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Outline. Minimum Spanning Tree Maximal Flow Algorithm LP formulation. 1. The Minimum Spanning Tree Problem 最小延展樹問題. 2. 8.1.2 The Minimum Spanning Tree Problem. application context nodes are cities （節點是城市） arcs are potential highways to be built to connect the cities （線段是連接城市的高速公路）

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  1. Outline Minimum Spanning Tree Maximal Flow Algorithm LP formulation 1

  2. The Minimum Spanning Tree Problem最小延展樹問題 2

  3. 8.1.2 The Minimum Spanning Tree Problem application context nodes are cities （節點是城市） arcs are potential highways to be built to connect the cities （線段是連接城市的高速公路） what is the cheapest way (i.e., shortest total distance) to have all cities connected? （最短的公路網將所有城市連接起來） 3

  4. 8.1.2 The Minimum Spanning Tree Problem general concept tree: a connected network such that the n nodes of the network are connected by n-1 arcs which is a tree? 4

  5. 8.1.2 The Minimum Spanning Tree Problem think about this the cheapest way to connect all cities should be a tree called spanning tree, as it connects all cities the only question: how to find the cheapest spanning tree? 只要細想，每個人都懂。 5

  6. 8.1.2 The Minimum Spanning Tree Problem question: would the cheapest arc always in the minimum spanning tree? must be an arc in the spanning tree an arc not in the spanning tree the cheapest arc, not in the spanning tree 6

  7. 8.1.2 The Minimum Spanning Tree Problem question: would we find the minimum spanning tree by always selecting the cheapest unselected arc? no, see, e.g., the cheapest three arcs. 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 7

  8. 8.1.2 The Minimum Spanning Tree Problem Let us grow the minimum spanning tree step by step, i.e., add the arcs one by one. （一步一步的養大這minimum spanning tree.） As argued before, arc (1, 2) must be in the minimum spanning tree. Thus, we have a small tree T{1, 2}. 4 26 22 4 26 2 22 6 13 18 27 24 2 6 13 18 20 1 33 27 24 5 30 20 1 33 32 18 5 3 30 32 18 7 3 49 7 49 8

  9. 8.1.2 The Minimum Spanning Tree Problem In the minimum spanning tree, which of the three arcs, (1, 3), (2, 3), and (2, 4), should be connected to the tree T{1, 2} grown so far? Arc (1, 3) of length 18 should be in the minimum spanning tree. Why? Considered a spanning tree without arc (1, 3). The cost of the spanning tree can be reduced by swapping an arc with higher cost with arc (1, 3). Thus, arc (1, 3) should be in the minimum spanning tree. 4 26 22 2 6 13 18 27 24 20 1 33 4 26 22 5 30 32 18 2 3 6 13 18 27 24 7 49 20 1 33 5 30 32 18 3 7 49 9

  10. 8.1.2 The Minimum Spanning Tree Problem to find the minimum spanning tree: start with the arc of minimum cost to from a tree （從最便宜的公路開始） grow the tree by adding an arc connecting to it such that （每次加一條公路） adding the arc maintains a tree （保持tree的形式）, and among possible arcs to add, the added arc is of the lowest cost （選能加上的公路中最便宜的） 10

  11. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 11

  12. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 12

  13. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 13

  14. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 14

  15. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 15

  16. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 16

  17. 8.1.2 The Minimum Spanning Tree Problem 4 26 22 2 6 13 18 27 24 20 1 33 5 30 32 18 3 7 49 17

  18. The Maximal Flow Problem最大流量問題 18

  19. Railway Map Each railway segment has a capacity (i.e., upper bound of tonnage (per day)). What is the maximal flow from, say, Paris to Moscow? http://www.worldchanging.com/archives/009028.html 19

  20. The Maximal Flow Problem The number beside an arc is the capacity of flow along the arc. （管線旁的數字是管線的容量） The movement can be of either direction. What is the maximal flow from node 1 to node 4? 2 6 10 4 1 2 3 6 3 20

  21. The Maximal Flow Problem At any moment, we need to keep track of the flow and the residual capacity (i.e., remaining capacity) of an arc. 2 0 0 6 10 0 0 0 4 1 2 0 0 3 6 3 number beside an arc: boxed = residual capacity, normal = current flow. 21

  22. The Maximal Flow Problem 2 2 0 6 6 0 0 6 10 4 6 0 6 0 0 0 4 4 1 1 2 2 0 0 0 0 3 3 6 6 3 3 The route 124 allows a flow of 6 units. 22

