Binary Tree - Free Essay Example

Binary Trees Page: 1 Binary Trees by Nick Parlante This article introduces the basic concepts of binary trees, and then works through a series of practice problems with solution code in C/C++ and Java. Binary trees have an elegant recursive pointer structure, so they are a good way to learn recursive pointer algorithms. Contents Section 1. Binary Tree Structure a quick introduction to binary trees and the code that operates on them Section 2. Binary Tree Problems practice problems in increasing order of difficulty Section 3. C Solutions solution code to the problems for C and C++ programmers Section 4. Java versions how binary trees work in Java, with solution code Stanford CS Education Library #110 This is article #110 in the Stanford CS Education Library. This and other free CS materials are available at the library (https://cslibrary. stanford. edu/). That people seeking education should have the opportunity to find it. This article may be used, reproduced, excerpted, or s old so long as this paragraph is clearly reproduced. Copyright 2000-2001, Nick Parlante, nick. [emailprotected] stanford. edu. Related CSLibrary Articles Linked List Problems (https://cslibrary. stanford. edu/105/) a large collection of linked list problems using various pointer techniques (while this binary tree article concentrates on recursion) Pointer and Memory (https://cslibrary. stanford. edu/102/) basic concepts of pointers and memory The Great Tree-List Problem (https://cslibrary. stanford. edu/109/) a great pointer recursion problem that uses both trees and lists Section 1 Introduction To Binary Trees A binary tree is made of nodes, where each node contains a left pointer, a right pointer, and a data element. The root pointer points to the topmost node in the tree. The left and right pointers recursively point to smaller subtrees on either side. A null pointer represents a binary tree with no elements the empty tree. The formal recursive definition is: a binary tre e is either empty (represented by a null pointer), or is made of a single node, where the left and right pointers (recursive definition ahead) each point to a binary tree. ttp://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 2 A binary search tree (BST) or ordered binary tree is a type of binary tree where the nodes are arranged in order: for each node, all elements in its left subtree are less-or-equal to the node (). The tree shown above is a binary search tree the root node is a 5, and its left subtree nodes (1, 3, 4) are 5. Recursively, each of the subtrees must also obey the binary search tree constraint: in the (1, 3, 4) subtree, the 3 is the root, the 1 3. Watch out for the exact wording in the problems a binary search tree is different from a binary tree. The nodes at the bottom edge of the tree have empty subtrees and are called leaf nodes (1, 4, 6) while the others are internal nodes (3, 5, 9). Binary Search Tree Niche Basically, binary search tree s are fast at insert and lookup. The next section presents the code for these two algorithms. On average, a binary search tree algorithm can locate a node in an N node tree in order lg(N) time (log base 2). Therefore, binary search trees are good for dictionary problems where the code inserts and looks up information indexed by some key. The lg(N) behavior is the average case its possible for a particular tree to be much slower depending on its shape. Strategy Some of the problems in this article use plain binary trees, and some use binary search trees. In any case, the problems concentrate on the combination of pointers and recursion. (See the articles linked above for pointer articles that do not emphasize recursion. ) For each problem, there are two things to understand The node/pointer structure that makes up the tree and the code that manipulates it The algorithm, typically recursive, that iterates over the tree When thinking about a binary tree problem, its often a good idea to draw a few little trees to think about the various cases. https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 3 Typical Binary Tree Code in C/C++ As an introduction, well look at the code for the two most basic binary search tree operations lookup() and insert(). The code here works for C or C++. Java programers can read the discussion here, and then look at the Java versions in Section 4. In C or C++, the binary tree is built with a node type like this struct node { int data; struct node* left; struct node* right; } Lookup() Given a binary search tree and a target value, search the tree to see if it contains the target. The basic pattern of the lookup() code occurs in many recursive tree algorithms: deal with the base case where the tree is empty, deal with the current node, and then use recursion to deal with the subtrees. If the tree is a binary search tree, there is often some sort of less-than test on the node to decide if the recursion sh ould go left or right. /* Given a binary tree, return true if a node with the target data is found in the tree. Recurs down the tree, chooses the left or right branch by comparing the target to each node. */ static int lookup(struct node* node, int target) { // 1. Base case == empty tree // in that case, the target is not found so return false if (node == NULL) { return(false); } else { // 2. ee if found here if (target == node-;data) return(true); else { // 3. otherwise recur down the correct subtree if (target ; node-;data) return(lookup(node-;left, target)); else return(lookup(node-;right, target)); } } } The lookup() algorithm could be written as a while-loop that iterates down the tree. Our version uses recursion to help prepare you for the problems below that require recursion. Pointer Changing Code There is a common problem with pointer intensive code: what if a function needs to change one of the pointer parameters passed to it? For example, the insert() function below ma y want to change the root pointer. In C and C++, one solution uses pointers-to-pointers (aka reference parameters). Thats a fine technique, but here we will use the simpler technique that a function that wishes to change a pointer passed to it will return the new value of the pointer to the caller. The caller is responsible for using the new value. Suppose we have a change() function https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 4 that may change the the root, then a call to change() will look like this / suppose the variable root points to the tree root = change(root); We take the value returned by change(), and use it as the new value for root. This construct is a little awkward, but it avoids using reference