How To Solve The Sum Of Square Roots Problem Over Polynomals

Transcription

1 On the Sum of Square Roots of Polynomals and Related Problems Neeraj Kayal Chandan Saha Abstract The sum of square roots problem over ntegers s the task of decdng the sgn of a non-zero sum, S = n =1 δ a, where δ {+1, 1} and a s are postve ntegers that are upper bounded by N (say). A fundamental open queston n numercal analyss and computatonal (n log N)O(1) geometry s whether S 1/2 when S 0. We study a formulaton of ths problem over polynomals: Gven an expresson S = n =1 c f (x), where c s belong to a feld of characterstc 0 and f s are unvarate polynomals wth degree bounded by d and f (0) 0 for all, s t true that the mnmum exponent of x whch has a nonzero coeffcent n the power seres S s upper bounded by (n d) O(1), unless S = 0? We answer ths queston affrmatvely. Further, we show that ths result over polynomals can be used to settle (postvely) the sum of square roots problem for a specal class of ntegers: Suppose each nteger a s of the form, a = X d + b 1 X d b d, d > 0, where X s a postve real number and b j s are ntegers. Let B = max({ b j },j, 1) and d = max {d }. If X > (B + 1) (n d)o(1) then a non-zero S = n =1 δ a s lower bounded as S 1/X (n d)o(1). The constant n the O(1) notaton, as fxed by our analyss, s roughly 2. We then consder the followng more general problem: gven an arthmetc crcut computng a multvarate polynomal f(x) and nteger d, s the degree of f(x) less than or equal to d? We gve a corp PP -algorthm for ths problem, mprovng prevous results of Allender, Bürgsser, Kjeldgaard-Pedersen and Mltersen (2009), and Koran and Perfel (2007). A prelmnary verson of ths paper appeared n the Proceedngs of the 26th IEEE Conference on Computatonal Complexty (CCC), Mcrosoft Research Inda, neeraka@mcrosoft.com Max Planck Insttute for Informatcs, csaha@mp-nf.mpg.de 1

2 1 Introducton The sum of square roots s the followng well-known problem: gven a set {a 1, a 2,..., a n } of postve ntegers and {δ 1, δ 2,..., δ n } { 1, +1} n, determne the sgn of the sum n δ a (1) =1 It was posed as an open problem by Garey, Graham and Johnson [GGJ76] n connecton wth the Eucldean travellng salesman problem. Eucldean TSP s not known to be n NP but s easly seen to be n NP relatve to the sum of square roots problem. The sum of square roots problem also arses naturally n some other problems nvolvng computatonal geometry (cf. [MR08]) and semdefnte progammng (cf. the survey by Goemans [Goe98]). Although t has been conjectured [Mal96] that the problem les n P, the best known result so far [ABKPM09] s contanment n the countng herarchy CH, whch s a subclass of PSPACE that contans the polynomal herarchy PH. For the related but easer problem of determnng whether the sum (1) s zero or not, a determnstc polynomal tme algorthm s known due to the work of Blömer [Blö91]. (The conference verson of Blömer s paper [Blö91] contans a randomzed polynomal tme algorthm, whch was later derandomzed n [Blö93]). A possble approach towards answerng the sum of square roots queston s to solve the followng number-theoretc problem whose current status s stll a conjecture. Problem 1.1 (Lower boundng a non-zero sgned sum of square roots of ntegers). Gven a sum S = n =1 δ a, where δ {+1, 1} and a s are postve ntegers upper bounded by N, fnd a tght lower bound on S n terms of n and N when S 0. Is t true that for a non-zero S, S 1/2 poly(n,log N) for some fxed polynomal poly( )? If the answer to the above queston s yes then computng the square roots up to poly(n, log N) bts of precson suffces to determne the sgn of the sum of square roots. Now, t s known (cf. [Bre76]) that square root of a number less than N can be computed up to m bts of precson n poly(m, log N) tme, whch together wth an affrmatve answer to the above queston would mply that the sum of square roots problem s n P. In fact, a stronger result s known n ths regard: t was shown by Ref and Tate [RT92] that TC 0 crcuts of polynomal sze can compute both terated sums and square roots up to polynomal bts of precson, whch means that the sum of square roots problem can even be shown to be n the complexty class TC 0 f we manage to solve Problem 1.1 postvely. Reducng the mmense gap between the known upper and lower bounds for Problem 1.1 s a challengng number-theoretc problem. Now, there s a well known analogy between ntegers and polynomals (cf. [EHM05]). We refer the reader to a survey by Immerman and Landau [IL93] for some algorthmc aspects of ths analogy. Whle the polynomal analog of a number-theoretc problem s typcally much easer than the nteger verson, nevertheless the nvestgaton of the polynomal verson can on occasson gve some nsght nto the nteger problem tself. For example, Allender et al. [ABKPM09] proved a hardness result of a closely related problem, whch they call BtSLP, by frst observng that the correspondng problem for polynomals s #P-hard. Ths motvates us to examne the polynomal analogue of the sum of square roots problem, the precse statement of whch s gven below as an nterestng problem n ts own rght. Problem 1.2 (Sum of square roots of polynomals). Gven an expresson S = n =1 c f (x), where c F (a feld of characterstc 0) and f (x) are unvarate polynomals wth degree bounded by d 2

