We present an idea to describe a polynomial with 2^n distinct integer zeros by an n-tuple of integers via a scheme of n recurring equations. 
We call such an n-tuple an exponential multiplication scheme of size n. Exponential multiplication schemes of size 1,2,3, and 4 are presented. 
Under the assumption that fast exponential multiplication scheme generators exist we suggest a fast randomized heuristic for the factorization problem

Borchert, Bernd

Reinhardt, Klaus

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Exponential Multiplication SchemesBernd BorchertKlaus ReinhardtWSI-2006-10Universita¨t Tu¨bingenWilhelm-Schickard-Institut fu¨r InformatikArbeitsbereich Theoretische Informatik/Formale SprachenSand 13D-72076 Tu¨bingenborchert/reinhard@informatik.uni-tuebingen.dec© WSI 2006ISSN 0946-3852Exponential Multiplication SchemesBernd Borchert Klaus ReinhardtUniversita¨t Tu¨bingen, Sand 13, 72076 Tu¨bingen, Germany{borchert,reinhard}@informatik.uni-tuebingen.deAbstractWe present an idea to describe a polynomial with 2n distinct integer zeros by an n-tupleof integers via a scheme of n recurring equations. We call such an n-tuple an exponentialmultiplication scheme of size n. Exponential multiplication schemes of size 1, 2, 3, and 4 arepresented. Under the assumption that fast exponential multiplication scheme generators existwe suggest a fast randomized heuristic for the factorization problem.1 Exponential Multiplication SchemesConsider the following value x3, defined on x via two intermediate values x1 and x2:x1 := x(x− 11)x2 := x1(x1 − 28)x3 := x2(x2 − 180)The term x3, seen as an polynomial in x, is of degree 8 and has 8 different integer zeros: 0, 1, 2, 4,7, 9, 10, 11. In other words:x3 = x(x− 1)(x− 2)(x− 4)(x− 7)(x− 9)(x− 10)(x− 11).This can be checked by comparing the two expansions of the two sides of the equation. Moreefficiently, this can be checked the following way: x3 = 0 implies by the last equation either case (A)x2 = 0 or case (B) x2 = 180. The case (A) together with the second equation leads to the casedistinction case (A.A) x1 = 0 or case (A.B) x1 = 28. Case (A.A) leads via the first equation tothe first two zeros: x = 0 and x = 11, and case (A.B) leads together with the first equation to thequadratic equation 28 = x(x− 11) which is solved by x = 11/2±√(112 − 4 ∗ 28)/4, which gives thenext two zeros: x = 4 and x = 7. Case (B) together with the second equation gives the quadraticequation 180 = x1(x1 − 28) which has the solutions x1 = 10 (call this case (B.A)) and x1 = 18 (case(B.B)). Together with the first equation case (B.A) gives the quadratic equation 10 = x(x − 11)solved by x = 1 and x = 10, and case (B.B) gives the quadratic equation 18 = x(x − 11) solved byx = 2 and x = 9. This way, we have found all 8 zeros of x3. The following picture shows the casedistinction tree.1x3 = 0x2 = 0 x2 = 180x1 = 0 x1 = 28 x1 = 10 x1 = 18x = 0 x = 11 x = 4 x = 7 x = 1 x = 10 x = 2 x = 9XXXXXXXXXXXXHHHHHHHHHHHH   @@@   @@@   @@@   @@@The interesting point about x3 is that eight integers – namely the zeros of x3 – are determined bythe three integers 11, 28, 180 from the recurring definition above. We will call the triple (11, 28, 180)an exponential multiplication scheme of size 3, see the formal definition later. This ”compression ofinformation” can be done one step higher, i.e. we will present a exponential multiplication schemeof size 4 as follows. The four constants 367, 33642, 125167320, 3665689134307200 determine thefollowing polynomial x4:x1 := x(x− 367)x2 := x1(x1 − 33642)x3 := x2(x2 − 125167320)x4 := x3(x3 − 3665689134307200)The polynomial x4 is of degree 16 and has 16 different integer zeros: 0, 4, 7, 12, 118, 133, 145,178, 189, 222, 234, 249, 355, 360, 363, 367. These can again be verified by a case distinctionprocess like above. As an example, we follow one path of the case distinction tree: x4 = 0 impliescase (A) x3 = 0 or case (B) x3 = 3665689134307200. Case (B) together with the third equationgives the quadratic equation 3665689134307200 = x2(x2 − 125167320). It has the solutions x2 =62583660 ±√(1251673202 − 4 ∗ 3665689134307200)/4 which are x2 = 46739880 (case (B.A)) andx2 = 78427440 (case (B.B)). Case (B.A) together with the second equation gives the quadraticequation 46739880 = x1(x1− 33642). It has the solutions x1 = 16821±√(336422 − 4 ∗ 46739880)/4which are x1 = 1452 (case (B.A.A)) and x1 = 32190 (case (B.A.B)). Case (B.A.B) together withthe first equation gives the quadratic equation 32190 = x(x − 367). It has the solutions x =183.5±√(3672 − 4 ∗ 32190)/4 which are x = 145 and x = 222. These are two of the 16 zeros of x4,the other 14 are found by continuing the other paths of the case distinction tree, as shown in thefollowing picture.2x4 = 0x3 = 0 x3 = 3665689134307200x2 = 0 x2 = 125167320 x2 = 46739880 x2 = 78427440x1 = 0 33642 4260 29382 1452 32190 2520 31122x = 0 367 178 189 12 355 118 249 4 363 145 222 7 360 133 234XXXXXXXXXXXXHHHHHHHHHHHH   @@@   @@@   @@@   @@@AAAAAAAAAAAAAAAAAAAAAAAACall an n-tuple (c1, ..., cn) an exponential multiplication scheme of size n if the polynomial xn definedby the recurrencex1 := x(x− c1)x2 := x1(x1 − c2)x3 := x2(x2 − c3)...xn := xn−1(xn−1 − cn)has 2n distinct integer zeros.We have presented (11,28,180) and (367, 33642, 125167320, 3665689134307200) as examples of ex-ponential multiplication schemes of size 3 and 4, resp. It is easy to verify that the 1-tuple (1) andthe 2-tuple (3,2) are exponential multiplication schemes of size 1 and 2, resp. For n ≥ 5 the authorsdo not know