Write a program to decipher messages encoded using a prefix code, gi.pdf
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Write a program to decipher messages encoded using a prefix code, given the encoding tree. Such codes are widely used in applications that compress data, including JPEG for images, MP3 for music, and DivX for video. using Huffman Trees (Prefix Binary Trees) c++ Solution #include #include #include #include // for CHAR_BIT #include #include const int UniqueSymbols = 1 << CHAR_BIT; const char* SampleString = \"this is an example for huffman encoding\"; typedef std::vector HuffCode; typedef std::map HuffCodeMap; class INode { public: const int f; virtual ~INode() {} protected: INode(int f) : f(f) {} }; class InternalNode : public INode { public: INode *const left; INode *const right; InternalNode(INode* c0, INode* c1) : INode(c0->f + c1->f), left(c0), right(c1) {} ~InternalNode() { delete left; delete right; } }; class LeafNode : public INode { public: const char c; LeafNode(int f, char c) : INode(f), c(c) {} }; struct NodeCmp { bool operator()(const INode* lhs, const INode* rhs) const { return lhs->f > rhs->f; } }; INode* BuildTree(const int (&frequencies)[UniqueSymbols]) { std::priority_queue, NodeCmp> trees; for (int i = 0; i < UniqueSymbols; ++i) { if(frequencies[i] != 0) trees.push(new LeafNode(frequencies[i], (char)i)); } while (trees.size() > 1) { INode* childR = trees.top(); trees.pop(); INode* childL = trees.top(); trees.pop(); INode* parent = new InternalNode(childR, childL); trees.push(parent); } return trees.top(); } void GenerateCodes(const INode* node, const HuffCode& prefix, HuffCodeMap& outCodes) { if (const LeafNode* lf = dynamic_cast(node)) { outCodes[lf->c] = prefix; } else if (const InternalNode* in = dynamic_cast(node)) { HuffCode leftPrefix = prefix; leftPrefix.push_back(false); GenerateCodes(in->left, leftPrefix, outCodes); HuffCode rightPrefix = prefix; rightPrefix.push_back(true); GenerateCodes(in->right, rightPrefix, outCodes); } } int main() { // Build frequency table int frequencies[UniqueSymbols] = {0}; const char* ptr = SampleString; while (*ptr != \'\\0\') ++frequencies[*ptr++]; INode* root = BuildTree(frequencies); HuffCodeMap codes; GenerateCodes(root, HuffCode(), codes); delete root; for (HuffCodeMap::const_iterator it = codes.begin(); it != codes.end(); ++it) { std::cout << it->first << \" \"; std::copy(it- >second.begin(), it->second.end(), std::ostream_iterator(std::cout)); std::cout << std::endl; } return 0; }.
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![Write a program to decipher messages encoded using a prefix code, given the encoding tree.
Such codes are widely used in applications that compress data, including JPEG for images, MP3
for music, and DivX for video. using Huffman Trees (Prefix Binary Trees) c++
Solution
#include #include #include #include // for CHAR_BIT #include #include const int
UniqueSymbols = 1 << CHAR_BIT; const char* SampleString = "this is an example for
huffman encoding"; typedef std::vector HuffCode; typedef std::map HuffCodeMap; class
INode { public: const int f; virtual ~INode() {} protected: INode(int f) : f(f) {} };
class InternalNode : public INode { public: INode *const left; INode *const right;
InternalNode(INode* c0, INode* c1) : INode(c0->f + c1->f), left(c0), right(c1) {}
~InternalNode() { delete left; delete right; } }; class LeafNode :
public INode { public: const char c; LeafNode(int f, char c) : INode(f), c(c) {} };
struct NodeCmp { bool operator()(const INode* lhs, const INode* rhs) const { return lhs->f
> rhs->f; } }; INode* BuildTree(const int (&frequencies)[UniqueSymbols]) {
std::priority_queue, NodeCmp> trees; for (int i = 0; i < UniqueSymbols; ++i) {
if(frequencies[i] != 0) trees.push(new LeafNode(frequencies[i], (char)i)); }
while (trees.size() > 1) { INode* childR = trees.top(); trees.pop();
INode* childL = trees.top(); trees.pop(); INode* parent = new
InternalNode(childR, childL); trees.push(parent); } return trees.top(); } void
GenerateCodes(const INode* node, const HuffCode& prefix, HuffCodeMap& outCodes) {
if (const LeafNode* lf = dynamic_cast(node)) { outCodes[lf->c] = prefix; }
else if (const InternalNode* in = dynamic_cast(node)) { HuffCode leftPrefix =
prefix; leftPrefix.push_back(false); GenerateCodes(in->left, leftPrefix,
outCodes); HuffCode rightPrefix = prefix; rightPrefix.push_back(true);
GenerateCodes(in->right, rightPrefix, outCodes); } } int main() { // Build
frequency table int frequencies[UniqueSymbols] = {0}; const char* ptr =
SampleString; while (*ptr != '0') ++frequencies[*ptr++]; INode* root =
BuildTree(frequencies); HuffCodeMap codes; GenerateCodes(root, HuffCode(),
codes); delete root; for (HuffCodeMap::const_iterator it = codes.begin(); it !=
codes.end(); ++it) { std::cout << it->first << " "; std::copy(it-
>second.begin(), it->second.end(), std::ostream_iterator(std::cout)); std::cout
<< std::endl; } return 0; }](https://image.slidesharecdn.com/writeaprogramtodeciphermessagesencodedusingaprefixcodegi-230707211456-a6424bab/85/Write-a-program-to-decipher-messages-encoded-using-a-prefix-code-gi-pdf-1-320.jpg)
Recommended
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
Sysprog17
The 17th session in eglug's system programming course, I only attend the session, the slides are not written by me
Implementing a basic directory-tree structure that is derived from a.pdf
Implementing a basic directory-tree structure that is derived from a base tree structure.