3 and f (0) 0 (for all ), can we show that unless S = 0, the mnmum exponent of x whch has a non-zero coeffcent n the power seres S s bounded by a fxed polynomal n n and d? The condton f (0) 0 n the above problem statement s smply to ensure that f(x) has a well defned power seres expanson around x = 0. Ths problem beng a close cousn of Problem 1.1, t seems reasonable to hope that solvng t mght shed some lght on the latter problem. At the least, one mght expect to solve Problem 1.1 for a nontrval class of ntegers startng from a soluton to Problem 1.2. We are not aware of any pror research work along ths lne. In ths work, we answer the queston posed n Problem 1.2 n the affrmatve. Usng ths result, we show that t s ndeed suffcent to keep polynomal amount of precson n computng the sgn of S n Problem 1.1 f the nput ntegers belong to a specal class that we call (by abusng termnology) the polynomal ntegers. We have mentoned earler that the sum of square roots problem les n the countng herarchy: CH = P P PP P PPPP..., where PP s the class of languages accepted by probablstc polynomal tme machnes. Ths result s due to Allender, Bürgsser, Kjeldgaard-Pedersen and Mltersen [ABKPM09]. In fact, they showed that the more general problem PosSLP, whch s the task of checkng f the nteger produced by a gven dvson-free straght-lne program s greater than zero, belongs to the complexty class P PPPPPP, the thrd level of CH. (The class P s taken to be the 0-th level of CH). The fact that PosSLP s a more general problem than the sum of square roots of ntegers follows from the work of Twar [Tw92] who used Newton Iteraton to construct a small (polynomal-szed) arthmetc crcut that computes a gven sum of square roots to exponentally many dgts of precson. The polynomal analog of the PosSLP problem s the task of comparng the degree of the polynomal computed by a gven arthmetc crcut wth a gven nteger. More precsely: Problem 1.3 (Degree Computaton). Let F be a feld (say the ratonal numbers Q). Gven an arthmetc crcut computng a multvarate polynomal f(x) over F and an nteger d, s the degree of f(x) at most d? Ths problem, whch Allender et al. [ABKPM09] refer to as DegSLP, was also studed by Koran and Perfel [KP07] and they put t n the second level of the countng herarchy. Here we gve a (slght) mprovement to the complexty theoretc upper bound for DegSLP. We show ts contanment n the class corp PP, whch s essentally the frst level of CH modulo the use of randomzaton. 1.1 Prevous work The work of Burnkel, Flescher, Mehlhorn and Schrra [BFMS00] consdered the problem of fndng the sgn of an arthmetc expresson E nvolvng the operatons addtons, subtractons, multplcatons and square root (n fact, dvson as well), and wth nteger operands. They showed that f u s the bound on the value of E when all the subtracton operatons n E are replaced by addtons and k s the number of dstnct square root operatons n E then E 1/u 2k 1 unless E = 0. Ths result mmedately gves an exponental bound on the bt sze of S n Problem 1.1: f S 0 then S 1/2 2n log(nn). (In ths regard, the work of Mehlhorn and Schrra [MS00] s also relevant). It s also noted n [BFMS00] that the bound obtaned for E s nearly optmal n general. For nstance, f E = (2 2k + 1) 1/2k 2 then u = (2 2k + 1) 1/2k and hence by the result n [BFMS00], E 1/5 2k 1. On the other hand, t was also shown that E 1/2 2k. However, Problem 1.1 s just a specal case of the problem studed n [BFMS00] where there s no occurrence of nested square 3

