Time due: 11:00 PM Tuesday, March 4 Wednesday, March 5
We anticipate that many people working on Project 3 will spend a lot of time debugging something that arises from a common novice misunderstanding. To save you that time later, we'll give you a chance to make that mistake in a simpler context, so you can work out that issue and how it manifests itself. (It may turn out that you don't have that misunderstanding, so you won't have any issues doing this problem. Still, keep this problem in mind, because you may still make the mistake in Project 3.)
Be especially sure to run your code for this problem under g32 to help ensure that there are no pointer/iterator violations or memory leaks which that common misunderstandings may lead to.
Material about vectors, lists, and iterators are in lecture09-updated.pptx in Carey Nachenberg's slides and David Smallberg's STL lecture and STL slides.
removeOdd function; you must use
list's erase member function; you must not use
list's remove or remove_if member
functions. Each int in the list must be examined for oddness no more than
once.
#include <list>
#include <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
using namespace std;
// Remove the odd integers from li.
// It is acceptable if the order of the remaining even integers is not
// the same as in the original list.
void removeOdd(list<int>& li)
{
}
void test()
{
int a[9] = { 5, 2, 8, 9, 6, 7, 3, 4, 1 };
list<int> x(a, a+9); // construct x from the array
assert(x.size() == 9 && x.front() == 5 && x.back() == 1);
removeOdd(x);
assert(x.size() == 4);
vector<int> v(x.begin(), x.end()); // construct v from x
sort(v.begin(), v.end());
int expect[4] = { 2, 4, 6, 8 };
for (int k = 0; k < 4; k++)
assert(v[k] == expect[k]);
}
int main()
{
test();
cout << "Passed" << endl;
}
For this problem, you will turn a file named oddlist.cpp with
the body of the removeOdd function, from its "void" to
its "}", no more and no less. Your function must compile and work
correctly when substituted into the program above, leaking no memory.
removeOdd function; you must use
vector's erase member function. Each int in the
vector must be examined for oddness no more than once.
#include <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
using namespace std;
// Remove the odd integers from v.
// It is acceptable if the order of the remaining even integers is not
// the same as in the original vector.
void removeOdd(vector<int>& v)
{
}
void test()
{
int a[9] = { 5, 2, 8, 9, 6, 7, 3, 4, 1 };
vector<int> x(a, a+9); // construct x from the array
assert(x.size() == 9 && x.front() == 5 && x.back() == 1);
removeOdd(x);
assert(x.size() == 4);
sort(x.begin(), x.end());
int expect[4] = { 2, 4, 6, 8 };
for (int k = 0; k < 4; k++)
assert(x[k] == expect[k]);
}
int main()
{
test();
cout << "Passed" << endl;
}
For this problem, you will turn a file named oddvector.cpp with
the body of the removeOdd function, from its "void" to
its "}", no more and no less. Your function must compile and work
correctly when substituted into the program above, leaking no memory.
removeBad function; you must use
list's erase member function; you must not use
list's remove or remove_if member
functions. Each Movie in the list must have its rating examined no more than
once.
#include <list>
#include <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
using namespace std;
vector<int> destroyedOnes;
class Movie
{
public:
Movie(int r) : m_rating(r) {}
~Movie() { destroyedOnes.push_back(m_rating); }
int rating() const { return m_rating; }
private:
int m_rating;
};
// Remove the movies in li with a rating below 50 and destroy them.
// It is acceptable if the order of the remaining movies is not
// the same as in the original list.
