tableau assignment

instructions are attached. the final project must be twbx format. I already gave the dataset so you just need to do the work with that

Python program to enable a user to perform image processing operations.

Write a Python program to enable a user to perform image processing operations. The program can use a GUI or a command line interface. Allow at least the following operations: brighten/darken, increase contrast, change color to grayscale and edge detection. You may choose to allow other operations, such as crop, resize, etc. Any group may use skimage, covered in the text. You may also use OpenCV, Mahotas, or Pillow.

discussion-10

 Describe three network security risks and how an administrator may be able to defend aganist them.
 

Business Intelligence

Examine the decision-making process. Please discuss the following areas of the decision-making process in your paper.
The following are the items to discuss in the paper:

  • Define and discuss what are the four (4) components of the decision-making process.
  • The decision-making process is comprised of four (4) components. What is the dependency between the different components?
  • Discuss at least two (2) advantages and disadvantages in using the decision-making process (minimum of 4 total).
  • Provide at least three (3) examples of where the decision-making process is used.

Paper requirements:

  • Please use section breaks defining each section above separately.
  • Minimum 1200 words (excluding title page, table of contents, abstract, and references pages)
  • Minimum of four (4) references
  • Format your paper consistent with APA guidelines

Read through all the provided source code to make sure that you understand the context. A class named BinarySearchTree with an add method

 

Instructions

  1. Read through all  the provided source code to make sure that you understand the context. A  class named BinarySearchTree with an add method plus various utility  methods is provided for you. You must not change any  provided method with a body that is already complete. Note that linked  nodes are used to implement the BinarySearchTree class. Note also that  the data type allowed in the BinarySearchTree is constrained to be a  class that implements the Comparable interface and thus has a natural  total order defined.
  2. The min method of the BinarySearchTree class currently has no body. You must provide a correct body for the min method.
  3. A  sample main method is provided to illustrate building a simple binary  search tree and then using the min method to search for particular  values.

Problem Description: Finding the Minimum Value in a Binary Search Tree

Complete the body of the min method so that it returns the minimum value in the binary search tree.

The  following table lists an example call to min and the expected return  value when called in the context of the binary search tree pictured  below.

 

public class BSTMin {

  /** Provides an example. */

  public static void main(String[] args) {

    BinarySearchTree iBst = new BinarySearchTree<>();

    iBst.add(10);

    iBst.add(12);

    iBst.add(8);

    iBst.add(2);

    iBst.add(6);

    iBst.add(4);

    Integer imin = iBst.min();

    // The following statement should print 2.

    System.out.println(imin);

    BinarySearchTree sBst = new BinarySearchTree<>();

    sBst.add(“W”);

    sBst.add(“A”);

    sBst.add(“R”);

    sBst.add(“E”);

    sBst.add(“A”);

    sBst.add(“G”);

    sBst.add(“L”);

    sBst.add(“E”);

    String smin = sBst.min();

    // The following statement should print A.

    System.out.println(smin);

  }

  /** Defines a binary search tree. */

  static class BinarySearchTree> {

    // the root of this binary search tree

    private Node root;

    // the number of nodes in this binary search tree

    private int size;

    /** Defines the node structure for this binary search tree. */

    private class Node {

      T element;

      Node left;

      Node right;

      /** Constructs a node containing the given element. */

      public Node(T elem) {

        element = elem;

        left = null;

        right = null;

      }

    }

    /*   >>>>>>>>>>>>>>>>>>  YOUR WORK STARTS HERE  <<<<<<<<<<<<<<<< */

    ///////////////////////////////////////////////////////////////////////////////

    //    I M P L E M E N T  T H E  M I N  M E T H O D  B E L O W     //

    ///////////////////////////////////////////////////////////////////////////////

    /**

     * Returns the minimum value in the binary search tree.

     */

    public T min() {

    }

    /*   >>>>>>>>>>>>>>>>>>  YOUR WORK ENDS HERE  <<<<<<<<<<<<<<<< */

    ////////////////////////////////////////////////////////////////////

    // D O  N O T  M O D I F Y  B E L O W  T H I S  P O I N T  //

    ////////////////////////////////////////////////////////////////////

    ////////////////////

    // M E T R I C S //

    ////////////////////

    /**

     * Returns the number of elements in this bst.

     */

    public int size() {

      return size;

    }

    /**

     * Returns true if this bst is empty, false otherwise.

     */

    public boolean isEmpty() {

      return size == 0;

    }

    /**

     * Returns the height of this bst.

     */

    public int height() {

      return height(root);

    }

    /**

     * Returns the height of node n in this bst.

     */

    private int height(Node n) {

      if (n == null) {

        return 0;

      }

      int leftHeight = height(n.left);

      int rightHeight = height(n.right);

      return 1 + Math.max(leftHeight, rightHeight);

    }

    ////////////////////////////////////

    // A D D I N G  E L E M E N T S //

    ////////////////////////////////////

    /**

     * Ensures this bst contains the specified element. Uses an iterative implementation.

     */

    public void add(T element) {

      // special case if empty

      if (root == null) {

        root = new Node(element);

        size++;

        return;

      }

      // find where this element should be in the tree

      Node n = root;

      Node parent = null;

      int cmp = 0;

      while (n != null) {

        parent = n;

        cmp = element.compareTo(parent.element);

        if (cmp == 0) {

          // don’t add a duplicate

          return;

        } else if (cmp < 0) {

          n = n.left;

        } else {

          n = n.right;

        }

      }

      // add element to the appropriate empty subtree of parent

      if (cmp < 0) {

        parent.left = new Node(element);

      } else {

        parent.right = new Node(element);

      }

      size++;

    }

  }

}