Friday, April 5, 2013
SLOG #9
This will be my last SLOG for the course. Although I was very afraid that I wasn't going to enjoy this course at the beginning, I can say that I have gained a lot of knowledge about a topic I was previous unfamiliar with. Danny did a great job teaching the material and Maria was thoroughly able to explain and clarify concepts during the tutorials. Overall, this course has been enjoyable. Lectures are officially over and now it's time to start preparing for exams. Good luck to everyone!
SLOG #8
Problem Solving Episode: Diagonals
Understanding the problem:
There is a rectangular grid with m rows and n columns. A diagonal line is drawn from the upper left corner to the lower right corner and it passes through certain number of grid squares. How many grid squares will the diagonal line cross through? This means to determine a formula for the number of squares using m and n.
Devise a plan:
Start small, and draw grids with different column lengths and rows to find a pattern between m and n, to figure out the number of diagonal squares by recording the number of squares the diagonal crossed through in a line. This was done by increasing the column lengths and rows by 1 and separating the odd, even, and mixed results.
Carry out the plan:
After drawing several grids, I was able to come up with the equation that the number of squares, x, is equal to the sum of the number of rows and the number of columns subtracted by the difference of the greatest common divisor of m and n. Algebraically, x = m + n - gcd(m,n). I determined this by observing the pattern for odd values of m and n, even values of m and n, and mixed values.
The calculations I used to determine the formula was:
d = m + n - s, where d was the missing difference.
Then I observed the values for d and realized that it was the gcd of m and n that resulted in the value of d.
Look back:
Although my plan wasn't the most efficient or organized, I was able to reach the final result after doing a few calculations using m and n.
Understanding the problem:
There is a rectangular grid with m rows and n columns. A diagonal line is drawn from the upper left corner to the lower right corner and it passes through certain number of grid squares. How many grid squares will the diagonal line cross through? This means to determine a formula for the number of squares using m and n.
Devise a plan:
Start small, and draw grids with different column lengths and rows to find a pattern between m and n, to figure out the number of diagonal squares by recording the number of squares the diagonal crossed through in a line. This was done by increasing the column lengths and rows by 1 and separating the odd, even, and mixed results.
Carry out the plan:
After drawing several grids, I was able to come up with the equation that the number of squares, x, is equal to the sum of the number of rows and the number of columns subtracted by the difference of the greatest common divisor of m and n. Algebraically, x = m + n - gcd(m,n). I determined this by observing the pattern for odd values of m and n, even values of m and n, and mixed values.
The calculations I used to determine the formula was:
d = m + n - s, where d was the missing difference.
Then I observed the values for d and realized that it was the gcd of m and n that resulted in the value of d.
Look back:
Although my plan wasn't the most efficient or organized, I was able to reach the final result after doing a few calculations using m and n.
Tuesday, April 2, 2013
SLOG #7
Unfortunately due to certain circumstances at home, I haven't been able to post as regularly as I'd like to. I have been reading other students' SLOGs and have come across some very interesting ones. I found the following blog very humorous but insightful at the same time: http://1d10terror.blogspot.ca/. I really enjoy how you can gain a sense of progress in the course through reading his SLOGs. As Michael said in his blog, I should have done the problem solving episode much earlier on, but I will complete it before the end of this week. Similarly, Assignment 3 isn't too challenging but manipulating the inequalities is what was taking a little bit of time. At first, using the proof of a limit that approaches zero was confusing me but gladly it was covered in the tutorial. Now I have completed the first four questions and the last one is the one that still confuses me, even after the quiz, which I know I did poorly on. I understand the structure and the proof by contradiction, but I'm still having a tough time coming up with the actual halt function itself. It seems to make sense when Maria explained it during the tutorial but when I attempt the problem myself, I can't seem to figure it out. I am going to review the tutorial exercises and the course notes until it becomes clearer. Another SLOG that interested me was the following: http://chargaslog.blogspot.ca/. It seems as if Charga and I tend to have the same problems in common, such as feeling off track, which is what makes it interesting to see that I am not the only one who feels this way.
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