Open guard edges and edge guards in simple polygons

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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, [email protected]

2013/06/10

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• Guard edges

• Assign. 2 - art gallery problem

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• Terms explanation

• Upper bound of open/closed guard

edges of Non-starshaped simple

polygon.

• Lemma

Outline

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Terms explanation

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Closed guard edges

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Open guard edges

Open guard edges not include end points

可以看到kernel

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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.

這裡我們講path(p,q) 就是Geodesic path(p,q)

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weakly visible

• A point is weakly visible to a set of

points

• this point is visible from some point in

that set.

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A starshaped n-gon P with kernel(P)

只有starshaped才有kernel

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Star-shaped polygon

https://en.wikipedia.org/wiki/Star-shaped_polygon

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.

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Non-starshaped simple polygon

No kernel

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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)

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Point p is visible from an open edge uv ⇔p is the only common vertex of path(p, u)

and path(p, v)

geodesics

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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).

Closed edge uv 看得到P，則path(p,u)和path(p,v)任何共同的節點，必是p,u,v其中

一個。

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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).

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OPEN GUARD EDGES

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A simple polygon with open

guard edges

• ≥ 2 ⇔ starshaped.

• ≤ 1 ⇔ Non-starshaped

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Lemma 2Given 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.

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• 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

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。

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• 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.

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。

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Lemma 3 Given ab, cd are open guard edges.

⇒The intersection point x = ac ∩ bd is in

the kernel of P

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• 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.

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。

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• 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.

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。

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Theorem 4Non-starshaped simple polygon ⇔ open

guard edge ≤ 1

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)

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Theorem 4• Non-starshaped simple polygon ⇒ open guard edge ≤ 1

• Open guard edge

≥ 2

• has kernel(P)

(Lemma 3)

• A starshaped polygon

• Contradiction

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CLOSED GUARD EDGES

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Theorem 5

non-starshaped simple polygon

⇔

closed guard edges ≤ 3

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

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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}.

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)。

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• 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}

Path(a,c)和path(b,d)在Q裡，所以任何path(a,c)和path(b,d)的內部節點，都會在Q

裡。如果path(a,c)和path(b,c)有共同的內部節點，則ab不能看到c。同樣地，

{a,b}和{c,d}上任兩點連成的路徑線，不能有任何的共同內部節點。因此，

path(a,c)和path(b,d)的所有內部節點都一定在四邊形{a,b,c,d}裡。

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Corollary 7

• If □{a, b, c, d} is convex ⇒ path(a,c)

and path(b,d) are straight line.

Figure 5: The convex hull of two closed guard edges, ab and cd, is either a

quadrilateral or a triangle

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Corollary 7convex({a, b, c, d})=△(abc)

⇒path(a,c) = (a,d,c) and path(b,d) = bd.

Figure 5: The convex hull of two closed guard edges, ab and

cd, is either a quadrilateral or a triangle

w.l.o.g

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Lemma 8 • The intersection point x = path(a, c) ∩

path(b, d) is in the kernel of P .

g2

Lemma 3 是直線

G1 2 3 4是guard edges

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• 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.

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。

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• 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)

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。

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• 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

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)

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Theorem 5

non-starshaped simple polygon

⇔

closed guard edges ≤ 3

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.

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TO FIND KERNEL(P)

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Left and right kernels