What is the complexity of this matrix transposition?

Question

I'm working on some exercises regarding graph theory and complexity. Now I'm asked to give an algorithm that computes a transposed graph of $G$, $G^T$ given the adjacency matrix of $G$. So basically I just have to give an algorithm to transpose an $N \times N$ matrix.

My first thought was to loop through all rows and columns and simply swapping values in each of the $M[i,j]$ place, giving a complexity of $O(n^2)$. But I immediately realized there's no need to swap more than once, so I can skip a column every time e.g. when I've iterated over row $i$, there's no need to start iteration of the next row at column $i$, but rather at column $i + 1$.

This is all well and good, but how do I determine the complexity of this? When I think about a concrete example, for instance a 6x6 matrix, this leads to 6 + 5 + 4 + 3 + 2 + 1 swaps (disregarding the fact that position $[i,i]$ is always in the right position if you want to transpose a $N \times N$ matrix, so we could skip that as well). This looks a lot like the well-known arithmetic series which simplifies to $n^2$, which leads me to think this is also $O(n^2)$. There are actually $n^2/2$ swaps needed, but by convention the leading constants may be ignored, so this still leads to $O(n^2)$. Skipping the $i,i$ swaps leads to $n^2/2 - n$ swaps, which still is $O(n^2)$, but with less work still.

Some clarification would be awesome :)

Answer 1

You can also "transpose" a matrix in $O(1)$ time and space. Maintain a bit $t$ for your matrix $M$. Whenever $M$ is transposed, flip $t$. Now, consider accessing an entry in $M$. If $t$ is set to true, return $M[j,i]$ instead of $M[i,j]$. This trick is arguably cheating, but not always is the matrix actually needed to be transposed. This might work nicely depending the situation.

Answer 2

The sequence you (correctly) identified sums up to $\frac{n(n-1)}{2}=\theta(n^2)$, which gives the runtime you were looking for.

I'm not sure what you are referring to by "leading constants". Do you mean you can ignore e.g. $1+2+3$ ? sure, but that will reduce $6$ from the complexity, which is clearly meaningless. On the other hand, you cannot ignore $f(n)$ leading constants for any $f(n)=\omega(1)$, so there is really no point in ignoring them altogether.