So Dr. Khovanova described Nim-Chomp to me at RSI more than a year ago, and I thought I solved it. Then a few months ago I found a flaw, thought it was hopeless, and gave up. Then a few minutes ago I realized that the flaw was in fact fixable. The point of this paragraph is that I’m not sure I’m to be trusted regarding this problem, but I’ll try.

I won’t repeat the problem statement here. Too lazy. Just read her post. Because I find it easier to think about this way, I’ll make a slight modification to the Chomp perspective on Nim-Chomp: on a given turn, each player may only eat squares from a single *column* (rather than a single *row*).

First let’s pretend the bottom left square is not actually poisoned. We can transform this easy-Nim-Chomp into regular Nim as follows: for a given position in easy-Nim-Chomp, suppose the number of squares remaining in each column is a_{1}, a_{2}, …, a_{n} from left to right. Let b_{1} = a_{1} – a_{2}, b_{2} = a_{3} – a_{4}, …, b_{[n/2]} = (a_{n-1} – a_{n}) or (a_{n}) depending on whether n is even or odd. Then this position is equivalent to regular Nim with piles of b_{1}, b_{2}, …, b_{[n/2]}. Basically, we’re splitting the chocolate columns into pairs of adjacent columns and considering the differences between the members of each pair to be the piles of our regular game of Nim. Because the piles are in nonincreasing order, this is a well-defined transformation.

It works as follows: suppose the loser of the Nim-game (b_{1}, b_{2}, …, b_{[n/2]}) eats some squares from the kth column where k is odd. This decreases the value of a_{k}, thereby decreasing one of the Nim-piles as in a regular Nim-game, so the winner just makes the appropriate response. Instead, the loser might try to dodge by eating squares from the kth column where k is even, thus decreasing the value of a_{k} but increasing one of the Nim-piles rather than decreasing it, which can’t be done in regular Nim. But the winner can simply decrease a_{k-1} by the same amount and leave the loser in the same position as before. There are a finite number of squares, so the loser can’t keep doing this. Eventually they must go back to decreasing piles, and lose.

This is how far I got at RSI. I didn’t realize the poisoned lower-left square was a significant issue, but it is. Thankfully, all it really does (I think) is turn the game into misère Nim rather than normal Nim. We make the transformation to Nim-piles in the same way as before, and the winner uses nearly the same strategy as in the previous paragraph, but they modify it to ensure that the loser is eventually faced with “b_{k}”s which are all 0 or 1 with an odd number of 1s. (Maybe one day if I get around to writing a basic game theory post I’ll explain why this is possible. Or you can check Wikipedia. Or just think about it.) When the loser increases some b_{k}, the winner eats squares in the corresponding column to decrease it back to 0 or 1; when the loser decreases a 1 to a 0, the winner decreases another 1 to a 0.

Eventually, the loser is forced to hand the winner a chocolate bar consisting of pairs of adjacent equal columns. At this point the winner takes a single square from any column for which this is possible, leaving a bunch of 0s with a single 1—i.e. another misère 2^{nd}-player win. This continues until we run out of squares, at which point we conclude that the loser of the new game of misère Nim is indeed the player who consumes the poisoned square in the original game of Nim-Chomp.

Question I’m too lazy to think about right now: can we still do this or something like this if we poison not only the bottom left square of the chocolate bar, but some arbitrary section at the bottom left?

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I have yet to find a satisfactory closed-form expression for even . I obtain an ugly series in terms of Stirling numbers of the second kind. However, I suspect that is asymptotically linear:

Is this a well-known problem?

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So, first of all, Damien Jiang, Anirudha Balasubramanian, and I have each uploaded the papers resulting from our RSI projects to arXiv. I’ve been discussing the story of my project on representation theory and the mathematics around it on my personal blog (see in particular here and here). There are others from the program who have placed their papers on arXiv as well (but are not involved in this blog).

I’d like to congratulate my friend and fellow Rickoid Yale Fan for winning the Young Scientist award at the International Science and Engineering Fair for his project on quantum computation (which deservedly earned him the title “rock star”). I also congratulate his classmate and fellow rock star Kevin Ellis (who did not do RSI, but whom I know from STS) for winning the (again fully deserved) award for his work on parallel computation. There is a press release here.

RSI 2010 is starting in just a few more days. I’m not going to have any involvement in the program myself (other than potentially proofreading drafts of kids’ papers from several hundred miles away), nor do I know much about what kinds of projects (mathematical or otherwise) will be happening there. I think I’d be interested in being a mentor someday—maybe in six years time. I’m going to be doing a project probably on something geometric this summer, but it remains to be seen on what.

