- Now OPT is at least d/2: else we would have covered less than d/2 positions, and hence missed some element.
- But the LP solution picks each x_S to a fraction 2/d: hence each element is fractionally covered to at least 1.
Wednesday, November 18, 2009
The deterministic integrality gap instance is super-simple: take all the d-bit strings that have at least d/2 1's in them. There are d sets, one for each dimension, with the set S_i containing all strings with the i-th bit 1.
Posted by Anupam at 1:52 PM