Polygon Partitioning O’Rourke, Chapter 2 Outline • Triangle Partitions • Convex Partitions Monotonicity A polygonal 푃 is monotone w.r.t. a line 퐿 if its boundary can be split into two polygon chains, 퐴 and 퐵, such that each chain is monotonic w.r.t. 퐿. 퐿 퐴 퐵 푃 퐿′ Monotonicity A polygonal 푃 is monotone w.r.t. a line 퐿 if its boundary can be split into two polygon chains, 퐴 and 퐵, such that each chain is monotonic w.r.t. 퐿. 퐿 A polygon 푃 is a monotone 푃 mountain w.r.t. 퐿 if it is 퐿′ monotone w.r.t. 퐿 and one of the two chains (the base) is a single segment. Note: Both endpoints of the base have to be convex Claim Every strictly convex vertex of a monotone mountain that is not on the base is an ear tip. WLOG, we will assume that 퐿 퐿 is vertical and that the base is to the left. 푃 퐿′ We will first show that the diagonal has to be interior and then that it cannot intersect the polygon. Proofs Assume ∃ 푣−, 푣, 푣+ ⊂ 푃, with 푣 strictly convex, s.t. 푣−푣+ is not an interior diagonal. + − 푣 푣 cannot be locally 퐿 exterior because: If 푣+ is not on the base, then 푣+푣++ must be below 푣+. 푣− Since 푣 is strictly convex, 푣 푣+푣− is to the left of 푣+푣. 푣+ 푣++ Proofs Assume ∃ 푣−, 푣, 푣+ ⊂ 푃, with 푣 strictly convex, s.t. 푣−푣+ is not an interior diagonal. + − 푣 푣 cannot be locally 퐿 exterior because: If 푣+ is on the base, then 푣+푣− is to the left of 푣+푣. But it must also be to the right 푣− of the base since 푣− is to the right of the base. 푣 푣+ Proofs Assume ∃ 푣−, 푣, 푣+ ⊂ 푃, with 푣 strictly convex, s.t. 푣−푣+ is not an interior diagonal. − + ⇒ If 푣 푣 is not an interior 퐿 diagonal then ∃푥 ∈ 푃, reflex, with 푥 interior to Δ푣+푣푣−. Interior ⇒ it cannot lie on the − chain 푣+푣푣−. 푣 푥 푣 Reflex ⇒ it is not on the base. + ⇒ The horizontal through 푥 푣 intersects 푃 in three points. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Note: We can induct because when we remove an ear, we do not violate the monotonicity of the chain. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. Monotone Mountain Triangulation Approach: 1. Find a convex vertex not on the base. 2. Remove it. 3. Go to step 1. To do this, we need to be able to quickly identify the next convex vertex. Recall: Strictly convex ⇔ interior angle < 휋. Monotone Mountain Triangulation MonotoneMountainTriangulation( 푃 ): – 퐵 ←FindBase( 푃 ) – 퐶 ← LinkConvexVertices( 푃 − 퐵 ) – while 퐶 ≠ ∅: • 푣 ← First( 퐶 ) • output( Δ푣−푣푣+ ) • 푃 ← 푃 − {푣} • if(푣+ ∉ 퐶 AND 푣+ ∉ 퐵 ) • if( ∠푣+ < 휋 ) 푣.addAfter( 푣+ ) • if( 푣− ∉ 퐶 AND 푣− ∉ 퐵 ) • if( ∠푣− < 휋 ) 푣.addBefore( 푣− ) • 퐶 ← 퐶 − {푣} Generating Monotone Mountains Approach: • Compute a trapezoidalization of 푃. • Connect supporting vertices that don’t come from the same side. Generating Monotone Mountains Claim: Such a partition generates pieces that are monotone mountains. Generating Monotone Mountains Proof: This algorithm removes all interior cusps, so the partition pieces are monotone. Generating Monotone Mountains Proof: Given a piece, we can find its topmost 푣 vertex 푣, with neighbors 푣− and 푣+. 푣− Assume 푣+ is lower than 푣−. 푣+ Generating Monotone Mountains Proof: Following the chain down 푣−, we get to the 푣 last vertex, 푤 ∈ 푃, above 푣+. 푣− There is a trapezoid supported by 푤 and 푣+. Since 푤 and 푣+ are on different sides of the trapezoid, the diagonal to 푣+ must be added. 푤 ⇒ 푣+ is a partition endpoint 푣+ Triangulation Triangulate( 푃 ): – Construct a trapezoidalization Triangulation Triangulate( 푃 ): – Construct a trapezoidalization – Connect supporting vertices from different sides. Triangulation Triangulate( 푃 ): – Construct a trapezoidalization – Connect supporting vertices from different sides. – Triangulate the mountains Triangulation Note: The algorithms for triangulating via monotone polygons and monotone mountains works for polygons with disconnected boundaries. Outline • Triangle Partitions • Convex Partitions Convex Partitions Definition: A convex partition by segments of a polygon 푃 is a decomposition of 푃 into convex polygons obtained by introducing arbitrary segments. Convex Partitions Definition: A convex partition by segments of a polygon 푃 is a decomposition of 푃 into convex polygons obtained by introducing arbitrary segments. A convex partition by diagonals of a polygon 푃 is a decomposition of 푃 into convex polygons obtained by only introducing diagonals. Convex Partitions Definition: A convex partition by segments of a polygon 푃 is a decompositionChallenge :of 푃 into convex polygons obtainedCompute by introducing a convex arbitrary partition segments. with the smallest number of pieces. 