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## CS 236501 Introduction to AI

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**CS 236501Introduction to AI**Tutorial 5 Automatic Generation of Heuristic Functions**Problem Definition**• Most search algorithms need a heuristic function to guide them in the search space • Writing a good heuristic function is hard: • Extensive knowledge about the problem is needed • The distance to the goal may be hard to estimate (especially when variables have nominal values) • It is hard to measure how informative a heuristic function is • We mostly want admissible heuristics • We sometimes want a heuristic function for an unknown problem, or a problem that changes with time Intro. to AI – Tutorial 5 – By Haggai Toledano**Proposed Solution**• Automatically generate a heuristic function • Should be a problem and domain independent method • Can be state specific • Should produce informative, preferably admissible heuristics • Should be fast in practice and preferably have low theoretical complexity • Such heuristics generally use a simplified version of the original problem, the (optimal) solution to which serves as the heuristic value for the original states • Most ideas come from the field of automated planning Intro. to AI – Tutorial 5 – By Haggai Toledano**A useful formal definition of search problems**• A search problem P=<S,A,I,G> can be defined in the following terms: • S – The search space. S can be defined as a set of variables V={v1,v2,…,vn} with value domains D={d1,d2,…,dn} such that each variable vi has an associated domain di, and a state s∊S is an element of the space defined by d1 X d2 X … X dn (alternatively this can be viewed as a complete assignment for the variables, i.e. s∊S={<vi,di>}i=1..n). • I – The initial state. I is an element of S. • G – The goal state. G is a partial element of S, i.e. G⊆{<vi,di>}i=1..n. When a state s∊S contains G, we say that s is in a goal state. Intro. to AI – Tutorial 5 – By Haggai Toledano**A useful formal definition of search problems**• A – The set of allowed actions (or operators). This is a function from S to S. An action <pre⊆{<vi,di>}i=1..n,eff⊆{<vi,di>}i=1..n>∊A takes as input a state s∊S, and if it meets the preconditions pre (i.e. the variables that pre specifies have the values that pre specifies), then the effect eff takes place (i.e. the variables that eff specifies are set to the values that eff specifies for them). Variables not specified in pre are “don’t cares” and the values of variables not in eff are copied from the input state to the output state. • The set A*(I) is the feasible state space. • Note: Each action a∊A is a symbolic manipulation on the values of the variables of a state s∊S. Intro. to AI – Tutorial 5 – By Haggai Toledano**Relaxation**• We will consider problem relaxation • A relaxation P’ of problem P is achieved by deleting some constraints present in P • Other methods include: Over constraining, Analogy, and Probabilistic models • Example: P = 8-Puzzle • Goal: Order tiles • Action: Swap(x, y) whentile x is H/V-adjacent totile y, and either x is emptyor y is empty 1 2 3 4 5 6 7 8 Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – The original problem**• The optimal solution of the original (non-relaxed) problem is: 2 4 1 5 3 h*=9 7 8 6 Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – Relaxation I**• Tiles can move to any location (whether occupied by another tile or not) • This is the number of misplaced tiles heuristic 2 4 1 5 3 h=5 7 8 6 Intro. to AI – Tutorial 7 – By Haggai Toledano**8-Puzzle – Relaxation II**• Tiles can move to any adjacent location (whether occupied by another tile or not) • Summing over all tiles, this is Manhattan Distance 2 4 1 5 3 h=7 7 8 6 Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – Relaxation III**• Tiles can move to the empty location (even if it is not adjacent to them) • Optimal solution: Sort (place a misplaced target tile in the empty place; prefer the target of that place) 2 4 1 5 3 h=5 7 8 6 Intro. to AI – Tutorial 5 – By Haggai Toledano**Relaxations – Usage**• Receive a problem P=<S,A,I,G> • Create P’=<S’,A’,I’,G’>, the relaxed version of P • Run your favorite heuristic algorithm on P • For each s∊S, starting with I, get the successors succs of s by applying all applicable actions from A. • For each t∊succs compute the optimal solution of P’ from t’ to G’. The solution value (length, cost etc.) will be denoted by h(t). • Use h(succs) as the heuristic function for the search algorithm. • Since P’ is only a symbolic manipulation of P, it is problem and domain independent Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Relaxation**• A general method for