The general selection algorithm with a O(n(logn)2)-time complexity
We write pi(a) for the permutation that orders the intercepts of l1,…, ln, in descending order at x=a. So ypi1(a)>…> ypin(a). We renumber l1,…, ln so that for a< t1, (a) is the identity. The permutation pi(t1) has one inversion (exactly one paire of lines crossed at t1), pi (t2) has two, etc…
In fact the function I( (x)) = number of inversions in pi(x), is a monotone step function in x with unit jumps at the ti's.
I(pi (x)) = j if and only if tj = max(ti:ti<=x).
For a given k, the problem of finding tk may be viewed as an usual sorting problem to which we apply Megiddo's technique of building algorithms from parallel ones. An implicit binary search over the ti's is performed, each step taking O(nlogn) time. This will an O(n(logn)2)-time algorithm.
Presentation of the algorithm
In seeking tk , we will attempt to sort y1(a*),…, yn(a*) at a* = tk + epsilon. We know that this sort may be achieved in O(nlogn) comparisons (cf Binary Search Tree), each answering a question Qij of the form "yi(a*)<=yj(a*)?". The O(nlogn) answers yield the permutation pi* that sorts these intercepts : ypi*1(a*)>…> ypi*n(a*). Once pi* has beed found, tk = max[upii*pii+1*:pii*>pii+1*]; the kth inversion must have just reversed a pair of adjacent intercepts in the permutation pi(tk-1).
The control structure for the sort will come from the O(logn)-depth sorting network of Ajtai, kmlos and Szemeredi (AKS-Network). At each level, n/2 questions are answered. The network is just a guide for blocking these comparisions into groups of size n/2. The sort is complete once the O(nlogn) answers are obtained and we have determined pi*.
Even though we do not know a*, we can answer the question Qij in time O(nlogn) as follows. We find uij the x-coordinate of , li inter lj , i<j, in constant time and then obtain its rank among the ti's.To find its rank, we sort the n intercepts at uij in decreasing order to get pi(uij). The rank uij is the number of inversions in pi(uij), I(pi (uij)).
If I(pi (uij))>k, we know that uij> tk so the answer is "no"; lines i and j have not yet crossed at tk.
If I(pi(uij))<k, we know that uij< tk so the answer is "yes".
If I(pi (uij))=k, uij=tk.
I may be computed in time O(nlogn) (merge sort of Knuth) : if pi(uij) = (r1,…,rn), we use merge-sort to sort the slopes mr1,…,mrn and count the number of inversions that were performed.
If we actually answered all n/2 questions Qi1j1,…, Qinjn on a level of the network by counting inversions, the complexity would be O(n2logn) for that level (n/2 questions*2nlogn for the sort of the n intercepts and the merge sort). And as there are O(logn) levels, we have a complexity of O((nlogn)2) overall.
The trick is to resolve the n/2 questions on a level by actually counting inversions only O(log) times.
As mentionned, each questions determines an intersection of a distinct pair of lines, and the answer is obtained by comparing the rank of that intersection point with k.
On a given level of the sequentialzed version of the sorting network, denote the x-coordinates of these points by zi1,…,zin/2. We can compute the median of these intersections points, zmed, in time O(n). Its cost time O(nlogn) to rank it, and answer half the questions. For example, if zmed < tk, then zik < tk for all the zik ≤ zmed. Continuing with the n/4 unresolved questions on this level, we again find the median z and rank it in time O(nlogn), etc…After O(logn) inversions counts, all n/2 questions on this level are resolved. Since each inversion count takes O(nlogn) steps and there are O(logn) levels, the algorithm has time complexity O(n(logn)3).
But, Cole shows how the result may be improved by a factor of logn, by considering the fact that in the network each question has two inputs. (see R.COLE, Slowing down sorting networks to obtain faster sorting algorithms, Journal of the.ACM, No34 (1987), pp200-208)
So we obtain an algorithm of O(n(logn)2)-time complexity