I was came across a problem during an exercise in a book and I don't really know how to solve it. I feel like something's missing.
"coverage, c = $NL/G$ (N=number of reads, L=read length, G=genome length) total expected gap length = $G(e^{-c})$ total number of gaps = $N(e^{-c})$
Question: You want to sequence a 4 Mb genome by the shotgun method, by assembling random fragments with read length 500. What coverage would you require, to expect no more than four gaps, assuming no complications arising from repetitive sequences or far-from equimolar base composition?"
The answer from the book is basically just: Solve $N(e^{-c})=4$ for $c$, but the question doesn't give what $N$ is?