#ifndef SRC_TREENODE_HPP_
#define SRC_TREENODE_HPP_
#include
#include
//#define USE_OSS
#ifndef USE_OSS
#define OUT std::cout
#else
#include
extern std::ostringstream oss;
#define OUT oss
#endif
namespace ensc251 {
static unsigned count = 0;
class TreeNode;
typedef TreeNode* (TreeNode::*action_func)(void);
typedef void (TreeNode::*traverse_func)(const TreeNode*);
class TreeNode
{
typedef std::vector TreeNodePVect;
protected:
TreeNodePVect childPVector;
public:
TreeNode() {};
TreeNode(const TreeNode& nodeToCopy) {
// ***** this needs work *****
if (nodeToCopy.childPVector.size())
add_childP(nodeToCopy.childPVector[0]->clone());
}
virtual TreeNode* clone() const { return new TreeNode(*this); };
virtual ~TreeNode() {
// do not modify this insertion on OUT
OUT << \"Destroying TreeNode with \" << childPVector.size() << \" children.\"<< std::endl;
// ***** Complete this member function *****
}
void swap(TreeNode & other) // the swap member function (should never fail!)
{
// ***** fill this in if desired *****
}
TreeNode & operator = (TreeNode other) // note: argument passed by value, copy made!
{
// ***** Complete this code *****
return *this; // by convention, always return *this
}
void add_childP(TreeNode* child) { /* ***** Complete this member function ***** */ }
void add_children(const TreeNodePVect& childPV) { childPVector.insert(childPVector.end(),
childPV.begin(), childPV.end()); }
const TreeNodePVect& get_children() const { return childPVector; }
TreeNode* traverse_children_post_order(traverse_func tf, action_func af)
{
for(auto childP : childPVector) {
(childP->*tf)(this); // traverse child\'s children using tf
}
return (this->*af)();
}
TreeNode* traverse_children_pre_order(traverse_func tf, action_func af)
{
// ***** Complete this member function *****
}
TreeNode* traverse_children_in_order(traverse_func tf, action_func af)
{
// ***** Complete this member function *****
}
TreeNode* count_action()
{
count++;
return nullptr;
}
void count_traverse(const TreeNode*)
{
traverse_children_post_order(count_traverse, count_action);
}
void reset_count() { count = 0; }
unsigned get_count() const { return count; }
};
}
#endif /* SRC_TREENODE_HPP_ */
#include \"makeDirectoryTree.hpp\"
using namespace std;
int doStuff()
{
// call function to build directory tree in memory
auto root_dir1P = make_directory_tree(\".\", \"test_dir\");
if (root_dir1P) {
root_dir1P->print_traverse(nullptr);
OUT << endl;
root_dir1P->reset_count();
root_dir1P->count_traverse(nullptr);
OUT << \"Count is \" << root_dir1P->get_count() << \"\ \" << endl;
auto root_dir2 = Directory(*root_dir1P); // copy construction
root_dir2.print_traverse(nullptr);
OUT << endl;
auto root_dir3P = make_directory_tree(\"test_dir\", \"Lectures\");
if (root_dir3P) {
root_dir2 = *root_dir3P; // overloaded copy assignment operation
OUT << endl;
delete root_dir3P;
OUT << endl;
}
root_dir2.print_traverse(nullptr);
OUT.
ANSimport java.util.Scanner; class Bina_node { Bina_node .pdf
ANS:
import java.util.Scanner;
class Bina_node
{
Bina_node l_1, r_1;
int dt;//here declare variables
public Bina_node()//constructor
{
l_1 = null;
r_1 = null;
dt = 0;
}
public Bina_node(int n)//here parameter constractor
{
l_1 = null;
r_1 = null;
dt = n;
}
public void nin_left(Bina_node n)//copy constractor
{
l_1 = n;
}
public void nin_right(Bina_node n)
{
r_1 = n;
}
public Bina_node get_lstf()
{
return l_1;
}
public Bina_node get_go()
{
return r_1;
}
public void set_data(int d)
{
dt = d;
}
public int get_data()
{
return dt;
}
}
class Bina_tree//class banary tree
{
private Bina_node rf;
public Bina_tree()
{
rf = null;
}
public boolean isEmpty()//here empty or not
{
return rf == null;
}
public void tee_insert(int dt)
{
rf = tee_insert(rf, dt);
}
private Bina_node tee_insert(Bina_node node, int dt)
{
if (node == null)
node = new Bina_node(dt);
else
{
if (node.get_go() == null)
node.r_1 = tee_insert(node.r_1, dt);
else
node.l_1 = tee_insert(node.l_1, dt);
}
return node;
}
public int cou_Node()
{
return cou_Node(rf);
}
private int cou_Node(Bina_node r)
{
if (r == null)
return 0;
else
{
int l = 1;
l += cou_Node(r.get_lstf());
l += cou_Node(r.get_go());
return l;
}
}
public boolean bin_serc(int tree_val)//function called
{
return bin_serc(rf, tree_val);
}
private boolean bin_serc(Bina_node r, int tree_val)
{
if (r.get_data() == tree_val)
return true;
if (r.get_lstf() != null)
if (bin_serc(r.get_lstf(), tree_val))
return true;
if (r.get_go() != null)
if (bin_serc(r.get_go(), tree_val))
return true;
return false;
}
public void in_order()
{
in_order(rf);
}
private void in_order(Bina_node r)
{
if (r != null)
{
in_order(r.get_lstf());
System.out.print(r.get_data() +\" \");
in_order(r.get_go());
}
}
public void pre_order()
{
pre_order(rf);
}
private void pre_order(Bina_node r)
{
if (r != null)
{
System.out.print(r.get_data() +\" \");
pre_order(r.get_lstf());
pre_order(r.get_go());
}
}
public void post_order()
{
post_order(rf);
}
private void post_order(Bina_node r)
{
if (r != null)
{
post_order(r.get_lstf());
post_order(r.get_go());
System.out.print(r.get_data() +\" \");
}
}
}
public class Bina_Tree//binary tree class
{
public static void main(String[] args)//here main methode
{
Scanner sc = new Scanner(System.in);//here scanner
Bina_tree bin_c = new Bina_tree();
System.out.println(\"Binary Tree \ \");
char ch;
do
{
System.out.println(\"\ here Binary Tree Operations\ \");
System.out.println(\"1 for insert 2 for search 3 for count 4 for check\");//here choice
numbers
int cho = sc.nextInt();
int k;
switch (cho)
{
k++;
case 1 :System.out.println(\"please enter insert ele\");
bin_c.tee_insert( sc.nextInt() );
break;
case 2 :System.out.println(\"enter bin_search\");//here search option
System.out.println(\"result\"+ bin_c.bin_serc( sc.nextInt() ));//result
break;
case 3 :System.out.println(\"Nodes\"+ bin_c.cou_Node());
break;
case 4 :System.out.println(\"Empty\"+ bin_c.isEmpty());
break;
default ://here default condition
System.out.println(\"enter 1 TO 4 only.
