COSC 320 Exam 2 Key Spring Part 1: Hash Functions

Transcription

1 COSC 320 Exam 2 Key Spring 2011 Part 1: Hash s 1. (5 Points) Create the templated function object lessthan, you may assume that the templated data type T has overloaded the < operator. template<typename T> class lessthan bool operator() (const T& x, const T& y) const return x < y; ; 2. (5 Points) Create the hash function object hfint2 which takes an integer, n, as its input parameter and returns an unsigned int defined as follows. First take the input n and convert it to an unsigned int, square the result, then take the upper half bits and lower half bits and add them, finally square that number. Assume that our machine uses 32 bits for an unsigned integer. Note that 2 16 = class hfint2 unsigned int operator()(int item) const unsigned int value = (unsigned int)item; ; value *= value; // square the value unsigned int ub = value / 65536; // upper 16 bits unsigned int lb = value % 65536; // lower 16 bits value = ub+lb; // add value *= value; // square the value return value; // return result

2 3. (10 Points) Consider a hash table that has 15 entries and stores the integers 27, 52, 16, 4, 25, 10, 37, 45, 44, 68, 32, 14, 1, 28 using the identity hash function. (a) What is the load factor of the table? 14/15 (b) If the table uses linear probing, i. Draw the table ii. What is the average number of probes needed to determine if a search is successful? Checking an empty array entry counts as a probe. 25/14 iii. What is the average number of probes needed to determine if a search fails? Checking an empty array entry counts as a probe. 120/15 = 8 (c) If the table uses chaining, i. Draw the table ii. What is the average number of probes needed to determine if a search is successful? Checking the end of a list counts as a probe. 9/7 iii. What is the average number of probes needed to determine if a search fails? Checking the end of a list counts as a probe. 29/15 Part 2: Trees 1. (10 Points) Draw the tree that results from inserting the following numbers in the given order27, 52, 16, 4, 25, 10, 37, 45, 44, 68, 32, 14, 1, 28

3 Part 3: Red-Black Trees: The implementation of the red-black tree is at the end of the exam. 1. (15 Points) Draw the Red-Black tree that results from inserting the following numbers in the given order21,, 5, 13, 14, 62, 18, 7, 25, 35, * 21 5* 21* * * 14* * 14* 62* 13* * 18* 13* * 7* 18* 13* * 62* 7* 18* 13* 25* * 18* 35* * 18* 27* 62*

4 2. (10 Points) Consider the following Red-Black tree (the asterisks represents a red node). Draw the Red-Black tree that results from deleting node * 21* 27* 48* 75* Erase * 27* 48* 75* 3. (10 Points) Consider the following Red-Black tree (the asterisks represents a red node). Draw the Red-Black tree that results from deleting node * 21* 27* 48* 75* Erase * 21* 27* 75*

5 Part 4: Inheritance and Polymorphism 1. (15 Points) Consider the following program. File: classes.h #include <iostream> #include <string> using namespace std; class basecl basecl(int aa, int bb, int cc) : x(aa), y(bb), z(cc) int x; virtual void print() cout << "x = " << x << " y = " << y << " z = " << z << endl; protected: int y; ; private: int z; class derivedcl : public basecl derivedcl(int aa, int bb, int cc, int dd, int ee, int ff) : basecl(aa, bb, cc), a(dd), b(ee), c(ff) int a; void print() basecl::print(); cout << "a = " << a << " b = " << b << " c = " << c << endl; protected: int b; ; private: int c; class derivedcl2 : public basecl derivedcl2(int aa, int bb, int cc, int dd, int ee, int ff) : basecl(aa, bb, cc), m(dd), n(ee), p(ff) ; int m; void print() basecl::print(); cout << "m = " << m << " n = " << n << " p = " << p << endl; protected: int n; private: int p;

6 File: main.cpp #include <iostream> #include "classes.h" int main() using namespace std; basecl bc(1,2,3); derivedcl dc(4,5,6,7,8,9); derivedcl2 dc2(10, 11, 12, 13, 14, 15); bc.print(); dc.print(); dc2.print(); bc = dc; bc.print(); basecl *arr[5]; for (int i = 0; i < 5; i++) basecl* b = new basecl(i, i+1, i+2); derivedcl* d = new derivedcl(i+3, i+4, i+5, i+6, i+7, i+8); derivedcl2* d2 = new derivedcl2(i+9, i+10, i+11, i+12, i+13, i+14); if (i % 3 == 0) arr[i] = b; else if (i % 3 == 1) arr[i] = d; else arr[i] = d2; for (int i = 0; i < 5; i++) arr[i]->print(); for (int i = 0; i < 5; i++) arr[i]->derivedcl2::basecl::print(); return 0; Write the output of the program. If there is a segment of code that will not compile, circle it, state why it would not compile and give the output of the program if that segment of code were removed. x = 1 y = 2 z = 3 a = 7 b = 8 c = 9 x = 10 y = 11 z = 12 m = 13 n = 14 p = 15 x = 0 y = 1 z = 2 a = 7 b = 8 c = 9 x = 11 y = 12 z = 13 m = 14 n = 15 p = 16 x = 3 y = 4 z = 5 x = 7 y = 8 z = 9 a = 10 b = 11 c = 12 x = 0 y = 1 z = 2 x = 11 y = 12 z = 13 x = 3 y = 4 z = 5 x = 7 y = 8 z = 9

7 2. (10 Points) Fill out the following access chart assuming that the inheritance is public inheritance. Put a check in each box where access is permitted and leave the box empty if assess would not be granted. Base Class Derived Class Base Object Base Private Base Protected Base Public Derived Private Derived Public Derived Object 3. (10 Points) Fill out the following access chart assuming that the inheritance is private inheritance. Put a check in each box where access is permitted and leave the box empty if assess would not be granted. Base Class Derived Class Base Object Derived Object Base Private Base Protected Base Public Derived Private Derived Public 4. (10 Points) Create a templated interface for a stack. The class name should be stackinterface, the methods must be as follows. Do not assume that the templated type T has a copy constructor. (a) push input of templated type T, no output. (b) pop no input or output. (c) top no input and type T output. (d) empty no input and boolean output. (e) size no input and integer output. template <typename T> class stackinterface virtual void push(const &T item) = 0; virtual void pop() = 0; virtual &T top() = 0; virtual bool empty() const = 0; virtual int size() const = 0; ;