A Faster Randomized Algorithm for Root Counting in Prime Power Rings

Transcription

1 A Faster Randomized Algorithm or Root Counting in Prime Power Rings Leann Kopp and Natalie Randall July 17, 018 Abstract Let p be a prime and Z[x] a polynomial o degree d such that is not identically zero mod p. We introduce a Las Vegas randomized algorithm to count the number o roots o in Z/(p k ) or k N with k which runs in time d 1.5+o(1) (log p) +o(1) 1.1 k. We compare the randomized algorithm to simple brute orce to see when we have practical time gains. In addition, we present an upper bound on the number o roots o (as a unction o p, k, and the degree o ) that is optimal or k =. 1 Introduction A deterministic algorithm or counting roots in Z/(p k ) in time (dlog(p) + t ) O(1) is given in []. Here we propose a Las Vegas randomized algorithm which runs in time d 1.5+o(1) (log p) +o(1) 1.1 k. By Las Vegas randomized, we mean that our algorithm undercounts roots with a ixed error probability but otherwise returns a correct root count and always correctly announces ailure. For instance, i we take our ixed error probability to be 1, we can get an overall 3 ailure probability o less than 1 by running the algorithm 100 times. Las Vegas randomized algorithms are common across algorithmic number theory; there are ast, widely accepted Las Vegas randomized algorithms or checking primality and or actoring polynomials over inite ields [1, 3, 4]. In our algorithm, we introduce randomization by using ast actorization (see [3]) to ind roots o in Z/(p). Prior to the deterministic algorithm in [] there was little inormation on counting the roots o a polynomial over prime power rings. We can easily count the number o roots o a polynomial in Z/(p) by taking the degree o gcd(x p x, ), but this method relies on Z/(p) being a unique actorization domain, and Z/(p k ) is not a unique actorization domain or k > 1. To overcome this issue, we consider the Taylor expansion o our polynomial about a root ζ o the mod p reduction o with a perturbation o pε, where ε {0,..., p k 1}. From this expansion, we can divide by certain powers o p in order to recursively isolate the roots o in the ring Z/(p k ). From a similar expansion, we also get an upper bound or the number o roots o in Z/(p k ) given by min{d, p}p k 1 and a sharp upper bound or k = given by min{ d, p}pk 1 + (d mod ). Background and Randomized Algorithm Lemma.1 (Hensel s Lemma). I Z[x] is a polynomial with integer coeicients, p is prime, and ζ J {0,..., p J 1 1} is a root o (mod p J ) and (ζ J ) 0 (mod p), then there is a unique ζ {0,..., p J+1 1} with (ζ) = 0 (mod p J+1 ) and ζ = ζ J (mod p J ). 1

2 We will see below that we can use Hensel s Lemma to determine the number o lits o a root ζ i with s(i, ζ i ) = 1. Consider the expansion o given by (ζ + pε) = (ζ) + (ζ)pε + + p min(d,k 1) ε min(d,k 1) mod p k, where ζ is a root o the mod p reduction o. Let s {1,..., k} be the maximal integer such that p s divides each o (ζ), (ζ)p,..., min(d,k 1)(ζ) p min(d,k 1). More precisely, s = min{ord p ((ζ)), ord p ( (ζ)p),..., ord p ( min(d,k 1)(ζ) p min(d,k 1) )}, where ord p (x) reers to the p-adic valuation o x. I (ζ + pε) = 0 (mod p k ), then we can write p s ( (ζ) + (ζ) ε + + ε min(d,k 1) ) = 0 mod p k, p s p s 1 (p s min(d,k 1) which is true i and only i (ζ) p s + (ζ) ε + + ε min(d,k 1) = 0 mod p k s. p s 1 (p s min(d,k 1) In the case where s = 1, i (ζ) = 0 mod p then we must have that p (ζ) and so ζ has no lits mod p k ; however, i (ζ) 0 mod p, then ζ lits to one unique root by Hensel s Lemma. In the case where s = k, the entire expression p s ( (ζ) + (ζ) ε+ + ε min(d,k 1) ) p s p s 1 (p s min(d,k 1) vanishes identically mod p k, so any ε {0,..., p k 1 } is a zero o (ζ + pε) and thereore we have that ζ has p k 1 lits. The key idea o the randomized algorithm is that counting the number o roots when s = 1 and s = k is simple, as described above, and we can reduce all the computations to these two cases using recursion. I s {,..., k 1}, we can reapply the algorithm to an + (ζ) (p s min(d,k 1) ε min(d,k 1) instance o counting roots or the polynomial (ζ) ε + + p s p s 1 in Z/(p k s ). Eventually this will reduce to the case where either s = 1 or s = k and the recursion will terminate, giving us that the root ζ o mod p has a total number o p s 1 (the number o roots o (ζ) + (ζ) ε + + ε min(d,k 1) mod p k s ) lits to roots p s p s 1 (p s min(d,k 1) in Z/(p k ). Algorithm 1 Randomized Prime Power Root Counting 1: unction count( Z[x] has degree d and is not identically 0 mod p, prime p, k N such that k ) : Factor as in [3] 3: count := number o distinct linear actors o multiplicity 1 These roots in Z/(p) can be lited uniquely to roots in Z/(p k ). 4: Push {ζ 0 {0,..., p 1} (ζ 0 ) = (ζ 0 ) = 0 mod p and (ζ 0 ) = 0 mod p } onto a stack S 5: while S 0 do 6: Pop a root ζ 0 rom the stack and deine s(0, ζ 0 ) := maximal integer such that p s(0,ζ 0) divides each o (ζ 0 ), (z 0 )pε,..., min(d,k 1)(ζ 0 ) p min(d,k 1) 7: i s(0, ζ 0 ) = k then 8: count count + p k 1 9: else 10: Deine ζ0 (x) := 1 (ζ p s(0,ζ 0 ) 0 + px) 11: count count + p s(0,ζ0) 1 count( ζ0 (x), p, k s(0, ζ 0 )) 1: end i 13: end while 14: return count 15: end unction