4 roots. So, t remans a concevable possblty that there s a better lower bound for S. Indeed, for a certan choce of parameters a better result s known due to the work of Cheng, Meng, Sun and Chen [CMSC10] (see also [Che06]). By connectng Problem 1.1 to the shortest vector problem over a certan nteger lattce, they showed that S 1/N 2O(N/ log N), whch s an mprovement over earler results when N c n log n for some constant c. However, a less desrable aspect of ths result s the doubly exponental dependency of the bt complexty of S on log N. On the other sde, one seeks to construct/prove the exstence of sets of ntegers {a 1, a 2,..., a n } for whch the sum S of Problem 1.1 s as small as possble n absolute value. In other words, what could be a good upper bound on S. Qan and Wang [QW06] gave an explct constructon that gves S = O(N n+3/2 ). The ntegers they construct are closely related to the specal class of ntegers that we look at (see Secton 1.2). For Qan and Wang, every nteger a s essentally of the form (X + ) (sutably scaled). They showed that f S = n =0 ( n ) ( 1) X + then S = O(X n+1/2 ) for fxed n 1. The complexty of the DegSLP problem was studed by Allender et al. [ABKPM09], who showed that DegSLP polynomal tme many-one reduces to PosSLP and hence also belongs to the thrd level of CH. Koran and Perfel [KP07] later mproved upon ther result over felds of characterstc p > 0 and obtaned a better bound of co-np ModpP, whch s contaned n the second level of CH. However, over felds of characterstc zero, the thrd level of CH remaned the best known upper bound on the complexty of DegSLP. 1.2 Our contrbuton Our frst contrbuton s an affrmatve answer to Problem 1.2. Usng ths, we prove the followng theorem. Theorem 1.4 (Sum of square root of polynomal ntegers ). Suppose S = n =1 δ a (δ {+1, 1}) such that every postve nteger a s of the form a = X d + b 1 X d b d (d > 0), where X s a postve real number and b j are ntegers. Let B = max({ b j },j, 1) and d = max {d }. If X > (B + 1) p1(n,d), where p 1 (n, d) s a fxed polynomal n n and d, then a non-zero S s lower bounded as, S 1/X p2(n,d), where p 2 (n, d) s another fxed polynomal n n and d. The polynomals p 1 (n, d) and p 2 (n, d) can be taken to be 12 dn 2 log 32d and 6 dn 2, respectvely. Note that the ntegers b j need not be postve. Expressng each a as X d + b 1 X d b d s nothng very unusual - t s lke a base-x representaton of a when X s a postve nteger. What makes the polynomal ntegers specal s the condton that X s exponentally large compared to the b j s; or n other words, all the dgts are small n X-ary representaton. Indeed, f one can prove Theorem 1.4 wthout ths condton then Problem 1.1 would stand solved n ts full generalty by takng X = 2. Fnally, we would lke to note that we have not made an attempt to fnd the best possble expressons for p 1 ( ) and p 2 ( ), our prmary ntenton beng to just show that the functons p 1, p 2 are some fxed polynomals n n and d. For the more general DegSLP problem, we show contanment n the frst level of the countng herarchy (modulo the use of randomzaton), thereby mprovng upon prevous best results [ABKPM09, KP07] for ths problem. More precsely, we show Theorem 1.5. DegSLP s n corp PP. 4

5 The above theorem holds over any underlyng feld. Organzaton - The rest of ths paper s organzed as follows. In Secton 2 we gve a soluton to Problem 1.2 and n Secton 3 we prove Theorem 1.4. The result on the complexty upper bound of DegSLP s presented n Secton 4. 2 Sum of square roots of polynomals In ths secton, we prove the followng theorem. Theorem 2.1. Gven a sum S = n =1 c g (x) f (x) where c F (a feld of characterstc 0), f and g are unvarate polynomals of degree at most d and f (0) 0 for all 1 n, ether S = 0 or the mnmum exponent of x whch has a nonzero coeffcent n the power seres S s bounded by dn 2 + n. The soluton to Problem 1.2 follows mmedately f we take g (x) to be 1 n the above theorem. But, we wll need the slghtly general form, that s, when the g s are not assumed to be 1, to prove Theorem 1.4. We wll see that the proof of Theorem 2.1 s not dffcult n hndsght - t uses a mathematcal object called the Wronskan, whch s used to study lnear dependence of functons. Let us spend some tme to brefly dscuss ths concept. 2.1 The Wronskan and lnear ndependence A set of n functons {h 1, h 2,..., h n } over a feld F s sad to be lnearly dependent f there exst elements c 1, c 2,..., c n F such that the functon (c 1 h 1 + c 2 h c n h n ) s dentcally zero. If each of the h s s n tmes dfferentable, then the Wronskan of ths set, denoted W(h 1,..., h n ) (or, W(h) for short) s defned as the followng determnant. h 1 h 2... h n W(h 1,..., h n ) def h (1) = det 1 h (1) 2... h (1) n.... h (n 1) 1 h (n 1) 2... h (n 1) n where h (j) s the j th dervatve of h. It s a well known functon used n the study of dfferental equatons. It s easy to observe that f the functons h 1,..., h n are F-lnearly dependent then ther Wronskan s dentcally zero. But, the converse need not be true n general. Bôcher [Bˆ00] showed that there are famles of nfntely dfferentable functons whch are lnearly ndependent and yet ther Wronskan vanshes dentcally. However, for analytc functons ths s not the case: a fnte famly of lnearly ndependent (real or complex valued) analytc functons has a non-zero Wronskan. More generally, ths property s true for any famly of formal power seres over any characterstc zero feld. Theorem 2.2 (Wronskan of a famly of power seres). Let F be a feld of characterstc zero. A fnte famly of power seres n F[[x]] has a zero Wronskan f and only f t s F-lnearly dependent. A short and smple proof of the above fact appears n [BD10]. Let us now see how to use ths result to prove Theorem 2.1., 5

