discrete mathematicsorder-theory
When drawing a Hasse diagram, I have seen that you can draw a bigraph for the poset and remove the reflexive and transitive edges of the poset. However, doing this for a poset with many elements can get tedious (by hand).
Is there a more efficient way of drawing a Hasse diagram?
To draw a Hasse diagram of a finite poset follow the algorithm below, but first notice that on any finite poset $P$, given $x,y\in P$ such that $x\neq y$, then $x$ and $y$ aren't comparable or there is a chain of covers from one to the other.
Now for the algorithm: Let $x,y,z\in P$
The whole idea of a Hasse diagram is just an efficient representation of your poset. If you think about the set of subsets of $\{1,2,3\}$ ordered by inclusion (that is, $\subseteq$), we have ordered pairs like $(\varnothing, \{1\}),\, (\varnothing \{1,2\}),\, (\{1\}, \{1,2\})$, and so on, reflecting the fact that $\varnothing \subseteq \{1\}$, etc.
But since we know that $\subseteq$ is transitive, then knowing $\varnothing \subseteq \{1\}$ and $\{1\} \subseteq \{1,2\}$ tells us that $\varnothing \subseteq \{1,2\}$; it would be silly for our efficient representation to waste time representing this, as long as we make sure to represent $\varnothing \subseteq \{1\}$ and $\{1\} \subseteq \{1,2\}$:
So for example, the fact that we can trace a path from $\{1\}$ up to $\{1, 2, 3\}$ using edges in the Hasse diagram means that $\{1\} \subseteq \{1,2,3\}$, or equivalently, that the ordered pair $(\{1\}, \{1,2,3\})$ is in our relation.
We only draw edges in the Hasse diagram to depict so-called covering relations: So, we drew a line from $\{1\}$ to $\{1,2\}$ because if we have an inequality like
$$\{1\} \subseteq \Delta \subseteq \{1,2\},$$ then we're forced to use $\Delta = \{1\}$ or $\Delta = \{1,2\}$; nothing "fits" between them properly. Thus, we say that $\{1\}$ is covered by $\{1, 2\}$, and draw a line in the Hasse diagram.
We generally use the symbol $\le$ any time we have a poset, even if the relation has nothing to do with the normal definition of $\le$ to compare numbers. So we could write $\varnothing \le \{1\}$ in the example above, if we wanted to. Some people, to prevent this confusion, use the symbol $\preceq$ for a generic poset, to help you realize that the relation might have nothing to do with the usual way to order real numbers.
When you're given an unlabeled Hasse Diagram as in your last example, just call all of the nodes by some name; $\{1, 2, 3,4\}$ or $\{a, b, c, d\}$, it doesn't really matter. Just that we can see the comparisons between them.
So we could pick 
and start getting ordered pairs $(a, c),\, (a, d),$ etc, since we can see that $a \le c$, $a \le d$, and so on. In this example, the only relations are covering relations, so we'll have as many ordered pairs as there are lines in the graph.
In your linear example with $1 \le 2 \le 3 \le 4$, we would have $4 + 3 + 2 + 1$ ordered pairs: $\{(i, j): i \le j\}$ even though there are only $3$ edges in the Hasse Diagram.
Best Answer
This problem is known as transitive reduction. It is shown equivalent to multiplication of binary matrices. A simple $n^3$ algorithm (so pretty good in regards of matrix multiplication) is as follow : Compute the transitive closure then find transitive edges by iterating over all sets of 3 vertices.