Termination of Linear Programs with Nonlinear Constraints

T ermination of Linear Programs with Nonlinear Constrain ts ⋆ Bican Xia, Zhihai Zhang School o f Mathematical Sciences, P eking U niversi ty xb c@math.pku .edu .cn, infzzh@gmail.co m Abstract. In [16] Tiw ari prov ed that termination of linear programs (loops with linear loop cond itions and up dates) o ver the reals is decid- able through Jordan forms and eigenve ctors computation. In [4] Brav er- man pro ved that it is also decidable ov er the integers. In this pap er, w e consider the t ermination of lo ops with p olynomial lo op conditions and linear up dates ov er the reals and integers. First, w e prov e th at the termination of such loops o ver the integers is u ndecidable. S econd, with an assumption, we pro vide an algorithm to decide the termination of a class of suc h programs o ver the reals. Our metho d is similar to that of Tiw ari in spirit but uses different tec hniqu es. Finally , we conjecture th at the termination of linear programs with p olynomial lo op conditions ov er the reals is undecidable in general by redu cin g the p roblem t o another decision prob lem related to num b er theory and ergodic theory , whic h w e guess undecidable. 1 In tr o duction T ermination analysis is an imp o rtant aspect of pro gram v erification. Guaranteed termination of program loops is necessary for man y applica tions, especially those for which unexp ected b ehavior can be ca tastrophic. F or a generic lo op while ( conditions ) { u pdates } , it is well known that the termina tion problem is undecida ble in g eneral, even for a simple class o f p olynomial pro grams [3]. In [2] Blo ndel et al. pro ved that, even when all the co nditions a nd updates are given as piecewise linear functions, the termination of the lo op remains undecidable. In [16] Tiwari prov ed that ter mination of the following pro grams is decidable ov er R (the real n umber s) P 0 : while ( B X > b ) { X := AX } , where A and B ar e resp ectively n × n a nd m × n matric es, B X > b represents a conjunction of linear inequalities ov er the state v ariables X and X := AX ⋆ This w ork is supp orted in part by NKBRPC-2004CB31800 3, NS F C-6057300 7, NSFC-907 18041 and NKBRPC-2005CB32 1902. represents a (deterministic) linear up date of e a ch v ar ia ble. Subsequen tly in [4] Brav er man prov ed that the ter mination of P 0 is dec idable ov er Z (the integers). In this paper , w e consider the pr oblem of termination of the following lo op: P 1 : while ( P ( X ) > 0) { X := AX } , where X = [ x 1 . . . x N ] T is the vector o f state v ar iables o f the pr o gram, P ( X ) = [ P 1 ( X ) P 2 ( X ) . . . P m ( X )] T are p olynomial co nstraints, each P i ( X ) (1 ≤ i ≤ m ) is a p olynomial in Q [ X ] and A is an N × N matrix ov er Q (the r ational num b ers). That is to say , we replace the linea r co nstraints in P 0 with p oly no mial cons traints and k eep linear updates unc hange d. There ar e some well known tec hniques for deciding termina tion o f some s p e- cial kinds of progra ms. Ra nking functions are most often used for this purp ose. A ra nking function for a lo op maps the v alues of the lo o p v aria bles to a well- founded domain; further, the v alues of the map decr ease on each iteratio n. A linear ranking function is a ranking function that is a linear combination of the loo p v aria bles and constants. Recently , the synthesis of ranking functions draws incr e asing a tten tio n, and some heuristics concerning how to automati- cally generate linear ra nk ing functions for linear programs hav e b een prop os ed in [8,9,12]. In [12] P o delski et al. provided a n efficien t and complete sy nthe- sis metho d based on linear progra mming to co nstruct linea r r anking functions. In [5] Chen et al. prop ose d a metho d to genera te non-linear ranking functions based on semi-algebraic system s olving. How ever, existence of ranking function is only a sufficient condition on the termination of a progra m. It is not difficult to construct prog rams that ter minate, but do not ha ve ra nking functions. T o s olve the pr oblem of termina tion of P 1 , we do not use the technique o f ranking functions. Our method is similar to that of Tiwari in spirit. O ur main contributions in this pap er a re as follows. First, we prov e that the termination o f P 1 ov er Z is undecidable. Then it is easy to prove that, if “ > ” is replaced with “ ≥ ” in P 1 , terminatio n of the resulted P 1 ov er Z is undecidable either. Seco nd, with an assumption, we provide an algorithm to decide the termination of P 1 ov er R . Finally , we conjecture tha t the termination of P 1 ov er R is undecidable in genera l by constructing a lo op and reducing the pr oblem to another decision problem related to n umber theory and ergo dic theory , whic h w e guess undecidable. The rest of the pap er is organized as follows. Section 2 pro ves the undecid- ability of P 1 and its v ariation over Z . Section 3 introduces our main algorithm. The main steps of the algorithm are o utlined first and some details of the steps are in tro duced separately in sev er al subsectio ns. With an ass umption, w e prove the correctnes s of our algor ithm at the end of Section 3. After presenting our conjecture that the termination of P 1 is generally undecidable in Section 4, we conclude the pap er in Section 5. 