I was reading about the $\mathcal{O}(\frac{1}{\log n})$ approximation algorithm for the Unique Coverage Problem from these notes.

The gist of the algorithm is as follows:

  • Partition the elements into $\log n$ classes according to their degrees, i.e., the number of sets that cover the element. So class $i$ contains all the elements which are covered with at least $2^i$ and at most $2^{i+1}$ sets.
  • Let $i$ be the class of the maximum cardinality.
  • Choose any set with probability $\frac1{2^i}$

Then in Lemma 3, they go on to show that

Lemma 3: The expected number of the elements uniquely covered from class $i$ is $\frac{1}{e^2}$ fraction of the element in class $i$.

I was able to follow the algorithm till here but I am completely stumped as to how they arrived at the statement immediately following lemma 3

Therefore, the total profit of uniquely covered elements is at least $\frac{1}{e^2 \log n} \times \text{opt}$

How did they arrive to this conclusion? They didn't even show what the upper bound is for this maximization algorithm. What am I overlooking?

It is trivial to prove that the expected number of unique elements covered by the algorithms is $\frac{n}{e^2}$ since it directly follows Lemma 3 but that is about all that I am able to observe :/


1 Answer 1


Suppose there are $m$ elements and $n$ sets (this is the meaning of your parameter $n$). The algorithm divides the set of elements into $\log n$ classes, so one of these classes, say class $i$, contains at least $m/\log n$ elements. Lemma 3 shows that if you pick each set with probability $1/2^i$, then in expectation you uniquely cover a $1/e^2$ fraction of the elements in class $i$. Thus, you uniquely cover at least $m/(e^2\log n)$ elements in expectation. On the other hand, the optimal cover uniquely covers at most $m$ elements. Thus the approximation ratio of this algorithm is at least $1/(e^2\log n)$.

  • $\begingroup$ But isn't this analysis lose? I mean, if we expect to cover $\frac{1}{e^2}$ elements from each class, if we take this sum over all the classes, by linearity of expectation, don't we expect to cover $\frac{m}{e^2}$ number of elements? This will give us a constant $\frac{1}{e^2}$ approximation scheme. $\endgroup$ Commented Sep 24, 2016 at 18:40
  • $\begingroup$ Not quite. Don't forget that we have to cover elements uniquely. $\endgroup$ Commented Sep 24, 2016 at 18:42
  • $\begingroup$ But the lemma says "The expected number of the elements uniquely covered from class $i$ is $\frac{1}{e^2}$ fraction of the element in class $i$" So from each class we expect to uniquely cover $\frac{1}{e^2}$ of the elements right? $\endgroup$ Commented Sep 24, 2016 at 18:45
  • $\begingroup$ Yes, but you will also cover elements from other classes. Take a couple of hours to think about it. $\endgroup$ Commented Sep 24, 2016 at 18:47
  • $\begingroup$ Oh I see so Lemma 3 was specifically for class $i$ with the maximum cardinality and wasn't used as an index variable since the inequalities won't carry through for the other classes. Did I get it right? $\endgroup$ Commented Sep 24, 2016 at 18:59

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