1. Lecture 4: Tree Search Strategies
主講人:虞台文

2. Content Introduction Depth-First Search Breadth-First Search
Heuristic Searches
Best-First Search/Hill Climbing
A* algorithm
Branch-and-Bound

3. Lecture 4: Tree Search Strategies
Introduction

4. Introduction
The procedure of solving many problems may be represented by trees.
Therefore the solving of these problems becomes a tree searching problem.

5. Example: n-Queen 1 2 3 4 5 6 7 8 x1 = 1 x2 = 5 x3 = 8 x4 = 6 x5 = 3

6. n = 4 Example: n-Queen x1 = 1 2 3 4 x2 = 2 3 4 1 3 4 1 2 4 1 2 3

7. n = 4 Example: n-Queen The size of search space is x!2 3 4
How to search effectively?
x2 = 2 3 4 1 3 4 1 2 4 1 2 3
x3 = 3 4 2 4 2 3 3 4 1 4 1 3 2 4 1 4 1 2 2 3 1 3 1 2
x4 = 4 3 4 2 3 2 4 3 4 1 1 4 4 1 2 1 3 2 3 1 2 1 3 2

8. Some Search Schemes Breadth-First Search Depth-First Search
Hill Climbing
Best-First Search
Branch-and-Bound Strategy
for optimization problems
Blind Search
Guide with
Heuristics

9. Nodes of Search Tree Live Node Dead Node Expanded Node
a node ready for being expanded
Dead Node
a node has been expanded or can not be expanded further
Expanded Node
a node that is currently selected for expansion

10. Example: Depth-First Search2 3 4
x2 = 2 3 4 1 3 4 1 2 4 1 2 3       2 4 2 3 1 3 2 4 2 3 1 3         3 3 2 2  

11. Example: Breadth-First Search2 3 4
x2 = 2 3 4 1 3 4 1 2 4 1 2 3       2 4 2 3 1 3 2 4 2 3 1 3         3 3 2 2  

12. Search Space S0: Initial State Solution Space Search Space
Solution Path
Sg: Goal State

13. Major Components in Tree Search
Tree-Node Generating Mechanism
List: Tree-Node Maintaining Mechanism
Stack, Queue, Priority Queue
Tree-Node Selection Mechanism
Branching
Answer Verification Mechanism
Bounding Mechanism
Pruning
Cost Estimation Function
Heuristic

14. Lecture 4: Tree Search Strategies
Depth-First Search

15. Depth-First Search Scheme
Step1: Form a stack consisting of the root node.
Step2: Test to see if the top element in the stack is a goal node. If it is, stop. Otherwise, go to step 3.
Step3: Remove the top element from the stack. Add the first element’s descendants, if any, to the top of the stack.
Step4: If the stack is empty, then signal failure. Otherwise, go to Step 2.

16. Example: Sum-of-Subset Problem
Given a set of positive integers, say S={s1, …, sn}, and an positive integer, say M, we want to determine whether there exists a subset, say S’S, whose element sum equals to M.
The problem is NP-Complete.

17. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
Solution Space
(I) 
{1}
{2}
{3}
{4}
{1,2}
{1,3}
{1,4}
{1,2,3}
{1,2,4}
{1,3,4}
{1,2,3,4}
{2,3}
{2,4}
{2,3,4}
x1=1
x1=2
x1=3
x1=4
x2=2
x2=3
x2=4
x3=3
x3=4
x4=4

18. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 9
Solution Space
(I) 
{1}
x1=1 7
{1,2}
x2=2
{1,3}
x2=3
{1,4}
x2=4  12 8 9
{1,3,4}
x3=4  10

19. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 9
Solution Space
(I) 
{1}
{2}
{3}
{4}
{1,2}
{1,3}
{1,4}
{1,2,3}
{1,2,4}
{1,3,4}
{1,2,3,4}
{2,3}
{2,4}
{2,3,4}
x1=1
x1=2
x1=3
x1=4
x2=2
x2=3
x2=4
x3=3
x3=4
x4=4

20. Example: Sum-of-Subset Problem
SumSubset_1(S[1..n], M)
x[1]1; depth 1; // generate the first tree node
loop
if (depth =0) return false; // solution space has been exhausted
// evaluate the current sum
sum  0;
for k  1 to depth do sum  sum + x[k];
if (sum = M) return true; // report success
if (sum > M or x[depth] = n) // backtracking
depth  depth1;
if (depth <> 0) x[depth]  x[depth] + 1;
else // dig into the tree
x[depth+1]  x[depth]+ 1;
depth  depth + 1;
repeat

21. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
Solution Space
(II)
x1=1
x1=0
x2=1
x2=0
x3=1
x3=0
x4=1
x4=0

22. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 13
Solution Space
(II) 7 12 18 22 16 13 17 11 5 15 9 6 10 4
x1=1
x1=0
x2=1
x2=0
x3=1
x3=0
x4=1
x4=0
x1=1
x1=0
x2=1
x2=0
x3=1
x3=0
x4=1
x4=0
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000

23. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 13
Solution Space
(II)
x1=1
x1=0
x2=1
x2=0
x3=1
x3=0
x4=1
x4=0
x1=1
x1=0 7
x2=1
x2=0
x2=1
x2=0 12 7 5 18
x3=1 12 16
x3=0
x4=1
1101
x3=1
x3=0
x3=1
x3=0
x3=1
x3=0 13 7 11 5 6
x4=1
x4=0
x4=0
x4=1
x4=0
x4=1
x4=0
x4=1
x4=0
x4=1
x4=0
x4=1
x4=0
x4=1
x4=0 22 18 12 17 13 11 7 15 11 9 5 10 6 4
1111
1110
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000

24. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 13
Solution Space
(II)
Lazy Version 7
x1=1 12
x2=1 18
x3=1 7
x2=0 13
x3=1 12 16
x3=0
x4=1  

25. Example: Sum-of-Subset Problem
n = 4
Example: Sum-of-Subset Problem
M = 13
Solution Space
(II)
Eager Version 7
x1=1 12
x2=1 7
x2=0  12
x3=0 13
x3=1 

26. Example: Sum-of-Subset Problem
Lazy Version
Example: Sum-of-Subset Problem
SumSubset_2(S[1..n], M, X[1..n]) // return true or false. If true, subset is in X
for i  1 to n do X[i] 0; // initialization
k  1; sum = 0; // start from the root node
loop
while k  n and sum < M do // extend left as deep as possible
x[k]  1; sum  sum + S[k]; k  k + 1;
if(sum = M) return true; // answer found
k  k  1; // overshooting
sum  sum  S[k]; // withdraw the last element
x[k]  0; // branch to right subtree
if(k = n) // if no more right subtree, backtrack one level
while(k > 0 and X[k] = 0) k  k  1;
if (k = 0) return false; // tree has been exhausted
else X[k]  0;
k  k + 1; // search on subtree
repeat

27. Example: Sum-of-Subset Problem
Eager Version
Example: Sum-of-Subset Problem
SumSubset_2(S[1..n], M, X[1..n]) // return true or false. If true, subset is in X
for i  1 to n do X[i] 0; // initialization
k  1; sum = 0; // start from the root node
loop
while k  n and sum + S[k]  M do // extend left as deep as possible
x[k]  1; sum  sum + S[k]; k  k + 1;
if(sum = M) return true; // answer found
if(k > n) // if no more right subtree, backtrack one level
while(k > 1 and X[k  1] = 0) k  k  1;
if (k = 1) return false; // tree has been exhausted
k  k + 1;
repeat

28. Discussion Sum-of-Subset problem is a constraint satisfaction problem.
Tree pruning by considering feasibility
Can we apply DFS scheme to solve optimization problem?
Do we need to inspect the costs (or profits) of all feasible solutions to determine the optimal one?
How to prune subtrees sure to with inferior feasible solutions only?

29. 0-1 Knapsack Problem

30. 0-1 Knapsack Problem Fractional Version
Fact: the profit for the optimal solution of a 0-1 knapsack problem cannot greater than that of its fractional version.

31. 0-1 Knapsack Problem xk+1 = 1 xk+1 = 0
We can compute the following items for each feasible node:
1. current weight
2. current profit
3. upper bound
xk+1 = 1
xk+1 = 0
Fact: a left child has the same upper bound as its root.

