Find intersection of two x-y tracks given time element

I have two tracks defined by x-y coordinates in which I would like to establish the crossing points. This appears to be relatively easy using several of the submissions on File Exchange such as Intersection and InterX . This submission a list of the x and y values of the crossing points. However it doesn’t take into account time. Let me explain....

Suppose my two tracks are for a cat and a mouse. The tracks explain where both the cat and mouse have been as a function of time using x-y coordinates. Let’s now suppose that I know that the cat always trails the mouse - that is to say that the mouse is always ahead of the cat as a function of time, never the other way round. Is there a quick way (like the File Exchange link I show) to calculate only crossing points where the cat crosses the mouses trail, not the mouse crossing the cats trail.

One solution would be to return x-y-time triplets which I could then filter to ensure the cat is behind the mouse. But I’m not sure how I could make a fast running function of this nature

[Adam Danz]

At time(t), the mouse is creating his track but it isn't until time (t+lag) that the cat's track catches up. So the only way for the mouse to cross the cat's track is if the mouse loops back across some old section of track that was established at time (t-x). Is that correct?

So your data looks something like this, is that correct?

t =
  2×n table
     xCat      yCat    xMouse  yMouse   time
    ______    ______    ____    ____    ____
     NaN       NaN       1       1       0  
       2         2       2       2       1  

[Mark Jackson]

Yes, that’s exactly right. The only way the mouse can cross the cat’s trail is by looping back. Hence my need for the time element. I want to be able to prevent an automated algorithm picking up the mouse crossing the cat’s trail 5 days later (for example).

Date-wise, yes very similar to what you’ve shown. The actual headings are xCat, yCat, timeCat, xMouse, yMouse, timeMouse

[Adam Danz]

Do you know the lag of the cat?

Is the catTime and mouseTime from the same timer? In other words, at time x we know where the mouse and cat was.

If not, I can't quickly think of a solution.

[Mark Jackson]

The lag of the cat behind the mouse is not constant but at any point in time the data is available to calculate the lag. One solution I came up with is to use the FEX functions to find all the crossing points then use the returned x-y values to look up timeCat and timeMouse and then filter the crossing points such that timeMouse > timeCat. However, this assumes that there are no repeated crossing points

[Adam Danz]

That's what I was going to suggest. The "intersection()" function you linked to has 4 outputs.

[x0,y0,iout,jout] = intersections(x1,y1,x2,y2)

The iout and jout will tell you which line segment contained the intersection. For example,

floor(iout)   % the index to the mouse coordinate just before the intersection
floor(jout)   % the index to the cat coordinate just before the intersection

Then you can look up the lag at that point in time and, as you mentioned, determine if the cat had already created that line segment.

"However, this assumes that there are no repeated crossing points"

why? Does that function only return unique crossing points?

[Mark Jackson]

I hadn’t thought about the iout and jout returns. The function returns all crossing points but I was assumed it only returned x-y so repeated crossing points would be indistinguishable (in terms of time). With iout and jout it should be easy to pick out the times and then check whether mouse is ahead of cat or vice-versa. Thanks for your help! I’ll try it today and let you know how I get on

[Adam Danz]

It's an interesting problem. Let me know how it works out.