6 2.2 Proof of Theorem 2.1 Let S = n =1 c g (x) f (x) be a gven non-zero sum. Assume wthout loss of generalty that f (0) = 1, for all. If ths s not the case then scale f (x) approprately: Express f (x) as f (x) = f (0) f (x)/f (0) and take f (x)/f (0) to be our new f (x) (by reusng symbols). Also, consder an approprate extenson feld that contans f (0) for every 1 n, ths extenson s our base feld F (agan, by reusng symbols). The condton f (0) = 1 s smply to ensure that f can be expressed as a formal power seres n x over F. Denote g f by h. We can also assume that h 1,..., h n are F-lnearly ndependent - f not, smply work wth an F-bass of h 1,..., h n. (Note that, we are not fndng a bass, we are only usng t for the sake of argument.) Suppose, x t dvdes the power seres S, where t s the maxmum possble. Pretend that, n c h = S (2) =1 s a lnear equaton n the varables c 1,..., c n. By takng dervatves of both sdes of Equaton 2 wth respect to x, we have the followng system of lnear equatons n c 1,..., c n, for 0 j n 1, n =1 c h (j) = S (j), (3) where h (j) and S (j) are the j th dervatves of h and S, respectvely. Let C be the coeffcent matrx of the above system of lnear equatons. That s, h 1 h 2... h n h (1) C = 1 h (1) 2... h (1) n..... h (n 1) 1 h (n 1) 2... h (n 1) n Observe that det(c) s the Wronskan W(h). The followng smple clam about W(h) s crucal to the proof. Clam 2.3. The Wronskan W(h) = n 2 det(m), where M s an n n matrx whose every entry s a polynomal n x of degree at most n d. =1 f 2n 3 Proof. Expandng h (j) we get the followng. (Superscrpts ndcate the order of the dervatves.) h (j) = j k=0 ( ) j k g (j k) ( f ) (k) f 2j 1 2 h (j) = j k=0 ( ) j g (j k) k f 2j 1 2 ( f ) (k), multplyng both sdes by f (2j 1)/2. Now notce that, f (2j 1)/2 ( f ) (k) s a polynomal of degree at most j d. Hence, f (2j 1)/2 h (j) s also a polynomal of degree at most (j+1) d, although ndvdually they are power seres n x. Snce j s at max n 1, the statement of the clam follows. Snce S 0, there must be one c whch s nonzero. Let t be c 1. Then, by applyng Cramer s rule, c 1 = det(m 1) W(h) = n =1 f 2n 3 2 det(m 1) det(m) 6 (by Clam 2.3),

7 where M 1 s the followng matrx, S h 2... h n S M 1 = (1) h (1) 2... h (1) n.... S (n 1) h (n 1) 2... h (n 1) n Note that, Cramer s rule apples here because W(h) 0, as h 1,..., h n are assumed to be lnearly ndependent, whch n turn mples that det(m) 0 (by Clam 2.3). Snce x t dvdes S, x t n+1 must dvde S (j) for every 0 j n 1 and hence x t n+1 dvdes det(m 1 ). Clam 2.4. The maxmum power of x dvdng det(m 1 ) and det(m) must be the same. Proof. Ths s because c 1 0 s an element of the feld F and n =1 f (0) 2n by assumpton. Therefore, t n must be less than the degree of det(m), whch s at most d n 2 (agan by Clam 2.3), and hence t d n 2 + n. Ths proves Theorem 2.1. Wth the polynomal verson of the sum of square roots problem at hand, one wonders as to what can be nferred about the correspondng problem over ntegers. In turns out that ndeed somethng nontrval can be shown about a specal class of ntegers that we have called before as the polynomal ntegers (see Theorem 1.4). Ths consttutes the content of the followng secton.. 3 Sum of square roots of polynomal ntegers Ths secton s devoted to the proof of Theorem 1.4. Let S = n =1 δ a ; δ {+1, 1}, be a gven nonzero sum, where each postve nteger a s of the followng form. a = X d + b 1 X d b d (4) where X s a postve real number and b j s are ntegers (not necessarly postve). Snce X d = ( X) 2d, by takng X to be our new X, we can assume henceforth that every d (1 n) s an even number. Overvew. The overall dea of the proof s to do a Taylor seres expanson for each a so that we get a Taylor expanson for the sum S overall. Usng Theorem 2.1 and the non-zeroness of S, we deduce that we must get a non-zero term very early n the Taylor expanson. That s, there must be some non-zero S l for l relatvely small (see Equatons 5 and 6 below for a precse defnton of S l ). We use the fact that l s small to deduce that such an S l s farly large n absolute value. We then use the fact that each b j s much smaller than X to upper bound each of the remanng terms and thereby deduce that the sum of the remanng terms cannot almost cancel out S l. More specfcally, the sum of the remanng terms s at most 1 2 S l n absolute value. Ths helps us deduce that S tself s farly large n absolute value. Dong the Taylor expanson. From (4) we have a = ( X) d 1 + b 1 X b d X d. 7

