C++ ProgrammingGivenWrite a closest Pair FunctionDeliverable.pdf

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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) ); }.

$C++ Programming Given: Write a closest Pair Function: Deliverables ClosestPair.h ClosestPair.cpp Solution code: Pair Closest_Pair( Point P[ ], int n) { float min = FLT_MAX; int indexone=0; int indextwo=0; for (int i = 0; i < n; ++i) for (int j = i+1; j < n; ++j) if (dist(P[i], P[j]) < min) { min = dist(P[i], P[j]); indexone=i; indextwo=j; } Struct Pair st ; st.indexPointOne = indexone; st.indexPointTwo = indextwo; return min; } float dist(Point p1, Point p2) { return sqrt( (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y) ); }$

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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 get a call from a customer who reinstalled Windows on his comput.pdf

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You get a call from a customer who reinstalled Windows on his comput.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..

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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; }.

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, gi.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..

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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:.

Write a class ArrayList that represents an array of integers. Init.pdf

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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..

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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..

Which of these conclusions is supported by plant phylogenies (evolut.pdf

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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.

Who are some other individuals who might want to use the informa.pdf

Who are some other individuals who might want to use the informa.pdf

Who are some other individuals who might want to use the informa.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).

Which account below is not a subdivision of owners equityAccumu.pdf

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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 the nucleoid region of a prokaryotic cell a membrane bound .pdf

What is the nucleoid region of a prokaryotic cell a membrane bound .pdf

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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..

What is heartwood What is sapwood What is the most abundant tissu.pdf

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What is heartwood What is sapwood What is the most abundant tissu.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 ---+-----+------+------+------.

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 .pdf

Use arrays to store data for analysis. Use functions to perform the .pdf

Using a telephone survey, a team of sociologists collect data on education and income for 154 individuals. Let xi be the highest grade completed by the ith respondent (example xi =14 means that ith respondent completed 2 years of college). Let yi be the income of the ith respondent. The sociologists have computed the following quantities after running the simple linear regression to predict the income of the individual by the highest grade completed: http://imgur.com/sdszLUM Solution.

Using a telephone survey, a team of sociologists collect data on edu.pdf

Using a telephone survey, a team of sociologists collect data on edu.pdf

Using a telephone survey, a team of sociologists collect data on edu.pdf

The “Tripwire” system calculates all the hashes for all executable files on a clean computer and then compares the hashes with the actual files, when the machine is scanned, to ensure nothing has changed and no malware is hiding as legitimate programs. 1) What kind of attacks will not be detectable by such a scheme? 2) What attack will render the above strategy useless (i.e. will not find infected files, even if they exist). Solution The near-ubiquitous UNIX system is an example of a file system where such monitoring is useful. Flaws and weaknesses in typical UNIX systems are well-documented (e.g., [8, 25, 19, 4, 9]). UNIX file systems are susceptible to threats in the guise of unauthorized users, intruders, viruses, worms, and logic bombs as well as failures and bugs. As such, UNIX system administrators are faced with prospects of subtle, difficult-to-detect damage to files, malicious and accidental. Tripwire is an integrity checking tool designed for the UNIX environment to aid system administrators to monitor their file systems for unauthorized modifications. First made available on November 2, 1992, it has proven to be a popular tool, being portable, configurable, scalable, flexible, manageable, automatable, and secure. It was written in response to repeated break-in activity on the Internet, and the difficulty experienced by affected administrators in finding all of the “backdoors” left by the intruders. To aid in the detection of the appropriate threats, system administrators would use an integrity checker to monitor file systems for added, deleted, and changed files. Meaningfully reporting changed files is difficult, because most files are expected to change: system log files are written to, program sources are updated, and documents are revised. Typically, these changes would not concern system administrators. However, changes to certain files, such as system binaries, might elicit a different reaction. The database used by the integrity checker should be protected from unauthorized modifications; an intruder who can change the database can subvert the entire integrity checking scheme. Although the system administrator can secure the database by storing it on some media inaccessible to remote intruders (e.g., paper printout), usability is sacrificed. A database stored in some machine readable format may risk unauthorized modification, but allows the integrity checking process to be automated. Storing the database on read-only media provides the best of both approaches, allowing machine access but preventing changes. Most available UNIX security tools fall into two categories: static audit tools and integrity checkers. Among the most prominent are COPS[8], TAMU[22], crc check[8], Hobgoblin[15], and ATP[28].5 A few commercial security tools also exist, but they are comparable to the user-community tools mentioned here. While many of these tools may be outstanding in their own right, most are mismatches for integrity checking in UNIX env.

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The “Tripwire” system calculates all the hashes for all executable f.pdf

The archaeal RNA viruses described in the introduction were single-stranded plus-sense viruses, like many of the viruses described in the textbook chapter (e.g., phage MS2, poliovirus, and coronaviruses). They had the genes for RNA replicase and some appeared to use polyproteins. Based on your knowledge of other viral life cycles, what are the most likely lifecycles for these archaeal RNA viruses? Solution RNA viruses are those viruses in which the sole nucleic acid present is RNA, mostly single stranded or some viruses have double-stranded RNA. The single stranded RNA can be (+) sense or (-) sense. The RNA viruses do not have DNA intermediates in their life cycle as compared to retroviruses. Most likely lifecycle for archaeal RNA virus.

