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5

The modulo operator keeps the result of the addition of $M$ and $K$ within the set $Z$. For example, if $m$ is 10, $M$ is 6 and $K$ is 5, $M + K$ would be 11 which is no longer in the set $Z$. Taking 11 mod 10 results in 1 which is in the set $Z$. As a help towards answering the question whether scheme $M + K$ mod $m$ is perfectly secure, when $m$ is 26 ...

4

If you use a concrete-security definition of security for a PRG, then this statement is true. The proof is a good exercise. If you know enough to pose the problem and to understand the definition of security for a PRG, you should be able to find the reduction proof without difficulty. Start by tracing out what the definition is saying. A general comment ...

4

A few definitions could assist: are $x$ and $x'$ random elements of $\{0,1\}^k$ and $\{0,1\}^l$? Is $\mathsf{G}$ a cryptographically-secure PRG or a PRG with some statistical properties? Let $\mathcal{D}$ be a PPT-distinguisher and $r$ be a uniformly random bitstring of length $n + \ell$. A common definition ($e.g.,$ from Katz-Lindell book) of a PRG ...

3

I'll expand my comments into a full answer. Start by examining that you know the value of $r^3$ - this is just $F(r)$. You know you can express $F(r+1)$ and $F(r+2)$ symbolically. You can do this for any term in the key sequence, but it will be useful to write down $F(r+3)$. Forget any constant terms in the computations because they can be added in at ...

3

Background: An infinite sequence $c_0,c_1,c_2,\dots$ is generated by a (linear feedback shift register (LFSR) with) polynomial $f(x) = \sum_{i=0}^n f_i x^i$ if for any $j$, $$\sum_{i=0}^n c_{j+i} f_{n-i} = 0 \text.$$ We can consider the sequence as a power series $\sum_{i=0}^{\infty} c_i x^i$. If we multiply this power series with the polynomial $f(x)$, we ...

3

The "million usernames" is a red herring, because the user name is used as "salt": the hash value is computed over the password and the user name. When the attacker tries a potential password, he must choose which user name he puts in the hash function; and if a match is found, then this will be for the hash value for this user name only. In other words, ...

2

An "encryption scheme" defines the encryption/decryption of data. A "message transmission scheme" is about securing transmission and defines both "privacy" and "authenticity" between a sender and a receiver. Since you haven't asked about the definition of CCA-secure (encryption) schemes and since you've been given this as an exercise, I won't mention ...

1

This is a classical example. Here is the proof system… Bob gives two gloves to Alice so that she is holding one in each hand. Bob can see the gloves at this point, but Bob doesn't tell Alice which is which. Alice then puts both hands behind her back. Next, she either switches the gloves between her hands, or leaves them be, with probability $1/2$ each. ...

1

AES-128 uses the full set $\{0, 1\}^{128}$ as keyspace, and for each key the blockcipher is defined for each input block in $\{0, 1\}^{128}$. The same goes for AES-256, but it uses a 256-bit keyspace (but still a 128-bit block). So the answer to 1 is yes. For 2, we have this equation: $$AES_K(AES_K^{-1}(x)) = x$$ We can decrypt both sides: ...

1

If you know the plaintext and the ciphertext, getting the key is trivial. $$c=m \oplus k$$ Try moving the terms around. Hint: $m$ and $k$ can be functionally swapped (change position with each other), ie. only one term needs to be secret. You already know one term. I've pretty much given you the answer. About the decoding procedure: The ASCII table ...

1

$\left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \:$ is the number of elements that the compression function can map to. If $\: m > \operatorname{log}_{\hspace{.01 in}2}\left(\hspace{-0.03 in}\frac{t^2}{2\cdot \epsilon}\hspace{-0.04 in}\right) \:$ then \$\;\; \left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \: = \: 2^m ...

1

None. No area of CS has been affected by fully homomorphic encryption yet, because it isn't practical (yet). If it becomes practical, it could have a significant effect on computer security and cryptography, but that remains speculative at this point. Read this answer for more details: http://crypto.stackexchange.com/a/628/351

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