Suppose that we are given three sequences $a1,a2$ and $a3$ each describing a total ordering on $N$ 'entities'. For example,
$$\langle a1\rangle=1<9<8<2<3<\cdots<N $$
means that entity #1 is worst and entity #N is the best according to $a1$.
I need to come up with a new total ordering, let us call it $\langle x \rangle$ that best approximates all given orderings.
One way I was thinking of doing this task with $N=4$:
$$<a1> = 1<3<4<2 $$ $$<a2> = 1<2<4<3 $$ $$<a3> = 4<2<3<1 $$
Look at each pair $x,y \in \lbrace 1,2,3,4 \rbrace, x\neq y $. For example let us look at the tuple $(1,4)$. Clearly the first two sequences claim $1<4$ while the third claims $4<1$. Pick out the majority winner. Here, $1<4$ wins.
Do this for each possible tuple. We then have $\binom{N}{2}$ such tuples. List them , for eg: $1<4$, $2<3$, ...
Use these now to combine them into a total ordering $<x>$.
ISSUE: This problem can be converted to the Hamilton Path problem. Let each number be a vertex. Put a directed edge between $x$ and $y$ iff $x<y$. Now find the directed path that convers each vertex exactly once, this is my $<x>$.
Is my argument that this problem is NP complete correct? (Since Ham path is NP complete). If so, in what other ways can I do this task?
The argument is both incorrect and wrong.
It's incorrect because reduction arguments have to go the other way: by solving your problem you can solve any instance of a known NP-complete problem (and the reduction is polynomial).
It's wrong because the graph defined in your problem is a tournament (Wikipedia, MathWorld) and a Hamiltonian path in a tournament on $n$ vertices can be found in $O(n \lg n)$ time. See e.g. Bang-Jensen and Gutin, On the complexity of hamiltonian path and cycle problems in certain classes of digraphs, for a survey.
Although it's not directly what you ask, I should note a problem with the proposed algorithm:
(my emphasis). There isn't necessarily a unique Hamiltonian path. Consider the minimal example where the three input orders are $$1 < 2 < 3 \\ 2 < 3 < 1 \\ 3 < 1 < 2$$ The graph is a cycle $1 \to 2 \to 3 \to 1$, so there are three Hamiltonian paths. Obviously in this case the three are equally good / bad approximations, but maybe a more complicated example which embeds them would have Hamiltonian paths which vary in how many of the original pairwise comparisons they contradict.