Homework solution

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 SS1S0 , 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.
2
2
1
F2
2
1
F1
1
2

• Each face is even in D(G)
• Each node is even in G ? then each cycle is even
• F1F2F1?F2 \ F1? F2 make a characteristic
  vector
• e1 e2 ... ei ... e10
• (0 0 1 0) C1 (01100100)
• C2 (01001000)

4
Homework solution

• Problem 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.

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 solutionspaths matching odd nodes
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The T-join Problem

• How to delete minimum-cost set of edges from
  graph G to eliminate odd cycles?
• Construct geometric dual graph Ddual(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
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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

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Planar graphs-Eulerian formula

• Given G planar
• V- number of nodes
• E- number of edges
• F- number of faces
• Then Eulerian formula
• V-EF2
• 4-642

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Planar graphs

• Given G planar
• Then E ?? 3V-6
• ( of edges is of the same order as of
  vertices, EO(V) )
• Proof Eulerian formula ? V-EF2
• 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
• 3F2E ? F2/3 E
• V-E2/3 E2 ? 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 SCP
1
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

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Analysis of Greedy Algorithm for SCP

• Theorem APR of the Greedy Algorithm is at most
  1ln k
• Proof

15
Analysis of Greedy Algorithm for SCP

• Theorem SCP cannot be approximated in polynomial
  time for any clt1 up to c?ln k, kX
• ? NP-hard to approx. SCP with approx. ratio
• (1-?) ln K for any ? gt0 ? Greedy algorithm
  is the best for this problem. We will prove
• Theorem Greedy
• Opt
• Let Si Si ? X
• 1/ S1? Opt/X X1X
• X2X\S1
• X3X\S1\S2

? 1 ln X
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Analysis of Greedy Algorithm for SCP

• 1/ S2? Opt/X2 ? 1/ Si ? Opt/Xi (1)
• XiXi-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

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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 ? (-last1) ln ((1-1/Opt)
• Now we use the inequality
• ln(1x) ? x
• ? ln X1 / Xlast ? (-last1)(-1/Opt)1/Opt(last-1)

yx
yln(1x)
yln x
1
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Analysis of Greedy Algorithm for SCP

• We got
• ln X / Xlast ? 1/Opt (last-1)
• lastGreedy
• Greedy-1
• Opt

?
? ln X/Xlast Xlast is at least 1