1. HEURISTIC SEARCH Heuristics:
Rules for choosing the branches in a state space that are most likely to lead to an acceptable problem solution.
Rules that provide guidance in decision making
Often improves the decision making:
Shopping: Choosing the shortest queue in a supermarket does not necessarily means that you will out of the market earlier
Used when:
Information has inherent ambiguity
computational costs are high
CS 331/531 Dr M M Awais

2. Finding Heuristics Tic - Tac - Toe Heuristic: which one to choose? X
calculate winning lines and move to state with most wining lines.
CS 331/531 Dr M M Awais

3. Calculating winning linesX X X
3 winning lines
4 winning lines
3 winning lines
Always choose the state with maximum heuristic value
Maximizing heuristics
CS 331/531 Dr M M Awais

4. Choosing city with minimum distance:
Always choose the city with minimum heuristic value
Minimizing heuristics (hi<hi-1)
CS 331/531 Dr M M Awais

5. Algorithms for Heuristic Search
Hill Climbing
Best first search
A* Algo
CS 331/531 Dr M M Awais

6. Simplified Algorithm:
Hill Climbing:
If the Node is better, only then you proceed to that Node
When is a node better?: Apply heuristic to compare the nodes
Simplified Algorithm:
1. Start with current-state (cs) = initial state
2. Until cs = goal-state or there is no change in the cs do:
(a) Get the successor of cs and use the EVALUATION FUNCTION to assign a score to each successor
(b) If one of the successor has a better score than cs then set the new state to be the successor with the best score.
CS 331/531 Dr M M Awais

7. Navigation Problem Choose the closest city to travel
CS 331/531 Dr M M Awais

8. Navigation Problem Choose the closest city to travel hi < hi-1
CS 331/531 Dr M M Awais

9. Navigation Problem GET STUCK No Backtracking
Choose the closest city to travel
CS 331/531 Dr M M Awais

10. Hill Climbing F:6 B:5 D:4 E:2 G:0 Goal/ Solution C:3 A:10
Node label: heuristic value of the node
A:10
A is node name,
10 is the heuristic evaluation
CS 331/531 Dr M M Awais

11. Hill Climbing F:6 B:5 D:4 E:2 G:0 Goal/ Solution C:3 A:10
Compare B with C
Node label: heuristic value of the node
A:10
A is node name,
10 is the heuristic evaluation
CS 331/531 Dr M M Awais

12. Hill Climbing F:6 B:5 D:4 E:2 G:0 Goal/ Solution C:3 A:10
C is better so move
F is poor than C
So gets stuck at C
Node label: heuristic value of the node
A:10
A is node name,
10 is the heuristic evaluation
CS 331/531 Dr M M Awais

13. Hill-climbing search
“is a loop that continuously moves in the direction of increasing value”
It terminates when a peak is reached.
Hill climbing does not look ahead of the immediate neighbors of the current state.
Hill-climbing chooses randomly among the set of best successors, if there is more than one.
Hill-climbing a.k.a. greedy local search
CS 331/531 Dr M M Awais

14. Hill-climbing search: Algo
function HILL-CLIMBING( problem) return a state that is a local maximum
input: problem, a problem
local variables: current, a node.
neighbor, a node.
current  MAKE-NODE(INITIAL-STATE[problem])
loop do
neighbor  a highest valued successor of current
if VALUE [neighbor] ≤ VALUE[current] then return STATE[current]
current  neighbor
CS 331/531 Dr M M Awais

15. Hill-climbing example
8-queens problem (complete-state formulation).
Successor function: move a single queen to another square in the same column.
Heuristic function h(n): the number of pairs of queens that are attacking each other (directly or indirectly).
CS 331/531 Dr M M Awais

16. Hill-climbing example
a) shows a state of h=17 and the h-value for each possible successor.
b) A local minimum in the 8-queens state space (h=1).
CS 331/531 Dr M M Awais

17. Drawback
Ridge = sequence of local maxima difficult for hill climbing to navigate
Plateaux = an area of the state space where the evaluation function is flat.
GETS STUCK 86% OF THE TIME.
CS 331/531 Dr M M Awais

