Question

Show that for any integer n that is a power of 2 , there is an instance of the set cover problem (Section 5.4) with the following properties:

1. There are n elements in the base set.
2. The optimal cover uses just two sets.
3. The greedy algorithm picks at least log n sets.

Thus the approximation ratio we derived in the chapter is tight.

Short answer

Any integer which count with 2 or count with power of 2 it gives elements in base set and optimal cover using two set in greedy algorithm.

Step-by-step solution

Elements of Even /Odd Numbers

Represent a set with n items that would be a positive integer of 2 . Permit the collection to be. $(1,2,...,2^k)$ for some $K\ge 2.$

So ideal number of matches would be two. • One has even numbers, while the other has odd numbers. As a result, the two sets will be localid="1658922602021" $X=\left\{1,3,...,2^k-3,2^k-1\right\}$ and$Y=2,4,...,2^k-2,2^k.$$Y=\left\{2,4,...,2^k-2,2^k\right\}.$.

Algorithm of number of 2 ’s in odd / even calculation

Now, Take, for example, a greedy algorithm.

Take into account everything. i ,

Let $D_i=2^{(k-1)},2^{(k-i)}+1,...,2^{k-i+1}$.

Clearly,

$A={\cup }_{{}_{i=1}}^k{D}_1$

As$n=2^k$

Thus,$\log n=k$

Its foundation setting is the combination of sets handled by greedy.

• Although X and Y do not cover and over half of both the elements, they provide the best solution.

• $D_1$, on the other hand, actually covers one additional vertex than X or Y .

As a result, there is a close approximated ratio among both greedy and optimum solutions.