Mathematics for Computer Science
Mathematics for Computer Science
MIT 6.042J/18.062J
MIT 6.042J/18.062J
Planar Graphs;
Bipartite Matching
Copyright © Albert R. Meyer, 2005.
Planar Graphs
lec 5f.1
October 7, 2005
Planar Graphs
Copyright © Albert R. Meyer, 2005.
lec 5f.2
October 7, 2005
Planar Graphs
A graph is planar if there
is a way to draw it in the plane
without edges crossing.
Copyright © Albert R. Meyer, 2005.
lec 5f.3
October 7, 2005
Copyright © Albert R. Meyer, 2005.
A Planar Graph
Planar Graphs
Maps are 2-connected planar graphs
with 43 faces (wait! also the outer face)
General connected planar graphs may have
dongles
lec 5f.4
October 7, 2005
1
2
4
3
cross bars
4
3
1
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.5
Copyright © Albert R. Meyer, 2005.
October 7, 2005
2
lec 5f.6
1
Planar Graphs
Planar Graphs
and record faces while drawing
draw it edge by edge:
graph
faces
(the outer face)
Copyright © Albert R. Meyer, 2005.
lec 5f.7
October 7, 2005
Planar Graphs
Copyright © Albert R. Meyer, 2005.
Planar Graphs
and record faces while drawing
and record faces while drawing
graph
graph
faces
faces
(the outer face)
Copyright © Albert R. Meyer, 2005.
(the outer face)
lec 5f.9
October 7, 2005
If you like curves…
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.10
“Planar Drawing” = Faces
With same faces, you can draw the
graph in the plane big or small,
curvy or straight:
An (abstract) planar drawing
is defined to be its set of faces.
The same planar graph may
have different drawings.
4
1
lec 5f.8
October 7, 2005
2
3
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.11
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.12
2
Euler's Formula
Team Problem
If a connected planar
drawing has v vertices,
e edges, and f faces,
then
Problem 1
v–e+f = 2
Copyright © Albert R. Meyer, 2005.
lec 5f.13
October 7, 2005
Euler's Formula
Copyright © Albert R. Meyer, 2005.
lec 5f.14
October 7, 2005
Adding an edge to a drawing
Inductive step: any n+1 edge drawing comes from adding an edge to some n
edge drawing.
(not a buildup error: it’s the definition of
drawing edge by edge)
So can assume Euler for n edge drawing
and see what happens to
v–e+f when 1 edge is added.
Proof by induction on # edges in drawing:
base case: no edges
connected,
so v = 1
outside face only, so f = 1
e=0
1−0+1 = 2
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.15
Adding an edge to a drawing
Copyright © Albert R. Meyer, 2005.
lec 5f.16
October 7, 2005
Face Creation Rules
1) choose face add edge to new vertex
Two cases for connected graph:
1) Attach edge from vertex on a
face to a new vertex.
2) Attach edge between vertices
on a face.
path x
w
v
old face vxv
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.17
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.18
3
Face Creation Rules
Face Creation Rules
1) choose face add edge to new vertex
1) choose face add edge to new vertex
path x
nothing else changes
w
v
new face is wvxvw
Copyright © Albert R. Meyer, 2005.
new face is wvxvw
lec 5f.19
October 7, 2005
Euler’s Formula
Copyright © Albert R. Meyer, 2005.
lec 5f.20
October 7, 2005
Face Creation Rules
2) choose face add edge across it
w
v increases by 1
e increases by 1
f stays the same
x
y
v
so v–e+f stays the same
old face: wxvyw
Copyright © Albert R. Meyer, 2005.
lec 5f.21
October 7, 2005
Face Creation Rules
Copyright © Albert R. Meyer, 2005.
lec 5f.22
October 7, 2005
Face Creation Rules
2) choose face add edge across it
w
2) choose face add edge across it
w
w
x
v
v
splits into 2 faces:
Copyright © Albert R. Meyer, 2005.
x
y
y
October 7, 2005
v
splits into 2 faces: wxvw, vywv
lec 5f.23
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.24
4
Face Creation Rules
Euler’s Formula
2) choose face add edge across it
v stays the same
e increases by 1
f increases by 1
so v–e+f stays the same
nothing else changes
splits into 2 faces: wxvw, vywv
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.25
Copyright © Albert R. Meyer, 2005.
Team Problems
Euler’s Formula
Problems
2 & 3
Inductive step:
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.26
October 7, 2005
lec 5f.27
Copyright © Albert R. Meyer, 2005.
lec 5f.28
October 7, 2005
Compatible Boys & Girls
Mathematics for Computer Science
MIT 6.042J/18.062J
Bipartite Matching:
Hall’s Theorem
B
G
compatible
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.29
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.30
5
Compatible Boys & Girls
Compatible Boys & Girls
B
G
match each girl to a
unique compatible boy
Copyright © Albert R. Meyer, 2005.
a matching
lec 5f.31
October 7, 2005
Copyright © Albert R. Meyer, 2005.
Compatible Boys & Girls
no match possible
lec 5f.33
October 7, 2005
Copyright © Albert R. Meyer, 2005.
No match possible
lec 5f.34
October 7, 2005
Bottleneck condition
B
G
B
G
suppose this edge was missing
Copyright © Albert R. Meyer, 2005.
lec 5f.32
October 7, 2005
Compatible Boys & Girls
B
G
B
G
B
G
N(S)
S
|N(S)|=2
|S|=3
because of bottleneck
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.35
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.36
6
Bottleneck Lemma
Hall’s Theorem
bottleneck: not enough boys for
some set of girls.
If there is a bottleneck,
then no match is possible.
S ⊆ G, N(S ) ::= {b | b adjacent to a g ∈ S } ,
| S | > | N(S ) |
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.37
Hall’s Theorem
There is a perfect match iff
there are no bottlenecks.
Proof in Notes: clever strong
induction on #girls.
(Better proof using duality principle
goes beyond 6.042)
Copyright © Albert R. Meyer, 2005.
Problem 4
Lots of elegant use in applications
& proofs
October 7, 2005
lec 5f.38
Team Problem
There is a perfect match iff
there are no bottlenecks.
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.39
Copyright © Albert R. Meyer, 2005.
October 7, 2005
lec 5f.40
7