whether exponential multiplication schemes of size n exist.The trees for the examples of exponential multiplication schemes confirm the following observation:An exponential multiplication scheme of size n exists if and only if there exists a complete binarytree of depth n labeled with integers such that all leaf labels are distinct, every label of a non-leafnode is the product of the labels of the two sons, the sums of the two labels of two brothers arethe same for every level of the tree, and the leftmost path is labeled with 0’s. The n integers forthe exponential multiplication scheme can be read as the labels of the right sons of the nodes of theleftmost path.The exponential multiplication scheme (11, 28, 180) was found ”manually” by the authors, the expo-nential multiplication scheme (367, 33642, 125167320, 3665689134307200) was found with the helpof a computer (a Java applet running some hours on a PC).The exponential multiplication scheme (11,28,180) is the lexicographically smallest one of size 3.3The following exponential multiplication schemes of size 3, in lexicographical order, are (13,42,360),(15,50,504), (16,63,720), (17,72,1260), (18,77,1440), (19,78,1080), (19,88,1260), (21,110,1800), and(22,112,2880), the last being a ”multiple” of the first: (22,112,2880) = (2∗11,4∗28,16∗180). Computersimulations suggest that for every n ≥ 22 there is an exponential multiplication scheme starting withc1 = n. The authors do not know a proof of this. It should be mentioned that the multitude seemson average to grow with n: for example, there are 9 exponential multiplication schemes of size 3 withc1 = 47.The quadruple (367, 33642, 125167320, 3665689134307200) is the lexicographically smallest expo-nential multiplication scheme of size 4. The next one is (474, 44933, 500669280, 58651026148915200).2 FactorizationThe factorization problem is the following computational problem:Input: a composite integer n ≥ 2 in binary representation.Output: an integer a with 2 < a < n which divides n.For the prime number problem (a ”yes/no” problem) there are fast algorithms known, the mostefficient of them are randomized. Although the factorization problem seems to be just a littlegeneralization of the prime number problem, no fast algorithm is known for it, see any book oncomplexity or cryptography, for example [Pa94] or [Sti95].Call a program which computes a function 0n → (c1, ..., cn) such that (c1, ..., cn) is a exponentialmultiplication schemes an exponential multiplication scheme generator. Note that the binary lengthof the last component cn will grow exponentially with n. Therefore, an exponential multiplicationscheme generator running in polynomial time does not exist, even if exponential multiplicationschemes of size n ≥ 5 do exist. We will call an exponential multiplication scheme generator fast iffor every n ≥ 2 the integers c1 mod n, ..., cn mod n are computable in polynomial time in |n|.Under the assumption that a fast exponential multiplication scheme generator exists we suggest thefollowing heuristic for the factorization problem.int factor(n);Input: a binary composite number n of length |n|.Do all following computations modulo n.Consider a fast exponential multiplication scheme (c1, . . . , cm) of size m = |n|.Randomly choose an integer x between 1 and n.Compute x1 := x(x− c1), x2 := x1(x1 − c2), x3 := x2(x2 − c3), . . .until the first i is found such that a :=gcd(xi, n) 6= 1 (or i = m).(use the well-known Euclid algorithm for gcd (greatest common divisor))If a > 1 output a, otherwise output ”don’t know” (and retry with another x).We believe that the probability that the algorithm reaches an a 6= 1 converges to 1 with growing nsince there are at least√n numbers < n which have a common factor with n. With the intuitive4assumption that the factors produced by the exponential multiplication scheme are randomly dis-tributed modulo n we expect that such a factor hits a number with a common factor with n alreadyfor i = m/2. The algorithm finds a non-trivial factor of n if not all factors of n are hit in the samestep (in this case a = 0 and we assume that the probability is independent for the next x).We call this randomized procedure only a heuristic (and not a randomized algorithm) because it isunverified in two ways: not only that it assumes the existence of a fast exponential multiplicationscheme generator but moreover we are unable to prove that in fact enough (=exponentially many)zeros remain when considering the polynomial as a polynomial modulo n.Note that in the case a = 0 we would be able to find a factorization for sure if we could also calculateany single integer in the tree and decend until factors come from different subtrees.3 Open Question and AcknowledgementsThe main open question is if there are exponential multiplication schemes of size n ≥ 5. And if theyexists, can they be computed fast modulo a given number?A way to find a solution may be to understand the size 3 case better. Is every integer a ≥ 22 thefirst component (a, , ) of an exponential multiplication scheme of size 3? Once this is understoodone may try to attack the size 4 case and higher size cases.Thanks to Andreas Krebs for some discussions.References[Pa94] Chr. Papadimitriou. Computational Complexity. Wiley, 1994.[Sti95] D. Stinson. Crytography - Theory and Practice. CRC Press, 1995.5

Exponential Multiplication Schemes

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Exponential Multiplication Schemes

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