creating a certain type of relaxations: • State variables accumulate values instead of merely changing their value – never forget achieved values • Simplifies the original problem – less constraints: • Allows “jumps” between previously unlinked states • In effect: Adds edges to the states transition graph • Preserves preconditions • Computation: Uninformed, complete search, e.g. BFS • Monotonic: The original actions are still applicable to the states they were applicable to beforehand • Admissible Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – h+**0 1 2 2 124X 12X 12X 2X 2X 4 4X 4 34X 4X 34X 2X 24X 24X X 24X 24X 1X 1 1X 1X 1 1 5 5 5 5 5 5 3 3 3 3X 3 3X 3 4 5 7 7 7 7 7 7 8 8 8 8 8 8 6 6 6 6 6 6 4 Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – h+**6 7 h+ = 7 h* = 9 4 124X 124X 124X 34X 34X 34X 24X 24X 24X 14X 14X 1X 5 5 5 36X 36X 36X 7 7 7 8 8 8 6X 6X 6X Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Relaxation – Other problems**• Lightz Out P P+ Goal! Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Relaxation – Other problems**• City Map • I = <Boston> • succ(I) = {<Quebec>, <New York>} • succ+(I)= {<Boston, Quebec>,<Boston, New York>} • succ+(<Boston, New York>) = <Boston, New York, Quebec>) Boston Boston Quebec New York Quebec New York Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Relaxation – Other problems**• Missionaries and Cannibals • State: <ML,CL,MR,CR,B>∊NxNxNxNx{right,left}, where N={0,1,2,3} • I = <3,3,0,0,L> ; G = <0,0,3,3> • A = {CCR = [move 2 cannibals from L to R] = <pre={CL≥2,B=left,MR≥CR+2}, eff={CL=CL-2,CR=CR+2,B=right}>, MMR, CML, … } • CCR(I=<3,3,0,0,L>) = <3,1,0,2,R> • CCR+ = <pre={CL≥2,B=left,MR≥CR+2}, eff={CL=CL U CL-2,CR=CR U CR+2,B=B U right}> • CCR+(I+=<{3},{3},{0},{0},{L}>)= <{3},{1,3},{0},{0,2},{L,R}> Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Relaxation: Complexity**• Although the +-relaxed problem seems simpler to solve, its complexity might still be too high • Proof (Bylander ’94): • Propositional planning is a type of search problem • PLANMIN(k) is the decision problem of determining whether an instance of a propositional planning problem is satisfiable in k or less actions • PLANMIN(k)+ is PLANMIN(k) restricted to positive postconditions (can only add facts, can’t delete them) • PLANMIN(k) is PSPACE complete • PLANMIN(k)+ is NP complete (reduction from 3SAT) • Solution: Approximate h+ Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Approximation I: hmax**• hmax – Find the longest path to any goal value for the relaxed problem • Build a layered planning graph: • Start from the initial state • At each layer apply all applicable actions • The first layer to hold all of the goal variables in their goal values is the approximated solution length • Properties: • Not very informative (too optimistic; no mutex detection) • Admissible Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – hmax**0 1 2 2 1245X 2X 1245X 1245X 4X 2345X 12345X 12345X 4 123458X 245X 123458X 245X X 1234578X 1 1X 1234578X 1 5X 123458X 5 123458X 123458X 1234568X 3 3 3X 1234568X 3 hmax = max(2(1), 1(2), 2(3), 3(4), 0(5), 3(6), 0(7), 0(8)) = 3 7 7X 7 7X 7 8 8X 12345678X 8 12345678X 6 6 6X 6 6X h+ = 7 h* = 9 hmax = 3 Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Approximation II: hadd**• hadd – Sum over the costs to achieve all goal variables values • Compute just like hmax but sum the costs instead of finding their maximum • Assumes total independence of goal variables • Properties: • Informative in problems where the goal variables have little interaction with one another • Not admissible – often over estimates Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – hadd**0 1 2 1245X 2 1245X 1245X 2X 4X 12345X 2345X 12345X 4 123458X 123458X 245X X 245X 1234578X 1 1234578X 1X 1 123458X 123458X 5 123458X 5X 1234568X 3 3 3X 1234568X 3 hadd = 2(1)+1(2)+2(3)+3(4)+0(5)+3(6)+0(7)+0(8) = 11 7 7X 7 7 7X 8X 8 12345678X 12345678X 8 6 6X 6 6 6X h+ = 7 h* = 9 hmax = 3 hadd = 11 Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Approximation III: hFF**• hFF – Fast Forward: The length of a non-optimal solution to the relaxed problem: • Build the planning graph • Backchain some solution (i.e. necessary actions) from the last (goal) layer to the first (initial state): • For each goal in layer i, set its preconditions as goals for layer i-1 • Select actions that were applied on layer i-1 and contain the goals from layer i as their combined postconditions • Repeat until reaching the initial graph layer • The solution’s length (i.e. the sum of the number of uniquely selected actions for each layer) is the heuristic value for the initial state Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ Approximation III: