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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
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Finding Heuristics
Tic - Tac - Toe
which one to choose?
X
X
X
Heuristic:
calculate winning lines and move to
state with most wining lines.
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Calculating winning lines
X
X
X
3 winning
lines
4 winning
lines
3 winning
lines
Always choose the state with maximum heuristic value
Maximizing heuristics
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Heuristic: Choosing city with minimum distance:
Always choose the city with minimum heuristic value
Minimizing heuristics (hi<hi-1)
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
Finding 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 ?
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8 Puzzle
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

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)=3+1+2+2+2+3+3+2=18
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Heuristics: Quality

Admissibility of Heuristics


Effective Branching factor


Heuristic function should never overestimate
the actual cost to the goal
Heuristic that has lower effective branching
factor is better
More Informed Heuristics

Heuristic that has a higher value is more
informed compared to the others
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Algorithms for Heuristic Search
Heuristic Search
Hill Climbing
Best first search
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A* Algo
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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.
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Navigation Problem
Choose the closest city to travel
10
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Navigation Problem
Choose the closest city to travel
hi < hi-1
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GET STUCK
No Backtracking
Navigation Problem
Choose the closest city to travel
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Hill Climbing
A:10
B:5
D:4
C:3
E:2
G:0
Goal/
Solution
F:6
Node label: heuristic value of the node
A:10
A is node name,
10 is the heuristic evaluation
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Hill Climbing
A:10
Compare B with C
B:5
D:4
C:3
E:2
G:0
Goal/
Solution
F:6
Node label: heuristic value of the node
A:10
A is node name,
10 is the heuristic evaluation
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Hill Climbing
A:10
C is better so move
B:5
D:4
C:3
E:2
G:0
Goal/
Solution
F:6
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
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Hill-climbing search

“is a loop that continuously moves in the
direction of increasing value (for maximizing
heuristic), or in the direction of decreasing value
(for minimizing heuristic)”




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
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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
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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).
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Hill-climbing example
a)
b)
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).
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Drawback
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


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.
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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.
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Reading Assignment on Simulated Annealing
Study the methods to get rid of the local minima
problem especially simulated annealing
You can consult the Text Books
(no need to turn in any report)
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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
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Summary: Best-first search

General approach:


Idea: evaluation function measures distance
to the goal.


Best-first search: node is selected for expansion
based on an evaluation function f(n)
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
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Evaluation function



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
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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
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Breadth - First
A:10
B:5
D:4
E:2
C:3
Depth First
F:6
Hill Climbing
G:0
Solution
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A:10
Best First Search Method
1.
Open [A:10] : closed []
Evaluate A:10;
2.
B:5
C:3
Open [C:3,B:5]; closed [A:10]
Evaluate C:3;
3.
Open [B:5,F:6]; closed [C:3,A:10]
Evaluate B:5;
4.
D:4
E:2
F:6
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]
G:0
Evaluate G:0;
Solution
the solution / goal is reached
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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
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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
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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
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Greedy search example
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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.
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Greedy search example
Open=[Fagaras:176 , RV:193 , Arad:366 , Or:380]
Arad
Sibiu
Arad
(366)

Rimnicu Vilcea
(193)
If Sibiu is expanded we get:


Fagaras Oradea
(176) (380)
Arad, Fagaras, Oradea and Rimnicu Vilcea
Greedy best-first search will select: Fagaras
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Greedy search example
Open=[Bucharest:0 ,… ] , goal achieved
Arad
Sibiu
Fagaras
Sibiu
(253)

If Fagaras is expanded we get:


Bucharest
(0)
Sibiu and Bucharest
Goal reached !!

Yet not optimal (see Arad, Sibiu, Rimnicu Vilcea, Pitesti)
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Greedy search, evaluation

Completeness: NO (DF-search)


Check on repeated states
With Oradea as GOAL, and start state lasi, what
would be the path
Lasi to Neamt to lasi and so on …
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8-Puzzle
f(N) = h(N) = number of misplaced tiles
3
5
3
4
3
4
2
2
4
3
1
3
4
5
4
0
Total nodes expanded 16
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8-Puzzle
6
f(N) = h(N) = S distances of tiles to goal
5
Savings 25%
2
2
4
1
3
5
4
6
5
0
Total nodes expanded 12
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Robot Navigation
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Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal
8
7
7
6
5
4
5
4
3
3
2
6
7
6
8
7
3
2
3
4
5
6
5
1
0
1
2
4
5
6
5
4
3
2
3
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Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal
8
7
7
6
5
4
5
4
3
3
2
6
77
6
8
7
3
2
3
4
5
6
5
1
00
1
2
4
5
6
5
4
3
2
3
4
5
6
Not optimal at all
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Greedy search, evaluation


Completeness: NO (DF-search)
m
O(b
)
Time complexity?
Worst-case DF-search
(with m is maximum depth of search space)
 Good heuristic
 can give dramatic
improvement.

