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Artificial Intelligence - Informed search algorithms

Artificial Intelligence - Informed search algorithms

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Informed search algorithms Chapter Sections 1-5
Informed= use problem-specific knowledge Which search strategies? —Best-first search and its variants Heuristic functions? —How to invent them Local search and optimization —Hill climbing, local beam search, genetic algorithms,... Local search in continuous spaces Online search agents
function TREE-SEARCH(problem,fringe) return a solution or failure fringe INSERT(MAKE-NODE(INITIAL-STATE problem]), fringe) loop do if EMPTY?(Fringe) then return failure node +- REMOVE-FIRST(fringe) if GOAL-TEST[pool/em] applied to STATE[node] succeeds then return SOLUTION(node) fringe INSERT-ALL(EXPAND(node, problem), fringe) A strategy is defined by picking the order of node expansion
General approach of informed search: - Best-first search • node is selected for expansion based on an evaluation function f(n) Idea: evaluation function measures distance to the goal. — Choose node which appears best Implementation: fringe is queue sorted in decreasing order of desirability. - Special cases: greedy search, A” search
—l›‹iil estimated cost of the cheapest path from node n to goal node. —If n is goal then li(i››=J — More information later
”, ••= .g ili ’é! * **** *" •° hS L D ——straight- line distance heuristic. hS‹D can NOT be computed from the problem description itself In this example f(n) — —h(n) §;;g “ ”-“ ,° ’ • - Expand node that is closest to goal = Greedy best-first search
Arad (366) Assume that we want to use greedy search to solve the problem of travelling from Arad to Bucharest. The initial state=Arad
Sibiu(253) Arad Timisoara (329) The first expansion step produces: - Sibiu, Timisoara and Zerind Greedy best-first will select Sibiu. Zerind(374)
Arad (366) Sibiu (380) (193) Fagaras (176) Arad Ora ea l i e u Vilcea If Sibiu is expanded we get: — Arad, Fagaras, Oradea and Rimnicu Vilcea Greedy best-firstsearch will select: Fagaras
Sibiu (253) Sibiu Fagaras ucharest (0) Arad If Fagaras is expanded we get: Sibiu and Bucharest Goal reached ! ! Yet not optimal (see Arad, Sibiu, Rim nicu Vil cea, Pitesti
Completeness: NO (e.g. DF-search) - Check on repeated states - Minimizing h(n) can result in false starts, e.g. lasi to Fagaras.
Completeness: NO (e.g. DF-search) Time complexity? - Worst-case DF-search @b fH) (with m is maximum depth of search space) —Good heuristic can give dramatic improvement.
Completeness: NO (e.g. DF- search) Time complexity: O(b”) Space complexity: o(bM —Keeps all nodes in memory
Time complexity: Completeness: NO (e.g. DF-search) O(b‘} Optimality? NO — Same as DF- search W) Space complexity: ( b
Best-known form of best-first search. Idea: avoid expanding paths that are already expensive. Evaluation function f(n) — —g(n) -i- h(n) —g(n) the cost (so far) to reach the node. —h(n) estimated cost to get from the node to the goal. —f(n) estimated total cost of path through n to goal.
A* search uses an admissible heuristic — A heuristic is admissible if it never overestimates the cost to reach the goal —Are optimistic Formally: t. h(n} < = h”(n) where is the true cost from n 2. h(n) > — 0 so h(G) — —0 for any goal G. e.g. hgtofn) never overestimates the actual road distance
U0 Tim I90e æ î0 J
360=0-›366 Find Buchareststarting at Arad I(Arad) - c(??,Arad) + h(Arad) -0+366= 366
4+z- 11s-›oze 440=75+27 4 Expand Arrad and determine f(n) for each node — f(5ibiu)=c(Arad,5ibiu)+h(5ibiu)=140+253=393 —f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329=447 —f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=449 Best choice is Sibiu
e46 @+300 415 3B+17B B71=æ1+2a0 413 1+1e3 Expand Sibiu and determine /¢nÿ for each node — f(Arad)=c(Sibiu,Arad)+h(Arad)=280+366=646 — f(Fagaras)-c(Sibiu,Fagaras)+h(Fagaras)=239+179—415 — f(Oradea) —c(Sibiu,Oradea)+h(Oradea) —291+380—671 — f(Rimnicu Vilcea)=c(Sibiu,Rimnicu Vìlcea)+ h(Rimnicu ViIcea)—220+ 192— 413 Best choice is Rimnicu Vilcea
+No 418 1?6 a71=&1+9W Expand Rimnicu Vilcea and determine f(n) for each node — f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)=360+ 160=526 - I(Pitesti)=c(Rimnicu Vilcea, Pitesti)-i- h(Pitesti)=317+100=417 - I(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)=300+253=553 Best choice is Fagaras
52'6W0+160 417-G17+10€i 553=3 €I+253 Expand Fagaras and determine f(n) for each node —f(Sibiu)=c(Fagaras, Sibiu)+h(Sibiu)—338+253= 591 —f(Bucharest) -—c(Fagaras, Bucharest) +h(Bucharest) -—450+0-— 450 Best choice is Pitesti !!!
Expand Pitesti and determine f(n) for each node f(Bucharest) ——c(Pitesti,Bucharest)+h(Bucharest) — —41 8+ 0 — —4 1 8 Best choice is Bucharest ! !! - Optimal solution (onIy if h(n) is admissable} Note values along optimal path !!
