Problem 1 Let \(G:\\{0,1\\}^{\lambda} \rig... [FREE SOLUTION] (2024)

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Chapter 5: Problem 1

Let \(G:\\{0,1\\}^{\lambda} \rightarrow\\{\theta, 1\\}^{\lambda+\ell}\) be aninjective (i.e., 1 -to-1) PRG. Consider the following distinguisher: $$ \begin{array}{l} \qquad \mathcal{A} \\ x:=\text { QUERY } 0 \\ \text { for all } s^{\prime} \in\\{\theta, 1\\}^{\lambda}: \\ \quad \text { if } G\left(s^{\prime}\right)=x \text { then return } 1 \\ \text { return } 0 \end{array} $$ (a) What is the advantage of \(\mathcal{A}\) in distinguishing\(\mathcal{L}_{\text {prg-real }}^{G}\) and \(\mathcal{L}_{\text {prg-rand }}^{G}?\) Is it negligible? (b) Does this contradict the fact that \(G\) is a PRG? Why or why not? (c) What happens to the advantage if \(G\) is not injective?

Short Answer

Expert verified

The advantage is near 1 and not negligible. It does not contradict the PRG property. If not injective, the advantage reduces.

Step by step solution

01

- Understand the problem statement

Given a pseudorandom generator (PRG) function \(G\) that is injective and a distinguisher \(\mathcal{A}\), determine the advantage of \(\mathcal{A}\) in distinguishing between two distributions \(\mathcal{L}_{\text{prg-real}}^{G}\) and \(\mathcal{L}_{\text{prg-rand}}^{G}\). Answer the related questions about the PRG and the behavior when the PRG is not injective.

02

- Define Advantage

The advantage of a distinguisher \(\mathcal{A}\) is defined as the absolute difference in probabilities of \(\mathcal{A}\) outputting 1 when given as input a sample from \(\mathcal{L}_{\text{prg-real}}^{G}\) versus \(\mathcal{L}_{\text{prg-rand}}^{G}\).

03

- Evaluate \(\mathcal{A}\)’s Request

When \(\mathcal{A}\) queries 0, \(x = G(s')\) if it is from \(\mathcal{L}_{\text{prg-real}}^{G}\). There may be an \(s'\) such that \(G(s') = x\). If such an \(s'\) is found, \(\mathcal{A}\) returns 1. Otherwise, \(\mathcal{A}\) returns 0.

04

- Calculate the probability for PRG-Real

For \(\mathcal{L}_{\text{prg-real}}^{G}\), since \(G\) is injective and given a random \(s\), it is highly likely that there exists a unique \(s'\) such that \(G(s') = x\). Therefore, the probability of \(\mathcal{A}\) outputting 1 is almost 1.

05

- Calculate the probability for PRG-Rand

For \(\mathcal{L}_{\text{prg-rand}}^{G}\), \(x\) is a uniformly random string. The probability that a specific string \(x\) matches any output of \(G\) applied to any \(s'\) is very low due to the length \(\lambda + \ell\) of the output.

06

- Evaluate the Advantage

The advantage is the difference in the probabilities determined: near 1 for \(\mathcal{L}_{\text{prg-real}}^{G}\) and near 0 for \(\mathcal{L}_{\text{prg-rand}}^{G}\). Thus, \(\mathcal{A}\) has a significant advantage.

07

- Assess Negligibility

Since the advantage is significant (near 1), it is not negligible.

08

- Check for Contradiction

It does not contradict the fact that \(G\) is a PRG. PRGs are not expected to be indistinguishable in distinguishing the real output from truly random sequences by every possible adversary but by those computationally bounded. \(\mathcal{A}\) takes advantage of the injectiveness, which is not a typical complexity feature.

09

- Consider Non-Injective PRG Scenario

If \(G\) is not injective, it means there could be multiple pre-images leading to the same output. This reduces the certainty of \(\mathcal{A}\) finding a unique \(s'\), thus reducing the advantage significantly.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Functions

An injective function, also known as a one-to-one function, is fundamental in understanding pseudorandom generators (PRGs) in cryptography.
When a function is injective, it means that every element in the function's domain maps to a unique element in its codomain. In other words, no two distinct inputs produce the same output.
For our PRG example, this characteristic ensures that each input value, represented as a binary string, translates into a unique pseudorandom output. This uniqueness is crucial when analyzing the distinguisher's advantage because it guarantees that if such a match exists, it is singular.
Injective functions offer stability and predictability in cryptographic algorithms, a vital property that helps prevent collisions and retain the integrity of encrypted messages.