I don’t really know what’s going to become of this blog as we all now finish high school and enter college. It looks like most of us will be in Cambridge, MA next year; this is hardly surprising given the RSI program’s location there. Also, just to annoy Yale, I’m going to further spread the word that he is going to Harvard.

If anyone from RSI 2010 wants to join/revive this blog, feel free to send an email to deltaepsilons [at] gmail [dot] com.

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The first two judges I had were mathematics judges. The first one asked me what I would do if I were giving a talk about my project at a colloquium. He asked me to explain one of my results, which I initially did incorrectly (having not looked through the older proof in quite some time) but fixed along the way. He asked me how I had learned algebraic geometry (or, more precisely, that rather small subset I can claim to vaguely understand). Interestingly, he referred to a specific result in my paper by number (3.10; I didn’t remember what that was for sure)—one of the differences between Intel and ISEF is that the judges read the papers.

The second mathematician asked me to give an overview of my project in detail, so I went into my usual spiel. She asked me a few questions along the way about how the results were proved. Finally, she asked where I was going to go to college. I said that I didn’t know yet. This was a somewhat longer interview.

There were others who wanted a brief overview and then left. A computer scientist who had asked me earlier about certain algorithms and an engineer that asked about the law of atmospheres chatted with me about extensions of those problems.

The exhibits were then opened to the public. I met a few RSI 2009 alumni from the D.C. area. Most people were not mathematicians, which made explaining my project (on representation theory in complex rank) a somewhat difficult task, though there were some that knew, e.g. group theory. I wasn’t envious of my neighbor Joshua Pfeffer with mobs of people craning to hear about the super Kahler-Ricci flow though owing to me extreme hoarseness despite my consuming two bottles of fluids. Also, my parents stopped by to say hello and see the other projects.

I’m somewhat tired now, and there’s not that much more I really can say about it without going into technical details.

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I had my third judging interview today at about 11:40 am. The judging panel included a computer scientist who pointed to the seven wrapped chocolates on the desk and informed me that to each was assigned a number, and that I needed to discover which contained the median. I didn’t have the actual numbers, but I could compare any two. In the end, I said something about sorting algorithms (e.g. heapsort). He then asked me about counting paths from (0,0) to (m,n) where one can move either right or up on each move; I stated a recursive formula, but got the wrong closed form expression (it’s a binomial coefficient, and I said it was a power of 2). I was then asked by the other judge about what change I would make if I had to design the human body. I suggested eliminating cognitive biases and improving rationality but that wasn’t legal; I then suggested various ideas such as removing vestigial organs and improving our ability to type, but settled on increasing the efficiency with which energy can be extracted from food.

I then had lunch and went to the National Academy of Sciences, where we set up our projects. My poster had developed a slight tear from being sent through the mail, but I fixed it with construction paper.

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**(9:45)** So I’m at the 2010 Intel Science Talent Search in Washington, D.C. Presumably many of the folk that come across this blog will have heard of it; it’s a science competition for high school seniors. There are seven people from RSI here this year. I’ve also met many interesting people among the other finalists for the first time, all of whom seem to be rather beastly. Everybody arrived yesterday–I took the train–but nothing competitive actually happened. Today, we will be judged by a panel of ten or eleven scientists and mathematicians who are going to ask us general questions about science in general, and not our projects. My first interview is in about half an hour, so I’m basically procrastinating by writing this entry, if there was anything that I could do to prepare :-).

In any case, it was pretty cool to find that I’m in a room that has a TV in the bathroom. The hotel is ridiculously fancy.

After this, I’m going to go back to random Wikipedia surfing about diverse scientific topics. They told us this morning that the judges want to see the scientific process rather than technical knowledge–perhaps this is a license for me to babble? I’ve always enjoyed idle pontification. In any case, I promise more later after my judging interview.

