A Variation of Minimum Latency Problem on Path, Tree and DAG

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A Variation of Minimum Latency Problemon Path, Tree and DAG

Author:

Yung-Hui Huang,

Yaw-Ling Lin,

Chuan-Yi Tang

2

Problem Definition

Given a graph G(V,E) with a distance matrix di,j and the weight function of each vertices vi

If the vertex vi wants to be completed, at least one ancestor of vi must be finished.

Let defined to be the distance traveled before first visiting vi

The problem is to find out the minimum latency tour traversal all vertices.

The objective function:

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Problem Sample

BA

C D

E

GF

Weight of vertices

Working Time of vertices t(vi) is zero.A ll edge distance d(ej) is 1

Result:T = {A,C,E,G,F,B,D}L(T) = 1*30+2*45+3*42+4*101+5*65+6*0+7*50

= 1325

A B C D E F G

w(vi) 30 0 45 50 42 65 101

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Compare with original MLP

The new problem is equivalent to the original MLP with no edge distance but vertex latency time and vertex weight.

In other words, we assume that the walk back latency of completed jobs be “0”.

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Our Result

Given a starting point of G– K-path : O(n log k) where k is a constant– Full k-ary tree with constant weight: O(n)– Tree : O(n log n)– DAG : NP-Hard (Reduce from 3-SAT)

Find the best starting point of G– Path : O(n) to find the best starting point.

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Inseparable Property

Let (vi)= w(vi)/d(vi) be the ratio cost of vi If u and v are inseparable, then there exist an o

ptimal solution containing uv sequence contiguously. (uv)= w(u)+w(v)/ d(u)+d(v)

Inseparable property:– If v is the only decedent of u, and (v) (u) then u

and v are inseparable.– If u have multiple decedents V. Let vV has the ma

ximum ratio cost among all others and (v) (u). u and v are inseparable.

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Inseparable Sample

3 4 2

Single decedent

Multiple decedents

2

5 4

7/2

4

11/3

2,5,4

Shrink 3 and 4

3 4 2

7/2

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Right Skew property

If a vertex sequence <v1,v2…vn> is right skew then (<v1…vi>) <(<vi+1…vn>) for all n i 1

The Inseparable is right skew.

2

5 4

2 5 4

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Decreasing right skew Property

G has decreasing property if for all u,v in G, u is the ancestor of v, then (u) (v) and u,v are both right skew.

4 8 5 3 9 3 7 6

4 8 5 3 9 3 7 6

12/2 17/3 16/3 Also, if G is a tree then G is a max-heap tree.

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Problem on K-path

We say a connected graph is the K-path if there exist a vertex has degree k and all other vertices have degree 2.

3-path

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Algorithm

1.Partition each paths into decreasing right skew.

2.Iterative select and output a maximum ratio cost from k candidate.

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Partition a path into several inseparable - O(n)

4 8 5 3 9 3 7 67 > 6

3 < 7

(3+7)/2 <6

4 8 5 3 9 3 7 6

4 8 5 3 9 3 7 6

4 8 5 3 9 3 7 6

4 8 5 3 9 3 7 6

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Partition a path into several inseparable - O(n)

After the partition steps, a path will be partition into several inseparable and the partition result will meet the decreasing right skew property.

We partition each paths in k-path graph individually. And get k decreasing right skew partition.

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Merge k decreasing right skew

S

13

1

10

5

6

2

S 13 10 6 5 2 1Result :

Time Complexity: n log(k)

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Problem on Tree

Observation: A node have maximum ratio cost is inseparable with its parent.

Algorithm:1.Build a Fibonacci heap

2.Select a maximum vertex v and shrink with its parent.

If v is the root, then output v.

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Example on Tree

5

3 10

8 11 2

Select 11

5

14/2 10

8 2

Select 10

15/2

14/2

8 2

Select 8

15/2

22/3 2

Select and output 15/222/3 2

Select and output 22/3

2Select and output 2

Result : 5,10,3,11,8,2

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Time Complexity

Construct a Fibonacci heap - O(n) Select maximum - O(1) Increasing Key - O(log n)

The time complexity is O(nlogn)

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Problem on full k-ary tree

If G is a full k-ary tree and external nodes have the constant weight we and the internal nodes also have the constant weight wi, then the problem can be solved in linear time.

1

1 2

2 2

Full k-ary tree

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Related Lemma

1. Any sub tree of full k-ary tree is inseparable

1

1 2

2 2

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Related Lemma

2. Let T1, T2 be the sub trees of full k-ary tree. If the T1 has more nodes, then (T2) (T1) and T2 will be visited before T1.

1

1 2

2 2(T1) 5/3

T1

(T2) 2/1

T2

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Algorithm

1.Compute the internal node number of each sub trees.

2. Use a radix sort to sort all nodes by the internal node number of sub tree rooted by each vertex.

3.According to the sorting result, we can build priority queues in each internal nodes.

4.Use a DFS to traversal the full k-ary tree.

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Example

A

B C

D E

F G 00

0

10

2

3 Sorting Result: G,F,D,C,E,B,A

G

A

B C

D E

F G

F

D

C

E

B

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Example

G

A

B C

D E

F G

F

D

C

E

B

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Problem on DAG

The problem on DAGs is NP-Hard. We can reduce from the 3-SAT. Given a 3-CNF F=C1 C2 …. Ck [k clauses]

with n literals {x1, x2,... xn} we can construct a DAG with 1+3n+4k vertices and 2n+6k edges.

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Construct a DAG

Given an instance of 3-SAT=(x1v x2v x3) (x3v x2v x4)...

S

x1 x1 x2 x2x3 x3 x4 x4

0 0 0

0 0 0

… n literals

… k clauses

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Constructing Step

1. If F has n literals.

S

x1 x1

x2 x2

x3 x3

x4 x4

… n literals

1 literal

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Constructing Step

2. If F has k clauses.

0 0 0

0 0 0

… k clauses

1 caluse

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Construct a DAG

Given an instance of 3-SAT=(x1v x2v x3) (x3v x2v x4)...

S

x1 x1 x2 x2x3 x3 x4 x4

0 0 0

0 0 0

… n literals

… k clauses

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The problem on DAG is NP-Hard

Let the w(xi) =0 If F is satisfiable the result of problem on DAG

is match the following format:

...S 0 0 0 0 0 0 ... 0 0 0...

2n vertices 2k vertices n+2k

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Find the best starting point on Path

The simplest way to find out the best starting point is perform the O(n) algorithm n times and select the starting point of the minimum result as the answer.

The simplest way cost O(n2). We propose an O(n) time algorithm to solve

the problem.

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Example to demonstration O(n) algorithm

4 8 5 3 9 3 7 6

3.Iterative select the minimum inseparable from Step 1 and 2 and remove it from 1 and 2

4 8 5 3 9 3 7 6

1.Find the partition starting from left point

4 8 5 3 9 3 7 6

2.Find the partition starting from right point

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Iterative select the minimum

4 8 5 3 9 3 7 6

4 8 5 3 9 3 7 6Result : 4

8 5 3 9 3 7 6

8 5 3 9 3 7 6Result : 358,4

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Iterative select the minimum

9 3 7 6

9 3 7 6Result : 376,358,4

9

9Result : 9,376,358,4

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Our Result

Given a starting point of G– K-path : O(n log k) where k is a constant– Full k-ary tree with constant weight: O(n)– Tree : O(n log n)– DAG : NP-Hard (Reduce from 3-SAT)

Find the best starting point of G– Path : O(n) to find the best starting point.

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Future Work

Find the best starting point of Tree Consider the on-line version. Consider the parallel version.

Q & A

Thank you!