Hm, ok, we have a relation here, where the value we pass is 2, and the value we get is 3. And if we pass 1, we get 1. Maybe t(n) = n+(n-1)? Let's keep going, two values are not enough.
It works. Now you see that the function 2^n - 1 applies to the relations of our previous values (t(1) = 1, t(2) = 3, t(3) = 7, t(4) = 15, t(5) = 31).
As to why 8 + 7, thanks to the previous exercise, we can now deduce that 2^(4-1) = 8 and therefore 2^(4-1) - 1 = 7. In the post, as soon as I got rid of t in 8t + 7, I tried to deduce the values of 8 and 7 and got that formula (2^(4-1) + (2^(4-1) - 1)).
Not quite... Solving for the given equation t(n) = 2^(n-1) + (2^(n-1) - 1) is straight forward. Nothing complicated about replacing n with an actual number. This is not the source of my confusion.
I am asking how you get to the given equation in the first place. As in, you start with 8+7 and a few steps later it's 2^(n-1) + (2^(n-1) - 1)
I'm going to use your example with t(4) from the previous comment, but reverse, because that is the process I am trying to figure out.
t(4) = 8+7 // factored recursive hanoi down to this. cool, makes sense. t(4) = 8 + (8 - 1) // ok I guess, but why? t(4) = 2^(3) + (2^(3) - 1) // Math makes sense, but why did we decide to represent 8 as an 2 to the 3rd? Is the goal to find exponential representations? t(4) = 2^(4-1) + (2^(4-1) - 1) // Why on earth did we decide 3 is better represented as 4-1??
From here, factoring 2^(4-1) + (2^(4-1) - 1) down to 2^n-1 makes perfect sense.
The source of my frustration here is not so much the math, but why the choices were made to eventually turn 8 into 2^(4-1)
I would really appreciate shedding some light on that part specifically.
t(4) = 2^(3) + (2^(3) - 1) // Math makes sense, but why did we decide to represent 8 as an 2 to the 3rd? Is the goal to find exponential representations? t(4) = 2^(4-1) + (2^(4-1) - 1) // Why on earth did we decide 3 is better represented as 4-1??
Hold that thought right there. The reason why we use 4-1 instead of 3 is to represent the value of n in the formula, so we can write n-1. Because in that example, n = 4.
If you notice in my previous comment, my goal was not simply to solve the equation with several values of n, but to show how the result and the parameters of the function relate each other.
The goal of the deduction process is to find the function that solves these relations: t(1) = 1 + 0 // same as 2^(1-1) + (2^(1-1) - 1)
t(2) = 2 + 1 // same as 2^(2-1) + (2^(2-1) - 1)
t(3) = 4 + 3 // same as 2^(3-1) + (2^(3-1) - 1)
t(4) = 8 + 7 // same as 2^(4-1) + (2^(4-1) - 1)
t(5) = 16 + 15 // same as 2^(5-1) + (2^(5-1) - 1)
That way you can find a function that solves for every value of n. Please let me know if I now made it clear.
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To understand that deduction process, let's solve this with more values of
n.Given the following equation:
t(n) = 2^(n-1) + (2^(n-1) - 1). We'll solvet(n)forn=1,n=2,n=3andn=4.t(1) = 2^(1 - 1) + (2^(1 - 1) - 1)t(1) = 2^(0) + (2^(0) -1)
t(1) = 1 + (1 - 1)
t(1) = 1 + 0 = 1
We can't deduce anything from
n=1only, let's try withn=2.t(2) = 2^(2 - 1) + (2^(2 - 1) - 1)t(2) = 2^(1) + (2^(1) - 1)
t(2) = 2 + (2 - 1)
t(2) = 2 + 1
t(2) = 3
Hm, ok, we have a relation here, where the value we pass is
2, and the value we get is3. And if we pass1, we get1. Maybet(n) = n+(n-1)? Let's keep going, two values are not enough.t(3) = 2^(3-1) + (2^(3 - 1) - 1)t(3) = 2^(2) + (2^(2) - 1)
t(3) = 4 + (4-1)
t(3) = 4 + 3
t(3) = 7
Wait, our relation (
t(n) = n+(n-1)) doesn't apply here. But the relation of3and7, seems familiar somehow. Let's keep going.t(4) = 2^(4-1) + (2^(4-1) - 1)t(4) = 2^(3) + (2^(3) - 1)
t(4) = 8 + (8 - 1)
t(4) = 8 + 7
t(4) = 15
Wait, this look like
2^n - 1. It applies to our previous values. If we try withn=5then...t(5) = 2^(5-1) + (2^(5-1) - 1)...
t(5) = 16 - 15
t(5) = 31
It works. Now you see that the function
2^n - 1applies to the relations of our previous values (t(1) = 1,t(2) = 3,t(3) = 7,t(4) = 15,t(5) = 31).As to why
8 + 7, thanks to the previous exercise, we can now deduce that2^(4-1) = 8and therefore2^(4-1) - 1 = 7. In the post, as soon as I got rid oftin8t + 7, I tried to deduce the values of8and7and got that formula (2^(4-1) + (2^(4-1) - 1)).Let me know if this helps to make myself clear.
Not quite... Solving for the given equation
t(n) = 2^(n-1) + (2^(n-1) - 1)is straight forward. Nothing complicated about replacingnwith an actual number. This is not the source of my confusion.I am asking how you get to the given equation in the first place. As in, you start with
8+7and a few steps later it's2^(n-1) + (2^(n-1) - 1)I'm going to use your example with
t(4)from the previous comment, but reverse, because that is the process I am trying to figure out.t(4) = 8+7// factored recursive hanoi down to this. cool, makes sense.t(4) = 8 + (8 - 1)// ok I guess, but why?t(4) = 2^(3) + (2^(3) - 1)// Math makes sense, but why did we decide to represent 8 as an 2 to the 3rd? Is the goal to find exponential representations?t(4) = 2^(4-1) + (2^(4-1) - 1)// Why on earth did we decide3is better represented as4-1??From here, factoring
2^(4-1) + (2^(4-1) - 1)down to2^n-1makes perfect sense.The source of my frustration here is not so much the math, but why the choices were made to eventually turn
8into2^(4-1)I would really appreciate shedding some light on that part specifically.
Hold that thought right there. The reason why we use
4-1instead of3is to represent the value ofnin the formula, so we can writen-1. Because in that example,n = 4.If you notice in my previous comment, my goal was not simply to solve the equation with several values of
n, but to show how the result and the parameters of the function relate each other.The goal of the deduction process is to find the function that solves these relations:
t(1) = 1 + 0 // same as 2^(1-1) + (2^(1-1) - 1)t(2) = 2 + 1 // same as 2^(2-1) + (2^(2-1) - 1)
t(3) = 4 + 3 // same as 2^(3-1) + (2^(3-1) - 1)
t(4) = 8 + 7 // same as 2^(4-1) + (2^(4-1) - 1)
t(5) = 16 + 15 // same as 2^(5-1) + (2^(5-1) - 1)
That way you can find a function that solves for every value of
n. Please let me know if I now made it clear.