18. Hill-climbing variations
Stochastic hill-climbing
Random selection among the uphill moves.
The selection probability can vary with the steepness of the uphill move.
First-choice hill-climbing
Stochastic hill climbing by generating successors randomly until a better one is found.
Random-restart hill-climbing
Tries to avoid getting stuck in local maxima/minima.
CS 331/531 Dr M M Awais

19. First Reading Assignment on Simulated Annealing
The method to get rid of the local minima problem
You can consult the Text Book
CS 331/531 Dr M M Awais

20. More on Heuristic functions
8-puzzle
Avg. solution cost is about 22 steps (branching factor +/- 3)
Exhaustive search to depth 22: 3.1 x 1010 states.
A good heuristic function can reduce the search process.
CAN YOU THINK OF A HEURISTIC ?
CS 331/531 Dr M M Awais

21. 8 Puzzle Two commonly used heuristics
h1 = the number of misplaced tiles
h1(s)=8
h2 = the sum of the distances of the tiles from their goal positions (manhattan distance).
h2(s)= =18
CS 331/531 Dr M M Awais

22. Heuristic quality Effective branching factor b*
Is the branching factor that a uniform tree of depth d would have in order to contain N+1 nodes.
A good heuristic should have b* as low as possible
This measure is fairly constant for sufficiently hard problems.
Can thus provide a good guide to the heuristic’s overall usefulness.
A good value of b* is 1.
CS 331/531 Dr M M Awais

23. Heuristic quality and dominance
h h2
h h2
Effective branching factor of h2 is lower than h1
If h2(n) >= h1(n) for all n then h2 dominates h1 and is better for search (value of h2 and h1 e.g. 18 vs 8)
CS 331/531 Dr M M Awais

24. Admissible heuristics
A heuristic h(n) is admissible if for every node n, h(n) ≤ h*(n), where h*(n) is the true cost to reach the goal state from n.
An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic
CS 331/531 Dr M M Awais

25. Navigation Problem
Shows Actual Road Distances
CS 331/531 Dr M M Awais

26. Heuristic: Straight Line Distance between cities and goal city (aerial)
Actual Road Distances
Is the new heuristic Admissible
CS 331/531 Dr M M Awais

27. Heuristic: Straight Line Distance between cities goal city (aerial)
Is the new heuristic Admissible
hsld(n)<=h*(n)
Consider n= sibiu
hsld(sibiu)=253
h*(sibiu)= =278 (actual cost through Piesti)
h*(sibiu)=99+211=310 (actual cost through Fagaras)
hsld<=h* Admissible (never overestimates the actual road distance)
CS 331/531 Dr M M Awais

28. Which Heuristic is Admissible?
h1 = the number of misplaced tiles
h1(s)=8
h2 = the sum of the distances of the tiles from their goal positions (manhattan distance).
h2(s)= =18
BOTH
CS 331/531 Dr M M Awais

29. New Heuristic: Permutation Inversions12 15 11 14 10 13 9 5 6 7 8 4 3 2 1
let the goal be:
Let nI be the number of tiles J < I that appear after tile I (from left to right and top to bottom)
h3 = n2 + n3 +  + n15 + row number of empty tile
n2 = 0 n3 = 0 n4 = 0
n5 = 0 n6 = 0 n7 = 1
n8 = 1 n9 = 1 n10 = 4
n11 = 0 n12 = 0 n13 = 0
n14 = 0 n15 = 0 12 15 11 14 6 13 9 5 10 7 8 4 3 2 1
 h3 = 7 + 4
IS h3 admissible
CS 331/531 Dr M M Awais

30. New Heuristic: Permutation Inversions12 15 11 14 10 13 9 5 6 7 8 4 3 2 1
IS h3 admissible
h3 = 7 + 4
h* = actual moves
required to achieve
the goal
If h3 <= h* then
Admissible,
Otherwise Not 12 15 11 14 6 13 9 5 10 7 8 4 3 2 1
Find out yourself
CS 331/531 Dr M M Awais

31. New Heuristic: Permutation Inversions5 4 7 1 2 6 8 3
STATE(Goal) 5 4 7 1 2 6 8 3
STATE(N)
Is h3 admissible here:
h3=3+1=4
3 (block 3 requires 3 jumps it should be before the empty block)
1(empty block is in row 1)
h*=2 (actual moves)
So h3 is not admissible
CS 331/531 Dr M M Awais

