The Life of a Mathematician

My sister has a card game called Set. Each card has a picture on it, which can be uniquely described in terms of four properties. Each property has three possible values. The deck has exactly one card with each combination of property values, giving a total of eighty-one cards. To play, you deal twelve cards face up, and then everyone looks for "Sets". A Set (capitalized to distinguish it from the math term) consists of a (lowercase) set of three cards which are either all the same or all different with regard to each of the four properties.
When someone identifies a Set, they remove those cards from the table and three more are dealt. If there are no Sets available, then you deal an extra three.

I had played this game once before, and thought I was competent, but no one comes even close to competing with my sister. She usually sees Sets before I have a chance to start thinking. Anyway, that's not what I'm posting about.

The instructions made claims about the probabilities of not having a Set when there are twelve or fifteen cards on the table. Over the past few days, I've tried to verify these figures mentally and have had much more difficulty than I anticipated in determining how many sets of a given size can be constructed which contain (or do not contain) a Set. I think I'm going to be obliged to work this out on paper.

A couple of interesting ideas:

For any given pair of cards, there is exactly one card which can be added to form a Set. Any given card is therefore a member of exactly forty Sets. This gives rise to an equivalence relation on pairs of cards, where two pairs are equivalent if they require the same card to make a Set. There are (81 choose 2) = 81*80/2 = 81*40 pairs of cards, which are thereby divided into 81 equivalence classes with 40 elements each.

I went to my sister's basketball game tonight. It was a very close game, although her team was slightly ahead for most of it. The score stayed tied for several minutes in the last quarter, and the game went into overtime. Each team scored one freethrow point, and it stayed tied again for a while. Her team finally won by two points, at least one of which was also from a freethrow.

I'm posting this from my mom's computer, since I'm visiting my family for a couple of weeks. Tomorrow my sister and I are planning to go finish our Christmas shopping together. Things are pretty much the same as I remember them here. The family cat is still the undisputed queen of the house. She's changed her habits slightly and will sometimes sit on someone's lap for as long as two minutes.

I'm just getting over a cold that I've had for most of a week. This prevented me from playing hockey, or at least that was my excuse, but maybe I was just being a wimp. After all, my roommate got hit in the lip with a rubber ball we were using for a puck, had to get five stitches in it, and went back and played again the next night. The nurse at the hospital was from Canada, and was disappointed to learn that we played without skates.

We still haven't decided when exactly we should open presents. Christmas morning has the disadvantage of being a Sunday morning, and we are supposed to go to our relatives' house (a ten minute drive) after church. I suggested either Saturday or Monday, so as not to be rushed either morning, but so far a concensus has not been reached.

Meanwhile, I'm looking at a movie poster for Holes which is attached to the side of my sister's locker (yes, she has a locker) so I think I'll end with a deep thought (pun intended): Hector Zeroni's nickname is "Zero", and the numeral 0 is shaped like a hole.

I've made some alterations to the left sidebar, over there to the left.

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Most notable are the blog links. If your blog is not on there, it's because I have to leave for work in a couple of minutes, and I just put up the blogs with addresses that I have memorized. The reason I have them memorized is because I don't get to them via links on some other page. I plan to expand the blog list later.

We played hockey last night. We do this at an outdoor skating rink, but just with tennis shoes, not skates. There were nearly thirty people by the time everyone showed up. After a while everyone not in our group had either left or been assimilated by us, so we played lengthwise and used the whole rink, since the teams were so big. I find it isn't that hard to keep my balance on the ice; what's hard is moving with so little traction. It takes a lot of effort to accelerate, and having accelerated, it is almost impossible to turn.

Work was interesting today. The computer that is normally used by the person with the drivethrough headset has officially died, which makes doing the drivethrough considerably more challenging. See, it normally works like this: A customer gives their order to whoever has the headset, and that person enters the order into that computer. It simultaneously appears on three other screens: One for the drivethrough window cashier, who is preparing napkins, sporks, and such, another for the people on the food line, who are making the food, and a third for the customer, so they can doublecheck that it is right. When the headset person hits the "Store" button, that order gets sent to the other computer, which is attached to the drivethrough cash register. It then prints a receipt, and brings that order up on its screen when all the orders before it have been paid for and handed out.