8 Addng these expressons together wth the approprate sgn, we get an expresson for S. n S = δ X d /2 1 + b 1 X b d X d. =1 Let y = 1/X and d = max {d }. Then, n S y d/2 = δ y (d d )/2 1 + (b 1 y) (b d y d ). =1 Now notce that, by pretendng that f (y) = 1 + b 1 y b d y d s a polynomal n the formal varable y, the sum S y d/2 s of the form n =1 δ g (y) f (y), where g (y) = y (d d)/2 (snce every d s even, by the adjustment dscussed before, (d d )/2 s an nteger). Therefore, n S y d/2 = δ n y j δ c j (by exchangng the summatons), =1 j 0 c j y j = j 0 =1 where c j s the coeffcent of y j comng from the th power seres g (y) f (y). Ths s the Taylor seres expanson of S that we wll work wth. We gve the name S j to each summand n the expresson above, namely S j = y j n =1 δ c j. Thus S y d/2 = j 0 S j. (5) The Proof Strategy ahead. Applyng Theorem 2.1, the mnmum exponent l of y wth n =1 δ c l 0 s such that l dn 2 + n. Suppose we could show that S l+t S l 1, (6) 2t+1 for every t 1, then from (5) t would follow that S y d/2 Sl S l+1... S l 1 2 S l = 1 2 S l S X d/2 1 2 S l. (7) Ths (potentally) gves us a lower bound on S va a lower bound of S l. But, to satsfy the condton gven by equaton (6), we also need an upper bound on S l+t for every t. Upper bound on S j : S j = y j n δ c j n y j max{ c j } =1 Let us upper bound the quantty c j. Fx any ndex. For the ease of presentaton, we wll avod wrtng the ndex whenever t s clear from the context that we have a specfc n mnd. For example, we wrte c j as the coeffcent of y j comng from the power seres, y d d b 1 y b d y d d d = y 2 8 u k (b 1 y b d y d ) k, k=0

9 where u 0 = 1 and u k = 1 2 ( 1 2 1)...( 1 2 (k 1)) k! = ( 1) k (2(k 1) 1) 2 k k!, for k > 0. Expressed dfferently, ( 2k ) u k = ( 1) k+1 (2k)! (2k 1) (k!) 2 2 2k u k k = (2k 1) 2 2k Now, notce that ( ) 2k k /(2k 1) = 2 Ck 1, where C k s the k th Catalan number 1/(k + 1) (2k ) k. Hence, u k = 2 C k 1 2 2k, and also u k 1. (8) For any j, the coeffcent c j s contrbuted to by those terms of k=0 u k (b 1 y b d y d ) k for whch k s n the range [(j (d d )/2)/d, j (d d )/2]. For any fxed k [j/d, j], the coeffcent of y j n (b 1 y b d y d ) k s exactly, ( ) k v kj = b k 1 1 bk bk d d. k 1 +2k d k d =j, k k d =k k 1,..., k d where k 1,..., k d are non-negatve ntegers. Then, assumng B = max({ b j },j, 1), ( ) k v kj b 1 k1 b 2 k 2... b d k d (B d ) k. k 1,..., k d k k d =k Snce c j+(d d )/2 = k [j/d,j] u k v kj, Lower bound on S j : c j k [0,j] (B d ) k (B d) j+1 (usng Equaton 8) S j n y j (B d) j+1 (9) S j = y j n δ c j Let us lower bound the sum n =1 δ c j. Notce that, n the prevous dscusson on upper boundng S j, the nteger v kj depends on the ndex whereas u k solely depends on k. So, to make the followng dscusson more precse, we swtch to the notaton v kj. For smplcty, the range of k s not specfed n the followng equatons - the approprate range of k s [(j (d d )/2)/d, j (d d )/2]. =1 n δ c j = =1 n δ u k v k(j (d d )/2) = k k =1 n u k δ v k(j (d d )/2). =1 Now notce that, the sum n =1 δ v k(j (d d )/2) s an nteger. Hence, f n =1 δ c j 0 then n =1 δ c j 1/2 2j 1 (by Equaton 8). Therefore, S j yj 2 2j 1 f S j 0. (10) 9

