Use LOGIC!

While going through the first test, in preparation for term-test #1, I found that Part B of the first question was pretty time consuming and a tad difficult. Since it is part of the first question and each sub-question(i, ii, iii, iv, v, vi) was only worth 1 mark, this indicated that these questions weren't supposed to be that difficult.

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The question:

Recall these python functions from lecture:

def quant1(L1, L2): return False in [x in L2 for x in L1]
def quant2(L1, L2): return True in [x in L2 for x in L1]
def quant3(L1, L2): return False not in [x in L2 for x in L1]
def quant4(L1, L2): return True not in [x in L2 for x in L1]


Part(B)

For each output (i)-(vi), either devise lists L1 and L2 so that the python expression:
[quant1(L1, L2), quant2(L1, L2), quant3(L1, L2), quant4(L1, L2)]
evaluates to that output, or else explain why it is impossible to devise such lists.


(i) [T,F,T,F]
(ii) [T,F,F,T]
(iii) [T,T,F,F]
(iv) [F,T,F,T]
(v) [F,F,T,T]
(vi) [F,T,T,F]


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I was wondering why I was taking so long with these questions that are assumed to be easy, so I spent some time trying to figure out what the problem was.

My first attempts at these questions, for each of the questions (i) - (vi), I didn't have an attack plan. I just jumped straight into the question, creating a small random list in my head, and then going through each of the 4 quants individually. Now that I think about it, that was kind of crazy. No wonder why I took so long. I was going through each function in my head, sometimes more than once for a question using a newly created list in my head, and I did this for each question the first time. Thus I ran over 4*6 functions using random made up lists in my head. That was such a waste of time! That definitely wasn't the method the prof had intended.

So then I kept trying to see whether there was a better method, an attack plan, something we've learned in lectures. Then I thought to myself that this is a logic course, and we've went over negations many times. So I went back to the quant functions and wrote comments to indicate what they predicate in English:


def quant1(L1, L2): return False in [x in L2 for x in L1]
## Means there exists an element in L1 that is not in L2

def quant2(L1, L2): return True in [x in L2 for x in L1]
## Means there exists an element in L1 that is also in L2

def quant3(L1, L2): return False not in [x in L2 for x in L1]
## Means all elements in L1 are in L2

def quant4(L1, L2): return True not in [x in L2 for x in L1]
## Means no elements in L1 are in L2

Now I can clearly see that quant1 and quant3 are negations of each other and quant2 and quant4 are negations of each other as well.

"What can I do with this information?" I thought to myself. Well, if some quants are negations of each other, then they both can't be True at the same time in each list. Now that was an easy way to find the impossible lists in (i) - (vi).

Now, thinking logically, it only makes sense that each list would have 2 Trues and 2 Falses in them. Also if any quant predicate is True, the negation quant predicate must be False, and vice versa. So now that eliminates having to go through each of the 4 quants in the list because you can trust that the negation of a quant is false if that quant is true.

So now, thanks to logic, I've developed an attack plan! All I had to do, after filtering out the impossible lists, was to find the quants corresponding to the Trues in the list and then generate a list that would satisfy the quants for them.

Now, with all this knowledge in mind derived from logic, all I had to do was eliminate the lists that have Trues for the quant and it's negation quant in the same list, then generate a list that would satisfy 2 quants.

Why didn't I think of this sooner?!?!

Since it took me a while to figure all this out, I'm guessing I didn't have a good grasp on logic and utilizing logic yet. This course is meant to teach you about logic and how to apply it and a lot of logic definitely applied in this question. Using logic, you can develop techniques for analyzing these types of questions swiftly. It was a good thing that I ran into these problems and figured out better ways to attack the problems because it really helped me with the actual term-test which had similar questions!



Also, for any TA or Professor reading my posts, I know they're pretty long and I apologize for that. I just like to ramble on and I feel that shorter posts aren't sufficient.

Sorry for making you guys read all this!

Venn Diagram Madness............

I came across something confusing for CSC165 Tutorial Exercise #1:

Drawing a Venn Diagram that Falsifies statement
d) "There is one of the three python programs that fails all three test suites."

So to falsify the above statement, you'd have to create a Venn Diagram portraying a situation where all python programs pass at least one test suite. If a python program passes at least 1 test suite, then it is not in the set of python programs that "fails all three test suites". Thus all we have to show on the Venn Diagram is that all of the 3 python programs in set Q passes at least one test suite. So technically, the programs can exist anywhere in Q, either in the overlapping Q and T area or just the Q area itself since the Q area itself just represents python programs, not stating how many test suites they pass or fail. So I figured the diagram for this would be:

(**Where Q is the set of 3 python programs, T is the set of universal python programs that pass all 3 test suites, and P is the universe of python programs. Also, O represents area must be occupied, X represents area that must not be occupied, and ? represents area could be empty or occupied.)

However, this diagram does not really say anything definite about making the statement false. For example, since ? represents may be empty or may be occupied, let's say for example that the overlapping T and Q area was empty. Now all the python programs in Q are in the Q only region. In this case, we may have a situation that falsifies the above statement (if all the programs in Q passed at least one test) or we may also have a situation that does not falsify it (any one of the python programs may fail all 3 tests and still be in this region). So this diagram wouldn't be enough to falsify the above claim.

During the tutorial, the Venn Diagram that was presented in answer to d) false situation was:

This Venn Diagram makes sense in that it presents a situation where the statement would be false for sure (if all of the 3 python programs passes all 3 test suites, then there can't be any that failed all 3), but there was just something still bothering me about this diagram. The area in Q may be occupied, not necessarily must not be. This diagram presents one situation where the statement would be false, but not a general diagram showing how to make the statement false. I don't know if I'm conveying myself properly, sorry for the confusion. For example, let's consider the Venn Diagram portraying a False situation for a) "All three python programs that passes all test suites." The correct diagram is:

However, another situation that would make the a) statement False would be this Venn Diagram:

This diagram would show a situation where a) would be False but it wouldn't be the correct most general Venn Diagram.

This is the feeling I'm getting for the tutorial provided Venn Diagram. I feel as if there is another region needed to indicate either programs that passed at least 1 test suite or a region indicating programs that failed all test suite.

The good thing about this experience is that after writing this blog, my head is a lot more clear about this whole situation!