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Int. J. Comm unications, Networ k and System Science s, 2010, 3, 639-644 doi:10.4236/ijcns.2010.38086 Published Online August 2010 (htt p:/ / www.SciRP.org/journal/ijcns) Copyright © 2010 SciRes . IJCN S Potential Vulnera bility of Encrypted Messages: Decomposability of Discrete Logarithm Problems B oris S. Verkhovsky Computer Science Department, Ne w J ersey Inst itute of Technology, Newark, USA E-mai l: verb@njit.edu Recei ved May 10, 2010; r evis ed June 15, 2010; accept ed Jul y 21, 2010 A bst ract This paper provides a framework that reduces the computational complexity of the discrete logarithm prob- lem. The paper describes how to decompose the initial DLP onto several DLPs of smaller dimensions. De- com posability of the DLP is an indicator of potential vulnerability of encrypted messages transmitted via op en cha nne ls of the Int ernet or wit hin cor porate net wor ks. S ever a l nu m er ica l examp l es il lus t r ate the fr a me - work and s how its computationa l efficiency. Keywords: Network Vulnerability, System Security, Discrete Logarithm, Integer Factorization, Multi-Level Decomposition, Complexity Analysi s 1. Introduction and Problem Stat ement The cryptoimmunity of numerous public key cryptogra- phi c protocols is based on the computational complexity of th e discrete logarithm pr oblems [1,2]. A DLP finds an integer x satisfying th e equation mod x g ph= . (1) Here 21gp≤≤− ; 11hp≤≤− (2) and p is a large prime. In (1) g, p and h are inputs, and the unknown integer x must be selected on the interval [ ] 1, 1p− . Two trivial cases: if h = 1, t hen x = p – 1; If h = g, t hen x = 1. If h is neither 1 nor g, then x must be selected on the int e rval [2, p – 2]. If g is a gen erator, th en (1) always has a solution, oth- erwise t he exi s tence of a sol ut ion is not guar anteed. For instance, if p = 7 and g = 2, then the DLP 2 mod75 x = doe s not have a s olution. Various a lgo rithms for solving the DLP were propose d an d th eir computati ona l compl exi ti es wer e an alyzed over the last forty years [3-15]. This paper provides the algorithmic framework that reduc es the com puta ti onal complexi ty of the DLP. The paper describes step-by-step procedure for deco- mposition of the initial DLP onto several DLPs with smaller dimensions. Several examples illustrate the dec- omposition algorithm and highlight its computational effi cienc y. Let 111 :; :; :ggh hxx== = ; 1 :1qp= − and 12 12p rr−= . (3) Her e it is a ssumed that integer fact or s 12 and rr in (3) are known or can be determined using existing algo- rithms for integer factor ization [5,16,17]. Proposition: Let ( ) 1: 1/Rp q= − ; (4) if ( ) |1qp− , then 1 R is an intege r (4). Let’s define 1 21 : mod R gg p= ; (5) 1 21 : mod R hh p= ; (6) If an integer 2 x is a solution of equation 2 22 mod x g ph= , where [ ] 2 0,xq∈ , (7) then q divides 12 xx− . Proof: Let’s multiply both sides of the Equation (1) by 2 1 mod x gp − [18], and fin d 2 x , such that 2 11 mod x hg p − (8) has a root of power q . By Euler ’s criterion [5] such a root exists if an d only if ( ) ( ) 2 1/ 11 mod 1 pq x hg p − − = (9) Using nota tions (4)-(6), rewr ite (8) as 2 22 mod 1 x hg p − = (10) or as Equation (7). Q. E.D. B. VERKHOVSKY Copyright © 2010 SciRes . IJCNS 640 Therefore, the unknown 1 x can be r epresent