Hmmmm

We did a class problem in lecture last Friday. The problem was to predict the ups and downs of paper after folding it from right to left a number of times. I thought about it but I'm not sure, but I tried. I will write about my guess at a solution.

First I thought that the middle crease is always down. My partner was writing out the ups and downs after folding it 1, 2, 3, 4, 5 times. He tried to find a pattern, and I tried to help, but when I thought about it, it was hard. But I noticed that there are always an odd number of creases.

Then professsor Heap said to think recursively and symmetrically. I'm still kind of confused about recursive, but I know what symmetric means. I thought I should define my functions and domain. My high school math teacher said that's a good place to start when you don't know what to do. So I stated my knowns and unknowns.

P is a function. P(m, n) is 1 when the nth crease from the left after m folds is up, and -1 when the nth crease from the left after m folds is down. m is the number of folds, starting from 1. n is the index of a crease, in [1, 2^m)

2^(m - 1) is the middle crease. Let's say k = 2^(m - 1) because it's easier. I think P(m, k) = -1. The first fold makes the middle crease which is always down.

Then I thought, for n in [1, k), we are looking at the creases on the left side. The left side is always flat on the table, so it is like folding a new piece of paper. So P(m, n) = P(m - 1, n).

Then I thought about the other side. For n in (k, 2^m), we are looking at the creases on the right side. When folded, the right side is like the left side, but upside down. While it is upside down, the right side has the same folds as the left side. But when you flip it back, all the ups become downs, and all the downs become ups. The creases on the left side become the creases on the right side. So P(m, n) = -P(m, 2^m - n)

I tried to write a python program to try it because sometimes professor Heap does python in class. But I kept getting a typeerror. But I wanted to write about my thoughts because I thought it was pretty cool. But I might be wrong. Together, I have

P(m, n) =
-1 if n = 2^(m - 1)
P(m - 1, n) if n in [1, 2^(m - 1))
-P(m, 2^m - n) if n in (2^(m - 1), 2^m)

Trying some simple values seems to work. P(1, 1) = -1. P(2, 1) = -1. P(2, 2) = -1. P(2, 3) = 1. P(3, 1) = -1, P(3, 2) = -1, P(3, 3) = 1, P(3, 4) = -1, P(3, 5) = -1, P(3, 6) = 1, P(3, 7) = 1.

Hmmm

Tutorial was interesting this week. The first interesting thing was question 2. The statement is "every course has a prerequisite." Someone said ∃x ϵ C, ∀ y ϵ C, P(x, y). Someone said that's wrong, and it should be the other way around (∀ y ϵ C, ∃x ϵ C, P(x, y)), and most people agreed. Then someone said that it's still true, and people said "wait, it does work." Because the first expression means that there exists one course that is a prerequisite for every course, and that means that every course has a prerequisite. But the two are not equivalent. x ϵ C, ∀ y ϵ C, P(x, y) ⇒ every course has a prerequisite, but every course has a prerequisite does not imply that there exists a course that is a prerequisite of every course. The second interesting thing was the last question in part 1. "Some courses have the same prerequisites." The first answer was ∃x, y, z ϵ C, x ≠ y, P(z, x) ∧ P(z, y). But this only means that some courses have a common prerequisite. ∃x, y ϵ C, x ≠ y, ∀ z ϵ C, P(z, x) ⇔ P(z, y) means some courses have the exact same prerequisites. The third interesting thing was that someone said in English, "only if" is the same as "if and only if". People said not to use English meaning in logic, but he continued. The English meaning is actually correct too. He said if I say "I will quit only if you report me to the media." He said that if I quit, you must have reported me to the media (P ⇒ Q). It's easy to think that if you report me to the media, I will quit (Q ⇒ P), because being reported to the media is scary. But that doesn't have to be true. But he insisted that was the case. Eventually someone said there will be rainbows only if it rains. There can't be rainbows if it doesn't rain, but if it rains, there don't have to be rainbows. Class discussion in tutorials brings up interesting things.

Hmmmm

Something new I learned in this class is that a statement about a set of elements is true if there are no counterexamples. Any claim about the empty set is true because there are no elements in the set that disprove it. We have to be careful in every day usage, because sometimes we do not know of counterexamples, but that doesn't mean there does not exist a counterexample. I might say "For all elements in the set of living creatures in the universe, none of them are unicorns." Most likely, other people won't have a counterexample to this claim, because there is no reputable evidence of at least one of the living creatures in the universe being a unicorn. However, there could be living creatures in the universe we don't know about, so there might be counterexamples to the claim in the set of all living creatures in the universe.

I feel confident about the material covered this week. Implication, the converse, and contrapositive were covered in Mat188 - Linear algebra. Burbulla was the lecturer. He was cool. The tutorial was good. We reviewed the homework questions. The quiz was interesting. It was like the tutorial preparation questions, but in the quiz question, we don't know if the sets are non empty. In the homework, we know there are three test programs. If T - P must be empty, then T ∩ P must be occupied, because there are three test programs. However, in the quiz, P - L does not imply that P ∩ L is occupied. P may be an empty set, and any claim about the elements of an empty set are true.