CS579 - Homework 2. Tu Phan. March 10, 2004

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1 I! CS579 - Hmewk 2 Tu Phan Mach 10, Review 11 Planning Pblem and Plans The planning pblem we ae cnsideing is a 3-tuple descibed in the language whse syntax is given in the bk, whee is the initial state, is the gal and is the dmain desciptin Plans We define a plan as a sequence f sets f actins, ie, "!$#% whee ', (*),+-),0/,1, is a set f actins That is, we allw actins t be executed paallelly Tansitin Functin Given a planning pblem with the set f actins 5 and set f fluents 6, the tansitin functin 7 f the planning dmain that maps sets f actins and states int states is defined as fllws ;= if 90 >@BA0 st is nt executable in 9 "CDC$E0FG BHJIKEM MN 73 9%- (1) 39*OP"CDCEQSRT BHJIKE thewise W E Z E Z whee VU,5 is sme set f actins, is a state, "CDC$EXY[ZJ\ "CDC, and ]H^IKE_Y[ZJ\ 3 BHJI It is wthwhile extending the definitin f f a plan as well Me pecisely, the tansitin functin f plans ae defined ecusively as fllws 7 ` bac 9%d73 9% 7e` 3f4ac 9%-g7hfS i73 9%e if f N j is called the initial state We say that can be achieved fm afte executing plan if 7kQ 9 l and 7kQ satisfies 12 SAT Encdings Given a planning pblem 2 in the language with the tansitin functin given in (1), the geneal pcedue f tanslating int the cespnding ppsitinal they m with time instants is sketched ut as fllws n Each fluent pap6 cespnds t the set f ppsitins q^ s in m, whee +tu(bvv n Each actin ]A05 cespnds t the set f ppsitins qj w, whee +xm(bvveg/y1 n The encding cnsists f the fllwing sentences ŒxŠ Œ " 4Ž 4Ž ZJ\} { \} - ZJ\} { \ E%ĩ š S \}ƒ% w e% ZJ\} { \ E%ĩ e% ZJ\} { \} - { \} ~ \} (2) (3) ]AP5p e+t (ŠvwP/ 1 (4) A 6 e+x (ŠvwP/ 1 (5) A 6 e+x (ŠvwP/ 1 (6) 1

2 I 13 Cectness We will shw that the encding m fm (2)-(6) is equivalent t the iginal planning pblem Pf f the Cectness T pve the cectness f the SAT-encding, we need pve the fllwing 1 [Sundness] If m has a mdel then we can extact fm a plan f that achieves fm the initial state 2 [Cmpleteness] If has a plan f length that achieves fm the initial state then m has at least ne mdel fm which can be extacted 21 Sundness Assume that is a mdel f m et qj Q is an actin such that, ` vv!$# a and W q^ A Nte that W might be empty By (2) and (3), it is easy t see that W is the initial state f and hlds in W! Befe pving the sundness f the encding, let us pve the fllwing lemmas A emma 1 F evey intege +, (l)y+ )y, a fluent liteal I hlds in a state W if and nly if MI Pf It is bvius fm the definitin f W emma 2 F evey intege +, (l)y+ )yg/y1, and f evey actin in, is executable in W Pf et I be a fluent liteal in H Z By the definitin f, we have u On the the hand, Hence, MI As a esult, I hlds in W (by emma 1) That is, is executable in W emma 3 F evey intege +, (l)y+ )yg/y1, 'CC$E%FP ]H^IKEQMN (by (4)) Pf Assume that 'CC$ESF0 ]H^IKE 2N Thus, thee exists a fluent A 6 st XA_"CDC$E F0 ]H^IKE This implies that thee exist tw actins ^ 'A st pap'cc and pap ]H^I By (5), we have Œx is tue in And by (6), we have Œ is tue in This is a cntadictin Theefe, "CDCDE%FB ]H^IKEQ N emma 4 F evey intege +, (l)y+ )yg/y1, we have 7 W W Œ Pf Accding t emma 2 and emma 3, evey actin in is executable in W and 'CCDEtF ]H^IKE' N f all ( )+)p/u1 Theefe, accding t (1), 73 W ' cw Op"CDCEetRŠ BHJIKEb W f sme W We will pve W Œx MW by shwing that, f evey A06, (1) if A WJŒ then pa0w (2) if X A W Œ then A0W Pf f (1) Assume that A W Œ Accding t emma 1, we have Œ Hence, Thee ae tw pssibilities a) j Ž ZJ\} { \} - It is easy t see that 2c Ž e Thus, pa WJ ZJ\} Š { \} d % ZJ\} Š { \ E%ĩ b) ZJ\} { \ E%ĩ % That is, thee must exist an actin such that and A 'CC Since, we have A ' Hence, A "CDC Uu"CDC E In additin, "CDC E %F0 ]H^I E -gn (emma 4), we have _ A ]H^I E As a esult pa cwj4op"cdc E esrt ]H^I E QMW Theefe, in bth cases, we have pa0w 2

