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CSCE 3110 Data Structures & Algorithm Analysis

CSCE 3110 Data Structures & Algorithm Analysis. Rada Mihalcea http://www.cs.unt.edu/~rada/CSCE3110 Graphs (I) Reading: Chap.9, Weiss. Network reliability. Distributed computing. Distributed computing. Many many more . . . Max flow and min cut. Two algorithmic problems.

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CSCE 3110 Data Structures & Algorithm Analysis

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  1. CSCE 3110Data Structures & Algorithm Analysis Rada Mihalcea http://www.cs.unt.edu/~rada/CSCE3110 Graphs (I) Reading: Chap.9, Weiss

  2. Network reliability. Distributed computing. Distributed computing. Many many more . . . Max flow and min cut. Two algorithmic problems. See Cormen et.al for more details Nontrivial applications / reductions. Network connectivity. Traffic control Data mining. Airline scheduling. Project selection. General scheduling Maximum Flow and Minimum Cut

  3. Max Flow Network • Max flow network: G = (V, E, s, t, u) . • (V, E) = directed graph, no parallel arcs. • Two distinguished nodes: s = source, t = sink. • u(e) = capacity of arc e. 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 Capacity 15 4 7 30

  4. Flows • An s-t flow is a function f: E   that satisfies: • For each e  E: 0  f(e)  u(e) (capacity) • For each v  V – {s, t}: (conservation) 0 2 5 9 4 0 0 10 15 15 0 10 4 4 0 4 4 5 s 3 6 t 8 10 0 0 15 0 4 0 6 10 Capacity 15 0 Flow 4 7 30 0

  5. Flows • An s-t flow is a function f: E   that satisfies: • For each e  E: 0  f(e)  u(e) (capacity) • For each v  V – {s, t}: (conservation) • MAX FLOW: find s-t flow that • maximizes net flow out of the source. 0 2 5 9 4 0 0 10 15 15 0 10 4 4 0 4 4 5 s 3 6 t 8 10 0 0 15 0 4 0 6 10 Capacity Value = 4 15 0 Flow 4 7 30 0

  6. Flows • An s-t flow is a function f: E   that satisfies: • For each e  E: 0  f(e)  u(e) (capacity) • For each v  V – {s, t}: (conservation) • MAX FLOW: find s-t flow that • maximizes net flow out of the source. 6 2 5 9 10 6 0 10 15 15 0 10 4 4 3 8 8 5 s 3 6 t 8 10 1 10 15 0 4 0 6 10 Capacity 15 Value = 24 11 Flow 4 7 30 11

  7. Flows • An s-t flow is a function f: E   that satisfies: • For each e  E: 0  f(e)  u(e) (capacity) • For each v  V – {s, t}: (conservation) • MAX FLOW: find s-t flow that • maximizes net flow out of the source. 9 2 5 9 10 9 1 10 15 15 0 10 4 0 4 9 8 5 s 3 6 t 8 10 4 10 15 0 4 0 6 10 Capacity 15 Value = 28 14 Flow 4 7 30 14

  8. Networks Flow Examples Network Nodes Arcs Flow communication telephone exchanges,computers, satellites cables, fiber optics,microwave relays voice, video,packets circuits gates, registers,processors wires current mechanical joints rods, beams, springs heat, energy hydraulic reservoirs, pumpingstations, lakes pipelines fluid, oil financial stocks, currency transactions money transportation airports, rail yards,street intersections highways, railbeds,airway routes freight,vehicles,passengers chemical sites bonds energy

  9. Cuts • An s-t cut is a node partition (S, T) such that s  S, t T. • The capacity of an s-t cut (S, T) is: • Min s-t cut: find an s-t cut of minimum capacity. 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 Capacity = 30 4 7 30

  10. Cuts • An s-t cut is a node partition (S, T) such that s  S, t T. • The capacity of an s-t cut (S, T) is: • Min s-t cut: find an s-t cut of minimum capacity. 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 Capacity = 62 4 7 30

  11. Cuts • An s-t cut is a node partition (S, T) such that s  S, t T. • The capacity of an s-t cut (S, T) is: • Min s-t cut: find an s-t cut of minimum capacity. 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 Capacity = 28 4 7 30

  12. Flows and Cuts • L1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. 6 2 5 9 10 6 0 10 15 15 0 10 4 4 4 8 8 5 s 3 6 t 8 10 0 10 15 0 4 0 6 10 15 10 4 7 30 Value = 24 10

  13. Flows and Cuts • L1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. 6 2 5 9 10 6 0 10 15 15 0 10 4 4 4 8 8 5 s 3 6 t 8 10 0 10 15 0 4 0 6 10 15 10 4 7 30 Value = 24 10

  14. Flows and Cuts • L1. Let f be a flow, and let (S, T) be a cut. Then, the net flow sent across the cut is equal to the amount reaching t. 6 2 5 9 10 6 0 10 15 15 0 10 4 4 4 8 8 5 s 3 6 t 8 10 0 10 15 0 4 0 6 10 15 10 4 7 30 Value = 24 10

  15. Flows and Cuts • Let f be a flow, and let (S, T) be a cut. Then, • Proof by induction on |S|. • Base case: S = { s }. • Inductive hypothesis: assume true for |S’| = k-1. • consider cut (S, T) with |S| = k • S = S'  { v } for some v  s, t, |S' | = k-1  cap(S', T') = | f |. • adding v to S' increase cut capacity by t t v v s s Before S' S After

  16. Flows and Cuts • L2. Let f be a flow, and let (S, T) be a cut. Then, | f |  cap(S, T). • Proof. • Corollary.Let f be a flow, and let (S, T) be a cut. If |f| = cap(S, T), then f is a max flow and (S, T) is a min cut. S T 4 8 t s 7 6

  17. Max Flow and Min Cut • Corollary.Let f be a flow, and let (S, T) be a cut. If |f| = cap(S, T), then f is a max flow and (S, T) is a min cut. 9 2 5 9 10 9 1 10 15 15 0 10 4 0 4 9 8 5 s 3 6 t 8 10 4 10 15 0 4 0 6 10 15 15 4 7 30 15 Flow value = 28 Cut capacity = 28

  18. Max-Flow Min-Cut Theorem • MAX-FLOW MIN-CUT THEOREM (Ford-Fulkerson, 1956): In any network, the value of the max flow is equal to the value of the min cut. 9 2 5 9 10 9 1 10 15 15 0 10 4 0 4 9 8 5 s 3 6 t 8 10 4 10 15 0 4 0 6 10 15 15 4 7 30 15 Flow value = 28 Cut capacity = 28

  19. Max-Flow Min-Cut Theorem • Intuitive Proof: • Lemma 2  | f |  cap(S, T). • Max for |f| is a certain cut C where Max|f| = C • Min cut for cap(S,T) is a certain flow F where |f| = Min cut • |f|  Max |f| = C  Min cut(S,T)  • Max |f|  Min cut(S,T) • L2  Max |f| Min cut(S,T) •  Max |f| = Min cut(S,T)

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