Given n line segments (p1, q1), (p2, q2), … (pn, qn), find if the given line segments intersect with each other or not.
Example:
Input: {{1, 5}, {4, 5}}, {{2, 5}, {10, 1}},{{3, 2}, {10, 3}},{{6, 4}, {9, 4}},{{7, 1}, {8, 1}}
Output: 2
Explanation: Lines {{1, 5}, {4, 5}}, {{2, 5}, {10, 1}} are intersecting, as well as {{2, 5}, {10, 1}},{{3, 2}, {10, 3}}.
We have discussed the problem to detect if two given line segments intersect or not. In this post, we extend the problem. Here we are given n line segments and we need to find out if any two line segments intersect or not.
Naive Algorithm A naive solution to solve this problem is to check every pair of lines and check if the pair intersects or not. We can check two line segments in O(1) time. Therefore, this approach takes O(n2).
Sweep Line Algorithm: We can solve this problem in O(nLogn) time using Sweep Line Algorithm. The algorithm first sorts the end points along the x axis from left to right, then it passes a vertical line through all points from left to right and checks for intersections. Following are detailed steps.
1) Let there be n given lines. There must be 2n end points to represent the n lines. Sort all points according to x coordinates. While sorting maintain a flag to indicate whether this point is left point of its line or right point.
2) Start from the leftmost point. Do following for every point
…..a) If the current point is a left point of its line segment, check for intersection of its line segment with the segments just above and below it. And add its line to active line segments (line segments for which left end point is seen, but right end point is not seen yet). Note that we consider only those neighbors which are still active.
….b) If the current point is a right point, remove its line segment from active list and check whether its two active neighbors (points just above and below) intersect with each other.
The step 2 is like passing a vertical line from all points starting from the leftmost point to the rightmost point. That is why this algorithm is called Sweep Line Algorithm. The Sweep Line technique is useful in many other geometric algorithms like calculating the 2D Voronoi diagram
What data structures should be used for efficient implementation?
In step 2, we need to store all active line segments. We need to do following operations efficiently:
a) Insert a new line segment
b) Delete a line segment
c) Find predecessor and successor according to y coordinate values
The obvious choice for above operations is Self-Balancing Binary Search Tree like AVL Tree, Red Black Tree. With a Self-Balancing BST, we can do all of the above operations in O(Logn) time.
Also, in step 1, instead of sorting, we can use min heap data structure. Building a min heap takes O(n) time and every extract min operation takes O(Logn) time (See this).
PseudoCode:
The following pseudocode doesn’t use heap. It simply sort the array.
sweepLineIntersection(Points[0..2n-1]):
1. Sort Points[] from left to right (according to x coordinate)
2. Create an empty Self-Balancing BST T. It will contain
all active line Segments ordered by y coordinate.
// Process all 2n points
3. for i = 0 to 2n-1
// If this point is left end of its line
if (Points[i].isLeft)
T.insert(Points[i].line()) // Insert into the tree
// Check if this points intersects with its predecessor and successor
if ( doIntersect(Points[i].line(), T.pred(Points[i].line()) )
return true
if ( doIntersect(Points[i].line(), T.succ(Points[i].line()) )
return true
else // If it's a right end of its line
// Check if its predecessor and successor intersect with each other
if ( doIntersect(T.pred(Points[i].line(), T.succ(Points[i].line()))
return true
T.delete(Points[i].line()) // Delete from tree
4. return False
Example:
Let us consider the following example taken from here. There are 5 line segments 1, 2, 3, 4 and 5. The dotted green lines show sweep lines.
Following are steps followed by the algorithm. All points from left to right are processed one by one. We maintain a self-balancing binary search tree.
Left end point of line segment 1 is processed: 1 is inserted into the Tree. The tree contains 1. No intersection.
Left end point of line segment 2 is processed: Intersection of 1 and 2 is checked. 2 is inserted into the Tree. Intersection of 1&2 Found (“Note that the above pseudocode returns at this point”). We can continue from here to report all intersection points. The tree contains 1, 2.
Left end point of line segment 3 is processed: Intersection of 3 with 1 is checked. No intersection. 3 is inserted into the Tree. The tree contains 2, 1, 3.
Right end point of line segment 1 is processed: 1 is deleted from the Tree. Intersection of 2 and 3 is checked. Intersection of 2 and 3 is reported. The tree contains 2, 3.
Left end point of line segment 4 is processed: Intersections of line 4 with lines 2 and 3 are checked. No intersection. 4 is inserted into the Tree. The tree contains 2, 4, 3.
Left end point of line segment 5 is processed: Intersection of 5 with 3 is checked. No intersection. 5 is inserted into the Tree. The tree contains 2, 4, 3, 5.
Right end point of line segment 5 is processed: 5 is deleted from the Tree. The tree contains 2, 4, 3.
Right end point of line segment 4 is processed: 4 is deleted from the Tree. The tree contains 2, 4, 3. Intersection of 2 with 3 is checked. Intersection of 2 with 3 is reported (Note that the intersection of 2 and 3 is reported again. We can add some logic to check for duplicates ). The tree contains 2, 3.
Right end point of line segment 2 and 3 are processed: Both are deleted from tree and tree becomes empty.