3 A A Ž { \ A W Ž { \ { \ E E E { A W Pf f (2) Assume that _ W^Œ Œ Accding t emma 1, is nt in, ie, Œ By (6), we have tw pssibilities a) ZJ\} E%ĩ In this case, we have Theefe, (by emma 1) On the the hand, since ZJ\} E%ĩ, we have ZJ\} E%ĩ e Thus, AP'CC Cnsequently, _ AXcWJ OG'CC e%rt ]H^I Q W b) ZJ\} \} - That is, thee must be exist an actin such that _A ]H^I and is tue in That is, A0 ]H^IKE and thus X AXcW OP"CDCE %RT ]H^IKEtW S, we can cnclude that if X W^Œ then _ The lemma is theefe pved Theem 1 The encding is sund Pf et 9 7 b v "3# 9 ^ f 1 ) +d) As we mentined abve, 9 MW By emma 4, using inductin, we can easily pve that W^b 94 f all (p)j+)j Hence, 7kQ 9 ^ 94!_ WJ! On the the hand, hlds W^! Thus, hlds in 73Q 94 That is, can be achieved fm 9 afte executing The encding is theefe pved t be sund 22 Cmpleteness Assume that has a plan!@#, whee, (_) +l) / 1, is a set f actins, that achieves gal fm 9 We dente 7 v # 9 by 9 Since is a plan that achieves gal, we have the fllwing ppeties (1) 9 ' (2) f evey intege, +, ([) +-) P/ 1, and f evey actin ]A0, is executable in 9 (3) 9 and ]H^IKE FG"CDCE- N f evey +, (l)y+-) (4) 9 Œ 239 OP"CDCEe RT ]H^IKE (5) f evey liteal ItAP, I hlds in 9! et m be the cespnding ppsitinal they f with time instants and f evey +, (P) +'), let dente the set f ppsitins q A09 We cnstuct an intepetatin f m as fllws j! cso qj BA0' š š!$# Fist f all, we will pve the fllwing lemmas and then use them t pve the cmpleteness f the encding emma 5 F evey intege +, (l)y+ )y, a fluent liteal I hlds in 9 iff MI Pf It is easy t see that I iff I is cntained in J, that is, I hlds in 9 emma 6 F evey intege +, (l)y+ )yg/y1, we have Pf It is bvius fm the cnstuctin f Theem 2 The encding is cmplete ]AP A0" Pf T pve that the encding is cmplete we need t pve (1) is a mdel f m and, (2) fm we can cnstuct the plan 3

4 Pving (2) is tivial We will shw (1) by checking all the ppsitinal sentences in m By emma 5, the sentence (2) and (3) ae appaently tue in F evey intege +, ( ) +d) 0/y1, and an actin GA0, since is executable in 9, evey fluent liteal ItA must hld in 9 By emma 5, we have I Thus, the sentence (4) is tue in Fm emma 6, we have ZJ\} { \} - ZJ\ E { \} - ZJ\} { \ E%ĩ ZJ\} { \ E%ĩ ZJ\} { \} - ZJ\ E { \ E%ĩ ZJ\ E { \ E%ĩ ZJ\ E { \} - S, t pve the sentences (5) and (6) ae tue in, we nly need t pve the fllwing sentences ae tue in Œ Š Œ b c 4Ž 4Ž ZJ\ E { \} - ZJ\ E { \ E%ĩ š % ZJ\ E { \ E%ĩ % ZJ\ E { \} d F each fluent A 6, and an intege +, ( ) +-)yg/y1, cnside the fllwing fu cases Case 1 hlds in bth 9 and 9 Œx Accding t emma 5, we have and Œ Since 9 Œ bj394 OG'CC E esrt BHJI E, X A0 ]H^I E Hence, we can educe (7) and () t Œ Š Œ " Ž ZJ\ E { \ E%ĩ ZJ\ E { \ E%ĩ (9) is tue in as bth and Œx ae tue in (9) is als tue in hand side ae false in Case 2 hlds in 9 but des nt hld in 9 Œx By emma 5, we have and M Œ (7) () (9) e (10) as bth the left hand side and the ight Since 94Œ 39 O "CDC E etr ]H^I E, A BHJI E S, thee exists ja such that A BHJI Z Thus, ZJ\ E { \} - is false in and [ZJ\ E { \} d ae tue in On the the hand, since BHJI3EQF "CDCE MN, we have X A0"CDCDE As a esult, bth sides f (7) ae false, wheeas bth sides f () ae tue That is, (7) and () ae tue in Case 3 des nt hld in 9 but hlds in 9 Œx In this case, we have Œ is tue in but is nt Since, 9 Œx 2 9 O"CDCE $R ]H^IKE, we have pap'cc$e Thus, A ]H^IKE Similaly t Case 2, (7) is tue in as its bth sides ae tue and () is als tue because its bth sides ae false Case 4 des nt hld in bth 9 and 9 Œ In this case, we have and Œ Since 9 Œ 94QO_"CDC E etr ]H^I E, we have eithe XA BHJI E u A 'CC E Hweve, A ]H^I E als implies that u A 'CC E Hence, in bth cases, we have _ A "CDC E H Z 4

5 (7) and () ae theefe equivalent t Œ Œ Ž ZJ\ E { \} - ZJ\ E { \} d (11) (12) It is easy t see that bth f them ae tue in is theefe a mdel f m S, the encding is cmplete 5