32. Example STATE(N) GOAL STATE1 4 7 5 2 6 3 8
STATE(N) x
GOAL STATE
Goal contains partial information, the sequence in the
first row is important.
CS 331/531 Dr M M Awais

33. Finding admissible heuristics
Relaxed Problem:
Admissible heuristics can be derived from the exact solution cost of a relaxed version of the problem:
Relaxed 8-puzzle for h1 : a tile can move anywhere As a result, h1(n) gives the shortest solution
Relaxed 8-puzzle for h2 : a tile can move to any adjacent square.
As a result, h2(n) gives the shortest solution.
The optimal solution cost of a relaxed problem is no greater than the optimal solution cost of the real problem.
CS 331/531 Dr M M Awais

34. Relaxed Problem: Example
By solving relaxed problems at each node
In the 8-puzzle, the sum of the distances of each tile to its goal position (h2) corresponds to solving 8 simple problems:
It ignores negative interactions among tiles
Store the solution pattern in the database 1 4 7 5 2 6 3 8
CS 331/531 Dr M M Awais

35. Relaxed Problem: Example1 4 7 5 2 6 3 8 8 5 6
h = h8 + h5 + h6 +…
CS 331/531 Dr M M Awais

36. Complex: Relaxed Problem
Consider two more complex relaxed problems:
h = d d5678 [disjoint pattern heuristic]
These distances could have been pre-computed in a database 1 4 7 5 2 6 3 8
CS 331/531 Dr M M Awais

37. Relaxed Problem Several order-of-magnitude speedups for
15- and 24-puzzle
Problem have been obtained through the application of relaxed problem
CS 331/531 Dr M M Awais

38. Finding admissible heuristics
Find an admissible heuristic through experience:
Solve lots of puzzles
Inductive learning algorithm can be employed for predicting costs for new states that may arise during search.
CS 331/531 Dr M M Awais

39. Summary: Admissible Heuristic
Defining and Solving Sub-problem:
Admissible heuristics can also be derived from the solution cost of a sub-problem of a given problem.
This cost is a lower bound on the cost of the real problem.
Pattern databases store the exact solution to for every possible sub-problem instance.
The complete heuristic is constructed using the patterns in the DB
Learning through experience
CS 331/531 Dr M M Awais

40. Robot Navigation Other ExamplesxN yN N yg xg
CS 331/531 Dr M M Awais

41. Robot Navigation Other ExamplesxN yN N yg xg
(Euclidean distance)
h2(N) = |xN-xg| + |yN-yg|
(Manhattan distance)
CS 331/531 Dr M M Awais

42. Best-First Search
It exploits state description to estimate how promising each search node is
An evaluation function f maps each search node N to positive real number f(N)
Traditionally, the smaller f(N), the more promising N
Best-first search sorts the nodes in increasing f [random order is assumed among nodes with equal values of f]
“Best” only refers to the value of f, not to the quality of the actual path. Best-first search does not generate optimal paths in general
CS 331/531 Dr M M Awais

43. Summary: Best-first search
General approach:
Best-first search: node is selected for expansion based on an evaluation function f(n)
Idea: evaluation function measures distance to the goal.
Choose node which appears best based on the heuristic value
Implementation:
A queue is sorted in decreasing order of desirability.
Special cases: greedy search, A* search
CS 331/531 Dr M M Awais

44. Evaluation function Heuristic Evaluation
Same as Hill Climbing
Heuristic Evaluation
f(n)=h(n) = estimated cost of the cheapest path from node n to goal node.
If n = goal then h(n)=f(n)=0
CS 331/531 Dr M M Awais

45. Best First Search Method
Algo:
1. Start with agenda (priority queue) = [initial-state]
2. While agenda not empty do:
(a) remove the best node from the agenda
(b) if it is the goal node then return with success Otherwise find its successors.
( c) Assign the successor nodes a score using the evaluation function and add the scored nodes to agenda
CS 331/531 Dr M M Awais

46. Breadth - First Depth First Hill Climbing F:6 B:5 D:4 E:2 G:0 Solution
CS 331/531 Dr M M Awais

47. Best First Search Method
G:0
Solution
C:3
A:10
1. Open [A:10] : closed []
Evaluate A:10;
Open [C:3,B:5]; closed [A:10]
Evaluate C:3;
Open [B:5,F:6]; closed [C:3,A:10]
Evaluate B:5;
Open [E:2,D:4,F:6]; closed [C:3,B:5,A:10].
Evaluate E:2;
5 Open [G:0,D:4,F:6]; closed [E:2,C:3,B:5,A:10]
Evaluate G:0;
the solution / goal is reached
CS 331/531 Dr M M Awais

