- #1
Shinjo
- 12
- 0
I've done 3 of my 5 problems, which took me 2 days and over 30-50 pieces of scrap paper. I'm serious, I didn't even eat dinner today because I spent straight hours just staring at questions, thinking I was coming close to solutions, then only to find out I've gotten no where. So in the end, I decided to come here and find out if anyone can lend me a hand.
Let T(n,m) = T(n,m-1) + T(n-1,m) and T(s,2) = T(2,s) = s for all natural numbers s. Use induction to prove that
[tex]T(n,m) \leq \left(\begin{array}{cc}n+m-1\\n-1\end{array}\right) \forall n,m \in N, n + m \geq 2[/tex]
Here's what I got so far
Basically if I expand the recurrence relation out a bit in a sort of "binary tree" form, so my top node will be T(n,m), then the next row is T(n,m-1) and T(n-1,m), then the next row will be T(n,m-2), T(n-1,m-1), T(n-1,m-1), and T(n-2,m), I started to notice something similar to Binomial Coefficients pascal's triangle thingy, which I also know that the left side of the inequality has a combination function.
I also noticed that as I go down each row on the tree, each sub node loses only an integer of one. For instance, the first root node is n+m, the next one is n+m-1, and the next is n+m-2, and so forth.
Someone told me to try and fix one variable then proving the other, which I think is how I'm supposed to tackle this problem. Unfortunately, for these kinds of proofs, there has to be some idea behind it, like a loop invariant, or a fixed form of the predicate. Without those, I can't even start the proving part.
If someone can help me, it would be much appreciated. Thank You.
Let T(n,m) = T(n,m-1) + T(n-1,m) and T(s,2) = T(2,s) = s for all natural numbers s. Use induction to prove that
[tex]T(n,m) \leq \left(\begin{array}{cc}n+m-1\\n-1\end{array}\right) \forall n,m \in N, n + m \geq 2[/tex]
Here's what I got so far
Basically if I expand the recurrence relation out a bit in a sort of "binary tree" form, so my top node will be T(n,m), then the next row is T(n,m-1) and T(n-1,m), then the next row will be T(n,m-2), T(n-1,m-1), T(n-1,m-1), and T(n-2,m), I started to notice something similar to Binomial Coefficients pascal's triangle thingy, which I also know that the left side of the inequality has a combination function.
I also noticed that as I go down each row on the tree, each sub node loses only an integer of one. For instance, the first root node is n+m, the next one is n+m-1, and the next is n+m-2, and so forth.
Someone told me to try and fix one variable then proving the other, which I think is how I'm supposed to tackle this problem. Unfortunately, for these kinds of proofs, there has to be some idea behind it, like a loop invariant, or a fixed form of the predicate. Without those, I can't even start the proving part.
If someone can help me, it would be much appreciated. Thank You.