Heuristics

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Heuristics
Some further elaborations of the art of
heuristics and examples.
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Goodness of heuristics
If a heuristic is perfect, search work is
proportional to solution length S O(bd)
b is average branching, d depth of
solution If h1 and h2 are two heuristics, and h1
lt h2 everywhere, then A(h2) will expand no more
nodes than A(h1) If an heuristic never
overestimates more than N of the least cost,
then the found solution is not more than N over
the optimal solution. h()0 is an admissible
trivial heuristic, worst of all In theory, we
can always make a perfect heuristics, if we
perform a full breadth first search from each
node, but that would be pointless.
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Example of good heuristics for the 8 puzzle
Transparency shows the heuristic f(n)g(n)
h(n) h(n) misplaced tiles On an A search
for the 8 - puzzle
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Graph Search
Original version function GRAPH-SEARCH(problem,
fringe) returns a solution, or failure closed lt-
an empty set fringe lt- INSERT(MAKE-NODE(INITIAL-S
TATEproblem),fringe) loop do if
EMPTY?(fringe) then return failure node lt-
REMOVE-FIRST(fringe) if GOAL-TESTproblem(STA
TEnode) then return SOLUTION(node) if
STATEnode is not in closed then add
STATEnode to closed fringe lt-
INSERT-ALL(EXPAND(node,problem),fringe)
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Heuristic Best First Search
f problem is the heuristic selection
function of the problem function
BEST-FIRST-SEARCH(problem)returns a solution,or
failure OPEN lt- an empty set //
P1 CLOSED lt- an empty set // P2 OPEN
lt- INSERT(MAKE-NODE(INITIAL-STATEproblem),OPEN)
// P3 repeat if EMPTY?(OPEN) then return
failure //P4 best lt- the
lowest f-valued node on OPEN // P5 remove
best from OPEN // P6 if GOAL-TESTproblem(ST
ATEbest) then return
SOLUTION(best) // P7 for all successors M of
best if STATEM is not in CLOSED then //
P8 OPEN lt-INSERT(successor,OPEN) // P9
add STATEbest to CLOSED // P10
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Heuristic Best First Search(A)Java Pseudocode
// Instantiating OPEN, CLOSED OPEN new
VectorltNodegt() // P1 CLOSED new
VectorltNodegt() // P2 // Placing
initial node on OPEN OPEN.add(0,
initialnode) // P3 // After
initial phase,we enter the main loop // of the
A algorithm while (true) // Check
if OPEN is empty if (OPEN.size() 0)
// P4 System.out.println("Failure ")
return // Locate next node on OPEN with
heuristic lowIndex 0
// P5 low OPEN.elementAt(0).f
for (int i 0 i lt OPEN.size() i)
number OPEN.elementAt(i).f if
(number lt low) lowIndex i low
number
// Move selected node from OPEN to n // P6 n
OPEN.elementAt(lowIndex) OPEN.removeElement(n
) // Successful exit if n is goal node // P8
if (n.equals(goalnode)) return //
Retrieve all possible successors of n M
n.successors() //Compute f-,g- and h-value for