parameters which confuse some C and C++ programmers, and Java does not have reference parameters at all. This allows us to focus on the recursion instead of the pointer mechanics. (For lots of problems that use reference parameters, see CSLibrary #105, Linked List Problems, https://cslibrary. stanford. du/105/). Insert() Insert() given a binary search tree and a number, insert a new node with the given number into the tree in the correct place. The insert() code is similar to lookup(), but with the complication that it modifies the tree structure. As described above, insert() returns the new tree pointer to use to its caller. Calling insert() with the number 5 on this tree 2 / 1 10 returns the tree 2 / 1 / 5 The solution shown here introduces a newNode() helper function that builds a single node. The base-case/recursion structure is similar to the structure in lookup() each call checks for the NULL case, looks at the node at hand, and then recurs down the left or right subtree if needed. /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* NewNode(int data) { struct node* node = new(struct node); // new is like malloc node-data = data; node-left = NULL; node- right = NULL; return(node); } 10 /* Give a binary search tree and a number, inserts a new node with the given number in the correct place in the tree. Returns the new root pointer which the caller should then use (the standard trick to avoid using reference parameters). */ struct node* insert(struct node* node, int data) { https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 5 // 1. If the tree is empty, return a new, single node if (node == NULL) { return(newNode(data)); } else { // 2. Otherwise, recur down the tree if (data data) node-left = insert(node-left, data); else node-right = insert(node-right, data); return(node); // return the (unchanged) node pointer } } The shape of a binary tree depends very much on the order that the nodes are inserted. In particular, if the nodes are inserted in increasing order (1, 2, 3, 4), the tree nodes just grow to the right leading to a linked list shape where all the left pointers are NULL. A similar thing happens if the nodes are inserted in decreasing order (4, 3, 2, 1). The linked list shape defeats the lg(N) performance. We will not address that issue here, instead focusing on pointers and recursion. Section 2 Binary Tree Problems Here are 14 binary tree problems in increasing order of difficulty. Some of the problems operate on binary search trees (aka ordered binary trees) while others work on plain binary trees with no special ordering. The next section, Section 3, shows the solution code in C/C++. Section 4 gives the background and solution code in Java. The basic structure and recursion of the solution code is the same in both languages the differences are superficial. Reading about a data structure is a fine introduction, but at some point the only way to learn is to actually try to solve some problems starting with a blank sheet of paper. To get the most out of these problems, you should at least attempt to solve them before looking at the solution. Even if your solution is not quite right, you will be building up the right skills. With any pointer-based code, its a good idea to make memory drawings of a a few simple cases to see how the algorithm should work. 1. build123() This is a very basic problem with a little pointer manipulation. (You can skip this problem if you are already comfortable with pointers. ) Write code that builds the following little 1-2-3 binary search tree 2 / 1 3 Write the code in three different ways a: by calling newNode() three times, and using three pointer variables b: by calling newNode() three times, and using only one pointer variable c: by calling insert() three times passing it the root pointer to build up the tree (In Java, write a build123() method that operates on the receiver to change it to be the 1-2-3 tree with the given coding constraints. See Section 4. ) struct node* build123() { 2. size() https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 6 This problem demonstrates simple binary tree traversal. Given a binary tree, count the number of nodes in the tree. int size(struct node* node) { 3. maxDepth() Given a binary tree, compute its maxDepth the number of nodes along the longest path from the root node down to the farthest leaf node. The maxDepth of the empty tree is 0, the maxDepth of the tree on the first page is 3. int maxDepth(struct node* node) { 4. minValue() Given a non-empty binary search tree (an ordered binary tree), return the minimum data value found in that tree. Note that it is not necessary to search the entire tree. A maxValue() function is structurally very similar to this function. This can be solved with recursion or with a simple while loop. int minValue(struct node* node) { 5. printTree() Given a binary search tree (aka an ordered binary tree), iterate over the nodes to print them out in increasing order. So the tree 4 / 2 / 1 3 5 Produces the output 1 2 3 4 5. This is known as an inorder traversal of the tree. Hint: For each node, t he strategy is: recur left, print the node data, recur right. void printTree(struct node* node) { 6. printPostorder() Given a binary tree, print out the nodes of the tree according to a bottom-up postorder traversal both subtrees of a node are printed out completely before the node itself is printed, and each left subtree is printed before the right subtree. So the tree 4 / 2 / 1 3 5 Produces the output 1 3 2 5 4. The description is complex, but the code is simple. This is the sort of bottom-up traversal that would be used, for example, to evaluate an expression tree where a node is an operation like + and its subtrees are, recursively, the two subexpressions for the +. https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 7 void printPostorder(struct node* node) { . hasPathSum() Well define a root-to-leaf path to be a sequence of nodes in a tree starting with the root node and proceeding downward to a leaf (a node with no children). Well say that an empty tree contains no root-to-leaf paths. So for example, the following tree has exactly four root-to-leaf paths: 5 / 4 / 11 / 7 2 8 / 13 4 1 Root-to-leaf paths: path 1: 5 4 11 7 path 2: 5 4 11 2 path 3: 5 8 13 path 4: 5 8 4 1 For this problem, we will be concerned with the sum of the values of such a path for example, the sum of the values on the 5-4-11-7 path is 5 + 4 + 11 + 7 = 27. Given a binary tree and a sum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Return false if no such path can be found. (Thanks to Owen Astrachan for suggesting this problem. ) int hasPathSum(struct node* node, int