2 Undecidabilit y of P 1 o ver Z Definition 1. A lo op with N variables is c al le d terminating over a ring R if for al l the inputs X ∈ R N , it is terminating; otherwise it is c al le d nonterminating . The undecidabilit y of P 1 is obtained by a reduction to Hilb ert’s 10 th problem. Consider the following lo o p: P 2 : while ( x m +1 − f ( x 1 , . . . , x m ) 2 > 0) { X := AX } where X =  x 1 . . . x m +1  T , A = diag(1 , ..., 1 , 1 / 2) is a diagona l ma trix and f ( x 1 , . . . , x m ) is a polyno mial with in teg er co efficients. Lemma 1. F or any input ( x 1 , . . . , x m +1 ) ∈ Z m +1 , P 2 terminates if and only if f ( x 1 , . . . , x m ) do es not have inte ger r o ots. Pr o of. ( ⇒ ) If f ( x 1 , . . . , x m ) has an in teg er ro ot, say ( y 1 , . . . , y m ) , ob viously P 1 do es not terminate with the input Y = ( y 1 , . . . , y m , 1) . ( ⇐ ) If f ( x 1 , . . . , x m ) has no integer roo ts, for an y given X ∈ Z m +1 , − f ( x 1 , . . . , x m ) 2 is a fixed negative n umber . Because ( x 1 , ..., x m ) will nev er b e changed and (1 / 2) n → 0 as n → + ∞ , the loo p will terminates after sufficien tly large n iterations. Theorem 1. T ermination of P 1 over Z is u nde cidable. Pr o of. Be c a use the existence of an int eger ro o t of a n a rbitrary Diophantine equation is undecidable, the termination of P 1 ov er Z is undecidable according to Lemma 1. If “ > ” is substituted with “ ≥ ” in P 1 , the loo p be comes P ′ 1 : while ( P ( X ) ≥ 0) { X := AX } . Analogously , we denote by P ′ 2 the lo o p o btained by substituting “ ≥ ” for “ > ” in P 2 . It is easy to see that Lemma 1 still holds for P ′ 2 . Then we get the following theorem. Theorem 2. T ermination of P ′ 1 over Z is u nde cidable. 3 Relativ ely Complete Algorithm for T ermination of P 1 o ver R T o decide whether P 1 terminates on a given input X ∈ R N , it is natura l to consider a gener a l expression of A n X , for instance, a unified for mu la expr essing A n X for any n . If o ne has suc h unified formula of A n X , o ne can express the v alues of P ( X ) after n iterations. Then, for each e le men t of P ( X ) (each constraint) , i.e. , P i ( X ), one ma y try to determine whether P i ( X ) > 0 as n → + ∞ b y guessing the dominant term o f P i ( X ) and deciding the sign of the term. That is the main idea of our algo r ithm w hich will b e describ ed formally in Subsection 3.2. A t several main steps of our algorithm, a few techniques and results in n umber theor y and erg o dic theo ry are needed. F o r the sake of clarity , the details are introduced subsequently in the next subsections. The first subsection is devoted to expr essing A n X in a unified formula. 3.1 General Expression for A n X by Generating F unction In this subsectio n the general expressio n of A n X will be deduced with generating function, not Jorda n for m. Lemma 2. [1 3] L et α 1 , . . . , α d b e a se quenc e of c omplex nu mb ers, d ≥ 1 and α d 6 = 0 . The fol lowing c onditions on a function f : N → C ar e e qu ivalent to e ach other: i. P n ≥ 0 f ( n ) x n = P ( x ) Q ( x ) , wher e Q ( x ) = 1+ α 1 x + . . . + α d x d and P ( x ) is a p olynomial in x of de gr e e less than d. ii. F or al l n ≥ 0 , f ( n + d ) + α 1 f ( n + d − 1) + α 2 f ( n + d − 2) + . . . + α d f ( n ) = 0 . iii F or al l n ≥ 0 , f ( n ) = k P i =1 P i ( n ) γ n i , and Q ( x ) = 1 + α 1 x + α 2 x 2 + . . . + α d x d = k Q i =1 (1 − γ i x ) d i , wher e t he γ i ’s ar e distinct, and P i ( n ) is a p olynomial in n of de gr e e less than d i . Corollary 1. A is a d × d squar e matrix with its entries in Q . Supp ose that the char acteristic p olynomial of A is D ( x ) = x d + α 1 x d − 1 + . . . + α d − s x s , wher e α d − s 6 = 0 and s ≥ 0 . Define f ( n ) = A n + s X and let f j ( n ) b e the j - th c omp onent of f ( n ) . Then for e ach j , f j ( n ) c an b e expr esse d as f j ( n ) = k X i =1 p j i ( n ) ξ n i , (1) wher e ξ i ’s ar e al l the distinct nonzer o c omplex eigenvalues of A and p j i ( n ) is a p olynomial in n of de gr e e less t han t he multiplicity of ξ i . Pr o of. Fir s t, A d + α 1 A d − 1 + · · · + α d − s A s = 0 b ecause D ( x ) is the characteristic po lynomial of A . So, for any n ≥ 0 , f ( n + d − s ) + α 1 f ( n + d − ( s + 1)) + · · · + α d − s f ( n ) = A n + d X + α 1 A n + d − 1 X + · · · + α d − s A n + s X = ( A d + α 1 A d − 1 + · · · + α d − s A s ) A n X = 0 . Thu s, for each j , f j ( n + d − s ) + α 1 f j ( n + d − ( s + 1)) + · · · + α d − s f j ( n ) = 0 . By Lemma 2, f j ( n ) = k P i =1 p j i ( n ) ξ n i and Q ( x ) = 1 + α 1 x + · · · + α d − s x d − s = k Q i =1 (1 − ξ i x ) d i where p j i ( n ) is a po lynomial in n o f degree less than d i . It’s easy to s ee that x = 0 is not a solution of Q ( x ) and P k i =1 d i = d − s . Because D ( x ) = x d Q ( 1 x ) = x d − s + α 1 x d − s − 1 + . . . + α d − s = x d k Y i =1 (1 − ξ i x ) d i = x s k Y i =1 ( x − ξ i ) d i , ξ i ’s are all the distinct nonz e r o complex eigenv alues of A and d i is the multiplicit y of ξ i . That completes the pro of. R emark 1. According to Cor ollary 1, we may compute the general expressio n of A n X as follows. First, compute all the complex eig env alues of A and their m ultiplicities. Second, supp ose each f j ( n ) of f ( n ) is in the for m of eq.