3 Since the non-degenerate roots o the mod p reduction o have a unique lit by Hensel s Lemma, we only need to keep track o the degenerate roots. Our recurrence takes a degenerate root ζ 0 as a point in a cluster o roots o in Z/(p k ) and recovers the other points in this cluster by expanding ζ 0 to more digits base-p. In this way, we count the number o roots o in Z/(p k ) by counting the number o lits rom each root ζ i o in Z/(p). 3 Discussion o Complexity Bound and Experimental Data d non-degenerate roots o mod p d 1 non-degenerate roots o ζ1 (ε) mod p ζ1 (ε) p s(i,ζ i) ζ1 identically zero mod p k p k 1 d non-degenerate roots o ζ (ε) mod p ζ (ε) p s(i,ζ i) ζ1 identically zero mod p k p k 1 d l non-degenerate roots o ζ1 (ε) mod p ζl (ε) Figure 1: Diagram o complexity tree In the Figure 1, we see the basic tree structure o the algorithm. We need only keep track o the degenerate roots o ; the degenerate roots o, denoted by ζ i, become the children 3

4 nodes rom which more branching occurs. The depth and branching o our recurrence tree is strongly limited by the value o k and the degree o. For the initial parent node, the total number o degenerate roots is less than or equal to d, and or each subsequent child node, the total number o degenerate roots is less than or equal to s(i,ζ i) d ζi. Non-degenerate roots have a unique lit by Hensel s Lemma, so non-degenerate roots require no additional computations and are thereore shown on the let o the tree. We also see that we have a maximum o s(1, ζ 1 ) s(l, ζ l ) nodes at the bottom level o the tree. We use Kedlaya-Umans ast Z/(p)[x] actoring algorithm ound in [3], which takes time d 1.5+o(1) (log p) 1+o(1) + d 1+o(1) (log p) +o(1) or a degree d polynomial, in order to actor the polynomials at each node in Z/(p). In simplest terms, we can consider our total complexity as being less than or equal to (the number o nodes in the recursion tree) (the complexity o actoring over Z/(p)[x]). Optimizing parameters, the worst case occurs when d e.7188 and the depth o the tree is k. The inal complexity o the randomized algorithm is given by e (d 1.5 log p) 1+o(1) + (d log p) 1+o(1) + [(min{d, k 1} 1.5 log p) 1+o(1) + (min{d, k 1} log p) 1+o(1) ](e/) k/e, where (e/) k/e 1.1 k. Based on this complexity bound, we expect to see time improvements even or p as small as when compared to brute-orce counting since brute-orce counting takes time approximately p k, giving us that brute-orce takes time approximately k or p =, while the randomized algorithm takes time approximately 1.1 k. More details regarding computational time with p = are given in Tables 1 and. We now present computational data which illustrates the advantages to using the randomized algorithm over the brute orce method. The brute orce method takes a polynomial, a prime p, and a power k, and evaluates at each value i rom 0 to p k 1. I (i) is identically equal to 0 (mod p k ), then that contributes to the total number o roots o Z/(p k ). We start by comparing the run times o the brute orce algorithm and the randomized algorithm or p =. Table 1 displays the average dierence in computation time or the number o roots o 100 random polynomials o degree less than or equal to 100 in Z/( k ) or the given k, between brute orce and the randomized algorithm (negative implies brute orce was aster). The times are shown in seconds. In general, a single computation took less than a second, so dierences in the milliseconds are not insigniicant. From the table, we see a switch rom brute-orce being more eicient to the randomized algorithm being more eicient at k = 10, and the dierence becomes more pronounced as k increases. k Avg Di (in seconds) Table 1: Average Dierence in Run Times or 100 random polynomials with p =, taken as (time o bruteorce)-(time o randomized algorithm) 4