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Greedy search, evaluation



Completeness: NO (DF-search)
m
O(b
)
Time complexity:
Space complexity: O(b m )

Keeps all nodes in memory


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Greedy search, evaluation




Completeness: NO (DF-search)
m
O(b
)
Time complexity:
Space complexity: O(b m )
Optimality? NO


Same as DF-search

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Robot Navigation Other Examples
N
yN
yg
xN
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Robot Navigation Other Examples
N
yN
yg
h1 (N) =
(xN -xg ) +(yN -yg )
2
2
h2(N) = |xN-xg| + |yN-yg|
xg
xN
(Euclidean distance)
(Manhattan distance)
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A* Alogorithm

Problems with Best First Search

It reduces the costs to the goal but


It is not optimal nor complete
Uniform cost
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A:10
Path Cost
2
2
Cost1 = 7 (2+3+2)
B:8
C:9
Cost2 = 14 (2+5+3+4)
5
3
Path for:
D:6
G:3
Hill Climbing: ABDEF
3
2
Best First: ABDEF
E:4
F:0
4
F:0
ABDEF
is it optimal / shortest pat. (NO)
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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)
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Coverage: g(n) >=
g*(n)
g(n)
Goal will
never be
reached
g*(n)
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Observations
h
g
h*
Remarks
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
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Example of A*
A:10
Path: (P1): Best First/Hill Climbing
2
ABDEF: Cost P1 = 14 (not optimal)
2
For A* algorithm
B:8
C:9
F(A)=0+10=10,
5
3
F(B)=2+8=10, f(C) = 2+9=11, Expand B D:6
F(D)=(2+5)+6=13, f(C)=11, Expand C 3
F(G)=(2+3)+3=8, f(D)=13, Expand G
G:3
E:4
F(f)=(2+3+2)+0=7, GOAL achieved
2
4
Path ACGF: Cost P2=7 (Optimal)
F:0
Path Admissibility
Cost P2 < Cost P1
hence P2 is admissible Path
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Explanation
A:10
For A* algorithm Path (P2)
Now lets start from A, should we
2
2
move from A to B or A to C.
B:8
C:9
Lets check the path cost SAME,
5
3
D:6
G:3
So lets see the total cost
(fb=hb+gb=8+2=10)
(fc=hc+gb=9+2=11), hence moving through
3
B is better
E:4
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
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F:0
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Explanation
A:10
2
2
B:8
C:9
Total Path cost via G is 2+3+2=7
5
3
And Total Path cost via D is 2+5+3+4=14
D:6
G:3
Now from G we move to the Goal node F
Hence moving through G is much better
will give the optimal path.
3
E:4
2
4
F:0
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For A*never throw away
unexpanded nodes:
Always compare paths through
expanded and unexpanded nodes
Avoid expanding paths that are already
expensive
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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.
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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
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Romania example
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A* search example

Starting at Arad

f(Arad) = c(Arad,Arad)+h(Arad)=0+366=366
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A* search example

Expand Arad and determine f(n) for each node




f(Sibiu)=c(Arad,Sibiu)+h(Sibiu)=140+253=393
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329=447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=449
Best choice is Sibiu
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A* search example
search

Previous Paths


f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329= 447

f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=
Expand Sibiu and determine f(n) for each node





449
f(Arad)=c(Sibiu,Arad)+h(Arad)=280+366=
f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)=239+179=
f(Oradea)=c(Sibiu,Oradea)+h(Oradea)=291+380=
f(Rimnicu Vilcea)=c(Sibiu,Rimnicu Vilcea)+
h(Rimnicu Vilcea)=220+192=
Best choice is Rimnicu Vilcea
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415
671
413
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A* search example






