We did not have a biology lecture this week. With regards to my math project, I organized my thoughts and came to the following claims: Let Tn be the triangular number corresponding to triangle with base n and let (n+1) be the number of nodes on our complete graph.
I proved the second claim. Using induction, I showed that 1 + 1/3 + 1/6 + ... + 1/Tn = 2/(1 - (-1/n)). What remains to be proven is the intermediate step, in that the geometric series gives the same "k" value as the DSD program. Parts of this seems intuitive as I developed the claim by calculating random walks by hand which is a part of the program, but the actual "a-ha" moment is yet to be found. Lenore suggested that I explain my work to the other students, so that I could get other inputs to help figure out a solution. In the meantime, she told me to start looking at paths (in the sense of a straight line or chain of nodes). With regards to this, I found another interesting pattern among endpoint DSD distances for length random walks on a graph of four nodes. The graphs looks like this: o - o - o - o. On a graph of four nodes, each time a random walk happens, two of the nodes are assigned weights and two have weight 0. Note: for all of my work, the random walks begin with initial node having weight 1. As the length of the random walk increases, the DSD distances between the first and fourth nodes are as follows: 2, 4, 4, 5, 4.5, 5.25, 4.625, ... . This pattern is alternating in the sense that the distance increases, then decreases, then increases, and so on and so forth. I found that the amount it increases or decreases by is equal to 2|x - y| where x and y correspond to the weights at the nodes (that are not 0). This will be described further in depth later on. Additionally, for our group meeting on Thursday, we heard a mini talk given by another student in our group. He briefly described what he was working which was function prediction. I was sick this week and had a slight fever, so I did not go out this weekend. |
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