Answer:
5.23 LAB: Adjust values in a list by normalizing When analyzing data sets, such as data for human heights or for human weights, a common step is to adjust the data. This can be done by normalizing to values between 0 and 1, or throwing away outliers.
Explanation:
Note: The matrix referred to in the question is: ![M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]](https://tex.z-dn.net/?f=M%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D)
Answer:
a) [5/18, 5/18, 4/9]'
Explanation:
The adjacency matrix is
To start the power iteration, let us start with an initial non zero approximation,
![X_o = \left[\begin{array}{ccc}1\\1\\1\end{array}\right]](https://tex.z-dn.net/?f=X_o%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To get the rank vector for the first Iteration:
![X_1 = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]\left[\begin{array}{ccc}1\\1\\1\end{array}\right] \\\\X_1 = \left[\begin{array}{ccc}5/6\\5/6\\4/3\end{array}\right]\\](https://tex.z-dn.net/?f=X_1%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2F2%261%2F3%260%5C%5C1%2F2%261%2F3%260%5C%5C0%261%2F3%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CX_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F6%5C%5C5%2F6%5C%5C4%2F3%5Cend%7Barray%7D%5Cright%5D%5C%5C)
Multiplying the above matrix by 1/3
![X_1 = \left[\begin{array}{ccc}5/18\\5/18\\4/9\end{array}\right]](https://tex.z-dn.net/?f=X_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2F18%5C%5C5%2F18%5C%5C4%2F9%5Cend%7Barray%7D%5Cright%5D)
An internal node is a node which carries at least one child or in other words, an internal node is not a leaf node.
Answer:
open-source
Explanation:
open-souce software allows any user to submit modifications of the source code
Answer:
Explanation:
If L(D1) = L(D2), the D has every state being final
If L(D1) = L¯(D2), the D has every state being final
If L(D1) = ∅, then L(D) = L(D2).
If L(D1)=Σ, L(D) = L(D2)
