Chapter 4 State-Space Planning

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1 Leture slides for Automted Plnning: Theory nd Prtie Chpter 4 Stte-Spe Plnning Dn S. Nu CMSC 722, AI Plnning University of Mrylnd, Spring

2 Motivtion Nerly ll plnning proedures re serh proedures Different plnning proedures hve different serh spes Two exmples: Stte-spe plnning Eh node represents stte of the world» A pln is pth through the spe Pln-spe plnning Eh node is set of prtilly-instntited opertors, plus some onstrints» Impose more nd more onstrints, until we get pln 2

3 Outline Stte-spe plnning Forwrd serh Bkwrd serh Lifting STRIPS Blok-stking 3

4 Forwrd Serh tke 3 tke 2 move r1 4

5 Forwrd-serh is sound Properties for ny pln returned y ny of its nondeterministi tres, this pln is gurnteed to e solution Forwrd-serh lso is omplete if solution exists then t lest one of Forwrd-serh s nondeterministi tres will return solution. 5

6 Deterministi Implementtions Some deterministi implementtions of forwrd serh: redth-first serh depth-first serh est-first serh (e.g., A*) greedy serh Bredth-first nd est-first serh re sound nd omplete But they usully ren t prtil euse they require too muh memory Memory requirement is exponentil in the length of the solution In prtie, more likely to use depth-first serh or greedy serh s 0 Worst-se memory requirement is liner in the length of the solution In generl, sound ut not omplete» But lssil plnning hs only finitely mny sttes 1 2 3» Thus, n mke depth-first serh omplete y doing loop-heking s 1 s 2 s s 4 s 5 s g 6

7 Brnhing Ftor of Forwrd Serh initil stte 3 gol Forwrd serh n hve very lrge rnhing ftor E.g., mny pplile tions tht don t progress towrd gol Why this is d: Deterministi implementtions n wste time trying lots of irrelevnt tions Need good heuristi funtion nd/or pruning proedure See Setion 4.5 (Domin-Speifi Stte-Spe Plnning) nd Prt III (Heuristis nd Control Strtegies) 7

8 Bkwrd Serh For forwrd serh, we strted t the initil stte nd omputed stte trnsitions new stte = γ(s,) For kwrd serh, we strt t the gol nd ompute inverse stte trnsitions new set of sugols = γ 1 (g,) To define γ -1 (g,), must first define relevne: An tion is relevnt for gol g if» mkes t lest one of g s literls true g effets()» does not mke ny of g s literls flse g + effets () = nd g effets + () = 8

9 Inverse Stte Trnsitions If is relevnt for g, then γ 1 (g,) = (g effets()) preond() Otherwise γ 1 (g,) is undefined Exmple: suppose tht g = {on(1,2), on(2,3)} = stk(1,2) Wht is γ 1 (g,)? 9

10 g 1 1 g 4 4 s 0 g 5 5 g 2 2 g 0 3 g 3 10

11 Effiieny of Bkwrd Serh initil stte Bkwrd serh n lso hve very lrge rnhing ftor E.g., n opertor o tht is relevnt for g my hve mny ground instnes 1, 2,, n suh tht eh i s input stte might e unrehle from the initil stte As efore, deterministi implementtions n wste lots of time trying ll of them 1 gol 11

12 Lifting foo(x,y) preond: p(x,y) effets: q(x) p( 1, 1 ) p( 1, 2 ) foo( 1, 1 ) foo( 1, 2 ) p( 1, 3 ) foo( 1, 3 ) q( 1 )... p(1, 50 ) foo( 1, 50 ) Cn redue the rnhing ftor of kwrd serh if we prtilly instntite the opertors this is lled lifting p( 1,y) foo( 1,y) q( 1 ) 12

13 Lifted Bkwrd Serh More omplited thn Bkwrd-serh Hve to keep trk of wht sustitutions were performed But it hs muh smller rnhing ftor 13

14 The Serh Spe is Still Too Lrge Lifted-kwrd-serh genertes smller serh spe thn Bkwrd-serh, ut it still n e quite lrge Suppose tions,, nd re independent, tion d must preede ll of them, nd there s no pth from s 0 to d s input stte We ll try ll possile orderings of,, nd efore relizing there is no solution More out this in Chpter 5 (Pln-Spe Plnning) d d s 0 d d gol d d 14

