Graph. Supply Vertices and Demand Vertices. Supply Vertices. Demand Vertices
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1 Partitioning Graphs of Supply and Demand Generalization of Knapsack Problem Takao Nishizeki Tohoku University
2 Graph Supply Vertices and Demand Vertices Supply Vertices Demand Vertices
3 Graph Each Supply Vertex has a number, called Supply. Each Demand Vertex has a number, called Demand Supply Demand Supply Vertices Demand Vertices
4 Desired Partition partition G into connected components so that
5 Desired Partition partition G into connected components so that
6 Desired Partition partition G into connected components so that (a) each component has exactly one supply vertex, (b) () supply is no less than the sum of demands in the component. Desired partition Sum :23 Sum : Sum: Sum:
7 Maximum Partition Problem (Max PP) No desired partition
8 Maximum Partition Problem (Max PP) A partition i of a graph must satisfy if (a) each component has at exactly most one supply vertex, (b) supply is no less than the sum of demands in the component. 20 no 4 supply vertex
9 Maximum Partition Problem (Max PP) finds a partition that maximizes the fulfillment. A partition i of a graph must satisfy if sum of demands in all components (a) each component has at most one supply vertex, with ihsupply vertices (b) supply is no less than the sum of demands in the component. if there is a supply vertex. Sum: Sum: Sum:12 Sum :
10 Maximum Partition Problem (Max PP) finds a partition that maximizes the fulfillment. The fulfillment of this partition = 41 Sum: Sum: Sum:12 Sum :
11 Maximum Partition Problem (Max PP) finds a partition that maximizes the fulfillment. The fulfillment of this partition = 54 Maximum fulfillment Sum: Sum: Sum:12 Sum :
12 Complexity Status Trees max subset sum problem (simple ver. of Knapsack) b = 13 NP-hard given set A Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C A which maximizes the sum of integers in C s.t. the sum does not exceed b.
13 Complexity Status Trees max subset sum problem (simple ver. of Knapsack) b = C NP-hard given set A Maximum Subset Sum Problem (NP-hard) instance: a set A of integers and an integer b find: a subset C A which maximizes the sum of integers in C s.t. the sum does not exceed b.
14 Related Result max subset sum problem (NP-hard) 13 Fully Polynomial-Time Approximation i Schemeh (FPTAS) Max PP for Stars with one supply at center [Ibarra and Kim 75] FPTAS: for any ε, 0 < ε < 1, the algorithm finds an approximation solution such that APPRO > (1 ε ε ) OPT in time polynomial in both n and 1/ε.
15 Related Result max subset sum problem (NP-hard) 13 Fully Polynomial-Time Approximation i Schemeh (FPTAS) Max PP for Stars with one supply at center [Ibarra and Kim 75] good approximation for larger classes?
16 Our Results (approximabiliy) General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP No FPTAS unless P=NP PTAS: for any ε, 0 < ε < 1, the algorithm finds an approximation solution such that APPRO > (1 ε ε ) OPT in time polynomial in n. (1/ε :regarded as a constant.)
17 Our Results (approximabiliy) General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard (2) FPTAS 5 15
18 Our Results (approximabiliy) General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard (2) FPTAS 5 15
19 (1) MAXSNP-hardness Max PP is MAXSNP-hard for bipartite graphs. L-reduction: preserves approximability error ratio: ε L-reduction error ratio: ε 3-occurrence MAX3SAT Max PP for bipartite graphs f each variable appears at most 3 times = ( x y z) ( x y z) ( x y z)
20 (1) MAXSNP-hardness 3-occurrence MAX3SAT (MAXSNP-hard) instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes #of satisfied clauses variables: v w x y z at most 3 times f = ( v w y) ( w x z) ( v w z) ( x y z)
21 (1) MAXSNP-hardness 3-occurrence MAX3SAT (MAXSNP-hard) instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes #of satisfied clauses variables: v w x y z at most 3 times f = ( v w y) ( w x z) ( v w z) ( x y z)
