Question

Estimate the probabilities of finding two messages with the same MD5 checksum, given total numbers of messages of \(2^{63}, 2^{64}\), and \(2^{65}\). Hint: This is the birthday problem again, as in Exercise 49 of Chapter 2, and again the probability that the \(k+1\) th message has a different checksum from each of the preceding \(k\) is \(1-k / 2^{128}\). However, the approximation in the hint there for simplifying the product fails rather badly now. So, instead, take the log of each side and use the approximation \(\log \left(1-k / 2^{128}\right) \approx-k / 2^{128}\).

Short answer

For \(2^{63}, P(\text{collision})\) \(\approx 1 - e^{-0.125} \). For \(2^{64}, P(\text{collision})\) \( \approx 1 - e^{-0.5} \ ). For \(2^{65}, P(\text{collision})\() \approx 1 - e^{-2} \ ).

Step-by-step solution

- Understanding the Birthday Problem

Recall that the problem is similar to the birthday problem. We are calculating the probability of at least two messages having the same MD5 checksum out of a given number of messages.

- Calculate the Probability of No Collision

For a given number of messages, the probability that the \(k+1\)th message has a different checksum from the previous \(k\) messages can be calculated using the formula: \[P(\text{no collision}) = \prod_{k=0}^{n-1} \left(1 - \frac{k}{2^{128}}\right)\] where \(n\) is the number of messages.

- Using Logarithms to Simplify

Take the natural logarithm of each side to simplify the product. Using the approximation \(\log (1 - \frac{k}{2^{128}}) \approx -\frac{k}{2^{128}}\), we get: \[\log \left(P(\text{no collision})\right) = \sum_{k=0}^{n-1} \log \left(1 - \frac{k}{2^{128}}\right) \approx \sum_{k=0}^{n-1} -\frac{k}{2^{128}}\]

- Summation of the Series

The summation of the series: \[\sum_{k=0}^{n-1} -\frac{k}{2^{128}}\] can be approximated using the sum of the first \(n-1\) integers: \[\sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}\] Therefore, \[\log \left(P(\text{no collision})\right) \approx -\frac{(n-1)n}{2 \cdot 2^{128}}\]

- Exponentiating to Find Probability

Exponentiate both sides to solve for \(P(\text{no collision})\): \[P(\text{no collision}) \approx e^{-\frac{(n-1)n}{2 \cdot 2^{128}}}\]

- Calculate for Given Values

Calculate \(P(\text{no collision})\) for \(n = 2^{63}, 2^{64}, 2^{65}\): \[\text{For } n = 2^{63}: P(\text{no collision}) \approx e^{-\frac{(2^{63} - 1)2^{63}}{2 \cdot 2^{128}}} = e^{-\frac{2^{126} - 2^{63}}{2^{129}}} = e^{-(2^{-3} - 2^{-65})} \approx e^{-0.125}\] \[\text{For } n = 2^{64}: P(\text{no collision}) \approx e^{-\frac{2^{128} - 2^{64}}{2^{129}}} = e^{-(0.5 - 2^{-65})} \approx e^{-0.5}\] \[\text{For } n = 2^{65}: P(\text{no collision}) \approx e^{-\frac{2^{130} - 2^{65}}{2^{129}}} = e^{-(2 - 2^{-64})} \approx e^{-2}\]

- Find Probability of Collision

The probability of at least one collision is: \[P(\text{collision}) = 1 - P(\text{no collision})\] Therefore, for the given values: \[\text{For } n = 2^{63}: P(\text{collision}) \approx 1 - e^{-0.125}\] \[\text{For } n = 2^{64}: P(\text{collision}) \approx 1 - e^{-0.5}\] \[\text{For } n = 2^{65}: P(\text{collision}) \approx 1 - e^{-2}\]

Key concepts

MD5 checksum collision

In cryptography, an MD5 checksum is a 128-bit hash value used to verify data integrity. It is designed such that if two different data inputs produce the same MD5 checksum, this is termed a collision. The birthday problem is used in this exercise to estimate the probability of such a collision. Given the size of the MD5 hash output (128 bits), the probability of finding at least one collision increases significantly when the number of hashed messages grows. As the hash space is finite (2^128 possible hash values), hashing around \(2^{64}\) unique messages will likely lead to collisions because we already have a significant portion of hash values covered.

Probability theory in cryptographic analysis

Probability theory helps in predicting the chances of certain outcomes, making it invaluable in cryptographic analysis. In this context, we utilize it to estimate the likelihood of checksum collisions within a given number of messages. The birthday problem's analogy provides a simplified way to understand these probabilities. Considering an event with many possible outcomes, such as generating a unique MD5 checksum for each message, the probability formula for no collision is \(P(\text{no collision}) = \prod_{k=0}^{n-1} \(1 - \frac{k}{2^{128}}\)\). This helps to calculate the odds of every new message having a unique checksum relative to the preceding ones. Hence, as more messages are hashed, the collision probability grows.

Logarithmic approximation

Logarithmic approximation simplifies complex multiplicative problems into additive ones, which are easier to manage. For instance, when calculating the product for no collision probability in the birthday problem, the product \(P(\text{no collision}) = \prod_{k=0}^{n-1} \(1 - \frac{k}{2^{128}}\)\) becomes unwieldy. By taking the natural log of both sides, we convert the multiplication into summation. Using the approximation \(\log(1-x) \approx -x\), the product simplifies to \(\sum_{k=0}^{n-1} -\frac{k}{2^{128}}\), further reduced to \(-\frac{(n-1)n}{2 \cdot 2^{128}}\). This logarithmic method makes dealing with large numbers more feasible in cryptographic analysis.