(1) where the co efficients of p j i are to b e computed. Third, compute f (1) , . . . , f ( d ) , and obtain a set of linear equations by comparing the coefficie nts of the resulted f j ( i ) (1 ≤ i ≤ d ) to those of eq.(1). Finally , b y so lving those linear equations, w e can obtain f j ( n ) and f ( n ) . Example 1. Let’s consider the following loop: while ( x 5 + x 2 1 + x 1 x 2 − x 2 3 − 2 x 3 x 4 − x 2 4 > 0) { X := AX ; } , where A = 2 6 6 6 6 4 1 − 2 5 0 0 0 2 1 5 0 0 0 0 0 0 2 0 0 0 − 1 2 − 1 0 0 0 0 0 − 1 2 3 7 7 7 7 5 . W e sha ll show how to compute the general ex pr ession of A n X by tak ing use of Corollar y 1. The characteristic po lynomial of A is D ( λ ) = λ 5 + 3 10 λ 4 + 7 10 λ 3 + 1 5 λ 2 + 9 10 λ + 1 2 = ( λ + 1 2 )( λ 2 − 6 5 λ + 1)( λ 2 + λ + 1) . The eigenv alues of A are ξ 1 = 3 + 4 i 5 , ξ 2 = 3 − 4 i 5 , ξ 3 = − 1 2 + √ 3 2 i , ξ 4 = − 1 2 − √ 3 2 i , ξ 5 = − 1 2 . Set f ( n ) = A n X = ˆ f 1 ( n ) f 2 ( n ) f 3 ( n ) f 4 ( n ) f 5 ( n ) ˜ T . F or an y n > 0 , we have f ( n + 5) + 3 10 f ( n + 4) + 7 10 f ( n + 3) + 1 5 f ( n + 2) + 9 10 f ( n + 1) + 1 2 f ( n ) = 0 . Because the multiplicities of ξ 1 , ξ 2 , ξ 3 , ξ 4 and ξ 5 are all 1, by Coro llary 1, we may assume for 1 ≤ j ≤ 5 f j ( n ) = ( 5 X i =1 a j i x i ) ξ n 1 + ( 5 X i =1 b j i x i ) ξ n 2 + ( 5 X i =1 c j i x i ) ξ n 3 + ( 5 X i =1 d j i x i ) ξ n 4 + ( 5 X i =1 e j i x i ) ξ n 5 . Let f ( 1) , f (2) , f (3) , f (4) and f (5) eq ua l to AX, A 2 X, A 3 X, A 4 X a nd A 5 X r esp ec- tively , and by solving some linear equations (see Remark 1) we can obtain f 1 ( n ) = „ 2 − i 4 x 1 + i 4 x 2 « ξ n 1 + „ 2 + i 4 x 1 − i 4 x 2 « ξ n 2 , f 2 ( n ) = „ − 5 i 4 x 1 + 2 + i 4 x 2 « ξ n 1 + „ 5 i 4 x 1 + 2 − i 4 x 2 « ξ n 2 , f 3 ( n ) = „ ( 1 2 − √ 3 i 6 ) x 3 − 2 √ 3 i 3 x 4 « ξ n 3 + „ ( 1 2 + √ 3 i 6 ) x 3 + 2 √ 3 i 3 x 4 « ξ n 4 , f 4 ( n ) = „ √ 3 i 6 x 3 + ( 1 2 + √ 3 i 6 ) x 4 « ξ n 3 + „ − √ 3 i 6 x 3 + ( 1 2 − √ 3 i 6 ) x 4 « ξ n 4 , f 5 ( n ) = x 5 ξ n 5 . 3.2 Main Algorithm According to subsection 3.1, the general expression of A n + m X is a p olynomial in x 1 , . . . , x N , n and ξ 1 , . . . , ξ q , the nonzero complex ro ots of D ( x ). If we substitute A n + m X for X in P ( X ) and deno te the resulted P j ( X )(1 ≤ j ≤ m ) ∈ P ( X ) b y P j ( X, n ), then P j ( X, n ) can b e written as P j ( X, n ) = p j 0 ( X, n ) + p j 1 ( X, n ) η n 1 + . . . + p j M ( X, n ) η n M , (2) where η k (1 ≤ k ≤ M ) is the pro duct of some ξ j ’s. T o determine whether P 1 terminates, we hav e to determine for each j whether P j ( X, n ) > 0 holds for all n . T o this end, it is sufficient to know whether the dominant term ( le ading term ) of eq.(2) is p ositive or not a s n → + ∞ . I n the following, we shall give a more deta ile d description of eq .(2 ) so that we can obtain the expression of the leading term of eq.(2). Let η k = r k e α k 2 π i , where i = √ − 1 and r k is the mo dulus of η k . Without loss of gene r ality , w e assume r 1 < r 2 < · · · < r M . F o r conv e nience, set η 0 = r 0 = 1. Rewrite P j ( X, n ) as P j ( X, n ) = p j 0 ( X, n ) r n 0 + p j 1 ( X, n ) e nα 1 2 π i r n 1 + · · · + p j M ( X, n ) e nα M 2 π i r n M . Suppo se T j is the common p erio d of all the e α q 2 π i (1 ≤ q ≤ M ) where α q is a rational n um be r . Definition 2. F or e ach j (1 ≤ j ≤ m P l =1 T l ) , if j = s − 1 P l =1 T l + i and 1 ≤ i ≤ T s , then define G j ( X, n ) , P s ( X, T s n + i − 1) . Notation 1 F or e ach j ( 1 ≤ j ≤ m P l =1 T l ) , exp and G j ( X, n ) , c ol le ct the r esult with r esp e ct to (w.r.t.) n l r n k , and let C j kl ( X, n ) denote the c o efficient of the term n l r n k . Then G j ( X, n ) can b e wr itten as C j 10 ( X, n ) r n 1 + C j 11 ( X, n ) nr n 1 + · · · + C j 1 d 1 ( X, n ) n d 1 r n 1 + · · · (3) + C j M 0 ( X, n ) r n M + C j M 1 ( X, n ) nr n M + · · · + C j M d M ( X, n ) n d M r n M , where d l (1 ≤ l ≤ M ) is the greatest degree of n in G j ( X, n ) w.r.t. r l . It can be deduced that if r i < r j , the order o f n l 1 r n i is less than the or der of n l 2 r n j for any l 1 and l 2 when n go es to infinit y . Similar ly , if l 1 < l 2 , the order of n l 1 r n i is less than the order of n l 2 r n i . So, it is natural to in tro duce an or dering on the terms n l r n j as follows. Definition 3. We define n l 1 r n i ≺ n l 2 r n j if r i < r j or r i = r j and l 1 < l 2 . A term C j kl n l r n j in e q. (3) is said to b e the leading term and C j kl the leading co efficient if n l r n j o c curring in C j kl is the lar gest one under that or dering ≺ . Suppo se G j ( X, n ) = P s ( X, T s n + t ) for some s a nd t . F or those e α q 2 π i ’s wher e α q ’s are rationa l n umber s, e ( T s n + t ) α q 2 π i = e tα q 2 π i bec ause T s is the commo n p e- rio d. Because there ma y b e so me e ( T s n + t ) α q 2 π i ’s with irrational α q ’s, each C j kl ( X, n ) can be div ided int o three parts, C j kl ( X, n ) = C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, n ) , where C j kl 0 ( X ) do es not contain an y e ( T s n + t ) α q 2 π i , C j kl 1 ( X ) co nt ains e ( T s n + t ) α q 2 π i with ratio nal α q and C j kl 2 ( X, n ) contains those e ( T s n + t ) α q 2 π i with irra tional α q 1 . F urther, C j kl 2 ( X, n ) can b e wr itten as C j kl 2 ( X, n ) = C j kl 2 ( X, sin( ( nT s + t ) α