Coverage required

Question

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?

Answer 1

According to the book you mentionned Introduction to Genomics SecondEdition by Arthur MLesk p.113

(a) What fraction of a genome could you expect to assemble from eightfold coverage? 
(b) What total gap length would you expect in an assembly of a 2 Mb target genome size from eightfold coverage? 
(c) How many gaps would you expect in an assembly of a 2 Mb target genome size from an eightfold coverage of fragments with a read length of 500? 
(d) 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?

I'm agreed that whitout knowing $ N $ it's not obvious to answer this last question so perhaps they're a mistake in the statement of this exercise. As Devon Ryan suggested, let's assume eightfold coverage means the coverage $ C $ = 8X

So we can calculate the number of reads:

$ N = \frac{GC}{L} $

$ N = \frac{4000000\times{8}}{500} = 64000 $ reads

Then we can calculate the required coverage to expect no more than four gaps:

$ N( e^{-C}) = 4 $

$ e^{C} = \frac{N}{4} $

$ C = log{\frac{N}{4}} $

$ C $ = 10X

(PS: I would like to thank @LordOfRacoons to help answer this question.)