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Report CopyRight/DMCA Form For : Public Key Cryptosystems With Secret Encryptor And Digital

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2 B VERKHOVSKY, Step3 5 Using the public key of the sender following Example4 a 35 and p 1 a 837 have binary. Bob computes weights 3 and 7 respectively Therefore D requires. fewer multiplications than d Hence it is faster to com. e u b mod p 3 4, pute D than d In general it is computationally advanta. where e is their mutual secret encryptor geous if both Alice and Bob select their secret keys a and. Remark3 1 Parameters a b and e must be distinct from b with smaller binary weights However these addi. q tional constraints provide clues for a potential intruder. cryptanalyst, 4 Several Ways to Hide Information Yet there is an alternative algorithm for modular. multiplicative inverse AAMMI proposed by the author. Suppose Alice wants to send a plaintext m which is rep. of this paper in 10 and analyzed in 11 Extensive, resented in a numeric for m where the message m satis. computer experiments demonstrated that the AAMMI, algorithm is computationally more efficient than 4 5.

2 m p 2 4 1 see Tables 1 3 below, Encryption Alice encrypts message m and sends it to In spite of computational simplicity applications of. Bob who decrypts it the encryptor e either in 4 2 or in 4 3 have negative. Remark4 1 Alice has several alternatives she can ei sides if at least one plaintext block mi becomes. ther create a ciphertext known an intruder can deduce every plaintext mk In. deed since 4 2 implies,c me mod p 4 2,e ci mi 1 mod p 4 9. which is an equivalent of the ElGamal scheme 3 or,consider then. C m e modp 4 3 mk ck mi ci 1 mod p 4 10, which is a poly alphabetic shift cipher 8 9 or she can Analogously the encryption 4 3 is vulnerable for. split the secret key e into two parts e11 and e12 and cryptanalysis if at least one plaintext block mi be. then consider comes known the intruder can compute every plaintext. c1 e11m e12 modp 4 4 mk Indeed since 4 3 implies,e ci mi mod p 4 11.

The encryption 4 4 is an equivalent of the poly al. phabetic affine algorithm 2 Yet another option is de then. scribed in Section 6,mk ck ci mi mod p 4 12, In general Bob can apply any ordered binary operator. B c mBe modp provided that the inverse of e ex In order to overcome these deficiencies the sender must. ists computable and it is unique for this operator compute dedicated encryptor e m for every plaintext. Below we describe Alice s and Bob s actions if 4 2. or 4 3 are used If 4 2 is applied Alice transmits the Table 1 Computation of MMI. ciphertext c to Bob via an open channel q 53 e 49 4 1. Decryption Using Alice s public key u and his private. Stack 1 12,key b Bob computes a decryptor,z 13 12 1 0. d u p 1 b mod p 4 5, and finally he recovers the plaintext Table 2 Computation of MMI of 195 modulo p 1 862. f dcmodp m 4 6 862 195 82 31 20 11 9 2 1,Stack 4 2 2 1 1 1 4. In 4 3 case the receiver computes his decryptor,z 389 88 37 14 9 5 4 1 0.

D u b mod p 4 7,and Table 3 Algorithm for MMI 1 a t 1. F C D mod p m 4 8 a0 t a1 a a2 an 1 an, Furthermore there are several ways see 4 5 and stack q1 q2 qn 1. 4 7 to compute the decryptors d and D respectively. b0 z b1 b2 bn 1 1 bn 0, One requires exponentiation p 1 a and another a In the. Copyright 2013 SciRes IJCNS,B VERKHOVSKY 3, block m and respectively the receiver must compute the secure transmission of information. dedicated decryptor d m to recover this plaintext Need First of all notice that there is no necessity to search. less to say that this requires additional computational for a generator in order to compute the hints and public. efforts i e it slows the transmission of information keys. Example1 Alice sends message m to Bob, 5 Dynamic Establishment of Secret Key Let p 47 g 4 b 25 x 33 and m 42.

between Sender and Receiver Encryption,Step6 1 1 Bob pre computes his public key. Step5 1 The sender Alice selects a large safe prime p. w g b 22b mod p 250 16 mod 47, and transmits it to the receiver Bob via an open chan. Step6 2 1 Alice computes her hint, Step5 2 Respectively Alice and Bob randomly select h g x 22 x 266 6 mod 47. large integers a and b as their private keys, Step6 3 1 Using Bob s public key w Alice computes her. Step5 3 Alice and Bob independently compute their,encryptor e and the ciphertext C.

public keys,e w x mod p 1633 36 mod 47,u p 22 a mod p 5 1. C m e mod p 42 36 31 mod 47, Step6 4 1 Alice sends the ciphertext and her hint to Bob. w p 22b mod p 5 2 C h 31 6,Decryption, Step5 4 The sender Alice selects an ephemeral secret. Step6 5 1 Using his private key b Bob computes his. key x and using the public key of the receiver she com. putes an ephemeral encryptor,e wx mod p 5 3,D hb 625 mod 47 36 g bx. and her hint Step6 6 1 Bob decrypts the ciphertext. F C D 31 36 mod 47 42 m,h p 22 x mod p 5 4,i e he recovers the plaintext m.

