不同内容同大小文件同时匹配SHA256与BLAKE256哈希的概率及多哈希对碰撞概率的影响

Let’s break down your questions with practical cryptographic hash function fundamentals:

1. Can two distinct, same-sized files collide on multiple different hash algorithms at once? #

The short answer: Theoretically possible, but practically impossible (for cryptographically secure hashes like SHA256 and BLAKE256).

Hash functions map arbitrary-length input to fixed-length output, so by the pigeonhole principle, collisions (distinct inputs producing identical outputs) must exist mathematically. However, secure modern hashes are engineered to make collisions computationally infeasible to find—this difficulty skyrockets when you require collisions across multiple independent algorithms.

2. What’s the probability that file B (distinct from A, same size) matches both SHA256 and BLAKE256 hashes of file A? #

Assuming SHA256 and BLAKE256 are cryptographically independent (a standard assumption for well-designed, unrelated hash functions), the probability is the product of their individual collision probabilities:

• A random distinct file matching A’s SHA256 hash has a probability of ~1/(2^256)
• The same file matching A’s BLAKE256 hash also has a probability of ~1/(2^256)
• Combined, this gives a total probability of 1/(2^512)

To contextualize this: 2^512 is an astronomically large number—far bigger than the total number of atoms in the observable universe. A natural, unforced collision is effectively impossible to encounter in real-world scenarios. Even intentional collision attacks (like birthday attacks) for a single 256-bit hash require 2^128 operations, which is already beyond current computational limits; pulling this off for two independent hashes would be exponentially harder.

3. Does using more hash algorithms (e.g., 5) reduce collision probability? #

Absolutely. Following the same independence assumption, each additional secure 256-bit hash multiplies the denominator by 2^256. For 5 algorithms, the probability drops to 1/(2^(256*5)) = 1/(2^1280)—an even more negligible likelihood.

A critical note: this only holds if the hash algorithms are truly independent. If two hashes share flawed design choices or underlying structures, their outputs might be correlated, and the combined collision probability wouldn’t be a pure product. But for modern, unrelated hashes like SHA256, BLAKE256, SHA3-256, Whirlpool, and RIPEMD-256, this independence assumption holds strong.

内容的提问来源于stack exchange，提问作者Akash