Merging two partial (jointly overdetermined) sets of ordering information

Question

I have a web app with data in a grid. The user can reorder the columns, and the server can change which columns exist. I would like to save the user\'s column order in a coo

Answer 1

One definition that might be reasonable is to minimize the number of inversions relative to the server's order, subject to the constraint that the restriction to the columns in common of the two orders be equal. I don't know offhand an algorithm to minimize this objective.

Answer 2

Let's say we have C columns, numbered from 1 to C. We have two sequences sequences of columns, U = u₁, u₂, ... u_n and S = s₁, s₂, ... s_m. We want to find a permutation P of S, such that P has no inversions with regard to U and a minimal number of inversions with regard to S.

We can show that there is such an optimal P which is an interleaving of U ∩ S and S \ U. By "interleaving" I mean that P has no inversions with regard to U ∩ S or S \ U.

We can apply dynamic programming to find the optimal interleaving: Let A = (a_i) = U ∩ S and B = (b_j) = S \ U. Let f(i,j) be the number of inversions w.r.t. S of the optimal interleaving of the prefixes a_1...i of A and b_1...j of B. The idea is very similar to the longest common subsequence DP algorithm. We have the recurrence

f(0,j) = 0 for all j >= 0
f(i,0) = f(i-1, 0) + sum(k=1 to i-1, [1 if A[i] appears before A[k] in S])
f(i,j) = min(f(i-1, j) + sum(k=1 to i-1, [1 if A[i] appears before A[k] in S])
                       + sum(k=1 to j, [1 if A[i] appears before B[k] in S]),
             f(i, j-1) + sum(k=1 to i, [1 if B[j] appears before A[k] in S])
                       + sum(k=1 to j-1, [1 if B[j] appears before B[k] in S]))

I used the notation [1 if X] here to denote the value 1, if X is true and 0, if X is false.

The matrix f can be built in time O(|A|^2 * |B|^2). The minimal cost (number of inversions w.r.t. S) will be f(|A|, |B|).

We can reconstruct the optimal permutation using the DP matrix as well: We build it from back to front. We start with the tuple (i,j) = (|A|, |B|) and at every step depending on which of the two options is minimum in the DP transition, we know whether we need to put A[i] or B[j] to the front of the permutation. Then we proceed with (i-1, j) or (i, j-1) depending on which we choose.

Here is an implementation of the algorithm, please excuse my lack of JS skills.

Answer 3

This relates to Niklas B's answer:

Theorem: Consider a sequence S = s₁, …, sₙ of some orderet set (e.g. integers). If i < j and sᵢ > sⱼ, then swapping sᵢ and sⱼ decreases the number of inversions—that is, let S' = s₁, …, sᵢ₋₁, sⱼ, sᵢ₊₁, …, sⱼ₋₁, sᵢ, sⱼ₊₁, …, sₙ; then S' has fewer inversions than S.

Intuitively stated: if two elements are out of order and you swap them, you're closer to having a sorted list.

Proof: Observe that the only elements that have a different relative order in S and S' are (sᵢ, sⱼ), (sᵢ, sₖ) and (sⱼ, sₖ) for each k where i < k < j. We know that (sᵢ, sⱼ) is an inversion in S but not S', so consider sₖ for some such k.

Either sₖ < sⱼ < sᵢ or sⱼ < sₖ < sᵢ or sⱼ < sᵢ < sₖ (we assume the elements of S to be unique).

In the first case, (sᵢ, sₖ) is an inversion in S and (sⱼ, sₖ) is an inversion in S'. In the second case, (sᵢ, sₖ) and (sⱼ, sₖ) are inversions in S but not in S'. In the third case, (sⱼ, sₖ) is an inversion in S and (sᵢ, sₖ). These are all the changes in inversions.

In each case, the number of inversions in S' is either the same as that in S or it is smaller. Recall that (sᵢ, sⱼ) got fixed from S to S', and we get the desired result. ■

Thus, if we have a₁, bᵢ, …, bⱼ, a₂ with each a ∈ S \ U and each b ∈ U ∩ S and a₁ > a₂ and we swap a₁ and a₂, getting a₂, bᵢ, …, bⱼ, a₁, the inversion count is lower. Since such swaps only rearrange elements of S \ U and not those of U ∩ S, any solution which has zero inversions on U ∩ S and (subject to that) a minimal number of inversions on S \ U must make all such swaps.

Ergo: the elements of S \ U must occur in order, and thus the solution is an interleaving of U ∩ S and S \ U.