We consider three different hash functions which produce outputs of lengths 64, 128 and 160 bit. After how many random inputs do we have a probability of ε = 0. 5 for a collision? After how many random inputs do we have a probability of ε = 0. 1 for a collision?

Answers

Answer 1

For ε = 0.1, approximately 2.147 random inputs are needed for a collision. The number of inputs required for the hash functions producing outputs of lengths 128 and 160 bits using the same formula.

To determine the number of random inputs needed to achieve a specific probability of collision, we can use the birthday paradox principle. The birthday paradox states that in a group of people, the probability of two individuals having the same birthday is higher than expected due to the large number of possible pairs.

The formula to calculate the approximate number of inputs required for a given probability of collision (ε) is:

n ≈ √(2 * log(1/(1 - ε)))

Let's calculate the number of inputs needed for ε = 0.5 and ε = 0.1 for each hash function:

For a hash function producing a 64-bit output:

n ≈ √(2 * log(1/(1 - 0.5)))

n ≈ √(2 * log(2))

n ≈ √(2 * 0.693)

n ≈ √(1.386)

n ≈ 1.177

For ε = 0.5, approximately 1.177 random inputs are required to have a probability of collision.

For ε = 0.1:

n ≈ √(2 * log(1/(1 - 0.1)))

n ≈ √(2 * log(10))

n ≈ √(2 * 2.303)

n ≈ √(4.606)

n ≈ 2.147

For ε = 0.1, approximately 2.147 random inputs are needed for a collision.

Similarly, we can calculate the number of inputs required for the hash functions producing outputs of lengths 128 and 160 bits using the same formula.

Please note that these calculations provide approximate values based on the birthday paradox principle. The actual probability of collision may vary depending on the specific characteristics of the hash functions and the nature of the inputs.

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