Wednesday, November 18, 2009

Lecture 21

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.
  • 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.
For d = log n, the universe has n elements. So this gives a Omega(log n) integrality gap. You should try to see whether you can get an integrality gap larger than this. Also, to fill in the details of the randomized construction I sketched in class.

No comments:

Post a Comment