48. Comments: Best First Search Method
If the evaluation function is good best first search may drastically cut the amount of search requested otherwise.
The first move may not be the best one.
If the evaluation function is heavy / very expensive the benefits may be overweighed by the cost of assigning a score
CS 331/531 Dr M M Awais

49. Romania with step costs in km
hSLD=straight-line distance heuristic.
hSLD can NOT be computed from the problem description itself
In this example f(n)=h(n)
Expand node that is closest to goal
= Greedy best-first search
CS 331/531 Dr M M Awais

50. Greedy search example
Open=[Arad:366]
Arad (366)
Assume that we want to use greedy search to solve the problem of travelling from Arad to Bucharest.
The initial state=Arad
CS 331/531 Dr M M Awais

51. Greedy search example The first expansion step produces:
Open=[Sibiu:253 , Tmisoara:329 , Zerind:374]
Arad
Zerind(374)
Sibiu(253)
Timisoara
(329)
The first expansion step produces:
Sibiu, Timisoara and Zerind
Greedy best-first will select Sibiu.
CS 331/531 Dr M M Awais

52. Greedy search example If Sibiu is expanded we get:
Open=[Fagaras:176 , RV:193 , Arad:366 , Or:380]
Arad
Sibiu
Arad
(366)
Fagaras
(176)
Rimnicu Vilcea
(193)
Oradea
(380)
If Sibiu is expanded we get:
Arad, Fagaras, Oradea and Rimnicu Vilcea
Greedy best-first search will select: Fagaras
CS 331/531 Dr M M Awais

53. Greedy search example Open=[Bucharest:0 ,… ] , goal achieved
Arad
Sibiu
Fagaras
Sibiu
(253)
Bucharest
(0)
If Fagaras is expanded we get:
Sibiu and Bucharest
Goal reached !!
Yet not optimal (see Arad, Sibiu, Rimnicu Vilcea, Pitesti)
CS 331/531 Dr M M Awais

54. Greedy search, evaluation
Completeness: NO (DF-search)
Check on repeated states
Minimizing h(n) can result in false starts, e.g. Iasi to Fagaras.
CS 331/531 Dr M M Awais

55. 8-Puzzle f(N) = h(N) = number of misplaced tiles3 4 3 4 5 3 3 4 2 4 4 2 1
Total nodes expanded 16
CS 331/531 Dr M M Awais

56. 8-Puzzle f(N) = h(N) = S distances of tiles to goal Savings 25%6 4 5 3
Savings 25% 2 5 4 2 1
Total nodes expanded 12
CS 331/531 Dr M M Awais

57. Robot Navigation
CS 331/531 Dr M M Awais

58. Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal 2 1 5 8 7 3 4 6
CS 331/531 Dr M M Awais

59. Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal 8 7 6 5 4 3 2 3 4 5 6 7 5 4 3 5 6 3 2 1 1 2 4 7 7 6 5 8 7 6 5 4 3 2 3 4 5 6
Not optimal at all
CS 331/531 Dr M M Awais

60. Greedy search, evaluation
Completeness: NO (DF-search)
Time complexity?
Worst-case DF-search
(with m is maximum depth of search space)
Good heuristic can give dramatic improvement.
CS 331/531 Dr M M Awais

61. Greedy search, evaluation
Completeness: NO (DF-search)
Time complexity:
Space complexity:
Keeps all nodes in memory
CS 331/531 Dr M M Awais

62. Greedy search, evaluation
Completeness: NO (DF-search)
Time complexity:
Space complexity:
Optimality? NO
Same as DF-search
CS 331/531 Dr M M Awais

63. A* Alogorithm Problems with Best First Search
It reduces the costs to the goal but
It is not optimal nor complete
Uniform cost
CS 331/531 Dr M M Awais