Recommended
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
Sysprog17
The 17th session in eglug's system programming course, I only attend the session, the slides are not written by me
Implementing a basic directory-tree structure that is derived from a.pdf
Implementing a basic directory-tree structure that is derived from a base tree structure.
#ifndef SRC_TREENODE_HPP_
#define SRC_TREENODE_HPP_
#include
#include
//#define USE_OSS
#ifndef USE_OSS
#define OUT std::cout
#else
#include
extern std::ostringstream oss;
#define OUT oss
#endif
namespace ensc251 {
static unsigned count = 0;
class TreeNode;
typedef TreeNode* (TreeNode::*action_func)(void);
typedef void (TreeNode::*traverse_func)(const TreeNode*);
class TreeNode
{
typedef std::vector TreeNodePVect;
protected:
TreeNodePVect childPVector;
public:
TreeNode() {};
TreeNode(const TreeNode& nodeToCopy) {
// ***** this needs work *****
if (nodeToCopy.childPVector.size())
add_childP(nodeToCopy.childPVector[0]->clone());
}
virtual TreeNode* clone() const { return new TreeNode(*this); };
virtual ~TreeNode() {
// do not modify this insertion on OUT
OUT << \"Destroying TreeNode with \" << childPVector.size() << \" children.\"<< std::endl;
// ***** Complete this member function *****
}
void swap(TreeNode & other) // the swap member function (should never fail!)
{
// ***** fill this in if desired *****
}
TreeNode & operator = (TreeNode other) // note: argument passed by value, copy made!
{
// ***** Complete this code *****
return *this; // by convention, always return *this
}
void add_childP(TreeNode* child) { /* ***** Complete this member function ***** */ }
void add_children(const TreeNodePVect& childPV) { childPVector.insert(childPVector.end(),
childPV.begin(), childPV.end()); }
const TreeNodePVect& get_children() const { return childPVector; }
TreeNode* traverse_children_post_order(traverse_func tf, action_func af)
{
for(auto childP : childPVector) {
(childP->*tf)(this); // traverse child\'s children using tf
}
return (this->*af)();
}
TreeNode* traverse_children_pre_order(traverse_func tf, action_func af)
{
// ***** Complete this member function *****
}
TreeNode* traverse_children_in_order(traverse_func tf, action_func af)
{
// ***** Complete this member function *****
}
TreeNode* count_action()
{
count++;
return nullptr;
}
void count_traverse(const TreeNode*)
{
traverse_children_post_order(count_traverse, count_action);
}
void reset_count() { count = 0; }
unsigned get_count() const { return count; }
};
}
#endif /* SRC_TREENODE_HPP_ */
#include \"makeDirectoryTree.hpp\"
using namespace std;
int doStuff()
{
// call function to build directory tree in memory
auto root_dir1P = make_directory_tree(\".\", \"test_dir\");
if (root_dir1P) {
root_dir1P->print_traverse(nullptr);
OUT << endl;
root_dir1P->reset_count();
root_dir1P->count_traverse(nullptr);
OUT << \"Count is \" << root_dir1P->get_count() << \"\ \" << endl;
auto root_dir2 = Directory(*root_dir1P); // copy construction
root_dir2.print_traverse(nullptr);
OUT << endl;
auto root_dir3P = make_directory_tree(\"test_dir\", \"Lectures\");
if (root_dir3P) {
root_dir2 = *root_dir3P; // overloaded copy assignment operation
OUT << endl;
delete root_dir3P;
OUT << endl;
}
root_dir2.print_traverse(nullptr);
OUT.
ANSimport java.util.Scanner; class Bina_node { Bina_node .pdf
ANS:
import java.util.Scanner;
class Bina_node
{
Bina_node l_1, r_1;
int dt;//here declare variables
public Bina_node()//constructor
{
l_1 = null;
r_1 = null;
dt = 0;
}
public Bina_node(int n)//here parameter constractor
{
l_1 = null;
r_1 = null;
dt = n;
}
public void nin_left(Bina_node n)//copy constractor
{
l_1 = n;
}
public void nin_right(Bina_node n)
{
r_1 = n;
}
public Bina_node get_lstf()
{
return l_1;
}
public Bina_node get_go()
{
return r_1;
}
public void set_data(int d)
{
dt = d;
}
public int get_data()
{
return dt;
}
}
class Bina_tree//class banary tree
{
private Bina_node rf;
public Bina_tree()
{
rf = null;
}
public boolean isEmpty()//here empty or not
{
return rf == null;
}
public void tee_insert(int dt)
{
rf = tee_insert(rf, dt);
}
private Bina_node tee_insert(Bina_node node, int dt)
{
if (node == null)
node = new Bina_node(dt);
else
{
if (node.get_go() == null)
node.r_1 = tee_insert(node.r_1, dt);
else
node.l_1 = tee_insert(node.l_1, dt);
}
return node;
}
public int cou_Node()
{
return cou_Node(rf);
}
private int cou_Node(Bina_node r)
{
if (r == null)
return 0;
else
{
int l = 1;
l += cou_Node(r.get_lstf());
l += cou_Node(r.get_go());
return l;
}
}
public boolean bin_serc(int tree_val)//function called
{
return bin_serc(rf, tree_val);
}
private boolean bin_serc(Bina_node r, int tree_val)
{
if (r.get_data() == tree_val)
return true;
if (r.get_lstf() != null)
if (bin_serc(r.get_lstf(), tree_val))
return true;
if (r.get_go() != null)
if (bin_serc(r.get_go(), tree_val))
return true;
return false;
}
public void in_order()
{
in_order(rf);
}
private void in_order(Bina_node r)
{
if (r != null)
{
in_order(r.get_lstf());
System.out.print(r.get_data() +\" \");
in_order(r.get_go());
}
}
public void pre_order()
{
pre_order(rf);
}
private void pre_order(Bina_node r)
{
if (r != null)
{
System.out.print(r.get_data() +\" \");
pre_order(r.get_lstf());
pre_order(r.get_go());
}
}
public void post_order()
{
post_order(rf);
}
private void post_order(Bina_node r)
{
if (r != null)
{
post_order(r.get_lstf());
post_order(r.get_go());
System.out.print(r.get_data() +\" \");
}
}
}
public class Bina_Tree//binary tree class
{
public static void main(String[] args)//here main methode
{
Scanner sc = new Scanner(System.in);//here scanner
Bina_tree bin_c = new Bina_tree();
System.out.println(\"Binary Tree \ \");
char ch;
do
{
System.out.println(\"\ here Binary Tree Operations\ \");
System.out.println(\"1 for insert 2 for search 3 for count 4 for check\");//here choice
numbers
int cho = sc.nextInt();
int k;
switch (cho)
{
k++;
case 1 :System.out.println(\"please enter insert ele\");
bin_c.tee_insert( sc.nextInt() );
break;
case 2 :System.out.println(\"enter bin_search\");//here search option
System.out.println(\"result\"+ bin_c.bin_serc( sc.nextInt() ));//result
break;
case 3 :System.out.println(\"Nodes\"+ bin_c.cou_Node());
break;
case 4 :System.out.println(\"Empty\"+ bin_c.isEmpty());
break;
default ://here default condition
System.out.println(\"enter 1 TO 4 only.
