Since we have already computed two major vanishing points for the detection of horizontal and vertical line segment at the beginning of this section, we can derive the line at infinity l ∞ =(l 1,l 2,l 3) T from the two vanishing points using l_{infty} = {vp}_{h} times {vp}_{v} (1) Fig. 7 Example cases of connected interceptions.
This expression can then be straightforwardly evaluated in O(n) time, given that we have already computed the aforementioned quantities.
Since we have already computed TC v) for every tip v∈ S, we can trivially evaluate ∑ v ∈ S TC v in O(n) time.
Hence, to compute quantity (V) it remains now to calculate the values SUM 4 (e ) = ∑ u ∈ S (e ) TC (u ) for every edge e∈ E. We can do this in O(n) time as follows: at each tip u∈ S we store the value TC u) that we have already computed.
The last expression can be computed in Θ |Ch e)|) time as well, if we have already computed the sum ∑ k ∈ Ch (e ) w k · s (k ) and the quantity T C s u b (e) for every edge e in the tree.
We explain now how we can compute the six quantities in (2) in O(n) time, assuming that we have already computed T C s u b (e) and s e) for every e∈ E. To make the description simpler, we show in detail how we can compute the second and fourth quantities that appear in the last expression; it is easy to show that the rest of the quantities in (2) can be calculated in a similar manner.
For any edge e∈ E it is easy to show that: (3) ∑ u ∈ S (e ) TC e (u ) = ∑ u ∈ S (e ) TC (u ) - TC (e ) Therefore, according to (3) we can compute the sum ∑ u ∈ S (e ) TC e (u ) for all edges e∈ E in linear time in total, given that we have already computed TC e) for every e∈ E, and TC u) for every u∈ S. Next we continue our description for computing QD e) using a divide-and-conquer approach.
Note that when the start point of a fragment f is computed, mincost0 has already been computed for each fragment that precedes f and each fragment that is visa(f) or visl(f).
