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- Search plays a key role in many parts of AI. These algorithms provide the conceptual backbone of almost every approach to the systematic exploration of alternatives.
- There are four classes of search algorithms, which differ along two dimensions:
- First, is the difference between uninformed (also known as blind) search and then informed (also known as heuristic) searches.
- Informed searches have access to task-specific information that can be used to make the search process more efficient.

- The other difference is between any solution searches and optimal searches.
- Optimal searches are looking for the best possible solution while any-path searches will just settle for finding some solution.

- First, is the difference between uninformed (also known as blind) search and then informed (also known as heuristic) searches.

- Graphs are everywhere; E.g., think about road networks or airline routes or computer networks.
- In all of these cases we might be interested in finding a path through the graph that satisfies some property.
- It may be that any path will do or we may be interested in a path having the fewest "hops" or a least cost path assuming the hops are not all equivalent.

- On holiday in Romania; currently in Arad.
- Flight leaves tomorrow from Bucharest.
- Formulate goal:
- be in Bucharest

- Formulate problem:
- states: various cities
- actions: drive between cities

- Find solution:
- sequence of cities, e.g., Arad, Sibiu, Fagaras, Bucharest

- However, graphs can also be much more abstract.

- A path through such a graph (from a start node to a goal node) is a "plan of action" to achieve some desired goal state from some known starting state.
- It is this type of graph that is of more general interest in AI.

- One general approach to problem solving in AI is to reduce the problem to be solved to one of searching a graph.
- To use this approach, we must specify what are the states, the actions and the goal test.
- A state is supposed to be complete, that is, to represent all the relevant aspects of the problem to be solved.
- We are assuming that the actions are deterministic, that is, we know exactly the state after the action is performed.

- In general, we need a test for the goal, not just one specific goal state.
- So, for example, we might be interested in any city in Germany rather than specifically Frankfurt.

- Or, when proving a theorem, all we care is about knowing one fact in our current data base of facts.
- Any final set of facts that contains the desired fact is a proof.

Vacuum cleaner

A problem is defined by four items:

- initial state e.g., "at Arad"
- actions and successor functionS: = set of action-state tuples
- e.g., S(Arad) = {(goZerind, Zerind), (goTimisoara, Timisoara), (goSilbiu, Silbiu)}

- goal test, can be
- explicit, e.g., x = "at Bucharest"
- implicit, e.g., Checkmate(x)

- path cost (additive)
- e.g., sum of distances, or number of actions executed, etc.
- c(x,a,y) is the step cost, assumed to be ≥ 0

- A solution is a sequence of actions leading from the initial state to a goal state

Example: The 8-puzzle

- states?
- actions?
- goal test?
- path cost?

Example: The 8-puzzle

- states?locations of tiles
- actions?move blank left, right, up, down
- goal test?= goal state (given)
- path cost? 1 per move

Tree search algorithms

- Basic idea:
- Exploration of state space by generating successors of already-explored states (i.e. expanding states)

- A state is a (representation of) a physical configuration
- A node is a bookeeping data structure constituting of state, parent node, action, path costg(x), depth
- The Expand function creates new nodes, filling in the various fields and using the SuccessorFn of the problem to create the corresponding states.

- The collection of nodes that have been generated but not yet expanded is called fringe. (outlined in bold)
- We will implement collection of nodes as queues. The operations on a queue are as follows:
- Empty?(queue) check to see whether the queue is empty
- First(queue) returns the first element
- Remove-First(queue) returns the first element and then removes it
- Insert(element, queue) inserts an element into the queue and returns the resulting queue
- InsertAll(elements, queue) inserts a set of elements into the queue and returns the resulting queue

public class Problem {

Object initialState;

SuccessorFunction successorFunction;

GoalTest goalTest;

StepCostFunction stepCostFunction;

HeuristicFunction heuristicFunction;

…

Successor-Fn[problem](State[node])

is in fact:

problem. Successor-Fn(node.State)

or

problem.getSuccessorFunction().getSuccessors( node.getState() );

The queue policy of the fringe embodies the strategy.

- A search strategy is defined by picking the order of node expansion
- Strategies are evaluated along the following dimensions:
- completeness: does it always find a solution if one exists?
- time complexity: number of nodes generated
- space complexity: maximum number of nodes in memory
- optimality: does it always find a least-cost solution?

