A Probabilistic Analysis of Propositional STRIPS. Planning. Tom Bylander. Division of Mathematics, Computer Science, and Statistics

Transcription

1 A Probabilistic Analysis of Propositional STRIPS Plannin Tom Bylander Division of Mathematics, Computer Science, and Statistics The University of Texas at San Antonio San Antonio, Texas USA April 24, 1995 to appear in Articial Intellience Abstract I present a probabilistic analysis of propositional STRIPS plannin. The analysis considers two assumptions. One is that each possible precondition (likewise postcondition) of an operator is selected independently of other pre- and postconditions. The other is that each operator has a xed number of preconditions (likewise postconditions). Under both assumptions, I derive bounds for when it is hihly likely that a plannin instance can be eciently solved, either by ndin a plan or provin that no plan exists. Rouhly, if plannin instances under either assumption have n propositions (round atoms) and oals, and the number of operators is less than an O(n ln ) bound, then a simple, ecient alorithm can prove that no plan exists for most instances. If the number of operators is reater than an (n ln ) bound, then a simple, ecient alorithm can nd a plan for most instances. The two bounds dier by a factor that is exponential in the number of pre- and postconditions. A similar result holds for plan modication, i.e., solvin a plannin instance that is close to another plannin instance with a known plan. Thus it appears that propositional STRIPS plannin, a PSPACE-complete problem, exhibits a easy-hard-easy pattern as the number of available operators increases with a narrow rane of hard problems. An empirical study demonstrates this pattern for particular parameter values. Because propositional STRIPS plannin is PSPACE-complete, this extends previous phase transition analyses, which have focused on NP-complete problems. Also, the analysis shows that surprisinly simple alorithms can solve a lare subset of the plannin problems. This paper is a revised and extended version of [4].

2 1 Introduction Lately, there has been a number of worst-case complexity results for plannin, showin that the eneral problem is hard and that several restrictions are needed to uarantee polynomial time [2, 3, 5, 6, 11, 12, 22]. A criticism of such worst-case analyses is that they do not apply to the averae case [8, 18]. Recent work in AI has shown that this criticism has some merit. Several experimental and theoretical results have shown that specic NP-complete problems are hard only for narrow ranes [7, 10, 19, 20, 27] and suests that even the instances within these ranes can usually be eciently solved [25, 26]. This paper extends these results by providin a probabilistic analysis of propositional STRIPS plannin [23]. In contrast to the above work, propositional STRIPS plannin is a PSPACE-complete problem [5], a much harder complexity class [14]. PSPACE problems are those that can be solved by alorithms with space requirements that are bounded by a polynomial on the size of the input. PSPACE-complete problems are the hardest problems in this class. Satisability of quantied Boolean expressions belons to this class. Naturally, because PSPACE-complete problems are harder than NP-complete problems, 1 they should also have reions of hard problems. See [14] for an introduction to NP- and PSPACEcompleteness. One way to think of the dierence between PSPACE-complete and NP-complete problems is that NP-complete problems are restricted to a polynomial number of nondeterministic choice points, while PSPACE-complete problems can have in eect an exponential number of choice points. This is because nondeterministic alorithms restricted to polynomial space (NPSPACE alorithms) can be reduced to PSPACE alorithms, and these nondeterministic alorithms can require exponential time. In propositional STRIPS plannin, this means that the shortest solution plan miht have exponential lenth. Thus, each choice in a PSPACEcomplete problem is potentially a much smaller component of a solution as compared to a NP-complete problem. Similarly, each modication to an instance of a PSPACE-complete problem miht have a small eect. This miht explain the smooth transitions between easy and hard instances in our empirical study. In common with previous work on NP-complete problems, I make stron independence assumptions about the distribution of instances. I assume that the probability that a iven operator is in a plannin instance is independent of what other operators are in the instance. Two variations of this theme are explored. One is that each possible precondition (likewise postcondition) is selected independently of the other pre- and postconditions in the operator. The other is that each operator has a xed number of preconditions (likewise postconditions). Under these assumptions, I derive bounds for when it is hihly likely that specic alorithms will eciently solve a plannin instance, either by ndin a plan or provin that no plan exists. Musick and Russell [21] also analyze a problem similar to plannin. They approximate a restricted kind of search (each operator has one postcondition) with Markov chains, which in turn leads to polynomial-time solutions on averae under certain conditions. In contrast, my analysis provides riorous probabilistic bounds in terms of ve parameters: the number of 1 Actually, P6=PSPACE has not yet been proven, but I shall make the standard assumptions that P6=NP and NP6=PSPACE. 1