each successor for (int i 0 i lt
M.size() i) Node s M.elementAt(i)
s.g n.g s.cost s.h s.estimate(goalnode)
s.f s.g s.h // Augmenting OPEN with
suitable nodes from M for (int i 0 i lt
M.size() i) // Insert node into OPEN if not
on CLOSED// P9 if (!(on CLOSED)) OPEN.add(0,
M.elementAt(i)) // Insert n into CLOSED
CLOSED.add(0, n) // P10
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AStarJava Code
See exercise 7 http//www.idi.ntnu.no/emner/tdt
4136/PRO/Astar.java
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ExampleMouse King Problem
(1,5) (5,5) ___________
X X
X M X C -----------
(1,1) (5,1) There is a 5X5
board. At (1,1) there is a mouse M which can move
as a king on a chess board. The target is a
cheese C at (5,1). There is however a barrier
XXXX at (3,1-4) which the mouse cannot
go through, but the mouse heuristics is to ignore
this.
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Heuristics for Mouse King
public class MouseKingState extends State
public int value public MouseKingState(int
v) value v public boolean equals(State
state) public String toString()
public VectorltStategt successors() public
int estimate(State goal) MouseKingState
goalstate (MouseKingState)goal int
goalarray goalstate.value int dx
Math.abs(goalarray0 - value0) int
dy Math.abs(goalarray1 - value1) return
Math.max(dx, dy) // End class
MouseKingState
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Behaviour of Mouse King
The mouse will expand the following nodes
1,1 2,1 2,2 2,3 1,2 1,3 2,4 1,4 2,5 3,5
4,4 4,3 4,2 5,1 Solution
Path start,f,g,h (1,1),4,0,4 (2,2),4,1,3 (2,3),5,
2,3 (2,4),6,3,3 (3,5),8,4,4 (4,4),8,5,3 (4,3),8,6,
2 (4,2),8,7,1 (5,1),8,8,0
(1,5) (5,5) _______________
910 8 7 11
6 4 12 5 3 13 1 2
14 ---------------- (1,1)
(5,1) Order of expansion
(1,5) (5,5) _______________
5 4 6
3 7 2 8 1
9 ---------------- (1,1)
(5,1) Solution Path
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Perfect Heuristics Behaviour
If the heuristics had been perfect, the
expansion of the nodes would have been equal to
the solution path. 1,1 2,2 2,3 2,4 3,5
4,4 4,2 4,2 5,1 This means that a
perfect heuristics encodes all the relevant
knowledge of the problem space. Solution
Path start,f,g,h (1,1),8,0,8 (2,2),8,1,7 (2,3),8,
2,6 (2,4),8,3,5 (3,5),8,4,4 (4,4),8,5,3 (4,3),8,6,
2 (4,2),8,7,1 (5,1),8,8,0
(1,5) (5,5)
_______________ 5 4
6 3 7 2 8
1 9 ----------------
(1,1)
(5,1) Order of expansion
(1,5) (5,5)
_______________ 5 4
6 3 7 2 8
1 9 ----------------
(1,1)
(5,1) Solution Path
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Monotone (consistent) heuristics
A heuristic is monotone if the f-value is
non-decreasing along any path from start to
goal. This is fulfilled if for every pair of
nodes n ? n f(n) lt f(n) g(n)
h(n) g(n) cost(n,n)
h(n) f(n) g(n) h(n) Which gives the
triangle inequality h(n) lt cost(n.n)
h(n)
cost(n,n)
h(n)
goal
h(n)
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Properties of monotone heuristics

