_max)};
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Originally posted by: SystemAdmin

In your very first formulation, you implicitly slice over the 6th component in Paths, named 'd' in both iterations.
That could explain the difference.

Tschau, Frank
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Originally posted by: UDOPS

You are correct. I did not notice that, since I ignore "d" otherwise.
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Originally posted by: UDOPS

I eliminated the duplicate "d", and it worked fine. Mark it up to bleary eyes and mental focus.
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