CSCI 4260MATH 4150

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CSCI 4260/MATH 4150

• Planarity

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Planar Graphs

• A graph is called planar if it can be drawn on
  the plane in a way that no two of its edges cross
  each other

A planar graph?
Yes!
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Why care?

• Many practical applications.
• Nodes gates
• Edges wires between the gates
• Find a circuit layout w/o intersecting wires

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Three houses and three utilities
Is K(3,3) planar?
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Todays question

• Given a graph G
• If we can find a drawing of G where no edges
  cross, then we know that G is planar
• What if we cant?
• Either G is non-planar, or
• Its our fault
• Our goal is to find properties to determine if a
  graph is planar.
• For example, we just saw that K(4) is planar. How
  about K(5)?

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Plane graphs

• A plane graph is a drawing of a planar graph
  where no edges cross.
• More common term planar embedding

Not a plane-graph
A plane-graph
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Formal Definitions

• Curve image of a continuous map from 0,1 to
  R2
• A u-v curve starts at u and ends at v
• Closed curve start point endpoint
• Simple curve a curve that does not intersect
  itself
• Polygonal curve a curve composed of finitely
  many line segments
• A drawing of a graph G(V,E) is a function f on
  V?? E that assigns
• Each vertex to a distinct point
• Each edge (u,v) to a u-v curve
• A point in f(e) ?? f(e) that is not a common end
  point is a crossing

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Jordans Curve Theorem

• Every simple closed curve partitions the plane
  into exactly two regions the interior and
  exterior
• In this course, we will take this theorem as an
  intuitive notion
• Formal proof requires advanced concepts from
  topology

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Lets revisit K(5) and K(3,3)

• Proposition K(5) and K(3,3) are not planar.
• Both graphs have spanning cycles
• A chord of a cycle is an edge that connects two
  non-consecutive vertices on the cycle
• Suppose K(5) is planar.
• Then the spanning cycle C is a closed curve
• Chords of C must be drawn inside or outside

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K(5) is not planar

• Two (u,v) and (u,v) conflict if u lt v but vgtv
• Eg (2,4) and (3,4)
• When two chords conflict we can not draw them
  both inside (or outside)
• We can put at most two conflicting chords inside
  (or outside)
• There are five conflicting chords

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5
2
4
3
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K(3,3) is not planar

• Similar argument
• 6-cycle in K(3,3) has three conflicting chords
• We can put at most one inside and one outside

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Regions

• Let us remove the drawing from the plane
• A plane graph divides the plane into connected
  pieces called regions

Boundary of a region the subgraph that contains
vertices and edges that define the region (those
that are incident with the region)
R1
R3
R2
R4
R4 exterior (note that it is unbounded)
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Some observations

• There is always one (and only one) unbounded face
  
• Also known as the outer face
• An edge can be on the boundary of at most ?
  faces.
• If e is a bridge, then e is on the boundary of
  only one face
• Can you prove this?

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Euler Identity

• For any connected plane graph G, n-mr
  2wheren order of Gm size of Gr number
  of regions of G

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Proof

• Case I G has no cycles (i.e. a tree)
• Let n be the number of vertices
• m ?
• r ?

n (n-1) 1 2 So the theorem holds for trees
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Proof (cont)

• Else (G has a cycle)
• Assume that the theorem is not true
• Let G be the smallest size graph where the
  theorem does not hold
• G is not a tree
• G has a cycle (hence a non-bridge edge e)
• n m r ? 2
• Let G G - e
• The euler identity holds for G n (m-1)
  r 2
• What is r?

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Theorem

• If G is a planar graph of order n ? 3 and size
  m, m ? 3n 6
• Interpretation the number of edges is small. It
  is O(n) as opposed to O(n2)

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Proof

• Suppose G is connected
• Lets take care of some special cases
• G has 3 vertices
• If m 2, the inequality holds
• m 2 ? 3n 6 3
• G is connected and n ?3 and m ? 3
• Suppose G has r regions R_1, , R_r
• Let m_i number of edges on the boundary of R_i
• m_i ? 3

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Proof (cont)

• Lets add all the edges region by region
• In M, every edge appears at most twice, so M ?
  2m
• Hence we have 3r ? M ? 2m ? 3r ? 2m
• Apply Euler
• n - m r 2 ? 6 3n3m3r ? 3n-3m2m 3n
  m

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Proof (cont)

• So far, we established that for connected planar
  graphs m ? 3n 6
• What about disconnected graphs ?

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So far

• We have established our first sufficient
  condition (for non-planarity).
• Count the number of edges, if it is more than 3n
  -6 then non-planar
• In other words, m ? 3n 6 is a necessary
  condition for planarity (but we dont know if it
  is sufficient i.e. are there non-planar graphs
  that satisfy this condition)
• Lets revisit K(5)
• n 5
• m 54/2 10
• 3n-6 9
• So it is not planar

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Corollary

• If G is planar, then G contains a vertex of
  degree 5 or less.
• Planar ? vertex with degree ? 5
• All vertices have degree ? 6 ? non-planar

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(All vertices have degree ? 6) ? non-planar
Hence, G is not planar
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Lets go over some of the properties we used again

• In the sum
• An edge appears once if it is a bridge, twice
  otherwise

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Back to the three houses and utilities problem
K(3,3)

• K(3,3)
• n ?
• m ?
• m 9 ? 3n 6 12
• Recall that this condition was necessary for
  planarity.
• Since K(3,3) is not planar, we know that this
  condition is not sufficient.
• Using similar ideas, we can come up with an
  alternative proof of the nonplanarity of K(3,3)

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Theorem

• K(3,3) is non-planar.
• Proof
• Assume, to the contrary that, it is.
• Apply Euler identity
• n-mr 2 ? r 5
• Let
• R1, , R5 be the five regions and
• m1,,m5 be the number of edges on the boundaries
  as before

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Proof (cont)

• K(3,3) has no triangles, so mi ?4 for each Ri
• So m ? 10
• A contradiction!

No bridges
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Moral of the story

• Our sufficient condition is not necessary.
• But we know that if a graph contains K(5) or
  K(3,3) as a subgraph then it is not planar.
• Unfortunately, this is also not a
  sufficient-and-necessary condition
• There exist non-planar graphs which contain
  neither K(5) nor K(3,3).
• See the book for an example.

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Subdivision

• A graph G is a subdivision of G if it is
  obtained by adding one or more vertices of degree
  2 into edges of G

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Subdivisions

• Every subdivision of a planar graph is planar
• Why?
• Every subdivision of a non-planar graph is also
  non-planar
• So if a graph contains a subdivision of K(5) or
  K(3,3) we know that it is not planar

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Theorem (Kuratowski)

• A graph G is planar if and only if G does not
  contain
• K(5) or a subdivision of it
• K(3,3) or a subdivision of it

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Discussion

• Proof of Kuratowskis theorem is beyond the scope
  of our course.
• Now we have a sufficient-and-necessary condition
• Note planarity can be checked effiently.