32. 0-1 Knapsack Problem
Weight/Profit
Upper Bound
xk+1 = 1
xk+1 = 0

33. Example: 0-1 Knapsack Problemi 1 2 3 4 5 6 7 8 pi 11 21 31 33 43 53 55 65 wi 23 45
W = 110

34. Example: 0-1 Knapsack Problemi 1 2 3 4 5 6 7 8 pi 11 21 31 33 43 53 55 65 wi 23 45
W = 110
Example: 0-1 Knapsack Problem
164.88
1/11
x1=1
155.11
x1=0 
12/32
x2=1
157.44
x2=0 
33/63
x3=1
159.76
x3=0
56/96
x4=1
160.22
x4=0
65/35
x4=1
154.88
x4=0 
89/139
x5=1
x5=0
162.44
66/106
x5=1
157.55
x5=0 
68/108
157.11 
164.66
x6=0
109/159
x6=1
159.79
x6=0
99/149
161.63
163.81
x7=0
162
105/151
160.18
158 
157.63  A
139
x8=0 B
149 C
151 D
159

35. Example: 0-1 Knapsack Problemi 1 2 3 4 5 6 7 8 pi 11 21 31 33 43 53 55 65 wi 23 45
W = 110
Example: 0-1 Knapsack Problem
164.88
Fact: the bounds can be integers.
1/11
x1=1
155.11
x1=0 
12/32
x2=1
157.44
x2=0 
33/63
x3=1
159.76
x3=0
56/96
x4=1
160.22
x4=0
65/35
x4=1
154.88
x4=0 
89/139
x5=1
x5=0
162.44
66/106
x5=1
157.55
x5=0 
68/108
157.11 
164.66
x6=0
109/159
x6=1
159.79
x6=0
99/149
161.63
163.81
x7=0
162
105/151
160.18
158 
157.63  A
139
x8=0 B
149 C
151 D
159

36. Example: 0-1 Knapsack Problemi 1 2 3 4 5 6 7 8 pi 11 21 31 33 43 53 55 65 wi 23 45
W = 110
Example: 0-1 Knapsack Problem
164.88
Fact: the bounds can be integers.
1/11
x1=1
155.11
x1=0 
12/32
x2=1
157.44
x2=0 
33/63
x3=1
159.76
x3=0 
56/96
x4=1
160.22
x4=0
89/139
x5=1
x5=0
162.44
66/106
x5=1
157.55
x5=0 
164.66
x6=0
109/159
x6=1
159.79
x6=0
99/149
161.63 
163.81
x7=0
162
105/151
160.18 A
139
x8=0 B
149 C
151 D
159

37. Lecture 4: Tree Search Strategies
Breadth-First Search

38. Breadth-First Search Scheme
Step1: Form a queue consisting of the root node.
Step2: Test to see if the first element in the queue is a goal node. If it is, stop. Otherwise, go to step 3.
Step3: Remove the first element from the queue. Add the first element’s descendants, if any, to the end of the queue.
Step4: If the queue is empty, then signal failure. Otherwise, go to Step 2.

39. Example: 8-Puzzle 2 8 3 1 6 4 7 5 1 2 3 8 6 4 7 5
Start
Goal

40. Example: 8-Puzzle    
Blank
Cell
Movement

41. Example: 8-Puzzle 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 1 7 6 4 5 2 8 3 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5

42. Exercise
Accompanying with a bounding function to solve the instance of 0-1 knapsack problem describe previously using the breadth-first search. Give your result graphically and compare it with the depth-first one.

43. Lecture 4: Tree Search Strategies
Heuristic Searches

44. Cost of Subtrees c1 c2 cn c2 cn c1

45. n = 4 Example: n-Queen 4  4  4  4  4 2 x1 = 1 3 4 x2 = 2 1 x3 = 3

46. Assume that we are completely informative, then search is trivial.
Example: n-Queen
Assume that we are completely informative, then search is trivial. 4 2
x1 = 1 3 4
x2 = 2 1
x3 = 3
x4 = 4  4  4  4  4

47. optimal path in the subtree
Cost Evaluation
root
c(x)
g(x)
current state (x)
h(x)
optimal path in the subtree
goal

48. Cost Evaluation root g(x) current state (x) c(x)
Usually can be computed precisely.
root
c(x)
g(x)
current state (x)
h(x)
optimal path in the subtree
Usually can be estimated only.
goal

49. Admissible Heuristic a How to make a heuristic?
generally, drop or relax constraints from the original problem.

50. Heuristic Search Algorithm
Step1: Form a priority queue with root node only.
Step2: Test to see if the first element in the queue is a goal node. If it is, stop. Otherwise, go to step 3.
Step3: Remove the first element from the queue, and add its descendants, if any, to the queue orderly by referencing to their estimated cost.
Step4: If the queue is empty, then signal failure. Otherwise, go to Step 2.