Suppose suboptimalgoal G2 in the queue. Let n be an unexpanded node on a shortest to optimal goal G. I(Gy) since h¿G — —0 — — g(Gy) > g(G) >= f(n) since Cl is suboptimal since h is admissible Since f(G,) > f(n), A* will never select G for expansion
A heuristic is consistent if (^ If h is consistent, we have ——g(n)+ c{n,a,n') + h{n') ?: g(n) + h(n) i.e. f(n) is nondecreasing along any path.
A” expands nodes in order of increasing I value Contours can be drawn in state space — Uniform -cost sea rch adds circ les. F-contours are g radually Aadea : 1) n Od es with I(n) < C 2) Some nodes on the goal ¿ Contour (I{n} — —C”’ ). Contour I nas a l l Nodes with f=f, vvnere
Completeness:YES —Since bands of increasing 7 are added —Unless there are infinitly many nodes with
Completeness: YESTime complexity: Number of nodes expanded is still exponential in the length of the solution.
Completeness: YES Time complexity: (exponential with path length) Space complexity: —It keeps all generated nodes in memory —Hence space is the major problem not time
Completeness: YES Time complexity: (exponential with path length) Space complexity:(all nodes are stored) Optimality: YES — Cannot expand f,+t until I, is finished. - A* expands all nodes with /°{n/ < C* — A ” expands some nodes with f(n) — —C ” - A* expands no nodes with f{n/ >C* Also optimally efficient (not including ties)
Some solutions to A” space problems (maintain completeness and optimality) —Iterative-deepening A* (IDA*) —Here cutoff information is the I-cost (g+h) instead of depth —Recursive best-first search(RBFS) —Recursive algorithm that attempts to mimic standard best-first search with linear space. — (simple)Memory-bounded A* ((S)MA*) — Drop the worst-leaf node when memory is full
8 function RECURSIVE-BEST-FIRST-SEARCH problem) return a solution or failure return RFBS(problem,MAKE-NODE(INITIAL-STATE problem]), ) function RFBS( prOblem, mode, /° /irnit) return a solution or failure and a new f- cost limit if GOAL-TEST(problem)(STATE[node]) then return node successors + — EXPAND(node, problem) if successors is empty then return failure, for each s in successors dn max(g(s) + h(s), f (node) j f (s) repeat best the lowest /-value node in successors if I [best) > I limit then return failure, I [test] alternative the second lowest I-value among successors RBFS¿prob/em, best, min(£ limit, alternative)) result, I best) if result failure then return result
Recursive best-first search Keeps track of the f-value of the best-alternative path available. — If current f-values exceeds this alternative f- value than backtrack to alternative path. - Upon backtracking change f-value to best f- value of its children. — Re-expansion of this result is thus
Path until Rumnicu Vilcea is already expanded Above node; /°-limit for every recursive call is shown on top. Below node: I(n) The path is followed until Pitesti which has a /°-value worse than the f-limit.
Ib ) un•*edmg toSJü° . ’ 1 41s 417 Unwind recursion and store best /-value for current best leaf Pitesti result, f (best) +- RBF5¿prOö/em, best, min(/°_//mit, alternative)) liest •is nowFagaras. Call RBFS for new best — best value is now 450
Unwind recursion and store best /°-value for current best leaf Fagaras result, I (best] +- RBFS¿prod/em, best, min(ñ_/imit, alternative)) best is now Rimnicu Viclea (again). Call RBFS for new best — Subtree is again expanded. Best alternative subtree is now through Timisoara. Solution is found since because 447 > 417.
RBFS is a bit more efficientthan IDA” —Still excessive node generation (mind changes) Like A”, optimal if h(n) is admissible Space complexity is 0(bd). —IDA” retains onIy one single number (the current f-cost limit) Time complexity difficultto characterize —Depends on accuracy if h(n) and how often best path changes. IDA” en RBFS suffer from too little memory.
E.g for the 8-puzzle knows two commonly used heuristics h — — the number of misplaced tiles — hp(s)=8 h — — the sum of the distances of the tiles from their goal positions (manhattan distance). — h2(*)= 3 -i-1--2 -i-2 -i-2 --3 -i-3 -i-2= 18
E.g for the 8-puzzle Avg. solution cost is about 22 steps (branching factor +/- 3) Exhaustive search to depth 22: 3.1 x 10‘0 states. A good heuristic function can reduce the search process.
Effective branching factor b* —Is the branching factor that a uniform tree of depth 4 would have in order to contain //+ 4 nodes. N + 1 = l + b * +(b*)2 + ...+ {b*)d —Measure is fairly constant for sufficiently hard problems. —Can thus provide a good guide to the heuristic's overall useful ness. —A good value of b” is 1
4 1 İ2 IO 47I27 I2 364d035 I4 - 16. 93' 539 J301 113 2IT ‹67f 1219 2.87 279 ?:J8 I.38 I:42 1,45 1.46 1:47 I.28 If hymn) >= h1(n) for all n (both admissible) then h2 dominates h1 and is better for search
Admissible heuristics can be derived from the exact solution cost of a relaxed version of the problem: a — Relaxed 8-puzzle for h1 tile can move anywhere As a result, h (n) gives the shortest solution — Relaxed B-puzzle for h a tile can move to any adjacent squa re. As a result, h (n) gives the shortest solution. The optimal solution cost of a relaxed problem is no greater than the optimal solution cost of the real problem. ABSo1veZ fOUnd a usefull heuristicfor the rubic cube.
Admissible heuristics can also be derived from the solution cost of a subproblemof a given problem. This cost is a lower bound on the cost of the real problem. Pattern databases store the exact solution to for every possible subproblem instance. — The complete heuristic is constructed using the patterns in the DB
Another way to find an admissible heuristic is through learning from experience: —Experience solving lots of 8-puzzles — An inductive learning algorithm can be used to predict costs for other states that arise during search.