SRC Technical Note June 17, Tight Thresholds for The Pure Literal Rule. Michael Mitzenmacher. d i g i t a l


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1 SRC Techical Note Jue 17, 1997 Tight Thresholds for The Pure Literal Rule Michael Mitzemacher d i g i t a l Systems Research Ceter 130 Lytto Aveue Palo Alto, Califoria Copyright cdigital Equipmet Corporatio All rights reserved
2 Abstract We cosider the threshold for the solvability of radom ksat formulas (for k 3) usig the pure literal rule. We demostrate how this threshold ca be foud by usig differetial equatios to determie the appropriate limitig behavior of the pure literal rule. 1 Itroductio We cosider the problem of the performace of the pure literal rule i solvig a radom kcnf satisfiability problem for k 3. The probability space k m, is the set of all formulas i cojuctive ormal form i variables with m clauses each cotaiig k literals. For example, the followig formula is a member of 4,3 : (x 1 x ) (x 3 x ) (x x 3 ) ( x 1 x 3 ). Whe we speak of a radom formula, or more specifically of radom k CNF formula with m clauses ad variables, we shall mea a formula chose uiformly at radom from k m,. Note that a alterative way of thikig of such a formula is that we radomly fill each of the mk holes with oe of the literals, each chose idepedetly ad uiformly at radom. The pure literal rule is a heuristic for satisfyig a CNF formula that works as follows. A pure literal is oe whose complemet does ot appear i the formula (ote that a literal ad its complemet ca both be pure, i our uderstadig). As log as there is a pure literal available, set a pure literal to the value 1 (true), remove all clauses cotaiig that literal, ad cotiue. The pure literal rule is the most coservative strategy, i that it oly assigs a value to a variable that will obviously maitai the satisfiability of the formula. Threshold behavior for the pure literal rule has bee studied by Broder, Frieze, ad Upfal [1], who foud that for sufficietly large a radom 3 CNF formula with (approximately) 1.63 clauses ca be solved by the pure literal rule with high probability, ad a radom 3CNF formula with 1.7 clauses is ot solvable by the pure literal rule with high probability. We expad upo their work here by fidig a exact threshold by cosiderig the limitig behavior as usig differetial equatios. These equatios ca also be thought of as describig the expected behavior of the system for large fiite values of. (For more o this approach, see also for example [3, 4, 5, 6, 7, 8].) The questio of the performace of the pure literal rule is related to the more geeral questio of fidig thresholds for the satisfiability of radom ksat formulas; see, for example, []. We ote that this prelimiary ote sketches the developmet of the appropriate differetial equatios ad their solutio. The justificatio that these differetial equatios accurately represet the behavior of the pure literal rule 1
3 is ot fully clarified, although it is easily checked usig argumets from [1] ad [4] or [8], for example. A full versio to be prepared i cojuctio with the authors of [1] will provide more complete details. The Equatios We shall thik of the pure literal rule i the followig maer: at each time step, if there is a pure literal available, a pure literal is chose uiformly at radom from all pure literals. That literal ad its egatio are the deleted (removed from cosideratio), ad all the clauses cotaiig the pure literal are deleted. I the followig, all the variables are scaled by a factor of, the umber of variables i the formula. This is useful i writig the appropriate differetial equatios. Hece (as will be see below) if we iitially have 10 clauses, we will represet this by a variable with value 10. We shall describe the pure literal rule as a process ruig from time 0 to 1. The variables we shall use are fuctios of time described as follows: L(t) : the scaled umber of udeleted literals remaiig X i (t) : the scaled umber of udeleted literals appearig i times i the formula C(t) : the scaled umber of clauses; that is, the umber of clauses remaiig divided by A(t) : the average umber of clauses i which a literal chose uiformly at radom from all literals appears We may drop the explicit depedece o t whe the meaig is clear. Note that X 0 (t) is the scaled umber of pure literals at time t. IfX 0 (t) = 0 while C(t) >0, the the pure literal rule fails to fid a solutio; the pure literals have ru out while clauses still remai. If, however, X 0 goes to 0 oly as C goes to 0 (ad ecessarily as t goes to 1), the the pure literal rule will succeed o a radom formula with high probability. Hece our goal is to determie how X 0 behaves as we vary the ratio m/. I particular, we shall show that for some costat c k, X 0 stays above 0 o the iterval t [0, 1) if m/ < c k ad it falls below 0 for some t < 1ifm/ > c k. We shall ow determie equatios that describe the limitig behavior as ad m/ is held fixed. The iitial values for C ad X i are easily determied: C(0) = m mk ad, lettig µ =, X i(0) = e µ µ i, sice i the i! limit as goes to ifiity, the distributio of the umber of times a literal approaches the Poisso distributio. To set up the differetial equatios, we assume for each block of time (which ca be though of as 1/) we choose a pure literal uiformly at radom ad remove it ad its egatio. Note that this assumes that X 0 (t) >0, ad the differetial equatios do ot hold oce X 0 (t) 0. I fact whe X 0 = 0 the system stops.