1. All monotone heuristics are admissible
2. Monotone heuristic is admissible at all nodes (if
   h(G)0)
3. If a node is expanded using a monotone heuristic,
   A has found the optimal route to that node
4. Therefore, there is no need to consider a node if
   it is already found.
5. If this is assumed, and the heuristic is
   monotone, the algorithm is still admissible
6. If the monotone assumption is not true, we risk
   a non optimal solution
7. However, most heuristics are monotone

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Some more notes on heuristics

• If h1(n) and h2(n) are admissible heuristics,
  then the following are also admissible heuristics
• max(h1(n),h2(n))
• ah1(n) ßh2(n) where aß1

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Monotone heuristic repair
Suppose a heuristic h is admissible but not
consistent, i.e. h(n) gt c(n,n)h(n),
which means f(n) gt f(n) (f is not
monotone) In that case, f(n) can be set to
f(n) as a better heuristic, (higher but still
underestimate), i.e. use h(n)
max(h(n),h(n)-c(n,n))
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An example of a heuristics
Consider a knight (horse) on a chess board. It
can move 2 squares in one direction and 1 to
either side. The task is to get from one square
to another in fewest possible steps (e.g. A1 to
H8). A proposed heuristics could be
ManHattanDistance/2 Is it admissible ? Is
it monotone ? (Actually, it is not
straightforward to find an heuristics that is
both admissible and not monotone)
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5
4
3
2
1
A B C D E F G H
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Relaxed problems
Many heuristics can be found by using a relaxed
(easier, simpler) model of the problem. By
definition, heuristics derived from relaxed
models are underestimates of the cost of the
original problem For example, straight line
distance presumes that we can move in straight
lines. For the 8-puzzle, the heuristics W(n)
misplaced tiles would be exact if we could
move tiles freely The less relaxed, (and
therefore better) heuristic P(n) ?
distance from home ( Manhattan distance) would
allow tiles to be moved to an adjacent square
even though there may already be a tile there.
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Generalized A

• f(n) ?g(n) ?h(n)
• ?1 A
• 0 Uniform cost
• 0 Greedy search
• lt 0, ? 0 Depth first
• gt ? Conservative (still
  admissible)
• lt ? Radical (not
  admissible)
• , ? depending on g,h Dynamic

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Learning heuristics from experience
Where do heuristics come from ? Heuristics can
be learned as a computed (linear?) combination of
features of the state. Example 8-puzzle
Features x1(n) number of misplaced
tiles x2(n) number of adjacent tiles
that are also adjacent in the goal
state Procedure make a run of searches from
100 random start states. Let h(ni) be the found
minimal cost. n1
h(n1) x1(n1) x2(n1)
.. .
n100 h(n100) x1(n100)
x2(n100) From these, use regression to
estimate h(n) c1x1(n)
c2x2(n)
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Learning heuristics from experience (II)
Suppose the problem is harder than the heuristic
(h1) indicates but that the hardness is assumed
to be uniform over the state space. Then , it is
an estimate to let an improved heuristics h2(x)
?h1(x).
S


n



G

Problem move a piece from S to G using ChessKing
heuristics h1(x) horisontal/vertical/diagonal
moves. h1(S)7 h1(n) 4 Assume problem
is actual harder (in effect Manhattan disance
h2(x), but we dont know that). It means
g(n) 6 We then estimate ? g(n)/(h1(s)-h1(n))
h2(n) g(n)/(h1(s)-h1(n)) h1(n)
6/(7 4) 4 8 (correct)
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Learning heuristics from experience (III)
Suppose the problem is easier than the heuristic
(h1) indicates but that the easiness is assumed
to be uniform over the state space. Then , it is
an estimate to let an improved heuristics h2(x)
?h1(x).
S


n



G

Problem move a piece from S to G using
Manhattan heuristics h1(x) horisontal/vertical
moves h1(S)14 h1(n) 8 Assume problem
is actual easier (in effect Chess King disance
h2(x), but we dont know that). It means
g(n) 3 We then estimate ? g(n)/(h1(s)-h1(n))
h2(n) g(n)/(h1(s)-h1(n)) h1(n)
3/(14 8) 8 4 (correct)
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Practical example(Bus world scenario)
Find an optimal route from one place to another
by public transport Nodes Bus passing events
A bus route passes a station at a time
Actions - enter bus -
leave bus - wait
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Bus scenario search space
Search space space(2) X time(1)
Time
Bus 3
wait
Space
Bus 5
Space
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Heuristics for bus route plannerwhich route is
best ?
T3 Z
N bus transfers A wait time 1. departure Z
wait time before arrival T sum transfer waiting
time K sum driving time
K3
T2
K2
Equivalent transfer
T1
K1
T0 A
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Planner discussion

1. (T1T2) T critical if rain
2. If Z is to be minimised, must search backwards
3. Many equivalent transfers (same T and K)
4. In practice, A is problematic
5. Waiting time A maybe unimportant

• Solution Relaxation
• Find trasees independent of time
• Eliminate equivalent transfers
• For each trasee, find best route plan
• Keep the best of these solutions