51. Heuristic Search Algorithm
Breadth-First Search
Best-First Search/Hill Climbing
A* algorithm

52. Example: 8-Puzzle 2 8 3 1 6 4 7 5 1 2 3 8 6 4 7 5
Start (x)
Goal

53. Example: 8-Puzzle 4 1+5 1+3 1+5 2+3 2+3 2+4 3+3 3+4 3+2 3+4 4+1 5+0 26 4 7 5 4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
1+5
1+3
1+5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
2+3
2+3
2+4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
3+3
3+4
3+2
3+4 2 8 3 1 6 4 7 5
4+1 2 8 3 1 6 4 7 5
5+0

54. Best-First Search
Best-first search is a search algorithm which optimizes depth-first search by expanding the most promising node chosen according to some rule.

55. Example: 8-Puzzle 2 8 3 1 6 4 7 5 4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 5 3 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 3 3 4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 3 4 2 4 2 8 3 1 6 4 7 5 1 2 8 3 1 6 4 7 5

56. Hill Climbing A simplification of Best-First Search The idea 
You don't keep the big list of states around, i.e., you just keep track of the one state you are considering, and the path that got you there from the initial state.
At every state you choose the state that leads you closer to the goal and continue from there.
In fact, this is a greedy algorithm
Local minima/maxima

57. A* Algorithm
Admissible

58. A* Algorithm 4 1+5 1+3 1+5 2+3 2+3 2+4 3+3 3+4 3+2 3+4 4+1 5+0 2 8 3 16 4 7 5 4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
1+5
1+3
1+5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
2+3
2+3
2+4 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5 2 8 3 1 6 4 7 5
3+3
3+4
3+2
3+4 2 8 3 1 6 4 7 5
4+1 2 8 3 1 6 4 7 5
5+0

59. Optimality of A* Algorithm
If A* returns a solution, that solution is guaranteed to be optimal as long as the cost estimation function is admissible.

60. Exercise
Use BFS, DFS, Hill-Climbing, Best-First Search and A* schemes to solve the following 8-puzzle problem with the evaluation function being the number of misplaced tiles. 2 3 1 8 4 7 6 5 1 2 3 7 6 4 8 5
Start
Goal

61. Branch-and-Bound Algorithm
Two main tools:
Branching
A Mechanism to jump to subtrees
Bounding
A Mechanism to prune subtrees

62. Example: Shortest Paths on a Multi-stage Graph1 2 3 5 4 7

63. Example: Shortest Paths on a Multi-stage Graph1 2 3 5 4 7
Search tree

64. Example: Shortest Paths on a Multi-stage Graph
Solved by B&B 1 3 2
Upper Bound 1 3 2 5 3 4 3 2 7  5     6 1 4 7 6 1 4 9 5 5

65. Example: Asymmetric TSP1 2 3 4

66. Example: Asymmetric TSP
n = 4
Example: Asymmetric TSP
Search Tree
(Static) 1 2 4 3 5 6 7 8 9 10 11 12 13 14 15 16
x1=2
x1=3
x1=4
x2=3
x2=4
x2=2
x3=4
x3=3
x3=2

67. Example: Asymmetric TSP
How to determine the lower-bound of each tree node?
n = 5

68. Example: Asymmetric TSP
How to determine the lower-bound of root node?
Example: Asymmetric TSP
10 2 2 3 4
21
n = 5
Row Reduced (21)

69. Example: Asymmetric TSP
How to determine the lower-bound of root node? 25
Example: Asymmetric TSP 1 3
Reduced Cost Matrix
(21 4 = 25)
Row Reduced (21)

70. Example: Asymmetric TSP25 1 2 4 3
x1=2
x1=3
x1=4 5
x1=5
How to determine the lower-bound of the children?
What are their cost reduced matrices?

71. Example: Asymmetric TSP
(i, j)
Given the cost reduced matrix of node, say R, how to obtain the cost reduced matrix of its child node, say S, generated by selecting edge (i, j)?