sum) { 8. printPaths() Given a binary tree, print out all of its root-to-leaf paths as defined above. This problem is a little harder than it looks, since the path so far needs to be communicated between the recursive calls. Hint: In C, C++, and Java, probably the best solution is to create a recursive helper function pr intPathsRecur(node, int path[], int pathLen), where the path array communicates the sequence of nodes that led up to the current call. Alternately, the problem may be solved bottom-up, with each node returning its list of paths. This strategy works quite nicely in Lisp, since it can exploit the built in list and mapping primitives. (Thanks to Matthias Felleisen for suggesting this problem. ) Given a binary tree, print out all of its root-to-leaf paths, one per line. void printPaths(struct node* node) { . mirror() Change a tree so that the roles of the left and right pointers are swapped at every node. So the tree 4 / 2 5 / https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 8 1 3 is changed to 4 / 5 2 / 3 1 The solution is short, but very recursive. As it happens, this can be accomplished without changing the root node pointer, so the return-the-new-root construct is not necessary. Alternately, if you do not want to change the tree nodes, you may construct and return a new mirror tree based on the original tree. void mirror(struct node* node) { 10. doubleTree() For each node in a binary search tree, create a new duplicate node, and insert the duplicate as the left child of the original node. The resulting tree should still be a binary search tree. So the tree 2 / 1 3 is changed to 2 / 2 3 / / 1 3 / 1 As with the previous problem, this can be accomplished without changing the root node pointer. void doubleTree(struct node* node) { 11. sameTree() Given two binary trees, return true if they are structurally identical they are made of nodes with the same values arranged in the same way. (Thanks to Julie Zelenski for suggesting this problem. int sameTree(struct node* a, struct node* b) { 12. countTrees() This is not a binary tree programming problem in the ordinary sense its more of a math/combinatorics recursion problem that happens to use binary trees. (Thanks to Jerry Cain for suggesting this problem. ) Suppose you are building an N n ode binary search tree with the values 1.. N. How many structurally different binary search trees are there that store those values? Write a recursive function that, given the number of distinct values, computes the number of structurally unique binary search trees that store those values. For example, https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 9 countTrees(4) should return 14, since there are 14 structurally unique binary search trees that store 1, 2, 3, and 4. The base case is easy, and the recursion is short but dense. Your code should not construct any actual trees; its just a counting problem. int countTrees(int numKeys) { Binary Search Tree Checking (for problems 13 and 14) This background is used by the next two problems: Given a plain binary tree, examine the tree to determine if it meets the requirement to be a binary search tree. To be a binary search tree, for every node, all of the nodes in its left tree must be the node. Consider th e following four examples a. 2 5 / 7 - TRUE b. 6 5 / 7 - FALSE, because the 6 is not ok to the left of the 5 c. 5 - TRUE / 2 7 / 1 d. 5 - FALSE, the 6 is ok with the 2, but the 6 is not ok with the 5 / 2 7 / 1 6 For the first two cases, the right answer can be seen just by comparing each node to the two nodes immediately below it. However, the fourth case shows how checking the BST quality may depend on nodes which are several layers apart the 5 and the 6 in that case. 3 isBST() version 1 Suppose you have helper functions minValue() and maxValue() that return the min or max int value from a non-empty tree (see problem 3 above). Write an isBST() function that returns true if a tree is a binary search tree and false otherwise. Use the helper functions, and dont forget to check every node in the tree. Its ok if your solution is not very efficient. (Thanks to Owen Astrachan for the idea of having this problem, and comparing it to problem 14) Returns true if a binary tree is a binary s earch tree. int isBST(struct node* node) { 4. isBST() version 2 https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 10 Version 1 above runs slowly since it traverses over some parts of the tree many times. A better solution looks at each node only once. The trick is to write a utility helper function isBSTRecur(struct node* node, int min, int max) that traverses down the tree keeping track of the narrowing min and max allowed values as it goes, looking at each node only once. The initial values for min and max should be INT_MIN and INT_MAX they narrow from there. * Returns true if the given tree is a binary search tree (efficient version). */ int isBST2(struct node* node) { return(isBSTRecur(node, INT_MIN, INT_MAX)); } /* Returns true if the given tree is a BST and its values are = min and left = lChild; root-right= rChild; return(root); } // call newNode() three times, and use only one local variable struct node* build123b() { struct node* root = newNode(2) ; root-left = newNode(1); root-right = newNode(3); return(root); } https://cslibrary. stanford. du/110/ BinaryTrees. html Binary Trees Page: 11 /* Build 123 by calling insert() three times. Note that the 2 must be inserted first. */ struct node* build123c() { struct node* root = NULL; root = insert(root, 2); root = insert(root, 1); root = insert(root, 3); return(root); } 2. size() Solution (C/C++) /* Compute the number of nodes in a tree. */ int size(struct node* node) { if (node==NULL) { return(0); } else { return(size(node-left) + 1 + size(node-right)); } } 3. maxDepth() Solution (C/C++) * Compute the maxDepth of a tree the number of nodes along the longest path from the root node down to the farthest leaf node. */ int maxDepth(struct node* node) { if (node==NULL) { return(0); } else { // compute the depth of each subtree int lDepth = maxDepth(node-left); int rDepth = maxDepth(node-right); // use the larger one if (lDepth rDepth) return(lDepth+1); else return(rDepth+1); } } 4. m inValue() Solution (C/C++) /* Given a non-empty binary search tree, return the minimum data value found in that tree. https://cslibrary. stanford. du/110/ BinaryTrees. html Binary Trees Page: 12 Note that the entire tree does not need to be searched. */ int minValue(struct node* node) { struct node* current = node; // loop down to find the leftmost leaf while (current-left ! = NULL) { current = current-left; } return(current-data); } 