20-13 Programming Project #05A Given the IntNode class- define the Fin.docx
20.13 Programming Project #05A
Given the IntNode class, define the FindMax() function to return the largest value in the list or -99 if the list is empty. Assume all values in the list are non-negative.
Ex: If the list contains:
FindMax(headNode) returns 191.
Ex: If the list contains:
FindMax(headNode) returns -99.
**Please make sure returns -99 and not -1**
Given code:
main.cpp
#include <iostream>
using namespace std;
class IntNode {
public:
// Constructor
IntNode(int dataInit);
// Get node value
int GetNodeData();
// Get pointer to next node
IntNode* GetNext();
/* Insert node after this node.
Before: this -- next
After: this -- node -- next
*/
void InsertAfter(IntNode* newNode);
private:
int dataVal;
IntNode* nextNodePtr;
};
// Constructor
IntNode::IntNode(int dataInit) {
this->dataVal = dataInit;
nextNodePtr = nullptr;
}
// Get node value
int IntNode::GetNodeData() {
return this->dataVal;
}
// Get pointer to next node
IntNode* IntNode::GetNext() {
return this->nextNodePtr;
}
/* Insert node after this node.
Before: this -- next
After: this -- node -- next
*/
void IntNode::InsertAfter(IntNode* newNode) {
IntNode* tempNext = this->nextNodePtr;
this->nextNodePtr = newNode;
newNode->nextNodePtr = tempNext;
}
// Return largest value in the list
int FindMax(IntNode* headNode) {
/* Type your code here. */
}
int main() {
IntNode* headNode = new IntNode(-1);
IntNode* currNode;
IntNode* lastNode;
int max;
// Initiaize head node
lastNode = headNode;
// Add nodes to the list
for (int i = 0; i < 20; ++i) {
currNode = new IntNode(i);
lastNode->InsertAfter(currNode);
lastNode = currNode;
}
max = FindMax(headNode);
cout << "Max is " << max;
return 0;
}
.
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Point.h header file #ifndef POINT_H #define POINT_H Po.pdf
//Point.h
//header file
#ifndef POINT_H
#define POINT_H
//Point class
class Point
{
//declare data members
private:
int x;
int y;
int z;
public:
Point();
//Mutator member functions
void setX(int) ;
void setY(int) ;
void setZ(int) ;
//Accessor member functions
int getX() const;
int getY() const;
int getZ() const;
};
#endif
--------------------------------------------------------------------------------
//Point.cpp
//Implementation of Point.h header file
#include \"Point.h\"
Point::Point()
{
x=0;
y=0;
z=0;
}
void Point::setX(int x)
{
this->x=x;
}
void Point::setY(int y)
{
this->y=y;
}
void Point::setZ(int z)
{
this->z=z;
}
int Point::getX() const
{
return x;
}
int Point::getY() const
{
return y;
}
int Point::getZ() const
{
return z;
}
--------------------------------------------------------------------------------
//Point.cpp
//header files
#include
//include Point.h
#include \"Point.h\"
using namespace std;
int main()
{
//declaration of Point object
Point p;
//Create a reference(pointer) that stores the address of the variable of Point class
Point *reference=&p;
//calling mutator methods to set x,y and z values
reference->setX(11);
reference->setY(7);
reference->setZ(9);
cout<<\"x = \"<
Solution
//Point.h
//header file
#ifndef POINT_H
#define POINT_H
//Point class
class Point
{
//declare data members
private:
int x;
int y;
int z;
public:
Point();
//Mutator member functions
void setX(int) ;
void setY(int) ;
void setZ(int) ;
//Accessor member functions
int getX() const;
int getY() const;
int getZ() const;
};
#endif
--------------------------------------------------------------------------------
//Point.cpp
//Implementation of Point.h header file
#include \"Point.h\"
Point::Point()
{
x=0;
y=0;
z=0;
}
void Point::setX(int x)
{
this->x=x;
}
void Point::setY(int y)
{
this->y=y;
}
void Point::setZ(int z)
{
this->z=z;
}
int Point::getX() const
{
return x;
}
int Point::getY() const
{
return y;
}
int Point::getZ() const
{
return z;
}
--------------------------------------------------------------------------------
//Point.cpp
//header files
#include
//include Point.h
#include \"Point.h\"
using namespace std;
int main()
{
//declaration of Point object
Point p;
//Create a reference(pointer) that stores the address of the variable of Point class
Point *reference=&p;
//calling mutator methods to set x,y and z values
reference->setX(11);
reference->setY(7);
reference->setZ(9);
cout<<\"x = \"<.
PVS-Studio 5.00, a solution for developers of modern resource-intensive appl...
PVS-Studio is a static analyzer that detects errors in source code of C, C++, C++11, C++/CX applications.
Describe 3 functions of the cellular membrane and how are these func.pdf
Describe 3 functions of the cellular membrane and how are these functions accomplished by the
chemical make up of the membrane.
