Partitioning Graphs of Supply and Demand Generalization of Knapsack Problem

Partitioning Graphs

of Supply and Demand

Generalization of Knapsack Problem

Takao Nishizeki

Tohoku UniversityTohoku University

Supply Vertices Demand Vertices

Supply Vertices and Demand Vertices

6 5 3 5

26 107

Each Supply Vertex has a number, called Supply.Each Supply Vertex has a number, called Supply.

Each Demand Vertex has a number, called DemandEach Demand Vertex has a number, called Demand

SupplySupply

DemandDemand

Supply Vertices Demand Vertices

Desired Partitionpartition G into connected components so that

6 5 3 5

26 107

Desired Partitionpartition G into connected components so that

6 5 3 5

26 107

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.

6 5 3 5

26 107

6 5 3 5

26 107

Desired partitionDesired partition

Sum ：

m ：13

Sum ：12

Maximum Partition Problem (Max PP)

6 5 3 5

26 107

No desired partitionNo desired partition

6 5 3 5

26 107

the component.

A partition of a graph must satisfy

6 5 3 5

26 107

no supply vertex no supply vertex

(a) each component has at most one supply vertex,(b) supply is no less than the sum of demands in

the component.

6 5 3 5

26 107

6 5 3 5

26 107

finds a partition that maximizes the “fulfillment.”

Sum ：

Sum ： 5

Sum ：12

the component if there is a supply vertex.

A partition of a graph must satisfy sum of demands in all components

with supply vertices

The fulfillment of this partition17 + 5 + 12 + 7 = 41

6 5 3 5

26 107

6 5 3 5

26 107

Sum ：

Sum ： 5

Sum ：12

Sum ：

Sum ： 8

Sum ：12

6 5 3 5

26 107

6 5 3 5

The fulfillment of this partition19 + 8 + 12 + 15 = 54

Maximum fulfillment

Complexity Status

2 3 9 4

25TreesTrees

NP-hard

Maximum Subset Sum Problem (NP-hard)instance: a set A of integers and an integer bfind: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b.

3 4 5 7 11

b = 13

max subset sum problem(simple ver. of Knapsack)

given set A

Complexity Status

TreesTrees

NP-hard

Maximum Subset Sum Problem (NP-hard)instance: a set A of integers and an integer bfind: a subset C ⊆ A which maximizes the sum of integers in C s.t. the sum does not exceed b.

2 3 9 4

3 4 5 7 11

b = 13

max subset sum problem(simple ver. of Knapsack)

given set A

Related Result

Max PP for Starswith one supply at center

Fully Polynomial-Time Approximation Scheme(FPTAS)

[Ibarra and Kim ’75]

max subset sum problem (NP-hard)max subset sum problem (NP-hard)

3 4 5 7 11

good approximationfor larger classesgood approximationfor larger classes??

2 3 9 4

6 5 3 5

26 107

Our Results (approximabiliy)

6 5 3 5

26 107

General graphs (1) MAXSNP-hard (APX-hard)

No PTAS unless P=NPNo PTAS unless P=NP

PTAS: for any , 0 < < 1, the algorithm finds an approximation solution such that

APPRO > (1–) OPTin time polynomial in n. (1/:regarded as a constant.)

PTAS: for any , 0 < < 1, the algorithm finds an approximation solution such that

APPRO > (1–) OPTin time polynomial in n. (1/:regarded as a constant.)

No FPTAS unless P=NPNo FPTAS unless P=NP

6 5 3 5

26 107

General graphs

(2) FPTAS

(1) MAXSNP-hard (APX-hard)

NP-hard

2 3 9 4

6 5 3 5

26 107

General graphs

(2) FPTAS

Trees NP-hard

2 3 9 4

(1) MAXSNP-hardnessMax PP is MAXSNP-hard for bipartite graphs.