void removeBad(list<Movie*>& li)
{
}
void test()
{
int a[9] = { 25, 85, 80, 30, 70, 20, 15, 90, 10 };
list<Movie*> x;
for (int k = 0; k < 9; k++)
x.push_back(new Movie(a[k]));
assert(x.size() == 9 && x.front()->rating() == 25 && x.back()->rating() == 10);
removeBad(x);
assert(x.size() == 4 && destroyedOnes.size() == 5);
vector<int> v;
for (list<Movie*>::iterator p = x.begin(); p != x.end(); p++)
{
Movie* mp = *p;
v.push_back(mp->rating());
}
// Aside: Since C++11, the above loop could be
// for (auto p = x.begin(); p != x.end(); p++)
// {
// Movie* mp = *p;
// v.push_back(mp->rating());
// }
// or
// for (auto p = x.begin(); p != x.end(); p++)
// {
// auto mp = *p;
// v.push_back(mp->rating());
// }
// or
// for (Movie* mp : x)
// v.push_back(mp->rating());
// or
// for (auto mp : x)
// v.push_back(mp->rating());
sort(v.begin(), v.end());
int expect[4] = { 70, 80, 85, 90 };
for (int k = 0; k < 4; k++)
assert(v[k] == expect[k]);
sort(destroyedOnes.begin(), destroyedOnes.end());
int expectGone[5] = { 10, 15, 20, 25, 30 };
for (int k = 0; k < 5; k++)
assert(destroyedOnes[k] == expectGone[k]);
for (list<Movie*>::iterator p = x.begin(); p != x.end(); p++)
delete *p;
}
int main()
{
test();
cout << "Passed" << endl;
}
For this problem, you will turn a file named badlist.cpp with
the body of the removeBad function, from its "void" to
its "}", no more and no less. Your function must compile and work
correctly when substituted into the program above, leaking no memory.
removeBad function; you must use
vector's erase member function. Each Movie in the
vector must have its rating examined no more than once.
#include <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
using namespace std;
vector<int> destroyedOnes;
class Movie
{
public:
Movie(int r) : m_rating(r) {}
~Movie() { destroyedOnes.push_back(m_rating); }
int rating() const { return m_rating; }
private:
int m_rating;
};
// Remove the movies in v with a rating below 50 and destroy them.
// It is acceptable if the order of the remaining movies is not
// the same as in the original vector.
void removeBad(vector<Movie*>& v)
{
}
void test()
{
int a[9] = { 25, 85, 80, 30, 70, 20, 15, 90, 10 };
vector<Movie*> x;
for (int k = 0; k < 9; k++)
x.push_back(new Movie(a[k]));
assert(x.size() == 9 && x.front()->rating() == 25 && x.back()->rating() == 10);
removeBad(x);
assert(x.size() == 4 && destroyedOnes.size() == 5);
vector<int> v;
for (int k = 0; k < 4; k++)
v.push_back(x[k]->rating());
sort(v.begin(), v.end());
int expect[4] = { 70, 80, 85, 90 };
for (int k = 0; k < 4; k++)
assert(v[k] == expect[k]);
sort(destroyedOnes.begin(), destroyedOnes.end());
int expectGone[5] = { 10, 15, 20, 25, 30 };
for (int k = 0; k < 5; k++)
assert(destroyedOnes[k] == expectGone[k]);
for (vector<Movie*>::iterator p = x.begin(); p != x.end(); p++)
delete *p;
}
int main()
{
test();
cout << "Passed" << endl;
}
For this problem, you will turn a file named badvector.cpp with
the body of the removeBad function, from its "void" to
its "}", no more and no less. Your function must compile and work
correctly when substituted into the program above, leaking no memory.
#include <iostream>
#include <vector>
#include <list>
using namespace std;
const int MAGIC = 11223344;
void test()
{
bool allValid = true;
vector<int> v1(5, MAGIC);
int k = 0;
for ( ; k != v1.size(); k++)
{
if (v1[k] != MAGIC)
{
cout << "v1[" << k << "] is " << v1[k] << ", not " << MAGIC <<"!" << endl;
allValid = false;
}
if (k == 2)
{
for (int i = 0; i < 5; i++)
v1.push_back(MAGIC);
}
}
if (allValid && k == 10)
cout << "Passed test 1" << endl;
else
cout << "Failed test 1" << endl;
allValid = true;
list<int> l1(5, MAGIC);
k = 0;
for (list<int>::iterator p = l1.begin(); p != l1.end(); p++, k++)
{
if (*p != MAGIC)
{
cout << "Item# " << k << " is " << *p << ", not " << MAGIC <<"!" << endl;
allValid = false;
}
if (k == 2)
{
for (int i = 0; i < 5; i++)
l1.push_back(MAGIC);
}
}
if (allValid && k == 10)
cout << "Passed test 2" << endl;
else
cout << "Failed test 2" << endl;
allValid = true;
vector<int> v2(5, MAGIC);
k = 0;
for (vector<int>::iterator p = v2.begin(); p != v2.end(); p++, k++)
{
if (k >= 20) // prevent infinite loop
break;
if (*p != MAGIC)
{
cout << "Item# " << k << " is " << *p << ", not " << MAGIC <<"!" << endl;
allValid = false;
}
if (k == 2)
{
for (int i = 0; i < 5; i++)
v2.push_back(MAGIC);
}
}
if (allValid && k == 10)
cout << "Passed test 3" << endl;
else
cout << "Failed test 3" << endl;
}
int main()
{
test();
}
Explain in a sentence or two what happens during the execution of test case 3 that eventually leads to test case 3 failing.