5. printTree() Solution (C/C++) /* Given a binary search tree, print out its data elements in increasing sorted order. */ void printTree(struct node* node) { if (node == NULL) return; printTree(node-left); printf(%d , node-data); printTree(node-right); } . printPostorder() Solution (C/C++) /* Given a binary tree, print its nodes according to the bottom-up postorder traversal. */ void printPostorder(struct node* node) { if (node == NULL) return; // first recur on both subtrees printTree(node-left); printTree(node-right); // then deal with the node printf (%d , node-data); } 7. hasPathSum() Solution (C/C++) /* Given a tree and a sum, return true if there is a path from the root down to a leaf, such that adding up all the values along the path equals the given sum. ttp://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 13 Strategy: subtract the node value from the sum when recurring down, and check to see if the sum is 0 when you run out of tree. */ int hasPathSum(struct node* node, int sum) { // return true if we run out of tree and sum==0 if (node == NULL) { return(sum == 0); } else { // otherwise check both subtrees int subSum = sum node-data; return(hasPathSum(node-left, subSum) || hasPathSum(node-right, subSum)); } } . printPaths() Solution (C/C++) /* Given a binary tree, print out all of its root-to-leaf paths, one per line. Uses a recursive helper to do the work. */ void printPaths(struct node* node) { int path[1000]; printPathsRecur(node, path, 0); } /* Recursive helper function given a node, and an array c ontaining the path from the root node up to but not including this node, print out all the root-leaf paths. / void printPathsRecur(struct node* node, int path[], int pathLen) { if (node==NULL) return; // append this node to the path array path[pathLen] = node-data; pathLen++; // its a leaf, so print the path that led to here if (node-left==NULL node-right==NULL) { printArray(path, pathLen); } else { // otherwise try both subtrees printPathsRecur(node-left, path, pathLen); printPathsRecur(node-right, path, pathLen); } } // Utility that prints out an array on a line. oid printArray(int ints[], int len) { int i; for (i=0; ileft); mirror(node-right); // swap the pointers in this node temp = node-left; node-left = node-right; node-right = temp; } } 10. doubleTree() Solution (C/C++) /* For each node in a binary search tree, create a new duplicate node, and insert the duplicate as the left child of the original node. The resulting tree should still be a binary search tree. So the tree 2 / https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 15 1 3 Is changed to / 2 3 / / 1 3 / 1 */ void doubleTree(struct node* node) { struct node* oldLeft; if (node==NULL) return; // do the subtrees doubleTree(node-left); doubleTree(node-right); // duplicate this node to its left oldLeft = node-left; node-left = newNode(node-data); node-left-left = oldLeft; } 11. sameTree() Solution (C/C++) /* Given two trees, return true if they are structurally identical. */ int sameTree(struct node* a, struct node* b) { // 1. both empty - true if (a==NULL b==NULL) return(true); // 2. oth non-empty - compare them else if (a! =NULL b! =NULL) { return( a-data == b-data sameTree(a-left, b-left) sameTree(a-right, b-right) ); } // 3. one empty, one not - false else return(false); } 12. countTrees() Solution (C/C++) /* For the key values 1 numKeys, how many structurally unique https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 16 binary search trees are p ossible that store those keys. Strategy: consider that each value could be the root. Recursively find the size of the left and right subtrees. */ int countTrees(int numKeys) { if (numKeys left! =NULL minValue(node-left) node-data) return(false); // false if the max of the right is right! =NULL maxValue(node-right) data) return(false); // false if, recursively, the left or right is not a BST if (! isBST(node-left) || !isBST(node-right)) return(false); // passing all that, its a BST return(true); } https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 17 4. isBST2() Solution (C/C++) /* Returns true if the given tree is a binary search tree (efficient version). */ int isBST2(struct node* node) { return(isBSTUtil(node, INT_MIN, INT_MAX)); } /* Returns true if the given tree is a BST and its values are = min and datadatamax) return(false); // otherwise check the subtrees recursively, // tightening the min or max constraint return isBSTUtil(node-left, min, node- data) isBSTUtil(node-right, node-data+1, max) ); } 15. TreeList Solution (C/C++) The solution code in C and Java to the great Tree-List recursion problem is in CSLibrary #109 https://cslibrary. stanford. edu/109/ Section 4 Java Binary Trees and Solutions In Java, the key points in the recursion are exactly the same as in C or C++. In fact, I created the Java solutions by just copying the C solutions, and then making the syntactic changes. The recursion is the same, however the outer structure is slightly different. In Java, we will have a BinaryTree object that contains a single root pointer. The root pointer points to an internal Node class that behaves just like the node struct in the C/C++ version. The Node class is private it is used only for internal storage inside the BinaryTree and is not exposed to clients. With this OOP structure, almost every operation has two methods: a one-line method on the BinaryTree that starts the computation, and a recursive method that wor ks on the Node objects. For the lookup() operation, there is a BinaryTree. lookup() method that the client uses to start a lookup operation. Internal to the BinaryTree class, there is a private recursive lookup(Node) method that implements the recursion down the Node structure. This second, private recursive method is basically the same as the recursive C/C++ functions above it takes a Node argument and uses recursion to iterate over the pointer structure. Java Binary Tree Structure To get started, here are the basic definitions for the Java BinaryTree class, and the lookup() and insert() methods as examples https://cslibrary. stanford. edu/110/ BinaryTrees. html Binary Trees Page: 18 // BinaryTree. ava public class BinaryTree { // Root node pointer. Will be null for an empty tree. private Node root; /* Node-The binary tree is built using this nested node class. Each node stores one data element, and has left and right sub-tree pointer which may be null. The node is a dumb neste d class we just use it for storage; it does not have any methods. */ private static class Node { Node left; Node right; int data; Node(int newData) { left = null; right = null; data = newData; } } /** Creates an empty binary tree a null root pointer. / public void BinaryTree() { root = null; } /** Returns true if the given target is in the binary tree. Uses a recursive helper. */ public boolean lookup(int data) { return(lookup(root, data)); } /** Recursive lookup given a node, recur down searching for the given data. */ private boolean lookup(Node node, int data) { if (node==null) { return(false); } if (data==node. data) { return(true); } else if (data false else return(false); } 12. countTrees() Solution (Java)

