Point covering problem

Question

I recently had this problem on a test: given a set of points m (all on the x-axis) and a set n of lines with endpoints [l, r] (again on the x-axis), find the minimum subset of n such that all points are covered by a line. Prove that your solution always finds the minimum subset.

The algorithm I wrote for it was something to the effect of: (say lines are stored as arrays with the left endpoint in position 0 and the right in position 1)

algorithm coverPoints(set[] m, set[][] n):
    chosenLines = []
    while m is not empty:
        minX = min(m)
        bestLine = n[0]
        for i=1 to length of n:
            if n[i][0] <= minX and n[i][1] > bestLine[1] then
                bestLine = n[i]
        add bestLine to chosenLines
        for i=0 to length of m:
            if m[i] <= bestLine[1] then delete m[i] from m
    return chosenLines

I'm just not sure if this always finds the minimum solution. It's a simple greedy algorithm so my gut tells me it won't, but one of my friends who is much better than me at this says that for this problem a greedy algorithm like this always finds the minimal solution. For proving mine always finds the minimal solution I did a very hand wavy proof by contradiction where I made an assumption that probably isn't true at all. I forget exactly what I did.

If this isn't a minimal solution, is there a way to do it in less than something like O(n!) time?

Thanks

Answer 1

Hint: first try proving your algorithm works for sets of size 0, 1, 2... and see if you can generalise this to create a proof by induction.

Answer 2

Your greedy algorithm IS correct. We can prove this by showing that ANY other covering can only be improved by replacing it with the cover produced by your algorithm.

Let C be a valid covering for a given input (not necessarily an optimal one), and let S be the covering according to your algorithm. Now lets inspect the points p1, p2, ... pk, that represent the min points you deal with at each iteration step. The covering C must cover them all as well. Observe that there is no segment in C covering two of these points; otherwise, your algorithm would have chosen this segment! Therefore, |C|>=k. And what is the cost (segments count) in your algorithm? |S|=k.

That completes the proof.

Two notes:

1) Implementation: Initializing bestLine with n[0] is incorrect, since the loop may be unable to improve it, and n[0] does not necessarily cover minX.

2) Actually this problem is a simplified version of the Set Cover problem. While the original is NP-complete, this variation results to be polynomial.