I am asking myself the question "Should I do PhD or should I leave academia and go for an industrial career?"

My life-goal is being a professor. And I love to do research. 

PhD is surely a bite that not everyone can chew.
But I wonder who can chew it?

I never was good at tests and exams. My BSc. GPA was 2.84/4.00 but finished my MSc. with 3.50/4.00

However, currently I am working on a conference paper and I feel like even that is too much for me. It has been nearly 3 months and still, the paper draft is to be improved (not the wording but the content).

I am surely a hard-worker but not always. Sometimes, I let go of my work and absorbed in other stuff (composing, amateur radio etc). If this period is too wide, I have to spend double effort to warm-up and remember where I left.

I don't know how things work in PhD. It usually is 5-6 years. It is the one of two most-challenging milestones in academic career (the other is getting the title Assoc. Prof).
Should I completely be a "nerd" and work on my thesis systematically (something I could never make in my entire life) or working periodically but with extra effort is still sufficient?

So, here's my question: If I say "I'm considering to do PhD" and ask your advice, what would you ask me? What kind of skills/characteristics do you look for a potential academician?

I know it is way too late for me to ask this kind of question, as a person who almost finished his master's degree. But better lose the saddle than the horse. 


Question: Hypergraph coloring
Answer: As far as I understand, your hyperedges are of the form $\{v+c,v+d,v+e\}$ for all suitable $v$. Hence you can just color the vertices from the left to the right. Color the vertices $u\leq e-c-1$ as you wish; then, when you consider some further vertex $u\geq e-c$, there is exactly one edge with the maximal element $u$; so you can color $u$ so that this edge is multichromatic. Proceeding in this way, you will finish with no troubles; each edge was considered on some step, so it is not monochromatic.

Surely this works for any bound instead of $2e-1$.

EDIT. Even simpler: just erase $v+d$ from each edge $\{v+c,v+d,v+e\}$. The remaining graph is obviously bipartite.