**(11:31**) OK, phew. My judging interview (and recounts thereof to many fellow finalists) is now complete. I don’t know if I’m allowed to say who the judges were, but two were doctors and one a mathematician. The doctor went first and asked me to explain why when one flies (at 35000 ft) 80% of the atmosphere (by mass) is below you. Uhhh….I was confused. I’m not really all that solid on physics (took AP Physics B freshman year, haven’t thought about the subject since with any success). Anyway, I have to figure out the problem by Sunday, where he’s going to ask me it again. Next the mathematician said that she was going to ask me LOTS of questions about my project (hopefully a good thing). She then asked me to explain why representation theory was interesting to her two nonmathematical colleagues. So, I started babbling at full speed. I started with the remarks that “representation theory is how algebraic objects act on vector spaces” and mentioned how it reveals information about the objects themselves (mentioning Burnside’s theorem). She asked me to go back to what I said, and I said that the tools of linear algebra are much easier than studying complicated algebraic objects directly. I said that the classification of finite simple groups was something like 10000 pages in length. I then started to say that group representations were useful in chemistry and Lie groups and Lie algebras in theoretical physcis (but freely confessed my ignorance of said subjects). I said that the representation theory of Lie algebras was interesting itself, which is where I was cut off. I was posed some fairly simple combinatorial problem about The next person asked me about stem cells. I told her that I knew about stem cells but couldn’t tell her the difference between adult and embryonic stem cells. Anyway, I was next questioned about the ethical and scientific issues in transporting stem cells from her to me. I suggested that there were potential incompatibilities (like blood types). Our time soon ended and she concluded by saying that I was on the right track. Probably not horrible, for a math person at least.

I then had my picture taken. All the finalists have to do this today. My next interview’s after lunch.

(**3:11) **I just got back from my second judging interview. The interviewers included a distinguished astronomer, who went first; she opened with a question about what latitude and longitude was. I said that they were a system of coordinates on a sphere minus the two poles. I said something about their being an injective immersion, which may or may not have piqued their interest (none of them was a mathematician). She then asked why astronomers would use them, to which i replied something about “objects being spherical.” Next, she asked me to tell me what I knew about the universe—which has to be by far the broadest question I have received all day. I babbled mildly about the four main forces and that they could be described by mathematical laws, sand omething about the age of the universe and its expansion. This is all very rudimentary, but I don’t know anything about astronomy. The next interviewer asked me why France would launch satellites from New Guinea; I suggested it was closer to the equator, but didn’t realize that the explanation for that was simply that the earth’s rotation provided additional speed. (Whoops.) I was then asked how one might determine whether switchgrass was an efficient fuel (including environmental concerns); I suggested that would one have to compute the total expenditure (i.e. of the land deforested, inputs of irrigation, fertilizer, fermentation, etc.) and compare the obtained energy per unit cost to the analog for, say, coal. My questioner pointed out that I should have determined the energy inputs of the tractor and taken them into accouont. Finally, he asked me the following math problem that I embarrassingly failed at: given 248 arm-wrestlers, how many matches would it take to determine the best one? I said something like given that an elimination process was optimal. This is wildly wrong, of course, because there are roughly **levels, **each of which (except the last) contains omre than 1 match. So he said there was a trivial solution and asked me to work out how many matches would be necessary. After a short computation I arrived at…247. Which is utterly trivial, because there must be one loser per match, and 247 losers. The judges laughed and said they had enjoyed the time; I am not sure whether that was necessarily a good thing.

(**9:24) **I’ve now had my third judging interview and a very enjoyable dinner. The first person at the third judging interview started by immediately asking me the probability that a function where has cardinality has no fixed points. The answer is (rather trivially) , but for some reason it didn’t occur to me over several minutes as I babbled about inclusion-exclusion and inductive proofs. Next the same judge asked me the same for continuous, which is an easy application of the intermediate value theorem; it’s also a special (and relatively easy) proof of the Brouwer fixed point theorem. He asked me whether the same was true for the circle; I said no, a rotation would do the trick. He asked whether it was true for (unit disk in ); I replied that it was the Brouwer theorem, which he then asked me to prove (in that case). I gave the usual proof using singular homology (which he said was allowed); I actually sort of half-alternated between talking about the fundamental group and that. (If you haven’t seen it, cf. any book on algebraic topology.)

The next judge then told me that the chair he pointed to in the back of the room was magical and could take me 1 million years in the past. I could take three things wit h me, and I had to decide. I said that I’d take a weapon to defend myself, fertilizer to help grow food, and textbooks to entertain myself (I really couldn’t resist the last part).

The third judge in the panel asked me about gases. He inquired about what would happen if a box of gas had its volume changed (temperature kept constant); I remembered enough basic chemistry from sophomore year to explain the ideal gas law. He asked me about the deviations from the ideal gas law. I said that they were caused by van der Waals forces and other intermolecular attractions, though I couldn’t remember enough to say it well. The 15 minutes were up, but they said it was a good interview. Dropping algebraic topology references was probably a good idea, though the brain-freeze on a baby problem was embarrassing.