3 propositions, the number of operators, the number of pre- and postconditions per operator, and the number of oals. These parameters characterize a full rane of plannin problems. The alorithms analyzed by this paper are all variations of simple hill-climbin, i.e., hill climbin with no backtrackin. Clearly, this type of search is far removed from the sophisticated partial plannin alorithms that are the subject of much current study, e.., TWEAK [6], SNLP [17]. and UCPOP [24], and is impoverished compared to almost any other plannin alorithm ever proposed. There are three reasons to consider alorithms of such simplicity: they permit formal analysis, they are ecient, and, as noted below, they cover a very lare portion of the plannin problem. The results for these alorithms should provide a baseline for analyzin and empirically comparin more sophisticated alorithms on random plannin instances. Specically, iven that randomly-enerated plannin instances have n propositions (round atoms) and oals, and that operators have r preconditions and s postconditions on averae, I derive the followin results. If the number of operators is at most ((2n?s)=s)(ln?ln ln 1=), then a simple, ecient alorithm can prove that no plan exists for at least 1? of the instances. If the number of operators is at least e r e s=n (2n=s + 1)(ln =), then a simple, ecient alorithm can nd a plan for at least 1? of the instances. If r, s, and are held constant as n and increase, then the bounds are O(n ln ) and (n ln ), respectively. They dier by a factor that is exponential in the number of pre- and postconditions. A similar result holds for plan modication. If the initial state or oals are dierent by one condition from that of another plannin instance with a known plan, and if there are at least e r+s (2n=s)(ln 1=) operators, then it is likely (1? ) that a sinle operator converts the old plan into a solution for the new instance. Thus, it appears that propositional STRIPS plannin is easy if the number of operators is below one threshold or above a somewhat hiher threshold. Conjecturin that some rane of problems between the thresholds are hard, then propositional STRIPS plannin exhibits a easy-hard-easy pattern similar to NP-complete problems. An empirical study demonstrates this pattern for particular parameter values. However, the empirical study shows smooth transitions between easy and hard instances, and so would not normally be considered a phase transition. Despite this, the theoretical analysis can be said to demonstrate an \asymptotic" phase transition. Larer random plannin instances are hard only if the number of operators is (n ln ). Outside this asymptote, larer instances become easy. In any case, future work is needed to narrow the ap between the bounds and to analyze more realistic distributional assumptions and more sophisticated alorithms. The rest of the paper is oranized as follows. First, denitions and key inequalities are presented. Then, the results of the analysis are derived. Finally, empirical results are displayed. 2 Preliminaries This section denes propositional STRIPS plannin, describes the distributions of instances to be analyzed, and presents key inequalities. 2

4 2.1 Propositional STRIPS Plannin An instance of propositional STRIPS plannin is specied by a tuple hp; O; I; Gi, where: P is a nite set of round atomic formula, called the propositions; a proposition is also called a positive condition; its neation is called a neative condition; a state is a satisable set of conditions of which each proposition or its neation is a member; O is a nite set of operators; the preconditions and postconditions of each operator are satisable sets of conditions; I is the initial state; and G, the oals, is a satisable set of conditions. If the preconditions of an operator are satised by a state, then the operator can be applied to that state, and the resultin state is determined by addin the postconditions, deletin those conditions that conict with the postconditions (cf. [13]). A solution plan is a sequence of operators that transforms the initial state into a oal state, i.e., a state that satises the oals. For example, a blocks-world instance can be represented usin propositions like clear(a) to represent \block A has a clear top," and on(a; B) to represent \block A is on top of block B." The set of preconditions of an operator to move A from on top of B to on top of C can be represented as: fclear(a); clear(c); on(a; B) That is, blocks A and C are clear, and block A is on top of block B. Its postconditions are: fclear(b); :clear(c); :on(a; B); on(a; C) If the preconditions are true before the operator is applied, then after the operator is applied, block B becomes clear, block C is no loner clear, block A is no loner on block B, and block A is now on block C. 2.2 Distributional Assumptions Let n be the number of propositions. Let o be the number of operators. Let r and s respectively be the expected number of pre- and postconditions within an operator. Let be the number of oals. For iven n, o, r, s, and, I assume that random plannin instances under the variable model are distributed by eneratin each operator as follows: For each proposition p 2 P, p is a precondition of the operator with probability r=(2n); alternatively :p is a precondition with probability r=(2n). These probabilities are independent of other pre- and postconditions. For postconditions, s=(2n) is the relevant probability. For each proposition p 2 P, p 2 I (the initial state) is as likely as :p 2 I. 3

5 For the oals, propositions are selected at random and are set to positive or neative so that no oal is satised in the initial state. This latter restriction is made for ease of exposition. The only dierence between the variable model and the xed model is that: Each operator has exactly r preconditions and s postconditions. Any leal set of r preconditions or s postconditions is equally likely. It must be admitted that these assumptions do not approximate certain aspects of plannin domains very well. For example, there are only b clear conditions for a blocks-world instance of b blocks compared to (b 2 ) on conditions. However, every blocks-world operator refers to one or more clear conditions, i.e., a iven clear condition appears more often within the set of round operators than a iven on condition. Also, there are correlations between the conditions, e.., clear(a) is more likely to appear with on(a; B) than with on(c; D). Similar violations can be found for any of the standard toy domains. Ultimately, the usefulness of these assumptions will depend on how well the threshold bounds of the analysis classify easiness and hardness of real plannin domains. Even so, it is worth notin that the xed model provides a uniform distribution over the set of instances dened by the parameters. Because the results show that such plannin instances are usually easy except for a narrow rane of the number of operators o, it follows that the vast majority of plannin instances are indeed easy. 2.3 Alorithm Characteristics Each alorithm in this paper is incomplete but sound, i.e., each alorithm returns correct answers when it returns yes or no, but miht answer \don't know." Specically, \success" is returned if the alorithm nds a solution plan, \failure" is returned if the alorithm determines that no plan exists, and \don't know" is returned otherwise. The performance of a iven alorithm is characterized by an accuracy parameter, 0 < < 1. Each result below shows that if the number of operators o is reater than (or less than) a formula on n, r, s,, and, then the accuracy of the alorithm on the correspondin distribution (see Distributional Assumptions section) will be at least 1?. 2.4 Inequalities I freely use the followin inequalities. For nonneative x and y: e?x=(1?x) 1? x for 0 x < 1 (1) 1? x e?x (2) xy 1 + xy 1? (1? x)y for 0 x < 1 (3) Inequalities (1) and (2) are easily derivable from [9]. Inequality (3) is derivable from inequalities (1) and (2). The loarithmic forms of these equalities are sometimes used. 4

