Euler Formula for planar graphs Sorry to post questions so many times...but I'm confused with what use Euler's formula for planar graphs has.
Because, say, $K_4$ the complete graph with 4 vertices is planar after some rearranging of one diagonal edge. But in its "original" state ->☒, it is not.
I thought this formula allows us to judge whether or not a given graph can be isomorphic to a planar graph. But it seems not.
In the original state, it has 5 regions, 4 vertices and 6 edges.
$e+2=r+v$(Euler) then the equation does NOT hold.
However, AFTER rearranging the ☒ so that a diagonal edge does not cross any other edge, then we have 4 regions, hence Euler's formula indeed, holds.
I am wondering...after the rearrangement, by observation, it's more than obvious that the graph is planar, why will we need the equation to verify it?
I thought it would be useful if we can tell planar/non-planar without trial and error and fiddling with the original graph.
But the formula only seems to say "yep you got it planar now" after I've already got the answer...Because it says "no that's not planar" to $K_4$ when it hasn't been altered with.
Then, why is this equation even useful? I don't need to check via the formula when I've gone through the effort to trasnform it to a planar graph and it is extremely obvious that it is.
Am I missing something here? The formula doesn't seem practical in terms of problem solving... 
 A: I think you are misunderstanding the definition of planar graphs. It means that you can draw it to the plane without it's edges crossing each other. $K_4$ is for sure a planar graph, for example:
About Euler's formula: You can only use that if you drew a graph to a plane without it's edges crossed. In that case, you can use it to determine whether or not it is a planar graph. In your case, you drew $K_4$ with two edges crossing, that's why it didn't work.
You can use the following inequalities for checking if it is planar graph or not:  
Let G be a connected planar simple graph with $n$ vertices, where $n \ge 3$ and m edges. Then $m \le 3n - 6.$ Let G be a connected planar simple graph with $n$ vertices and $m$ edges, and no triangles. Then $m \le 2n - 4 $.
For $K_4$, you can see, that $n=4$, $m=6$, and if you check: $6 \le 3*4-6=6$, it works, so $K_4$ must be a planar graph.
A: Indeed, the faulty assumption seems to be that "Euler's Formula is used to test whether a graph is planar." As you've pointed out, this line of reasoning doesn't really make sense - mostly, without a planar drawing, the number $f$ isn't well defined! This is exactly what trouble you ran into ($f$ not being well defined) when looking at different drawings of $K_4$, unless you used a planar drawing. So in one sense, you could say that Euler's Formula guarantees that $f$ is well defined for all planar drawings of a graph $G$, if a single planar drawing $G$ exists.
Instead, Euler's Formula is really more of a topological fact: it tells us that, no matter what planar graph we draw, then $v - e + f = 2$, no questions asked.
Here's a classic puzzle, the water, gas, electricity puzzle:

Suppose there are three cottages on a plane (or sphere) and each needs to be connected to the gas, water, and electricity companies. Without using a third dimension or sending any of the connections through another company or cottage, is there a way to make all nine connections without any of the lines crossing each other?

Graphically, we can draw what looks like the complete bipartite graph $K_{3,3}$:

And our puzzle can be "solved" if we can find a planar drawing of this graph. By a famous theorem (Kuratowski), we know no such solution exists - on the plane! But if we draw our graph on a torus, suddenly we can do it without crossings. Even more, on the torus, such a graph would have $v - e + f = 0$.
So, the one way to think about Euler's Formula is that it tells us more about the Euclidean plane $\Bbb R^2$ itself, than about graphs: We're just using graphs to study space, from this viewpoint.
(Caveat: Euler's formula can also be used to provide bounds on $v, e$, or $f$, for example by considering so-called maximal planar graphs, and that's another alternative).
A: Let us look into a tetrahedron. It has 4 faces, 4 vertices, 6 edges. It does satisfy the Euler's Formula.
The point is, Euler's Formula is a theorem about polyhedron, but not about graph drawn on paper. A planar graph satisfies Euler's is just because polyhedrons can be "stretched" to planar graphs and vice versa.
Come back to the question. What you miss is, ☒ is not "stretched" (by Euler's way) from a polyhedron, consequently Euler's Formula not works to it.
A: Let me add one way Euler formula can be used. For example, one of the former answers mentions the famous inequality about the number of edges and vertices in a planar graph: $m\leq 3n-6$.

Let G be a connected planar simple graph with n vertices, where n≥3 and m edges. Then m≤3n−6. Let G be a connected planar simple graph with n vertices and m edges, and no triangles. Then m≤2n−4. For K4, you can see, that n=4, m=6, and if you check: 6≤3∗4−6=6, it works, so K4 must be a planar graph.

Where does this inequality come from? Here is an informal explanation that uses the Euler formula.
First let's define that a graph is maximal planar if it is a planar graph and adding one more edge would destroy that property. Now, take a piece of paper and draw some maximal planar graphs that you get a feeling about how such a graph looks. Also you can search some images (Apollonian networks look great).
You can observe that each face  (both inner and outer) is bound by a triangle. For this reason maximal planar graphs are sometimes called a triangulation.
Now, each face is bound by $3$ edges, thus on edge 'gives' $\frac{1}{3}$ of a face. But every edge bounds two faces so each edge is responsible for $\frac{2}{3}$ faces. In any maximal planar graph it, therefore, holds that $\frac{2}{3}m = r$. Every edge 'gives' us $\frac{2}{3}$ of face.
That's good, but the moment when this becomes really powerful is when we use the Euler's formula $n - m + r = 2$. We plug the earlier result into it so $n - m + \frac{2}{3}m = 2$.
In any maximal planar graph the following holds
$$
3n - 6 = m.
$$
When dealing with planar graphs in general, the $\frac{2}{3}m = r$, becomes inequality $\frac{2}{3}m \geq r$ and we have
$$
3n - 6 \geq m.
$$
Let's see how this can be used. Prove that $K_5$ is non-planar. Observe that is has $5$ vertices but $10$ edges. And
$$
15 - 6 \ngeq 10
$$
so $K_5$ can't be planar.
Also note, that the lemma that any maximal planar graph is a triangulation can be proven in a rigorous way. But I just don't remember it offhand.
