Correction to problem 4.

Hi guys,

Problem 4 on the latest homework turns out to be a lot harder than we anticipated. (Thanks to Jiri and Min for pointing this out).

Here is a modification that makes it much more tractable. The separation penalty between vertices i and j will now depend on the distance between the labels of i and j. That is, label pairs (Bac, Foc) and (Foc, For) still result in penalty p_i,j, but the label pair (Bac,For) now results in penalty 2*p_i,j

Sorry about this. Grading will be lenient, and extra credit if you solve the original problem.

Homework 3 Question 2

Hi,

For homework 3 question 2, you may assume the graph is connected (aka. k>0 and the algorithm can always find some edge to contract). Also, part 5 of that question should have "at most $O(|V|^{2\alpha})$" instead of $\Omega(|V|^{2\alpha})$.

Sorry about this.

Richard

Office Hours

Hi Class,

I'll be out of town this coming week, so I won't be holding my regularly scheduled office hours. I should still be in frequent email contact though, so feel free to email questions about the homework if you're getting started early (It should be posted by later today).

I'll also hold extra office hours by appointment, in addition to my regular office hours next week, so if you can save up your questions until then, we can meet when it is convenient for you.

-- Aaron

Wednesday office hours moved

Hi Class,

I'll be out of town during my regularly scheduled Wednesday office hours, so I'll be holding office hours Thursday instead. (Same time, 10:00-11:00 am)

Modification to Homework 2 Question 4

Hi,

Question 4 of homework 2 has been changed to remove the degenerate case of multiple points having the same y-coordinate. You make assume the input points have been perturbed to avoid this.

Thanks to Henry DeYoung for pointing this out and sorry for not realizing this issue earlier.

Richard,

Hints for Problem 2

Here are some hints for problem 2, for those of you who are stuck. Even though the solutions are all simple, for these problems its easy to think for awhile without getting anywhere. This should get you on the right track; feel free to email or visit office hours for further hints.

2a)
See if you can mentally divide the pizza into two sets in such a way that you can select every slice in one or the other of the sets, no matter what your opponent does. If you can do this, you are done, because one of the two sets has to have at least half of the pizza.

2b)
This problem doesn't depend on the specifics, like the exact size of the pepperonis. Think symmetry. Suppose we're playing on an annulus (a disk with a circle cut out of the middle). If I make the first move, what is some move you are guaranteed to be able to make? If you are the first player on a pizza, how can you guarantee that you will be the second player on an annulus?

For 2c:
It's a long standing open problem to actually come up with a winning strategy for the 1st player, so don't try to do that. Just prove one must exist. Suppose it were the case that the 2nd player had a winning strategy. What this means is that even if he were to tell you exactly what his strategy was, he could still win. Is this possible, or if he were to tell you his strategy in the chocolate bar game, could you use it to win yourself? An important property of this game is that many moves wipe out all sign of the previous history of the game in the current game state: for example, if I go first and select square (1,1), and you go second and select square (3,3), the state of the game is exactly the same as if you had gone first and selected square (3,3). The solution to this problem is a nice example of a nonconstructive proof.

Homework 1 Question 4

Thanks to several people who pointed this out during Aaron's office hours, what we thought was a minor detail on question 4 is a lot more complicated. The rearrangement of the 4 rank k subtrees causes changes to the parent pointers of their siblings, which are required to determine and fix violations. There exists a fairly complicated set of extra invariants on the violations that keeps this process at O(1) worst case, but that is beyond what is reasonable for this assignment. Sorry for this oversight.

As a result, your solutions may assume that there is an oracle which can support the rearrangement of rank-k subtrees and queries for parents in O(1) worst case time. You are also welcome to work towards the full solution for up to 20 bonus points.