CSTBC: Theory Bridge CourseInstructor: Kevin Milans (milans@uiuc.edu) Newsgroup: news.cs.uiuc.edu/class.i2cs.theorybridge 

Photo by Bryan Clark 
[(n1)/3n] * k < k/3of the troublesome pairs eat together. By the time the exam is due, I think we'll have the tools we need to prove this stronger statement. I'll go over it in the exam 2 discussion lecture.
For example, if you are given a situation with 30 pirates and 60 troublesome pairs, then you are only allowed to have 20 bad pairs eat at the same time. If you try to "assume the worst case" and look at the situation where all (30 choose 2) = 435 pairs have fought, you are allowed to have 435/3 = 145 bad pairs eat together. Although we have more bad pairs in this situation, we are also allowed to have more bad pairs eat together. It is not clear how solving the problem in this case leads to a solution where at most 20 of the original 60 bad pairs eat together.
The second point of confusion is that we want to assign pirates to dinner times so that the total number of bad pairs that share a time, summed up over all 3 time slots, is k/3. If you multiply the number of bad pairs below by 3, you'll get k/3.
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