4 It is clear that dl =, sice at each step, two literals are removed. Hece, as L 0 =, we have L = t. Whe a radom pure literal is chose, the expected umber of times it appears i the formula is simply the average umber of clauses a radom variable appears i (up to a O( 1 ) additive error). Oe ca see this by otig that ay give pure literal is equally likely to be ay of the remaiig variables; the fact that its egatio appears 0 times does ot affect the coditioal distributio of its umber of appearaces, give the curret state (X 0 (t), X 1 (t),...). (The O( 1 ) discrepacy is caused by the fact that a pure literal is slightly less likely that a radom literal to appear 0 times, as we kow that oe literal, its egatio, appears 0 times; this, however, oly chages thigs by a O( 1 ) term, which ca be safely dismissed i the limit as. From ow o, we igore this discrepacy i establishig the differetial equatios.) Hece dc i 0 = A = ix i. L Makig use of the idetity = i 0 ix i, which expresses the total umber of remaiig variables i the formula i two differet ways, ad our kowledge of the form of L, we may rewrite this as dc = t, from which it is easily derived that C = C 0 (1 t) k/. The equatios describig the behavior of the X i are slightly more complex. First, ote that the pure literal deleted durig a time step appears i times with probability Xi. Now, suppose the pure literal occurs j times. The we L lose a literal that appears i times wheever oe of j clauses cotaiig that variable cotais a literal that appears exactly i times. Note that there are j (k 1) variables deleted, as there are k 1 variables per clause (1 variable for each clause is take by the pure literal!). The probability that each such variable is oe that appears i times is ixi. (Agai, ote that we have here igored additive O( 1 ) terms, such as whe a two appearaces of a literal are deleted.) Hece the expected loss of literals of size i is Xi Ai Xi L. Oe ca similarly determie the expected gai i X i durig a time step from all literals that appear i + 1 times ad have 1 appearace deleted. The result yields: dx i = A(k 1)iX i + A(k 1)(i + 1)X i+1 X i L for i 1. Note the case of X 0 is special, sice we always remove the egatio of a pure literal, which by defiitio appears 0 times, at each step: dx 0 = A(k 1)X 1 X 0 L 1. 3
5 3 The Solutio Recall that, oce X 0 = 0, the process stops. Hece our goal is to determie a explicit equatio for X 0, ad use it to determie what values of m guaratee that X 0 > 0 for t [0, 1). Oce we have solved this determiistic case give by the differetial equatios, we ca use this iformatio to make statemets regardig the limitig case of the radom process as. (Note that, for techical reasos, we also require k 3; see Lemma 4.4 of [1].) For the equatios below, we use c = m which is a fixed costat. Oe may check that the solutios for the X i, i 1, are give by the followig formulas: where X i (t) = ( ) C j (k 1)/k λ i, j (1 t) 1/, c j=i λ i, j = ( ck ) j ( ( 1) i+ j j) i. j! X 0 ca be solved for explicitly, or by otig that X 0 = L i 1 X i, yieldig X 0 (t) = t i=1 ( ) C j (k 1)/k λ i, j (1 t) 1/. c j=i We ow fid a coveiet form for X 0 (t): ( ) C j (k 1)/k X 0 (t) = t λ i, j (1 t) 1/ c i=1 j=i ( ck ) j = (1 t) [(1 ( 1/ t) 1/ ( 1) i+ j j) ] i (1 t) (k 1) j/ j! i=1 j=i ( = (1 t) 1/ j ( ) ) ( ) j ck(1 t) (k 1)/ j (1 t) 1/ ( 1) i+ j i j! j=1 = (1 t) 1/ (1 t) 1/ + j=1 i=1 ( ) ck(1 t) (k 1)/ j ( (1 t) = (1 t) [(1 1/ t) 1/ (k 1)/ ) ] ck 1. Hece, to determie whe X 0 (t) >0, it suffices to examie the expressio ( (1 t) (1 t) 1/ (k 1)/ ) ck 1, 4 j!
