What is the reverse of greedy algorithm for setcover?
The approximation guarantee will be significantly worse.

Assume you want to cover the set $U=\{1,\ldots,2n\}$.
For every $i=1,\ldots,n$ define a set with n+1 elements by $S_i=\{i,n+1,\ldots,2n\}$.
Assume we want to cover U with the sets
$C=\{S_1,\ldots,S_n,U\}$, where $c(U)=2, c(S_i)=1$.

We would now start with $C$ ans see that the effectivity of the sets $S_i$ is 
$$\rho(S_i)=\frac{|S_i|}{c(S_i)}=\frac{n+1}{1}=n+1.$$

In contrast to this the effectivity of $U$ is
$$\rho(U)=\frac{|U|}{c(U)}=\frac{2n}{2}=n&lt;n+1.$$

Thus the reverse greedy would throw out the less effective $U$ leaving us with a solution of cost $n$ while there is a solution of cost $2$.