  23. The Maximal Flow Problem 2 6 6 0 4 6 6 0 4 1 2 0 0 3 6 3 The route 1234 allows a flow of 2 units. 2 6 8 6+2 6+2 0 2 6 6 2 4 1 0 0 2 3 4 3 23

  24. The Maximal Flow Problem 2 6 8 6+2 6+2 0 2 6 6 2 4 1 0 0 2 3 4 3 The route 134 allows a a flow of 3 units. 2 6+2+3 6+2+3 6 8 6+2 6+2 0 2 6 6 2 4 1 0 3 2+3 0 1 3 24

  25. The Maximal Flow Problem The maximal flow from node 1 to node 4 is 11 units. The flow pattern is shown below, with the residual capacities of arcs boxed. 2 6 8 0 2 11 11 2 4 1 0 3 5 0 1 3 25

  26. The Maximal Flow Problem The method seems all right: We first find a route of positive flow from the source to the sink. Find the residual capacities of arcs by subtracting added flows from their (residual) capacities. Keep on repeating the above two steps until no route with positive flow can be added. However, the method can run into trouble. 26

  27. The Maximal Flow Problem 2 6 10 2 0 0 4 1 2 6 10 0 0 0 4 1 3 6 2 0 0 3 3 6 3 27

  28. The Maximal Flow Problem 2 2 2 0 0 0 6 4 10 10 0 2 0 2 0 2 4 4 1 1 0 2 2 0 0 0 1 3 6 6 3 3 The route 1324 allows a flow of 2 units. 28

  29. The Maximal Flow Problem 2 2 2 6 4 0 0 4 10 6 6 2 6 2 2 2 4 4 1 1 0 0 2 2 0 0 1 1 6 6 3 3 The route 124 allows a flow of 4 units. 29

  30. The Maximal Flow Problem 2 2 6 6 4 4 0 0 6 6 7 6 7 6 2 2 4 4 1 1 0 0 2 3 0 1 0 1 5 6 3 3 The route 134 allows a flow of 1 unit. 30

  31. The Maximal Flow Problem There seems to be no more flow to node 4, because residual capacity of arc (1, 3) = 0, arc (2, 3) = 0, arc (2, 4) = 0 2 6 4 0 6 7 7 2 4 1 0 3 1 0 5 3 31

  32. The Maximal Flow Problem However, this does not make sense, because the flow is found to be 11 units before. What is wrong? 2 6 4 7 7 2 4 1 3 1 3 32

  33. The Maximal Flow Problem Compare the two flow patterns. Any observations? In the second case, the flow in the vertical arc is in the wrong direction. 2 2 6 4 6 8 0 6 7 7 2 0 2 4 1 11 11 2 4 1 0 3 0 1 3 5 0 5 0 3 1 3 33

  34. The Maximal Flow Problem In a large network, how can we be sure which direction to follow when we add flows one by one? There should be ways to revert the direction of a wrong flow. 2 4 4 2 7 7 2 4 1 2 1 1 3 34

  35. The Maximal Flow Problem Suppose that we send 2 units from node 1 to node 2 to take up the capacity of node 2 to node 4 of the red flow. The red flow is along arc (3, 2) is then re-directed to arc (3, 4). 2 4 4 2 4 2 7 7 4 2 4 1 6 2 2 7+2 0 2 7+2 1 4 1 0 1 2 2 3 0 5 1 1 3 35

  36. The Maximal Flow Problem The result: Two more units of flow can be sent along the route 1234 2 2 6 6 6+2 6 2 4 9 9+2 0 0 9 9+2 2 0 4 4 1 1 0 2 3 3 3 5 1 3 0 0 3 3 36

  37. The Maximal Flow Problem There is a labeling algorithm basically keeping track of the flows, the residual capacities, and the re-direction of flows. The algorithm works for networks with capacity limits on bi-directional flows on arcs. It is possible to have different limits for the two directions of the same arc. （容許i到j跟j到i的capacity不一樣） 3 5 j i the capacity of i j flow the capacity of j i flow 37

  38. The Maximal Flow Problem Find the maximal flow from node 1 to node 4 for the following network. The number beside an arc is the capacity limit of the arc. All arcs are bi-directional. 2 6 10 4 1 2 3 6 3 38

