Sections 2.9-2.11, due on September 17

Difficult: Wow! This was by far the most difficult section so far. If I had to pick a most difficult part, I suppose it would be section 2.11, about Linear Feedback Shift Register Sequences. I think the main problem started because I do not understand the equivalence describing the first relation. It seems to say that the n+5th term is given by adding the nth term to the n+2nd term, but then how do you find the nth term? The nth term would have to be the n+5th term minus the n+2nd term, but you haven't gotten those yet, because you are trying to get the nth term! Possibly this is the misunderstanding that carries over so that I do not understand the know plaintext attack.

Reflective: Earlier, I mentioned that I did not see how anyone could make a code that was "unbreakable". Now that the one time pad system has been explained, and I know a bit more, I can see how this would work. It seems that would be a great idea in general, but the great foil to the method seems to be the necessity of the courier to transport the one time use key. That is expensive (because you have to pay the guy) and dangerous (he could be intercepted). Other than that, it doesn't seem like it would be too hard to create a one time pad cipher. It seems to me that even with the weakness of random number generators and LFSR sequences, if you had one of these generate a key that was as long as your message and then used, say, a Vigenere style cypher, you would still not have enough patterns to mount an attack on your cyphertext.