ed a s 12 3 xx qx= + (11) where the int ege r x3 must be on the interval [ ] [ ] 33 0,(1) /0,xpq q∈−= (12) After 2 x is determined, we need to find an integer 3 x , for which the following equation holds 23 11 mod x qx g ph + = . (13) This equation can be rewritten as ( ) ( ) 32 1 11 mod xx q g hgp − = (14) where in contrast with the BSGS algorithm, the value of 2 x is already known. Let ( ) 3 1/ 31 : mod pq gg p − =; (15) and 2 3 11 : mod x h hgp − = . (16) 2. Divide-and-Conquer Decomposition: Illustrative Example-1 Let’s solve 1 2mod947 273 x = , (17) i.e., here 11 2; 947;273gp h= == , and [ ] 1 1,946x∈ . Let 1 :1qp= − . Si nce 1 12 22 1143q rr= =×× , select ( ) 201 minmax ,(1)/43 zp qzp z ≤≤− = −= . Then 1 12 : /22R qq= = ; 1 22 21 :mod2 mod947 R gg p= = 41= ; and 1 22 21 :mod273 mod947283 R hh p= == . Ther efore we n eed to solve t he DLP(2): 2 41 mod947283 x = (7), (18) where [ ] 2 1, 42x∈ . Remark1: Noti ce tha t the int er val of un cer ta int y [ 1, 42] for 2 x is much smaller than the corresponding interval of uncert ainty [1, 946] for 1 x . Equation (18) can be solved using any algorithm for the DLP [3,6,8-10,12]. In this example 2 x = 39 a nd 2 43q= . Therefore 13 39 43xx= + , where ( ) [ ] 32 0, 1/0,22x pq∈− = . To fin d 3 x solve th e DLP(3): ( ) ( ) 3 43 39 2273 2mod947 x− = × , wh ich is equivalent to ( ) 3 367273111946 mod947 x=×= . (19) Therefore 3 11x= . Verification: 11 367 mod947946= . (20) Finally, 1 394311512x=+×= . 3. Multi-Level Decomposition: Illus trative Example-2 Initial DLP(1): Find an integer 1 x , such that 1 30 mod9999145636 x= , (21) where [ ] 11,99990x∈. Because 99990=303*330, select 2 330q= and represent the unknown 1 x as 12 3 330xx x= + . Si nce ( ) 12 :1 /303Rp q=−= ; then 303 21 :mod99991 151gg= = ; and 303 21 :mod99991 64099hh== . Remark2: To better describe th e concept of decompo- sition, a more suitable system of notations is considered bel ow in th e fol lowi ng Table 1. These notati ons are us ed to de scribe th e process of sol vi ng three D LPs . DLP(2): S olve 2 22 mod99991 x gh= , i.e., 2 151 mod9999164099 x = , where [ ] 2 0,330x∈ . (22) The so lut ion is 2 115x= ; inde ed 115 151mod99991 64099= . Therefore 3 1 115 330 3030mod99991 45636 x x+ = = . Consider the equation ( ) ( ) 3 330 115 303045636 mod99991 x− = × . Let 330 3: 30mod99991g= = 2593; and ( ) 115 3 115 : 3045636 9665845636mod99991 49845 h − = × = × = Ther efore, we need to solve DLP(3): 3 2593 mod9999149845 x= , where [ ] 3 0,303x∈ . (23) It is easy to ver ify that 3 47x= . Finally, 12 xx= 23 115330 4715625qx+= + ×= . Decomposit ion of DLP(2): Solve 2 22 mod x g ph= , (24) where [ ] 22 0,xq∈ = [0,330]. B. VERKHOVSKY Copyright © 2010 SciRes . IJCNS 641 Tabl e 1. Solu tions of DL P(1) via th e dec ompos ition of DLP (2) and DL P(3). DLP(1): 1 11 mod x g ph= Proble m A Problem B Problem C Inputs { } 11 ;;gph {2; 9 47; 273} {2 ; 947; 641} {30; 99991; 45636} 1 12 :12... t qprr r= −= 2 11 43×× 2 11 43×× 2 2311 101××× DLP(2): ( ) 21 min max,/ z qzq z= 2 q = 43 2 q = 43 2 q = 330 ( ) 22 : 1/Rp q= − 2 R = 22 2 R = 22 2 R = 303 2 21 : mod R gg p= 241g= 241g= 303 2 30 mod99991151g= = 2 21 : mod R hh p= 2283h= 2283h= 303 245636 mod99991h= = 64099 2 22 mod x g ph= , [ ] 22 0,xq∈ [ ] 2 0,43x∈ ; 239x= [ ] 2 0,43x∈ ; 223x= [ ] 20,330x∈ ; 2115x= DLP(3): 1 23 q qq= , ( ) 33 : 1/Rp q= − 3 R = 43 3 R = 43 3 R = 330 3 