15 Pruning the Serh Spe I ll sy lot out this lter, in Prt III of the ook For now, just two exmples: STRIPS Blok stking 15

16 π the empty pln STRIPS do modified kwrd serh from g insted of γ -1 (s,), eh new set of sugols is just preond() whenever you find n tion tht s exeutle in the urrent stte, then go forwrd on the urrent serh pth s fr s possile, exeuting tions nd ppending them to π repet until ll gols re stisfied π = 6, 4 s = γ(γ(s 0, 6 ), 4 ) g 6 stisfied in s 0 g 4 g5 g urrent serh pth g 1 g 2 g g 16

17 Quik Review of Bloks World unstk(x,y) Pre: on(x,y), ler(x), hndempty Eff: ~on(x,y), ~ler(x), ~hndempty, holding(x), ler(y) stk(x,y) Pre: holding(x), ler(y) Eff: ~holding(x), ~ler(y), on(x,y), ler(x), hndempty pikup(x) Pre: ontle(x), ler(x), hndempty Eff: ~ontle(x), ~ler(x), ~hndempty, holding(x) putdown(x) Pre: holding(x) Eff: ~holding(x), ontle(x), ler(?x), hndempty 17

18 The Sussmn Anomly Initil stte gol On this prolem, STRIPS n t produe n irredundnt solution Try it nd see 18

19 The Register Assignment Prolem Stte-vrile formultion: Initil stte: Gol: Opertor: {vlue(r1)=3, vlue(r2)=5, vlue(r3)=0} {vlue(r1)=5, vlue(r2)=3} ssign(r,v,r',v') preond: vlue(r)=v, vlue(r')=v' effets: vlue(r)=v' STRIPS nnot solve this prolem t ll 19

20 How to Hndle Prolems like These? Severl wys: Do something other thn stte-spe serh» e.g., Chpters 5 8 Use forwrd or kwrd stte-spe serh, with dominspeifi knowledge to prune the serh spe» Cn solve oth prolems quite esily this wy» Exmple: lok stking using forwrd serh 20

21 Domin-Speifi Knowledge A loks-world plnning prolem P = (O,s 0,g) is solvle if s 0 nd g stisfy some simple onsisteny onditions» g should not mention ny loks not mentioned in s 0» lok nnot e on two other loks t one» et. Cn hek these in time O(n log n) If P is solvle, n esily onstrut solution of length O(2m), where m is the numer of loks Move ll loks to the tle, then uild up stks from the ottom» Cn do this in time O(n) With dditionl domin-speifi knowledge n do even etter 21

22 Additionl Domin-Speifi Knowledge A lok x needs to e moved if ny of the following is true: s ontins ontle(x) nd g ontins on(x,y) - see elow s ontins on(x,y) nd g ontins ontle(x) - see d elow s ontins on(x,y) nd g ontins on(x,z) for some y z» see elow s ontins on(x,y) nd y needs to e moved - see e elow e d initil stte d gol 22

23 Domin-Speifi Algorithm loop if there is ler lok x suh tht x needs to e moved nd x n e moved to ple where it won t need to e moved then move x to tht ple else if there is ler lok x suh tht x needs to e moved then move x to the tle else if the gol is stisfied then return the pln else return filure d repet e d initil stte gol 23

24 Esily Solves the Sussmn Anomly loop if there is ler lok x suh tht x needs to e moved nd x n e moved to ple where it won t need to e moved then move x to tht ple else if there is ler lok x suh tht x needs to e moved then move x to the tle else if the gol is stisfied then return the pln else return filure repet initil stte gol 24

25 The lok-stking lgorithm: Properties Sound, omplete, gurnteed to terminte Runs in time O(n 3 )» Cn e modified to run in time O(n) Often finds optiml (shortest) solutions But sometimes only ner-optiml (Exerise 4.22 in the ook)» Rell tht PLAN LENGTH for the loks world is NPomplete 25

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