This document discusses open and closed guard edges in simple polygons. It defines key terms like open guard edges, closed guard edges, and star-shaped polygons. It presents two lemmas: 1) A point p is visible from an open/closed edge depending on the common vertices of paths from p to the edge's endpoints. 2) Given open guard edges ab and cd, their associated paths are disjoint and straight lines. The document proves that a non-starshaped polygon has at most 1 open guard edge and at most 3 closed guard edges. It also shows the intersection of paths associated with closed guard edges is in the polygon's kernel.
Influencing policy (training slides from Fast Track Impact)
Open Guard Edges and Edge Guards in Simple Polygons
1. Open Guard Edges and Edge
Guards in Simple Polygons
,with Csaba D. Tóth and Godfried T. Toussaint, Proceedings of 23rd
Canadian Conference on Computational Geometry, 449-454, 2011.
http://www.eecs.tufts.edu/~awinslow/
Presenter : Oscar, openguards@olife.org
2013/06/10
7. Geodesic path(p,q)
• Geodesic path(p,q) →
path(p,q)
• Shortest directed path from p to q that lies
entirely in P.
• path(p,q) is straight line p and q see each⇔
other.
8. weakly visible
• A point is weakly visible to a set of points
• this point is visible from some point in
that set.
12. Lemma 1(1) open edge
• Let p be a point inside a simple polygon P.
1. Point p is visible from an open edge uv ⇔
p is the only common vertex of path(p, u)
and path(p, v)
13. Point p is visible from an open edge uv p is⇔
the only common vertex of path(p, u) and
path(p, v)
14. Lemma 1(2) closed edge
• Let p be a point inside a simple polygon P.
2.p is visible from a closed edge uv {p, u,⇔
v} are only three possible common
vertices of path(p, u) and path(p, v).
15. p is visible from a closed edge uv ⇔ {p, u, v}
are only three possible common vertices of
path(p, u) and path(p, v).
17. A simple polygon with open
guard edges
• ≥ 2 starshaped.⇔
• ≤ 1 Non-starshaped⇔
18. Lemma 2
Given ab, cd are open guard edges.
⇒path(b, c) and path(a, d) are disjoint.
⇒path(a, c) and path(b, d) are straight line.
19. • quadrilateral Q {← ab, path(b,c), cd, and path(a,d)}
• Assume interior vertex q path(b,c) path(a,d)∈ ∩
• a or b is not visible from the open edge cd (Lemma 1)
• cd is not a guard edge (contradiction)
• Conclude
path(b,c) and path(a,d) are disjoint
Q is a simple polygon
20. • Assume an interior vertex of path(a,c) in path(b,c)
• c is not visible from ab
• ab is not a guard edge (contradiction)
• Conclude
path(a,c) has no interior vertices.
path(b,d) has no interior vertices as well.
21. Lemma 3
Given ab, cd are open guard edges.
⇒The intersection point x = ac ∩ bd is in
the kernel of P
22. • To show that an arbitrary point p is visible from x.
• ac and bd are diagonals of P (lemma 2)
• assume p (abx) (cdx)∈ △ ∪ △
• px lies in the same triangle
• p can be seen by x.
23. • Assume p is outside of both triangles.
• ∵ ab and cd are open guard edges
• p sees q and o (relative interiors)
• quadrilateral Q = (o, p, q, x) is simple and inside P
• diagonal px lies inside Q
• ∴p can be seen by x.
28. Lemma 6
• Given g1,g3 are closed guard edges.
the path(b, c) and path(a, d) are disjoint⇒
path(a, c) and path(b, d) in⇒ □{a, b, c, d}.
29. • path(a,c) and path(b,d) lie in Q
• Any interior vertex of path(a,c) and path(b,d) is in Q
• Assume path(a,c) and path(b,c) have a common interior vertex
• c is not visible from ab
• ab is not a guard edge (contradiction)
• Conclude
all interior vertices of path(a,c) and path(b,d) are in {a,b,c,d}
30. Corollary 7
• If □{a, b, c, d} is convex path(a,c) and⇒
path(b,d) are straight line.
32. Lemma 8
• The intersection point x = path(a, c) ∩
path(b, d) is in the kernel of P .
g2
33. • To show that an arbitrary point p is visible from x.
• △ (abx) , (cdx) are diagonals in P△ (Corollary 7)
• Assume p (abx) (cdx)∈ △ ∪ △
• px lies in the same triangle
• p can be seen by x.
34. • Assume p is outside of both triangles.
• w.l.o.g. assume, p is on the right side of the
directed path(a,c) and path(b,d).