k 1 2 π ) , cos(( nT s + t ) α k 1 2 π ) , . . . , sin(( nT s + t ) α ks k 2 π ) , cos(( nT s + t ) α ks k 2 π )) , where { α k 1 2 π , . . . , α ks k 2 π } is a maximum r ational ly indep endent g roup 2 . Example 2. W e contin ue to use the lo op in Exa mple 1 to illustra te the ab ov e concepts and notations. Because | ξ 1 | = | ξ 2 | = 1 , let ξ 1 = e α 1 2 π i and ξ 2 = e − α 1 2 π i , where α 1 2 π is the argument of ξ 1 . It’s not difficult to chec k that α 1 is an irr a tional num b er. 3 F or the sake of clarity , in the following we firstly reduce the expr e s sions of f 1 ( n ) , f 2 ( n ) , f 3 ( n ) , f 4 ( n ) and f 5 ( n ) , a nd then substitute them in the lo op g uard. Let α 2 = 1 3 , then ξ 3 = e α 2 2 π i , ξ 4 = e − α 2 2 π i , and f 1 ( n ) , f 2 ( n ) , f 3 ( n ) , f 4 ( n ) and f 5 ( n ) can be r ewritten as f 1 ( n ) = x 1 cos( nα 1 2 π ) + x 1 − x 2 2 sin( nα 1 2 π ) , f 2 ( n ) = x 2 cos( nα 1 2 π ) + 5 x 1 − x 2 2 sin( nα 1 2 π ) , f 3 ( n ) = x 3 cos( nα 2 2 π ) + √ 3 x 3 + 4 √ 3 x 4 3 sin( nα 2 2 π ) , f 4 ( n ) = x 4 cos( nα 2 2 π ) − √ 3 x 3 + √ 3 x 4 3 sin( nα 2 2 π ) , f 5 ( n ) = x 5 ( − 1 2 ) n . Substituting f 1 ( n ) , f 2 ( n ) , f 3 ( n ) , f 4 ( n ) and f 5 ( n ) for x 1 , x 2 , x 3 , x 4 and x 5 resp ec- tively in the lo op guar d, w e get that the res ulted lo op guard is L 1 r n 1 + ( L 21 + L 22 + L 23 ) r n 2 , where r 1 = 1 2 , r 2 = 1 , and L 1 = ( − 1) n x 5 , L 21 = − x 2 3 + 2 x 3 x 4 + 4 x 2 4 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 , L 22 = 2 x 2 4 − 2 x 3 x 4 − x 2 3 2 cos( n 2 α 2 2 π ) − ( √ 3 x 3 x 4 + √ 3 x 2 4 ) sin( n 2 α 2 2 π ) , L 23 = − x 2 1 + x 2 2 − 6 x 1 x 2 4 cos( n 2 α 1 2 π ) + 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 sin( n 2 α 1 2 π ) . 1 Later it will b e proved th at C j kl 0 ( X ) , C j kl 1 ( X ) and C j kl 2 ( X, n ) are reals for any n . 2 “Rationally independ ent” will be d escrib ed later. 3 A general alg orithm for chec king whether an argumen t is a rational multiple of π will b e stated in detail in sub section 3.4. Since α 2 = 1 3 , the perio d of ξ 2 3 = e 2 α 2 2 π = cos(2 α 2 2 π ) + sin(2 α 2 2 π ) i is 3. Then we compute G 1 ( X, n ) = ( ( − 1) 3 n x 5 ) r 3 n 1 + C 120 r 3 n 2 , G 2 ( X, n ) = ( ( − 1) 3 n +1 x 5 ) r 3 n +1 1 + C 220 r 3 n +1 2 , G 3 ( X, n ) = ( ( − 1) 3 n +2 x 5 ) r 3 n +2 1 + C 320 r 3 n +2 2 . T ake G 1 ( X, n ) as an ex a mple. C 120 = C 1200 + C 1201 + C 1202 , C 1200 = − x 2 3 + 2 x 3 x 4 + 4 x 2 4 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 , C 1201 = 2 x 2 4 − 2 x 3 x 4 − x 2 3 2 , C 1202 = − x 2 1 + x 2 2 − 6 x 1 x 2 4 cos(3 n 2 α 1 2 π ) + 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 sin(3 n 2 α 1 2 π ) . Notation 2 We denote by C j kl ( X, n ) ≻ 0 (and c al l C j kl ( X, n ) “p ositive”) if min { C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 ) } > 0 subje ct t o { y 2 11 + y 2 12 = 1 , · · · , y 2 s k 1 + y 2 s k 2 = 1 } . R emark 2. It is not difficult to see C j kl ( X, n ) ≻ 0 iff ∀ ( y 11 , y 12 , . . . , y s k 1 , y s k 2 ) ( ( y 2 11 + y 2 12 = 1) ∧ · · · ∧ ( y 2 s k 1 + y 2 s k 2 = 1) → C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 ) > 0) bec ause { ( y 11 , y 12 , . . . , y s k 1 , y s k 2 ) : y 2 i 1 + y 2 i 2 = 1 , i = 1 , ..., s k } is a bounded closed set. If ≻ and > ar e repla ced with  and ≥ resp ectively in the notatio n, w e get the notation of C j kl ( X, n )  0 (“no nnegative”). Roughly sp ea k ing, fo r a ny G j ( X, n ) , if its leading co efficient C j kl ( X, n ) ≻ 0 , there ex is ts an integer N 1 such that for a ll n > N 1 , G j ( X, n ) > 0 . If all the leading co efficients of all the G j ( X, n ) ’s are “p o sitive”, there exists N ′ such that for all n > N ′ , a ll the G j ( X, n ) ’s ar e p ositive. Ther efore, P 1 is non terminating with input X ′ := A N ′ X . On the o ther hand, if P 1 is nonterminating, do es there exist an input X s uch that the leading co efficients of a ll the G j ( X, n ) ’s a re “ p o sitive”? W e do not know the answer yet . How ever, w ith an assumption describ ed b elow, the answer is yes. Assump tion for the mai n algorithm : F or any X ∈ R n and any C j kl ( X, n ) , C j kl 2 ( X, n ) being not identically zero implies min( C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 )) 6 = 0 sub ject to y 2 i 1 + y 2 i 2 = 1 ( i = 1 , . . . , s k ) . It is not difficult to se e that the as sumption is e q uiv a lent to the following formula: [ ∀ X ∀ Y ( C j kl 2 ( X, n ) ≡ 0 ∨ V 1 ≤ i ≤ s k y 2 i 1 + y 2 i 2 = 1 → C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, Y ) > 0) i W [ ∀ X ∃ Y ( C j kl 2 ( X, n ) ≡ 0 ∨ V 1 ≤ i ≤ s k y 2 i 1 + y 2 i 2 = 1 → C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, Y ) < 0) i . Because C j kl 2 ( X, n ) can b e written as P i ∈ I f i 1 ( X ) sin( nα i 2 π ) + f i 2 ( X ) cos( nα i 2 π ) , where I is an index set, C j kl 2 ( X, n ) ≡ 0 is equiv a lent to V i ∈ I f i 1 ( X ) = f i 2 ( X ) = 0 . Thu s, the assumption can b e chec ked with r eal q ua nt ifier elimination tec hniques. Example 3. F or those C ij k ’s in E xample 2, let’s chec k whether they satisfy the assumption. F or clar ity , w e present a clear pro of here rather than make use of any to o l for real quantifier elimination. T ake G 1 ( X, n ) for exa mple. It’s clear that C 1202 ( X, n ) ≡ 0 if and only if x 