31 : mod R gg p= 3 367g= 3 367g= 330 330 mod999912593g= = 2 3 11 : mod x h hgp − = 3946h= 3643h= 1 30 mod96658fp − = = , 2 3 96658 mod9381 x hp== 3 33 mod x g ph= , [ ] 33 0,xq∈ [ ] 3 0,22x∈ ; 3 11x= [ ] 3 0,22x∈ ; 314x= [ ] 3 0,303x∈ ; 347x= Solution of DLP(1): 1 2 23 x x qx= + 1 3943 11 512 x= + ×= 1 2343 14 625 x=+×= 1 115 330 47 15625 x= +×= Remark3: Notice that the interval of uncertainty in DLP(2) is n ot [1, p – 1], but [ ] 22 1,xq∈ , which is much smaller than [1, p – 1]. In stead of solvin g (24) dir ectly usin g an existing DLP algorithm, we can again apply the meth od of dec om posi- tion described above. Consider a factor 4 q of 2 q that i s c los e to the s quare root of 2 q = 330: ( ) ( ) 2 42 0 minmax , / minmax,330/30 zq z qzqz zz ≤≤ = = = (25) Let’s represent the unknown in (24) as 2445 xx qx= + , (26) [] [] [ ] [ ] 44 55 24 where 1,1,30 and 1,:/1,11 xq xq qq ∈= ∈== . (2 7) Let us now investigate whether 2 h has an integer root of powe r 30 modulo p. By Euler ’s criterion, such a root exists if and only if ( ) 4 1/ 2mod 1 pq hp −= . (28) However, if ( ) 4 1/ 2mod 1 pq hp −≠ , find an integer 4 x , which s atis fi es the equat ion ( ) ( ) 4 4 1/ 22 mod 1 pq x hg p − − = . (29) Let ( ) 4 1/ 42 : mod pq ggp − =; (30) and ( ) 4 1/ 42 : mod pq hhp − = . (31) Now we need t o sol ve th e equat ion 4 44 mod x g ph= , (32) where [ ] 40,30x∈ . An d aga i n, th e Equation (32) i tself is also a DLP with a much smaller interval (27) for x4 than the interval for 2 x in (24), and so on. 4. Mu lti-Level Decomposition: Illustrative Example-3 First level: Let’s s olve the equation 1 11 mod x g ph=, where g = 2, p = 4,000,000,003,231; and h = 3,024,336,139,227. Then p – 1 = 863*2310*2006491, where 863 and 2,006,491 ar e pr imes. In this case the initial DLP(1) 1 11 mod x g ph= ; is de- composable into two sub-problems: DLP(2) a nd DLP(3). DLP(2): Compute ( ) 2 1/ 21 1993530 : 2mod4000000003231 pq gg − = = =3278213345371; and ( ) 2 1/ 21 : pq hh − = =302433613922719 935 30 mod 4000000003231 =2084778340641. Solve 2 22 mod4000000003231 x gh= , where 22 0xq≤≤ 2006491= ; It is easy to verify the solution 2 1853979 2006491x= ≤ . DLP(3):Compute ( ) 3 1/ 2006491 31 :2 mod4000000003231 pq gg − = = = 3767306619080; and B. VERKHOVSKY Copyright © 2010 SciRes . IJCNS 642 2 3 11 1853979 30243361392272000000001616 mod 4000000003231 : 3024336139227 629308445687 mod4000000003231 2623468766941. x h hg − × = = ⋅ = ×⋅ = Solve ( ) 3 33 mod x gh p= , where ( ) 3 32 0146221 /1993530;xqp q≤=≤=− = and 1 23 q qq= . Then 1223 =1,853,9792,006,49114,622 =29,340,765,381. xx qx ∗ = + + It is easy to verify that the solution 3 14622 1993530x= ≤ . Comparison of complexities: While the size of the required memory/storage for DLP(1) equals 1 1 2000000Tp = −= ; the corresponding memory requirement for DLP(2) and DLP(3) are res pe ctively 22 12006491 1416Tq = −== and 33 11993530 1411Tq = −== . Ther efore the spe ed -up ratio ( ) ( ) 1 23 /2000000 /14161411707.ST TT=+=+ = Thus the decomposition algor ithm for solving DLP(1) via DLP(2) and DLP(3) is 707 times faster than a direct solution of t he original DLP(1). 