Solution
The main functions of the cell membrane are:to maintain the physical integrity of the cell,that is
to mechanically enclose the contents of the cell,also to control the movement of particles e.g.
ions or molecules,into and out of the cell.It also isolates the cytoplasm from the external
environment.
The hydrophilic head and hydrophobic tails of the phospholipid molecules that form the
phospholipid bilayer of the cell membrane forming its physical/mechanical structure.The ion
channels responsible for the selective permeability of the cell membrane.Membrane pumps that
enable active transport of specific molecules across the cell membrane.Carrier proteins and
receptor proteins that transport specific substance across the cell membrane and receive
(chemical) signals from outside the cell that tell the cell to take a specific
action,respectively.Cholesterol molecules and carbohydrate chains that support the structure of
the membrane and its attachment to other tissues,respectively.Surface protein markers that
identify the cell.Membrane proteins acts as a recptor to combine with specific substance such as
hormones while some form narrow passage ways or channels through which various molecules
and ions can cross the cell membrane.Enzymatic proteins carry out metabolic reactions..
Discuss the three main types of intellectual capitalSolutionth.pdf
Discuss the three main types of intellectual capital
Solution
the main types of intellectual capitals are: Copy rights, patents, trade marks and trade secrets.
copy rights: gives the right to the inventor or author or a book or artistick work to his invention.
book publishing and movie distribution comes under this example.
patents: gives the inventor to sell the right to use his invention for commercial and for that he
will charge something from the user. the user will get licence to use it and will be benefited.
manufacturing of McDonald and Dominos pizzas comes under this action.
trademark: is facilitate the seller or producer to differentitate his prodcut from his competitors
product and will be benefited. is like giving licence to sell the products in a region for a
particular period. the producer will allows the seller to sell the product and will get some benefit..
More Related Content
Similar to Write a program to decipher messages encoded using a prefix code, gi.pdf
20-13 Programming Project #05A Given the IntNode class- define the Fin.docx
20.13 Programming Project #05A
Given the IntNode class, define the FindMax() function to return the largest value in the list or -99 if the list is empty. Assume all values in the list are non-negative.
Ex: If the list contains:
FindMax(headNode) returns 191.
Ex: If the list contains:
FindMax(headNode) returns -99.
**Please make sure returns -99 and not -1**
Given code:
main.cpp
#include <iostream>
using namespace std;
class IntNode {
public:
// Constructor
IntNode(int dataInit);
// Get node value
int GetNodeData();
// Get pointer to next node
IntNode* GetNext();
/* Insert node after this node.
Before: this -- next
After: this -- node -- next
*/
void InsertAfter(IntNode* newNode);
private:
int dataVal;
IntNode* nextNodePtr;
};
// Constructor
IntNode::IntNode(int dataInit) {
this->dataVal = dataInit;
nextNodePtr = nullptr;
}
// Get node value
int IntNode::GetNodeData() {
return this->dataVal;
}
// Get pointer to next node
IntNode* IntNode::GetNext() {
return this->nextNodePtr;
}
/* Insert node after this node.
Before: this -- next
After: this -- node -- next
*/
void IntNode::InsertAfter(IntNode* newNode) {
IntNode* tempNext = this->nextNodePtr;
this->nextNodePtr = newNode;
newNode->nextNodePtr = tempNext;
}
// Return largest value in the list
int FindMax(IntNode* headNode) {
/* Type your code here. */
}
int main() {
IntNode* headNode = new IntNode(-1);
IntNode* currNode;
IntNode* lastNode;
int max;
// Initiaize head node
lastNode = headNode;
// Add nodes to the list
for (int i = 0; i < 20; ++i) {
currNode = new IntNode(i);
lastNode->InsertAfter(currNode);
lastNode = currNode;
}
max = FindMax(headNode);
cout << "Max is " << max;
return 0;
}
.
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Point.h header file #ifndef POINT_H #define POINT_H Po.pdf
//Point.h
//header file
#ifndef POINT_H
#define POINT_H
//Point class
class Point
{
//declare data members
private:
int x;
int y;
int z;
public:
Point();
//Mutator member functions
void setX(int) ;
void setY(int) ;
void setZ(int) ;
//Accessor member functions
int getX() const;
int getY() const;
int getZ() const;
};
#endif
--------------------------------------------------------------------------------
//Point.cpp
//Implementation of Point.h header file
#include \"Point.h\"
Point::Point()
{
x=0;
y=0;
z=0;
}
void Point::setX(int x)
{
this->x=x;
}
void Point::setY(int y)
{
this->y=y;
}
void Point::setZ(int z)
{
this->z=z;
}
int Point::getX() const
{
return x;
}
int Point::getY() const
{
return y;
}
int Point::getZ() const
{
return z;
}
--------------------------------------------------------------------------------
//Point.cpp
//header files
#include
//include Point.h
#include \"Point.h\"
using namespace std;
int main()
{
//declaration of Point object
Point p;
//Create a reference(pointer) that stores the address of the variable of Point class
Point *reference=&p;
//calling mutator methods to set x,y and z values
reference->setX(11);
reference->setY(7);
reference->setZ(9);
cout<<\"x = \"<
Solution
//Point.h
//header file
#ifndef POINT_H
#define POINT_H
//Point class
class Point
{
//declare data members
private:
int x;
int y;
int z;
public:
Point();
//Mutator member functions
void setX(int) ;
void setY(int) ;
void setZ(int) ;
//Accessor member functions
int getX() const;
int getY() const;
int getZ() const;
};
#endif
--------------------------------------------------------------------------------
//Point.cpp
//Implementation of Point.h header file
#include \"Point.h\"
Point::Point()
{
x=0;
y=0;
z=0;
}
void Point::setX(int x)
{
this->x=x;
}
void Point::setY(int y)
{
this->y=y;
}
void Point::setZ(int z)
{
this->z=z;
}
int Point::getX() const
{
return x;
}
int Point::getY() const
{
return y;
}
int Point::getZ() const
{
return z;
}
--------------------------------------------------------------------------------
//Point.cpp
//header files
#include
//include Point.h
#include \"Point.h\"
using namespace std;
int main()
{
//declaration of Point object
Point p;
//Create a reference(pointer) that stores the address of the variable of Point class
Point *reference=&p;
//calling mutator methods to set x,y and z values
reference->setX(11);
reference->setY(7);
reference->setZ(9);
cout<<\"x = \"<.