GMAT Tip of the Week The Overly Specific Question Stem

For most of our lives, we ask and answer relatively generic questions: Hows it going? What are you up to this weekend? What time do the Cubs play tonight? And think about it, what if those questions were more specific: Are you in a melancholy mood today? Are you and Josh going to dinner at Don Antonios tonight and ordering table-side guacamole? Do the Cubs play at 7:05 tonight on WGN? If someone is asking those questions instead, youre probably a bit suspicious. Why so specific? Whats your angle? The same is true on the GMAT. Most of the question stems you see are relatively generic: What is the value of x? Which of the following would most weaken the authors argument? So when the question stem get a little too specific, you should become a bit suspicious. Whats the test going for there? Why so specific? The overly-specific Critical Reasoning question stem is a great example. Consider the problem: Raisins are made by drying grapes in the sun. Although some of the sugar in the grapes is caramelized in the process, nothing is added. Moreover, the only thing removed from the grapes is the water that evaporates during the drying, and water contains no calories or nutrients. The fact that raisins contain more iron per food calorie than grapes do is thus puzzling. Which one of the following, if true, most helps to explain why raisins contain more iron per calorie than do grapes? (A) Since grapes are bigger than raisins, it takes several bunches of grapes to provide the same amount of iron as a handful of raisins does. (B) Caramelized sugar cannot be digested, so its calories do not count toward the food calorie content of raisins. (C) The body can absorb iron and other nutrients more quickly from grapes than from raisins because of the relatively high water content of grapes. (D) Raisins, but not grapes, are available year-round, so many people get a greater share of their yearly iron intake from raisins than from grapes. (E) Raisins are often eaten in combination with other iron-containing foods, while grapes are usually eaten by themselves. Look at that question stem: a quick scan naturally shows you that you need to explain/resolve a paradox, but the question goes into even more detail for you. It reaffirms the exact nature of the paradox its not about iron, but instead that that raisins contain more iron per calorie than grapes do. By adding that extra description into the question stem, the testmaker is practically yelling at you, Make sure you consider caloriesdont just focus on iron! And therefore, you should be prepared for  the  correct answer B, the only one that addresses calories, and deftly avoid answers A, C, D, and E, which all focus only on iron (and do so tangentially to the paradox). Strategically speaking, if a Critical Reasoning question stem gets overly specific, you should pay particular attention to the specificity thereits most likely directing you to the operative portion of the argument. Overly specific questions are most helpful in Data Sufficiency questions (and that same logic will help on Problem Solving too, as youll see). The testmaker knows that youve trained your entire algebraic life to solve for individual variables. So how can a question author use that lifetime of repetition against you? By asking you to solve for a specific combination that doesnt require you to find the individual values. Consider this example, which appears courtesy the Official Guide for GMAT Quantitative Review: If x^2 + y^2 = 29, what is the value of (x y)^2? (1) xy = 10 (2) x = 5 Two major clues should stand out to you that you need to Leverage Assets on this problem. For one, using both statements together (answer choice C) is dead easy. If xy = 10 and x = 5 then y = 2 and you can solve for any combination of x and y that anyone could ever ask for. But secondly and more subtly, the question stem should jump out as a classic way-too-specific, Leverage Assets question stem. They asked for a really, really specific value: (x y)^2. Now, immediately upon seeing that specificity you should be thinking, Thats too specifictheres probably a way to solve for that exact value without getting x and y individually. That thought process alone tells you where to spend your time you want to really leverage Statement 1 to try to make it work alone. And if youre still unconvinced, consider what the specificity does: the squared portion removes the question of negative vs. positive from the debate, removing one of the most common reasons that a seemingly-sufficient statement just wont work. And, furthermore, the common quadratic (x y)^2 shares an awful lot in common with the x^2 and y^2 elsewhere in the question stem. If you expand the parentheses, you have What is x^2 2xy + y^2? meaning that youre already 2/3 of the way there (so to speak), since theyve spotted you the sum x^2 + y^2. The important strategy here is that the overly-specific question stem should scream LEVERAGE ASSETS and You dont need to solve for x and ytheres probably a way to solve directly for that exact combination. Since you know that youre solving for the expanded x^2 2xy + y^2, and you already know that x^2 + y^2 = 29, youre really solving for 29 2xy. Since you know from Statement 1 that xy = 20, then 29 2xy will be 29 2(10), which is 9. Statement 1 alone is sufficient, even though you dont know what x and y are individually. And one of the major signals that you should recognize to help you get there is the presence of an overly specific question stem. So remember, in a world of generic questions, the oddly specific question should arouse a bit of suspicion: the interrogator is up to something! On the GMAT, you can use that to your advantage an overly specific Critical Reasoning question usually tells you exactly which keywords are the most important, and an overly specific Data Sufficiency question stem begs for you to leverage assets and find a way to get the most out of each statement. Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on  Facebook, YouTube,  Google+  and Twitter! By Brian Galvin.

Sustainability and Its Impact on the Four Managerial...