22 (1) MAXSNP-hardness 3-occurrence MAX3SAT (MAXSNP-hard) instance: variables and clauses s.t. each clause has exactly 3 literals; and each variable appears at most 3 times in the clauses find: a truth assignment which maximizes #of satisfied clauses variables: v w x y z f = ( v w y) ( w x z) ( v w z) ( x y z)
23 (1) MAXSNP-hardness variable gadget x = true variable x 7 7 x 4 4 x x 4 4 x x =false 4+4 = 8 > 7 7 x 4 4 x
24 (1) MAXSNP-hardness variable x variable y variable z x x y y z z f = ( x y z ) ( x y z ) ( x y z )
25 (1) MAXSNP-hardness variable x variable y variable z x x y y z z f = ( x y z ) ( x y z ) ( x y z )
26 (1) MAXSNP-hardness variable x variable y variable z x x y y z z f = ( x y z ) ( x y z ) ( x y z )
27 (1) MAXSNP-hardness 7 4=3 3 variable x enough variable power y variable z x x y y z z at most f = ( x y z ) ( x y z ) ( x y z )
28 (1) MAXSNP-hardness variable x variable y variable z x x y y z z f = ( x y z ) ( x y z ) ( x y z )
29 (1) MAXSNP-hardness x = true y = false z = false x x y y z z f = ( x y z ) ( x y z ) ( x y z )
30 (1) MAXSNP-hardness variable x variable y variable z x x y y z z f = ( x y z ) ( x y z ) ( x y z )
31 (1) MAXSNP-hardness x = false y = false z = false x x y y z z not satisfied f = ( x y z ) ( x y z ) ( x y z )
32 (1) MAXSNP-hardness Max PP is MAXSNP-hard for bipartite graphs. L-reduction: preserves approximability 3-occurrence MAX3SAT L-reduction Max PP for bipartite graphs f each variable appears at most 3 times = t t 3 ti ( x y z ) ( x y z ) ( x y z ) 1 1 1
33 Our Results (approximabiliy) General graphs (1) MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard (2) FPTAS Pseudo-poly.-time algorithm
34 Pseudo-Polynomial-Time Algorithm Max PP is NP-hard even for trees. Max PP can be solved for a tree T in time O(F 2 n) if the supplies and demands are integers. F = min sum of all demands sum of all supplies = = F= min{43,40}= max fulfillment F
35 Pseudo-Polynomial-Time Algorithm Dynamic Programming root root
36 Pseudo-Polynomial-Time Algorithm T optimal partition of T v optimal partition of T 5 5 T v v 10 9 T v v fulfillment = 18 fulfillment = 20
37 Pseudo-Polynomial-Time Algorithm Dynamic Programming marginal power 20 - (10+5+3) = 2 v =10 v 10 5 T v 10 9 T v fulfillment: fulfillment: 10
38 Pseudo-Polynomial-Time Algorithm Dynamic Programming 20-(10+3)= 7 20-(10+5)=5 5 5 T v v 10 9 T v v fulfillment: fulfillment: 15
39 Pseudo-Polynomial-Time Algorithm optimal partition of T T v T 9 v v10 v T v v fulfillment: 18 fulfillment: T v T v v10 9 v fulfillment: fulfillment: 15
40 Pseudo-Polynomial-Time Algorithm staircase,non-increasing marginal power T v T 9 v v10 v fulfillment: fulfillment: T v T v v10 9 v fulfillment: 13 fulfillment: F F points fulfillment
41 Pseudo-Polynomial-Time Algorithm dfii deficient power 5+4 = 9 5 optimal partition of T 5 T v v T v v fulfillment: 9
42 Pseudo-Polynomial-Time Algorithm T v optimal partition of T v T v 15 v fulfillment: 5 fulfillment: T v v 5 15 T v v T v v fulfillment: 9 fulfillment: 12
43 Pseudo-Polynomial-Time Algorithm staircase, non-decreasing deficient power T v v 5 15 T v v fulfillment: T v 5 15 v fulfillment: 9 fulfillment: T 15 v 5 v fulfillment: F F points fulfillment
44 Pseudo-Polynomial-Time Algorithm leaves
45 Pseudo-Polynomial-Time Algorithm marginal power deficient power F F fulfillment fulfillment 20 O(F 2 ) time marginal power deficient power marginal power deficient power fulfillment 0 F F F F 12 3 fulfillment fulfillment fulfillment
46 Pseudo-Polynomial-Time Algorithm marginal power deficient power root t fulfillment max fulfillment F F fulfillment Dynamic Programming
47 Pseudo-Polynomial-Time Algorithm Computation time for each vertex O(F 2 ) There are n vertices. root Dynamic Programming
48 Pseudo-Polynomial-Time Algorithm Computation time for each vertex O(F 2 ) There are n vertices. Computation time O(F 2 n) Th l ith t k l i lti if The algorithm takes polynomial time if F is bounded by a polynomial in n.