PVS-Studio 5.00, a solution for developers of modern resource-intensive appl...
PVS-Studio is a static analyzer that detects errors in source code of C, C++, C++11, C++/CX applications.
Similar to Write a program to decipher messages encoded using a prefix code, gi.pdf (20)
Automatically Describing Program Structure and Behavior (PhD Defense)
Automatically Describing Program Structure and Behavior (PhD Defense)
20-13 Programming Project #05A Given the IntNode class- define the Fin.docx
20-13 Programming Project #05A Given the IntNode class- define the Fin.docx
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Data Structures and Agorithm: DS 14 Binary Expression Tree.pptx
Point.h header file #ifndef POINT_H #define POINT_H Po.pdf
Point.h header file #ifndef POINT_H #define POINT_H Po.pdf
PVS-Studio 5.00, a solution for developers of modern resource-intensive appl...
PVS-Studio 5.00, a solution for developers of modern resource-intensive appl...
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Describe 3 functions of the cellular membrane and how are these func.pdf
Describe 3 functions of the cellular membrane and how are these functions accomplished by the
chemical make up of the membrane.
Solution
The main functions of the cell membrane are:to maintain the physical integrity of the cell,that is
to mechanically enclose the contents of the cell,also to control the movement of particles e.g.
ions or molecules,into and out of the cell.It also isolates the cytoplasm from the external
environment.
The hydrophilic head and hydrophobic tails of the phospholipid molecules that form the
phospholipid bilayer of the cell membrane forming its physical/mechanical structure.The ion
channels responsible for the selective permeability of the cell membrane.Membrane pumps that
enable active transport of specific molecules across the cell membrane.Carrier proteins and
receptor proteins that transport specific substance across the cell membrane and receive
(chemical) signals from outside the cell that tell the cell to take a specific
action,respectively.Cholesterol molecules and carbohydrate chains that support the structure of
the membrane and its attachment to other tissues,respectively.Surface protein markers that
identify the cell.Membrane proteins acts as a recptor to combine with specific substance such as
hormones while some form narrow passage ways or channels through which various molecules
and ions can cross the cell membrane.Enzymatic proteins carry out metabolic reactions..
Discuss the three main types of intellectual capitalSolutionth.pdf
Discuss the three main types of intellectual capital
Solution
the main types of intellectual capitals are: Copy rights, patents, trade marks and trade secrets.
copy rights: gives the right to the inventor or author or a book or artistick work to his invention.
book publishing and movie distribution comes under this example.
patents: gives the inventor to sell the right to use his invention for commercial and for that he
will charge something from the user. the user will get licence to use it and will be benefited.
manufacturing of McDonald and Dominos pizzas comes under this action.
trademark: is facilitate the seller or producer to differentitate his prodcut from his competitors
product and will be benefited. is like giving licence to sell the products in a region for a
particular period. the producer will allows the seller to sell the product and will get some benefit..
Describer in detail the innate immune response to both an intercellu.pdf
Describer in detail the innate immune response to both an intercellular infection and an
extracellular infection.
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Innate immunity is the type of immunity which is developed by birth..! The defense against the
extracellur infection are from the barrier which are present on the body like skin, hairs in the
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Then the intra cellular immunity like cytokines, inflammatory barriers like histamine which
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Describe the difference between tropic and non-tropic hormones.S.pdf
Describe the difference between tropic and non-tropic hormones.
Solution
TROPIC HORMONES : Tropic hormones are those hormones which act on other endocrine
hormonesthey are named as respective tropin.
Ex: adreno cartico tropic hormone (acth) is a tropic hormone which acts on the cells of adrenal
cortex and stimulate the synthesis of glucocorticoid hormones.similarly,Thyroid stimulating
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C++ ProgrammingGivenWrite a closest Pair FunctionDeliverable.pdf
C++ Programming
Given:
Write a closest Pair Function:
Deliverables
ClosestPair.h
ClosestPair.cpp
Solution
code:
Pair Closest_Pair( Point P[ ], int n)
{
float min = FLT_MAX;
int indexone=0;
int indextwo=0;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
{
min = dist(P[i], P[j]);
indexone=i;
indextwo=j;
}
Struct Pair st ;
st.indexPointOne = indexone;
st.indexPointTwo = indextwo;
return min;
}
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
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The learning curve equation is defined as follows : -
Tn = T1 nb
Tn is the production time for the nth unit
T1 is the production time for the first unit
n = number of units
b is slope of the learning curve and is defined as b = log(learning rate) / log2
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Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
fusion glycoproteins. Experimental exposure to low pH (5.5) buffer will inactivate IFV, but not
MV. Based on this information, What can you infer about the entry pathways and mechanisms of
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Both influenza virus and (IFV), and Measles Virus (MV) are enveloped viruses with class I
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Solution
Influenza virus is an enveloped, single-stranded negative-sense RNA virus belonging to the
Orthomyxoviridae family. Influenza A virions have three integral membrane proteins: an ion
channel protein (M2) and two glycoproteins, hemagglutinin (HA), required for entry into host
cells, and neuraminidase (NA), involved in the release of progeny virions from the host cell.
Underneath the membrane, most virions have a layer of M1 matrix protein enclosing the viral
ribonucleoprotein complexes (RNPs). On exposure to low pH of around 5 the viral RNP\'s are
released. It is followed by the M2-mediated pumping of protons into the virion, this induces the
change in the intermolecular interactions of the M1 matrix protien and its binding to the viral
membrane, this leads to membrane fusion and conformational changes in the hemagglutinin
causing them to be inactive because this conformational change causes exposure of a
hydrophobic part of the molecule. Fusion of the virion with target membrane occurs at low pH
where a transition from a relaxed conformation to an active conformation of HA occurs, this
transition at low pH in absence of a target membrane leads to inactivation.
