Number of adjacent swaps Given an initial sequence of $n$ integers $(1,2,\ldots,n)$ and an ending sequence (which is a permutation of those $n$ integers), what is the minimum number of adjacent swaps involving the $n/2$ integers? (Assume $n \geq 2$ is a power of $2$).
Note that the operation to get the final sequence is only by adjacent swaps.  
For example when $n = 4$, if the ending sequence is $(4,3,1,2)$, then there have been a minimum of $3$ adjacent swaps involving the first $4/2 = 2$ integers.
Since we go
$(1,2,3,4) \rightarrow(1,2,4,3)$ ($4$ swaps with $3$, but the count is $0$ since we consider the first $4/2 = 2$ integers
$(1,2,4,3)\rightarrow (1,4,2,3)$  ($4$ swaps with $2$, and count is now $1$, since $4$ was in the first $2$ integers)
$(1,4,2,3) \rightarrow(4,1,2,3)$ ($4$ swaps with $1$, and count is now $2$, since $1$ is in the first $2$ integers)
$(4,1,2,3)\rightarrow(4,1,3,2)$ ($3$ swaps with $2$ but count is still $2$)
$(4,1,3,2) \rightarrow (4,3,1,2)$ ($3$ swaps with $1$ and count is incremented to $3$).  
Let me know if any clarification is needed!  
I've tried doing some recursion since it involves only the first $n/2$ integers, but no ideas from there. I thought I could use some combinatorial argument too but it gets messy when I test with $n\geq 8$.  
Response to possible duplicate: I think mine is different. Mine only counts the number of operations in the first $n/2$ integers, whereas that gives a method for the entire string of integers. I'm having trouble finding a formula or function for the number of minimum operations.

It is different from previous question, but answer is similar: call "wrongness" of permutation the number of pairs of elements in wrong order in the first half multiplied by $2$, plus number of pairs (element from the first half, element from the second half).
If wrongness is zero (so the first half is already as we want it to be), we need to just rearrange second half, and we don't need any swaps in the first half to do it.
If wrongness is not zero, we can always decrease it by either $1$ swapping the $n/2$-th and the $n/2+1$ elements, or by $2$ by swapping some elements in the first half. Indeed, take some $p_i$ in the first half and $p_j$ s.t. $i < j$ but $p_i$ should be in the right of $p_j$. There is some adjacent out-of-order pair in $p_i, p_{i + 1}, \ldots, p_j$. If this pair intersects first half - we can swap this elements, reduce wrongness by $1$ or $2$ and pay $1$ or $2$ correspondingly. If both elements are in the second part - swap them (and pay nothing). As it decreases number of out-of-order pairs in the second part, we will eventually get a pair that intersects the first part.
So wrongness is upper bound on number of swaps we need. It's easy to see that any operation decreases wrongness by at most amount of swaps in the first half, so it's also lower bound.