49 (2) FPTAS Let all demands and supply be positive real numbers. For any ε, 0< ε <1, the algorithm finds a partition of a tree T such that OPT APPRO < ε OPT in time polynomial in both n and 1/ε. O 5 n 2 ε n : # of vertices
50 (2) FPTAS The algorithm is similar to the previous algorithm. marginal power 0 t sampled fulfillment
51 (2) FPTAS The algorithm total error is similar < error/merge to the previous # of algorithm. merge < 2t n marginal power OPT APPRO < 2nt OPT 0 t sampled fulfillment APPRO
52 (2) FPTAS Error OPT APPRO < 2nt m d : max demand t = ε m 2nn d m d = 9
53 (2) FPTAS Error OPT APPRO < 2nt m d = ε ε OPT m d : max demand t = ε m 2nn d m d OPT d OPT APPRO < ε OPT error ratio OPT APPRO OPT < ε
54 (2) FPTAS Error m d : max demand OPT APPRO < ε OPT Computation time marginal power 2 F O n t t ε m 2n d =, 5 n = O 2 ε F nm d 0 t F t Sampled fulfillment points
55 Conclusions General graphs (1) () MAXSNP-hard (APX-hard) No PTAS unless P=NP Trees NP-hard (2) FPTAS 5 15
56 Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand d vertices that t are supplied power. 10 3, 50 7, 30 8, 200 demand profit
57 Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand d vertices that t are supplied power , 50 7, 30 8, 200 3, 50 7, 30 8, 200 Sum demands = 8 profits = 200 max Sum demands = 3+7 = 10 profits = = 80
58 Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand d vertices that t are supplied power. 10 2, 10 7, 30 5, 45 8, 20 3, 50 7, 30 8, 200 FPTAS for KP [Ibarra and Kim 75] 10 6, FPTAS for trees
59 Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand d vertices that t are supplied power. 10 2, 10 7, 30 5, 45 8, 20 3, 50 7, 30 8, 200 FPTAS for KP [Ibarra and Kim 75] extendable to partial k-trees (graphs with bounded treewidth) if there is exactly one supply 10 6, FPTAS for trees 20 2, 10 7, 8 6, 12 5, 25 8, 3 3, 5 8, 13
60 Variant (Generalization of Knapsack Problem) find a partition which maximizes sum of profits of all demand d vertices that t are supplied power. 10 2, 10 7, 30 5, 45 8, 20 3, 50 7, 30 8, 200 FPTAS for KP [Ibarra and Kim 75] extendable to partial k-trees (graphs with bounded treewidth) if there is exactly one supply 10 6, FPTAS for trees 20 2, 10 7, 8 6, 12 5, 25 8, 3 3, 5 8, 13
61 Thank You!
62
63 Series-Parallel Graphs (recursively) (1) A single edge is a SP graph. (2a) Series connection :terminal G G G1 1 1 G 2 G2 (2b) Parallel connection G G 1 G 1 2 G 2 G 2
64 SP Graph & Decomposition Tree e 2 e 1 e 3 S P e 4 e 6 e 5 S P e 7 S S S :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
65 SP Graph & Decomposition Tree e 2 e 1 e 3 e 6 e 4 e 7 e 5 :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
66 SP Graph & Decomposition Tree e 2 e 1 e 3 e 6 e 4 e 7 S e 5 Series connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
67 SP Graph & Decomposition Tree e 2 e 1 e 3 e 6 e S S e 4 e 7 S S e 5 Series connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
68 SP Graph & Decomposition Tree e 2 e 1 e 3 e 6 e 4 e 7 S S e 5 P Parallel connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
69 SP Graph & Decomposition Tree e 2 e 1 e 3 S P e 6 e 4 e 7 S S e 5 S Series connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
70 SP Graph & Decomposition Tree e 2 e 1 e 3 S S P e 6 e 4 e 7 S S e 5 S Series connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
71 SP Graph & Decomposition Tree e 2 e 1 e 3 S P S P e 6 e 4 e S S 7 e 5 S Parallel connection :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
72 SP Graph & Decomposition Tree e 2 e 1 e 3 S P e 4 e 6 e 6 e e 4 5 e 7 e 5 S P e 7 e 7 S S S :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
73 SP Graph & Decomposition Tree DP algorithm P e 2 e 1 e 3 S e 4 e 6 e 5 S P e 7 S S S :terminal e 1 e 2 e 3 e 4 e 5 e 6 e 7 leaves Decomposition tree
74 Pseudo-Polynomial-Time Algorithm Suppose that a SP graph has exactly one supply. Max PP for such a SP graph can be solved in time O(F( 2 n) ) if the demands and the supply are integers. F : sum of all demands n: # of vertices 10 5 pseudo-polynomialtime algorithm F max = fulfillment = F29 (2) FPTAS
75 Pseudo-Polynomial-Time Algorithm What should we store in the table for Max PP? DP G table 3 2 P S S 7 G 1 G Decomposition tree DP table Corresponding SP subgraphs DP table
76 Pseudo-Polynomial-Time Algorithm
77 Pseudo-Polynomial-Time Algorithm
78 Pseudo-Polynomial-Time Algorithm
79 Pseudo-Polynomial-Time Algorithm
80 Pseudo-Polynomial-Time Algorithm If G had more than one supply, Max PP would ldfind a partition. G
81 Pseudo-Polynomial-Time Algorithm Max PP finds only one component with the supply G
82 Pseudo-Polynomial-Time Algorithm two terminals are supplied power and contained in same component G
83 Pseudo-Polynomial-Time Algorithm Max PP finds only one component with the supply G