Remark5 1 Parameters a b e and x must be distinct,Example2 Let now p 863 g 4 a 35 y 21 m 754. from 1 q and p 1, and suppose Bob wants to send a secret message m to. Step5 5 The sender computes the ciphertext,c me mod p 4 2 Encryption. and transmits c h to the receiver Step6 1 2 Alice pre computes her public key. Step5 6 Using his public key the receiver computes an u g a 270 mod 863 660. ephemeral decryptor,Step6 2 2 Bob computes his hint. d h p 1 b mod p 5 5,h g y 242 mod 863 392,and then computes.

Step6 3 2 Bob computes the ephemeral secret encryptor. f dcmodp m 4 6 e and ciphertext C, Remark5 2 As an alternative the sender computes the e u y 66021 mod 863 171. ciphertext,C m e 754 171 mod863 62 5 1,C m e modp 4 3. Step6 4 2 Bob sends the ciphertext C and his hint h to. and sends C h to the receiver, In this case the receiver computes his ephemeral de. Decryption,cryptor D hb mod p 4 7 and then computes. Step6 5 2 Using her private key a and hint h Alice. F C D mod p m 4 8 i e he recovers the,computes her ephemeral decryptor D.

D h a 39235 mod 863 171 g ay,original plaintext,Step6 6 2 Alice decrypts the ciphertext. 6 Numeric Illustrations 1,f C D 62 171 mod863 754 m. In the Examples 1 and 2 modifications that simplify the. computation are introduced and as a result accelerate i e she recovers the plaintext m. Copyright 2013 SciRes IJCNS,4 B VERKHOVSKY, Remark6 3 If the plaintext is intelligible Alice accepts it ed 1 p 1 k 7 10. as legitimate i e this protocol also provides a digital. signature Therefore 7 8 and the FLT imply that,f 2 c2d med. 7 Encryption via Exponentiation with Secret 7 11,Encryptor EvESE m m p 1 m 1k mod p m.

The following encryption decryption scheme requires Remark7 2 Although 7 8 and 7 9 resemble the RSA. fewer exponentiations than the schemes described above protocol 4 there are three distinct features. System design 1 the encryptor e is a secret not public key. a Users Alice and Bob select their private keys a and b 2 modulo reduction is executed with the public prime p. respectively and then compute their public keys rather than with the product of two secret primes indi. vidually selected for each user,u2 g a mod q 7 1, 3 this scheme does not require additional computations. and for sender identification in other words it automatically. provides the digital signature,w2 g b mod q 7 2, b each user computes his her common secret encryptor 8 Numeric Illustrations 2. e u w mod p,2 7 3 Example3 Let p 107 g 4 a 33 b 28 and m 42. System design either a or b must be selected in such. c if e and p 1 are relatively prime a way that gcd e q 1 otherwise the MMI of e mo. i e if dulo q does not exist u2 g a mod q 433 mod 53 7. gcd e p 1 1 7 4 and, then the users compute an integer d that satisfies the w2 g b mod q 428 mod 53 16. e u2b 728 w2a 1633 mod 53 49,edmod p 1 1 7 5,d eq 2 mod q 4951 mod 53 13 8 1.

else if gcd e p 1 v, they consider Remark8 1 If q is a very large integer the computation. of the multiplicative inverse in 8 1 requires many mul. tiplications even if the square and multiply algorithm. and compute d in 7 5 To find d decryptor both users is used Yet computation of the MMI is much simpler. compute using the algorithm described in 10 as it is shown in the. d eq 2 mod p 1 7 7 Since the number of columns in the Table 1 is even. Remark7 1 As an alternative the users can apply the al then d z. gorithm for MMI 10 11 see Examples 3 4 and Table Verification demodq 13 49mod53 1. 3 Explanations are provided in Section 9, Encryption Encryption A sender computes the ciphertext c2 and. d A sender of message m computes the ciphertext transmits it to the receiver. c2 me mod p 7 8 c2 me mod p 4249 mod107 48, e the ciphertext c2 is sent to a receiver via an open Decryption using the decryptor d the receiver recovers. communication channel notice that no any hint is neces the plaintext. sary f 2 c2d mod p 4813 mod107 42 Therefore the me. Decryption ssage m is correctly recovered by the receiver. f The receiver computes Example4 Let now p 863 g 4 a 35 b 49 and. f 2 c2d mod p m 7 9 System design, Validation of EvESE Algorithm u2 g a mod q 435 mod 863 660. Since gcd e p 1 1 i e e is odd and distinct from, q then 7 5 implies the existence of an integer k such.