447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=
f(Arad)=c(Sibiu,Arad)+h(Arad)=280+366=
f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)=239+179=
f(Oradea)=c(Sibiu,Oradea)+h(Oradea)=291+380=
449
646
415
671
Expand Rimnicu Vilcea and determine f(n) for each node




f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329=
f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)=360+160=
f(Pitesti)=c(Rimnicu Vilcea, Pitesti)+h(Pitesti)=317+100=
f(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)=300+253=
Best choice is Fagaras
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A* search example
f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329=
447
f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=

f(Arad)=c(Sibiu,Arad)+h(Arad)=280+366=

f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)=239+179=

f(Oradea)=c(Sibiu,Oradea)+h(Oradea)=291+380=
Expand Rimnicu Vilcea and determine f(n) for each node

f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)=360+160=

f(Pitesti)=c(Rimnicu Vilcea, Pitesti)+h(Pitesti)=317+100=

f(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)=300+253=
Expand Fagaras and determine f(n) for each node

f(Sibiu)=c(Fagaras, Sibiu)+h(Sibiu)=338+253=

f(Bucharest)=c(Fagaras,Bucharest)+h(Bucharest)=450+0=
Best choice is Pitesti !!!
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646
415
671





526
417
553
591
450
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A* search example

Expand Pitesti and determine f(n) for each node


Best choice is Bucharest !!!


f(Bucharest)=c(Pitesti,Bucharest)+h(Bucharest)=418+0=418
Optimal solution (only if h(n) is admissable)
Note values along optimal path !!
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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 )
> g(G)
>= f(n)
Since f(G2) > f(n), A* will
since h(G2 )=0
since G2 is suboptimal
since h is admissible
never select G2 for expansion
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BUT … graph search

Discards new paths to repeated state.


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)
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Consistency

A heuristic is consistent if

If h is consistent, we have
h(n)  c(n,a,n')  h(n')
f (n')  g(n')  h(n')
 g(n)  c(n,a,n')  h(n')

 g(n)  h(n)
 f (n)
i.e. f(n) is non decreasing along any path.

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Optimality of A*(more useful)


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.
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A* search, evaluation

Completeness: YES


Since bands of increasing f are added
Unless there are infinitely many nodes with
f<f(G)
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A* search, evaluation


Completeness: YES
Time complexity:

Number of nodes expanded is still
exponential in the length of the solution.
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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
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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)
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Quality of Heuristics


Admissibility
Effective Branching factor
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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
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Navigation Problem
Shows Actual Road Distances
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Heuristic: Straight Line Distance between cities and
goal city (aerial)
Actual Road Distances
Is the new heuristic Admissible
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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)=80+97+101=278 (actual cost through Piesti)
h*(sibiu)=99+211=310 (actual cost through Fagaras)
hsld<=h* Admissible (never overestimates the actual road distance)
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
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)=3+1+2+2+2+3+3+2=18



BOTH
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


New Heuristic: Permutation Inversions
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
1
3
4
5
6
7
8
13 14 15
4
5 10 7
8
6
2
9 10 11 12
3
9
2
1
11 12
13 14 15
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 h3 = 7 + 4
IS h3 admissible
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New Heuristic: Permutation Inversions
IS h3 admissible
h3 = 7 + 4
h* = actual moves
required to achieve
the goal
If h3 <= h* then
Admissible,
Otherwise Not
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
2
3
4
5
10
7
8
9
6
11
12
13
14
15
Find out yourself
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New Heuristic: Permutation Inversions
1
2
3
1
2
4
5
6
4
5
3
7
8
7
8
6
STATE(N)
STATE(Goal)
Is h3 admissible here:
h3=6+1=7
6 (block 3 and 6 requires 2 jumps each)
1(empty block is in row 1)
h*=2 (actual moves)
So h3 is not admissible
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Example
5
8
1
2
3
x
4
2
1
x
x
7
3
6
x
x
STATE(N)
GOAL STATE
Goal contains partial information, the sequence in the
first row is important.
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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.
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

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:
5


8
1
2
3
6
4
2
1
4
5
7
3
6
7
8
It ignores negative interactions among tiles
Store the solution pattern in the database
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Relaxed Problem: Example
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5
8
1
2
3
6
4
2
1
4
5
7
3
6
7
8
8
8
5
5
6
6
h = h8 + h5 + h6 +…
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
Complex: Relaxed Problem
Consider two more complex relaxed problems:
5
2
3


1
1
2
3
6
4
2
1
4
5
7
3
6
7
8
1
4
8
2
3
5
8
4
5
7
6
7
6
8
h = d1234 + d5678 [disjoint pattern heuristic]
These distances could have been pre-computed in a
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Relaxed Problem
Several order-of-magnitude speedups
for
15- and 24-puzzle
Problem have been obtained through the
application of relaxed problem
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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.
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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
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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 ?
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8 Puzzle
search


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)=3+1+2+2+2+3+3+2=18
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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.
N  1  b * (b*)2  ...  (b*)d


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.
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

Heuristic quality and dominance
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)
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