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This implies that 7 is a primitive root mod Therefore k is a multiple of , but is not a multiple of Of these, 16 are multiples of The remaining 16 are possibilities for k. If a is odd, then the output of step 2 is the same as the input, hence of the desired form. If a is even, step 3 does nothing, so the output still has the desired form. The new c is still j y2.
Therefore step 5 outputs y x , as desired. Therefore, the output of step 2 always has the desired form, as claimed. As pointed out above, the output of step 5 is then y x mod n. Therefore x cannot exist. Therefore, several random selections of these square roots should include the meaningful message m. By induction, we have d ri for all i. By induction, rk ri for all i. Since rk is a common divisor, it is the largest.
But part b says that this is not the case. Therefore, the order of 3 must be , and 3 is a primitive root mod If it factors, it must be divisible by a degree one polynomial. Clearly X does not divide it. Since none of the polynomials from part a divide it, it must be irreducible. Therefore it is irreducible. The first keeps repeating 1, 2. The second keeps repeating 1, 1, 1, 4. Similarly for the multiples of q.
Therefore, m is a multiple of p, q, or pq, hence a multiple of p or of q possibly both. By part c , these two sets of numbers do not overlap. Chapter 4 - Exercises 1. The switch the left and right of the final output. Verification is the same as that on pages Therefore 3 rounds is very insecure! If you also know the plaintext, you know M1 are therefore can deduce K. If someone discovers the fixed key and obtains the encrypted password file, this person can easily decrypt by the usual decryption procedure.
However, knowing the ciphertext and the plaintext does not readily allow one to deduce the key. Also, clearly the new left side is the complementary string. So each round of DES gives the complementary string, so this is true for the final result.
Decryption is ac- complished by reversing the order of the keys to K16 ,. Since the Ki are all the same, this is the same as encryption, so encrypting twice gives back the plaintext. Let m, c be a plaintext-ciphertext pair. There should be a small number of such pairs. For each such pair, try it on another plaintext and see if it produces the corresponding ciphertext. This should eliminate most of the incorrect pairs. So, make two lists. The left list consists of encryptions using the second encryption E 2 with different choices for K2.
Similarly, the right side contains decryptions using different keys for the first encryption algorithm. Note: The two lists need not be the same size, as the different algorithms might have different key lengths, and hence different amount of keys see part b. Now, look for matches between yj and zl. The composition of these two gives the affine cipher.
The total computation needed involves producing 26 encryptions for E 2 and 12 decryptions for E 1. The total is It suffices to look at an arbitrary round of the encryption process. This is of the form of the decryption algorithm specified in the problem. However, note that by the end of the third round, the register X4 is no longer corrupted. Thus, the fourth step of decryption is uncorrupted.
All subsequent decryption steps also will be free of errors. Now start decrypting. Let K be the key we wish to find. Use the hint. Now, suppose we start a brute force attack by encrypting M1 with different keys. However, when we use Kj we can eliminate another key. Here is how. Hence, if this happens, we know the key is Kj since Kj would decrypt C2 to get M1. We are effectively testing two keys for the price of one!
Hence, the key space is cut in half and we only have to search an average of Chapter 5 - Exercises 1. In the notation in Subsection 5. Since all steps in this modified AES algorithm have the equal difference property, the composition of all the steps has the property.
Let Ej x represent the result after j steps there are 30 such steps. We start with x1 and x2. If the jth step is MixColumn, then everything is multiplied on the left by a matrix M. By XORing with x1 , Eve obtains x2. Therefore, BS does not satisfy the equal difference property.
By 3 a , affine maps satisfy this property, so BS is not affine. Chapter 6 - Exercises 1. The two possible plaintexts are 8 and 9. Hence, the correct plaintext is 8. The number e is maba1 mod n. This will be m.
Eve divides by 2 mod n to obtain m. Therefore, this double encryption is the same as single encryption with encryption exponent e1 e2. So the security is at the same level as single encryption. Fractional exponents must be avoided.