64. ABDEF is it optimal / shortest pat. (NO)
Path Cost
Cost1 = 7 (2+3+2)
Cost2 = 14 ( )
Path for:
Hill Climbing: ABDEF
Best First: ABDEF
A:10
B:8 C:9
5 3
D:6 G:3
3 2
E:4 F:0 4
F:0
ABDEF is it optimal / shortest pat. (NO)
CS 331/531 Dr M M Awais

65. Path cost to node n + heuristic cost at n
A* Search
Evaluation function:
f(n) = g(n) +h(n)
Path cost to node n + heuristic cost at n
Constraints:
h(n) <= h*(n) (Admissible: Studied earlier)
g(n) >= g*(n) (Coverage)
CS 331/531 Dr M M Awais

66. Coverage: g(n) >= g*(n)
Goal will never be reached
CS 331/531 Dr M M Awais

67. Observations h g Remarks
h* Immediate convergence, A* converges to goal (No Search)
0 0 Random Search
0 1 Breath - First Search
>=h* No Convergence
<=h* Admissible Search
<=g* No Convergence
CS 331/531 Dr M M Awais

68. Example of A* A:10 Path: (P1): Best First/Hill Climbing 2 2
B:8 C:9
5 3
D:6 G:3 3 E: 4
F:0
Path: (P1): Best First/Hill Climbing
ABDEF: Cost P1 = 14 (not optimal)
For A* algorithm
F(A)=0+10=10,
F(B)=2+8=10, f(C) = 2+9=11, Expand B
F(D)=(2+5)+6=13, f(C)=11, Expand C
F(G)=(2+3)+3=8, f(D)=13, Expand G
F(f)=(2+3+2)+0=7, GOAL achieved
Path ACGF: Cost P2=7 (Optimal)
Path Admissibility
Cost P2 < Cost P1 hence P2 is admissible Path
CS 331/531 Dr M M Awais

69. Explanation A:10 2 2 B:8 C:9 5 3 D:6 G:3 3 E:4 2 4 F:0
B:8 C:9
5 3
D:6 G:3 3 E: 4
F:0
For A* algorithm Path (P2)
Now lets start from A, should we
move from A to B or A to C.
Lets check the path cost SAME,
So lets see the total cost
(fb=hb+gb=8+2=10)
(fc=hc+gb=9+2=11), hence moving through
B is better
Next move to D, total path cost to D is 2+5 = 7, and heuristic cost is 6, total is 7+6=13.
On the other side If you move through C to G, then the path cost is 2+3=5, and heuristic cost is 3, total = 3+5=8, which is much better than moving through state D. So now we choose to change path and move through G
CS 331/531 Dr M M Awais

70. Hence moving through G is much better will give the optimal path.
Explanation
A:10
B:8 C:9
5 3
D:6 G:3 3 E: 4
F:0
Now from G we move to the Goal node F
Total Path cost via G is 2+3+2=7
And Total Path cost via D is =14
Hence moving through G is much better will give the optimal path.
CS 331/531 Dr M M Awais

71. For A*never throw away unexpanded nodes:
Always compare paths through expanded and unexpanded nodes
Avoid expanding paths that are already expensive
CS 331/531 Dr M M Awais

72. A* search Best-known form of best-first search.
Idea: avoid expanding paths that are already expensive.
Evaluation function f(n)=g(n) + h(n)
g(n) the cost (so far) to reach the node.
h(n) estimated cost to get from the node to the goal.
f(n) estimated total cost of path through n to goal.
CS 331/531 Dr M M Awais

73. A* search A* search uses an admissible heuristic
A heuristic is admissible if it never overestimates the cost to reach the goal
Are optimistic
Formally:
1. h(n) <= h*(n) where h*(n) is the true cost from n
2. h(n) >= 0 so h(G)=0 for any goal G.
e.g. hSLD(n) never overestimates the actual road distance
CS 331/531 Dr M M Awais

74. Romania example
CS 331/531 Dr M M Awais

75. A* search example Starting at Arad
f(Arad) = c(Arad,Arad)+h(Arad)=0+366=366
CS 331/531 Dr M M Awais

76. A* search example Expand Arad and determine f(n) for each node
f(Sibiu)=c(Arad,Sibiu)+h(Sibiu)= =393
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)= =447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=449
Best choice is Sibiu
CS 331/531 Dr M M Awais

