So what is wrong with NYCIML? For a large portion of the math team, absolutely nothing. However, for the people who actually have the potential to make our A team, there is a lot wrong with NYCIML. The biggest is that it is completely unrepresentative of the contests that they will have to face if they get selected for the team. For example, freshman year I qualified to be one of the members of TJ A at PUMaC. I got a 0 on number theory, what I considered at the time to be my best subject. That's right. I didn't get a single number theory question correct.
What went wrong? Well, it was a combination of PUMaC being crazy hard and a severe lack of preparation for that kind of math. To put it simply and bluntly, the contests that we do simply do not prepare people for contests, and they furthermore test for something completely different than what PUMaC and HMMT test for.
After three years of experience at PUMaC and HMMT, I think I have a good idea of what characteristics makes one successful at each of them. PUMaC rewards knowledge and an ability to work through standard computations, regardless of how ugly they get. Many PUMaC problems are not theoretically difficult, but finding the actual numerical answer is the bulk of the work. HMMT, on the other hand, focuses much more on problem solving. Most of their problems require little prior knowledge; almost everythign necessary can be derived on the spot. Some exceptions exist: the fact that
has been used, for example. But by and large, HMMT is a contest based around how well you can handle things you've never seen before.So what is NYCIML? NYCIML is just a hard math test. Practically everything can be seen instantly. There is no creativity involved. In fact, the last NYCIML this year essentially repeated two of the problems from the previous one. Contests that do that are generally bad. They become formulaic; anybody who has seen enough problems will do well, regardless of how good they are at math. As a result, for the people competing for the top spots on our math team, NYCIML is not about solving problems. It's about eliminating stupid mistakes. While I am all for stupid mistakes being considered in rankings, it should not be the determining factor as it is here.
This is why I wrote more than 50 problems over the summer. I wanted to create contests that would prepare the team for what they will face at PUMaC and HMMT. So far, the team has not met my expectations, but I believe that they will grow now that they have been faced with the challenge and the knowledge that there are three more to come. The biggest difference between the contest that I wrote and the NYCIMLs is the average. Out of 6 points, the NYCIML top 15 averages have been around 5. Out of 19, the top 15 average on my contest was a little over 6. What does this mean? It means that you have more than one route to victory. The first is to do the same thing as NYCIML: do the problems you know how to do and do them right. The problem is that I ramped the difficulty up quickly, so many of these people probably got a 4, or possibly a 2 because they still missed one of the first two problems (the contest was six problems weighted 2,2,3,3,4,5 respectively). The other method to victory is to be more ambitious, solving more problems, and then maybe you make a mistake or two that drops your score from 11 to 8. You'll still be ahead of most of the people. However, making a single mistake on a NYCIML costs you dearly, even if you know how to do all the problems.
At HMMT, the top score is not 50. If you know how to do all the problems, you will win, even if you make stupid mistakes. I speak from experience: On the 2009 Calculus test, I solved all 10 problems. I made some stupid errors on 9 and 10, dropping my score down to 35. As a result, I tied with Kee Young for first, but lost the tiebreaker. Regardless, the fact that I could make two mistakes and still tie for first demonstrates how different HMMT and NYCIML are. The two mistakes probably cost me the overall first place, but I was second, not eighth. One mistake is much more acceptable at HMMT than on NYCIML.
So why do we choose our team based mostly on a contest where problem solving is minimal? And why do we pass it off as good practice? It's good practice for most of the team, but not A team. But A team is the team that represents us at competitions, so it seems weird that we train the rest of the team more. Hopefully these performance contests will help jolt people into realization that being good at NYCIML does not mean that they will do well at every contest and will continue to work to improve in order to conquer the challenges I have given them.
The performance contest (to be done in 35 minutes):
And as an amusing sidenote:
time, whereas an algorithm that uses geometric properties of spaces can compute homology groups in 
time. This is a huge gap, although the big issue is going to be how to handle torsion, since the paper that computes homology groups in 
, which implies that there is no torsion (so the homology groups are all isomorphic to 
for some k). Well let's look at why Smith Normal Form is hard to compute. To do that, I should explain what it is.
). Then we can find invertible square matrices P and Q with entries in the PID such that PAQ is diagonal and 
if i < j. This matrix need not be square, 
for varying i. Here is how I think you should think about it:
. Each column represents one generator, and each row represents one relation.
and 
to get 
and 
we want to use 
. Q lets us do that.
. However, this may cause the elements on the diagonal to get larger. This is probably not obvious to you, but when I explain the algorithm for computing the normal form it should become more clear. The divisibility condition is also one of the weirder things, but it is essentially just removing all the ambiguities caused by the Chinese Remainder Theorem: 
. Smith normal form uses the divisibility condition to give the latter, although in the form 
.
and 
.
. We then do row and column operations to eliminate entries: subtracting two times the first row from the second row gives 
. Then subtracting twice the first column from the second column and one times the first column from the third column, we get 
, which is our Smith normal form, and we now know that our module can be better written as 
. The generators corresponding to each of these three components can be obtained from the column operations that we did. In this case, the generating set is 
, and our relations become 
and 
.
. To make it simpler, let's pretend it just looks like 
. It might take a while, but you should be able to see that it doesn't matter what the dimensions of the stretched version is, as long as we multiply by square matrices on either side. So now say we want to eliminate c. How can we do that? Well, what we do is we write 
where g is the greatest common divisor of a and c. Then 
, so we know that 
is invertible because the determinant is 1. Furthermore, when we multiply we get 
, and we have eliminated c, just as we wanted. The problem is: this might have undone our elimination of some of the columns! Well, never fear! We can alternate between eliminating the rows and eliminating the columns, leading to a sequence of top left elements where each divides the previous. Therefore at each step the top left element gets smaller, meaning it terminates eventually. (For general PIDs, we create an increasing sequence of ideals 
and then use the fact that our ring is a PID to show that it is Noetherian, and therefore the sequence becomes constant eventually). So once both the first row and first column are eliminated, we ignore the element that's now in the top left and proceed to diagonalize the rest.