1. Introducing Underestimates
OPTIMAL Search
When cost of TRAVERSING the path should be minimized (even at expense of more complicated SEARCHING) :
Uniform Cost
Branch and Bound
Introducing Underestimates
Path Deletion A*

2. Re-introduce the costs of arcs in the NETB E C F G S 3 4 5 2 S A D B E C F G 3 2 4 5

3. Uniform cost algorithm = uniformed best-first3 4 5 2 7 8 9 6 10 11 12 13
At each step, select the node with the lowest accumul-ated cost.

4. Uniform cost algorithm:
1. QUEUE <-- path only containing the root;
2. WHILE QUEUE is not empty
AND goal is not reached
DO remove the first path from the QUEUE;
create new paths (to all children);
reject the new paths with loops;
add the new paths and sort the entire QUEUE;
3. IF goal reached
THEN success;
ELSE failure;
(BY ACCUMULATED COST)

5. Problem: NOT always optimal!C 1 5 B 1 2 D 5 10
Uniform cost returns the path with cost 102, while there is a path with cost 25. G
100 E 5 15 F 5 20 5

6. The Branch-and-Bound principle
Use any (complete) search method to find a path.
Remove all partial paths that have an accumulated cost larger or equal than the found path.
Continue search for the next path.
Iterate.
3.5 S B D C G A 5 E 6
First
goal
reached 2 3
0.5 G X
ignore

7. A weak integration of branch and bound in uniform cost:
100 B 5 S A C 1 2
102
Change the termination condition:
only terminate when a path to a goal node HAS BECOME THE BEST PATH. F 5 D E 10 15 20 25

8. Optimal Uniform cost version:
1. QUEUE <-- path only containing the root;
2. WHILE QUEUE is not empty
AND first path does not reach goal
remove the first path from the QUEUE;
create new paths (to all children);
reject the new paths with loops;
add the new paths and sort the entire QUEUE;
3. IF goal reached
THEN success;
ELSE failure;
(by accumulated cost)

9. Example: 3 4 4 7 8 7 8 9 6 7 8 9 11 10 A D S S A D A B D S A B D D A EF 11 10

10. S A D B 8 E 9 F 11 10 B C E 11 12 S A D B E 9 F 11 10 C 12 D E 10 S A D B E F 11 10 C 12 A B 13 S A D B E F 11 10 C 12 13 E B F 15 14

11. S A D B E F 11 C 12 13 15 14 F G 13
Don’t stop yet! S A D B E F C 11 12 13 15 14 G B A C 15 S A D B E F C 13 15 14 G E F D 14 16
STOP!

12. Properties of extended uniform cost :
Optimal path:
If there exists a number  > 0, such that every arc has cost  , and if the branching factor is finite,
Then extended uniform cost finds the optimal path (if one exists).
Memory and speed:
In the worst case, at least as bad as for breadth-first:
needs additional sorting step after each path- expansion !
How to improve ?

13. Extension with heuristic estimates:
Replace the ‘accumulated cost’ in the ‘extended uniform cost’ algorithm by a function:
f(path) = cost(path) + h(endpoint_path)
where:
cost(path) = the accumulated cost of the partial path
h(T) = a heuristic estimate of the cost remaining
from T to a goal node
f(path) = an estimate of the cost of a path extending
the current path to reach a goal. +

14. Example: Reconsider the straight-line distance:B C F 4 3 2 5
Example: Reconsider the straight-line distance:
h(T) = the straight-line distance from T to G D E G S A B C F 4
6.7
10.4 11
8.9
6.9 3

15. A D
= 13.4
= 12.9 S S A D
13.4 D A E
= 19.4
= 12.9 S A D E
13.4
19.4 E B F
= 13.0
= 17.7 S A D E B F
13.4
19.4
17.7
STOP! F G
= 13.0

16. Estimate-extended Uniform cost algorithm:
1. QUEUE <-- path only containing the root;
2. WHILE QUEUE is not empty
AND first path does not reach goal
remove the first path from the QUEUE;
create new paths (to all children);
reject the new paths with loops;
add the new paths and sort the entire QUEUE;
3. IF goal reached
THEN success;
ELSE failure;
(by f = cost + h)

17. Optimality
With the same condition on the arcs-costs and on the branching factor:
IF for all T: h(T) is an UNDERestimate of the remaining cost to a goal node
THEN estimate-extended uniform cost is optimal.
Intuition:
includes
underestimated
remaining cost S A D E B F
13.4
19.4
17.7 G 13 G
with real remaining
cost, this path must
at least be 13.4

18. More on underestimates:
Example:
Real
remaining
costs 3 2 1 S A D F B E H C G I 1 3 2 1 5 4
Over-
estimate
If h is NOT an underestimate: S A D F
1+3
1+5
1+4 B
2+2 A B C
3+1 C G 4
Not the optimal path!

19. More on underestimates:
Example: 3 2 S A D F B E H C G I 1
Real
remaining
costs 1 2 3
Under-
estimate
If h is an underestimate: S A D F
1+1
1+2
1+3 A B
2+0 C
3+0 B G 4 C
2+1 E
3 ! D E
Bad underestimates always get cor-
rected by increasing accumulated cost

20. Speed and memory
In the worst case: no improvement over ‘branch and bounded extended uniform cost’
Just take h = 0 everywhere.
For good heuristic functions: search may expand much less nodes !
See our running example.
BUT: the cost of computing such functions may be high
Trade-off

21. An orthogonal extension: path deletion
Discard redundant paths:
in the ‘branch and bound extended uniform cost’ : S A D 4 3 A B D 7 8
Accumulated distance X
discard !
Principle:
the minimum distance from S to G via I = (min. dist. from S to I) + (min. dist. from I to G)

22. More precisely: 7 8 9 6 Q P X IF the QUEUE contains: THEN S A D B E
a path P terminating in I, with cost cost_P
a path Q containing I, with cost cost_Q
cost_P  cost_Q
THEN
delete P

23. A D 3 4 S S A D 4 B 7 8 X S A D B 7 E 9 6 X S A D B 7 E F 11 10 X

24. S A D B E F 10 C 11 12 X S A D B E F C 11 G 13 X
Note how this optimization reduces the number of expansions VERY much, compared to ‘branch and bound extended uniform cost’.
5 expansions less !

25. A* search
IS:
Branch and bound extended,
Heuristic Underestimate extended,
Redundant path deletion extended,
Uniform Cost Search.
Note that redundant path deletion is based on the accumulated costs only, so that there is no problem combining it with heuristic underestimates.

26. A* algorithm: 1. QUEUE <-- path only containing the root;
2. WHILE QUEUE is not empty
AND first path does not reach goal
remove the first path from the QUEUE;
create new paths (to all children);
reject the new paths with loops;
add the new paths and sort the entire QUEUE;
IF QUEUE contains path P terminating in I, with cost cost_P, and path Q containing I, with cost cost_Q AND cost_P  cost_Q
THEN delete P
3. IF goal reached THEN success;
ELSE failure;
(by f = cost + h)

27. STOP! A D S S A D E X Only difference is here. X X 3 + 10.4 = 13.4
= 12.9 S S A D E
13.4
= 19.4
= 12.9 X
Only difference is here. S A D E B F
13.4
= 17.7
= 13.0 X S A D E B F
13.4
17.7 G
= 13.0
STOP! X