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\"Molecular clock\" studies by Sarich and Wilson have suggested that humans and the African apes diverged from a common ancestor around 2 to 3 million years ago. 5 to 8 million years ago. 15 to 20 million years ago. 50 million year ago. 10 to 15 million years ago. Which of the following is true of the gibbon? they are the smallest of the apes. they live mainly in central Africa. they live in multi-male, multi-female groups in the rainforest. all of the above Old World monkeys are characterized by all of the following EXCEPT bilophodont molars. quadrupedal molars quadrupedal locomotion, with all four limbs around the same length. a 2-1- 2-3 dental formula. a grasping, or prehensile, tail. Solution Q. 19. 4 to 5 million years ago Q. 20 they are the smallest of the apes - true They live mainly in Central Africa - false. They live in Southeast Asia They live in multi male and multi female groups in rainforest - false. They live in small families in the rainforest- (monogamous) Q 21. They do not have a grasping and prehensile tail.

Molecular clock studies by Sarich and Wilson have suggested that .pdf

Molecular clock studies by Sarich and Wilson have suggested that .pdf

Molecular clock studies by Sarich and Wilson have suggested that .pdf

Limestone rocks, coral reefs, and forests are examples of Select one a. carbon sources b. carbon sinks c. carbon credits d. geo-sequestration. Differences in the climates of Venus, Earth, and Mars can be best explained by Select one: a. the distances of each planet from the Sun b. the chemical make-up of their atmospheres c. the percentage of carbon dioxide in their atmospheres d. the distances of each planet from the Sun and the chemical make-up of their atmospheres Which of the following statements is supported by the data presented in the graph? Select one a. The planet closest to the Sun has the warmest temperature b. Mars is too far from the Sun to support life c. Distance from the Sun alone is not sufficient to explain the variation in temperatures of the planets d. Because it is the warmest of the planets, Venus is likely to support life Solution 1. carbon sinks 2. the distances of each planet from Sun and their chemical makeup of their atmospheres. Because of its thin atmosphere and greater distance from Sun, the surface temperature is colder for Mars. however, it experiences temperature fluctuations. 3. distance from Sun alone is not sufficient to explain the variation in temperatures of the planets. 3..

Limestone rocks, coral reefs, and forests are examples of Select one.pdf

Limestone rocks, coral reefs, and forests are examples of Select one.pdf

Limestone rocks, coral reefs, and forests are examples of Select one.pdf

I have the first program completed (not how request, but it works) and I am trying to get the second portion together. First Program requested (complete)********************************************** (1) Create three files to submit: ItemToPurchase.h - Class declaration ItemToPurchase.cpp - Class definition main.cpp - main() function Build the ItemToPurchase class with the following specifications: Default constructor Public class functions (mutators & accessors) SetName() & GetName() (2 pts) SetPrice() & GetPrice() (2 pts) SetQuantity() & GetQuantity() (2 pts) Private data members string itemName - Initialized in default constructor to \"none\" int itemPrice - Initialized in default constructor to 0 int itemQuantity - Initialized in default constructor to 0 (2) In main(), prompt the user for two items and create two objects of the ItemToPurchase class. Before prompting for the second item, call cin.ignore() to allow the user to input a new string. (2 pts) Ex: (3) Add the costs of the two items together and output the total cost. (2 pts) Ex: Second Portion I am on now************************************************************************** ******* 7.19 Program: Online shopping cart (continued) (C++) This program extends the earlier \"Online shopping cart\" program. (Consider first saving your earlier program). (1) Extend the ItemToPurchase class per the following specifications: Parameterized constructor to assign item name, item description, item price, and item quantity (default values of 0). (1 pt) Public member functions SetDescription() mutator & GetDescription() accessor (2 pts) PrintItemCost() - Outputs the item name followed by the quantity, price, and subtotal PrintItemDescription() - Outputs the item name and description Private data members string itemDescription - Initialized in default constructor to \"none\" Ex. of PrintItemCost() output: Ex. of PrintItemDescription() output: (2) Create three new files: ShoppingCart.h - Class declaration ShoppingCart.cpp - Class definition main.cpp - main() function (Note: main()\'s functionality differs from the warm up) Build the ShoppingCart class with the following specifications. Note: Some can be function stubs (empty functions) initially, to be completed in later steps. Default constructor Parameterized constructor which takes the customer name and date as parameters (1 pt) Private data members string customerName - Initialized in default constructor to \"none\" string currentDate - Initialized in default constructor to \"January 1, 2016\" vector < ItemToPurchase > cartItems Public member functions GetCustomerName() accessor (1 pt) GetDate() accessor (1 pt) AddItem() Adds an item to cartItems vector. Has parameter ItemToPurchase. Does not return anything. RemoveItem() Removes item from cartItems vector. Has a string (an item\'s name) parameter. Does not return anything. If item name cannot be found, output this message: Item not found in cart. Nothing removed. ModifyItem() Modifies an it.

I have the first program completed (not how request, but it works) a.pdf

I have the first program completed (not how request, but it works) a.pdf

I have the first program completed (not how request, but it works) a.pdf

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How does chemotherapy function as a cancer treatmentA.It kills ce.pdf

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