As a result of this exchange, both sides now share a secret encryption key. Public-key cryptography, or asymmetric cryptography, is a cryptographic system that uses pairs of keys: public keys, which may be disseminated widely, and private keys, which are known only to the owner. Public Key Encryption from a Hardcore Predicate, ElGamal (1984) and Pointcheval-Stern (1996). These actions are passive in nature, as they neither affect information nor disrupt the communication channel. Parse pk=(param,y). Then, the verifier sends a challenge c to the prover, who answers with a response t. Compute the sequence xn-j=fi-1(xn-j+1) for j=1,...,n using the trapdoor ti. Thus, 727(mod30)=727(mod8)(mod30)7^{27} \pmod{30} = 7^{27 \pmod 8} \pmod{30}727(mod30)=727(mod8)(mod30). Therefore, k′=12k' = 12k′=12 because 8+12(mod20)=08 + 12 \pmod{20} = 08+12(mod20)=0. The other key is known as the private key. Let's walk through the Diffie-Hellman key exchange using q=353q = 353q=353 and α=3\alpha = 3α=3. Proposition 3. Padlock icon from the Firefox Web browser, which indicates that TLS, a public-key cryptography system, is in use. Parse sk=(x). Choose a random x∈Zq and compute the group element y=gx∈G. Choose a random x∈Zp-1 and compute y=gx mod p. A single secret key per user for all communications......but the public keys must be reliably distributed (see below). Soundness: Proposition 1. Both Alice and Bob must have information about the key, in order to perform the encryption and decryp-tion. Because the numbers in this example are quite small, an attacker can feasibly compute the discrete log to retrieve either XBX_BXB​ or XAX_AXA​. Output (pk,sk)=((param,y),x), where param=(g,p). The intermediate CAs can certify other CAs' public keys, or they can directly certify the public keys of the final users. This assumption is notknown to be true, but is widely believed. Now A can compute a digital signature for a message m∈Mpk by computing s=Sig(sk,m). or open a Choose a generator g of a cyclic subgroup of order q in the multiplicative group Zp×. Public Key Cryptography: Notes and Links Last updated: Oct 22 20:06:57 2018 Contents 1 Public Key Encryption 1.1 Cryptography in the Multiuser Setting 1.2 Public Key Encryption Syntax 1.3 Man-in-the-Middle Attack 1.4 Public Key Certification 1.5 Hybrid Encryption 1.5 Diffie-Hellman Key Exchange 1.6 Practical Constructions That is, the challenge is now computed as c=H(pk,a,m), and the signature is the pair s=(a,t). The resulting ciphertext is decrypted to obtain the number of positive votes. 1.Asymmetric algorithms rely on one key for encryption and a different but related key for decryption. The RSA cryptosystem is based on the assumption that factoringlarge integers is computationally hard. •Encryption and decryption are carried out using two different keys. A root CA can issue certificates for the intermediate CAs, by signing their signing public keys. Public key cryptography typically uses a pair of keys to secure communications—a private key that is kept secret, and a public key that can be widely distributed. If y=1(modϕ(n))y = 1 \pmod{\phi(n)}y=1(modϕ(n)), then xy(modn)=x(modn)x^y \pmod n = x \pmod nxy(modn)=x(modn). The public key is the two numbers (n,e) and the (public) encryption transformation E(M) is E(M) = Memod n. The secret key is the number d (as well as p,q and φ(n)) and the secret decryption transformation D(C) is D(C) = Cdmod n. T. Johansson (Lund University) 15 / 44 Verify that decrypting a ciphertext returns the encrypted plaintext. For example, everyone can publish their public key to a trusted, public repository instead of sending it and risking interception and forgery. Choose a secure hash function H : {0,1}* → Zq. Homomorphic encryption has many applications, being homomorphic talling in electronic voting the most intuitive one. The two ciphertexts are identically distributed, conditioned to the values of r1 and r2, because r0 and r3 are uniformly distributed. The presence of an additive inverse means that modular addition is reversible. Observe that the previously described attack is no longer possible, since now the attacker must give a preimage of xe mod n by the hash function, which seems to be an infeasible task. For example, 2+8(mod10)=02 + 8 \pmod{10} = 02+8(mod10)=0, because 10÷1010 \div 1010÷10 divides evenly whereas 3+8(mod10)=13 + 8 \pmod {10} = 13+8(mod10)=1 because 11÷1011 \div 1011÷10 yields a remainder of 111. A public key encryption scheme (KeyGen,Enc,Dec) is called (weakly) homomorphic with respect to the operations (M,+) and (C,*) if If someone finds an efficient way to factor a large number into primes, then the security of RSA is effectively broken. For deterministic encryption schemes, the two notions are equivalent, with Rerand(pk,c)=c, since the ciphertexts contain no randomness. There exists some security proofs in the Random Oracle Model that require some extra assumptions. Public key Encryption is important because it is infeasible to determine the decryption key given only the knowledge of the cryptographic algorithm and encryption key. cryptographic algorithm and the public key. Output (pk,sk)=((n,e,H),(p,q,d)). This is just a toy example of a electronic voting system, because there is no verification mechanism