The files Sequence.h and Sequence.cpp contain the definition and implementation of Sequence implemented using a doubly-linked list. A client who wants to use a Sequence has to change the type alias declaration in Sequence.h, and within one source file, cannot have two Sequences containing different types.
Eliminate the using statement defining the type alias, and
change Sequence to be a class template, so that a client can say
#include "Sequence.h"
#include <string>
using std::string;
...
Sequence<int> si;
Sequence<string> ss;
si.insert(30);
ss.insert("30 For 30");
...
Also, change subsequence and zipper to be function
templates.
(Hint: Transforming the solution based on a type alias is a mechanical task that takes five minutes if you know what needs to be done. What makes this problem non-trivial for you is that you haven't done it before; the syntax for declaring templates is new to you, so you may not get it right the first time.)
(Hint: Template typename parameters don't have to be named with single
letters like T; they can be names of your choosing. You might
find that by choosing the name ItemType, you'll have many
fewer changes to make.)
(Hint: The Node class nested in the Sequence class can talk about the template parameter of the Sequence class; it should not itself be a template class.)
The declarations and implementations of your Sequence class
template and the subsequence and zipper template
functions must be in just one file, Sequence.h, which is all that you will
turn in for this problem. Although the implementation of a non-template
non-inline function should not be placed in a header file (because of linker
problems if that header file were included in multiple source files), the
implementation of a template function, whether or not it's declared inline,
can be in a header file without causing linker problems, and in fact
the header file is the normal place to put it in most C++ environments.
There's a pre-C++20 language technicality that relates to a type declared
inside a class template, like N below:
template <typename T>
class S
{
...
struct N
{
...
};
N* f();
...
};
The technicality affects how we specify the return type of a function
(such as S<T>::f) when that return type uses a type defined
inside a template class (such as S<T>::N). If we attempt to
implement f this way:
template <typename T>
S<T>::N* S<T>::f() // Error! Won't compile in C++17 or earlier.
{
...
}
the pre-C++20 technicality requires the compiler to not recognize
S<T>::N as a type name; it must be announced as a type name
this way:
template <typename T>
typename S<T>::N* S<T>::f() // OK in all C++ versions
{
...
}
Giving g32 the -std=c++20 option will cause it to use C++20.
We will test your code with C++17, unless it doesn't compile, in which case
we'll test it with C++20 instead.
For you to not get a score of zero for this problem, this test program
that we will try with your Sequence.h must
build and execute successfully under both g32 and either Visual C++ or
clang++, with no Sequence.cpp file on the command line (for
g32) or as part of the project (for Visual C++ or Xcode):
#include "Sequence.h"
#include <iostream>
#include <string>
#include <cassert>
using namespace std;
void test()
{
Sequence<int> si;
assert(si.empty());
assert(si.size() == 0);
assert(si.insert(0, 20) == 0);
assert(si.insert(10) == 0);
assert(si.find(20) == 1);
assert(si.remove(10) == 1);
int i;
assert(si.get(0, i));
assert(si.set(0, 30));
assert(si.erase(0));
Sequence<int> si2(si);
si2.swap(si);
si2 = si;
assert(subsequence(si,si) == -1);
zipper(si,si2,si);
Sequence<string> ss;
assert(ss.empty());
assert(ss.size() == 0);
assert(ss.insert(0, "Hello") == 0);
assert(ss.insert("Goodbye") == 0);
assert(ss.find("Hello") == 1);
assert(ss.remove("Goodbye") == 1);
string s;
assert(ss.get(0, s));
assert(ss.set(0, "Aloha"));
assert(ss.erase(0));
Sequence<string> ss2(ss);
ss2.swap(ss);
ss2 = ss;
assert(subsequence(ss,ss2) == -1);
zipper(ss,ss2,ss);
}
int main()
{
test();
cout << "Passed all tests" << endl;
}
Consider this program:
#include "Sequence.h" // class template from problem 2
class Coord
{
public:
Coord(int r, int c) : m_row(r), m_col(c) {}
Coord() : m_row(0), m_col(0) {}
double r() const { return m_row; }
double c() const { return m_col; }
private:
double m_row;
double m_col;
};
int main()
{
Sequence<int> si;
si.insert(50); // OK
Sequence<Coord> sc;
sc.insert(0, Coord(50,20)); // OK
sc.insert(Coord(40,10)); // error!