Okay, so with the first computer out of commission, it is necessary to ring up the orders on the second computer. This has always been possible, because people sometimes add items to their order after they pull up to the window. If just one person is working over there, there's not much difference. The only time it is not possible to enter items to the new order is when you are taking cash for the one currently at the window. This is the case even when both computers are working, because it's not possible to stand at the window and at the other computer at the same time. One person can do both jobs, although things move a bit slower because almost every customer gets put on hold at least for a few seconds when it's busy.

The trouble is that when you introduce a second person, the computer is still tied up while cash is being taken. So the only way to speed things up is for the order taker (which was me this morning) to write down the orders on paper, so that the cashier can ring them in when the register is free. I got a fresh roll of register tape, and basically produced handwritten receipts for about two hours. These got replaced with printed receipts once the orders were entered into the computer. The system got more efficient once I figured out what order number I should be on and started writing the numbers at the top.

The other difficulty is that until the orders get entered into the computer, they don't appear on any of the screens, and with that computer they never appear on the customer's screen. This meant that I had to keep repeating orders back to the food line people so that they could make the correct items, as well as repeat each order back to the customer since they weren't able to look at it themselves.

It actually was kind of fun to do it this way, if a bit more stressful for everyone. It was hard for me to get all the drinks made since I had to be standing with my back to the dispenser while writing the manual receipts. I was rather pleased when the delivery truck arrived, and I had to give someone else the headset so I could go put everything away.

Okay, I'm not in this semester's Modern Geometry class, but I have become mildly obsessed with one of their homework problems. The statement is simple enough: Given that the circumcenter of a triangle is the same as the Gergonne point, prove that the triangle is equilateral. The circumcenter is the center of the circumcircle, which is the circle that circumscribes the triangle, i.e., intersects all three vertices. It is also the intersection point of the perpendicular bisectors of the sides. The Gergonne point is defined as the point of intersection of the lines connecting the vertices with the points of tangency for the incircle, where the incircle is the circle inscribed in the triangle. The approach I've been trying is to show that when the Gergonne point and the circumcenter are the same, they must also be the same as the incenter, which is the center of the incircle. But I haven't had success yet. Apparantly no one in the class, including the teacher, has been able to solve this.

Today the 2000th consecutive Schlock Mercenary comic strip was posted. I've been reading this comic ever since my friend Matt convinced me to read the archives, which took me about a semester to get through. It has become my second favorite comic that's still active, right after Foxtrot. It would probably be my favorite if it had more math jokes. Furthermore, the cartoonist has promised to post a list of sites mentioning this historic event. So, by making this blog post and emailing him about it, I should be able to get my blog in his list of links.

I had my oral exam today, and I passed. So I now have a master's degree. It will be nice to be able to put that on job applications. They didn't ask me anything about Numerical Analysis, or Partial Differential Equations. Not too surprising, I guess. Those were pretty much applied classes and I picked theory people to be on the committee. I was glad that they did ask about Representation Theory. That was a fun class (in retrospect) and I enjoyed reviewing the material.

Most of the test focused on the three courses that I took full year sequences in; namely, Real Analysis, Topology, and Modern Algebra. Most of the topology questions were on point-set topology, because Dr. Stowe didn't realize that we did a lot of algebraic topology second semester. As a result, he asked me about fundamental groups but not about homology groups, since he was just asking stuff off the top of his head once he realized that algebraic topology had been covered. The most involved proof they asked me to do was in point-set topology; namely, to explain why any infinite sequence of points in a compact metric space must have a cluster point. It's not a complicated proof, given some preliminary stuff that we had already talked about, but given the fact that I didn't actually look at this proof while studying, it was rather involved to come up with extemporaneously. Then he asked me the difference between a cluster point and a limit point, which was ironic because the whole time we were talking about cluster points, I was trying to think how the two concepts were different, and hadn't figured it out. After thinking about just that for a few moments, I realized that I was reading a condition into the definition of cluster point that wasn't there. Hmm, I just realized that I failed to answer one question that he asked in that regard. I answered a related question, and then he said something like, "That's close enough; let's move on." Huh. I still don't know the answer. What he actually asked was whether something like the Bolzano-Weirstrauss Theorem applies to an arbitrary compact metric space. He asked it in such a way that implied the answer is no. I'll have to think about that one.

When we got to Modern Algebra, Dr. Hill pointed out that all four questions I skipped on the written exam were about Field Theory, and then proceeded to ask all Field Theory questions. That part of the exam didn't go so well. At least she didn't ask about Galois Theory.

After it was all over, the outside committee member, who was from the business department, commented that the only thing she understood the entire time was the term "function of x".