Week 5
During the fifth week, we decided to try some new approaches to the problem. We had been focusing on bagging algorithms, where bags were filled with items in a strategic fashion until an agent valued it at more than 2/3 and claimed it. We decided that since we kept running into the same problem when trying to construct algorithms using bagging, we should try something else. We explored both round-robin and LP. With round robin, each agent takes turns (in a predetermined constant order) picking the best item, and since we assume ordinality in the problem, essentially the first agent will get the 1st, n+1st, 2n+1st, and so on items, while agent n will get the nth, 2nth, 3nth, and so on items, all of which are worse than the first agents. However, this does not matter as long as we can prove that since we know agent 1 had value greater than 1, then maybe agent n is only off by a factor of 2/3. This method works great in the case of 1/2-MMS, but the issue with 2/3-MMS is it lacks the guarantee that every item is less than 1/3, because after we perform something called a two-good reduction, there are n items that may have value greater than 1/3. Thus we cannot guarantee that each of agent n’s items are at most 1/3 less than each of agent 1’s items as needed. We though maybe we could improve round robin slightly by turning it into a sort of round-robin/bagging algorithm, where the agents aren’t in predetermined order and assigned a bag, but they can claim a good bag by choice. However, we ran into the same issue we had with bagging before, and decided to move on to LP. We defined the linear program and familarized ourselves with a process that has been proved to work for the case of 1/2. Essentially, the LP solution would provide each agent with at most k items with value 1 (guaranteed by assignment of 1/nth of each of the top nk item to each agent) and then through reassignment of the items, a binary solution can be found that guarantees 1/2 value for each agent. We want to proceed to prove this for 2/3, but again have to tackle the issue of the n items that might not have value less than 1/3.