5. Second-Level Decomposition: Solution of DLP(3) Remark4: The second problem, DLP(2), cannot be sol- ved by decomposition since q2 = 2,006,491 is a prime integer. However, the third problem, DLP(3), is decom- posable, therefore the speed-up ratio S can be further increas ed. Inde ed, s elect ( ) 3 63 0 :minmax/ , zq qq zz ≤≤ = = 2310. Let’s repr es ent 3 x as 36 67 xxqx= + , where 66 0 2310xq<<= and 77 0 863xq<<= , and solve DLP(3) by decomposition into DLP(6) and DLP(7). DLP(6): Compute ( ) 6 1/ 63 : mod pq gg p − = ; and ( ) 6 1/ 63 : mod pq hh p − =; where 67 3 1993530qqq= = ; and solve ( ) 6 66 mod 1993531 x gh= ; { 66 0 2310xq<<= }. DLP(7): Compute ( ) 7 1/ 73 : mod pq gg p − = ; and 6 7 33 : mod x h hgp − = ; and solve ( ) 7 77 mod 1993531 x gh= ; { 77 0 863xq<<= }. Then 66 48Tq = = and 77 29Tq = = . Therefore ( ) ( ) 1 267 / =2000000/14164829 =2000000/1493 =, ST TTT= ++ ++ 1339.6 wh ich implies that by decomposing th e original problem DLP(1) into three sub-problems {DLP(2), DLP(6) and DLP(7)}, we can solve the initial DLP(1) 1340 times faster than if we directly solve it without employing de- composition. In general, the speed-up increases as the size of p in- creas es . 6. Computational Considerations It is quite re as ona ble to ask under what c onditions should we stop t he de compositi on of a DLP(k) and tr y to solve it direct ly. Here are th e major iss ue s that must be taken in to the consideration: 1) Feasibility of factoring 2 21kkk q qq+ = in such a way that ( ) 2 1/ 2 :mod 1 k pq kk ggp − = ≠± . (33) For instance, if ( ) 24 |2 1qq p− , then ( )()( ) ( )( ) ( ) 4 42 24 1/ 1/ 1/ 42 1 1 /2 2 1/ 1 : 1mod pq pq pq p p qq ww w wp − −− − − == == ± , (34) where w = {g, h}. In such a ca se Equation (32) ha s on ly trivial solutions { 0 or 1} or no solution if 44 1 and 1gh==− . 2) Magnitude of the overhead computations required to fin d 2 21 and kk gg + an d t h en t o s olve t h es e t wo DLP s, provided that these intermediate computations do not B. VERKHOVSKY Copyright © 2010 SciRes . IJCNS 643 be come too “ c ostly”. Re mark 4: Analogously, we can solve DLP(3) by de- composing it into two DLPs with smaller intervals of uncertainty for the corresponding unknown s. 7. Algorithmic Decomposition of DLP(k) Suppose t hat we need to solve DLP(k) mod k u kk g ph= , (33) where [ ] 0, kk uq∈ . If k q is a prime or if factors of k q are unknown, then (33) can be solved by a n algori thm for DLP such as: BSGS, Pollard’s rho-algorithm, Lenstra’s number field algorithm etc. However, if k q cd= , where both c and d are integers, then the DLP(k) can be reduced to solving two less complex DLPs: DLP(2k) and DLP(2k + 1). Let 2 21kkk q qq + = ; DLP(2k): Solve 2 22 mod k ukk g ph= ; (34) where 2: k qc= and [ ] 20, k uc∈ ; (35) ( ) : 1/ kk Rpq= − ; (36) 2 : mod k R kk gg p= ; (37) and 2 : mod k R kk hh p= ; (38) DLP(2k+1): Solve 21 21 21 mod k ukk g ph + ++ = ; (39) where [ ] 21 0, / kk u qc + ∈ , (40) ( ) 21 21 : 1/ kk R pq ++ = − ; (41) 21 21 : mod k R kk gg p + + = ; (42) and 2 21 : mod k u k kk h hgp − + = . (43) 8. Conclusions Provi de d t hat we know how to fa c tor p – 1, we can reduce the initial DLP(1) to two discrete logarithm problems: DLP(2) and DLP(3), for solut i on of whi ch the bes t kn own algorithms can be implemented. The decomposition can be im plem ent ed r ecursi vel y for solution of th e DLP(k) by reducing it to a pair of DLP(2k) and DLP (2k + 1). 9. Acknowledge ment s I express my appreciation to R. Rubino and P. Fay for their comments and suggestions that improved the style of this paper. 