Conclusion - Slog #10

   
As we wrap up CSC165, I would like to thank the profs, TA, and students for making this course very insightful and useful. In particular, in most mathematics oriented courses, I would just solve the problem using the techniques learned in class without really understanding why,  but now I have largely sharpened up my knowledge about limits and inequalities.


In the beginning, other students who have previously taken CSC165 have warned me about the difficulty thus painted an alarming picture of this course. However, I am honestly happy to say that this perception is not what it makes out to be. Initially, I have stumbled on logical notation but picked up the pace for proof structures. In all, I have learned an abundance of concepts such as vacuous truth, conjunction, implication, contradiction and proofs that would be very useful in my other courses and in reality.

Once again, it's been a pleasure taking this course and I wish everyone the best for exams! :)

Problem Solving - Diagonals - Slog #9

Understanding the problem:
Well we have a rectangular grid made out of side length m and n. When a diagonal line is draw from the upper left corner to the bottom right corner, the diagonal will pass through some number of squares.

Problem: How many squares will the line cross through?

Devise a plan:
Depending on the orientation of the rectangle, side length (m,n) can be looked at as (n,m).
To simplify the problem, m will be the vertical side length and n will be the horizontal side length.
Also, we know that when m = n then the number of shaded sides would be m number.

Carry out the Plan:
After devising multiple scenarios, I noticed that when m is even and:

·         n is odd, the formula is n // 2 * m + 2

·         n is even, the formula is n//2 *m

(3//2 * 2)+ 2 = 4

4//2 * 2 = 4

(5//2 *2) + 2 = 6

2//2 * 4 = 4

(3//2 *4) + 2 = 6

Surely, this seems true for the above cases but as m and n becomes larger, this is obviously not true.

Looking Back:Unfortunately, this method was unsuccessful. After drawing a whole bunch of diagrams, there seems to be some relationship to the remainder (longest length % shortest length) since an equal grid would result in
 m % n = 0 + m or n = number of shaded squares. Ultimately at this moment, this is just a loose end and possibly a lead to the actual solution. .