10 Puttng everythng together. Wth the upper and the lower bounds on S j at hand, we are now ready to pnpont the requrement that ensures that condton (6) s satsfed. We want S l+t / S l 1/2 t+1, for all t 1. Hence, by combnng equatons (9) and (10), t s suffcent f, n y l+t (B d) l+t+1 y l /2 2l t+1 X t n 2 2l+t (B d) l+t+1 Therefore, t suffces f X (B + 1) 3 dn2 log 16d (takng nto consderaton that l dn 2 + n). And f ths happens then condton (6) s satsfed and by equaton (7), S X d/2 1 2 S l 1 2 2l X l d/2 (by Equaton 10) Ths mples that S 1/X 6 dn2. Recall that at the begnnng of ths secton, we have adjusted X (by choosng X to be our X) so that d can be assumed to be even whch n turn s used to argue that (d d )/2 s n fact an nteger. Replacng X n place of X and 2d n place of d n the above bounds on X and S, we obtan the bounds X (B + 1) 12 dn2 log 32d and S 1/X 6 dn2 as promsed n Theorem The complexty of DegSLP In ths secton, we consder the algorthmc complexty of the followng problem: gven a polynomal f(x) as an arthmetc crcut, and an nteger d n bnary determne f deg(f) d. Towards ths end, we need to defne another natural computatonal problem. CoeffSLP : Gven an arthmetc crcut computng a polynomal f(x) over ntegers, a monomal X α and a prme p, determne the coeffcent of X α n f(x) modulo p. (The addtonal nput p s to ensure that the output s not too large). Our defnton of problem CoeffSLP s slghtly dfferent from that of [ABKPM09, KP07] and more talored towards our needs. We wll need the followng theorem from [KP07] to prove our result. A smpler, self-contaned proof of the theorem s gven below. Theorem 4.1. [KP07] CoeffSLP s #P-complete. We frst gve some warm-up lemmas. Lemma 4.2. For any m 7, the lcm of the frst m numbers s at least 2 m. Lemma 4.3. For any nteger t Z 1 and prme p, there s a prme r = O(t 2 log p) such that the rng R def = F p [z]/ zr 1 z 1 s the drect sum of fnte felds of sze q > p t. 10

11 Proof. Consder the nteger M := (p 1) (p 2 1)... (p t 1). Then M < p t2. By lemma 4.2, there exsts a prme r < log M such that r does not dvde M. Ths s the prme r that we seek. Let m denote the order of p modulo r,.e. m s the smallest postve nteger such that p m = 1 (mod r). Snce r does not dvde M = [t] (p 1), therefore r does not dvde any (p 1) for 1 t and therefore m > t. Let φ r (z) denote the r-th cyclotomc polynomal, that s φ r (z) def = zr 1 z 1. It s known (cf. Theorem 2.47 of [LN94]) that over F p, φ r (z) factors nto r 1 m rreducble polynomals each of degree m. Thus, R def = F p [z]/ φ r (z) s the drect sum of fnte felds of sze p m > p t. Our next two warmup lemmas pertan to the coeffcent of a monomal x d of a gven polynomal f(x), denoted Coeff(x d, f(x)). The frst lemma s vald when the characterstc of the feld s larger than d and expresses the coeffcent of x d n any shft of f(x) as the evaluaton of the d-th order dervatve of f at a sutable pont. When the sze of the underlyng feld s large enough, our second lemma below expresses Coeff(x d, f(x)) as an exponental sum (wth each summand beng computable n polynomal-tme). Ths wll help us n showng that the computaton of Coeff(x d, f(x)) s n #P. Lemma 4.4. Over a feld of characterstc larger than d, Coeff(x d, f(x + β)) (the coeffcent of x d n f(x + β)) s precsely 1 d! f (d) (β). Proof. By the lnearty of dervatves and substtutons, t s suffcent to show ths for monomals. So let f(x) = a x e. If e < d then f (d) (x) s the zero polynomal and we are done. So let e d. Expandng (x + β) e usng bnomal theorem we get that coeffcent of x d s a (e d) β e d. On the other hand f (d) (x) = a e (e 1)... (e d + 1)x e d. It s now easly verfed that the coeffcent of x d n f(x + β) s 1 d! f (d) (β). Lemma 4.5. For any fnte feld F q and any polynomal f(x) F q [x], we have β d f(β 1 ) = β F q In partcular f (q 1) > deg(f) then d (mod (q 1)) Coeff(x d, f(x)) = β F q Coeff(x, f(x)). β d f(β 1 ). Proof. The proof s a routne computaton n the manner of dong Fourer transforms. In partcular we use the fact that for any nteger k we have { β k 1 f k 0 (mod (q 1)) = (11) 0 otherwse β F q 11

12 Consder the lhs of the dentty we wsh to prove. We have β d Coeff(x, f(x)) β 0 β d f(β 1 ) = β F q β F q = Coeff(x, f(x)) β d β F q 0 = Coeff(x, f(x)) β d 0 β F q = 0 Coeff(x, f(x)) β F q β d = 0 Coeff(x, f(x)) ( 1) δ (d ) 0 (mod (q 1)) (usng (11)) = d (mod (q 1)) Coeff(x, f(x)) As a corollary we get an exact expresson for Coeff(x d, f(x)) as an exponental when the sze of the underlyng fnte feld s larger than the degree of f. Corollary 4.6. Let F q be a fnte feld and f(x) F q [x] be a polynomal. Suppose that d deg(f) < (q 1). Then Coeff(x d, f(x)) = β d f(β 1 ). β F q Note that ths corollary mples that when the sze of the underlyng fnte feld s large enough Coeff(x d, f(x)) can be computed n P #P. To extend ths for arbtrary fnte felds, we need to go to a sutably large extenson of the underlyng fnte feld. However constructng extensons of felds determnstcally s not known but we get around the problem by gong to a sutably rng extenson and ths suffces for our purpose. We collect all ths together nto a proof of theorem 4.1 below. Proof of Theorem 4.1: The #P-hardness of ths problem s well-known and a proof can be found for example n [ABKPM09]. It s suffcent to show ths for unvarate polynomals (by replacng each ndetermnate x by an exponentally ncreasng sequence of monomals, f necessary). That s, our problem now becomes the followng: gven a crcut of sze s computng a unvarate polynomal f(x) and a d Z 0 gven n bnary, compute the coeffcent of x d n f(x). Notce that D def = 2 s s an upper bound on deg(f(x)). Usng lemma 4.3, we obtan an extenson rng R of the form R = F p [z]/ zr 1 z 1 such that r (log D)2 (log p) and R = F q... F q, wth q 1 > D. Combnng ths structure of the rng R wth corollary 4.6 va Chnese remanderng we have Coeff(x d, f(x)) = β R β d f(β 1 ). 12