2 A convex partitionChallenge by diagonals: of a polygonCompute 푃 is a decomposition it efficiently of 푃 into convex polygons obtained by only introducing diagonals. Convex Partitions (by Segments) Claim (Chazelle): Assume the polygon 푃 has 푟 reflex vertices. If Φ is the fewest number of polygons required for a convex partition by segments of 푃 then: 푟/2 + 1 ≤ Φ ≤ 푟 + 1 Convex Partitions (by Segments) Proof (Φ ≤ 푟 + 1): For each reflex vertex, add the bisector. Because the segment bisects, the reflex angle splits into two convex angles. (Angles at the new vertices have to be < 휋.) Doing this for each reflex vertices, gives a convex partition with 푟 + 1 pieces. Convex Partitions (by Segments) Proof ( 푟/2 ≤ Φ): Each reflex vertex needs to be split and each introduced segment can split at most two reflex vertices. Convex Partitions (by Diagonals) Definition: A diagonal in a convex partition is essential for vertex 푣 ∈ 푃 if removing the diagonal creates a piece that is not convex at 푣. Convex Partitions (by Diagonals) Claim: If 푣 is a reflex vertex, it can have at most two essential diagonals. Convex Partitions (by Diagonals) Proof: Given a reflex vertex 푣, let 푣− and 푣+ be the vertices immediately before and after 푣 in 푃. There can be at most one essential segment in the half-space to the right of 푣푣+. (If there were two, we could 푣+ remove the one closer to 푣푣+.) 푣 푣− Convex Partitions (by Diagonals) Proof: Given a reflex vertex 푣, let 푣− and 푣+ be the vertices immediately before and after 푣 in 푃. Similarly, there can be at most one essential segment in the half-space to the right of 푣푣−. 푣+ 푣− Convex Partitions (by Diagonals) Proof: Given a reflex vertex 푣, let 푣− and 푣+ be the vertices immediately before and after 푣 in 푃. Similarly, there can be at most one essential segment in the half-space to the right of 푣푣−. Since the two half-spaces 푣+ cover the interior of the vertex there are at most two essential vertices at 푣. 푣− Convex Partitions (by Diagonals) Algorithm (Hertl & Mehlhorn): Start with a triangulation and remove inessential diagonals. Claim: This algorithm is never worse than 4 × optimal in the number of convex pieces. Convex Partitions (by Diagonals) Proof: When the algorithm terminates, every remaining diagonal is essential for some (reflex) vertex. Each reflex vertex can have at most two essential diagonals. ⇒ There can be at most 2푟 + 1 pieces in the partition. Since at least 푟/2 + 1 are required, the result is within 4 × optimal. Convex Partitions Why do we care? Convex polygons are easier to intersect against. For a polygon with 푛 vertices: • Testing if a point is inside is 푂(log 푛). • Testing if a line intersects is 푂(log 푛). • Testing if two polygons intersect is 푂(log 푛). Point/Convex-Polygon Intersection Algorithm: Recursively split the polygon in half and test the half the point could be in.푣7 푣6 푣8 푣5 푣1 푣4 푞 푣 2 푣3 Point/Convex-Polygon Intersection Algorithm: Recursively split the polygon in half and test the half the point could be in. 푣1 푣4 푞 푣 2 푣3 Point/Convex-Polygon Intersection Algorithm: Recursively split the polygon in half and test the half the point could be in. 푣1 푣4 푞 푣3 Point/Convex-Polygon Intersection InConvexPolygon( 푞 , {푣1, … , 푣푛} ) if( 푛 == 3 ) »return InTriangle( 푞 , {푣1, 푣2, 푣3} ); if( Left( 푣1, 푣푛 2 , 푝 ) ) »return InConvexPolygon( 푞 , {푣푛 2, … , 푣1} ); else »return InConvexPolygon( 푞 , {푣1, … , 푣푛 2} ); Line/Convex-Polygon Intersection Note: Given a convex polygon 푃, a line segment 퐿, and * vertices 푣푖, 푣푗 ∈ 푃 on the same side of 퐿.