L-reduction: preserves approximability

L-reductionMax PP for

bipartite graphs

3-occurrence MAX3SAT

each variable appears at most 3 times

zyxzyxzyxf

error ratio: error ratio: ’

(1) MAXSNP-hardness3-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 clausesfind: a truth assignment which maximizes # of satisfied clauses

variables: v w x y z

zyxzwvzxwywvf

at most 3 times

(1) MAXSNP-hardness

zyxzwvzxwywvf

at most 3 times

3-occurrence MAX3SAT (MAXSNP-hard)

(1) MAXSNP-hardness

zyxzwvzxwywvf

3-occurrence MAX3SAT (MAXSNP-hard)

(1) MAXSNP-hardnessvariable gadgetvariable gadget

variable x

x = false

x = true

4+4 = 8 > 7

(1) MAXSNP-hardness

zyxzyxzyxf

variable x variable y variable z

(1) MAXSNP-hardness

zyxzyxzyxf

(1) MAXSNP-hardness

zyxzyxzyxf

(1) MAXSNP-hardness

x x y y z z

zyxzyxzyxf

at most 3

7 – 4 = 3enough power

(1) MAXSNP-hardness

x x y y z z

zyxzyxzyxf

(1) MAXSNP-hardness

x x y y z z

x = true y = false z = false

zyxzyxzyxf

(1) MAXSNP-hardness

x x y y z z

zyxzyxzyxf

(1) MAXSNP-hardness

x x y y z z

zyxzyxzyxf

x = false y = false z = false

not satisfied

(1) MAXSNP-hardnessMax PP is MAXSNP-hard for bipartite graphs.

L-reduction: preserves approximability

3-occurrence MAX3SAT

Max PP for bipartite graphs

L-reduction

each variable appears at most 3 times

zyxzyxzyxf

6 5 3 5

26 107

General graphs

(2) FPTAS

Trees NP-hard

Pseudo-poly.-time algorithm

2 3 9 4

Max PP is NP-hard even for trees.

Max PP can be solved for a tree T in time O(F 2n)

if the supplies and demands are integers.

Max PP can be solved for a tree T in time O(F 2n)

if the supplies and demands are integers.

2+3+4+ ･･･ +4+5= 43

15+25= 40

F= min{43,40}=40

max fulfillment ≦ Fmax fulfillment ≦ F

• sum of all demands• sum of all supplies

F = min

Pseudo-Polynomial-Time Algorithm

8 2 7 5

4 3 6 9

2 7 2 420 25

rootroot

rootroot3

2 8 2 4 3 6

5 2 3 9 4

Dynamic ProgrammingDynamic Programming

optimal partition of Tv optimal partition of T

fulfillment = 18 fulfillment = 20

fulfillment: 10

fulfillment: 18

20 - (10+5+3) = 2marginal power

1020-10=

fulfillment: 15

20-(10+3)= 7

fulfillment: 13

20-(10+5)=5

fulfillment: 15

fulfillment: 13

fulfillment: 10

fulfillment: 18

optimal partition of T

marginal power

F1 2 3 fulfillment

fulfillment: 15

fulfillment: 13

fulfillment: 10

fulfillment: 18

staircase,non-increasingstaircase,non-increasing

F points

Tv 15v

fulfillment: 9

5+4 = 9deficient power

fulfillment: 8

fulfillment: 12

fulfillment: 9

fulfillment: 5

4Tv 15v

Tv 15v

fulfillment: 8

fulfillment: 12

fulfillment: 9

fulfillment: 5

deficient power

F1 2 3 fulfillment

staircase, non-decreasingstaircase, non-decreasing

F points

2 8 2 4 3 6

5 2 3 9 4

25leaves

2 8 2 4 3 6

5 2 3 9 4

2 8 2 4 3 6

5 2 3 9 4

25marginal power

F12 3 fulfillment

deficient power

F123 fulfillment

deficient power

F123 fulfillment

marginal power

F12 3 fulfillment

O(F 2) time

marginal power

F1 2 3 fulfillment

deficient power

F123 fulfillment

rootroot3

2 8 2 4 3 6

5 2 3 9 4

Dynamic Programming

marginal power

F1 2 3 fulfillment

deficient power

F1 2 3 fulfillment

max fulfillment

rootroot3

2 8 2 4 3 6

5 2 3 9 4

Computation timeComputation time

for each vertex ･･････････････ O(F 2)

Dynamic Programming

There are n vertices.