Sustainability and Its Impact on the Four Managerial Functions Nowadays people all over Australia use their own environmentally friendly shopping bags when they go to the Supermarket. Why have so many people changed their every day habit and stopped using the convenient plastic bags? The transition from plastic bags to recycled ones is only one aspect of a global attitude change towards environmental topics. During the last few decades, businesses have been focusing on lowering their production costs, with no considerations on the consequences for the upcoming years. Today, organizations all over the globe are starting to change the way they operate in order to prevent an environmental crisis. Sustainability is the new approach that†¦show more content†¦In prior years, while managers used to set goals for their organization, their only consideration was how to maximize revenues, without paying attention to the created waste during the manufacturing process and the operation of the organization. Nowadays, managers need to ensure that the manufacturing process is sustainable. The Lean Manufacturing (Womack, Jones Roos, 1990) term is a managerial philosophy which tries to eliminate waste as much as possible during the manufacturing process. For example maximize use and reuse of raw materials, decrease the amount of excess supply and defective products, effective shipping handling and consecutive manufacturing process. The second factor managers must relate to is reduction in the amount of waste in the organizations building itself. For instance total consumption of water and energy should be minimized as well. Then, the organizations board has to structure its vision, and set specific goals relating to environment issues, for the upcoming years. For illustration, for more than half a century The Coca Cola Company had put its emphasis on protecting the environment. In 1956, The Coca Cola Company set a goal of protecting our world and environment. Since then the company conducts researches about environmental issues consistently, and is considered as a revolution maker in regards to environment and Sustainability topics. In addition, in 1991 the first beverage bottle was made fromShow MoreRelatedThe Modern Era Of Social Responsibility1523 Words   |  7 Pagesbetter understand the relationship between CSR and ethics. Literature Review The primary CSR functions are as follows: to cause no harm, prevent harm, and do good. These functions can be implemented through several dimensions such as ethics, legal transparency and accountability, socio-economic developments, and environmental sustainability. Heinz, for example, has demonstrated environmental sustainability efforts including a reduction in gas emissions, solid waste, energy and water consumption, andRead MoreManaging A Successful Business Management1244 Words   |  5 Pagesthem out to achieve organizational goals. Managers have four main functions they carry out in the management world. Managers plan which requires setting goals, and decide how to accomplish them. They need to organize which means they arrange tasks, people, and other resources to accomplish the work. They also control which means they monitor performance, compare it with goals, and take corrective action as needed. Another function managers do is lead. They motivate, direct, and influenceRead MoreManaging A Successful Business Management1244 Words   |  5 Pagesthem out to achieve organizational goals. Managers have four main functions they carry out in the management world. Managers plan which requires setting goals, and decide how to accomplish them. They need to organize which means they arrange tasks, people, and other resources to accomplish the work. They also control which means they monitor performance, compare it with goals, and take corrective action as needed. Another function managers do is lead. They motivate, direct, and influenceRead MoreCorporate Ethics And Social Responsibility1528 Words   |  7 Pagesqualitative paradigm needs to be used in order to ensure in-depth knowledge and understanding of the issues and challenges among business practices and how they can be handled. 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Associational Thinking Innovators use associational thinking, where they â€Å"cross-pollinate ideas in their own heads and in others. They connect wildly different ideas, objects, services, technologies, and disciplines to

Mass Incarceration A New Form Of Slavery Essay - 1555 Words

Mass Incarceration: A New Form of Slavery in the United States Lorena P. Ambriz History 12A Abstract Starting in the 1970s, the rising rate of imprisonment came to be known as Mass Incarceration. What was once an average of 100 people getting imprisoned for every 100,000 adults, prior to the 1970s, has now grown to become more than 600 individuals per every 100,000 adults imprisoned. With only five present of the total world population, The United States holds an astonishing 25 percent of the world’s prisoners. From 1980 to 2013, the number of people who are in United States prisons and jails has risen from a little over five hundred thousand to over two million people (Kilgore, 2015). The amount of people in prisons has risen more than 450% than what it was prior to the 1980s, despite the fact that crime in the United States has declined steeply since the 1990s. Why and how did this occur? What role does Mass Incarceration play in our Society? Mass Incarceration: A New Form of Slavery in the United States The Bigger Picture In 1865, the United Sates abolished slavery with the Implementation of the 13th Amendment to the United States Constitution. While it brought a major victory to many Americans at the time, it was not without its flaws. Stating that â€Å"Neither Slavery nor involuntary servitude, except as punishment for crime whereof the party shall have been duly convicted, shall exist within the United State, nor any place subject to their jurisdiction,† the 13thShow MoreRelatedHow Mass Incarceration Has Become The New Form Of Jim Crow And Slavery Essay1422 Words   |  6 Pagesnation there has always been a racial caste systems due to slavery, money, and greed. The End of slavery was after the civil war and enfourced through the 13th Amendment. The loophole that was created that was the exception that criminals can be treated as a involuntary servitude, which was noted in the U.S constitution. To speed things along you have the slavery which transferred to convicted leasing to Jim Crow Er a and now Mass Incarceration which all has striped millions of the people, whom are inRead MoreThe New Jim Crow By Michelle Alexander1313 Words   |  6 Pages The New Jim Crow Michelle Alexander’s the new Jim Crow Mass Incarceration in the Age of Colorblindness examine the Jim Crow practices post slavery and the mass incarceration of African-American. The creation of Jim Crows laws where used as a tool to promote segregation among the minority and white American. Michelle Alexander’s the new Jim Crow Mass takes a look at Jim Crow laws and policies were put into place to block the social progression African-American from the post-slavery to the civilRead MoreThe New Jim Crow By Michelle Alexander1316 Words   |  6 Pages The New Jim Crow Michelle Alexander’s the new Jim Crow Mass