Question

Have been scatching my head about this – and I reckon it’s simple but my geometry/algebra is pretty rubbish, and I can’t remember how to do this stuff from my school days!

EDITED:
I have a list of coordinates with people stood by them – I need an algorithm to order people from top left to bottom right in a list (array), and the second criteria demands that coordinates that are closer towards the top left origin take prescendance over all others – how would you do this?

The code should show the order as:

1. Tom
2. Harry
3. Bob
4. Dave

See diagram below:

Answer 1

From your ordering, it looks like you are putting y position at a higher priority than x position, so something like this would work when comparing two people:

if (a.y > b.y)
// a is before b
else if (a.x < b.x)
// a is before b
else
// b is before a

edit for update
This comparison still works with your new criteria. Y position is still has precedence over the X position. If the Y values are equal, the point closest to the top left corner would be the one with the smaller X value. If you want to make your object a comparator, implementing this as your comparator function would allow you to do ArrayList.sort() where negative means the first person is before the second:

public int compareTo(person a, person b) {
    if (a.y == b.y)
       return a.x-b.x
    else
       return b.y-a.y
}

//compareTo(Tom, Harry) == -50 (tom is before harry)
//compareTo(Tom, Bob) == -25 (tom is before bob)
//compareTo(Dave, Bob) == 30 (dave is after bob)