Quantitative and Qualitative Study Design Examples

Recent questions in Study design
High school statisticsAnswered question
djo57bgfrqn djo57bgfrqn 2022-10-26

What I have found is that I love theoretical discussions of mathematics and proving theorems, though the education system focuses on doing exercises and applications, and that is what grades are determined on. I have a hard time doing exercises mainly because I don't find them motivating, although, I admit, doing the exercises does cultivate one's intuition to a certain degree.
Up until now my main technique of learning was to take meticulous notes from textbooks(not in class as all they do is mundane exercises), think about those concepts and then practice exercises before exams, and it worked fine for discrete math, precalculus, and even the first course of calculus, which was mainly focused on derivatives. Further. I as able to maintain A's in all my classes with enough energy left over to even do very well in other classes.
However, this hasn't gone too well in the second course of calculus, which is mostly about integrals of various functions and their applications and techniques. The instructor usually does problems on the board step by step without any discussion for what we are doing. For example, both the fundamental theorems were introduced and given about 15 minutes of time to discuss. But, of course, this could be bad teaching style.
My question is this, should I learn math theoretically and understand the material well even though on tests this would mean getting about B's that I have gotten or should I only care about the grades and just do what I am told, like a human calculator, practice exercises and just focus on getting A's? What is the better for me in the long run if I want to get into a decent grad school? And have any of you ever had to decide between getting an A or spending more energy and time actually learning the theory, which is more satisfying? I could be looking at this completely wrong and perhaps there is a way to do both? Thank you and your input will be appreciated.

High school statisticsAnswered question
Bairaxx Bairaxx 2022-10-18

Reminder : Given a set S of n elements (we will use [n] in the following for simplicity), a Latin square L is a function L : [ n ] × [ n ] S, i.e., an n × n array with elements in S, such that each element of S appears exactly once in each row and each column. For example,
Latin square
1, 2, 3
3, 1, 2
3, 1, 2
Let L 1 and L 2 be two Latin squares over the ground sets S 2 respectively. They are called orthogonal if for every ( x 1 , x 2 ) S 1 × S 2 there exists a unique ( i , j ) [ n ] × [ n ] such that L 2 ( i , j ) = x 2 . For example, the following are two orthogonal Latin squares of order 3.
1, 2, 3 2, 3, 1
3, 1, 2 3, 1, 2
1, 2, 1 1, 2, 3
It is known that there at most n−1 mutually orthogonal Latin squares of order n, and that the bound is achieved if and only there exist an affine plane of order n.
Graph : I'm building a graph G n with vertex set the latin squares of order n and two vertices are adjacent iff the Latin squares are orthogonal.
I want to understand some properties of this graph. For simplicity I consider the squares up to permutation of [n], hence w.l.o.g. all my squares have for first line { 1 , 2 , , n }. Indeed if I call H n the graph not up to permutations, then H n is the n! graph blowup of G n , or using the Tensor product
H n = G n × K n !
As I'm mainly interested in the chromatic number of my graph, and we know that χ ( H n ) min { χ ( G n ) ; n ! }, I will study only G n
For instance G 3 = K 2
I know that :
It's trivial that G n is not complete.
If there exist an affine plane of order n then G n contains K n 1 as a subgraph, and χ ( G n ) n 1
I wonder the following :
What is the maximum degree of G n ? We know that we have at most n−1 mutually orthogonal latin squares, but to how many squares can one square be orthogonal (still up to permutation)?
Do we have any other info on the chromatic number, not coming from the property χ ( G n ) Δ + 1
Can G n contains an induce k-cycle with k>3 (i.e. chordless cycle)?
Can it be conjectured that
Conjecture : for any n, G n is the disjoint union of complete subgraphs (of different sizes).
Edit After some simple Brute force and some additional reading, I can tell that
G 4 is made of 2 disjoint K 3 and 18 isolated vertices, for a total of 24 Latin squares up to permutations.
G 5 is made of 36 disjoint K 4 and 1200 isolated vertices, for a total of 1344 Latin squares up to permutation.
The case n=6 would be the first interesting case, as there are no affine plance of order 6, hence we will find no K 5 in G 6 . It is known since 1901 (from Tarry hand checking all Latin squares of order 6) that no two were mutually orthogonal. So G 6 is made of only isolated vertices.
It is also know that the case n=2 and n=6 are the only one with only isolated vertices. (see design theory by Beth, Jingnickel and Lenz)
From the article "Monogamous Latin Square by Danziger, Wanless and Webb, available on Wanless website here. The authors show that for all n>6, if n is not of the form 2p for a prime p 11, then there exists a latin square of order n that possesses an orthogonal mate but is not in any triple of Mutually Orthogonal Latin Squares. Therefore our graph G n will have some isolated K 2

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