to avoid voting more than once (e.g., by encrypting a number greater than one or a negative number), there is no way to prevent voting in a way related to other voter (e.g., rerandomizing other's vote as the own vote), to mention a few examples of all possible attacks. This tutorial covers the basics of the science of cryptography. Similarly, AAA computes k=YBXA(mod353)=24897(mod353)=160k = {Y_B}^{X_A} \pmod{353} = 248^{97} \pmod{353} = 160k=YB​XA​(mod353)=24897(mod353)=160. Bob sends αb(modq)\alpha^b \pmod qαb(modq) to Alice, which is equivalent to 515(mod23)=195^{15} \pmod{23} = 19515(mod23)=19. Notice a tyop typo? While RSA is the most widely used public-key algorithm for encryption and digital signature, a competing system known as Elliptic-Curve Cryptography (ECC), has recently begun to challenge RSA. Output (pk,sk)=((n,e),(p,q,d)). Lecture notes on Cryptography by Boaz Barak. Sig(pk,sk,m): Suppose we have plaintext p=3p = 3p=3, key k=2k = 2k=2 and encryption algorithm p+k(mod10)p + k \pmod{10}p+k(mod10). As A, if we sign with person B’s public key only they can decrypt the message. The scheme is clearly insecure because an attacker can generate an arbitrary pair (m,s)=(xe mod n,x) for a random x, that is accepted by the verification algorithm. A public key is used for encryption and a private key is used for decryption. We assume that an attacker can access YAY_AYA​, YBY_BYB​, qqq, and α\alphaα. Parse pk=(n,e,H). The bible for people who want to implement cryptograpy. Then she publishes pk and keeps sk secret as her long term key. Dec(pk,sk,c): A MODEL FOR NETWORK SECURITY A message is to be transferred from one party to another across some sort of internet. The values of XAX_AXA​ and XBX_BXB​ are private while α\alphaα, qqq, YAY_AYA​, and YBY_BYB​ are public. The value of local secret XAX_AXA​ is equal to the discrete logarithm dlog(α,q)(YA)dlog(\alpha,q)(Y_A)dlog(α,q)(YA​). Then the prover only computes (a,t) with the (interactive) sigma protocol but using as challenge c=H(pk,a). ∀(pk,sk)∈K, ∀m1,m2∈Mpk, Dec(sk,Enc(pk,m1)*Enc(pk,m2)) = m1+m2. Therefore, a more involved strategy would be necessary to build a general proof. Both public and private key are related to each other and unique for User ID. Output 1 if gm ≡ rtyr (mod p), and 0 otherwise. Public-key cryptography, or asymmetric cryptography, is a cryptographic system that uses pairs of keys: public keys, which may be disseminated widely, and private keys, which are known only to the owner. Choose a prime number q of λ bits that divides p-1. We select 7 as our public key eee, as 7 and 160 are relatively prime. ElGamal signature scheme is based on the difficulty of computing discrete logarithms on some cyclic groups. RSA is homomorphic with respect to the product modulo n. Proof. public key PK and computes c = E(PK, m), and sends it to Bob. Public-key cryptography refers to a class of cryptographic systems in which each actor uses two keys: a public key that is known to all, and a corresponding private key that is known only to the actor. If t=0, repeat the procedure by choosing a new random k. Always use standard libraries, as they have been reviewed and tested by experts in the field. The public key typically contains some auxiliary information such the message space Mpk (or simply M) and the signature space Spk (or simply S), corresponding to the particular choice of λ and pk. 1Introduction Public Key Cryptography is a solid tool which ensures the transfer of confidential data upon insecure channels. On receipt of ciphertext CCC, Alice can use her private key, d,n{d, n}d,n, and compute Cd(modn)C^d \pmod nCd(modn) to recover mmm. The RSA alg… KeyGen(λ): The generation of such keys depends on cryptographic algorithms based on mathematical problems t But this implies that P can compute x from the two conversations. On the other hand, Enc(pk,m1+m2) = r3ngm1+m2. An Intensive Introduction to Cryptography. Parse pk=(n,e). With XAX_AXA​ in hand, they can compute SSS and effectively eavesdrop on communication between AAA and BBB. However, a better solution is applying the hash-and-sign paradigm to the plain RSA signature, because this solves two problems at once: preventing the descibed attack and allowing messages of arbitrary length to be signed. The equivalence between the interactive and the non-interactive protocols can only be proven in the Random Oracle Model. If nnn is prime, ϕ(n)=n−1\phi(n) = n - 1ϕ(n)=n−1, because every number smaller than nnn is relatively prime to nnn because nnn itself is prime. Bob can verify that YAY_AYA​ is really from Alice using her RSA public key. Given a modulus MMM, only the numbers that are relatively prime to MMM have multiplicative inverses in (modM)\pmod M(modM). One issue with RSA is that the algorithm is deterministic. To decrypt ccc, we compute m=cd(modn)m = c^d \pmod nm=cd(modn). Upon receiving YAY_AYA​, BBB computes k=YAXB(mod353)=40233(mod353)=160k = {Y_A}^{X_B} \pmod{353} = 40^{233} \pmod{353} = 160k=YA​XB​(mod353)=40233(mod353)=160. n=p∗q=11∗3=33n = p * q = 11 * 3 = 33n=p∗q=11∗3=33 and ϕ(n)=(p−1)∗(q−1)=2∗10=20\phi(n) = (p - 1) * (q - 1) = 2 * 10 = 20ϕ(n)=(p−1)∗(q−1)=2∗10=20. Next, AAA sends 40 to BBB and BBB sends 248 to AAA. First, user AAA selects a random number, XA