Polya problem solving

Question (From week 10 lecture):

If L1 is the set of binary strings with an even number of zeroes, and L2 is

the set of binary strings with an odd number of 1s, can you express L1 /\ L2 as a regex?

1) Understand the problem

Data: L1 is defined, L2 is defined, L1 /\ L2 is defined.

Unkown: Regular Expression for L1 /\ L2

I know it is possible to come up with one because Prof Heap said somebody did it on the bulletin board.

Seperate: First write L1s regex, then write L2s regex then combine.

2) Plan

I will:

i) Find regular expressions for L1 and L2

ii) Break up the regular expression for L1 and insert pieces of the regular expression of L2 into it to create the cases of the intersection

iii) Combine the cases with + (ORs)

iv) Factor to reduce size.

This was devised by noticing the similarity between this and straight forward algebra. If we first examine L1, the 1s in L1's regex can be seen as variables holding the place of a regex to satisfy L2's condition also. The solution to this problem could possibly be applied to creating regular expressions to denote unions of other languages.

3) Carry out the plan

i) A regular expression to denote L1: (1*01*01*)*

A regular expression to denote L2: 0*10*(0*10*10*)*

ii) If we examine L1, we need to have an odd number of 1s in the 1*s. To do this, there are a few cases. 1) The first, second and third 1* contains an odd number of 1s (then odd + odd + odd = odd). 2) The first 1* contains an odd number of 1s (then odd + even + even = odd).  3) The second 1* contains an odd number of 1s (then even + odd + even = odd). 4) The third 1* contains an odd number of 1s (then even + even + odd = off).

1) 1(11)*[01(11)*01(11)*]*

2) 1(11)*[0(11)*0(11)*]*

3) [(11)* 0 1(11)*0(11)*][(11)*01(11)*0(11)*01(11)*0(11)*]* 

# 3 is an interesting case because we cannot split up the 0s. Thus we must have at least 1 repetition and then subsequent repititions will contain two center pieces (1(11)*). This is interesting because in this case, 0s can only increment by 4 rather than 2

4) [(11)*0(11)*0]*1(11)*

iii) {1(11)*[01(11)*01(11)*]*} + {1(11)*[0(11)*0(11)*]*} + {[(11)* 0 1(11)*0(11)*][(11)*01(11)*0(11)*01(11)*0(11)*]* } + { [(11)*0(11)*0]*1(11)*}

iv) Notice how two of the cases start with the same regex? We can combine these with +

{1(11)*[[01(11)*01(11)*]+[0(11)*0(11)*]]*} + {[(11)* 0 1(11)*0(11)*][(11)*01(11)*0(11)*01(11)*0(11)*]* } + { [(11)*0(11)*0]*1(11)*}

Finally, I am sure this works because of the means of construction. All cases are accounted for.

4) Looking Back.

Although this is a complex way of doing this, the regex is right based on the way it is constructed (including each possible case then logically combining them). 

We learned a better way of doing this in lecture, using FSAs and then converting them to regex. I think that is a more practical way since it is more systematic and will work on all problems. However, in problems where you are doing different things with the 0s and 1s, this solution should work.

Week 13

The test wasn't too bad. The state question was very easy, and I kept looking at the transitions to make sure there wasn't a trick somewhere in there (like making the state invariant of q1 be x equiv 2%3 rather than 1%3). I couldn't find any tricks. The other questions were less straightforward. I almost forgot that I had to prove Language equivelancy both ways and I ended up rewriting my proof in the last 5-10 minutes.

All in all, I study the wrong things. I expected the test to have questions on the pumping lemma, turning NFSAs into DFSAs into regex and things of that nature.

Week 12

The assignment went by smoothly once again with the help of my group. Working in groups is great as everybody has different insights into the problems, so the thing I will get stuck on my partner has a quick solution to and vice versa. Despite this, I obviously need to practice proving Regex equivelancy and will make sure to review it before the test. I can create my own equivelant regular expressions and practice proving their equivelancy.

The lecture: AAAAHH! There's almost no notes on the bulletin board and I wasn't feeling great and had to miss class. The pumping lemma, from the lecture notes, is very easy and I grasped it quickly. My only question that arises is why there needs to be a u v and a w. Although its obvious there are uses for it, there are situations that it seems pointless. For example, the question of having a string with the same amount of 1s as 0s, it could easily be solved with just a v and a w. The u is just dead weight. Oh well, I'll do it the way im taught!

Now I need to read the textbook to find out what I missed in lecture!

Week 11

Epsilon transitions are very weird.  The way NFSAs are defined with the epsilon transitions is very strange and I need to review this to get a better grasp on it. Same thing goes for turning NFSAs into DFSAs. However, the cartesian product is very easy, although tedious. Listing states over and over again changing 1 tiny thing each time is not my definition of fun. Despite their usefulness I can see myself growing to dislike both NFSAs and the cartesian product. Time will tell.

Week 10

DFSAs. These are the coolest things ever. You can draw a regular expression evaluator. It's a simple idea put together very interestingly. There are quite a few things to remember; states, transitions, start state, etc. but they are pretty obvious. I have to remember not to forget to mention the start state and the end state. All in all DFSAs are cool, easy, and seemingly useful. :)

I liked the example of the "difficult" regex to construct and how I was able to have a solution very quickly (despite being told it was hard). However on review, my solution had a hole that I need to fix, which is why I did not put it on the bulletin board. That is the problem I will do the full Polya method of problem solving on. 

Week 9

The test was not extremely difficult but I was unprepared due to my lack of understanding about loop invariants. The question involving the loop was the most difficult for me only for the fact that I did not know how to include an invariant that will help me in my proof. It was easy to prove that the function terminated but my proof was definitely not in the pattern of what we were taught (and was probably wrong). The rest of the questions were easy.

Finally a change in the flavor of the material, the regular expressions are a wonderful break from induction (unless were going to use that for this too...). They seemed very confusing at first (especially the Kleene star but by the end of a lecture, they made sense.) The proof was very bizare but I am sure that with practice these things will become easy.

Week 8

I did not attend the lecture and after writing the test I wish I had. It was very difficult to figure out the proofs for functions with loops simply from the lecture notes. The concept of a loop invariant, although seemingly useful, does not differ from a postcondition in my mind?

The assignment went by very well even with my limited amount of time to work through and understand it. The partner system helped very much and I am upset I never utilized partners before. The questions were very similar to those done in the lecture and it was not difficult to extend those ideas to the assignment.