1. Homework solution
Problem 2. The number of odd degree vertices in a graph is even.
(recom. book: G. Harary: Graph Theory)
Solution: Let G=(V,E,w)
S=vVdeg(v) = 2, =|E|
Let S=S1+S0 , where S1= v odd deg(v)
S0= v even deg(v)
S is even, S0 is even (sum of even numbers) 
S1 is even. Since S1 is a sum of odd numbers, it has
to be an even numbers of odd degree vertices in G.

2. Homework solution Problem 1. Graph G is Eulerian  D(G) is bipartite.
Solution: definitions:
Face of a planar graph = cycle without “diagonals”
Given: graph G=(V,E) with the set of faces F.
G’ is dual to G if G’=(F’,E’) s.t. for any face fF there is a vertex f’F’ of graph G’, for any edge eE there is an edge e’E’ (1-1 correspondence) such that if ef1, f2, then e’=(f1’,f2’).
1st face (cycle goes counterclockwise)
2nd face (cycle goes clockwise)

3. Homework solution Problem 1. F2 F1 Each face is even in D(G)
Each node is even in G  then each cycle is even
F1+F2=F1F2 \ F1 F2 make a “characteristic vector”:
e1 e2 ... ei ... e10
( ) C1 ( )
C2 ( )

4. Homework solution Problem 1. F1 (0001110001) F2 (1111000000)
C=( )
if our face is even  then every cycle is even
Bipartite graph
Theorem:
Graph G is bipartite iff G does not have odd cycles.
 Graph will not have odd cycles because the dual is Eulerian.

5. Planar bipartite graph
Theorem: Graph G is bipartite  G does not have odd cycles.
Problem (The T-join problem):
Given a planar, not bipartite, graph G = (V,E).
Find the set H with the minimum number of edges such that G’= (V, E \ H) is bipartite.
Solution: Construct the dual graph D(G). G not bipartite  D(G) is not Eulerian  D(G) has at least 2 odd-degree vertices. Take 1 edge from the odd degree vertex, then delete the corresponding edge in the original graph.

6. Planar bipartite graph
Solution: we are deleting edges in D(G) to get Eulerian graph. D(G) Eulerin  G bipartite.
Any solution consists of paths in G matching odd-degree vertices.
We need to find minimum size such paths that match odd-degree vertices.
All solutions=paths matching odd nodes

7. The T-join Problem
How to delete minimum-cost set of edges from graph G to eliminate odd cycles?
Construct geometric dual graph D=dual(G)
Find odd-degree vertices T in D
Solve the T-join problem in D:
find min-weight edge set J in D such that
all T-vertices have odd degree
all other vertices have even degree
Solution J corresponds to desired min-cost edge set in graph G

8. Optimal Odd Cycle Elimination
conflict graph G
dual graph D
T-join odd degree nodes in D

9. T-join Problem in Sparse Graphs
Reduction to matching
construct a complete graph T(G)
vertices = T-vertices
edge costs = shortest-path cost
find minimum-cost perfect matching
Typical example = sparse (not always planar) graph
note that conflict graphs are sparse
#vertices = 1,000,000
#edges  5  #vertices
# T-vertices  10% of #vertices = 100,000
Drawback: finding all shortest paths too slow and memory-consuming
#vertices = 100,000 ® #edges = 5,000,000,000

10. Planar graphs-Eulerian formula
Given: G planar
V- number of nodes
E- number of edges
F- number of faces
Then: Eulerian formula
V-E+F=2
4-6+4=2

11. Planar graphs Given: G planar Then: E  3V-6
(# of edges is of the same order as # of vertices, E=O(V) )
Proof: Eulerian formula  V-E+F=2
In maximal planar graph each face is a triangle (triangulation)
-each edge incident with 2 faces, we count each edge twice 
3/2 F = E
3F=2E  F=2/3 E
V-E+2/3 E=2  1/3 E = V-2 (for max. graph)  in any graph: E  3V-6

12. The Set Cover Problem Sets Ai cover a set X if X is a union of Ai
Weighted Set Cover Problem
Given:
A finite set X (the ground set X)
A family F of subsets of X, with weights w: F  +
Find:
sets S  F, such that
S covers X, X = {s | s  S} and
S has the minimum total weight  {w(s) | s  S}
If w(s) =1 (unweighted), then minimum # of sets

13. Greedy Algorithm for SCP1 2 6 3 4 5
Greedy Algorithm:
While X is not empty
find s  F minimizing w(s) / |s  X|
X = X - s
C = C + s
Return C

14. Analysis of Greedy Algorithm for SCP
Theorem: APR of the Greedy Algorithm is at most 1+ln k
Proof:

15. Analysis of Greedy Algorithm for SCP
Theorem: SCP cannot be approximated in polynomial time for any c<1 up to cln k, k=|X|
 [ NP-hard to approx. SCP with approx. ratio
(1-) ln K for any  >0 ]  Greedy algorithm is the best for this problem. We will prove:
Theorem: Greedy
Opt
Let Si=| Si  X|
1/ S1 Opt/|X| X1=|X|
X2=|X\S1|
X3=|X\S1\S2|
 1+ ln |X|

16. Analysis of Greedy Algorithm for SCP
1/ S2 Opt/|X2|  1/ Si  Opt/|Xi| (1)
Xi=Xi-1-Si-1
(1)  Si  Xi/Opt
 Xi  Xi-1- Xi-1/Opt= Xi-1(1-1/Opt)
Xi  Xi-1(1-1/Opt)
Xlast  Xlast-1 (1-1/Opt) 
Xi-1/ Xi  (1-1/Opt)-1
X1 / X2  (1-1/Opt)-1
X2 / X3  (1-1/Opt)-1 
Xlast-1 / Xlast  (1-1/Opt)-1

17. Analysis of Greedy Algorithm for SCP
By multiplying all these terms, we get:
X1 / Xlast  (1-1/Opt)-(last-1) / take a logarithm
ln X1 / Xlast  (-last+1) ln ((1-1/Opt)
Now we use the inequality:
ln(1+x)  x
 ln X1 / Xlast  (-last+1)(-1/Opt)=1/Opt(last-1)
y=x
y=ln(1+x)
y=ln x 1

18. Analysis of Greedy Algorithm for SCP
We got:
ln X / Xlast  1/Opt (last-1)
last=|Greedy|
Greedy-1
Opt 
 ln X/Xlast Xlast is at least 1