I next went to dinner, which included the other 39 finalists, the judges, various SSP people, and a couple of Intel alumni who had distinguished themselves. One of them was Roger Tsien, who said that his obsession with producing pretty colors in chemistry experiments was part of his motivation to study fluorescent proteins. Tsien gave an amusing recount of his life story and major discoveries. The other was an aerospace engineer from MIT who told us that each of the finalists get an asteroid named after him or her.

I’ve got one more (nonmathematical) interview tomorrow. On Sunday, we’re actually going to talk to the judges about our projects.

(**10:38) **A conversation with David Liu just made me realize that I omitted a part of the third judging interview. In particular, a doctor asked me to name a letter from A-Z (which, upon asking, was inclusive of the endpoints). I picked “A” and was asked what the adaptations of aardvarks are; I said something about their fur and warm-bloodedness.

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In an unrelated note, if you haven’t already seen it, you ought to watch the MAA’s great debate. And that’s regardless what you think of the holiday “Pi Day”–I’m mostly in agreement with what John Armstrong has to say on this subject. Also, cf. this MO thread for some good alternatives to memorizing digits.

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Since we have been too lazy to post lately (and the so-not-lazy Akhil posts mostly elsewhere now), I’m going to post some problems that I probably should be able to solve, but haven’t.

First, an interesting (and apparently classic) problem given to me by Tim Chu, USA IMO alternate:

Find all such that f is multiplicative (for relatively prime positive integers), monotonically increasing, and f(2) = 2.

A second from an old IMO longlist:

Show that the union of all segments with endpoints in includes every point in .

EDIT: Well apparently this is wrong. However, part b of the problem asks if the convex hull of is equal to .

And a third from Russia 2005:

A quadrilateral ABCD without parallel sides is circumscribed around a circle with center O. Show that O is the barycenter of ABCD iff (OA)(OC) = (OB)(OD).

(Looks quite vector/complex-number friendly, but… I probably didn’t try hard enough.)

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The book *Real and Complex Analysis*, by Christopher Apelian and Steve Surace, was recently released.

It’s mainly for an introductory upper-level undergraduate course in real and complex analysis, especially at small liberal-arts colleges. In this post, I’ll describe this book and how I was involved in its production.

At the start of my freshman year, my analysis teacher, Professor Surace, asked me to check over the drafts of a book he and his colleague (and my former teacher) Prof. Apelian were working on. It was the textbook for the course. At the time, if I remember correctly, there were six chapters: on the real and complex spaces, basic topology, limits, continuity, convergence of functions, and derivatives. The complex analysis part of the book was in its infancy (e.g., there was only a rudimentary outline of one chapter, which had been written some time back and was typeset in Word—they wrote it well before before they had switched to LaTeX).

Anyway, since one of my hobbies that year was playing around with LaTeX and trying to figure out all the cool formatting tricks used in books, and since they hadn’t yet done much to change the (somewhat bland) default book style, I pointed out a few tweaks that, in my opinion, would make it look better. They agreed, and I ended up being put in charge of the layout. It turned out that I would need to learn a lot more about LaTeX though (or at least, learn to look up things a lot more). As you can see from the sample pages, the authors—or to be precise, one of them :-)—had fairly detailed ideas of how the book should appear. I don’t know if I learned LaTeX properly, but I sure learned a lot of hacks.

I also ended up sketching the figures, which enabled me to pick up the useful (and definitely nontrivial) skill of using Adobe Illustrator. In fact, I’ve used it in making some of these blog posts. (Though I would recommend interested passers-by also to consider, say, Tikz depending on your aims; I’m not familiar with it, but it has the benefit of being free.)

I don’t know all the details of how the book got started–I think the idea goes back many years, and certainly before I started helping them. The authors, who were faculty members at Drew University, a small liberal arts college, found it inconvenient to have separate courses for real and complex analysis. So they taught them together, i.e. covered topics such as continuity and convergence for real and complex functions simultaneously. They couldn’t find a textbook to use though, so they decided to write one.

I had noticed that a lot of basic complex analysis textbooks (e.g. Ahlfors, Hille) covered much logically redundant material. Then again, at least when I first got interested in analysis, it was Cauchy’s theorem and the residue theorem—not rigorizing integration—that piqued my curiosity. So I first learnt about compactness and continuity from a book on complex analysis—if I remember correctly, an old volume by Herb Silverman that I found in a public library.