13 The number of terms n the above summaton s exponentally large but notce that each summand n the above expresson, (β d f(β 1 )), s polynomal-tme computable so that overall ths sum s computable n P #P. Theorem 4.7. DegSLP corp T CoeffSLP. Here we frst prove ths theorem assumng the underlyng feld to be the feld of ratonal numbers. A smlar proof wll go through over other felds of zero characterstc. The theorem s vald even f the underlyng feld has small characterstc. We relegate the proof for small characterstc felds to Secton 4.1. Proof. We frst reduce the multvarate to the unvarate problem by makng a random substtuton of the form g(z) = f(a 1 z, a 2 z,..., a n z). The unvarate verson of DegSLP s n fact equvalent (va determnstc reductons as well) to the multvarate verson. For a proof cf. [ABKPM09]. Clam 4.8. Wth hgh probablty over a random choce of the vector a = (a 1,..., a n ) we have: deg(g(z)) = deg(f(x)). Proof of Clam 4.8: Let f have degree d. We can wrte the polynomal f(x) as d =0 f (X), where each f (X) s a homogeneous polynomal of degree. Applyng the substtuton x := a z, we get that g(z) = f 0 + z f 1 (a) + z 2 f 2 (a) z d f d (a). By the Schwartz-Zppel lemma, f d (a) s non-zero wth hgh probablty so that deg(g(z)) = deg(f(x)) also wth hgh probablty. Our problem thus s the followng: Gven an arthmetc crcut computng a unvarate polynomal g(z) and an nteger d n bnary, we want to determne f deg(g(z)) d. The most natural thng to do s to use the CoeffSLP oracle to determne whether the coeffcent of z d n g(z) s nonzero. (For the oracle call to CoeffSLP, we choose the prme p at random. Ths ensures that wth hgh probablty, the coeffcent α of z d n g(z) s zero as a ratonal number f and only f α s zero modulo p.) If the coeffcent of z d n g(z) happens to be nonzero we have a certfcate that the degree of g s at least d. The converse s easly seen to be false: the coeffcent of z d n g can be zero and yet the degree of g can be larger than d. To fx ths, we take a random shft of g and compute ts degree nstead. Specfcally, we look at the polynomal g(z + β), where β s chosen unformly at random from a large enough subset of F. Clearly, deg(g(z)) = deg(g(z + β)). It suffces then to prove the followng clam: Clam 4.9. The coeffcent of z d n g(z + β) s zero for more than (deg(g) d) choces of β F f and only f deg(g(z)) < d. Proof of Clam 4.9: By lemma 4.4, the coeffcent of z d n g(z + β) s 1 d! g(d) (β), where g (d) (z) denotes as usual the d-th order dervatve of g. Snce the characterstc of the feld s zero, g (d) (z) has degree precsely deg(g) d. In partcular, g (d) (z) s dentcally zero f and only f deg(g) < d. Now the clam follows from an applcaton of the Schwarz-Zppel lemma. Ths completes the proof of the theorem. Combnng Theorems 4.1 and 4.7, we mmedately get: 13