Computation timeO(F

2n)Computation time

O(F 2n)

The algorithm takes polynomial time if F is bounded by a polynomial in n.The algorithm takes polynomial time if F is bounded by a polynomial in n.

There are n vertices.

for each vertex ･･････････････ O(F 2)

(2) FPTAS

For any , 0< <1, the algorithm finds a partition of a tree T such that

OPT – APPRO < OPT

in time polynomial in both n and 1/ .

For any , 0< <1, the algorithm finds a partition of a tree T such that

OPT – APPRO < OPT

Let all demands and supply be positive real numbers.

n : # of vertices

The algorithm is similar to the previous algorithm.

0fulfillment

marginal power

t sampled fulfillment

(2) FPTAS

0fulfillment

marginal power

(2) FPTAS

OPT – APPRO < 2nt

total error < error/merge × # of merge

< 2t × n

sampled fulfillment

OPTOPT

APPROAPPRO

ErrorError md: max demand

2 3 9 4

md = 9

(2) FPTAS

OPT – APPRO < 2nt

ErrorError md: max demand

(2) FPTAS

OPT – APPRO < 2nt

dm md OPT≦ ≦ OPT

OPT – APPRO < OPTOPT – APPRO < OPT

error ratio

APPROOPT

dd nmF

md: max demand

marginal power

0 t Sampled fulfillment

pointst

(2) FPTASErrorError

OPT – APPRO < OPT

Conclusions

6 5 3 5

26 107

General graphs(1) MAXSNP-hard (APX-hard)

(2) FPTAS

NP-hard

2 3 9 4

Variant (Generalization of Knapsack Problem)

find a partition which maximizes sum of profits of all demand vertices that are supplied power.

3, 50 7, 30 8, 200

demand profit

3, 50 7, 30 8, 200

demands = 3+7 = 10

profits = 50+30 = 80

SumSum

3, 50 7, 30 8, 200

demands = 8

profits = 200

SumSum

FPTAS for KP [Ibarra and Kim ’75]

FPTAS for trees

7, 30 5, 45

3, 50 7, 30 8, 200

FPTAS for trees

7, 30 5, 45

extendable to partial k-trees(graphs with bounded treewidth) if there is exactly one supply

7, 8 5, 25

3, 5 8, 13

3, 50 7, 30 8, 200

FPTAS for trees

7, 30 5, 45

extendable to partial k-trees(graphs with bounded treewidth) if there is exactly one supply

3, 50 7, 30 8, 200

7, 8 5, 25

3, 5 8, 13

Thank You!Thank You!

(2b) Parallel connection

:terminal

Series-Parallel Graphs (recursively)(1) A single edge is a SP graph.

(2a) Series connection

:terminal

SP Graph & Decomposition Tree

e1 e2 e3 e4 e5 e6 e7

leavesDecomposition tree

e1 e2 e3 e4 e5 e6 e7

:terminal

e1 e2 e3 e4 e5 e6 e7

:terminal

Series connection

:terminal

Series connectione1 e2 e3 e4 e5 e6 e7

:terminal

Parallel connection

e1 e2 e3 e4 e5 e6 e7

:terminal

Series connection

e1 e2 e3 e4 e5 e6 e7

:terminal

Series connection

e1 e2 e3 e4 e5 e6 e7

:terminal

Parallel connection

e1 e2 e3 e4 e5 e6 e7

:terminal

e1 e2 e3 e4 e5 e6 e7

DP algorithmDP algorithm

:terminal

e1 e2 e3 e4 e5 e6 e7

Pseudo-Polynomial-Time AlgorithmSuppose that a SP graph has exactly one supply.

Max PP for such a SP graph can be solved in time O(F 2n) if the demands and the supply are integers. Max PP for such a SP graph can be solved in time O(F 2n) if the demands and the supply are integers.

F : sum of all demands

F = 3+7+6+2+6+5 = 29

pseudo-polynomial-time algorithm

(2) FPTASmax fulfillment ≦ Fmax fulfillment ≦ F

n: # of vertices

Decomposition tree Corresponding SP subgraphs

DP table

What should we store in the table for Max PP ?

