Looking for the minimum takes more work than looking for one that’s exactly 100, and you’re right that this is binary optimization, which is a giant pain.
I think I was thinking of it as including m binary variables, which I’ll call z_j, which indicate whether or not the u_j are nonzero, and then an L0 minimization is that the cost function is the sum of the z_j or the L0 constraint is a constraint that their sum must equal 100. (Standard caveat than L0 is not actually a norm, which I should edit in to the previous post.)
The SAT problem Eliezer mentioned is binary (well, boolean), and it felt awkward to claim that it’s a SAT problem when that could be mistaken as a question on the otherSAT, so I decided to switch names without moving too far in concept-space. Your explanation seems much neater than mine, though.
[edit]I should be clear that I mean that looking for the L0 minimization directly is at least as hard as doing it the binary optimization way, but that the L1 minimization problem, which indirectly solves the L0 minimization problem only sometimes (read: almost always), is polynomial time, i.e. much easier to solve, because it’s linear.
I should be clear that I mean that looking for the L0 minimization directly is at least as hard as doing it the binary optimization way, but that the L1 minimization problem, which indirectly solves the L0 minimization problem only sometimes (read: almost always) is polynomial time, i.e. much easier to solve, because it’s linear.
Why binary optimization?
I think I was thinking of it as including m binary variables, which I’ll call z_j, which indicate whether or not the u_j are nonzero, and then an L0 minimization is that the cost function is the sum of the z_j or the L0 constraint is a constraint that their sum must equal 100. (Standard caveat than L0 is not actually a norm, which I should edit in to the previous post.)
The SAT problem Eliezer mentioned is binary (well, boolean), and it felt awkward to claim that it’s a SAT problem when that could be mistaken as a question on the other SAT, so I decided to switch names without moving too far in concept-space. Your explanation seems much neater than mine, though.
[edit]I should be clear that I mean that looking for the L0 minimization directly is at least as hard as doing it the binary optimization way, but that the L1 minimization problem, which indirectly solves the L0 minimization problem only sometimes (read: almost always), is polynomial time, i.e. much easier to solve, because it’s linear.
Ok.