- Time and space complexity are measured in terms of
- b: maximum branching factor of the search tree
- d: depth of the least-cost solution
- m: maximum depth of the state space

- Uninformed do not use information relevant to the specific problem.
- Breadth-first search
- Uniform-cost search
- Depth-first search
- Depth-limited search
- Iterative deepening search

- TreeSearch(problem, FIFO-QUEUE()) results in a breadth-first search.
- The FIFO queue puts all newly generated successors at the end of the queue, which means that shallow nodes are expanded before deeper nodes.
- I.e. Pick from the fringe to expand the shallowest unexpanded node

- Expand shallowest unexpanded node
- Implementation:
- fringe is a FIFO queue, i.e., new successors go at end

- Expand shallowest unexpanded node
- Implementation:
- fringe is a FIFO queue, i.e., new successors go at end

- Expand shallowest unexpanded node
- Implementation:
- fringe is a FIFO queue, i.e., new successors go at end

- Complete?
- Yes (if b is finite)

- Time?
- 1+b+b2+b3+… +bd + b(bd-1) = O(bd+1)

- Space?
- O(bd+1) (keeps every node in memory)

- Optimal?
- Yes (if cost is a non-decreasing function of depth, e.g. when we have 1 cost per step)

Suppose b=10, 10,000 nodes/sec, 1000 bytes/node

- Expand least-cost unexpanded node.
- The algorithm expands nodes in order of increasing path cost.
- Therefore, the first goal node selected for expansion is the optimal solution.
- Implementation:
- fringe = queue ordered by path cost (priority queue)

- Equivalent to breadth-first if step costs all equal
- Complete? Yes, if step cost ≥ ε(I.e. not zero)
- Time? number of nodes with g ≤ cost of optimal solution, O(bC*/ ε) where C* is the cost of the optimal solution
- Space? Number of nodes with g ≤ cost of optimal solution, O(bC*/ ε)
- Optimal? Yes – nodes expanded in increasing order of g(n)

Try it here

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Expand deepest unexpanded node
- Implementation:
- fringe = LIFO queue, i.e., put successors at front

- Complete? No: fails in infinite-depth spaces, spaces with loops
- Modify to avoid repeated states along path
complete in finite spaces

- Modify to avoid repeated states along path
- Time?O(bm): terrible if m is much larger than d
- but if solutions are dense, may be much faster than breadth-first

- Space?O(bm), i.e., linear space!
- Optimal? No

DepthLimitedSearch (int limit)

{

stackADT fringe;

insert root into the fringe

do {

if (Empty(fringe)) return NULL; /* Failure */

nodePT = Pop(fringe);

if (GoalTest(nodePT->state))

return nodePT;

/* Expand node and insert all the successors */

if (nodePT->depth < limit) {

insert into the fringe Expand(nodePT)

} while (1);

}

IterativeDeepeningSearch ()

{

for (int depth=0; ; depth++) {

node=DepthLimitedtSearch(depth);

if ( node != NULL )

return node;

}

}

- Number of nodes generated in a depth-limited search to depth d with branching factor b:
NDLS = b0 + b1 + b2 + … + bd-2 + bd-1 + bd

- Number of nodes generated in an iterative deepening search to depth d with branching factor b:
NIDS = (d+1)b0 + d b1 + (d-1)b2 + … + 3bd-2 +2bd-1 + 1bd

- For b = 10, d = 5,
- NDLS = 1 + 10 + 100 + 1,000 + 10,000 + 100,000 = 111,111
- NIDS = 6 + 50 + 400 + 3,000 + 20,000 + 100,000 = 123,456

- Overhead = (123,456 - 111,111)/111,111 = 11%

- Complete? Yes
- Time?(d+1)b0 + d b1 + (d-1)b2 + … + bd = O(bd)
- Space?O(bd)
- Optimal? Yes, if step cost = 1

- Failure to detect repeated states can turn a linear problem into an exponential one!

- You have three jugs, measuring 12 gallons, 8 gallons, and 3 gallons, and a water faucet.
- You can fill the jugs up, or empty them out from one another or onto the ground.
- You need to measure out exactly one gallon.