In this problem we will crack RSA. Suppose the parameters for an instance of the RSA cryptosystem are N = 31313, e = 7. We have obtained some ciphertext y = = a) Factor N = 31313 into its constituent primes p and 9. min(p, q) = max(p, q) b) Compute (N). (31313) c) Compute d, the decryption exponent. d = = = X = = d) Decrypt y 1412. = 1412 to find the plaintext.
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- ONLY THE TAP BITS. The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m=4m=4.The plaintext given by 01001011=x0x1x2x3x4x5x6x701001011=x0x1x2x3x4x5x6x7 when encrypted by the LFSR produced the ciphertext 11010110=y0y1y2y3y3y5y6y711010110=y0y1y2y3y3y5y6y7. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3=0,p2=1,p1=0,p0=1p3=0,p2=1,p1=0,p0=1).RSA cryptosystem Let p = 29 and q = 31. Then the RSA modulus n = pq = 899. We set the length of plaintext blocks to be 2 and the length of ciphertext blocks 3. Let the encryption exponent b = 29. If, for example, we want to encrypt the word live, we first divide the word into blocks of length 2, li and ve, and then find its corresponding number blocks 11 8 and 21 4. We consider these number blocks as numbers in base 26. Then we find the corresponding integers in base 10, that is 11 8 = 11 × 26 + 8 = 294 and 21 4 = 21 × 26 + 4 = 550. The integers 294 and 550 are regarded as modular numbers in Z899 and encrypted as 29429 (mod 899) = 91 and 5502⁹ (mod 899) = 492. Now write 91 and 492 in base 26 of length 3 we get 91 = 0 x 26² +3 x 26 +13 and 492 = 0 x 262 +18 × 26 +24 and the corresponding number blocks of length 3 are 0 3 13 and 0 18 24. Hen the ciphertext for live is ADNASY as 0 3 13 = ADN and 0 18 24 = ASY.For affine cipher, the key is a pair (a,b) for a,b<26. The encryption algorithm is E(x)=(ax+b) mod 26. And the decryption algorithm is D(y)=a¹(y-b) mod 26. Note that we map A, B, C, D, ... to numbers 0, 1, 2, 3, ... respectively. Assume that K=(7,3), then E(x)=(7x+3) mod 26, and 7-1-15. Thus D(y)=15(y-3) mod 26. 1. For the key K=(7,3), computer the cipher text for "GOLUNCH" is (please use all capital letters) 2. For the key K=(7,3), try to decrypt the cipher text "YHQQFPGHJF" is (please use all capital letters)
- Recall that the sponge construction describes how to turn a public random permutation into a cryptographic hash function. Once the data to be hashed has been padded, the next step is to absorb the data r bits at a time. blk = <0>while not done: xor the next r bits of data into the first r bits of blk blk = p(blk) As a convenient (but not secure because it's too small) public random permutation, lets use the AES S-box for p. What is blk after absorbing each of the following when r=4 bits. Give your answer as exactly either 2 hexadecimal digits or 8 binary digits with no spaces (eg, 0A or 00001010). Note that the data is already padded and you should restart at blk=<0> each for each computation Data: 1001 Resulting blk:____ Data 10011001 Resulting blk:_____ Data 100110011001 Resulting blk:____2. Recall the concept of Physically Unclonable Functions, which are an importantcomponent of hardware based security. Here is a simple scenario and simple protocol usingPUFs for authenticating a tag. In the protocol below, the Reader stores ALL possiblecombinations of Challenge-Response pairs for every tag it needs to authenticate. Duringauthentication, the Reader broadcasts the ID of the tag, and a greeting (Gn ). Every tag maysee this message, but only the one with the right ID will process the message. The right tagwith the ID in the message will then feed the greeting to its PUF and compute the responseSn. The response is then sent to the Reader. This should be clear from the Figure below. Forthe next round of authenticating the same Tag, the process repeats with a new Greeting(G n+1 ). Since the Reader has the responses, authentication is straightforward. Recall thatChallenges and Responses are typically 128 or 256 bit strings for RFIDs.From the protocol, and class, it should be…Many forms of cryptography are based on the following discrete log problem. Choose a cyclic group G (produced by one generator g) with finitely many elements. Alice and Bob want to set up a secret communication and need a secret key for that, which becomes a group element from G. They set up the following exchange (the so-called Diffie-Hellman protocol): 1. Alice chooses a secret group element a and sends unsecured g^a to Bob 2.Bob chooses a secret group element b and sends unsecured g^b to Alice 3. Alice calculates (g^b )^a and keeps this group element secret 4.Bob calculates (g^a )^b and keeps this group element secret Since a cyclic group is abelian, Alice and Bob now share the secret key (g^b )^a=(g^a )^b. The blue group elements are publicly known, the red elements are secret. The discrete log problem presented to an eavesdropper is computing say a from the publicly known data G,g,g^a. The naming of the problem comes from the solution a= glog(ga ) which is generally difficult to…
- Consider a very simple symmetric block encryption algorithm in which 32-bits blocks of plaintext are encrypted using a 64-bit key. Encryption is defined as C = (PK₁) K₁ where C = ciphertext, K = secret key, Ko = leftmost 64 bits of K, K₁ = rightmost 64 bits of K,+ = bitwise exclusive OR, and is addition mod 264. a. Show the decryption equation. That is, show the equation for P as a function of C, Ko, and K₁. b. Suppose and adversary has access to two sets of plaintexts and their correspond- ing ciphertexts and wishes to determine K. We have the two equations: C = (PK) K₁; C = (PK) K₁ First, derive an equation in one unknown (e.g., Ko). Is it possible to proceed fur- ther to solve for Ko?If an encryption function er is identical to the decryption function dg, then the key K is said to be an involutory key. (a) Let K = (a, b) be a key in an affine cipher over Z/nZ. Show that K is an involutory key if and only if a- = a (mod n) and b(a+1) = 0 (mod n).Use a coding matrix A of your choice. Use a graphing utility to find the multiplicative inverse of your coding matrix. Write a cryptogram for each message. C perform all necessary matrix multiplications. DGE OU T AHEAD BR 2 18 4 75 0 15 21 20 0 1 8 5 1 4 .... 1 1 -12 32 -15 Using the coding matrix the coded message is.····☐☐☐☐☐☐☐☐☐☐ 2 2 1 5 4 4 4 7 (Simplify your answers.) 61