Cfrgtkky

I'm reading about MinHash technique to estimate the similarity between 2 sets: Given set A and B, h is the hash function and $h_min(S)$ is the minimum hash of set S, i.e. $h_min(S) = min(h(s))$ for s in S. We have the equation:
$$
p(h_min(A) = h_min(B)) = frac{|A cap B|}{|A cup B|}
$$
Which means the probability that minimum hash of A equals to minimum hash of B is the Jaccard similarity of $A$ and $B$.

I am trying to prove above equation and come up with a proof, which is:
for $a in A$ and $b in B$ such that $h(a) = h_min(A)$ and $h(b) = h_min(B)$. So, if $h_min(A) = h_min(B)$ then $h(a) = h(b)$. Assume that hash function h can hash keys to distinct hash value, so $h(a) = h(b)$ if and only if $a = b$, which is $frac{|A cap B|}{|A cup B|}$. However, my proof is not complete since hash function can return the same value for different keys. So, I'm asking for your help to find a proof which can be applied regardless the hash function. Thanks.

L

It seems like it is an underlying assumption in this formula that different elements map to distinct hashes. Otherwise, just have your hash function map everyone to 0. The probability you are looking at would be 1 regardless of any sets.

There is a remarkable connection between minhashing and Jaccard similarity
of the sets that are minhashed.

• The probability that the minhash function for a random permutation of
rows produces the same value for two sets equals the Jaccard similarity
of those sets.

To see why, we need to picture the columns for those two sets. If we restrict
ourselves to the columns for sets S1 and S2, then rows can be divided into three
classes:

1. Type X rows have 1 in both columns.


2. Type Y rows have 1 in one of the columns and 0 in the other.


3. Type Z rows have 0 in both columns.

Since the matrix is sparse, most rows are of type Z. However, it is the ratio
of the numbers of type X and type Y rows that determine both SIM(S1, S2)
and the probability that h(S1) = h(S2). Let there be x rows of type X and y
rows of type Y . Then SIM(S1, S2) = x/(x + y). The reason is that x is the size
of S1 ∩ S2 and x + y is the size of S1 ∪ S2.

Now, consider the probability that h(S1) = h(S2). If we imagine the rows
permuted randomly, and we proceed from the top, the probability that we shall
meet a type X row before we meet a type Y row is x/(x + y). But if the
first row from the top other than type Z rows is a type X row, then surely
h(S1) = h(S2). On the other hand, if the first row other than a type Z row
that we meet is a type Y row, then the set with a 1 gets that row as its minhash
value. However the set with a 0 in that row surely gets some row further down
the permuted list. Thus, we know h(S1) 6= h(S2) if we first meet a type Y row.
We conclude the probability that h(S1) = h(S2) is x/(x + y), which is also the
Jaccard similarity of S1 and S2. ( from Mining of Massive Datasets book )