1 = x 2 = 0 . It is not difficult to compute D = inf n ≥ 1 C 1202 ( X, n ) = − s „ − x 2 1 − x 2 2 + 6 x 1 x 2 4 « 2 + „ 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 « 2 . If x 1 = x 2 = 0 does not ho ld, 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 + D < 0 b ecause ( 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 ) 2 − D 2 = − 1 16 (5 x 2 1 + x 2 2 − 2 x 1 x 2 ) 2 < 0 . Let M = min y 2 11 + y 2 12 =1 ( C 1200 ( X ) + C 1201 ( X ) + C 1202 ( X, y 11 , y 12 )) , then M = − ( x 3 + x 4 ) 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 + D < 0 if x 1 = x 2 = 0 do es not hold. Consequently , G 1 ( X, n ) satisfies the assumption. Similarly it can b e prov ed G 2 ( X, n ) and G 3 ( X, n ) satisfy the assumption to o . Thu s the lo op in Example 1 satisfies the assumption. W e s hall sho w in nex t section how ha rd it is to deal with the case tha t the assumption do es not hold. Now, w e are re a dy to describ e our main algo rithm. F or the sake of br evity , the algorithm is describ ed as a nondeterministic alg orithm. The basic idea is to g ue s s a leading ter m for ea ch G j ( X, n ) fir s t. Then, setting its co efficient be “p o s itive” a nd the co efficients of the terms with higher or der b e “nonnegative”, we can g et a semi-algebraic system (SAS). If o ne of our guess is satisfiable, i.e. , one of the SASs ha s so lutions, P 1 is nonterminating. Otherwise, it is terminating. Algorithm Terminati on Step 0 Co mpute the general expressio n of A n + m X . Step 1 Substitute A n + m X for X in P ( X ) , and compute a ll G j ( X, n ) (finite ma ny , say , j = 1 , ..., L ). Step 2 Gues s a leading term for each G j ( X, n ) , sa y C j k j l j n l j r n k j . Step 3 Co nstruct a semi-algebra ic sy stem S as follows. S j = C j k j l j ≻ 0 ∧ V ( k>k j ) ∨ ( k = k j ∧ l>l j ) C j kl ( X, n )  0 , S = V L j =1 S j . Step 4 If one of these systems is satisfia ble, return ”nonterminating”. Otherwise return ”terminating”. R emark 3. If t he assumpt ion for the main algor ithm do es not hold, then Termin ation is incomplete. That is, if it returns “nonterminating”, the lo o p is nonterminating. O therwise, it tells nothing. Example 4. F or the lo op in Example 1, we ha ve computed the G i ( X, n ) ’s in Example 2 and verified that it s atisfies the as sumption o f the main algo rithm in E xample 3. W e shall finish its ter mination decision in this example, following the steps in Termin ation . By Steps 2 and 3 of Termin ation , we sho uld guess leading terms and con- struct SASs according ly . T o be concrete, let’s take as an example one certain guess and suppo se w e hav e the following se mi-algebra ic system 8 < : C 120 ( X, n ) ≻ 0 , C 220 ( X, n ) ≻ 0 , C 320 ( X, n ) ≻ 0 . According to No ta tion 2 a nd Remark 2, the ab ov e inequa lities are equiv alent to ( ∀ y 11 , y 12 ) y 2 11 + y 2 12 = 1 → 8 > < > : − ( x 3 + x 4 ) 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 − x 2 1 + x 2 2 − 6 x 1 x 2 4 y 11 + 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 y 12 > 0 , − ( x 4 − x 3 2 ) 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 − x 2 1 + x 2 2 − 6 x 1 x 2 4 y 11 + 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 y 12 > 0 , − ( x 4 + x 3 2 ) 2 + 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 − x 2 1 + x 2 2 − 6 x 1 x 2 4 y 11 + 7 x 2 1 − x 2 2 − 2 x 1 x 2 4 y 12 > 0 . In E xample 3, we hav e shown that 5 x 2 1 + x 2 2 − 2 x 1 x 2 4 + D ≤ 0 . Th us, the ab ov e pred- icate for mula do es not hold. In fac t, none of the fo rmulas obta ined in Step 3 holds. Th us, the lo op in Example 1 is terminating. R emark 4. It is w ell known that real qua ntifier elimination is de c ida ble from T arski’s work [15]. Ther efore, the semi-a lgebraic sy stems in Step 3 can be s olved. F or the to ols for solving semi-alg ebraic systems, please be referred to [6,7,10,18]. R emark 5. There are some techniques to decrea se the a mount of computation of the algorithm Terminati on . F or example, we can use Lemma 3 when guessing leading terms for each G j ( X, n ) . Lemma 3. [4 ] L et ξ 1 , ξ 2 , . . . , ξ m ∈ C b e a c ol le ction of distinct c omplex num b ers such that | ξ i | = 1 and ξ i 6 = 1 for al l i . L et α 1 , α 2 , . . . , α m b e any c omplex numb ers and z n = α 1 ξ n 1 + . . . + α m ξ n m . Then one of the fol lowing is true: 1. the r e al p art Re ( z n ) = 0 for al l n ; or 2. ther e is c < 0 such t hat Re ( z n ) < c for infinitely many n ’s. According to Lemma 3, if C j kl 0 = 0 and C j kl ( X, n ) is not identically zero w.r.t. n , then C j kl ( X, n ) can no t b e alw ays nonneg a tive. According to the former discussion { G j − 1 P k =1 T k +1 ( X, n ) , . . . , G j − 1 P k =1 T k + T j ( X, n ) } are obtained fro m P j ( X, n ) . Then the lea ding term with the greatest order among all the leading terms of the above set should not b e o f the form C j k j l j n l j r n k j where C j k j l j 0 = 0. Thus, this should b e a voided when guessing leading ter ms for each G j . In the follo wing s ubsections, we shall ex plain the details of the main steps of Termin ation and prove its correctness . 