cs 312: algorithm analysis lecture #34: branch and bound design options for solving the traveling...

CS 312: Algorithm Analysis

Lecture #34: Branch and Bound Design Options for Solving the Traveling Salesman

Problem: Tight Bounds

This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License.

Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, and Sean Warnick

Announcements

Homework #24 due now

Homework #25 due Friday

Project #7: TSP ASAP: Read the helpful “B&B for TSP Notes” linked from

the schedule Read Project Instructions Today: We continue discussing main ideas Next Wednesday: Early day Week from Friday: due

Objectives

Review the Traveling Salesman Problem (TSP)

Develop a good bound function for the TSP

Reason about Tight Bounds Augment general B&B algorithm

Traveling Salesman (Optimization) Problem

Rudrata or Hamiltonian Cycle Cycle in the graph that passes through each vertex

exactly once

+ Find the least cost or “shortest”

cycle1

2

3 4

58

67

5

4

3

2

19

1012

Distinguish from theTSP search problem and theTSP decision problem

How to solve?

If with B&B, what do we need?

How to solve?

If with B&B, what do we need?

Initial BSSF

1

2

3 4

5

8

6

75

4

3

2

19

10

12

How to compute?

Should be quick.

What if you have a complete graph?

What if you don’t?

Simple-Minded Initial BSSF

1

2

3 4

5

8

6

75

4

3

2

19

10

12

Cost of BSSF= 9+5+4+12+1 = 31

A Bound on Possible TSP Tours

We need a bound function. Lower or Upper?How to compute?

1

2

3 4

58

67

5

4

3

2

19

1012

A Bound on Possible TSP Tours

We need a bound function. Lower or Upper?How to compute?

1

2

3 4

58

67

5

4

3

2

19

1012

A Bound on Possible TSP Tours

1

2

3 4

5

8

6

75

4

3

2

19

10

12

What’s the cheapest way to leave each vertex?

Bound on Possible TSP Tours

1

2

3 4

5

8

6

75

4

3

2

19

10

12

Save the sum of those costs in the bound (as a rough draft).

Rough draft bound= 8+6+3+2+1 = 20

Bound on Possible TSP Tours

1

2

3 4

5

8-8=0

6

74

4

3

2

19-8=1

10

12

For a given vertex, subtract the least cost departure from each edge leaving that vertex.

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

12

1

0

0

01

9

6

Repeat for the other vertices.What do the numbers on the edges mean now?

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

12

1

0

0

01

9

6

Now, can we find a tighter lower bound?

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

12

1

0

0

01

9

6

Does that set of edges now having 0 residual cost arrive at every vertex?

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

12

1

0

0

01

9

6

In this case, those edges never arrive at vertex #3.

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

12

1

0

0

01

9

6

We have to take an edge to vertex 3 from somewhere. Assume we take the cheapest.

Rough draft bound= 20

Bound on Possible TSP Tours

1

2

3 4

5

0

0

01

1

0

0

01

9

6

Subtract its cost from other edges entering vertex 3 and add the cost to the bound.

We have just tightened the bound.

Bound = 21

This Bound

It will cost at least this much to visit all the vertices in the graph. There’s no cheaper way to get in and out of each

vertex. Each edge is now labeled with the extra cost of

choosing that edge.

The bound is not a solution; it’s a bound!

Why are tight bounds desirable?

Bound on Possible TSP Tours

1

2

3 4

5

8

6

74

4

3

2

19

10

12

Our algorithm can do this reasoning using a cost matrix.

999 9 999 8 999999 999 4 999 2999 3 999 4 999999 6 7 999 12

1 999 999 10 999

To:1 2 3 4 5From:

1

2

3

4

5

Bound on Possible TSP Tours

999 1 999 0 999999 999 2 999 0999 0 999 1 999999 0 1 999 6

0 999 999 9 999

1

2

3 4

50

01

2

1

0

0

01

9

6

Reduce all rows.

To:1 2 3 4 5From:

1

2

3

4

5

Bound on Possible TSP Tours

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

1

2

3 4

50

01

2

1

0

0

01

9

6

Then reduce column #3. Now we have a tighter bound.

To:1 2 3 4 5From:

1

2

3

4

5

Search

Let’s start the search Arbitrarily start at vertex 1

Why is this OK? Focus on:

the bound function and the reduced cost matrix representation of

states

Using this bound for TSP in B&B

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21 BSSF=31

Start at vertex 1 in graph (arbitrary)

What should our state expansion strategy be?

Using this bound for TSP in B&B

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21

1-2 1-3 1-4 1-5

bound = 21+1

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

BSSF=31

Start at vertex 1 in graph (arbitrary)

bound = 21

Focus: going from 1 to 2

999 999 999 999 999999 999 1 999 0999 999 999 1 999999 999 0 999 6

0 999 999 9 999

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21

1-2

bound = 22

1

2

3 4

50

00

1

1

0

0

01

96

1

2

3 4

5

01

1

0

01

96

Add extra cost from 1 to 2, exclude edges from 1 or into 2.