Incarceration in the Age of Colorblindness examine the Jim Crow practices post slavery and the mass incarceration of African-American. The creation of Jim Crows laws were used as a tool to promote segregation among the minority and white American. Michelle Alexander’s the new Jim Crow Mass takes a look at Jim Crow laws and policies were put into place to block the social progression African-American from the post-slavery to the civilRead MoreWacquant - From Slavery to Mass Incarceration - Critique and Reflection1394 Words   |  6 PagesFrom Slavery to Mass Incarceration: Necessary Extremes Of the supplementary readings provided, I found â€Å"From Slavery to Mass Incarceration† by Loà ¯c Wacquant the most intriguing. This particular article is based on â€Å"rethinking the ‘race question’ in the US† and the disproportionate institutions set apart for African Americans in the United States. The volatile beginnings of African Americans presented obvious hardships for future advancement, but Wacquant argues that they still suffer from a formRead MoreThe New Jim Crow?919 Words   |  4 PagesAlexander, the author of The New Jim Crow, did not see the prison systems as racially motivated until doing further research. After researching the issue, Alexander found the prison system was a way to oppress African Americans and wrote the novel The New Jim Crow. The New Jim Crow follows the history of the racial caste system and in the novel Alexander comes to the conclusion that the mass incarceration of African American is the New Jim Crow, or in othe r words a new system of black oppression.Read MoreMass Incarceration : The Color Of Justice Essay1352 Words   |  6 PagesMass Incarceration: The Color of Justice (DRAFT) Racial discrimination in the United States has been a radical issue plaguing African Americans from as early as slavery to the more liberal society we see today. Slavery is one of the oldest forms of oppression against African Americans. Slaves were brought in from Africa at increasingly high numbers to do the so-called dirty work or manual labor of their white owners. Many years later, after the abolishment of slavery came the Jim Crow era. In theRead MoreIs The Mass Incarceration Of Blacks The New Jim Crow?1540 Words   |  7 PagesIs the Mass Incarceration of Blacks the new Jim Crow? American has a legacy of the mistreatment and disenfranchisement of African Americans. The same bad treatment that many think only took place in the past is in fact still intact, it’s just presented in a new way. The mass incarceration of blacks in the Unites States can be attributed to the â€Å"racial hierarchy† that has always existed. The U.S contributes to about 5% of the worlds overall population, and about 25% of the worlds prison populationRead MoreThe New Jim Crow : Mass Incarceration1361 Words   |  6 PagesBook Review Michelle Alexander, The New Jim Crow: Mass Incarceration in the Age of Colorblindness The premise of the ‘The New Jim Crow: Mass Incarceration in the Age of Colorblindness’ by Michelle Alexander, is to refute claims that racism is dead and argue that the War on Drugs and the federal drug policy unfairly targets communities of color, keeping a large majority of black men of varying ages in a cycle of poverty and behind bars. The author proves that racism thrives by highlighting theRead MoreThe United States Criminal Justice System Essay1132 Words   |  5 PagesConstitution which outlawed slavery unless you are being punished for a crime. The film focuses on racism in the United States’ criminal justice system. According to DuVernay, the part of the 13th Amendment that says â€Å"unless you are being punished for a crime† is a loophole that has been used to allow slavery to continue in the early days during reconstruction and even now. This loophole coupled with the criminalization of the black man has led to mass incarceration of minorities. The United StatesRead MoreThe New Jim Crow : Mass Incarceration Essay795 Words   |  4 PagesThe video we were asked to write a reflection on discussed The New Jim Crow: Mass Incarceration in the Age of Colorblindness which is a book written by Michelle Alexander a highly acclaimed civil rights lawyer, advocate and Associate Professor of Law at Oh io State University. Michelle Alexander states that although we made tremendous progress with Civil Rights Movement in the 1950s by unifying as a race and fought to seemingly ended the old Jim Crow era by the passing of laws such as the 1965 voting

12 Monkeys and Societys Perception of Drugs Essay

Illegal Drugs vs. Psych Ward Drugs The definition of drugs ranges from supposedly positive to automatically classified as negative. But who is to say which drug is bad and which drug is considered good. In the movie entitled 12 Angry Men the director Terry Gilliam used this film to bring about an idea that I had to rethink. This movie was released on the 27 of December in 1995, this movies genre is often questionable but for the most part it is an Action, Thriller, Sci-Fi. The two characters that the drugs affected the most were the main character James Cole played by Bruce Willis and another important character Jeffrey played by Brad Pit. Our society often contradicts themselves when it comes to distributing and using drugs†¦show more content†¦While the scientists on the cast question his trip one of the questions I would like to point out is Cole is asked, How was it? Lots of drugs? The definition of a Drug according to dictionary.com is A chemical substance, such as a narcotic or hallucinogen, that affects the central nervous system, causing changes in behavior and often addiction and A substance used in the diagnosis, treatment, or prevention of a disease or as a component of a medication. Being a controlled substance we automatically classify drugs as an illegal act that is not often accepted. In our society today you can go to jail for possession and possession with intent to supply. Our generation is receiving years in jail or the misuse and overdose of drugs. Drugs are abused everyday by teens and adults and often looked down upon. Numerous accidents have occurred do to the abuse of some drugs. Drugs in our society today are initially thought of as bad, and then we come to the understanding that there are other types of drugs that may in fact not be bad but used for good. Illegal drugs are portrayed in our society as really bad, but the use of drugs is abused in psych wards and other facilities also. From previous research I learned that Thorazine is an antipsychotic drug, used as a major tranquilizers. So it is ok for our government and other higher authority to allow the use of tranquilizers on humans andShow MoreRelatedStephen P. Robbins Timothy A. Judge (2011) Organizational Behaviour 15th Edition New Jersey: Prentice Hall393164 Words   |  1573 PagesManager, Production: Lisa Rinaldi Full-Service Project Management: Christian Holdener, S4Carlisle Publishing Services Composition: S4Carlisle Publishing Services Printer/Binder: Courier/Kendallville Cover Printer: Courier/Kendalville Text Font: 10.5/12 ITC New Baskerville Std Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on the appropriate page within text. 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Analysis Of The Movie The Great Gatsby By F. Scott...