6 A particular form of Bonferroni's inequality shall be used. If E 1 ; E 2 ; : : : ; E m are m events and the probability of each event is reater than or equal to 1? =m, then: P (E 1 ^ E 2 ^ : : : ^ E m ) 1? (4) Finally, the followin recurrence relation is useful in analyzin the xed model. Its justication is described in the proof for Lemma 4. Let f(s; n; k) be the probability that s conditions that are randomly enerated from n propositions are consistent with some particular set of k conditions. Then, for nonneative inteers n, s n, and k n: f(s; n; k) = n? k k f(s? 1; n? 1; k) + f(s? 1; n? 1; k? 1) (5) n 2n The base cases are f(0; n; k) = 1, f(s; n; 0) = 1, f(s; n; n) = 2?s, and f(n; n; k) = 2?k. In the appendix, the followin inequalities are demonstrated: e?sk=n f(s; n; k) 2n 2n + sk 3 Eciently Provin Plan Nonexistence If there are few operators, it becomes unlikely that the postconditions of the operators cover all the oals, i.e., it is likely that some oal is not a postcondition of any operator. Recall that random plannin instances are dened so that no oal is true of the initial state. So if some oal is not a postcondition of any operator, then the instance has no solution plan. This leads to the followin simple alorithm: Posts-Cover-Goals for each oal if the oal is not in the postconditions of any operator then return failure return don't know While Posts-Cover-Goals miht be considered a trivial alorithm, the followin two theorems show that Posts-Cover-Goals works for a substantial rane of the plannin problem. Theorem 1 For random plannin instances under the variable model, if o 2n? s (ln? ln ln 1=) s then Posts-Cover-Goals will determine that no plan exists for at least 1? of the instances. Proof: The probability that there exists a oal that is not a postcondition of any operator can be developed as follows. Consider a particular oal and operator: s=2n probability that the oal is a postcondition of the operator 2 1? s=2n probability that the oal is not a postcondition of the operator (1? s=2n) o probability that the oal is not a postcondition of any operator 1? (1? s=2n) o probability that the oal is a postcondition of some operator (1? (1? s=2n) o ) probability that every oal is a postcondition of some operator 5 (6)

7 The inequality of the theorem implies that the above probability is less than or equal to. Suppose that the inequality of the theorem is true: o 2n? s (ln? ln ln 1=) s This is equivalent to: and: os 2n? s ln? os 2n? s ln 1= ln ln 1= ln(1? s=2n)?s=(2n? s) by Inequality (1), which implies: ln 1= o ln(1? s=2n) ln This is equivalent to: and: ln (1? s=2n) o 1=?(1? s=2n) o ln ln(1? (1? s=2n) o )?(1? s=2n) o by Inequality (2), which implies: ln(1? (1? s=2n) o ) ln This is equivalent to: and nally: ln(1? (1? s=2n) o ) ln (1? (1? s=2n) o ) which is the desired inequality. Thus, if the inequality of the theorem is satised, then the probability that some oal is not a postcondition of any operator is at least 1?. 2 For arithmetic expressions within this paper, multiplication has hihest precedence, followed by division, loarithm, subtraction, and addition. E.., 1? s=2n is equivalent to 1? (s=(2n)). 6

8 Theorem 2 For random plannin instances under the xed model, if o 2n? s (ln? ln ln 1=) s then Posts-Cover-Goals will determine that no plan exists for at least 1? of the instances. Proof: The derivation in the previous proof holds for the xed model to the point where 1? (1? s=2n) o is the probability that a particular oal is a postcondition of some operator. However, if this oal is a postcondition of some operator, then this reduces the probability that other oals will be postconditions of that operator, i.e., the number of \available" postconditions is reduced from s to s? 1. Althouh (1? (1? s=2n) o ) is not the probability that every oal is a postcondition of some operator, this expression does remain an upper bound. Thus, the same inequality holds for the xed model. For xed and increasin n and, the above bound approaches (2n? s)(ln )=s. If s is also xed, the bound is O(n ln ). In eneral then, plannin instances with a number of operators linear in n (or linear in n times loarithmic in times a small constant) will not have plans. Fortunately thouh, it is usually easy in such cases to prove that a plan does not exist. Naturally, more complex properties that are ecient to evaluate and imply plan nonexistence could and should be used, e.., the above alorithm does not look at preconditions or consider how postconditions conict with the oals. Nevertheless, the analysis of Posts- Cover-Goals provides a stron bound on when it is easy to prove plan nonexistence. 4 Eciently Findin Plans With a sucient number of operators, then it becomes likely that some operator will make proress towards the oal. In this section, I consider four alorithms. One is a simple forward search from the initial state to a oal state, at each state searchin for an operator that decreases the number of oals to be achieved. The second is a backward search from the oals to the initial state. The third is also a backward search from the oals, but tries nds a plan that will reach the oals from any initial state. The fourth is a very simple alorithm for when the initial state and oals dier by just one condition. To illustrate the alorithms the followin instance is used: P = fa 1 ; a 2 ; a 3 ; a 4 O = fa 1 ^ a 2 ) :a 3 ^ a 4 ; a 2 ^ a 4 ) a 3 ; :a 1 ^ a 2 ) a 3 ^ a 4 ; a 2 ^ :a 4 ) :a 1 ; :a 2 ) a 3 ^ a 4 ; ) :a 2 I = fa 1 ; a 2 ; :a 3 ; :a 4 G = fa 3 ; a 4 7