C++14
#include <bits/stdc++.h>using namespace std;// A point in 2D planestruct Point{ int x, y;};// A line segment with left as Point// with smaller x value and right with// larger x value.struct Segment{ Point left, right;};// An event for sweep line algorithm// An event has a point, the position// of point (whether left or right) and// index of point in the original input// array of segments.struct Event { int x, y; bool isLeft; int index; Event(int x, int y, bool l, int i) : x(x), y(y), isLeft(l), index(i) {} // This is for maintaining the order in set. bool operator<(const Event& e) const { if(y==e.y)return x<e.x; return y < e.y; }};// Given three collinear points p, q, r, the function checks if// point q lies on line segment 'pr'bool onSegment(Point p, Point q, Point r){ if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) && q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y)) return true; return false;}// To find orientation of ordered triplet (p, q, r).// The function returns following values// 0 --> p, q and r are collinear// 1 --> Clockwise// 2 --> Counterclockwiseint orientation(Point p, Point q, Point r){ // for details of below formula. int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y); if (val == 0) return 0; // collinear return (val > 0)? 1: 2; // clock or counterclock wise}// The main function that returns true if line segment 'p1q1'// and 'p2q2' intersect.bool doIntersect(Segment s1, Segment s2){ Point p1 = s1.left, q1 = s1.right, p2 = s2.left, q2 = s2.right; // Find the four orientations needed for general and // special cases int o1 = orientation(p1, q1, p2); int o2 = orientation(p1, q1, q2); int o3 = orientation(p2, q2, p1); int o4 = orientation(p2, q2, q1); // General case if (o1 != o2 && o3 != o4) return true; // Special Cases // p1, q1 and p2 are collinear and p2 lies on segment p1q1 if (o1 == 0 && onSegment(p1, p2, q1)) return true; // p1, q1 and q2 are collinear and q2 lies on segment p1q1 if (o2 == 0 && onSegment(p1, q2, q1)) return true; // p2, q2 and p1 are collinear and p1 lies on segment p2q2 if (o3 == 0 && onSegment(p2, p1, q2)) return true; // p2, q2 and q1 are collinear and q1 lies on segment p2q2 if (o4 == 0 && onSegment(p2, q1, q2)) return true; return false; // Doesn't fall in any of the above cases}// Find predecessor of iterator in s. set<Event>::iterator pred(set<Event> &s, set<Event>::iterator it) { return it == s.begin() ? s.end() : --it;}// Find successor of iterator in s.set<Event>::iterator succ(set<Event> &s, set<Event>::iterator it) { return ++it;}// Returns true if any two lines intersect.int isIntersect(Segment arr[], int n){ unordered_map<string,int> mp; // to note the pair for which intersection is checked already // Pushing all points to a vector of events vector<Event> e; for (int i = 0; i < n; ++i) { e.push_back(Event(arr[i].left.x, arr[i].left.y, true, i)); e.push_back(Event(arr[i].right.x, arr[i].right.y, false, i)); } // Sorting all events according to x coordinate. sort(e.begin(), e.end(), [](Event &e1, Event &e2) {return e1.x < e2.x;}); // For storing active segments. set<Event> s; int ans=0; // Traversing through sorted points for (int i=0; i<2*n; i++) { Event curr = e[i]; int index = curr.index; // If current point is left of its segment if (curr.isLeft) { // Get above and below points auto next = s.lower_bound(curr); auto prev = pred(s, next); // Check if current point intersects with // any of its adjacent bool flag=false; if (next != s.end() && doIntersect(arr[next->index], arr[index])){ string s=to_string(next->index+1)+" "+to_string(index+1); if(mp.count(s)==0){mp[s]++;ans++;} //if not already checked we can increase count in map } if (prev != s.end() && doIntersect(arr[prev->index], arr[index])){ string s=to_string(prev->index+1)+" "+to_string(index+1); if(mp.count(s)==0){mp[s]++;ans++;} //if not already checked we can increase count in map } // if same line segment is there then decrease answer as it got increased twice if(prev != s.end() && next != s.end() && next->index==prev->index)ans--; // Insert current point (or event) s.insert(curr); } // If current point is right of its segment else { // Find the iterator auto it=s.find(Event(arr[index].left.x, arr[index].left.y, true, index)); // Find above and below points auto next = succ(s, it); auto prev = pred(s, it); // If above and below point intersect if (next != s.end() && prev != s.end()) { string s=to_string(next->index+1)+" "+to_string(prev->index+1); string s1=to_string(prev->index+1)+" "+to_string(next->index+1); if (mp.count(s)==0&&mp.count(s1)==0&&doIntersect(arr[prev->index], arr[next->index])) ans++; mp[s]++; } // Remove current segment s.erase(it); } } //print pair of lines having intersection for(auto &pr:mp){ cout<<"Line: "<<pr.first<<"\n"; } return ans;}// Driver codeint main() { Segment arr[] = { {{1, 5}, {4, 5}}, {{2, 5}, {10, 1}},{{3, 2}, {10, 3}},{{6, 4}, {9, 4}},{{7, 1}, {8, 1}}}; int n = sizeof(arr)/sizeof(arr[0]); cout<<"Number of intersection points: "<<isIntersect(arr, n); return 0;} |
Line: 2 3 Line: 1 2 Number of intersection points: 2
Time Complexity: The first step is sorting which takes O(nLogn) time. The second step process 2n points and for processing every point, it takes O(Logn) time. Therefore, overall time complexity is O(nLogn)
Auxiliary Space: O(N)
References:
http://www.cs.uiuc.edu/~jeffe/teaching/373/notes/x06-sweepline.pdf
http://courses.csail.mit.edu/6.006/spring11/lectures/lec24.pdf
http://www.youtube.com/watch?v=dePDHVovJlE
http://www.eecs.wsu.edu/~cook/aa/lectures/l25/node10.html
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