LP for set cover:
================

min Sum_S x_S c(S)
s.t.
    Sum_{S containing e} x_S >= 1, for all e in U
    x_S >= 0

LP-based greedy algorithm for set cover:  (Hooman's algorithm)
=======================================

1. Solve LP

2. Sort the x_S in decreasing order.

3. Select the smallest prefix of this sequence of sets (in the sorted
order) that covers all of the elements.

Counterexample for the above algorithm:
======================================

The approach is to build a system in which any solution will have to
select at least one set from a subcollection of high-cost sets since
these sets are the only ones that contain a particular element.  Now,
the optimal LP solution may pick these sets fractionally.  We set the
system up in such a way that the fractional values for these sets in
the LP solution exceed the fractional values chosen for a large number
of other sets, which are the ones that primarily cover the other
elements that belong to the expensive set collection.

Define A = {a}, B = {b_1, b_2, ..., b_n}, C = {c_1, c_2, ...,
c_{2n-2}}

U = A U B U C = {a, b_1, b_2, ..., b_n, c_1, c_2, ..., c_{2n-2}}

Collection of sets = X U Y U Z

X = {{a,b_1}, {a,b_2}, ..., {a,b_n}}
Y = {{c_1}, {c_2}, ..., {c_{2n-2}}}
Z =  {{b_1,c_1}, ..., {b_1,c_{2n-2}},
     {b_2,c_1}, ..., {b_2,c_{2n-2}},
     ...,
     {b_n,c_1}, ..., {b_n,c_{2n-2}}}

c(S) = L for S in X, 1 for S in Y and 2 for S in Z, L >> 1

Here is a solution x for the set cover LP.

x_S = 1/n for all S in X
x_S = 1/2 for all S in Y
x_S = 1/(2n) for all S in Z
     
Feasibility of x:
================

-- For the element a, Sum_{S containing a} = n*(1/n) = 1

-- For an element e in {b_1,..,b_n}, Sum_{S containing e} = 1/n +
(2n-2)/(2n) = 1

-- For an element e in {c_1, c_2, ..., c_{2n-2}}, Sum_{S containing e}
= 1/2 + n/(2n) = 1

Optimality of x:
===============

The cost of x = L + (2n-2)/2 + 2*n(2n-2)/(2n) = L+n-1+2n-2 = L+3n-3.

Recall that the dual LP is:

max Sum_{e in U} y_e
s.t.
    Sum_{e in S} y_e <= c(S)
    y_e >= 0

Consider the following solution y to the dual LP.

y_e = L - 1 if e = a
    = 1 for e in B U C

The solution is feasible since the sum of y_e is L for each set in X,
1 for each set in Y and 2 for each set in Z.

The cost of y = L-1+3n-2 = L+3n-3.

Since the cost of x equals the cost of y, x and y are both optimal for
their respective LPs.

Performance of the LP-based greedy algorithm:
============================================

The above algorithm will select X U Y for a total cost of nL+2n-2.  A
good integral solution is to select one set in X, all sets in Y, and n
sets in Z (one for each c_i), yielding a total cost of L + 2n-2 + 2n =
L+4n-2.  If we make L large, the approximation ratio of the algorithm
tends to n.