מיכל מבקשת שמי שמעוניין להגיע לשעת הקבלה שלה ביום ראשון הקרוב (5/6) ישלח לה מייל לפני כן, על מנת לחלק את הזמנים בין הסטודנטים באופן הוגן ויעיל.

As Michal said in class, you should submit a proposal for your intended project topic. The proposal should include a list of articles you plan to look at, a short description of the topic (few paragraphs), and a possible research question(s).

It is highly recommended to try to have a research project.
Thus, in general, reseach projects will gain a higher grade.

Projects should be submitted in groups of 2-3 students.

אני מצטערת אם לא הייתי ברורה.
העניין הוא שיש לך שאלות והערות רבות על כל הרצאה. אם הפסדת כמה הרצאות, כנראה שכדאי להשתמש בספר על מנת להשלים את החומר, בנוסף לסיכומים (או במקום). שימוש במקור נוסף יפתור את הבעיה של חוסר דיוקים ושגיאות בסיכומים.
אין לי יכולת לענות על כל כך הרבה שאלות על כל הרצאה במסגרת הפורום. הפורום נועד בעיקר להבהרות לגבי תרגילי בית (אני רק בודקת תרגילים, ולא מתרגלת).
אם יש שאלה ספציפית אני משתדלת לעזור, אבל בכמות הזאת זה מעבר לscope של העבודה שלי.

Thanks a lot

3. The function in Brouwer fixed-point theorem is from domain C to range C, when C is a subset of R^t. (So any element in C can be viewed as a vector of size t). The domain and range used in the proof of the existence of NE are not a subset of some R^t. (Any element in that domain is a "vector of vectors"). So the implementation of Brouwer fixed-point theorem seems to be not very trivial.
Anyway, I was asking: can we interpret a profile as a vector of size t=|S_1|+..+|S_n| (and then the using of Brouwer theorem will be more trivial)?

4. in page 10, after the minmax theorem (lemma 15) it said "Immediate implication: with mixed strategies, the order of the turns does not matter - the players will take the same strategy sets."
How does it implies from that lemma? (The lemma above is an inequality (unlike the minmax theorem in lecture 2) and it seems that the only thing implies from it is that each player is better playing second).

Thank you

1.Is that the correct definition?
2.h can be anything, like h = sizeof(v_i)?
3. Does 5b should be proven for any h?

About definition 8 - you are right.
For the others, I didn't understand your intention.

(meaning each player has its own Delte(S_i), possibly with different vector sizes then the others)?
2. In example 1 p.8 - We assume we get NE when player 1 plays B with prob' p and O with prob' 1-p.
We can calculate p using the fact that each choice of player 2 leads to the same utility expectation, meaning E[u_2| player 2 plays B] = E[u_2| player 2 plays O]. So we have 2p+0(p-1) = 0p+3(p-1) leading to p=3/5.
Where is the mistake here?
3. Using the Brouwer fixed-point theorem - the set in the theorem is a subset of R^t. The set used to prove NE theorem is a bit more complicated. Should we look at it as a subset of R^t when t=|S_1|+..+|S_n|?
4. How is the immediate implication implies from lemma 15?
Thank you very much

התרגילים הבדוקים נמצאים במדפים של התרגילים בשרייבר (כשיורדים למרתף, מצד שמאל).
המדף שלנו הוא בפינה השמאלית התחתונה.

Besides, do we need to show also membership in NP? Or only NP-hardness?

(Binary search works because you know the set of critical values, but that's not a complexity\optimization course, so it really doesn't matter).

Given S, S1 and S2, are S1 and S2 a partition of S that minimizes the makespan?

is in coNP. A verifier for S1 and S2 not being a partitioning that minimizes the makespan is easy to formulate - just accept a different partitioning of S and calculate its makespan.

Since it's in coNP I don't think we can prove that it's NP-complete.

The following problem is in NP:

Given S and k, is there a partitioning of S with makespan less than k?

One may try to think of a scheme to verify the following problem by applying the above verifier repeatedly:

Given S, what is the minimal makespan of any partition of S?

But I really can't see how. Binary search won't work since S's elements can be arbitrarily large.

I'm confused about what you expect to see in this question.
If the the current formulation is changed to say "NP hard" at the end, then I'm fine with it.

For the left side:
ln(1-eps) = - eps - (eps^2)/2 - O(n^4) >= - eps - eps^2.
(note that esp^2 is divided by 2).

For the right side:
ln(1-eps*A) = - eps*A - O((eps*A)^2) <= - eps*A.

Does that fit?