Edit distance between two partitions

Question

I have two partitions of $[1 \ldots n]$ and am looking for the edit distance between them.

By this, I want to find the minimal number of single transitions of a node into a different group that are necessary to go from partition A to partition B.

For example the distance from {0 1} {2 3} {4} into {0} {1} {2 3 4} would be two

After searching I came across this paper, but a) I am not sure if they are taking into account the ordering of the groups (something I don't care about) in their distance b) I am not sure how it works and c) There are no references.

Any help appreciated

Answer 1

This problem can be transformed into the assignment problem, also known as maximum weighted bipartite matching problem.

Note first that the edit distance equals the number of elements which need to change from one set to another. This equals the total number of elements minus the number of elements which do not need to change. So finding the minimum number of elements which do not change is equivalent to finding the maximum number of vertices that do not change.

Let $A = \{ A_1, A_2, ..., A_k \}$ and $B = \{ B_1, B_2, ..., B_l \}$ be partitions of $[1, 2, ..., n]$. Also, without loss of generality, let $k \ge l$ (allowed because $edit(A, B) = edit(B, A)$). Then let $B_{l+1}$, $B_{l+2}$, ..., $B_k$ all be the empty set. Then the maximum number of vertices that do not change is:

$\max_f \sum_{i=1}^k |A_i \cap B_{f(i)} |$

where $f$ is a permutation of $[1, 2, ..., k]$.

This is exactly the assignment problem where the vertices are $A_1$, ..., $A_k$, $B_1$, ..., $B_k$ and the edges are pairs $(A_i, B_j)$ with weight $|A_i \cap B_j|$. This can be solved in $O(|V|^2 \log |V| + |V||E|)$ time.

Answer 2

Look at this paper's PDF

http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.0030160

The definition of edit distance in there is exactly what you need I think. The 'reference' partition would be (an arbitrary) one of your two partitions, the other would simply be the other one. Also contains relevant citations.

Best, Rob

Answer 3

Cranky Sunday morning idea that might or might not be correct:

Wlog, let $P_1$ be the partition with more sets, $P_2$ the other. First, assign pairwise different names $n_1(S) \in \Sigma$ to your sets $P_1$. Then, find a best naming $n_2(S)$ for the sets $P_2$ by the following rules:

• $n_2(S) := n_1(S')$ for $S \in P_2$ with $S \cap S'$ maximal amongst all $S' \in P_1$; pick the one creating the least conflicts if multiple choices are possible.
• If now $n_2(S) = n_2(S')$ for some $S \neq S'$, assign the one that shares less elements with $S'', n_1(S'') = n_2(S)$, the name of the set in $P_1$ it shares the second most elements with, i.e. have it compete for that set's name.
• If the former rule can not be applied, check for both sets wether they can compete for the name of other sets they share less elements with (they might still have more elements from some $S'' \in P_1$ than the sets that got assigned its name!). If so, assign that name to the one of $S, S'$ that shares more elements with the respective set whose name they can compete for; the other keeps the formerly conflicting name.
• Iterate this procedure until all conflicts are resolved. Since $P_1$ does not have less sets than $P_2$, there are enough names.

Now, you can consider the bit strings of your elements wrt either partition, i.e. $w_1 = n_1(1) \cdot \dots \cdot n_1(n)$ and $w_2 = n_2(1) \cdot \dots \cdot n_2(n)$ (with $n_j(i) = n_j(S), i \in S \in P_j$). Then, the desired quantity is $d_H(w_1, w_2)$, i.e. the Hamming distance between the bit strings.