9 The notation pre ) post is used for operators; the preconditions are represented by a conjunction of conditions before the arrow; the postconditions are after the arrow. This instance would possible under the variable model for n = 4, o = 6, = 2, and r and s set to any positive number between 1 and 3, inclusive, thouh it would be an especially unlikely instance for r = 3 or s = Forward Search Consider the followin alorithm: Plan-Forward(S) if G is satised by S then return success else if some operator can be applied to S and satises more oals then let S 0 be the result of applyin the operator to S return Plan-Forward(S 0 ) else return don't know end if end if If Plan-Forward(I) is called, then it searches for an operator that increases the number of satised oals. If there is such an operator, the current state S is updated. Plan- Forward succeeds if it reaches a oal state and is noncommittal otherwise. For the example instance, a plan of the rst two operators miht be enerated. The rst operator achieves the oal a 4 from the initial state, leavin the other propositions unchaned. The second operator achieves the remainin oal a 3. Plan-Forward just performs simple hill-climbin. I do not claim that this is a practical alorithm for plannin in eneral, but the analysis is reatly simplied by avoidin backtrackin and partial plans. The probability that the alorithm will succeed can be bound by considerin the probability that an additional oal can be satised by some operator. Certainly, a more systematic search alorithm that eciently includes Plan-Forward would exceed its probability of success. 3 Despite its handicaps, Plan-Forward is surprisinly robust under certain conditions. First, I demonstrate two lemmas for the number of operators that need to be considered to increase the number of satised oals. One lemma is for the variable model, and the other, the xed model. Lemma 3 Consider random plannin instances under the variable model except that d of the oals are not satised. If o e r e s(?d)=n 2n sd + 1 ln 1 3 One such alorithm would be A* search from the initial state usin a heuristic equal to the number of unsatised oals times any constant reater than 1. The constraint on the constant ensures that the number of oals achieved are considered more valuable than the number of operators applied. However, this heuristic miht not lead to optimal plans. 8

10 then, for at least 1? of the instances, applyin some operator will increase the number of satised oals. Proof: The expression for the probability that some operator will increase the number of satised oals can be developed as follows: (1? r=2n) n probability that a state satises the preconditions of an operator, i.e., each of n propositions is not a precondition with probability 1? r=n; alternatively, a proposition is a matchin precondition with probability r=2n (1? s=2n)?d probability that the postconditions of an operator are consistent with the? d oals already achieved (1? s=2n) d probability that the postconditions do not achieve any of the d remainin oals, i.e., for each oal, it is not a postcondition with probability 1? s=2n 1? (1? s=2n) d probability that the postconditions achieve at least one of the d remainin oals Thus, the probability p that a particular operator can be applied, will not clobber any satised oals, and will achieve at least one more oal is: p = 1? r n 1? s?d 1? 1? s! d 2n 2n 2n 1? p is the probability that the operator is missin one or more of these properties, and (1? p) o is the probability that o operators are unsatisfactory. If (1? p) o, then there will be some satisfactory operator with probability at least 1?. This inequality is satised if o (1=p)(ln 1=) because in such a case: (1? p) o e?po e? ln 1= = All that remains then is to determine an upper bound on 1=p, i.e., a lower bound on p. For each term of p: (1? r=2n) n e?rn=(2n?r) e?r (1? s=2n)?d e?s(?d)=(2n?s) e?s(?d)=n 1? (1? s=2n) d sd 2n + sd Invertin these terms leads to the bound of the lemma. Lemma 4 Consider random plannin instances under the xed model except that d of the oals are not satised. If at least e r e s(?d)=n 2n sd + 1 ln 1 operators are considered, then, for at least 1? of the instances, Plan-Forward will nd an operator that increases the number of satised oals. 9

11 Proof: Just as in the previous Lemma, the approach is to show that the probability p that a particular operator can be applied, will not clobber any satised oals, and will achieve at least one more oal satises: p e?r e?s(?d)=n sd 2n + sd The probability that a particular operator can be applied is 2?r e?r. The probability that the postconditions of an operator are consistent with the? d oals already achieved can be described with a recurrence equation. Let f(s; n; k) be the probability that s conditions can be randomly enerated from n propositions so that they are consistent with some particular set of k conditions. If a condition is randomly enerated from the n propositions, there is a (n? k)=n probability that it is neither identical to nor the neation of one of the k conditions; this leaves s? 1 conditions to be enerated from n? 1 propositions and to be consistent with k conditions. Alternatively, there is a k=2n probability that it is identical to one of the k conditions; this leaves s? 1 conditions to be enerated from n? 1 propositions to be consistent with k? 1 conditions. The remainin k=2n probability is when it conicts with one of the k conditions. This leads to the followin recurrence equation: f(s; n; k) = n? k k f(s? 1; n? 1; k) + f(s? 1; n? 1; k? 1) n 2n which was introduced as equation (5). In the base cases, f(0; n; k) = 1, f(s; n; 0) = 1 (the probability is 1 if there are no conditions to select or no conditions to be consistent with), f(s; n; n) = 2?s (each condition to select must have a particular sin), and f(n; n; k) = 2?k (each condition to be consistent with must be selected). For this recurrence equation and these base cases, inequality (6) holds, in this case, f(s; n;? d) e?s(?d)=n. The probability of achievin at least one more of the d remainin oals is one minus the probability that none of the d oals are achieved, i.e., 1? f(s; n; d). Inequality (6) implies that f(s; n; d) 2n=(2n + sd), from which 1? f(s; n; d) sd=(2n + sd) follows. The probability that the postconditions of an operator are consistent with the? d oals already achieved is not independent of the probability of achievin at least one more of the d remainin oals. Fortunately, the two events have an additive dependency relationship, i.e., either event \increases" the probability of the other event. For example, if the postconditions of an operator are consistent with the oals already achieved, then this increases the probability of achievin one of the remainin oals because there are fewer conditions to choose from, and because any of the oals can still be chosen. This completes the proof. The maximum value of the expression in the lemmas can be used to describe Plan- Forward, which leads to the followin theorem: Theorem 5 For random plannin instances under either the variable or the xed model, if o e r e s=n 2n s + 1 ln then Plan-Forward will nd a plan for at least 1? of the instances. 10

