Introduction to Artificial Intelligence Class 1 Planning & Search Henry Kautz Winter 2007

Outline of Course Heuristic search Constraint satisfaction Automated planning Propositional logic First-order logic and logic programming Knowledge engineering Probabilistic reasoning: directed and undirected graphical models Learning graphical models Decision trees and ensemble methods Neural networks

3 Planning Input Description of initial state of world Description of goal state(s) Description of available actions –Optional: Cost for each action Output Sequence of actions that converts the initial state into a goal state May wish to minimize length or cost of plan

Classical Planning Atomic time Deterministic actions Complete knowledge No numeric reward (just goals) Only planner changes world

Route Planning State = intersection Operators = block between intersections Operator cost = length of block

Blocks World Control a robot arm that can pick up and stack blocks. Arm can hold exactly one block Blocks can either be on the table, or on top of exactly one other block State = configuration of blocks { (on-table G), (on B G), (clear B), (holding R) } Operator = pick up or put down a block (put-down R)put on table (stack R B)put on another block

State Space pick-up(R) pick-up(G) stack(R,B) put-down(R) stack(G,R) Planning = Finding (shortest) paths in state graph

STRIPS Representation (define (domain prodigy-bw) (:requirements :strips) (:predicates (on ?x ?y) (on-table ?x) (clear ?x) (arm-empty) (holding ?x))

Problem Instance (define (problem bw-sussman) (:domain prodigy-bw) (:objects A B C) (:init (on-table a) (on-table b) (on c a) (clear b) (clear c) (arm-empty)) (:goal (and (on a b) (on b c)))) goal may be a partial description

Operator Schemas (:action stack :parameters (?obj ?under_obj) :precondition (and (holding ?obj) (clear ?under_obj)) :effect (and (not (holding ?obj)) (not (clear ?under_obj)) (clear ?obj) (arm-empty) (on ?obj ?under_obj))) delete effects – make false add effects – make true

Blocks World Blackbox Planner Demo

Search Algorithms Today: Space-State Search Depth-First Breadth-First Best-First A* Next Class: Local Search Constraint Satisfaction

A General Search Algorithm Search( Start, Goal_test, Criteria ) Open = { Start }; Closed = { }; repeat if (empty(Open)) return fail; select Node from Open using Criteria; if (Goal_test(Node)) return Node; for each Child of node do if (Child not in Closed) Open = Open U { Child }; Closed = Closed U { Node }; Closed list optional

Breadth-First Search Search( Start, Goal_test, Criteria ) Open: fifo_queue; Closed: hash_table; enqueue(Start, Open); repeat if (empty(Open)) return fail; Node = dequeue(Open); if (Goal_test(Node)) return Node; for each Child of node do if (not find(Child, Closed)) enqueue(Child, Open) insert(Child, Closed) Criteria = shortest distance from Start

Depth-First Search Search( Start, Goal_test, Criteria ) Open: stack; Closed: hash_table; push(Start, Open); repeat if (empty(Open)) return fail; Node = pop(Open); if (Goal_test(Node)) return Node; for each Child of node do if (not find(Child, Closed)) push(Child, Open) insert(Child, Closed) Criteria = longest distance from Start

Best-First Search Search( Start, Goal_test, Criteria ) Open: priority_queue; Closed: hash_table; enqueue(Start, Open, heuristic(Start)); repeat if (empty(Open)) return fail; Node = dequeue(Open); if (Goal_test(Node)) return Node; for each Child of node do if (not find(Child, Closed)) enqueue(Child, Open, heuristic(Child)) insert(Child, Closed) Criteria = shortest heuristic estimate of distance to goal

Properties Depth First Simple implementation (stack) Might not terminate Might find non-optimal solution Breadth First Always terminates if solution exists Finds optimal solutions Visits many nodes Best First Always terminates if heuristic is “reasonable” Visits many fewer nodes May find non-optimal solution

Best-First with Manhattan Distance (  x+  y) Heuristic 52 nd St 51 st St 50 th St 10 th Ave 9 th Ave 8 th Ave 7 th Ave6 th Ave5 th Ave4 th Ave 3 rd Ave 2 nd Ave S G 53 nd St

Non-Optimality of Best-First 52 nd St 51 st St 50 th St 10 th Ave 9 th Ave 8 th Ave 7 th Ave6 th Ave5 th Ave4 th Ave 3 rd Ave 2 nd Ave SG 53 nd St

A* Criteria: minimize (distance from start) + (estimated distance to goal) Implementation: priority queue f(n) = g(n) + h(n) f(n) = priority of a node g(n) = true distance from start h(n) = heuristic distance to goal

Optimality of A* Suppose the estimated distance is always less than or equal to the true distance to the goal heuristic is a lower bound heuristic is admissible Then: when the goal is removed from the priority queue, we are guaranteed to have found a shortest path!

Maze Runner Demo

Observations on A* Perfect heuristic: If h(n) = h*(n) (true distance) for all n, then only the nodes on the optimal solution path will be expanded. Null heuristic: If h(n) = 0 for all n, then this is an admissible heuristic and A* acts like breath-first search. Comparing heuristics: If h1(n)  h2(n)  h*(n) for all non-goal nodes, then h2 is as least as good a heuristic as h1 Every node expanded by A* using h2 is also expanded by A* using h1 if h1(n)<h1(n) for some n, then h2 is stronger than h1 Combining heuristics: if h1(n) and h2(n) are admissible, then h3(n) = MAX(h1(n),h2(n)) is admissible Why?

Search Heuristics “Optimistic guess” at distance to a solution Some heuristics are domain specific Manhattan distance for grid-like graphs Euclidean distance for general road maps Rubik’s Cube –Admissible, but weak: # cubits out of place / 8 –Better: MAX( Sum( Manhattan distance edge cubits )/4, Sum( Manhattan distance corner cubits )/4 )

Planning Heuristics A useful non-admissible heuristic for planning is the number of goals that need to be achieved Why not admissible? Good admissible heuristics for planning can be created by relaxing the operators, e.g.: Eliminate preconditions, or Eliminate negative preconditions & effects Use the length of the solution to the relaxed problem as a heuristic for the length of the solution to the original problem

Homework Shakey the robot has to bring coffee to Prof. Kautz. In order to make the coffee, Shakey will need to gather coffee filters, coffee, and Prof. Kautz's mug and bring them to the coffee maker in the kitchen. The coffee and filters are in the supply room, but it is locked. To unlock the supply room, Shakey will need to get the key from Prof. Kautz's office. Represent this problem in STRIPS notation What is the true value of the start state? What is the heuristic value of the start start, based on rhe relaxed problem with no preconditions on actions?