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Due to the graphic nature of the following slides, user discretion is advised.

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Goal StateRepeated StateIllegal StateSearch Tree for Farmer, Wolf, Duck, CornLast time we saw

A farm hand was sent to a nearby pond to fetch 8 gallons of water. He was given two pails - one 11, the other 6 gallons. How can he measure the requested amount of water?

Sliding Tile PuzzleYou can slide any of the numbered tiles into the blank space.Can you arrange the numbers into order?Can you place 8 queens on a chessboard such that no piece is attacking another?Find a route from LAX to the Golden Gate bridge that minimizes the driving time, ...that minimizes the mileage, ...that minimizes the number of Taco Bells you must pass. Which tree shows the correct relationship between gorilla, chimp and man?When you have just 3 animals, there are only three possible trees...

Last time we saw

Problem Solving using SearchA Problem Space consists of

The current state of the world (initial state) A description of the actions we can take to transform one state of the world into another (operators). A description of the desired state of the world (goal state), this could be implicit or explicit.

A solution consists of the goal state*, or a path to the goal state.

* Problems were the path does not matter are known as constraint satisfaction problems.

Initial StateGoal StateOperatorsSlide blank square left.Slide blank square right..FWDCFWDCMove FMove F with W.Distributive property Associative property... Add a queen such that it does not attack other, previously placed queens.A 4 by 4 chessboard with 4 queens placed on it such that none are attacking each other4 Queens

Representing the statesA state space should describe Everything that is needed to solve the problem. Nothing that is not needed to solve the problem.

In general, many possible representations are possible, choosing a good representation will make solving the problem much easier.For the 8-puzzle 3 by 3 array 5, 6, 7 8, 4, BLANK 3, 1, 2

A vector of length nine 5,6,7,8,4, BLANK,3,1,2

A list of factsUpper_left = 5Upper_middle = 6 Upper_right = 7Middle_left = 8.

Choose the representation that make the operators easiest to implement.

Operators I Single atomic actions that can transform one state into another. You must specify an exhaustive list of operators, otherwise the problem may be unsolvable.

Operators consist of Precondition: Description of any conditions that must be true before using the operator. Instruction on how the operator changes the state.

In general, for any given state, not all operators are possible.

Examples:In FWDC, the operator Move_Farmer_Left is not possible if the farmer is already on the left bank.

In this 8-puzzle, The operator Move_6_down is possibleBut the operator Move_7_down is not.

Operators IIExample: For the eight puzzle we could have... Move 1 left Move 1 right Move 1 up Move 1 down Move 2 left Move 2 right Move 2 up Move 2 down Move 3 left Move 3 right Move 3 up Move 3 down Move 4 left There are often many ways to specify the operators, some will be much easier to implement... Move Blank left Move Blank right Move Blank up Move Blank downOr

A complete example: The Water Jug Problem Two jugs of capacity 4 and 3 units. It is possible to empty a jug, fill a jug, transfer the content of a jug to the other jug until the former empties or the latter fills. Task: Produce a jug with 2 units. A farm hand was sent to a nearby pond to fetch 2 gallons of water. He was given two pails - one 4, the other 3 gallons. How can he measure the requested amount of water? Abstract away unimportant details State representation (X , Y) X is the content of the 4 unit jug. Y is the content of the 3 unit jug.Initial State (0 , 0)Goal State (2 , n)Operators Fill 3-jug from faucet (a, b) (a, 3) Fill 4-jug from faucet (a, b) (4, b) Fill 4-jug from 3-jug (a, b) (a + b, 0) ...Define a state representationDefine an initial stateDefine an goal state(s)May be a description rather than explicit stateDefine all operators

Once we have defined the problem space (state representation, the initial state, the goal state and operators) are we are done?

We start with the initial state and keep using the operators to expand the parent nodes till we find a goal state.

but the search space might be large

really large

So we need some systematic way to search.

The average number of new nodes we create when expanding a new node is the (effective) branching factor b. The length of a path to a goal is the depth d.

A Generic Search Treebb2bdSo visiting every the every node in the search tree to depth d will take O(bd) time.