14 Theorem DegSLP s n corp PP for felds of characterstc zero. 4.1 DegSLP over felds of small characterstc In ths secton, we gve the proof of Theorem 4.7 n the general case,.e even when the underlyng feld has small characterstc. We wll need a lemma orgnally due to Edouard Lucas. Lemma [vl99, p.55] Let n, m be postve nte gers whose p-ary representaton s the followng: Then m = m 0 + m 1 p m d p d, : 0 m p 1 n = n 0 + n 1 p n d p d : 0 n p 1. ( ) n = m ( n0 m 0 ) ( n1 m 1 )... ( nd m d ) (mod p) In partcular, for any ntger n 1, the bnomal coeffcent ( n p ) s dvsble by p f and only f the n, the -th dgt n the p-ary representaton of n s zero. Proof of Theorem 4.7: The proof of theorem 4.7 needs to be refned carefully for the low characterstc stuaton because clam 4.9 s no longer true. As before t does hold that deg(g(z)) = deg(g(z + β)) for every β F. In one drecton the clam does ndeed hold true over any feld. Indeed f deg(g(z)) < d then the coeffcent of z d n g(z + β) s zero for every β F. The converse drecton s unfortunately not true. The dea s to prove that a partal converse of clam 4.9 and use t to derve a randomzed reducton from DegSLP to CoeffSLP. Clam Let F be a feld of characterstc p, F ts algebrac closure and g(z) F[z] be a polynomal. Let t 0 be an nteger. If the coeffcent of z pt n g(z + β) s zero for more than (deg(g(z)) p t ) dstnct choces of β F then ether 1. deg(g(z)) < p t or, 2. deg(g(z)) p t+1 Proof of Clam 4.12: g(z + β) s gven by Let the coeffcent of z n g(z) be a F. Then the coeffcent of z d n h(β) = d m deg(g) a m ( ) m β m d. d Note that h(β) s a polynomal functon of β of degree at most (deg(g) d). Snce h(β) s zero for more than (deg(g) d) values of β F t means that h(β) must be dentcally zero as a polynomal functon of β. Thus a m ( ) m = a m d Now from Lucas s lemma 4.11 t follows that ( ) m p t 0 m [p t..(p t+1 1)]. ( ) m p t = 0 m d. (12) 14

15 Combned wth equaton (12) t means that a m = 0 m [p t..(p t+1 1)]. Ths means that ether deg(g(z)) < p t or that deg(g(z)) p t+1, provng the clam. Ths partal converse allows us to modfy the earler proof sutably n the small characterstc stuaton as follows. Proceedng as before, we can assume wthout loss of generalty that the gven crcut computes a unvarate polynomal g(z). Our problem then s the followng: gven an arthmetc crcut computng a unvarate polynomal g(z) and an nteger d n bnary, we want to determne f deg(g(z)) d. Frst observe that multplyng g(z) wth a sutable power of z, we may assume wthout loss of generalty that d s a power of p, say d = p t. Let the sze of the crcut computng f be s. Then the formal degree of f s bounded by 2 s. Let r = 1 + s log 2 log p. If the underlyng feld s fnte we can go to a sutable extenson n randomzed polynomal tme and thereby ensure that F p 2r. Wth ths preprocessng behnd us, let us see how the partal converse of clam 4.12 allows us to test f deg(g(z)) p t usng an oracle for CoeffSLP. 1. Pck a β F unformly at random from a subset of F of sze at least 2 2s. 2. Usng the CoeffSLP oracle, for each [t..r] compute a p := Coeff(z p, g(z + β)). 3. Accept f and only f a p 0 for some [t..r]. To see why ths works note that f deg(g(z)) < p t, then each a p s zero and therefore the above test always rejects. For the other drecton assume deg(g(z)) p t log deg(g(z)). Let := log p so that p deg(g(z)) < p +1. Then by clam 4.12 a p wll be nonzero (wth hgh probablty over the random choce of β) and hence our test accepts wth hgh probablty. Ths completes the proof of the theorem. Combnng Theorems 4.1 and 4.7, we mmedately get: Theorem DegSLP s n corp PP. 5 Dscusson We have seen that for the class of polynomal ntegers t s possble to compare two sums of square roots by keepng precson of up to polynomally many bts (durng square root computatons). Although, polynomal ntegers form a nontrval class, the condton that X s suffcently large also makes them very restrctve at the same tme. The hope s that t may be possble to explot results smlar to that of Theorem 2.1 to show somethng stronger for the case of ntegers. As a next step, we would be nterested n a smlar result where X s constraned as X poly(n, d) B c (c s a constant), nstead of X (B + 1) poly(n,d) as s the case n our analyss. Could encodng the ntegers as multvarate polynomals be useful n ths regard? 15

16 Nonetheless, polynomal ntegers are perhaps nterestng from one perspectve. A plausble way to make the sum S very small s to assume that the number of + and sgns n S = n =1 δ a are equal and all the ntegers a s are somewhat very close to each other. Notce that, because of the assumpton that X s large, all ntegers of the form X d + b 1 X d b d are reasonably close to X d. Our analyss shows that at least for ths case any non-zero sum S s stll suffcently large. As a fnal remark on the sum of square roots problem, we would lke to note that the proofs and the results presented here generalze n a straghtforward manner to general sums of radcals - lke sums of cube roots or fourth roots of ntegers. We feel that the complexty of the problem DegSLP s not understood well enough. In partcular no hardness results are known for t. We conclude by posng the followng problem: Problem 5.1. (Hardness of DegSLP ): Does there exst an effcent randomzed algorthm for DegSLP?... or, wll the exstence of such an algorthm for DegSLP lead to a collapse of the polynomal herarchy? Acknowledgements We thank Kurt Mehlhorn and Pyush Kurur for some nsghtful dscussons on ths work. Thanks also to the anonymous revewers whose comments have helped us mprove the presentaton of ths paper. 16

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