Absolutely fascinating that a sequence can be represented as a permutation graph, which means that the problem of finding the longest subsequence morphs into the problem of finding the largest clique!

Thus, essentially, the question which I was trying to solve – prove that a sequence from 1…101, will contain a increasing or decreasing sub-seq of length 10, morphs into a problem of proving that the ramsey number is 10.

1. Define new arrays. Length & pointer – length contains the length of the longest subsequence starting with a[i], and pointer points to the next element of that particular subsequence.
2. (for i = lastElementIndex -1 ; i >= 0 ; i–)

1. If (i is last element index)

i. Length[i] = 1, pointer[i] = null

2. Else

i. Go through all elements from j = i to j = lastElement, appropriately setting pointer & length – i.e. if a[i] > a[j], then clearly no sequence possible. Else, just choose the one with the longest length.

Constructing the sequence from the length/pointer arrays is a straightforward task then.

An intuitive feeler to Algebraic Geometry: http://rigtriv.wordpress.com/ag-from-the-beginning/

A nice introduction to what Algebraic Topology actually does: http://blog.mikael.johanssons.org/archive/2006/01/introduction-to-algebraic-topology-and-related-topics-i/

This is something that I never really understood well until actually working out copious amounts of basic topology.

1st approach was to try out a recursive algorithm, starting with {a_n-1,a_n} in {a_1…a_n}, and build up the sequence from there.

However, this doesn’t always work. Take for example if a_n-1 = a high number forwhen we want a monotonically decreasing subsequence.

Then that number ends up blocking out other numbers that come later but could be part of the subsequence.

Take for example the sequens {8,6,12,4,2}. Here, the algorithm yields {12,4,2}, which is clearly suboptimal.

Proofs use a tedious amount of notation manipulation, but thinking of it in terms of indicator functions is interesting.

What’s required is focus.

So what I plan to do for 2010 is focus exclusively on one topic – probability. I shall work on 3 aspects of probability theory.

1. Random walks/stochastic processes (non-measure theoretic).

2. Measure & Probability

3. Applications of Probability – Statistical learning & Information theory

Things have been a little busy with work so I haven’t had as much time as I would like for math. I’ve been pottering about here and there – mostly around finite fields and trying to prove , but didn’t get anywhere I’m afraid.

I came pretty close though – at least by the proof in Apostol’s book. However, failed to make one simple but crucial connection.

Anyways, I started reading this book – Fearless Symmetry – and thought this could be  good point to start learning a little bit about quadratic reciprocity and it’s connections to Galois theory. So the first problem that comes out is ‘What are the solutions for .

One small conclusion that immediately comes about is that for any solution , is also a solution.

.

This is very easily proved by squaring both sides, and cancelling the common terms.

One gets the following on doing this:

.

For every pair of indices i & j, we can take the two same terms from the RHS and put them onto the LHS, creating a sequence of sum of squares whose generic term is:

.

The sum of squares of the above form for every i,j is clearly positive.

Hence the inequality is true.

If , then for some .

Now to evaluate , we know that for some .

Thus, we have .

Thus, we have . This implies that , for some m.  Thus, we have . Therefore, since the GCD has to be the maximal divisor, , otherwise, a new maximal divisor could be created either for (a,b) or (ka,kb).

I’ve posted my first question here: http://mathoverflow.net/questions/840/the-core-question-of-topology