Interesting problem handout from lecture today. It is called the Penny Pile Problem. Basically you have two boxes, the left box contains 64 pennies and the right contains nothing. Now, there are two operations that you are allowed to use in order to move pennies from one box to the other.
L: If the left box has an even number of pennies, you can transfer half of them to the right box. If the left box has odd numbers of pennies, you can not use this operation.
R: If the right box has an even number of pennies, you can transfer half of them to the left box. If the right box has odd numbers of pennies, you can not use this operation.
Now, the problem is to arrange things so that one of the boxes has 48 pennies using the operations from above.
So, let's just carry out the operations.
(64,0)
(32,32)
(48,16) - Done!
Well, that was easy.
Now, are we able to choose any number from a range of 0 to 64 and arrange it so that we are able to achieve that number? I think it is definitely possible.
Let's begin from (64,0), the only step is to divide the pennies into two piles each containing 32 pennies. Therefore, I think I will tackle the problem from 32 since once we reach 32, we are able to get back to 64. From here on, my intuition is to work backwards from the chosen number and reach 32. If the chosen number is bigger than 32, then the other number can be arranged to become 32 and reach 64.
Friday, 24 October 2014
Proofs
Yesterday was my tutorial quiz day and we had to prove one statement in the quiz. Before the quiz, I was pretty confident in myself with the quiz because proving a statement is basically making assumptions, proving the consequent and then unwraping the assumptions. I was pretty stupid not to solve some extra questions and practice proving statements. The question was similar to the one on the tutorial exercise:
For all x that is in whole numbers, there exists a y in the whole numbers that makes x = 3y + 1. Then there exists a y in the whole numbers that makes x^2 = 3y + 1.
It's pretty simple right?
Square the antecedent x = 3y +1 to x^2 = 9y^2 + 6y + 1
Factor out 3 to make x^2 = 3(3y^2 +2y) + 1
Then the 3y^2 + 2y is equivalent to the consequent's y.
And then unwrap all the assumptions and the proving is done.
The question that the class received in the quiz was a little altered and when I squared the antecedent and factored out a common number, the numbers inside the bracket did not equal to the consequent's y. I was going insane on why I was not able to solve it and in the end I could not figure it out.
CSC165 is a strange course for me because half the course is more about logics and the way you think and tackle each problems in an interesting way. I was suprised that I did so poorly on my assignment #1 because it was very confusing for me but the average was 80% which is insanely high in my opinion. I need to start studying more for CSC165 and up my game in the assignments and quizzes.
For all x that is in whole numbers, there exists a y in the whole numbers that makes x = 3y + 1. Then there exists a y in the whole numbers that makes x^2 = 3y + 1.
It's pretty simple right?
Square the antecedent x = 3y +1 to x^2 = 9y^2 + 6y + 1
Factor out 3 to make x^2 = 3(3y^2 +2y) + 1
Then the 3y^2 + 2y is equivalent to the consequent's y.
And then unwrap all the assumptions and the proving is done.
The question that the class received in the quiz was a little altered and when I squared the antecedent and factored out a common number, the numbers inside the bracket did not equal to the consequent's y. I was going insane on why I was not able to solve it and in the end I could not figure it out.
CSC165 is a strange course for me because half the course is more about logics and the way you think and tackle each problems in an interesting way. I was suprised that I did so poorly on my assignment #1 because it was very confusing for me but the average was 80% which is insanely high in my opinion. I need to start studying more for CSC165 and up my game in the assignments and quizzes.
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