(One could of course ask—why not just talk about metric spaces, or topological spaces more generally? But the curriculum at Drew University does not typically cover that, and this is an introductory textbook.)

I also learned that while having two authors definitely does not halve the time for completion, it can improve the quality.

The book also has a website, here. Soon I am going to write up partial solutions to the exercises to be posted there. It’s now been more than three years that I’ve been helping out, and as fun as it’s been, I’m glad to see the book in print!

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The Hahn-Banach theorem is a basic result in functional analysis, which simply states that one can extend a linear function from a subspace while preserving certain bounds, but whose applications are quite manifold.

**Edit (12/5): **This material doesn’t look so great on WordPress. So, here’s the PDF version. Note that the figure is omitted in the file.

**The Hahn-Banach theorem**** **

**Theorem 1 (Hahn-Banach)** *Let be a vector space, ** a positive homogeneous (i.e. ) and sublinear (i.e. ) function. *

*Suppose is a subspace and is a linear function with for all . *

*Then there is an extension of to a functional with *

I’ll omit the proof; I want to discuss why it is so interesting. One of its applications lies in questions of the form “are elements of this form dense in the space”? The reason is that if is a normed linear space and a **closed** subspace, the quotient vector space is a norm with the norm (The closedness condition is necessary because otherwise there might be nonzero elements of with zero norm.) So:

* *

Corollary 2Let be a normed linear space, a subspace. Then if and only if:Every bounded linear functional vanishing on vanishes on .

One implication is immediate; linear functionals are continuous. For the other, note that the condition is equivalent to “any bounded linear functional on satisfies , where denotes the image of .”In particular, we are reduced to showing that if is a normed linear space is nonzero, there is a bounded linear functional that does not annihilate . To do this, we apply the Hahn-Banach theorem to and the subspace , with as some multiple of the norm. Choose any nonzero linear functional on this subspace, and then extend to all of .

**An approximation result for powers of **A spectacular application of this is the following:

Theorem 3 (Muntz)If is a sequence of positive real numbers with , then the linear combinations of the functions are dense in , the space of real continuous functions on the unit interval with the sup norm.

The proof starts by employing the above corollary: for a sequence with the powers not dense in , there is a nonzero continouus functional with The elegant idea in this proof is to consider the *complex function* defined in the right half-plane . This is an analytic function in , because When , the function tends uniformly in (in the unit interval) to .These are the basic properties of :

- . Indeed, is a bounded functional and the function has norm 1.
- . Indeed, otherwise for , implying vanishes on polynomials. But the Weierstrass theorem implies these are dense in , so , contradiction.

There are general theorems characterizing the roots of a bounded analytic function in some disk (in some, boundedness can be weakened to a condition). Since the right half-plane is conformally equivalent to the disk via the transformation we can consider the bounded analytic function on the unit disk, which has roots at Now I quote the following theorem:

Theorem 4Let be a bounded analytic function. If the zeros of in are , then

This can be proved using Blaschke products; see Rudin’s *Real and Complex Analysis*.So for our function , we find which shows that , and which proves the theorem.It’s in fact true that the condition is necessary, and the can be made complex. Cf. Rudin.

**Geometric applications **Let be a closed convex plane region, and :Then there exists a line through such that lies on one side of —i.e., a **support line**. This can be proven using a bit of elementary geometry, but it also follows from fancier arguments as well.Basically, the assertion of the theorem can be rephrased as follows: the line corresponds to a linear functional , such that and .

Theorem 5Let , be disjoint convex sets in a topological vector space such that contains an interior point. Then there is a nonzero continuous linear functional with

The geometric fact above follows from taking , .For a convex set , we define the **Minkowski functional** Then:

- is positive homogeneous—this is evident.
- . Indeed, if , then by convexity. So , which proves subadditivity.

If has an interior point at zero, is always finite.Now consider . This is a convex set, and I will show that there is a continuous functional with , which will prove the theorem.Choose and consider the set ; if is chosen appropriately, this will contain a neighborhood of . Now (since otherwise and ). There is a linear functional with bounded by (this is seen for with by positive homogeneity; for with , is negative), which we can extend to all of with the same bound. Then where I have abused notation by writing for when every element of is greater than or equal to every element of .

Corollary 6If are closed convex subsets of a locally convex space , we can get a nonzero continuous as above with even if .

Indeed, we can find a small convex neighborhood of the origin such that are disjoint. Then apply the above result to and note that .

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