Drawing a Planar Graph on a Grid

1
Drawing a Planar Graph on a Grid
2
Last Week

• Any planar graph has a Straight-line drawing
• But it might take a lot of space

3
Grid

• Grid point a point with integer coordinates
• m by n grid the set of grid points (x, y) with
• 0 x m
• 0 y n
• Map each vertex in the graph to a grid point
• Goal - minimize the size of the needed grid

4
Theorem

• Any planar graph with n vertices has a
  straight-line embedding on the 2n-4 by n-2 grid.

n 6
n - 2
2n - 4
5
Proof

• G triangulation
• Exterior face of G u v w
• v1 u, v2 v, v3, ... , vn w canonical
  labeling of the vertices of G
• Gk the sub-graph of G induced by v1, v2, ... ,
  vk

G
G6
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Each graph Gk can be embedded on the 2k-4 by k-2
Grid

• By induction on k, show that Gk can be embedded
  on the 2k-4 by k-2 grid (k 3)
• k 3

f3(v3) (1, 1)
2k 4 2 K 2 1 2 by 1 grid

f3(v1) (0, 0)
f3(v2) (2, 0)
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Induction Hypothesis

• Assume we already foundfk-1(vi) (xk-1(vi),
  yk-1(vi)) (1 i k-1), (4 k)
• Where
• fk-1(v1) (0, 0), fk-1(v2) (2k 6, 0)
• Exterior face monotonic in x (cyclic order)
• The segments fk-1(ui) , fk-1(ui1) all have
  slopes 1 or -1

K 7, k-1 6 2k 4 10 K 2 5 10 by 5
grid
u3
u4
u2
u1
u6
fk-1(v1) (0, 0)
fk-1(v2) (2k 6, 0)
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Show for k

• Wed like to define fk(vk) (draw vk)
• up, up1, ... , uq the neighbors of vk in Gk
• Good Candidate P(up, uq)
• We might have a problem...

vks neighbors in Gk are u3, u4, u5





u3
u4
u2
u1
u6
f6(v1) (0, 0)
f6(v2) (8, 0)
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Problem

• Assume vks neighbors in Gk are u2, u3, u4, u5

How do we connect vk to u2 with a straight line?





u3
u4
u2
u1
u6
f6(v1) (0, 0)
f6(v2) (8, 0)
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Modify the previous embedding

• Modify fk-1 first
• Move fk-1(ui) where p1 i q-1 one unit to the
  right
• Move fk-1(uj) where q j m two units to the
  right

vks neighbors in Gk u2, u3, u4, u5





u2
u1
f6(v1) (0, 0)
f6(v2) (8, 0)
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Another Problem

• What about the interior vertices?

Insert v8 with neighbors u2, u3, u4, u5 in G8






u3
v7
u5
u4
u3
v6
u2
v4
u4
v5
u5
u1
v1
v2
u6
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Define a New Order

• Define a total order on v1, v2, ... , vn
  recursively
• v1 v3 v2
• Assume the order has already been defined for
  v1, v2, ... , vk-1
• Insert vk just before up1

f3(v3) (1, 1)

f3(v1) (0, 0)
f3(v2) (2, 0)
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Example

• In our example

v7s Neighbors in G7 u2, u3, u4, u5
v7





u5
u3
v6
u2
v4
u4
v5
u1
u6
v1
v2
v7
... v4 (u2)
v6 (u3) ...
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Solution continued

• Now modifing fk-1 will be defined as
• Move fk-1(ui) where up1 ui uq-1 one unit
  to the right
• Move fk-1(uj) where uq uj um two units to
  the right

Insert v8 with neighbors u2, u3, u4, u5 in G8






u3
v7
u4
v6
u2
v4
v5
u5
v2
u1
v1
... v4 v7 v6 ...
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Prove fk is a Straight Line Embedding

• Assume that fk-1 is a straight line embedding of
  Gk-1
• fk-1(vi) (xk-1(vi), yk-1(vi))
• a1, a2, ... ,am 0
• For any 1 i k-1 , 1 j m
• x(vi) xk-1(vi) a1 a2 ... aj if uj
  vi uj1
• y(vi) yk-1(vi)
• Then f k-1 (x, y) is also a straight line
  embedding of Gk-1

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Induction Base

• k 4
• a1, a2, a3, a4
• x(v1) xk-1(v1) a1 , since u1 v1 u2
• x(v4) xk-1(v4) a1 a2 , since u1 v4
  u2 ,u2 v4 u3
• x(v3) xk-1(v3) a1 a2 a3
• x(v2) xk-1(v2) a1 a2 a3 a4

v4 u2
v3
v2 u3
v1 u1
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Show For Gk

• u1, ... , up, vk, uq, ... , um the exterior
  face of Gk
• Fix a(u1), a(u2), ... , a(up), a(vk), a(uq), ...
  , a(um)
• Take Gk-1, and a1 a(u1), ... , ap a(up),
  ap1 a(vk) 1, ap2 0, ... , aq a(uq)
  1, aq1 a(uq1), ... , am a(um)
• Apply the induction hypothesis the restriction
  of f k to Gk-1 is a straight line embedding.
• Also vks edges dont create any crossing

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Questions?