hFF**• Properties: • Takes into account positive interactions • Not admissible • Rarely over estimates • Uses the fact that while PLANMIN(k)+is NPC, PLANSAT+ (relaxed satisfiablity) is in P (so if the goal cannot be reached – h+ is unsolvable; if the goal is reached – h+ is solvable and a witness can be constructed) • Note that solving for the minimal number of actions in each layer is NPC (reduction from SC) • The original FF algorithm uses enforced hill-climbing (use BFS to escape local minima) Intro. to AI – Tutorial 5 – By Haggai Toledano**h+ and its approximations**• Problem: • Variables: x∊{x1,x2,x3}, y∊{y1,y2,y3} , z∊{z1,z2} • Actions (<prepost>): a1=<x1x2>, a2=<x2x3,y2>, a3=<x2x3>, a4=<x2y2>, a5=<y2y3>, a6=<z1z2> • Initial state: {x=x1, y=y1, z=z1} • Goal state: {x=x3, y=y3, z=z2} x2x3 x1x2 y2y3 x2y2 x2x3,y2 z1z2 hmax=max{2(x),3(y),1(z)}=3 hFF=2(a1,a6)+2(a3,a4)+1(a5)=5 hadd=sum{2(x),3(y),1(z)}=6 h+=1(a1)+1(a2)+1(a5)+1(a6)=4 = h* Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – hFF**0 1 2 1245X 2 1245X 1245X 2X 4X 12345X 2345X 12345X 4 123458X 123458X 245X 245X X 1234578X 1 1 1234578X 1X 123458X 5 123458X 5X 123458X 1234568X 3 3 3X 1234568X 3 hFF = by a certain choice of actions = 9 h+ = 7 h* = 9 hmax = 3 hadd = 11 hFF = 9 7 7 7X 7X 7 8X 8 12345678X 12345678X 8 6 6X 6 6 6X Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – hFFbackchaining**action noop bad Unique non-noop actions: 3+4+2 hFF = 9 good Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**• The Pattern Database (PDB) relaxation method maps a problem P to a relaxed problem P’ by ignoring some aspects of the states • Option I: Variables deletion: Use only a subset of the variables from P (but take care to always include a goal variable in this subset) • Option II: Values domain abstraction: Use only a subset of the domain for some variables • PDB only clusters states that share some properties (rather than arbitrary states) • We now have a set of patterns of states • The new state(+actions) space is homomorphic to the original state space Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**Use this pattern Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**• How to solve using the state patterns: • For each state, optimally solve the abstracted problem and use the solution length as the heuristic value of the original state • Construct a database of solutions for all abstracted states in advance (rather than on the fly) • This method guarantees admissibility: All of the abstracted solutions are shorter than the optimal solution of the original problem Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**h* = 3 Use this pattern h = 2 Intro. to AI – Tutorial 5 – By Haggai Toledano**8-Puzzle – PDB Relaxation**1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 2 1 2 4 4 7 7 7 8 8 8 2 9 9 9 4 1 1 5 3 6 7 7 7 8 8 8 6 6 Example action: <5(5), X(2) → 5(2)> <X(2) → Ø> =eff Ø <2(5), X(6) → 2(6)> <2(5), X(6) → 2(6)> Goal h = 6 Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**• Representation is important for understanding the patterned state space • In the last example, the representation was: “Where is tile X located”. Eliminating tiles 3 and 5 adds two extra empty places on the board • Using the representation “What tile does place X hold” deletes places, creating strange actions like “place y holds x ↔ place y holds empty”, allowing multiple instances of a tile and allowing the empty board – all of these do not impact admissibility • However “What tile does place X hold” is a more intuitive representation for values domain abstraction: We could place the empty tile on several placeholders simultaneously Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**• PDB is not necessarily surjective: The new state space often contains states that have no mapping from the original state space (may affect search speed but not admissibility) • Example: 3-Puzzle 1 1 1 2 2 2 3 3 3 4 4 4 1 2 1 3 2 1 2 This state has no source in the original state space Intro. to AI – Tutorial 5 – By Haggai Toledano**Pattern Database Relaxation**• Very important subject (which we will not discuss): How to choose which variables to delete • Always keep at least one goal variable • Possible strategy: Try to decompose the problem into as many independent subproblems as possible while retaining as many variables as possible Intro. to AI – Tutorial 5 – By Haggai Toledano**Summary**• The solution to a relaxed problem can be used as a heuristic for the original problem • It is preferable to use an admissible heuristic when possible • Some (crucial) aspects of the automated process of heuristic functions generation still need to be taken care of manually Intro. to AI – Tutorial 5 – By Haggai Toledano