Homework solutionProblem 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=vVdeg(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.

Homework solutionProblem 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 fF there is a vertex f’F’ of graph G’, for any edge eE there is an edge e’E’ (1-1 correspondence) such that if ef1, f2, then e’=(f1’,f2’).

1st face (cycle goes counterclockwise)

2nd face (cycle goes clockwise)

Homework solution

Each face is even in D(G)

Each node is even in G then each cycle is even

F1+F2=F1F2 \ F1 F2 make a “characteristic vector”:

e1 e2 ... ei ... e10

(0 0 1 0) C1 (01100100)

C2 (01001000)

1 2

1 2

2

12

F1

F2

Problem 1.

Homework solutionProblem 1.

F1 (0001110001)

F2 (1111000000)

C=(1110110001)

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.

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.

Planar bipartite graphSolution: 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

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

Optimal Odd Cycle Elimination

conflict graph G

dual graph D

T-join odd degree nodes in D

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

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

Planar graphsGiven: 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=2In maximal planar graph each face is a triangle (triangulation)

-each edge incident with 2 faces, we count each edge twice 3/2 F = E3F=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

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

Greedy Algorithm for SCP

• Greedy Algorithm:– While X is not empty

• find s F minimizing w(s) / |s X| • X = X - s• C = C + s

– Return C

6

3 4 5

2

1

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

Proof:

iii nXS ||

11 iii nkk

ii

i

k

opt

n

w

opt

kwn iii

opt

wkk

opt

wkk i

iii

ii 1111

|| 1 ii Sk

opt

wkk i

last 10

opt

w

opt

w

k

k ii

last

1ln 0xx )1ln(

opt

approx

k

k

last

0ln1

Analysis of Greedy Algorithm for SCPTheorem: SCP cannot be approximated in polynomial

time for any c<1 up to cln 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|

Analysis of Greedy Algorithm for SCP1/ 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

Analysis of Greedy Algorithm for SCPBy 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)

1

y=x

y=ln x

y=ln(1+x)

Analysis of Greedy Algorithm for SCPWe got:

ln X / Xlast 1/Opt (last-1)

last=|Greedy|Greedy-1

Opt ln X/Xlast Xlast is at least 1