Not necessarily O(bd) space.Fringe (Frontier)Set of nonterminal nodes without childrenI.e nodes waiting to be expanded.

Branching factors for some problemsThe eight puzzle has a branching factor of 2.13, so a search tree at depth 20 has about 3.7 million nodes. (note that there only 181,400 different states).

Rubiks cube has a branching factor of 13.34. There are 901,083,404,981,813,616 different states. The average depth of a solution is about 18. The best time for solving the cube in an official championship was 17.04 sec, achieved by Robert Pergl in the 1983 Czechoslovakian Championship.

Chess has a branching factor of about 35, there are about 10120 states (there are about 1079 electrons in the universe).

Detecting repeated states is hard.

We are going to consider different techniques to search the problem space, we need to consider what criteria we will use to compare them. Completeness: Is the technique guaranteed to find an answer (if there is one). Optimality: Is the technique guaranteed to find the best answer (if there is more than one). (operators can have different costs) Time Complexity: How long does it take to find a solution. Space Complexity: How much memory does it take to find a solution.

General (Generic) Search Algorithm function general-search(problem, QUEUEING-FUNCTION) nodes = MAKE-QUEUE(MAKE-NODE(problem.INITIAL-STATE)) loop do if EMPTY(nodes) then return "failure" node = REMOVE-FRONT(nodes) if problem.GOAL-TEST(node.STATE) succeeds then return node nodes = QUEUEING-FUNCTION(nodes, EXPAND(node, problem.OPERATORS)) endA nice fact about this search algorithm is that we can use a single algorithm to do many kinds of search. The only difference is in how the nodes are placed in the queue.

Breadth First SearchEnqueue nodes in FIFO (first-in, first-out) order. Complete? Yes. Optimal? Yes. Time Complexity: O(bd) Space Complexity: O(bd), note that every node in the fringe is kept in the queue.Intuition: Expand all nodes at depth i before expanding nodes at depth i + 1

Uniform Cost SearchEnqueue nodes in order of cost Complete? Yes. Optimal? Yes, if path cost is nondecreasing function of depth Time Complexity: O(bd) Space Complexity: O(bd), note that every node in the fringe keep in the queue.Intuition: Expand the cheapest node. Where the cost is the path cost g(n) 252517251745Note that Breadth First search can be seen as a special case of Uniform Cost Search, where the path cost is just the depth.

Depth First SearchEnqueue nodes in LIFO (last-in, first-out) order. Complete? No (Yes on finite trees, with no loops). Optimal? No Time Complexity: O(bm), where m is the maximum depth. Space Complexity: O(bm), where m is the maximum depth.Intuition: Expand node at the deepest level (breaking ties left to right)

Depth-Limited SearchEnqueue nodes in LIFO (last-in, first-out) order. But limit depth to L Complete? Yes if there is a goal state at a depth less than L Optimal? No Time Complexity: O(bL), where L is the cutoff. Space Complexity: O(bL), where L is the cutoff.Intuition: Expand node at the deepest level, but limit depth to LL is 2 in this examplePicking the right value for L is a difficult, Suppose we chose 7 for FWDC, we will fail to find a solution...

Iterative Deepening Search IDo depth limited search starting a L = 0, keep incrementing L by 1. Complete? Yes Optimal? Yes Time Complexity: O(bd), where d is the depth of the solution. Space Complexity: O(bd), where d is the depth of the solution.Intuition: Combine the Optimality and completeness of Breadth first search, with the low space complexity of Depth first search

Iterative Deepening Search II1+10+100+1000+10,000+100,000 = 111,111

11+101+10+1001+10+100+10001+10+100+1000+10,0001+10+100+1000+10,000+100,000= 123,456Consider a problem with a branching factor of 10 and a solution at depth 5... Iterative deepening looks wasteful because we reexplore parts of the search space many times...

Bi-directional SearchIntuition: Start searching from both the initial state and the goal state, meet in the middle. Complete? Yes Optimal? Yes Time Complexity: O(bd/2), where d is the depth of the solution. Space Complexity: O(bd/2), where d is the depth of the solution.Notes Not always possible to search backwards How do we know when the trees meet? At least one search tree must be retained in memory.