that w2 g b mod q 449 mod863 213,Copyright 2013 SciRes IJCNS. B VERKHOVSKY 5, e u2b 66049 w2a 21335 mod 863 195 ever using the carrousel DHKE we can generalize the. proposed algorithm for communication among several. Each user computes the MMI of 195 modulo 862 de users. tails are shown in the Table 2, Since the number of columns in the Table 2 is odd 11 Algorithm EvESE Modification. then z p 1 d therefore,If e is a power of 2 or e s 2t where s 3 and s is. d 2q 389 473, a small integer then the algorithm does not work since.

Encryption the decryptor either does not exist or the encryptor is too. c2 m e mod p 756195 mod 863 166 small One option is to select new private keys a and b in. hope that new e will be more suitable However this is a. Decryption non deterministic procedure Besides it requires many. f 2 c2d mod p 166 473 mod 863 756 m multiplications of multidigit long integers. There are other simple options if p 11 e is even and. 9 Algorithm for MMIazmodt 1,a if e q 1 then assign e e 1. In this section Table 3 briefly describes the algorithm b if e q 1 then assign e e 1. for the MMI For more details see 10 and 11 Finally if e 2 then e q 2. In the following Table 3 both integers a and t are the. inputs and integer z is the output 12 Conclusions, Assign a0 t a1 a Notice that the ElGamal algorithm is just one of several. for k 1 2 n 1 iterate and store in a stack the quo constructive demonstrations how to dynamically apply a. tients secret key for secure communication Indeed the inverse. qk ak 1 ak value of the decryptor is the same as encryptor There. fore both parties are dynamically establishing the com. ak 1 ak 1 qk ak mon secret key encryptor e and then use it to hide the. repeat until an 0 or an 1 message m by multiplying it on the encryptor Another. if an 0 then the MMI does not exist possibility instead of multiplying they add the encryptor. if an 1 then assign bn 0 and bn 1 1 for k e to m or can consider exponentiation m e mod p There. from n 1 down to 1 iterate bk 1 qk bk bk 1 fore in the case of addition in 4 3 they eliminate two. if n is odd then z b0 else z t b0 multiplications for every plaintext since D e. However both protocols require twice more exponen, 10 Complexity Analysis of EvESE tiations than the EvESE algorithm described in the Sec. Cryptosystem tion 7 and demonstrated in Section 8 As the analysis. shows the most efficient is to use the exponentiation 7 5. On the system design level each user performs two ex for the encryption. ponentiations to compute their public key 7 1 and 7 2 I wish to express my appreciation to Dr Roberto. and the secret encryptor 7 3 Rubino for his comments that improved the style of this. For the encryption it is necessary to perform only one paper. exponentiation 7 5 Analogously for decryption every. receiver performs only one exponentiation 7 7 Al, though for the purpose of maintaining a high security REFERENCES. level we need periodically select new private keys and 1 W Diffie and M E Hellman New Directions in Cryp. re compute the encryptor and decryptor we do not need tography IEEE Transactions on Information Theory. to send the hints with every block of transmitted message Vol 22 No 6 1976 pp 644 654. doi 10 1109 TIT 1976 1055638, like it is done in the ElGamal algorithm Since this algo.

rithm is based on the difficulty of the DLP it has certain 2 A J Menezes P C van Oorschot and S A Vanstone. Handbook of Applied Cryptography CRC Press Boca, advantages over the RSA algorithm based on factoriza Raton 1997. tion One of them is that the encryptor and decryptor pro. 3 T ElGamal A Public Key Crypto System and a Signa. vide a digital signature sender identification since they ture Scheme Based on Discrete Logarithms IEEE Trans. are computed for communication between the specific actions on Information Theory Vol 31 No 4 1985 pp. pair of users Alice and Bob On the other hand in the 469 472 doi 10 1109 TIT 1985 1057074. described form the algorithm cannot be applied for 4 R L Rivest A Shamir and L M Adleman A Method. broadcasting of secret messages to several users How of Obtaining Digital Signature and Public Key Crypto. Copyright 2013 SciRes IJCNS,6 B VERKHOVSKY, systems Communication of ACM Vol 21 No 2 1978 8 J Katz and Y Lindell Introduction to Modern Cryptog. Public Key Cryptosystems with Secret Encryptor and Digital Signature Boris S Verkhovsky Computer Science Department New Jersey Institute of Technology Newark USA Email verb73 gmail com Received October 20 2012 revised November 28 2012 accepted December 25 2012 ABSTRACT This paper describes and compares a variety of algorithms for secure transmission of information via open

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