10. R eferen c es [1] W. Diffie and M. E. Hellman, “New Directions in Cry- ptography,” IEEE Transactions on Information Theory, Vol. 22, No. 6, 1976, pp. 644-654. [2] T. ElGamal, “A Public Key Cryptosystem and a Digital Signature Scheme Based on Discrete Logarithms,” IEEE Transactions on Information Theory, Vol. 31, No. 4, 1985, pp. 469-472. [3] L. M. Adleman and J. DeMarrais, “A Sub-Exponential Algorithm for Discrete Logarithms over all Finite Fields,” Mathematics of Computation, Vol. 61, No. 203, 1993, pp. 1-15. [4] E. Ba ch, “ Di scr ete Logar it hms and Fact oring, ” Technical Repor t: CSD-84-186, University of California, Berkeley, 1984. [5] R. Crandall and C. Pomerance, “Pri m e N umbers: A Com- put ati onal Perspe ctive,” The Qua drat ic Sieve Fact ori z ation Method, Springer, Berlin, 2001, pp . 227- 244. [6] A. Enge and P. 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Shanks, “Class Number, a Theory of Factorization and Genera,” Proceedings of Symposium in Pure Mathematics, Vol. 20, American Mathematical Society, Providence, 1971, pp. 415-440. [12] J. Silverman, “The xedni Calcul us a nd t he Elliptic Cur ve Di scr ete Loga ri thm Pr oblem,” Designs, Codes and Cryp- tography, Vol. 20, N o. 1, 2000, pp. 5-40. [13] D. C. Terr, “A Modifi cat i on of Sha nks’ Ba by-Step Giant- Step Algorithm,” Mathematics of Computation, Vol. 69, No . 230, 2000, pp. 767-773. [14] B. Ver khovsky, “Generalized Baby-Step Giant-Step Alg- orithm for Discrete Logarithm Problem,” Advances in Decision Technology and Intelligent Information Systems , International Institute for Advanced Studies in Systems Research and Cybernetics, Baden-Baden, 2008, pp. 88- 89. [15] R. Zucche rato, “The Equivalence between Elliptic Curve and Quadratic Function Field Discrete Logarithms in Characteristic 2,” Algorithmic Number Theory Seminar B. VERKHOVSKY Copyright © 2010 SciRes . IJCNS 644 ANTS-III, Lecture Notes in Computer Science, Springe r, Berlin, Vol. 1423,1998, pp . 621-638. [16] J. P. Pollard, “A Mont e Carlo Method for F act ori zation,” BIT Numerical Mathematics, Vol. 15, No. 3, 1975, pp. 331-334. [17] C. Pomerance, J. W. Smith and R. Tuler, “A Pipeline Architecture for Factoring Large Integers with the Qua- dratic Sieve Algorithm,” S IAM Journal on Computing, Vol. 17, No. 2, 1988, pp. 387-403. [18] B. Verkhovsk y, “Multiplicative Inverse Algorithm and its Complexity,” Proceedings of International Conference on System Research, Informatics & Cyber netics , Baden- Baden , 28-30 July 1999, pp. 62-67. APPENDIX Numeric example as an exercise Let p = 5,000 ,491 ; t he n 1990 5051p−= × Let 11 2 and 1020305gh= = . In this ca se DLP(1) is ( ) 1 21020305mod5000491 x = , where the unknown [ ] 1 1, 1xp∈− . The DLP(1) is decomposable into two sub-problems: DLP(2): ( ) 2 22 mod x ghp = {see ( 4)-(6)}, wher e [] [] 22 1, 1,5051xq∈= ; and DLP(3): ( ) 3 33 mod x gh p= {see (15) and (16)}, where [ ] [ ] 33 1,1,990xq∈= . Therefore 12 23 xx qx= + . Remark5: The r eader n ow has an opportunity to solve this problem himself since values required for the de- composition are purposely omitted. From DLP(2) and DLP(3) we find that 2 1947 5051x= < ; and 3 470 990x= < . Finally, 1 194750514702375917.x=+×= Overall complexity: the storage requirement for DLP (2) and DLP(3) equal to 71 and 31 respectively, yet the size of required storage for the DLP(1) is 2236, i.e. al- most 32 times larger . |