Distinguishers in Cryptography

Distinguishers in cryptography are algorithms intended to differentiate between two distributions. In the context of PRGs, a distinguisher \( \mathcal{A} \ \) is used to determine whether a given output is truly random or pseudorandom.
The distinguisher works by querying certain inputs and analyzing the corresponding outputs, looking for patterns or statistical anomalies.
In our exercise, the distinguisher queries an input of 0 and checks if the output matches any potential pseudorandom strings generated by the PRG. If an exact match is found, the distinguisher returns 1, suggesting the output is from the PRG and not random.
The core idea is to measure the 'advantage' of the distinguisher, which is the absolute difference between the probabilities of correctly identifying the PRG output and the random output.
This analysis helps cryptographers to evaluate the strength of the PRG and ensure it meets the necessary security standards.

Probability and Statistical Analysis in Cryptography

Cryptographic security often relies on rigorous probability and statistical analysis. This helps in validating whether the outputs of cryptographic algorithms like PRGs behave as expected under pseudorandom conditions.
In our exercise, the advantage of the distinguisher hinges on the probability that it correctly identifies whether an output is from \( \mathcal{L}_{\text{prg-real}}^{G} \ \) or \( \mathcal{L}_{\text{prg-rand}}^{G} \ \).
By assessing these probabilities, cryptographers can calculate the extent to which \( \mathcal{A} \ \) can differentiate between real and random distributions.
When the PRG is injective, there is a high probability that the distinguisher will find a matching input, making the advantage significant. Conversely, if the PRG is not injective, multiple inputs could map to the same output, reducing the certainty and the distinguishers' advantage.
Therefore, these statistical measures are crucial in certifying the robustness of cryptographic functions against potential attackers who might exploit predictable patterns.

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Problem 1 Let \(G:\\{0,1\\}^{\lambda} \rig... [FREE SOLUTION] (3)

Most popular questions from this chapter

Let \(G_{1}\) and \(G_{2}\) be deterministic functions, each accepting inputs oflength \(\lambda\) and producing outputs of length \(3 \lambda\) (a) Define the function \(H\left(s_{1} \| s_{2}\right)=G_{1}\left(s_{1}\right)\oplus G_{2}\left(s_{2}\right) .\) Prove that if either of \(G_{1}\) or \(G_{2}\)(or both) is a secure \(\mathrm{PRG},\) then so is \(\mathrm{H}\). (b) What can you say about the simpler construction \(H(s)=G_{1}(s) \oplusG_{2}(s),\) when one of \(G_{1}, G_{2}\) is a secure PRG?(a) Let \(f\) be any function. Show that the following function \(G\) is not asecure PRG, no matter what \(f\) is. Describe a successful distinguisher andexplicitly compute its advantage: $$ \frac{G(s):}{\operatorname{return} s \| f(s)} $$ (b) Let \(G:\\{\theta, 1\\}^{\lambda} \rightarrow\\{0,1\\}^{\lambda+\ell}\) be acandidate PRG. Suppose there is a polynomial-time algorithm \(V\) with theproperty that it inverts \(G\) with non-negligible probability. That is, $$ \underset{s \leftarrow\\{0,1\\}^{\lambda}}{\operatorname{Pr}}[V(G(s))=s] \text{ is non-negligible. } $$ Show that if an algorithm \(V\) exists with this property, then \(G\) is not asecure PRG. In other words, construct a distinguisher contradicting the PRG-security of \(G\) and show that it achieves non-negligible distinguishingadvantage. Note: Don't assume anything about the output of \(V\) other than theproperty shown above. In particular, \(V\) might very frequently output the"wrong" thing.
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Problem 1 Let \(G:\\{0,1\\}^{\lambda} \rig... [FREE SOLUTION] (2024)
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