• p and the guard edge g4 are on opposite sides
• If path(p,x) = px, then p is visible from x (done)
35. • path(p,x) is not a straight line
• w.l.o.g. assume, path(p,x) makes a right turn at its last
interior vertex q
• = path(p,d) also makes a right turn at q
• ∵ p is visible from the guard edge cd
• ∴ q = c (Lemma 1b)
• ∵any paths from g4 to p make a right turn at c
• ∴ p is not visible from g4 (Lemma 1b) (contradiction)
• ∵ from g4 to p is straight line
• ∴ path(p,x) is a straight line
Open guard edges not include end points 可以看到 kernel
這裡我們講 path(p,q) 就是 Geodesic path(p,q)
只有 starshaped 才有 kernel
a polygon that contains a point from which the entire polygon boundary is visible The set of all points z with this property (that is, the set of points from which all of P is visible) is called the kernel of P.
ab, path(b, c), cd, and path(a, d) 形成四邊形 Q , ab 上任一點和 cd 上任一點所連結的路徑都在 Q 裡。如果 path(b,c) 和 path(a,d) 有共同內部節點 q ,則 open edge cd 看不到 a 或 b 其中一個 (Lemma 1) ,所以 cd 就不能當 guard edge 。我們得到結論, path(b,c) 和 path(a,d) 不相連,且 Q 是 simple polygon 。
Path(a,c) 和 path(b,d) 在 Q 中,所以任何 path(a,c) 和 path(b,d) 的內部節點是 Q 的節點。如果 path(a,c) 的內部節點在 path(b,c) 裡,則 ab 看不到 c 。同樣地,如果 path(a,c) 的內部節點在 path(a,d) 裡,則 cd 看不到 a 。因此, path(a,c) 沒有內部節點,同樣 path(b,d) 也不會有內部節點。
圖中證明在多邊形 P 內, x 可以看到任意的點 p 。 基於 lemma 2 , ac 及 bd 是對角線 (ad, cb 不交集 ) 。三角形 (abx) 及 (cdx) 在 P 內部。如果 p 屬於 (abx) 或 (cdx) ,則 px 線段在同一個三角形裡。
假設 p 在兩個三角形之外,由於 ab 和 cd 是 open guard edges , p 看到一些點在它的相對內側,例如 o 屬於 ab 和 q 屬於 cd 。四邊形 Q {o,p,q,x} 是 simple ,而且它一樣在 P 裡。注意 Q 的 o 和 q 是 convex 節點。無論 Q 是 convex 或非 convex 四邊形,它的對角線 px 在 Q 裡 (x 看得到 p) ,所以也在 P 裡。
If a simple polygon has two open guard edges, then it has a nonempty kernel by Lemma 3, and thus is starshaped. So every non-starshaped simple polygon has at most one open guard edge. No kernel(P)
We proceed by contradiction, and show that the presence of four closed guard edges implies that the polygon is starshaped. Let P be a simple polygon where g1, g2,g3, and g4, in counterclockwise order, are guard edges. Let g1 = ab and g3 = cd such that a, b, c, and d are in counterclockwise order along P . Note that the vertices a, b, c, and d are distinct. In fect, it is equal to 3
all vertices of the geodesics path(a, c) and path(b, d) are in {a, b, c, d}. geodesic quadrilateral Q 重點 :g3,g1 一定要能看到彼此 ------------------- 考慮由 ab, path(b,c), cd, path(a,d) 形成的四邊形 Q ,所有由 ab 中任一點及 cd 中任一點所連起來的路徑必在 Q 之內。假設 path(b,c) 的一內部節點 q 是 path(a,d) 的一個節點。如果 q 是 a 或 path(a,d) 中任一內部節點,則 closed edge cd 無法看到 b (Lemma 1) 。同樣地,如果 q 是 d ,則 closed edge ab 看不到 c 。我們得到結論, path(b,c) 和 path(d,a) 是不相連 ( 因為 g3 和 g1 要能完全看到彼此,因為他們是 guard edges) ,而且 Q 是 simple polygon( 沒有洞在裡面擋住 ) 。
Figure 5: The convex hull of two closed guard edges, ab and cd, is either a quadrilateral or a triangle
Figure 5: The convex hull of two closed guard edges, ab and cd, is either a quadrilateral or a triangle w.l.o.g
Lemma 3 是直線 G1 2 3 4 是 guard edges
我們要證明任意的 point p 在多邊形 P 裡,都可以被 x 看到。 1. 根據推論 7 ,三角形 (adx) (cdx) 是位在 P 裡,所以如果 p 是在這兩個三角形裡,則 px 線段也必在三角形裡。
Generality 2. 如果 p 是在三角形之外,不失一般性的假設 p 在 path(a,c) 和 path(b,d) 右邊,所以 p 和 guard edge g4 在不同邊。如果 path(p,x) 是直線,則 x 看得到 p 。
Generality 我們假設 path(p,x) 不是直線, path(p,x) 在最後一個內部節點 q 右轉,則 path(p,d) 也在 q 右轉。由於 guard edge cd 可以看到 p ,故必 q=c (Lemma1b) 。重提一下, p 到 g4 的路徑跨越 path(a,c) 和 path(b,d) 。由於 path(p,x) 在 c 右轉,每條從 p 到 g4 的路徑都在 c 右轉。但是 c 和 g4 不相連,而且 q4 看不到 p(lemma1b) ,和我們最初的假設 contradict 。我們得到結論, path(p,x) 是直線,所以 x 看得到 p 。 (g4 也是 closed guard edge , g4 必需要能看到 p 及所有多邊形內部 )
Proof If a simple polygon has four closed guard edges, then it has a nonempty kernel by Lemma 8, and thus is starshaped. So every nonstarshaped simple polygon has at most three closed guard edges.