2. Goodness of heuristics
If a heuristic is perfect, search work is proportional to solution length
S = O(b*d) b is average branching, d depth of solution
If h1 and h2 are two heuristics, and h1 < 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.
3. 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
4.
5. Graph Search
%% Original version
function GRAPH-SEARCH(problem,fringe) returns a solution, or
failure
closed <- an empty set
fringe <- INSERT(MAKE-NODE(INITIAL-STATE[problem]),fringe)
loop do
if EMPTY?(fringe) then return failure
node <- REMOVE-FIRST(fringe)
if GOAL-TEST[problem](STATE[node]) then return SOLUTION(node)
if STATE[node] is not in closed then
add STATE[node] to closed
fringe <- INSERT-ALL(EXPAND(node,problem),fringe)
6. 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 <- an empty set // P1
CLOSED <- an empty set // P2
OPEN <- INSERT(MAKE-NODE(INITIAL-STATE[problem]),OPEN) // P3
repeat
if EMPTY?(OPEN) then return failure //P4
best <- the lowest f-valued node on OPEN // P5
remove best from OPEN // P6
if GOAL-TEST[problem](STATE[best])
then return SOLUTION(best) // P7
for all successors M of best
if STATE[M] is not in CLOSED then // P8
OPEN <-INSERT(successor,OPEN) // P9
add STATE[best] to CLOSED // P10
7. Heuristic Best First Search
(A*)Java Pseudocode
// Instantiating OPEN, CLOSED // Move selected node from OPEN to n // P6
OPEN = new Vector<Node>(); // P1 n = OPEN.elementAt(lowIndex);
CLOSED = new Vector<Node>(); // P2 OPEN.removeElement(n);
// Successful exit if n is goal node // P8
// Placing initial node on OPEN if (n.equals(goalnode)) return;
OPEN.add(0, initialnode); // P3 // Retrieve all possible successors of n
M = n.successors();
// After initial phase,we enter the main loop //Compute f-,g- and h-value for each successor
// of the A* algorithm for (int i = 0; i < M.size(); i++) {
while (true) { Node s = M.elementAt(i);
// Check if OPEN is empty s.g = n.g + s.cost;
if (OPEN.size() == 0) { // P4 s.h = s.estimate(goalnode);
System.out.println("Failure :"); s.f = s.g + s.h;
return; // Augmenting OPEN with suitable nodes from M
// Locate next node on OPEN with heuristic for (int i = 0; i < M.size(); i++)
lowIndex = 0; // P5 // Insert node into OPEN if not on CLOSED// P9
low = OPEN.elementAt(0).f; if (!(on CLOSED))
for (int i = 0; i < OPEN.size(); i++) { OPEN.add(0, M.elementAt(i));
number = OPEN.elementAt(i).f; // Insert n into CLOSED
if (number < low) { CLOSED.add(0, n); // P10
lowIndex = i;
low = number;
8. AStar
Java Code
See exercise 7
http://www.idi.ntnu.no/emner/tdt4136/PRO/Astar.java
9. Example
Mouse 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.
10. 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 Vector<State> successors() {…}
public int estimate(State goal) {
MouseKingState goalstate = (MouseKingState)goal;
int[] goalarray = goalstate.value;
int dx = Math.abs(goalarray[0] - value[0]);
int dy = Math.abs(goalarray[1] - value[1]);
return Math.max(dx, dy);
}
} // End class MouseKingState
12. 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 (1,5) (5,5) (1,5) (5,5)
_______________ _______________
(1,1),8,0,8 | | |5 | | | | | |5 | | |
(2,2),8,1,7 | | 4| |6 | | | | 4| |6 | |
(2,3),8,2,6 | | 3| |7 | | | | 3| |7 | |
| | 2| |8 | | | | 2| |8 | |
(2,4),8,3,5
| 1| | | |9 | | 1| | | |9 |
(3,5),8,4,4 ---------------- ----------------
(4,4),8,5,3 (1,1) (5,1) (1,1) (5,1)
(4,3),8,6,2
Order of expansion Solution Path
(4,2),8,7,1
(5,1),8,8,0
13. 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) <= 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) <= cost(n.n’) + h(n’)
cost(n,n’)
h(n’)
goal
h(n)
14. 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
15. 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))
• α*h1(n) + β*h2(n) where α+β=1
16. Monotone heuristic repair
Suppose a heuristic h is admissible but not consistent, i.e.
h(n) > c(n,n’)+h(n’), which means
f(n) > 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’))
17. 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)
8 | | | | | | |
|*|
7 | | | | | | |
| |
6 | | | | | | |
| |
5 | | | | | | |
| |
4 | | | | | | |
| |
3 | | | | | | |
| |
2 | | | | | | |
| |
1 |*| | | | | |
| |
A B C D E F G H
18. 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.
20. 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) = c1*x1(n) + c2*x2(n)
21. 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| | | | | | | | Problem: move a piece from S to G using ChessKing
| | | | | | | | | heuristics h1(x) = # horisontal/vertical/diagonal moves.
| | | | | | | | | h1(S)=7 h1(n) = 4
| | | |n| | | | |
| | | | | | | | | 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))
| | | | | | | | G|
h2(n) = g(n)/(h1(s)-h1(n)) * h1(n)
= 6/(7 – 4) *4 = 8 (correct)
22. 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| | | | | | | | Problem: move a piece from S to G using Manhattan
| | | | | | | | | heuristics h1(x) = # horisontal/vertical moves
| | | | | | | | | h1(S)=14 h1(n) = 8
| | | |n| | | | |
| | | | | | | | | 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))
| | | | | | | | G|
h2(n) = g(n)/(h1(s)-h1(n)) * h1(n)
= 3/(14 – 8) *8 = 4 (correct)
23. 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
24. Bus scenario search space
Search space space(2) X time(1)
Time
Bus 3
wait
Space Bus 5
Space
25. Heuristics for bus route planner
which route is best ?
T3 = Z N # bus transfers
K3 A wait time 1. departure
Z wait time before arrival
T2 T sum transfer waiting time
K2 K sum driving time
Equivalent
transfer
T1
K1
T0 = A
26. Planner discussion
1. (T1+T2) = 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
a) Find trasees independent of time
b) Eliminate equivalent transfers
c) For each trasee, find best route plan
d) Keep the best of these solutions