BSSF=31

Before After

999 999 999 999 999999 999 1 999 0999 999 999 1 999999 999 0 999 6

0 999 999 9 999

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21

bound = 22+1

1

2

3 4

50

00

1

1

0

0

01

96

1

2

3 4

5

01

1

0

01

96

No edges into vertex 4 w/ 0 reduced cost.

Focus: going from 1 to 2

BSSF=31

Before After

1-2

999 999 999 999 999999 999 1 999 0999 999 999 0 999999 999 0 999 6

0 999 999 8 999

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21

bound = 21+1+1

1

2

3 4

5

01

0

0

01

86

Add cost of reducing edge into vertex 4.

Focus: going from 1 to 2

BSSF=31

1-2

Bounds for other choices

999 999 999 999 999999 999 1 999 0999 999 999 0 999999 999 0 999 6

0 999 999 8 999

999 1 999 0 999999 999 1 999 0999 0 999 1 999999 0 0 999 6

0 999 999 9 999

bound = 21

bound = 23

999 999 999 999 999999 999 1 999 0999 0 999 999 999999 0 0 999 6

0 999 999 999 999

bound = 21

1-2(23),1-4(21)BSSF=31

1-2 1-3 1-4 1-5

Agenda:

Leaving Vertex 4

999 999 999 999 999999 999 1 999 0999 0 999 999 999999 0 0 999 6

0 999 999 999 999

bound = 21

1

2

3 4

50

00

1 0

0

0

6

999 999 999 999 999999 999 0 999 0999 999 999 999 999999 999 999 999 999

0 999 999 999 999

999 999 999 999 999999 999 999 999 0999 0 999 999 999999 999 999 999 999

0 999 999 999 999

999 999 999 999 999999 999 0 999 999999 0 999 999 999999 999 999 999 999

0 999 999 999 999

1-4-2 1-4-3 1-4-5

bound = 22 bound = 21 bound = 28

BSSF=311-4

Leaving Vertex 4

999 999 999 999 999999 999 1 999 0999 0 999 999 999999 0 0 999 6

0 999 999 999 999

bound = 21

1

2

3 4

50

00

1 0

0

0

6

999 999 999 999 999999 999 0 999 0999 999 999 999 999999 999 999 999 999

0 999 999 999 999

999 999 999 999 999999 999 999 999 0999 0 999 999 999999 999 999 999 999

0 999 999 999 999

999 999 999 999 999999 999 0 999 999999 0 999 999 999999 999 999 999 999

0 999 999 999 999

bound = 22 bound = 21 bound = 28

1-4-2(22), 1-4-3(21)1-4-5(28),1-2(23)

BSSF=311-4

1-4-2 1-4-3 1-4-5

Agenda:

Leaving Vertex 3

999 999 999 999 999999 999 999 999 0999 0 999 999 999999 999 999 999 999

0 999 999 999 999

bound = 21

1

2

3 4

50

00

0

0

999 999 999 999 999999 999 999 999 0999 999 999 999 999999 999 999 999 999

0 999 999 999 999

1-4-3-2

bound = 21

BSSF=311-4-3

Leaving Vertex 3

999 999 999 999 999999 999 999 999 0999 0 999 999 999999 999 999 999 999

0 999 999 999 999

bound = 21

1

2

3 4

50

00

0

0

999 999 999 999 999999 999 999 999 0999 999 999 999 999999 999 999 999 999

0 999 999 999 999

bound = 21

4-2(22), 3-2(21)4-5(28), 1-2(23),

BSSF=31

1-4-3-2

Agenda:

1-4-3

Search Tree for This Problem

b=21

b=23 b=21

b=22 b=21 b=28

b=21

1-to-2 1-to-4

4-to-2 4-to-3 4-to-5

3-to-2

2-to-5

Termination Criteria for a B&B Algorithm

Repeat until Agenda is empty Or time is up Or BSSF cost is equal to original LB

Update: Branch and Boundfunction BandB(v)

BSSF quick-solution(v) // BSSF.cost holds costAgenda.clear()v.b bound(v)Agenda.add(v, v.b)while !Agenda.empty() and time remains and BSSF.cost != v.b do

u Agenda.first()Agenda.remove_first()children = generate_children_ascending(u)

for each w in children doif ! time remains then breakw.b bound(w)

if (w.b < BSSF.cost) thenif criterion(w) then

BSSF wAgenda.prune(BSSF.cost)

else if partial_criterion(w) thenAgenda.add(w, w.b)

return BSSF

Assignment

HW #25: Compute bound for TSP instance using

today’s method Reason about search for TSP solution