77. A* search example Expand Sibiu and determine f(n) for each node
Previous Paths
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)= = 447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374= 449
Expand Sibiu and determine f(n) for each node
f(Arad)=c(Sibiu,Arad)+h(Arad)= = 646
f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)= = 415
f(Oradea)=c(Sibiu,Oradea)+h(Oradea)= = 671
f(Rimnicu Vilcea)=c(Sibiu,Rimnicu Vilcea)+
h(Rimnicu Vilcea)= = 413
Best choice is Rimnicu Vilcea
CS 331/531 Dr M M Awais

78. A* search example
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)= = 447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374= 449
f(Arad)=c(Sibiu,Arad)+h(Arad)= = 646
f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)= = 415
f(Oradea)=c(Sibiu,Oradea)+h(Oradea)= = 671
Expand Rimnicu Vilcea and determine f(n) for each node
f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)= = 526
f(Pitesti)=c(Rimnicu Vilcea, Pitesti)+h(Pitesti)= = 417
f(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)= = 553
Best choice is Fagaras
CS 331/531 Dr M M Awais

79. A* search example
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)= = 447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374= 449
f(Arad)=c(Sibiu,Arad)+h(Arad)= =
f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)= = 415
f(Oradea)=c(Sibiu,Oradea)+h(Oradea)= = 671
Expand Rimnicu Vilcea and determine f(n) for each node
f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)= = 526
f(Pitesti)=c(Rimnicu Vilcea, Pitesti)+h(Pitesti)= = 417
f(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)= = 553
Expand Fagaras and determine f(n) for each node
f(Sibiu)=c(Fagaras, Sibiu)+h(Sibiu)= = 591
f(Bucharest)=c(Fagaras,Bucharest)+h(Bucharest)=450+0= 450
Best choice is Pitesti !!!
CS 331/531 Dr M M Awais

80. A* search example Note values along optimal path !!
Expand Pitesti and determine f(n) for each node
f(Bucharest)=c(Pitesti,Bucharest)+h(Bucharest)=418+0=418
Best choice is Bucharest !!!
Optimal solution (only if h(n) is admissable)
Note values along optimal path !!
CS 331/531 Dr M M Awais

81. Optimality of A*(standard proof)
Suppose suboptimal goal G2 in the queue.
Let n be an unexpanded node on a shortest to optimal goal G.
f(G2 ) = g(G2 ) since h(G2 )=0
> g(G) since G2 is suboptimal
>= f(n) since h is admissible
Since f(G2) > f(n), A* will never select G2 for expansion
CS 331/531 Dr M M Awais

82. BUT … graph search Discards new paths to repeated state. Solution:
Previous proof breaks down
Solution:
Add extra bookkeeping i.e. remove more expsive of two paths.
Ensure that optimal path to any repeated state is always first followed.
Extra requirement on h(n): consistency (monotonicity)
CS 331/531 Dr M M Awais

83. Consistency A heuristic is consistent if If h is consistent, we have
i.e. f(n) is non decreasing along any path.
CS 331/531 Dr M M Awais

84. Optimality of A*(more usefull)
A* expands nodes in order of increasing f value
Contours can be drawn in state space
Uniform-cost search adds circles.
F-contours are gradually
Added:
1) nodes with f(n)<C*
2) Some nodes on the goal
Contour (f(n)=C*).
Contour I has all
Nodes with f=fi, where
fi < fi+1.
CS 331/531 Dr M M Awais

85. A* search, evaluation Completeness: YES
Since bands of increasing f are added
Unless there are infinitely many nodes with f<f(G)
CS 331/531 Dr M M Awais

86. A* search, evaluation Completeness: YES Time complexity:
Number of nodes expanded is still exponential in the length of the solution.
CS 331/531 Dr M M Awais

87. A* search, evaluation Completeness: YES
Time complexity: (exponential with path length)
Space complexity:
It keeps all generated nodes in memory
Hence space is the major problem not time
CS 331/531 Dr M M Awais

88. A* search, evaluation Completeness: YES
Time complexity: (exponential with path length)
Space complexity:(all nodes are stored)
Optimality: YES
Cannot expand fi+1 until fi is finished.
A* expands all nodes with f(n)< C*
A* expands some nodes with f(n)=C*
A* expands no nodes with f(n)>C*
Also optimally efficient (not including ties)
CS 331/531 Dr M M Awais