Informally, the Avalanche effect says that two similar, but not identical, inputs should produce radically different outputs when fed through a hash function.

I've seen, at least, two formal definitions of the effect. The first is that if we flip one bit of an input, the two respective outcomes should typically be very different and it's guaranteed that they're not the same. The other definition is that if we flip one bit of an input, the outcomes are statistically independent. I find this confusing: the two definitions are contradictory.

What is the proper definition? And practically, does modern hash functions like sha256 have this property?

To set ideas. Let $H(\cdot)$ be some hash function with $N$ possible outputs. Let $X_1$ be an (integer) random variable, uniformly distributed on some interval. (Assume that $\Pr(H(X_1) = x) = N^{-1}$ for each $x$.) Let $X_2$ be a RV such that it differs exactly in one bit with respect to the first RV (i.e., the Hamming distance between $X_1$ and $X_2$ is exactly one). What does the Avalanche effect says about $\Pr(H(X_2) = H(X_1)\,|\, H(X_1))$?

The first definition says that $\Pr(H(X_2) = H(X_1)\,|\, H(X_1)) = 0$. The second says that $\Pr(H(X_2) = H(X_1)\,|\, H(X_1)) = N^{-1}$. Obviously, both cannot be true.

  • $\begingroup$ Do you mean outcomes should typically be very different and outcomes are statistically independent are contradictory? They mean the same to me.. $\endgroup$
    – Susilo
    Commented Sep 27, 2016 at 7:56
  • $\begingroup$ @Susilo Consider setting $H(X_2)$ to the bitwise complement (i.e., bitwise NOT) of $H(X_1)$. In that case, the two hashes are very different (they don't have a single bit in common), but they are as far from statistically independent as one can get: each hash is a degenerate RV conditional on the other hash. $\endgroup$ Commented Sep 27, 2016 at 8:03

1 Answer 1


The proper definition of avalanche effect is defined in the paper of Webster, A. F. "On the design of S-boxes". Advances in Cryptology - Crypto '85 as :

For a given transformation to exhibit the avalanche effect, an average of one half of the output bits should change whenever a single input bit is complemented.

It is also seen as each bit should have 50% chances to change if you change 1 bit of the input.

This basically mean that in order to test an avalanche effect, you have to apply an algorithm similar to the following.

# result array
array res[n] = {0, ..., 0}

# compute all inputs
for each inputs x of size n
   ref_val = H ( x )

   # compute all possible 1 bit change
   for i in 0..n-1
      test_val = H ( X ^ (1 << i) )

      # compute statistics per bit
      for j in 0 .. n-1
         res[j] += ((test_val ^ ref_val) >> j) & 1

# in the end forall k, res[k] / (n * 2**n) should be around 50%.

Due to the size of the inputs, you can test with a relative precision whether or not $\operatorname{SHA256}$ does have a strict avalanche effect. But by doing a sampling over a large number of inputs where you flip each bit once, you can have a measure.But in most cases we usually analyze the components of the primitive in order to assert whether or not the avalanche effect is respected.

  • $\begingroup$ Thank you, @Biv, the paper was helpful. The first definition you state is consistent with both of the definitions I asked about (and your test would not be able to differentiate between the two). Your second definition (which Webster & Tavares refer to as "the strict avalanche criterion") is different from your first. This definition is more inline with my second definition, but it is much weaker. The difference is that W&T only require pairwise independence between the bits in each hash, while my definition require mutual independence. $\endgroup$ Commented Sep 27, 2016 at 16:01
  • $\begingroup$ @Biv It's actually quite easy to test any size hash function for avalanche effect. I've been testing a hash function that I wrote against Sha-512 to see if they have identical avalanche effects. I've explained the 6 step method in my answer to crypto.stackexchange.com/questions/34269/…. $\endgroup$
    – Paul Uszak
    Commented Sep 28, 2016 at 2:17

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