5 p k Brute Force Randomized Algorithm 71x 4 + 1x 3 84x 47x ns 0 ns 1x 5 66x 4 4x 3 88x 17x ns µs 75x 6 + 8x 5 93x 4 19x 3 + 3x µs µs x 7 + x 6 + 6x 5 3x 3 58x µs µs 48x 8 3x x 5 19x x ms µs 80x 8 37x 7 89x x 3 + 3x ms 0 ns 5x x 6 75x 5 + 3x 3 7x 38x ms 3.00 ms 61x 10 80x 9 17x 6 90x x ms.00 ms 18x x x x 5 64x ms.00 ms 89x 1 56x x 5 x x x ms 6.00 ms 93x 10 36x x 5 78x 4 67x ms.00 ms Table : Run times or 5 k 15, d < k 1 Table shows the dierence in computational time with speciic examples, giving an idea o the overall time it takes or both the randomized algorithm and brute-orce to run when p =. The dierence in computational run time becomes more noticeable when we introduce larger primes. p k Brute Force Randomized Algorithm 44x x 83 17x 67 75x 49 10x sec 11.00ms 10x x 8 51x x x 48 + x min µs 15x 99 59x 74 96x 9 + 7x 8 87x x hours µs Table 3: Run times or p with at least 3 digits Table 3 illustrates the advantages o using the randomized algorithm over brute-orce or computations involving large prime numbers. Table 4 displays the run times or the randomized algorithm or prime numbers with at least our digits. We begin to see a very signiicant dierence between brute orce and the randomized algorithm or large primes; it took the brute orce method almost 4 hours to count the number o roots o a polynomial in Z/(p k ) when p was a 4 digit prime number, while the randomized algorithm counted the roots o a polynomial in Z/(p k ) when p was a 9 digit prime in approximately a minute and a hal. p k t 56x x 64 x x x x ms 53x 94 78x 37 67x x 6 5x 9 36x ms 55x 98 49x x 60 3x x x ms 35x x 84 14x 56 9x 54 90x 7 3x ms 6x 78 31x x x 16 80x 6 51x s 40x 90 10x x 69 40x 41 8x 36 8x s 80x 87 7x x x x x s Table 4: Run times or the randomized algorithm when p has 4 digits 5