If G had more than one supply, Max PP would find a partition.

Max PP finds only one component with the supply

two terminals are supplied power and contained in same component

two terminals will be supplied power, but are contained in different components

this terminal will be supplied power

this terminal is never supplied power

two terminals are never supplied power

Pseudo-Polynomial-Time Algorithmboth are/will besupplied powerboth are/will besupplied power

one is/will be supplied, but the other is never suppliedone is/will be supplied, but the other is never supplied

both are never suppliedboth are never supplied

same component

different components

If the component has the supply vertex, then the component may have the “marginal” power.

marginal power

10 – (3+2) = 5

If the component has no supply vertex, the component may have the “deficient” power.

deficient power

3+7+2 = 12

same component

F1 2 3 fulfillment

marginal power

fulfillment

marginal power

staircase,non-increasingstaircase,non-increasing

deficient power

fulfillment

deficient power

F1 2 3 fulfillment

staircase, non-decreasingstaircase, non-decreasing

same component

marginal power

F12 3 fulfillment

deficient power

F123 fulfillment

deficient power

F123 fulfillment

marginal power

F12 3 fulfillment

marginal power

F1 2 3 fulfillment

deficient power

F123 fulfillment

marginal power

F12 3 fulfillment

deficient power

F123 fulfillment

Decomposition tree Corresponding SP subgraphs

DP table DP table

DP table

sizeO(F )

timeO(F

2)time

O(F 2)

e1 e2 e3 e4 e5 e6 e7

Decomposition tree

O(F 2)O(F 2)

# of nodes = O(n)

O( F 2 n )O( F 2 n )

n : # of vertices in a given SP graph

Pseudo-Polynomial-Time AlgorithmSuppose that a SP graph has exactly one supply.

Max PP for such a SP graph can be solved in time O(F 2n) if the demands and the supply are integers. Max PP for such a SP graph can be solved in time O(F 2n) if the demands and the supply are integers.

F : sum of all demands

F = 3+7+6+2+6+5 = 29

pseudo-polynomial-time algorithm

(2) FPTASmax fulfillment ≦ Fmax fulfillment ≦ F

n: # of vertices

(2) FPTAS

For any , 0< <1, the algorithm finds a component with the supply vertex such that

APPRO > (1–) OPT

For any , 0< <1, the algorithm finds a component with the supply vertex such that

APPRO > (1–) OPT

Let all demands and supply be positive real numbers.

n : # of vertices

(2) FPTAS

0fulfillment

marginal power

t sampled fulfillment

0fulfillment

marginal power

(2) FPTAS

APPRO > (1–) OPT

md: max demand n

Open ProblemIs there an approximation algorithm for SP graphs having more than one supply?

• FPTAS or PTAS ?• constant-factor ?

Open Problemboth are/will besupplied powerboth are/will besupplied power

same component

Open Problemboth are/will be supplied powerboth are/will be supplied power

different componentsThe two components must become ONE component.

deficient power= j + k

deficient power

1 2 3 fulfillment

j + k(1D)

Open Problem

deficient power= j + k

If a given graph has more than one supply.

Our Results

Max subset sum problem

FPTAS [Ibarra and Kim, 1975]

Max PP

our FPTAS

straightforward ?

5 6 62

Our Results

Max subset sum problem

FPTAS [Ibarra and Kim, 1975]

Max PP

our FPTAS

straightforward ?

5 6 632

Sum: 10margin: 2

Sum: 8

The graph structure plays crucial role.The graph structure plays crucial role.

Every demand vertex must be supplied from exactly one supply vertex.

Assumption of the Problem

Max PP is applied to Power delivery networks.

It is difficult to synchronize two or more sources.

Power delivery network

Application of Max PPPower delivery problem

feeder

supply vertex=

demand vertex

= cable segment

Power delivery network

Power delivery problem

Determine whether there exists a switching so that all loads can be supplied.

If not all loads can be supplied, we want to maximize the sum of loads supplied power.

Application of Max PP

Max Partition Problem

Partitioning Graphs of Supply and Demand Generalization of Knapsack Problem

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