12 Proof: For oals, the number of satised oals will be increased at most times. If each increase occurs with probability at least 1? =, then increases (the most possible) will occur with probability at least 1? (this follows from Bonferroni's inequality). Thus, Lemmas 3 and 4 can be applied usin = instead of. Maximizin over the oals leads to: max d=1 e r e s(?d)=n 2n sd + 1 ln e r e s=n 2n s + 1 ln The bound is exponential in the expected numbers of pre- and postconditions. Naturally, as operators have more preconditions, it becomes exponentially less likely that they can be applied. Similarly, as operators have more postconditions, it becomes exponentially less likely that the postconditions are consistent with the oals already achieved. Note thouh that if n=s, then e s=n e, so the expected number of postconditions s is not as important a factor if the number of oals are small. For xed, r, and s, and increasin n and, the above bound is (n ln ). Takin into account the result for Posts-Covers-Goals, it is clear that two sides of the random plannin problem are easy. Below a O(n ln ) bound on the number of operators, it is usually easy to prove that a plan does not exist; above a (n ln ) bound on the number of operators, it is usually easy to nd a plan. Remainin is a ap of a constant between the two bounds, which is exponential in the number of pre- and postconditions. It is a safe conjecture that some rane of instances within the ap is hard, so I conclude that random plannin instances exhibit the easy-hard-easy pattern of other NP-hard problems, with the hard problems occupyin a narrow rane of the number of operators. The empirical study in a followin section displays the results of usin this alorithm. 4.2 Backward Search One could also perform a backward search from the oals to the initial state. Consider the followin alorithm: Plan-Backward(G) if G is consistent with I, then return success else if there is a an operator with R and S as its pre- and postconditions such that G is consistent with S, R is consistent with G? S, and jg? Ij > j((g? S) + R)? Ij then return Plan-Backward((G? S) + R) else return don't know Initially, Plan-Backward(G) is invoked. Plan-Backward then chooses an operator if it can achieve a oal state from another state that has fewer conicts with the initial state I. The postconditions S of such an operator must be consistent with the current set of oals 11

13 G, its preconditions R must be consistent with the oals not achieved by the postconditions, and the new set of oals (G? S) + R must have fewer conditions that are not in the initial state. In the example instance, the third operator :a 1 ^a 2 ) a 3 ^a 4 achieves the oals fa 3 ; a 4, leavin f:a 1 ; a 2 as the new oals. a 2 is consistent with the initial state, so this reduces the number of unsatised oals by one. Now the fourth operator a 2 ^ :a 4 ) :a 1 achieves the unsatised oal :a 1, and the new set of oals will be fa 2 ; :a 4, which is consistent with the initial state. The disadvantae of Plan-Backward is that the number of current oals G can increase steadily to the number of propositions n because each new set of oals (G? S) + R can be r? 1 larer than the previous set of oals (under the xed model). First, I present an analysis for the eneral case, then I consider two special cases in which the performance of Plan-Backward will be more comparable to Plan-Forward. Lemma 6 Consider random plannin instances under the variable or xed model except that d oals are not satised. If o e r+s 2n sd + 1 ln 1 then, for at least 1? of the instances, some operator will achieve a oal state from another state that has fewer conicts with the initial state. Proof: The expression for the probability for the variable model can be developed as follows: (1? r=2n) n a lower bound on the probability that the preconditions of an operator are consistent with the oals minus the postconditions, and all preconditions not in the oals minus the postconditions are consistent with the initial state 1? (1? s=2n) d probability that the postconditions achieve at least one of the d remainin oals (1? s=2n) n a lower bound on the probability that the postconditions of an operator are consistent with the oals, i.e., each of at most n oals is not neated with probability 1? s=2n The last two probabilities are not independent, but their interaction is additive, i.e., an operator that is consistent with the remainin oals is more likely to achieve one of the remainin oals. Thus, the probability p that some operator will reduce the number of oals that are not true of the initial state is bounded by: p (1? r=2n) n (1? s=2n) n (1? (1? s=2n) d ) 1? p is the probability that the operator is unsatisfactory, and (1? p) o is the probability that o operators are unsatisfactory. If o (1=p)(ln 1=), then there will be some satisfactory operator with probability at least 1?. For each term of p: (1? r=2n) n e?rn=(2n?r) e?r 12

14 (1? s=2n) n e?sn=(2n?s) e?s 1? (1? s=2n) d sd 2n + sd Invertin these terms leads to the bound under the variable model. Under the xed model, let f be recurrence equation (5). Then, the probability of suitably consistent pre- and postconditions is at least f(r; n; n) = 2?r e?r and f(s; n; n) = 2?s e?s, respectively. The probability that the postconditions achieve at least one of the d remainin oals is 1? f(s; n; d) sd=(2n + sd). The probability that the postconditions are consistent with the oals is not independent of the probability that the postconditions achieve at least one of the d remainin oals, but their interaction is additive. Thus, the probability p that some operator reduces the number of oals not true of the initial state under the xed model is also bounded by: p e?r e?s sd 2n + sd which leads to the inequality of the theorem. As for Plan-Forward, to determine a bound for Plan-Backward, the maximum value of the expression in the above lemma needs to be determined. This is done to prove the followin theorem: Theorem 7 For random plannin instances under either the variable or the xed model, if o e r+s 2n s + 1 ln then Plan-Backward will nd a plan for at least 1? of the instances. Proof: For oals, the number of unsatised oals will be decreased at most times. If each decrease occurs with probability at least 1? =, then decreases (the most possible) will occur with probability at least 1?. Thus, the expression in the previous lemma can be used substitutin = instead of. Maximizin over the decreases leads to: max d=1 e r+s 1 + 2n sd ln e r+s 1 + 2n s ln The dierence between this bound for Plan-Backward and the bound for Plan- Forward is that the Plan-Backward bound has an e s term instead of e s=n. This makes the Plan-Backward bound larer by a factor of e s(n?)=n, so if the number of oals is small relative to the number of propositions n, and if the expected number of postconditions s is lare, then the increase is substantial. This suests that Plan-Forward will outperform Plan-Backward when is small relative to n. This performance dierence should become more pronounced for larer s. However, if r, s, and remain constant as and n increases, the order of the Plan- Backward bound will be (n ln ), which is identical to the order of the Plan-Forward bound. Also, the followin analysis suests that the dierence will be smaller than suested by Theorem 7. 13

