CSCI 4260MATH 4150

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CSCI 4260-MATH 4150

• Lecture 16

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Looking ahead

• In the next few lectures, we will study some of
  the most difficult (from a computational
  viewpoint) problems in graph theory.
• Colorings, cliques, independent sets
• In this lecture, I would like to motivate the
  study of one such problem (coloring) with an
  important problem in computational geometry

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The Art Gallery Problem

• How many stationary guards are necessary to guard
  an art gallery?
• Lets make this precise

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Problem formulation

• Polygon
• V 0,,n-1 ? n points on the plane
• E (i,i1 mod n) 0?i?n
• Edges form a cycle
• Constraint edges dont intersect

Not allowed! (the interior is not well-defined)
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Note that

• We are not allowing holes inside the polygon
• Such polygons with no holes are sometimes
  referred to as simply-connected
• Essentially they are 2-cell regions on the plane

Not allowed!
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Art Gallery

• An art gallery will be represented by a polygon
• Now we need to define what it means to guard

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Visibility

• A polygon separates the plane into two regions
  exterior and interior
• For convenience, we will include the boundary in
  the interior as well
• We say two points a and b see each other if the
  line segment connecting them lies in the interior

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Guards

• A set S of points guard a polygon P if every
  point inside the polygon is visible from some
  point in S
• For this lecture, we will allow guards to be
  placed only on the vertices

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The guarding number

• g(P) the minimum-number of (vertex) guards to
  guard polygon P
• We can now state the art gallery problem
  mathematically

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The Art Gallery Problem

• What is the sufficient and necessary number of
  guards to guard a polygon with n vertices?
• k is the necessary number means that there is a
  polygon with n vertices that require k guards
• sufficient means that k is enough to guard any
  polygon with n vertices

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The Art Gallery Problem

• What is the sufficient and necessary number of
  guards to guard a polygon with n vertices?
• Is this number independent of n?

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Empirical ExplorationG(n) sufficient and
necessary number for n-vertex polygons

• G(n) gt 1
• G(n) lt n
• G(n) gt n/3

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The Art Gallery Theorem

• Floor(n/3) guards are sufficient and sometimes
  necessary to guard a polygon with n vertices.

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Proof of Sufficiency

• First proof by Chvatal in 1975, proof by
  induction
• Ill present Fisks proof (1978)

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Proof Step 1

• Step 1 Every polygon can be triangulated

Vertices of the triangles are chosen from the
vertices of the polygon
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Colorings

• A k-coloring of graph is an assignment of
  colors/labels to the vertices such that no two
  adjacent vertices get the same color and at most
  k colors are used.

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Proof Outline

• Step 1 Every polygon can be triangulated
• Step 2 The triangulation graph (whose vertices
  and edges edges correspond to the vertices and
  edges of triangles) can be 3 colored

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Proof Outline

• Step 1 Every polygon can be triangulated
• Step 2 Triangulaton Graph can be 3 colored
• After coloring, if we pick any color and place
  guards on the nodes that have that color, we
  cover the polygon. Why?
• If n objects are placed in k holes , at least one
  hole must contain no more than n/k objects. Why?

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Hence

• If we can prove our claims, we will be done
• Polygons can be triangulated
• Triangulation graph can be 3-colored

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Triangulation

• A diagonal A line segment that connects two
  non-consecutive vertices which can see each other
• That is, it lies inside the polygon

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Lemma

• Every polygon with 4 or more vertices has a
  diagonal

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Existence of a Diagonal

• Let a and b be two neighbors of a vertex v
• Case I a-b is a diagonal
• Case 2

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From diagonals to triangulations

• By induction on the number of vertices of the
  polyon P
• Basis P has 3 vertices. Triangulated.
• Inductive step
• Find a diagonal
• This splits P into two smaller polygons P1 and
  P2
• Apply the inductive hypothesis to P1 and P2
• Combine the triangulations

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Properties of Triangulations

• Every triangulation of P with n vertices has
  (n-3) diagonals and (n-2) triangles
• Corollary Sum of internal angles (n-2)?

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Triangulation Dual
The dual T of a triangulation is a
graph Vertices triangles (u,v) is an edge iff
corresponding triangles share a side
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Lemma

• Triangulation dual is a tree

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Proof

• By contradiction.
• Suppose we had a cycle C v1, , vk
• Start at v1 (located inside a triangle)
• Follow the edges of C looking right keeping
  track of the boundary of the polygon
• How can you come back at v1 and keep observing
  the boundary?

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3-coloring of the triangulation graph(not the
dual)

• Again by induction on the number of vertices of
  the polygon
• Basis n 3 (3 colorable)
• Inductive step Since the dual is a tree it must
  have at least 2 leaves
• That is, there is triangle T that is adjacent to
  only one other triangle
• T (a, b, c) T (b,c,d)
• Remove a from the triangulation and apply the
  ind. hyp.
• To color a, pick the unused color that was not
  assigned to b or c
• Done!

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Recap Proof Outline

• Step 1 Every polygon can be triangulated
• Step 2 Triangulaton Graph can be 3 colored
• After coloring, if we pick any color and place
  guards on the nodes that have that color, we
  cover the polygon.
• If n objects are placed in k holes , at least one
  hole must contain no more than n/k objects.