3.3 Compute the Minim al P olynomials of α + β , α − β , α · β and α β A t Step 1 of the main alg o rithm, we ma y compute η k (1 ≤ k ≤ M ) which are the pro ducts of so me ξ j ’s after substituting the general expre s sion of A n + m X for X in P ( X ) . In order to describ e η k , we need the minimal polynomia ls of η k . In this subsection a method is present ed to solv e a more general problem. In [14] Strzeb onski gav e an a lgorithm to compute { α + β , α − β , α · β , α β } nu merically where α and β are given algebr aic num b er s. Here we wan t to present another more in tuitive metho d based on symbolic computation. In the following let α ⋄ β denote one of { α + β , α − β , α · β , α β } . Without loss of generality let’s assume that the minimal p olynomial o f α is f 1 ( x ) whos e degree is d 1 and the minimal po lynomial of β is f 2 ( x ) who se degree is d 2 . W e can b ound α and β in W 1 and W 2 , respectively , where W 1 , W 2 are “b oxes”, by isolating the complex zer os of f 1 ( x ) and f 2 ( x ). Since the degree of α is d 1 and the degree of β is d 2 , the degr ee of α ⋄ β is at most d 1 · d 2 . Then there must ex ist x 1 , . . . x d 1 · d 2 +1 in Z such that x 1 + x 2 α ⋄ β + . . . + x d 1 · d 2 +1 ( α ⋄ β ) d 1 · d 2 = 0 . Thu s we c a n design a n algor ithm to enumerate a ll the ( x 1 , . . . , x d 1 · d 2 +1 ) ∈ Z d 1 · d 2 +1 and chec k whether it is a solution. Since there e x ists one solution this algorithm must terminate a nd output a solution. Assume that its output is a 0 , a 1 , . . . , a d 1 · d 2 and f ( x ) = a 0 + a 1 x + . . . + a d 1 · d 2 x d 1 · d 2 . Because f ( α ⋄ β ) = 0 , the minimal p olyno mia l of α ⋄ β is an irreducible factor of f ( x ). F actor f ( x ) in Q . Without loss o f generality , we assume f ( x ) = g 1 ( x ) m 1 · g 2 ( x ) m 2 . . . g d ( x ) m d . W e can chec k whether g j ( x )(1 ≤ j ≤ d ) is the minimal p olynomial of α ⋄ β b y solving the following semi- algebraic system (SAS): { g j ( x ⋄ y ) = 0 , f 1 ( x ) = 0 , f 2 ( y ) = 0 , x ∈ W 1 , y ∈ W 2 } . If it is s a tisfiable, g j ( x ) is the minimal polyno mial of α ⋄ β ; otherwise it is not. Thu s the minimal po lynomial of α ⋄ β can b e obtained. 3.4 Chec k Whe ther the Argument of α Is a Rational Mul ti ple of π A t Step 1 of the main algo r ithm, T s is necessar y for de fining G j ( X, n ) . Th us for a given η k we have to chec k whether its argument is a rational mu ltiple o f π and if it is, we need to know the p e r io d o f η k | η k | . This subsectio n a ims at this problem. Suppo se the minimal polynomia l of α is p ( x ) whose degree is d . Without loss of genera lit y s uppo se α = r e β 2 π i . W e can b ound α in W by isolating all the complex ro o ts of p ( x ). Since the degree of α is d , the degr ee of α must b e d . Then the degree o f α · α = r 2 is a t most d 2 . Thus the degree of r is at mo st 2 d 2 . The degr ee of α − 1 is at most 2 d 2 bec ause the degree of α − 1 is the same as the degree of α . Since the degree of α is d the degree of α · r − 1 = e iβ 2 π is at most 2 d 3 . If β is a rational num b er , α must b e a unit ro o t and its minimal polynomial m ust b e a cyclo tomic po lynomial. As a result if β is a rational nu mber the mini- mal p o ly nomial of α m ust be a c y clotomic polyno mia l whose degree is less tha n or equal to 2 n 3 . All the cyclotomic p olynomia ls can b e computed explicitly a c- cording to the theor y of cyclotomic field. Th us let C P j ( x ) denote the cyclo to mic po lynomial who se deg ree is j . Then, β is a rational n um be r if and only if the following is sa tis fia ble ∃ r (( r 6 = 0) ^ ( 2 d 3 _ j =1 C P j ( x/r ) = 0) ^ ( p ( x ) = 0) ^ ( x ∈ W )) . Because C P j ( x/r ) = 0 ⇐ ⇒ r j C P j ( x/r ) = 0, the ab ov e q uantifier formula is decidable. If it’s satisfiable the minimal p olynomial of α can b e computed by chec king whether ∃ r (( r 6 = 0) ^ ( C P j ( x/r ) = 0) ^ ( p ( x ) = 0) ^ ( x ∈ W )) is satisfiable one b y one. 3.5 Chec k Ratio nal Indep endence Given a set of algebr aic num b ers, α 1 = e β 1 2 π i , . . . , α d = e β d 2 π i , wher e β 1 , . . . , β d are irrationa l n umber s. In this subsection we present a metho d to c heck whether β 1 , . . . , β d are r ational ly indep endent . Definition 4. Irr ational numb ers β 1 , . . . , β d ar e rationally indep endent if ther e do es not exist r ational numb ers a 1 , . . . , a d such that d P j =1 a j β j ∈ Q . Obviously , β 1 , . . . , β d are rationally indep endent if and only if 1 , β 1 , . . . , β d are linea rly indepe ndent in Q . It ca n b e deduced that β 1 , . . . , β d are r ationally independent if and only if ∀ ( b 1 , . . . , b d ) ∈ Z d , d P j =1 b j β j / ∈ Z . Lemma 4. [1 ] L et λ 1 , . . . , λ m with m ≥ 2 b e line arly dep endent lo garithms of algebr aic numb ers. Define α j = e λ j (1 ≤ j ≤ m ) . F or 1 ≤ j ≤ m , let log A j ≥ 1 b e an upp er b ound for max { h ( α j ) , | λ j | D } wher e D is the de gr e e of the numb er field K = Q ( α 1 , . . . , α m ) over Q and h ( α ) denotes t he absolute lo garithmic heig ht of α . Then ther e exist r ational inte gers n 1 , . . . , n m , not al l of which ar e zer o, such that n 1 λ 1 + . . . + n m λ m = 0 and | n k | < (11( m − 1) D 3 ) m − 1 (log A 1 ) ... (log A m ) log A k for 1 ≤ k ≤ m . R emark 6. Baker is the fir s t one to use his tr a nscendence ar g ument s to establish such an estima te. Ho wever, the description here follows Lemma 7.19 in [17]. Sequence { β 1 , . . . , β d , 1 } ar e linearly indep endent in Q iff { β 