The measles virus is a paramyxovirus, is 120–250 nm in diameter, with a core of single-stranded
RNA. It has Two membrane envelope proteins are important in pathogenesis. They are the F
(fusion) protein, responsible for fusion of virus and host cell membranes, viral penetration, and
hemolysis, and the H (hemagglutinin) protein, which is responsible for adsorption of virus to
cells. measles virus could not be inactivated at pH range of 5.5 to 9.0. It is inactivated at pH of
<3 or >7Attachmentnt to host cell happens through binding of virus to the cellular receptor and
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Solution
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A cancer biology topic: Describte how ChIP (chromatin immunoprecipitation), DNA
methylation array, and gene expression arrays are performed. What are the similarities?
Solution
DNA methylation is main epigenetic mechanism of gene expression regulation.Alteration in this
leads to tumor. Various stimulant factors stimulate DNA methylation leads to generation of
canceroua cells.
DNA methylation process do down regulation of gene.
Methylation of CpG sites blocks the binding of regulatory proteins resulting transcription
modulation.
In CHIP DNA methylation stauninig of CHIP done by various stains. Stained CHIP show
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ratio between methylated and unmethylated group.
Value of ratio is 0, 1, 0.5 represents non methylation, total methylation and one copy methylated
other not respectively.
DNA methylation measure by following other method also
Bisulfite treatment
Whole genomic DNA amplification
Hybridization and single base extension
In Gene expression two following procees take important part:-
TRANSCRIPTION:- DNA to RNA by RNA polymerase which use DNA as template and
produce RNA.
TRANSLATION:- RNA to Protein translation (tRNA charged with amino acid enter in ribosome
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In cancer, gene promoter contain CpG island with numerous CpG sites. When CpG
sitesethylated gene become silence.
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You get a call from a customer who reinstalled Windows on his comput.pdf
You get a call from a customer who reinstalled Windows on his computer after a crash.
Fortunately, he had recently backed up all this data to an external USB drive and was able to
restore everything on his NTFS-formatted file system. However, he had a number of encrypted
files that he had backed up along with the rest of his data that he can no longer access. What
happened?
A. NTFS-encrypted files cannot be backed up, and when the computer crashed, those files were
lost forever. When the customer tried to restore these files, only corrupted versions were
restored.
B. NTFS-encrypted files are not restored with the original encryption key used before the backup
process and thus become inaccessible.
C. When the customer reinstalled Windows, even though he used his old username, the computer
assigned him a completely different security identifier (SID). The restored NTFS-encrypted files
are looking for his original SID.
D. When the customer reinstalled Windows, he didn’t use his original username. The restored
NTFS-encrypted files are looking for the original username.
Solution
C. NTFS encryption used the customer’s SID to associate with the encrypted file. Now that the
customer reinstalled Windows, a completely new SID was assigned to the customer. Without the
original SID, he can’t access the encrypted files. A third-party decryption application may help
him access his encrypted data..
You are running the Windows installation setup for Windows 7 on a co.pdf
You are running the Windows installation setup for Windows 7 on a computer. When the
Windows Setup screen appears, what do you notice?
A. The Setup screen has been drastically redesigned from Windows Vista.
B. The Setup screen is almost exactly the same as Windows Vista.
C. The Setup screen hasn’t changed since Windows XP.
D. Unlike Windows Vista, the Setup screen immediately launches into the GUI.
Solution
B. The Windows 7 Setup screen is virtually the same as the screen in Windows Vista, both
beginning in a black character-mode screen and then launching the GUI..
With the biological species concept, the process of speciation is fre.pdf
With the biological species concept, the process of speciation is frequently defined as the
evolution of between populations. reproductive isolation genetic drift gene flow reproductive
inclusion geographic separation The phylogenetic species concept looks at __. environmental
adaptations and morphology species response to adverse conditions and genetic data genetic data
and environmental adaptations morphology and environmental adaptations morphology and
genetic data A phylogenetic species comprise __. interbreeding population that are
reproductively isolated individuals that share measurable traits that distinguish them populations
that share a recent evolutionary history individuals that share a recent evolutionary history one
large gene pool subdivided into local populations
Solution
Reproductive isolation
According to biological species concept, species is a group of actually or potentially
interbreeding individuals who are reproductively isolated from other such groups..
Write a class ArrayList that represents an array of integers. Init.pdf
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
Write a class ArrayList that represents an array of integers. Initially it has a capacity of 1 and is
“empty”.
At any time, it can be partially full, so it keeps track of its current occupied size, as well as its
capacity
(true size).
However the output for the prime numbers 1 -50 is incorrect. Displaying numbers like 0, 625 for
prime numbers.
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a: Default constructor used to unitialize the array with capacity=1
ArrayList::ArrayList()
{
asize = 0;
capacity = 1;
array = new int[capacity];
}
// This function returns the actual size of array
int ArrayList::size()
{
return asize;
}
// 1.b:The overloaded [] oprtaor return element at ith position in array
int & ArrayList::operator [](unsigned int i)
{
if(i >= asize)
{
cout << \"invalid reference \";
exit(0);
}
return *(array + i);
}
//1.c: This function push m at the end of the array
void ArrayList::push_back(int m)
{
// this condition checks if size is beyond capacity or not. if so double the capacity.
if(asize == capacity)
{
capacity = 2 * capacity;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
array[asize] = m;
asize = asize + 1;
}
void ArrayList::erase(int m){
int pos = -1;
for(int i = 0; i < asize; i++){
if(array[i] == m){
pos = i;
break;
}
}
if(pos == -1){
cout << \"Element not present\ \";
}
else{
for(int i = pos; i < asize - 1; i++){
array[i] = array[i + 1];
}
}
asize--;
if(asize == capacity / 2)
{
capacity = capacity / 2;
int *temp = new int[capacity];
for(int i = 0; i < asize; i++)
temp[i] = array[i];
array = new int[capacity];
for(int i = 0; i < asize; i++)
array[i] = temp[i];
}
}
string ArrayList::toString(){
string str = \"[\";
for(int i = 0; i < asize; i++){
ostringstream oss;
oss << array[i];
str += oss.str();
if(i != asize - 1){
str += \", \";
}
}
str += \"]\";
return str;
}
//---------------------------
//You can test above class by using main() below
int main()
{
ArrayList arr; // call ro default constructor
for (int i = 1; i <= 50; i++)
{
arr.push_back(i); // call to push_back()
}
cout << \"Should contain numbers 1..50, is \";
cout << arr.toString();
cout << \"\ \";
return 0;
}
This is my code with the main method below
#include
#include
#include
using namespace std;
class ArrayList
{
private:
int *array;
int capacity, asize;
public:
int size();
ArrayList();
int & operator [](unsigned int);
void push_back(int);
void erase(int m);
string toString();
};
// 1.a:.
what is the relationship between lnx and 1xSolutionThe relat.pdf
what is the relationship between lnx and 1/x?