6 ad to determie the supremum of the set of all c such that this expressio is positive for all t [0, 1). This ca be foud by fidig the values of c ad t such that the above expressio is 0 at t ad its derivative is 0 at t. This poit must satisfy: ( (1 t) (1 t) 1/ (k 1)/ ) ck 1 = 0 (1 t) 1/ ( (1 t) (k 1)/ ) ck ck(k 1) (1 t) (k 3)/ = 0 4 We use the first equatio above to remove ( the expoetial ) expressio from the secod by otig that it implies exp (1 t) (k 1)/ ck = 1 (1 t) 1/ ad substitutig accordigly. The equatios ca the be solved for c to yield: c = k(k 1)[(1 t) (k )/ (1 t) (k 1)/ ]. This, i tur, yields a coditio o t based solely o k: ( (1 t) 1/ 1 (k 1)((1 t) 1/ 1) ) 1 = 0. This ca easily be solved umerically for the correct t [0, 1) ad i tur for the correct value of c. Usig this framework, we derive Table 1 of values c k, where c k is the appropriate threshold for ksat formula. That is, c k is the umber such that for ay fixed ɛ>0, if we have a radom ksat formula with variables ad (c k ɛ) clauses, with high probability we may fid a solutio usig the pure literal rule, while if we have (c k + ɛ) clauses with high probability the pure literal rule fails to fid a solutio. Refereces [1] A.Z. Broder ad A. M. Frieze ad E. Upfal. O the satisfiability ad maximum satisfiability of radom 3CNF formulas. Joural of Algorithms 0 (1996) pp [] A.M. Frieze ad S. Sue. Aalysis of two simple heuristics o a radom istace of ksat. I Proceedigs of the Fourth Aual ACMSIAM Symposium o Discrete Algorithms (1993) pp [3] B. Hajek. Asymptotic aalysis of a assigmet problem arisig i a distributed commuicatios protocol. I Proceedigs of the 7th Coferece o Decisio ad Cotrol (1988) pp
7 k c k Table 1: The thresholds for the pure literal rule for ksat. These values match simulatios quite well eve for a very small umber of clauses (i the tes of thousads). [4] R. M. Karp ad M. Sipser. Maximum matchigs i sparse radom graphs. I Proceedigs of the d IEEE Symposium o Foudatios of Computer Sciece (1981) pp [5] T. G. Kurtz. Approximatio of Populatio Processes. SIAM (1981) [6] M. Mitzemacher. Load balacig ad desity depedet jump Markov processes. I Proc. of the 37 th IEEE Symp. o Foudatios of Computer Sciece (1996) pp. 13. [7] M. Mitzemacher. The Power of Two Choices i Radomized Load Balacig Ph.D. thesis, Uiversity of Califoria, Berkeley. (September 1996) [8] N.C. Wormald. Differetial equatios for radom processes ad radom graphs. The Aals of Applied Probability 5 (1995) pp
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