15 4.3 Backward Search with Few Goals The above analysis of Plan-Backward assumes the worst case reardin the size of the current set of oals G, namely that jgj is always close to n. However, if the number of oals is small enouh, specically n=r, then jgj can never reach n under the xed model, and is very unlikely to reach n under the variable model. The analysis for the xed model is much more tractable, and is iven below. Lemma 8 Consider random plannin instances under the xed model except that d oals are not satised, and there is no more than r? dr + d n oals. If o e r e s(r?dr+d)=n 2n sd + 1 ln 1 then, for at least 1? of the instances, some operator will achieve a oal state from another state that has fewer conicts with the initial state. Proof: The only dierence from the proof for Lemma 6 is that the postconditions must be consistent with at most r? dr + d oals rather than at most n oals. The probability of this is bounded by f(s; n; r? dr + d) e?s(r?dr+d))=n, where f is recurrence equation (5), and the inequality follows from inequality (6). Substitutin e?s(r?dr+d)=n for e?s in the proof for Lemma 6 leads to the bound of this lemma. Theorem 9 For random plannin instances under the xed model, with r n and r 1, if o e r e sr=n 2n s + 1 ln then Plan-Backward will nd a plan for at least 1? of the instances. Proof: For oals, the number of unsatised oals will be decreased at most times. If each decrease occurs with probability at least 1? =, then decreases (the most possible) will occur with probability at least 1?. Also, satisfyin a oal results in increasin the total number of oals by at most r? 1. Thus, when there d oals left to achieve, there are at most + (? d)(r? 1) = r? dr + d oals. Thus, the expression in the previous lemma can be used substitutin = instead of. Maximizin over the decreases leads to: max d=1 e r e s(r?dr+d)=n 2n sd + 1 ln e r e sr=n 2n s + 1 ln In the case where r = 1, this bound for Plan-Backward is the same as for Plan- Forward. For xed r and s and increasin and n, if sr remains small relative to n, then there is little dierence between the two bounds, but the Plan-Forward bound will still be smaller for r > 1, thouh it should be noted that both bounds are (n ln ). 14

16 4.4 Backward Search Independent of the Initial State Consider the followin variation of Plan-Backward: Plan-Backward2(G) if G = ; then return success else if there is a an operator with R and S as its pre- and postconditions such that G is consistent with S, and j(g? S) + Rj < jgj then return Plan-Backward2((G? S) + R) else return don't know Like the previous alorithm, Plan-Backward2 looks for operators that reduce the number of oals, but unlike Plan-Backward, Plan-Backward2 does not depend on the initial state, and it repeatedly looks for an operator that reduces the number of oals until there are no oals left. If Plan-Backward2 succeeds, then it will have discovered a sequence of operators that achieves a oal state from any initial state, althouh note that the rst operator in this sequence (the last operator selected by Plan-Backward) must not have any preconditions; otherwise j(g? S) + Rj would be non-zero. Havin such an operator is probably unrealistic; it is impossible under the xed model if r 1. Nevertheless, the analysis below suests that reducin a set of oals into a much smaller set of oals is often possible, which, of course, can then be followed by forward search or a more eneral backward search. In the example instance, the fth operator :a 2 ) a 3 ^ a 4 achieves the oals leavin the sinle new oal :a 2. The sixth operator ) :a 2 achieves the new oal without any preconditions. I rst introduce a lemma for the number of operators needed to nd one operator that reduces the number of oals. Lemma 10 For random plannin instances under the variable model, with r n=2 and s n=2, if: o e 2r e s=n! 3n s + 1 ln 1 then, for 1? of the instances, some operator reduces the number of oals. Proof: The preconditions should not refer to any condition that is not a oal or the neation of a non-oal. This has probability (1?r=n) n?. It does not matter what the postconditions do to these conditions. The preconditions and postconditions should be consistent with the oals. This has probability 1? r 1? s = 1? r 2n 2n 2n? s 2n + rs 4n 2 15

17 However, the case in which every oal equal to a postcondition is also equal to a precondition must be avoided. The probability that this occurs for a iven oal is a sum of the followin: (1? r=n)(1? s=n) probability that the oal and its neation is not in the preand postconditions (r=2n)(1? s=n) probability that the oal is in the preconditions, but it (as well as its neation) is not in the postconditions (r=2n)(s=2n) probability that the oal is in both the pre- and postconditions Usin the sum of these probabilities, the probability of this case happenin for all oals is: 1? r 2n? s n + 3rs 4n 2 Thus, the probability p that a random operator will satisfy the stated requirements is: p = (1? r=n) n? 1? r 2n? s 2n + rs? 1? r 4n 2 2n? s n + 3rs 4n 2 Now recall from the previous proofs that if the number of operators considered exceeds (1=p)(ln 1=), then the probability that some operator is satisfactory is at least 1?. What remains then is to demonstrate an upper bound on 1=p (lower bound on p). For the rst term of p, (1? r=n) n? e?r(n?)=(n?r) Because r n=2, e?r(n?)=(n?r) e?2r(n?)=n Reardin the second term of p: 1? r 2n? s 2n + rs? 1? r 4n 2 2n? s n + 3rs 4n 2!! = 1? r 2n? s 2n + rs 1? 4n2? 2rn? 4sn + 3rs 4n 2 4n 2? 2rn? 2sn + rs 2rn + 2sn? rs 2sn? 2rs = 1? 1? 1? 4n 2 4n 2? 2rn? 2sn + rs 2rn + 2sn? rs 2sn? 2rs exp? 4n 2? 2rn? 2sn + rs 4n 2? 2rn? 4sn + rs + 2sn? 2rs = exp (? (2n? s)r + (2n? r)s + rs (2n? r)(2n? s) ) 2s(n? r) (n? r)(2n + 2s? s) + n(2n? s) 16