Solution
The relation between ln x and (1/x) is explained in terms of definite and indefinte integrals.
Indefinite integral:
Int (1/x) dx + C = lnx +C.
Definite integral:
{Int (1/t) dt from t = 0 to t = x} = lnx.
Or in other words ln (x) is the area under the curve f(x) = ln(t) between the ordinates t= 0 to t =
x, which could be calculated by evaluation of the definite integral Int (1/t)dt from t = 0 to t =x..
Which of these conclusions is supported by plant phylogenies (evolut.pdf
Which of these conclusions is supported by plant phylogenies (evolutionary trees)?
Question 4 options:1)It is likely that the transition to land occurred directly from red algae.2)In
all of the trees generated so far, land plants trace back to more than one ancestral group.3)The
oldest lineages of land plants are descended from some green algae (Charophytes)4)The closest
relative to land plants is fungi
Solution
Answer is option 3) The oldest lineages of land plants are descended from some green algae
(Charophytes)
Land plants evolbed from green algae approximately 510 million years ago. The Charophytes are
still present..
Who are some other individuals who might want to use the informa.pdf
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Who are some other individuals who might want to use the information in the financial
statements
Solution
Some Other Individuals who might use the information in financial statements
1) Suppliers - Suppliers are interested in the information of financial Statements. Suppliers get to
know about solvency and liquidity of company to pay its short term obligations.
2) Employees- Employees use the financial statement information to analyse about the
company\'s ability to pay salary and benefits. Others reasons of using financial statement
information are career development, Financial performance.
3) Customers - Customers use the financial statement information for assessing the company
expansion possibilities, its stability of operations, Its compliances review, for investment
decisions.
Which account below is not a subdivision of owners equityAccumu.pdf
Which account below is not a subdivision of owner\'s equity?
Accumulated Comprehensive Income.
Bond Sinking fund.
Addition Paid-in-Capital.
Retained Earnings.
Accumulated Comprehensive Income.
Bond Sinking fund.
Addition Paid-in-Capital.
Retained Earnings.
Solution
Bonk Sinking Fund not a subdivision of owner\'s equity
Parts of Stockholder Equity are
Capital Stock, Addition Paid-in-Capital, Retained Earnings., Accumulated Comprehensive
Income, Treasury Stock, NONCONTROLLING INTEREST (MINORITY INTEREST).
What is the nucleoid region of a prokaryotic cell a membrane bound .pdf
What is the nucleoid region of a prokaryotic cell? a membrane bound organelle that contains
the DNA a region within the nucleus where parts of ribosomes are made a place within the cell
where DNA is more abundant a place where you\'d expect to find a proton and a neutron
Solution
A place with the cell where DNA is more abundant
The nucleoid refers to an irregularly shaped region within a prokaryotic cell that contains the
genetic material..
What is heartwood What is sapwood What is the most abundant tissu.pdf
What is heartwood? What is sapwood? What is the most abundant tissue in the three-year-old
stem? What are three contrasting differences in structure between the sunflower stem and the
TNia stem? What are two techniques that can be used to determine the age of a woody twig?
Solution
1. Heart wood is the dead and inner wood tissue produced by cork cambium. It is dark in colr
and is distinguished from sap wood by its color and texture.
Sapwood is the outermost living wood tissue. It is also produced by cambium but is composed of
different types of cells. It is lightin color. It is involved conduction of water and minerals from
roots to all parts of tree.
2. In a three year old stem, the secondary xylem and phloem are bundant than primary xylem and
phloem. These vascular tissues conspicuously appear as annual rings by which the age of the tree
is known.
3.
4.
The age of a woody twig can be determined by counting the number of groups of bud scale scars
on it and he the number of terminal bud scale scars present.SunflowerTiliaAnnualperennialherb
exhibits no secondary growthtree exhibits secondary growthVascular bundles form a ring. Xylem
lies inside the vascular bundles. Phloem lies outside and phloem fibers are tough. The center of
stem is filled with ground tissue known as pithOne year old stem exhibits no secondary growth
and contains a layer of xylem and two later meristems called vascular and cork cambium.
During second year growth 2 rings of xylem is observed. During third year secondary xylem is
obseved and secondary phloem towards inside to xylem..
Use arrays to store data for analysis. Use functions to perform the .pdf
Use arrays to store data for analysis. Use functions to perform the analysis described below. All
output is to be written to an output file.
Solution
import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
public class DNA{
private String dna;
public DNA(){
}
public void read_store(String filename){
/*try {
Scanner in;
//in = new Scanner(new FileReader(filename));
dna = in.next();
}catch(IOException e){
System.out.println(\"file not there \" + e);
return;
}*/
}
public char compliment_base(char c){
if(c == \'A\') return \'T\';
if(c == \'T\') return \'A\';
if(c == \'G\') return \'C\';
if(c == \'C\') return \'G\';
return \' \';
}
public void DrawDNA(String seq){
System.out.println(\"---+-----+------+------+------\");
for (char c : seq.toCharArray() ) {
System.out.print(c + \" \");
}
System.out.println(\"DNA Sequence \");
System.out.println(\"................................\");
for (char c : seq.toCharArray() ) {
System.out.print(compliment_base(c) + \" \");
}
System.out.println(\" Complement \");
System.out.println(\"---+-----+------+------+------\");
System.out.println(\"\");
}
public void Prefix(int n, String base){
String output = base.substring(0,n);
DrawDNA(output);
}
public static void main(String[] args) {
DNA d = new DNA();
d.DrawDNA(\"ATGC\");
d.Prefix(4,\"ACGTTGCA\");
}
}
---+-----+------+------+------
A T G C DNA Sequence
................................
T A C G Complement
---+-----+------+------+------
---+-----+------+------+------
A C G T DNA Sequence
................................
T G C A Complement
---+-----+------+------+------.
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