  39. The Maximal Flow Problem Since node 1 is the source, flow can only be possible from node 1 to node 2. Thus, arc (1, 2) is practically a directed arc. Similar, flow can only be possible from node 1 to node 3, node 2 to node 4, and node 3 to node 4. Thus, practically arcs (1, 3), (2, 4), and (3, 4) are directed arcs. The flow on (2, 3) can be on either direction. 2 6 0 10 2 0 4 1 3 0 2 0 3 6 39

  40. The Maximal Flow Problem 2 6 0 10 2 0 0 0 4 1 3 0 2 0 3 6 4 2 6 0 2 10 2 4 2 0 2 0 0 4 1 the residual capacity of flowing from node 1 to node 3 3 0 0 2 1 0 3 6 the residual capacity of flowing from node 3 to node 1 2 The route 1324 allows a flow of 2 units. 2 40

  41. The Maximal Flow Problem 4 2 6 0 2 10 2 4 2 0 2 0 0 4 1 3 0 0 2 1 0 3 6 2 The route 124 allows a flow of 4 units. 0 4 4 6 2 6 0 6 2 6 6 10 2 4 2 0 2 0 0 4 1 3 0 0 2 1 0 3 6 41 2

  42. The Maximal Flow Problem 0 4 4 6 2 6 0 6 2 6 6 10 2 4 2 0 2 0 0 4 1 3 0 0 2 1 2 0 3 6 The route 134 allows a flow of 1 unit. 0 4 4 6 2 7 6 7 0 6 2 6 6 10 2 4 2 0 2 0 0 4 1 3 0 0 2 1 1 2 0 0 3 6 5 42 2 3

  43. The Maximal Flow Problem 0 4 4 6 2 7 6 7 0 6 2 6 6 10 2 4 2 0 2 0 0 4 1 3 0 0 2 1 1 0 0 3 6 5 2 3 The route 1234 allows a flow of 4 unit. 0 8 11 4 4 6 2 7 11 2 6 7 0 6 2 6 6 10 2 4 2 0 2 0 0 0 4 1 3 0 4 0 2 1 1 0 5 0 3 6 5 43 2 1 3

  44. The Maximal Flow Problem How to get the flows? Compare with the original. The differences in (residual) capacities give the actual flows. 2 6 0 10 2 0 0 0 4 1 3 0 2 0 3 6 2 0 8 2 0 6 11 11 4 1 0 5 4 3 3 1 13: 3 units 12: 8 units 23: 2 units 24: 6 units 34: 5 units maximal flow from node 1 to node 4 : 11 units 44

  45. A Linear Program Formulation of the Maximal Flow Problem 45

  46. LP Formulation of the Maximal Flow Problem Oil is pumped out at node 1, temporarily stored at nodes 2, 3, and 4, and eventually sent the refinery at node 5 through the pipe lines. In the following, arc represents a pipe line, with its capacity (i.e., barrels of oil per day) written beside it. Formulate a linear program to find the maximal amount of oil sent from node 1 to node 5. A maximal flow problem can be formulated as a linear program (and solved by an optimizer). 2 50 45 25 4 20 1 40 5 15 35 10 3 46

  47. LP Formulation of the Maximal Flow Problem Let xij be the number of barrels of oil pumped from node i to node j; xij 0. Because node 1 is the source, x21 = x31 = 0. Similarly, x52 = x53 = x54 = 0 because node 5 is the sink. Because we do not know the direction of flow in arc (2, 3), we are not sure which of x23 and x32 should be equal to zero. Similarly, we have no idea of fixing which of x24 and x42, and of x34 and x43, to zero. 2 50 45 25 4 20 1 40 5 15 35 10 3 47

  48. LP Formulation of the Maximal Flow Problem We further introduce a variable x51. Effectively we act an artificial arc from node 5 to node 1. 2 50 45 25 4 20 1 40 5 15 35 10 3 x51 48

  49. LP Formulation of the Maximal Flow Problem Each LP constraint matches with a physical relationship of the problem. There are two types of constraints: conservation of flow: the amount of oil into a node must be the same as the amount out of the node. bounds on flow: 0 xij capacity of i to j flow 49

  50. LP Formulation of the Maximal Flow Problem objective function: max x51, conservation of flow node 1: x51 = x12 + x13, node 2: x12 + x32 + x42 = x23 + x24 + x25, node 3: x13 + x23 + x43 = x32 + x34 + x35, node 4: x24 + x34 = x42 + x43 + x45, node 5: x25 + x35 + x45 = x51, 2 50 45 25 4 20 1 40 5 15 35 10 3 50

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