1 2 π i , . . . , β d 2 π i , 2 π i } are linearly independent in Q . If n 1 β 1 2 π i + . . . + n d β d 2 π i + n d +1 2 π i = 0 , then e n 1 β 1 2 π i · · · e n d β d 2 π i = 1 . That is α n 1 1 · · · α n d d = 1 . According to Lemma 4 w e can decide whether { β 1 , . . . , β d } are rationally independent by enumerating n k from ⌊− (11 dD 3 ) d (log A 1 ) · · · (log A d +1 ) log A k ⌋ , . . . , ⌈ (11( d ) D 3 ) d (log A 1 ) · · · (log A d +1 ) log A k ⌉ for k = 1 , . . . , d and chec king whether α n 1 1 · · · α n d d = 1 . F or an y g iven ( n 1 , . . . , n d ) , whether α n 1 1 · · · α n d d = 1 ca n b e determined b y c hecking whether the fo llowing SAS has solutions. { x n 1 1 · · · x n d d = 1 , q j ( x j ) = 0 , x j ∈ W j , j = 1 , ..., d } , where q j is the minima l poly nomial of α j and W j contains only one complex ro ot of q j for j = 1 , . . . , d . 3.6 Compute the Infim um of C j kl 2 T o prov e the correctness of our main algor ithm, we need some further results. First, let’s introduce a lemma in ergo dic theory . Let S be the unit circumfer- ence. Usually any p oint on S , sa y ( a, b ) , is denoted as a co mplex n umber a + b i . Define π : R m → S m as ( x 1 , . . . , x m ) → ( e x 1 2 π i , . . . , e x m 2 π i ) . Define m-torus T m = { ( e x 1 2 π i , . . . , e x m 2 π i ) | x j ∈ R } and L π ( α ) : T m → T m as ( e y 1 2 π i , . . . , e y m 2 π i ) → ( e ( y 1 + α 1 )2 π i , . . . , e ( y m + α m )2 π i ) . Lemma 5. [1 1] If α ∈ R m , the t r anslation L π ( α ) ( X ) is er go dic iff for al l K ∈ Z m , ( K, α ) / ∈ Z wher e ( K, α ) stands for the inner pr o duct of K and α . According to Lemma 5 w e know tha t if α 1 is a n irra tional num b er, the clo- sure of { e nα 1 2 π i } n ≥ 1 is the unit cir c umference. Also, if α 1 , . . . , α m are rationally independent, L π ( α ) ( X ) is erg o dic. Th us the clos ure of { L n π ( α ) (0) } n ≥ 1 is T m . Lemma 6. If α k 1 , . . . , α ks k ar e ra t ional ly indep endent and C j kl 2 is of the form C j kl 2 ( X, sin( nα k 1 2 π ) , cos( nα k 1 2 π ) , . . . , sin( nα ks k 2 π ) , cos( nα ks k 2 π )) for a fixe d X , t hen inf n ≥ 1 { C j kl 2 ( X, sin( nα k 1 2 π ) , cos( nα k 1 2 π ) , . . . , sin( nα ks k 2 π ) , cos( nα ks k 2 π )) } is e qual t o min { C j kl 2 ( X, x 1 , y 1 , . . . , x s k , y s k ) } subje ct to { x 2 i + y 2 i = 1 , 1 ≤ i ≤ s k } . Pr o of. Acco rding to Lemma 5 for an y ( x 1 , y 1 , . . . , x s k , y s k ) there exists a subse- quence, say { n i } i ≥ 1 , such that lim i → + ∞ (sin( n i α k 1 2 π ) , cos( n i α k 1 2 π ) , . . . , sin( n i α ks k 2 π ) , cos( n i α ks k 2 π )) = ( x 1 , y 1 , . . . , x s k , y s k ) . Theorem 3. L et γ i = ( nT j ′ + j ′′ ) α ki 2 π ( 1 ≤ i ≤ s k ) and supp ose that C j kl 2 = C j kl 2 ( X, sin ( γ 1 ) , cos( γ 1 ) , . . . , sin( γ s k ) , cos( γ s k )) . Then inf n ≥ 1 { C j kl 2 ( X, sin ( γ 1 ) , cos( γ 1 ) , . . . , sin( γ s k ) , cos( γ s k )) } is e qual t o min { C j kl 2 ( X, x 1 , y 1 , . . . , x s k , y s k ) } subje ct to { x 2 i + y 2 i = 1 , 1 ≤ i ≤ s k } . Pr o of. Acco rding to Lemma 5 a nd Lemma 6, it’s sufficien t to prov e that T s k is the clo sure of { ( e γ 1 i , . . . , e γ s k i ) } n ≥ 1 . Beca use { α 1 , . . . , α s k } ar e rationally indepen- dent , { T j ′ α 1 , . . . , T j ′ α s k } are rationa lly independent, to o. Thus, T s k is the closure of { ( e nT j ′ α k 1 2 π i , . . . , e nT j ′ α ks k 2 π i ) } n ≥ 1 . The result of rota ting ( e nT j ′ α k 1 2 π i , . . . , e nT j ′ α ks k 2 π i ) by ( j ′′ α k 1 2 π , . . . , j ′′ α ks k 2 π ) is ( e γ 1 i , . . . , e γ s k i ) . Conseque ntly , T s k is the closure of { ( e γ 1 i , . . . , e γ s k i ) } n ≥ 1 . That completes the pro of. 3.7 Correctness F or each j , P j ( X, n ) can b e wr itten as D j 10 ( X, n ) r n 1 + D j 11 ( X, n ) nr n 1 + . . . + D j 1 d 1 ( X, n ) n d 1 r n 1 + · · · D j H 0 ( X, n ) r n H + D j H 1 ( X, n ) nr n H + . . . + D j H d H ( X, n ) n d H r n H . The D j kl ’s in the ab ov e are real b ecause P j ( X, n ) ∈ R and those n l r n k ’s are of different o rders. Just lik e C j kl , D j kl can be div ided int o three parts, D j kl = D j kl 0 ( X ) + D j kl 1 ( X, n ) + D j kl 2 ( X, n ) . Because D j kl 0 ( X ) contains no e ( nT j ′ + j ′′ ) α q 2 π i ’s, the e ( nT j ′ + j ′′ ) α q 2 π i ’s c o ntained in D j kl 1 ( X, n ) a re p erio dic and the e ( nT j ′ + j ′′ ) α q 2 π i ’s contained in D j kl 2 ( X, n ) are not, D j kl 0 , D j kl 1 and D j kl 2 are all rea l. Since C j kl ( X, n ) results from D j kl ( X, n ) , we get the following lemma . Lemma 7. F or al l n ∈ N and e ach C j kl ( X, n ) = C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, n ) , C j kl 0 ( X ) ∈ R , C j kl 1 ( X, n ) ∈ R and C j kl 2 ( X, n ) ∈ R . Lemma 8. Consider C j kl ( X, n ) = C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, n ) . 1. inf n ≥ 1 { C j kl ( X, n )) } > 0 if and only if min { C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 ) } > 0 subje ct t o { y 2 t 1 + y 2 t 2 = 1 , 1 ≤ t ≤ s k } . 