4 rod Tower of Hanoi proof

In summary: This would complete the proof for our original inequality.In summary, you can use the given inequality and the fact that $2^{n-1} \leq 2^{n}$ to simplify the expression and reduce the problem to a simpler one. Then, you can use mathematical induction to prove the simpler inequality. I hope this helps!
  • #1
lemonthree
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I am having trouble solving part 2, for

$ W_{\frac{n(n+1)}{2}} \leq 2^{n} (n-1) + 1 , n \geq 0 $

I know that $W_{m} \leq 2*W_{m-k} + 2^{k} – 1, 0 \leq k \leq m$

Let $m = \frac{n(n+1)}{2}$

So now $W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n+1)}{2} - k} + 2^{k} - 1, 0 \leq k \leq \frac{n(n+1)}{2}$

Let k = n (just a hunch, I can't really explain why except that because the equation has an n inside)

$W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n-1)}{2}} + 2^{n} – 1$

From here, I'm not very sure, just trying to manipulate and force things in

$W_{\frac{n(n+1)}{2}} -1 \leq 2*W_{\frac{n(n-1)}{2}} + 2^{n} – 2$
And I'm lost. I tried letting $W_{\frac{n(n+1)}{2}} -1$ be $Q_{n}$ and manipulating from there such that

$Q_{n} \leq 2*Q_{n-1} + 2^{n}$ but to no avail.
So I re-worked everything and tried mathematical induction too, which I also ended up getting stuck halfway.

I think my working here is closer to the answer (and also less tedious than the induction), so I am posting this instead. Please, if there is any guidance at all, I would really appreciate it.
 
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  • #2


Hi there,

I can see that you have made some good progress in trying to solve this inequality. I would suggest trying to simplify the expression on the right side of the inequality by using the fact that $2^{n-1} \leq 2^{n}$ for all $n \geq 0$. This will help you to get rid of the $2^{n}$ term and make the inequality easier to work with.

Also, instead of trying to manipulate the expression $W_{\frac{n(n+1)}{2}}$ directly, you can try to use the given inequality $W_{m} \leq 2*W_{m-k} + 2^{k} – 1$ to reduce the problem to a simpler one.

For example, let's say we want to show that $W_{\frac{n(n+1)}{2}} \leq 2^{n-1}(n-1) + 1$ is true. We can use the given inequality with $m = \frac{n(n+1)}{2}$ and $k = n-1$ to get:

$W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n+1)}{2} - (n-1)} + 2^{n-1} – 1$

$W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n-1)}{2} + 1} + 2^{n-1} – 1$

Now, we can use the fact that $W_{m} \leq 2*W_{m-1} + 2^{m} – 1$ to simplify the expression further:

$W_{\frac{n(n+1)}{2}} \leq 2*(2*W_{\frac{n(n-1)}{2}} + 2^{n-1} – 1) + 2^{n-1} – 1$

$W_{\frac{n(n+1)}{2}} \leq 2^{n-1}*(2*W_{\frac{n(n-1)}{2}} + 1)$

Now, we can use mathematical induction to show that $2*W_{\frac{n(n-1)}{2}} + 1 \
 

1. What is the "4 rod Tower of Hanoi proof"?

The "4 rod Tower of Hanoi proof" is a mathematical problem that involves moving a stack of disks from one rod to another, using four rods instead of the traditional three. It is a variation of the classic Tower of Hanoi puzzle.

2. What is the objective of the "4 rod Tower of Hanoi proof"?

The objective of the "4 rod Tower of Hanoi proof" is to move all the disks from the starting rod to the destination rod, using the fewest number of moves possible. The disks must be moved one at a time and can only be placed on top of a larger disk or an empty rod.

3. How many moves are required to solve the "4 rod Tower of Hanoi proof"?

The minimum number of moves required to solve the "4 rod Tower of Hanoi proof" is 2^(n+1) - 2, where n is the number of disks. For example, if there are 4 disks, the minimum number of moves required is 2^(4+1) - 2 = 30 moves.

4. What is the significance of the "4 rod Tower of Hanoi proof"?

The "4 rod Tower of Hanoi proof" is significant because it is a variation of a well-known mathematical problem and requires a different approach to solve it. It also has real-world applications in computer programming and data storage systems.

5. Is there a general formula for solving the "4 rod Tower of Hanoi proof" with any number of disks?

Yes, there is a general formula for solving the "4 rod Tower of Hanoi proof" with any number of disks. It is 2^(n+1) - 2, where n is the number of disks. This formula can be applied to any variation of the Tower of Hanoi puzzle with any number of rods.

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