6 We expect the randomized algorithm to take the longest when a polynomial has many degenerate roots because a polynomial o this type will require many recursive calls. Polynomials with many degenerate roots do take longer than a random polynomial, but overall the randomized algorithm still outperorms other methods. For instance, counting roots o the 55 degree polynomial (x 1)(x ) (x 10) 10 in Z/(31 10 ) took 6.4 seconds using the randomized algorithm, while counting roots in the same ring with a random polynomial o the same degree took only 1 millisecond. Despite this slowdown or polynomials with very degenerate roots, the randomized algorithm still outperorms other methods; counting roots o a polynomial in just Z/(31 6 ) using brute orce took.7 hours. 4 Bound on Number o Roots Lemma 4.1. I a root ζ o the mod p reduction o has multiplicity j, then s ζ j, where s ζ is the greatest integer such that p s ζ divides each o (ζ),..., (k 1) (ζ) p k 1 ε k 1. (k 1)! Proo. I ζ has multiplicity j, then (ζ) = = j 1 (ζ) = 0 (mod p), but (j) (ζ) 0 (mod p). So j (ζ) j! p j is divisible by p j but not p j+1 and thereore s ζ j. Theorem 4.. Let p be a prime, Z[x] a polynomial o degree d, and k N such that d k. Then N (p, d, k) min{d, p}p k 1, where N (p, d, k) denotes the number o roots o in Z/(p k ). Proo. Let ζ i {0,..., p 1} be any root o the mod p reduction o, and let s(i, ζ i ) be the greatest integer such that p s(i,ζi) divides each o (ζ i ),..., min(d,k 1) (ζ i ) p k 1. Set ζi (x) = 1 (ζ p s(i,ζ i i + px). Clearly, we have that N (p, d, 1) min{d, p}. We know rom Lemma 4.1 that i ζ i Z/(p) is a root o multiplicity J, then J s. Let δ 1 denote the number o non-degenerate roots o mod p. From this, we see that N (p, d, k) δ 1 + min(d,k 1) J= ζ i with s(i,ζ i )=J ps(i,ζ i) 1 N ζi (p, k 1, k s(i, ζ i )) + ζ i with s(i,ζ i )=k pk 1. Considering that N ζi (p, k 1, k s(i, ζ i )) p k s(i,ζi), we get N (p, d, k) min(d,k 1) J=1 ζ i with s(i,ζ i )=J ps(i,ζ i) 1 p k s(i,ζi) + ζ i with s(i,ζ i )=k pk 1, N (p, d, k) min(d,k 1) J=1 ζ i with s(i,ζ i )=J pk 1 + ζ i with s(i,ζ i )=k pk 1, N (p, d, k) k J=1 ζ i with s(i,ζ i )=J pk 1. Since the number o distinct roots o the mod p reduction o is less than min{d, p}, we get that N (p, d, k) min{d, p}p k 1, as desired. Examples o polynomials with more than d k pk 1 roots are given below. These examples show that our bound is within a actor o k o optimality when d p. Example 4.3. (x ) 7 (x 1) 3 with p = 17, k = 7 has 4, 1, 090 roots, which is greater than d k pk 1 = 4, 137, 569. Example 4.4. (x 1) k x has p k roots when d = k + 1 p. The ollowing examples show that we can have p k roots when d p. Example 4.5. (x p x) k is a polynomial o degree pk with p k roots in Z/(p k ). Example 4.6. (x pk p k 1 1)x k has degree p k p k 1 + k and also vanishes on all o Z/(p k ). 6

7 Theorem 4.7. Let p be a prime and Z[x] a polynomial o degree d such that d. Then the number o roots o in Z/(p ) is less than or equal to min{ d, p}p + (d mod k), and this bound is sharp. Proo. Let ζ i {0,..., p 1} be any root o the mod p reduction o, and let s(i, ζ i ) be the greatest integer such that p s(i,ζi) divides each o (ζ i ),..., min(d,k 1) (ζ i ) p k 1. Let δ 1 denote the number o roots o in Z/(p) with s(i, ζ i ) = 1, and let δ denote the number o roots o in Z/(p) with s(i, ζ i ) =. We know that δ 1 + δ d and that δ d by Lemma 4.1. Using this, N (p, d, ) δ 1 + pδ, N (p, d, ) (d δ ) + pδ, N (p, d, ) (d d ) + d p, N (p, d, ) d p + (d mod ). To show that this bound is sharp, we give several examples below or which this bound equals the number o roots o in Z/(p ). Example 4.8. With p = 5, the degree 3 polynomial (x 1) x has (3 mod ) = 6 roots in Z/(p ). Example 4.9. In general, or i, j Z/(p) such that i j, the polynomial (x i) (x j) has d p + (d mod ) roots in Z/(p ) when d and d p. 5 Acknowledgements We would like to thank our advisor, Dr. J. Maurice Rojas, or his assistance and guidance; we would also like to thank Yuyu Zhu or all the suggestions and advice she gave us. We also thank Texas A&M University or hosting and the National Science Foundation (NSF) or unding this program. Reerences [1] Eric Bach and Je Shallit, Algorithmic Number Theory, Vol. 1: Eicient Algorithms, MIT Press, Cambridge, MA, [] Qi Cheng; Shuhong Gao; J. Maurice Rojas; and Daqing Wan, Counting Roots or Polynomials Modulo Prime Powers, Proceedings o ANTS XIII (Algorithmic Number Theory Symposium, July 160, 018, University o Wisconsin, Madison), to appear. [3] Kiran Kedlaya and Christopher Umans, Fast polynomial actorization and modular composition, SIAM J. Comput., 40 (011), no. 6, pp [4] Qi Cheng, Primality Proving via One Round in ECPP and One Iteration in AKS, Journal o Cryptology, July 007, Volume 0, Issue 3, pp [5] Ivan Niven; Herbert S. Zuckerman; and Hugh L. Montgomery, An Introduction to the Theory o Numbers. John Wily & Sons, Inc.,