2. If I = C j kl 0 ( X ) + C j kl 1 ( X ) + D < 0 , then ther e is c < 0 s uch that C j kl ( X, n ) < c for infinitely many n ’s, wher e D = min { C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 ) } subje ct t o { y 2 t 1 + y 2 t 2 = 1 , 1 ≤ t ≤ s k } . Pr o of. 1. It has b een prov ed that inf n ≥ 1 C j kl 2 ( X, n ) = min { C j kl 2 ( X, y 11 , y 11 , . . . , y s k 1 , y s k 2 ) } sub ject to { y 2 t 1 + y 2 t 2 = 1 , 1 ≤ t ≤ s k } . Consequently inf n ≥ 1 { C j kl ( X, n )) } = min { C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, y 11 , y 12 , . . . , y s k 1 , y s k 2 ) } . sub ject to { y 2 t 1 + y 2 t 2 = 1 , 1 ≤ t ≤ s k } . 2. Let Y = ( y 11 , y 12 , . . . , y s k 1 , y s k 2 ) and Y ′ = ( y ′ 11 , y ′ 12 , . . . , y ′ s k 1 , y ′ s k 2 ) . D can b e at- tained b ecause { y 2 t 1 + y 2 t 2 = 1 , 1 ≤ t ≤ s k } is a b ounded clo sed s et and C j kl 2 ( X, Y ) is a co nt inu ous function of Y . Assume that D = C j kl 2 ( X, Y ′ ) . Since I < 0 there exists a neighbor ho o d o f Y ′ , say U , such that ∀ Y ∈ U ⇒ C j kl 0 ( X ) + C j kl 1 ( X ) + C j kl 2 ( X, Y ) < I / 2 . Let c = I / 2 and γ i = ( nT j ′ + j ′′ ) α ki 2 π . Because of the density o f { ( e γ 1 i , . . . , e γ s k i ) } n ≥ 1 , there are infinitely many n ’s such that (cos γ 1 , sin γ 1 , . . . , cos γ s k , sin γ s k ) lies in U . Thus there a re infinitely many n ’s such that C j kl ( X, n ) < c . If the main algor ithm, Termin ation , finds one solution, X 0 , the leading co ef- ficient o f G j ( X 0 ) , say C j kl ( X 0 , n ) , satisfies C j kl ( X, n ) ≻ 0 . Accor ding to the defini- tion of ≻ there exist c j > 0 ( j = 1 , . . . , L ) such tha t C j kl ( x, n ) > c j for all n . Thus P 1 is nonterminating. This means tha t if the a lgorithm outputs “ no nt erminat- ing”, then P 1 is nonterminating indeed. On the other hand, if the main algo rithm o utputs “ terminating”, then for any { C j k j l j ( X, n ) , j = 1 , . . . , L } there is a subset V ⊆ { 1 , . . . , L } such that ^ j ∈ V C j k j l j ( X, n ) ≻ 0 is not satisfiable sub ject to V j / ∈ V C j k j l j ≻ 0 ∧ V ( k>k j ) ∨ ( k = k j ∧ l>l j ) C j kl ( X, n )  0 According to th e assump tion for the mai n algorithm , we get that with the ab ov e constraints ∀ j ∈ V , inf n ≥ 1 C j k j l j ( X, n ) ≤ 0 . Thu s by Lemma 8 , ∀ j ∈ V , C j k j l j ( X, n ) is identically zero or there are infinitely many n ’s and so me c < 0 such tha t C j k j l j ( X, n ) < c . Tha t means P 1 is terminating. Therefore, we g et the follo wing theorem. Theorem 4. U nder the assumption of t he main algo rithm, Termi natio n r e- turns “terminating” if and only if P 1 is t erminating. 4 Conjecture In this section, we s hall discuss the general case of P 1 wherein our assumption for the main algor ithm may not hold. Suppo se p ( X ) = p ( x 11 , x 12 , . . . , x m 1 , x m 2 ) ∈ Q [ X ] , and one of the loop condi- tions is p ( X ) 2 > 0 . F ro m the discussio n in Section 3 , we kno w that if we substi- tute A n + m X for X in the co nditions, ther e must b e a p olyno mial q such that the condition bec o mes q ( X , sin( nα 1 2 π ) , cos( nα 1 2 π ) , . . . , sin( nα m 2 π ) , cos( nα m 2 π )) 2 > 0 . Because p is arbitra r y , q c a n b e arbitrar y . F urther, α 1 , ..., α m can be made ratio- nally indep endent because A can b e arbitrary . It’s not har d to see that we can construct a progra m Q such that it is terminating if and only if S q,α , { n : q (sin( nα 1 2 π ) , cos( nα 1 2 π ) , . . . , sin( nα m 2 π ) , cos( nα m 2 π )) = 0 } contains infinitely ma n y elemen ts. F or an y p ( X ) ∈ Q [ X ] the dec is ion pr oblem “whether { X : p ( X ) = 0 } T Z 2 m = ∅ ” is undecidable. Z 2 m is a “regular” set while E = { (sin( nα 1 2 π ) , cos( nα 1 2 π ) , . . . , sin( nα m 2 π ) , cos( nα m 2 π )) } n ≥ 1 is a chaotic set when { α 1 , . . . , α m } a r e rationa lly independent acco rding to the er- go dic theory . Intuitiv ely , deciding whether S p,α = { X : p ( X ) = 0 } T E = ∅ is more difficult than deciding whether { X : p ( X ) = 0 } T Z 2 m = ∅ . So, we intuit ively guess the decision pr oblem “whether S p,α is empty” is undecida ble. F ollowing the same idea, we guess the decision problem “whether S p,α contains infinitely many elemen ts” is m uch mo re difficult and thus undecidable. Thus, we ma ke the following co njecture: Conjec ture. The decis ion problem “whether the lo op P 1 is terminating ov er R ” is undecidable. 5 Conclusion In this pap er we hav e proved tha t termination of P 1 ov er Z is undecidable. Then w e give a relatively complete algor ithm, with an assumption, to determine whether P 1 is terminating o ver R . If the assumption holds, P 1 is terminating iff o ur a lgorithm outputs “terminating”. If the assumption do es not hold, P 1 is nonterminating if the algor ithm outputs “ nonterminating”. W e demo nstrate the main steps of our a lgorithm b y a n example. Finally we show how har d it is to determine the termination of P 1 by reducing its ter mination to the problem of “ whether S p,α has infinite many elements”. W e conjecture the la tter problem is undecidable. Thus, if our conjecture holds, the termination of P 1 ov er R is undecidable. Ac knowledgem en t The authors would like to thank P rof. Lu Y ang for prop os ing the problem to us a nd encoura g ing us to work o n it. W e thank Pro f. Mic hel W aldschmidt for his introduction to Baker’s work. W e a re also gr ateful to Prof. Bob Caviness, Prof. Ho o n Hong a nd Pr of. Daniel Richardson for their insightful suggestio ns . Finally w e w o uld like to thank Prof. Chao chen Z ho u, Pro f. 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