Programming Bits

Red Black tree deletion involves following cases: If node to delete is a RED leaf node then just delete it. If node to delete is a BLACK node and has one red child then, just replace it with child and retain color of node to be deleted. If node is having more than two children then just replace the node with in-order predecessor and call the delete routine at in-order predecessor recursively. But the problem arises when node to delete is a BLACK node and it has only one BLACK child or node to be deleted is a BLACK leaf node. Then DOUBLE BLACK problem arises. Note: Points form 1 to 3 are easy to implement and can be understood using the code given below. To resolve DOUBLE BLACK problem we have following four cases: Why double black problem: Suppose we have deleted a BLACK node and the number of nodes in the path going through it is reduced by 1. So the BLACK replacement will carry one extra BLACK color to retain property 5 of RB Tree that in any path from root to leaf...

Red Black Tree is a binary search tree. With extra property with each node having color either RED or BLACK. It has following properties: Every node is either RED or BLACK Root element will have BLACK color. Every RED node has BLACK children. Every path from root to leaf will have same number of BLACK nodes. We have few theorems over RB tree: Note: Height of RB tree is θ(logn). Insertion: Insertion in RB tree is little tricky as compared to BST, we have following cases to insert a node in RB tree: When tree is null just insert the node and it will be root, color will be BLACK. To insert an element first find its position: if key <= node.key then move to left else move to right Once we find the correct place to insert, we insert the node with color RED. But if the parent of the node just inserted has color RED then it will violate 3 property. So we have to fix this issue. We call it double RED problem, we resolve it using following few cases. DOUBLE R...

Meaning of Goodness of fit test: We find out which distribution fits the sample data the most. And this is achieved using chi-square distribution (Snedecor and Cochran, 1989). How to apply: There are 4 steps to follow: State the hypothesis:  Data follows a distribution or not Criteria to reject null hypothesis: if Χ 2  > Χ 2 (k,1-α) then reject null hypothesis. Analyze sample data: Compute the chi-square value using below formula: ∑(Oi- Ei) 2 /Ei        : Oi is observed frequency and Ei is expected frequency Interpret the results: Declare the results after comparing the values of Χ 2 and Χ 2 (k,1-α), where k is degree of freedom and  α is significance level. Degree of Freedom: It is  =  n - 1 - m m: number of parameter in the distribution. So in case of normal distribution m is 2 ( μ ,α) and in case of poisson dist. m is = 1 ( λ). Example 1: Goodness of fit test for Normal Distribution Year wise d...

RUDP provides  a solution where UDP  is too primitive because guaranteed-order  p acket  delivery is desirable, but  T CP  adds too much overhead.  In order for RUDP to gain higher Quality of Service , RUDP implements features that are similar to TCP with less overhead. Implementations: In order to ensure quality, it extends UDP by means of adding the following features: Acknowledgment of received packets Flow control Re-transmission of lost packets Over buffering (Faster than real-time streaming) RUDP is not currently a formal standard, however it was described in an n 1999. It has not been proposed for standardization. Working Example: One way to think about how RUDP types of transport work is to use a basic model where all the data is sent in UDP format, and each missing packet is indexed.  Once the main body of the transfer is done, the recipient sends the sender the index list and the sender re-transmits ...