18 The rst term of this expression can be further simplied usin r n=2 and s n=2: exp ( (2n? s)r + (2n? r)s + rs? (2n? r)(2n? s) ( = exp? r 2n? r? s 2n? s? rs (2n? r)(2n? s) exp? r n? 2s 3n? 4rs 9n 2 exp? r n? 2s 3n? 2s 9n e?r=n e?s=n At this point, the followin lower bound for p has been derived: p e?2r(n?)=n e?r=n e?s=n 2s(n? r) (n? r)(2n + 2s? s) + n(2n? s) This can be further simplied usin e?2r(n?)=n e?r=n e?2r. Finally, an upper bound for (1=p)(ln 1=) is derived: 1 p ln 1 e2r e s=n e 2r e s=n! 2n + 2s? s n(2n? s) + ln 1 2s 2s(n? r) n + s s + 2n s = e 2r s=n 3n + s e ln 1 s which proves the lemma.! ln 1 ) Similar to previous theorems, the maximum of this expression needs to be determined. This is done to prove the next theorem. Theorem 11 For random plannin instances under the variable model, with r n=2 and s n=2, if o e 2r 5n s + 3es s + es=n ln 1 ) o e 2r n s (4es(+1)=n + 3e ln ) ln then Plan-Backward2 will nd a plan that achieves the oals from any initial state for at least 1? of the instances. 17

19 Proof: For oals, the number of remainin oals will be decreased at most times. If each decrease occurs with probability 1? =, then decreases (if necessary) will occur with probability at least 1?. Thus, Lemma 10 can be applied usin = instead of. Maximizin over the decreases leads to: max d=1 e 2r e sd=n 3n sd + 1 ln e 2r ln e s=n + 3 max d=1 e sd=n n sd e sd=n n=sd has one minimum for positive d, i.e., when d = n=s. So: max d=1 e sd=n n sd As a result: e 2r ln e s=n + 3 = max which proves the theorem. ( max d=1 e s=n n s ; es=n n s e sd=n n sd ) max e 1=2ns es ; s e 2r ln e s=n 5ns 3es + + s 5n 3s + e s s Comparin the two bounds for Plan-Forward and Plan-Backward2, the bound for Plan-Backward2 is worse in that it has a larer constant and has a e 2r term as opposed to a e r term for the Plan-Forward bound. Because Plan-Backward2 does not use the initial state, some increase would be expected. However, the Plan-Backward2 bound is better in that one component is additive, i.e., O(e s + n=s); whereas the correspondin subexpression for the Plan-Forward bound is O(e s=n n=s). The reason is that in the maximization of the Plan-Backward2 bound, e sd=n is maximum when d is at its maximum, while the maximum value for n=sd is when d is at its minimum. In the maximization for the Plan-Forward bound, e s(?d)=n attains its maximum at the same time as n=sd does, when d is at its minimum. However, for xed r, s, and, and increasin n and, both bounds are (n ln ). 4.5 Plan Modication So far I have considered the problem of eneratin a plan from scratch. In many cases, however, the current plannin instance is close to a previously solved instance, e.., [15, 16]. Consider a simplied version of plan modication, specically, when the initial state or set of oals of the current plannin instance diers by one condition from a previously solved instance. In this case, the new instance can be solved by showin how the new initial state can reach the old initial state, or how the old oal state can reach a new oal state. Within the framework of random plannin instances then, I shall analyze the problem of reachin one state from another when the two states dier by one condition, i.e., there are n oals, and all but one oal is true of the initial state. The worst-case complexity of this problem, like the problem of plannin from scratch, is PSPACE-complete [22]. However, the followin theorem shows that ecient plan modication does not appear to require as many operators as ecient plannin from scratch. 18

20 Theorem 12 For random plannin instances under either the variable or the xed model in which there are n oals, where n? 1 oals are true of the initial state, if: o e r e s 2n s ln 1 then, for at least 1? of the instances, some operator solves the instance in one step. Proof: First, I develop the probability p that a random operator solves a random instance for the variable model. The probability that the preconditions are consistent with the initial state is (1? r=2n) n. The probability that the postconditions are consistent with the n? 1 oals already achieved is (1?s=2n) n?1. In addition, the probability that the oal is achieved by a postcondition is s=2n. Thus: 4 p = (1? r=2n) n (1? s=2n) s n?1 2n Lower bounds for p are: p e?rn=(2n?r) e s?sn=(2n?s) 2n e?r e?s s 2n The probability that none of o operators solves the instance is (1? p) o. If o satises the inequality stated in the theorem, then: (1? p) o e?po e? ln 1= = which proves the bound for the variable model. For the xed model, the probability that the preconditions are true of the initial state is 2?r e?r. The probability that the oal is achieved by a postcondition is s=2n (it is selected with probability s=n and has the riht sin with probability 1=2). The probability that the remainin s? 1 postconditions are consistent with the n? 1 oals already achieved is f(s? 1; n? 1; n? 1) 2?s+1 e?s, where f is dened by recurrence equation (5). Thus, the probability p that a random operator solves a random instance in the xed model has a lower bound of e?r e?s s=2n, and the inequality of the theorem follows. Thus, for xed r, s, and, (n) operators suce to solve plannin instances that dier by one condition from previously solved instances. So, for at least the distributions of plannin instances considered here, the number of operators needed for ecient plan modication appears to be a factor of O(ln ) lower than that needed for ecient plannin from scratch. As a corollary, consider any state sequence of lenth + 1 from the initial state to a oal state, where each successive state in the sequence diers by just one condition. If = is substituted for in the theorem, and the inequality is satised, then any particular transition in the sequence can be accomplished with probability 1? =. Consequently, all transitions can be performed with probability 1?. Thus, (n ln ) plannin can be accomplished even if the transitions from the initial state to a oal state are chosen in advance. 4 This does not scale up to the case of attainin oals by a sinle operator. The probability that the postconditions of a random operator contain the oals is (s=2n), i.e., exponentially small in the number of oals. 19