72. Example: Asymmetric TSP1 2 i j n … R S
(i, j)
Reduced Cost Matrix of R
Non-leaf

73. Example: Asymmetric TSP
To obtain the reduced cost matrix of S, set the surrounded entries to , then reduce it.
Example: Asymmetric TSP 1 2 i j n … R S
(i, j)
Reduced Cost Matrix of R
Non-leaf

74. Example: Asymmetric TSP
Upper Bound  1 2 4 3
x1=2
x1=3
x1=4 5
x1=5 25 10 35 2
Reduced

75. Example: Asymmetric TSP
Upper Bound  1 2 4 3
x1=2
x1=3
x1=4 5
x1=5 25 10 17 35 53 3
11

76. Example: Asymmetric TSP
Upper Bound  1 2 4 3
x1=2
x1=3
x1=4 5
x1=5 25 35 53 3 25 31 6 8 7
x2=2
x2=3
x2=5 28 50 36 9 10
x3=3
x3=5 32 28 11
x4=3 28

77. Example: Asymmetric TSP
Upper Bound  1 2 4 3
x1=2
x1=3
x1=4 5
x1=5 25 28  35  53  3 25 31 6 8 7
x2=2
x2=3
x2=5 28 50  36  9 10
x3=3
x3=5 32  28 11
x4=3 28

78. Example: Asymmetric TSP
Search Tree
(Dynamic)
Tours
include (i, j)
Tours
exclude (i, j)

79. Example: Asymmetric TSP
Given the cost reduced matrix of node, say r, and how to obtain the cost reduced matrices of its child nodes, say x,and y, and
and
by considering (i, j)? r
Tours
include (i, j)
Tours
exclude (i, j) x y

80. Example: Asymmetric TSP
Obtain the cost reduced matrix of the right child by set (i, j)th entry to  and then reduce it. The low-bound of the right child can be obtained by accumulation.
Example: Asymmetric TSP r
Tours
include (i, j)
Tours
exclude (i, j)  x y
Cost Reduced Matrix of r.

81. Example: Asymmetric TSP
Obtain the cost reduced matrix of the left child by set the entries to  and then reduce it. The low-bound of the left child can be obtained by accumulation.
Example: Asymmetric TSP r
Tours
include (i, j)
Tours
exclude (i, j) x y
Cost Reduced Matrix of r.

82. Example: Asymmetric TSP
At each stage, the edge to be considered is determined based on heuristic.
Possible heuristics:
Choose (i, j) such that
as large as possible. r
Tours
include (i, j)
Tours
exclude (i, j) x y or

83. Example: Asymmetric TSP
We will use the first one in the following example.
At each stage, the edge to be considered is determined based on heuristic.
Possible heuristics:
Choose (i, j) such that
as large as possible.
Tours
include (i, j)
Tours
exclude (i, j) x y or

84. Example: Asymmetric TSP
How to determine the lower-bound of each tree node?
n = 5

85. Example: Asymmetric TSP
How to determine the lower-bound of root node?
Example: Asymmetric TSP
10 2 2 3 4
21
n = 5
Row Reduced (21)

86. Example: Asymmetric TSP
How to determine the lower-bound of root node? 25
Example: Asymmetric TSP 1 3
Reduced Cost Matrix
(21 4 = 25)
Row Reduced (21)

87. Example: Asymmetric TSP25 1 2 3 ?
You should choose one of zero entries that reduces most cost for the right child.
Why?

88. Example: Asymmetric TSP25 1 2 3 0 ?  1 1
You should choose one of zero entries that reduces most cost for the right child.

89. Example: Asymmetric TSP25 1 2 3 0 ? 1 2
You should choose one of zero entries that reduces most cost for the right child.  2

90. Example: Asymmetric TSP25 1 2 3 ?
11 1
You should choose one of zero entries that reduces most cost for the right child. 2  0
11

91. Example: Asymmetric TSP25 1 2 3 0 ? 1
You should choose one of zero entries that reduces most cost for the right child. 2  0
11

92. Example: Asymmetric TSP25 1 2 3 0 ? 1
You should choose one of zero entries that reduces most cost for the right child. 2
11 3  3

93. Example: Asymmetric TSP25 1 2 3 3 ? 1
You should choose one of zero entries that reduces most cost for the right child. 2
11 3  0 3

94. Example: Asymmetric TSP25 1 2 3
11 ? 1
You should choose one of zero entries that reduces most cost for the right child. 2
11 3  0 3
11

95. Example: Asymmetric TSP25 1 2 3
11 ? 1
You should choose one of zero entries that reduces most cost for the right child. 2
11 3 0 3
11

96. Example: Asymmetric TSP25 1 2 3 ? 25 36
already reduced

97. Example: Asymmetric TSP1 3 2 4 5 25 1
Optimal
TSP Tour 2 3 ? 25 36  4 5 28 36  6 7 28 37 

98. Exercise
Using B&B algorithm to solve the TSP instance given in the above table.
Using static search tree
Using dynamic search tree.