受限字符串SHA1哈希的碰撞概率计算及最小化n的技术问询

Let's break this down step by step, since we're dealing with a specific input space, SHA1 hashing, and the goal of minimizing n (likely the number of hash bits you'll use) while keeping collision risk in check.

First, let's lock in our key variables:

• x: Total possible input strings. Your inputs are 64 characters long, with 62 options per character (10 digits + 26 lowercase + 26 uppercase letters). That means x = 62^64—this is an enormous number, way bigger than SHA1's full 160-bit hash space (2^160).
• n: The number of bits we'll use from the SHA1 output (we want the smallest n possible while keeping collisions acceptable). The hash space for n bits is 2^n.

The Core Idea: Birthday Paradox #

Hash collision probability boils down to the birthday problem. Here's the practical, simplified version: if you have k items picked from a space of size S, the probability of at least one collision can be approximated (when k is much smaller than S) with this formula:

Collision Probability ≈ k*(k-1)/(2*S)

For most real-world cases where k is large, k-1 is almost identical to k, so we can simplify this to:

Collision Probability ≈ k²/(2*S)

Applying This to Your Setup #

Now let's map this to your x and n values:

Scenario 1: Input set size x ≤ hash space 2^n #

If you're using every unique input once, plug k = x and S = 2^n into the approximation:

P ≈ x²/(2*2^n) = x²/(2^(n+1))

If you're only sampling a smaller subset of m inputs (where m ≤ x), just swap x with m in the formula.

Scenario 2: Input set size x > hash space 2^n #

Here, collisions are guaranteed by the pigeonhole principle—you have more unique inputs than unique hash values. But if you're only working with a subset m of inputs (where m ≤ 2^n), you still use the birthday approximation with k = m and S = 2^n.

Minimizing n for Acceptable Collision Risk #

To find the smallest n that keeps collision probability below your desired threshold (say, P_max), rearrange the approximation formula:

n ≥ log2(x²/(2*P_max)) = 2*log2(x) - log2(2*P_max)

Since x = 62^64, log2(x) ≈ 64 * 5.954 = 381.056 (because log2(62) is roughly 5.954). For example, if you want a collision probability of ≤ 1e-9, plugging that in gives:

n ≥ 2*381.056 - log2(2*1e-9) ≈ 762.11 - (-29.9) ≈ 792 bits

That's way larger than SHA1's 160 bits, which makes sense given how massive your input set is. If you're only dealing with a smaller subset of inputs, replace x with your subset size m in the calculation to get a smaller, more feasible n.

内容的提问来源于stack exchange，提问作者mrhijack