Friendly Carrie was more of a quiet person who never did anything to anyone. She worked hard on her grades. She wasn’t a very outgoing person. Carrie’s hair was dyed pink, she usually wore ripped jeans and plain t-shirts. Alison on the other hand was the total opposite of Carrie. Alison was known as the school bully. She never did her homework, and she was loud and outgoing and not to mention very mean. Alison had black hair and usually wore shorts and t-shirts. Both Carrie and Alison had blue eyes. One day Carrie was walking down the hallway, they hallway had white walls and white tiled floor with one colorful tiled tile. Alison was walking towards her. When Alison got closer to Carrie she started bullying her by saying very mean and offensive things to her like â€Å"You’re stupid, you’re ugly and no one likes you!’’ Carrie didn’t say anything to her just ignored her. When Alison saw Carrie had no reaction she pushed Carrie into a wall. Carrie hit the back of her head on the locker and fell to her knees afterwards. She cried because of the paint it caused. Months went by and Alison’s bullying got worse verbally and physically. One day it got to Carrie and she finally stood up for herself, she wasn’t going to get bullied into silence. The next day Alison came up to her and before she could say anything Carrie said in a firm voice â€Å"What is your problem? I’ve never done anything to you. Why are you bullying me? If something is going on at home talk to someone about it.’’Show MoreRelatedAnalysis Of The Movie The Great Gatsby By F. Scott Fitzgerald1186 Words   |  5 PagesLong beaten out by the glaring sun, Noctis and his friends stood on outside of the car. Prompto, who was leaning on the front wheel, gawked at the young woman coming up to them. 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Research Project for Earthworms in Earthworm- myassignmenthelp

Question: Discuss about theResearch Project for Earthworms in Earthworm Farming. Answer: Feasibility of reusing coffee powder waste as feed for earthworms in earthworm farming The most significant issues encountered by todays world are the ability to manage waste that is affecting the public health and the environment. As commented by Murthy Naidu (2012), coffee powder is the most widely produced waste that is produced on a daily basis. This is because of an increased rate of coffee consumption among the population in Singapore. It is estimated that there are hundreds of coffee shops around Singapore that produce an average of 15 kg coffee waste daily. Therefore, tonnes of coffee powders are wasted per month in Singapore. Using natural fertilizers for the crops is essential to promote growth and level of nutrition within them. Disposing of the coffee powder in the natural environment adversely affects the public health and environment. Therefore, it is feasible to reuse the coffee powder waste for vermin composting, as worms enjoy similar food habits as humans (Pathma Sakthivel, 2012). As mentioned by Yuan (2016), worms recycle the coffee powders into nutrients for the plants. Regular composting produces healthy food for the plants whereas compost by the earthworms helps in enriching the compost even more. After feeding on the coffee powders, the earthworms excrete castings that help in providing sufficient nutrients to the garden. Coffee powders help in attracting worms similarly as traditional composting piles used to do. The coffee powders are considered as a tasty addition to the products that are used for compost production. Thus, coffee powder helps in digestion for the earthworms while vermin composting. An adequate amount of coffee powder needs to be added while compost production because an excessive amount of coffee powder makes the compost acidic. The best environment for the worms ranges from 6.0-8.0. Excessive acidic nature of the bed can burn the skin of the worms thereby, hampering the compost formation (Makkar et al. 2014). The research suggests that the use of coffee powder helps in enriching the property of the compost used for agriculture. However, the feasibility of reusing the coffee powder needs attention, as a huge amount of coffee powder is produced as waste daily in Singapore. The practicality of the objective is presented by SMART objective below: Specific- The objective is specific because reusing the coffee powder waste will enhance the quality of the compost used for agriculture. Enriched quality of the vermin compost will help in better plant growth, as the plants will have access to better nutritional values. Additionally, reusing the wasted coffee powders will also help in decreasing the adverse environmental impact. Therefore, the reusing coffee powder will help in mitigating the adverse environmental impact and enhancing the quality of crops. Measurable- The objective is measurable because the use of the amount of waste coffee powder in Singapore can be compared between the past and the present. Comparing the amount of coffee powder used in vermin compost in the past and in the present will give an idea of the feasibility of reusing the waste for earthworm culture. Tools such as SQL data compare and various statistical tools can measure the data (Sanchez-Hernandez Dominguez, 2017). Additionally, studying the reports of the company that uses coffee powder as vermin compost will help in measuring the feasibility. Attainable- The objective is attainable because successful completion of the objective contributes largely to the environment and the agricultural field in Singapore. The practicality of the objective will help in reducing the environmental pollution caused by general disposal of coffee powder. Realistic- The objective is realistic because it is appropriate for the caffeine industry as well as the environment. Judging the feasibility of reusing coffee powder will help in evaluating the available resources that can be used in attaining the objectives. It has to be seen whether there are sufficient and adequate resources available in Singapore that can be used for collecting the daily coffee powder waste. This will provide an overall idea of the amount of coffee powder that is used in vermin compost thereby, enriching the quality of the compost (Chen et al., 2015). Time-specific- The objective is time specific because reusing the wasted coffee powder 1 month of the collection will increase the feasibility of the objective. Therefore, the wasted coffee powder produced daily needs to be utilized within 1 month of collection by the compost producing companies. Time Frame Activities 1st week Collecting the coffee powder waste 2nd week Mixing the coffee powder with other composts and creating the bed for the earthworms 3rd week Wait for the earthworms to use the coffee waste 4th week Production of enriched compost through vermin compost References Chen, Y., Zhang, Y., Zhang, Q., Xu, L., Li, R., Luo, X., ... Tong, J. (2015). Earthworms modify microbial community structure and accelerate maize stover decomposition during vermicomposting. Environmental Science and Pollution Research, 22(21), 17161-17170. Makkar, H. P., Tran, G., Heuz, V., Ankers, P. (2014). State-of-the-art on use of insects as animal feed. Animal Feed Science and Technology, 197, 1-33. Murthy, P. S., Naidu, M. M. (2012). Sustainable management of coffee industry by-products and value additionA review. Resources, Conservation and recycling, 66, 45-58. Pathma, J., Sakthivel, N. (2012). Microbial diversity of vermicompost bacteria that exhibit useful agricultural traits and waste management potential. SpringerPlus, 1(1), 26. Sanchez-Hernandez, J. C., Domnguez, J. (2017). Vermicompost derived from spent coffee grounds: assessing the potential for enzymatic bioremediation. Yuan, Y. (2016). Vermicomposting, waste recycling and plant growth (Doctoral dissertation, Lincoln University).