84 Pseudo-Polynomial-Time Algorithm two terminals will be supplied power, but are contained din different components G
85 Pseudo-Polynomial-Time Algorithm Max PP finds only one component with the supply G
86 Pseudo-Polynomial-Time Algorithm this terminal will be supplied power this terminal is never supplied power G
87 Pseudo-Polynomial-Time Algorithm Max PP finds only one component with the supply G
88 Pseudo-Polynomial-Time Algorithm two terminals are never supplied power G
89 Pseudo-Polynomial-Time Algorithm both are/will be supplied power one is/will be supplied, but the other is never supplied same component both are never supplied different components
90 Pseudo-Polynomial-Time Algorithm If the component has the supply vertex, then the component may have the marginal power
91 Pseudo-Polynomial-Time Algorithm If the component has the supply vertex, then the component may have the marginal power (3+2) = 5 marginal power
92 Pseudo-Polynomial-Time Algorithm If the component has no supply vertex, the component may have the deficient i power
93 Pseudo-Polynomial-Time Algorithm If the component has no supply vertex, the component may have the deficient i power = 12 deficient power
94 Pseudo-Polynomial-Time Algorithm both are/will be supplied power one is/will be supplied, but the other is never supplied same component both are never supplied different components
95 Pseudo-Polynomial-Time Algorithm both are/will be supplied power fulfillment staircase,non-increasing marginal power 5 marginal power F fulfillment
96 Pseudo-Polynomial-Time Algorithm both are/will be supplied power staircase, non-decreasing deficient power fulfillment deficient power 1 23 F fulfillment
97 marginal Pseudo-Polynomial-Time power deficient power marginal power Algorithm deficient power both are/will be F supplied power F fulfillment fulfillment one is/will be supplied, but F the other is never supplied F fulfillment fulfillment same component different components 0 marginal power marginal power dfii deficient power both are never supplied deficient F power 0 F 12 3 fulfillment 0 F 12 3 fulfillment fulfillment F 12 3 fulfillment
98 Pseudo-Polynomial-Time Algorithm S P time O(F 2 ) S G DP table G 1 G Decomposition tree DP size table Corresponding DP size table O(F ) SP subgraphs O(F )
99 Pseudo-Polynomial-Time Algorithm O(F 2 ) S P P S # of nodes = O(n) S S e 1 e 2 e 3 e 4 e 5 e 6 e 7 O( F 2 n ) Decomposition tree n : # of vertices in a given SP graph
100 Pseudo-Polynomial-Time Algorithm Suppose that a SP graph has exactly one supply. Max PP for such a SP graph can be solved in time O(F( 2 n) ) if the demands and the supply are integers. F : sum of all demands n: # of vertices 10 5 pseudo-polynomialtime algorithm F max = fulfillment = F29 (2) FPTAS
101 (2) FPTAS Let all demands and supply be positive real numbers. For any ε, 0< ε <1, the algorithm finds a component with the supply vertex such that APPRO > (1 ε ) OPT in time polynomial in both n and 1/ε. O 5 n 2 ε n : # of vertices
102 (2) FPTAS The algorithm is similar to the previous algorithm. marginal power 0 t sampled fulfillment
103 (2) FPTAS The algorithm is similar to the previous algorithm. marginal power APPRO > (1 ε ) OPT 0 m d : max demand t t ε m = 4n d fulfillment
104
105 Open Problem Is there an approximation algorithm for SP graphs having more than one supply? FPTAS or PTAS? constant-factor?
106 Open Problem both are/will be supplied power one is/will be supplied, but the other is never supplied same component both are never supplied different components
107 Open Problem both are/will be supplied power different components The two components must become ONE component. j k deficient power = j + k j + k (1D) deficient power F fulfillment
108 Open Problem If a given graph has more than one supply j k deficient power = j + k
109
110 Our Results Max subset sum problem Max PP FPTAS [Ibarra and Kim, 1975] straightforward? our FPTAS
111 Our Results Max subset sum problem FPTAS [Ibarra and Kim, 1975] Max PP our FPTAS straightforward? margin: 2 Sum: 10 Sum: The graph structure plays crucial role.
112
113 Assumption of the Problem Every demand vertex must be supplied from exactly one supply vertex. Max PPi is applied to Power delivery networks. It is difficult to synchronize two or more sources.
114
115 Application of Max PP Power delivery problem load = demand vertex cable segment = edge feeder = supply vertex Power delivery network
116 Application of Max PP Power delivery problem Determine whether there exists a switching so that t all loads can be supplied. If not all loads can be supplied, we want to maximize the sum of loads supplied power. Max Partition ProblemPower delivery network
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