21 5 Empirical Study The formal analysis provides riorous probabilistic bounds on when random plannin instances can be eciently solved, either by provin that no plan exists or by ndin a solution plan. However, the derivation of the bounds in the above theorem depends on rather crude inequalities. In this section, I display the empirical results when two of the above alorithms, Posts-Cover-Goals and Plan-Forward, are applied to randomly-enerated plannin instances. Note that there are ve parameters to choose (the number of operators o, the number of propositions n, the number of pre- and postconditions r and s, and the number of oals ) as well as the choice of variable model (on averae, operators have r and s pre- and postconditions) or xed model (each operator has exactly r and s pre- and postconditions). As a result, it is not feasible to empirically cover many of the possibilities. Any choice of values will be arbitrary to some extent. I chose the followin values for this study: n = 100 and n = 1000, r = 2 and s = 2 under the xed model, ranin from low values up to n, and o varyin over where the alorithms' performance chanes. The two values for n allow to vary over many values, and also show how the transitions chane from a lower value to a hiher value. The values of r and s are the minimum values that make propositional STRIPS plannin PSPACE-complete [5]. The xed model ensures eciency in eneratin an operator. 5 Dierent values for will test the ln asymptote. For each trial, it is assumed that there is an unbounded stream of randomly-enerated operators that can be used. For the Posts-Cover-Goals alorithm, it can be determined when the stream of operators covers all the oals. At this number of operators (call the number a), Posts-Cover-Goals fails to solve the problem. In my implementation of Plan-Forward, whenever an additional oal is achieved, the alorithm reverts to the beinnin of the stream, attemptin to achieve remainin oals with previously enerated operators. For this implementation, it can be determined how much of the stream was used by Plan-Forward to solve the problem. At this number of operators (call this number b), Plan-Forward solves the problem trials were performed for each value of n and considered. This will, as shown below, ive a ood indication of where these alorithms solve from 1% to 99% of the instances. I.e., if 99% of the a values are reater than 100, then Posts-Cover-Goals empirically solves at least 99% of the instances when o 100. If 1% of the b values are less than or equal to 1000, then Plan-Forward empirically solves at least 1% of the instances when o Fiure 1 displays the empirical results for Posts-Cover-Goals when n = 100. The x-axis is the number of oals (even numbers from 2 to 100) on a loarithmic scale, and the y-axis is the number of operators. Each point on the raph indicates the number of operators where Posts-Cover-Goals solves a certain percentae of instances for a iven number of oals. The ve kinds of points correspond to ve isolevels. The diamond points indicate 99% eectiveness; the plus points 90% eectiveness; the square points 50%; the x points 10%; and the trianle points 1%. The Posts-Cover-Goals alorithm has hiher 5 It takes constant time to enerate two random numbers. For the variable model, eneratin an operator would take time linear in n because each condition must be independently considered. 20

22 % 10% 50% 90% 99% Operators Goals Fiure 1: Empirical eectiveness of Posts-Cover-Goals for 100 propositions. The x-axis is loarithmically scaled. eectiveness for lower number of operators. For a iven level of eectiveness, the loarithmic scalin makes it clear that the number of operators varies loarithmically with, which is consistent with the theoretically derived bound. Also, the theoretical bound is remarkably close to the empirical results, e.., for = 100 and = 0:01, the theoretical bound ives 305 operators; the empirical result is 311. This closeness is also true for n = Fiure 2 displays the results for Plan-Forward when n = 100. The x-axis is displayed on a linear scale in this raph. In this case, Plan-Forward has hiher eectiveness for hiher number of operators. In apparent contradiction to the analysis for Plan-Forward, the eect of the number of oals appears to be linear. For example, the 99% eectiveness level appears to vary linearly from about 2000 for = 2 to about 6000 for = 100. The apparent contradiction can be resolved by lookin closer at the theoretical bound of Theorem 5, i.e., e r e s=n (2n=s + 1)(ln =), where for this data r = s = 2 and n = 100. As the number of oals increases from 2 to 100, ln = with = 0:01 only doubles, but note that e s=n increases by more than a factor of 7, with most of the increase occurrin for > 50. This is more than sucient to account for the three-fold increase observed for the 99% eectiveness level. Because e s=n more than doubles when > n=2 while ln does not even increment by 1 (no matter what n is), a similar eect should be present for hiher values of n. In this case, the theoretical bound (which is over for = 100 and = 0:01) exceeds the empirical values by a very pessimistic marin. This is also true for n = Fiure 3 shows the combined performance of the two alorithms for n = 100. For this raph, the y-axis is iven a loarithmic scalin and is cut o at 100 to increase readability. There is more than 99% eectiveness at the top and bottom of the raph with minima in the middle. The minimum eectiveness of the alorithms steadily decreases as more oals are 21