Therefore, we can factor n1 and n2 and break the systems. This would contradict the assumption in part a , and hence n must not be prime. Since Eve can calculate this last quantity, she can calculate m. Make a list of 1e , 2e ,. For each block of ciphertext, look it up on the list and write down the corresponding letter. The message given is hello. The verification is the same as the one for RSA. This can be easily verified by hand. Let cA and cB be the outputs of the two machines.
Eve makes two lists: 1. Another way: Eve divides c1 by e mod n and then uses the short message attack from Section 6. If all of the factors of f x are in B, then the register is 0 if the prime factors are distinct.
In general, the register will contain a sum of logs of some primes from B. These will tend to be small. Moreover, the multiple prime factors of f x will tend to be the small primes in the factor base for example, it is much more likely that 22 or 32 divides f x than divides f x. Therefore, the register will tend to be small.
Subtracting is a much faster operation than dividing. Also, the subtraction can be done in floating point, while the division is done with large integer arithmetic. Chapter 7 - Exercises 1. By the proposition in Section 3. Instead of using the proposition in Section 3. This means we can stop. Chapter 8 - Exercises 1. However, it is fairly quickly computed though not fast enough for real use , and it is preimage resistant.
This is computationally equivalent to factoring see page Taking different numbers of 0-blocks yields collisions, so it is not strongly collision-free. The other inequality follows similarly. Each such step requires an evaluation of f , which takes time a constant times n. Use the same formulas as in Section 8. This gives 5A in hexadecimal. The first: DK1 c for N random keys K1.
The second: EK2 m for N random keys K2. By hypothesis, K3 is probably the only such key, so we have probably found the key K3. The attack makes two lists. The first: the ciphertext shifted by 6 random shifts.
The second: the plaintext shifted by 6 random shifts. A match gives a shift that sends the plaintext to the ciphertext, as desired. This last term dominates, so the sum has approximately bits.
Chapter 9 - Exercises 1. This will probably be a small number. So the signature will probably not be valid. We must solve for a, b, and c. These are the same. They each send their respective result to the person to their right. The next round, they take what they received and raise it to their private exponent and pass it to their right. Thus Eve can calculate the key. The hacker can then perform the six steps needed to create a valid coin, since the hacker knows x, which is needed in step 5.
The fact that c1 , c, x, w are related as in step 5 is the key to the security. If that equation can be satisfied and the hacker performs all the computations required by the Spender , all the verification equations work.
All the verification equations work. There is no way to identify the Spender. The Sender uses these to obtain z and b. If we are ignoring z and b, the sender never needs the information provided by the bank to produce a legitimate coin. Therefore, any value of I, hence of u, can be used. Chapter 12 - Exercises 1. To approach this problem, one should use a 2, 4 threshold scheme.
In this problem, the 2, 30 scheme requires solving for a line that inter- polates the points 1, 13 and 3, The slope can be calculated to be 50, and the intercept to be Hence the secret is 2, No information is obtained on the secret until there are 5 people, so there are still possibilities. Only AB is correct, so if they try all three, they will find the correct combination within three tries. If the three incorrect secret are not all the same, then the correct secret can be obtained by seeing which secret occurs most often among the 6 pairs.
This latter possibility is not allowed, since it would mean that C has the share 0, y for some y. This type of share is never given out since 0, f 0 is the secret. Take a 10, 30 scheme and give the general 10 shares, the colonels 5 shares each, and the clerks 2 each. Then each of the desired groups, and no smaller groups, have the 10 shares needed to launch the missile. Therefore A, B, D lie on a line.
C is not on this line, so C is the foreign agent. Split the launch code into three equal components using a 3 party se- cret splitting scheme. One component will be given to each of A, B, and C. For the component that belongs to Government A, use a 3, 10 secret sharing scheme to give shares to the delegates of Government A.
Take the shares xk , yk and evaluate p x at xk. The adversary is the agent that The last stage is modern cryptography. Its central feature is the reliance on mathematics and Introduction. Its central feature is the reliance on mathematics and Where can i find "Introduction To Modern Cryptography Solutions Manual" If you have,can you send me?
In fact, it was so good that we used it as a starting point for our TSS-based solution. However, after reading this paper, suggested to us by Omer Shlomovits, we found a. Introduction to modern cryptography 2nd edition solution manual.
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