}
Explain in a sentence or two why the call to the one-argument form of
Sequence<Coord>::insert causes at least one compilation
error. (Notice that the call to the one-argument form of
Sequence<int>::insert is fine, as is the call to the
two-argument form of Sequence<Coord>::insert.) Don't
just transcribe a compiler error message; your answer must indicate you
understand the ultimate root cause of the problem and why that is
connected to the call to Sequence<Coord>::insert.
A class has a name (e.g., Actor) and zero or
more subclasses (e.g., the class with name Player or
the class with name Enemy).
The following program reflects this structure:
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Class
{
public:
Class(string nm) : m_name(nm) {}
string name() const { return m_name; }
const vector<Class*>& subclasses() const { return m_subclasses; }
void add(Class* d) { m_subclasses.push_back(d); }
~Class();
private:
string m_name;
vector<Class*> m_subclasses;
};
Class::~Class()
{
for (size_t k = 0; k < m_subclasses.size(); k++)
delete m_subclasses[k];
}
void listAll(string path, const Class* c) // two-parameter overload
{
You will write this code.
}
void listAll(const Class* c) // one-parameter overload
{
if (c != nullptr)
listAll("", c);
}
int main()
{
Class* d1 = new Class("Koopa");
listAll(d1);
cout << "====" << endl;
d1->add(new Class("SoopaKoopa"));
Class* d2 = new Class("Enemy");
d2->add(new Class("Fireball"));
d2->add(d1);
Class* d3 = new Class("Goodie");
d3->add(new Class("Garlic"));
d3->add(new Class("Beans"));
d3->add(new Class("ExtraLife"));
listAll(d3);
cout << "====" << endl;
Class* d4 = new Class("Actor");
d4->add(d2);
d4->add(new Class("Player"));
d4->add(d3);
listAll(d4);
delete d4;
}
This main routine should produce the following output (the first line
written is Koopa, not an empty line):
Koopa ==== Goodie Goodie=>Garlic Goodie=>Beans Goodie=>ExtraLife ==== Actor Actor=>Enemy Actor=>Enemy=>Fireball Actor=>Enemy=>Koopa Actor=>Enemy=>Koopa=>SoopaKoopa Actor=>Player Actor=>Goodie Actor=>Goodie=>Garlic Actor=>Goodie=>Beans Actor=>Goodie=>ExtraLife
Each call to the one-parameter overload of listAll produces a
list, one per line, of the inheritance path to each class in the
inheritance tree rooted at listAll's argument. An
inheritance path is a sequence of class names separated by "=>". There is
no "=>" before the first name in the inheritance path.
You are to write the code for the two-parameter overload of
listAll to make this happen. You must not use any additional
container (such as a stack), and the two-parameter overload of
listAll must be recursive. You must not use any global
variables or variables declared with the keyword static, and
you must not modify any of the code we have already written or add new
functions. You may use a loop to traverse the vector; you must not use
loops to avoid recursion.
Here's a useful function to know: The standard library string class has a + operator that concatenates strings and/or characters. For example,
string s("Hello");
string t("there");
string u = s + ", " + t + '!';
// Now u has the value "Hello, there!"
It's also useful to know that if you choose to traverse an STL container
using some kind of iterator, then if the container is const, you must use
a const_iterator:
void f(const list<int>& c) // c is const
{
for (list<int>::const_iterator it = c.begin(); it != c.end(); it++)
cout << *it << endl;
}
(Of course, a vector can be traversed either by using some kind of iterator, or by using operator[] with an integer argument).
For this problem, you will turn a file named list.cpp with
the body of the two-parameter overload of the listAll
function, from its "void" to its "}", no more and no less. Your function
must compile and work correctly when substituted into the program above.
We introduced the two-parameter overload of listAll. Why could you
not solve this problem given the constraints in part a if we had only a
one-parameter listAll, and you had to implement it as
the recursive function?
Suppose we have a list of N world cities, numbered from 0 to N-1.
The two-dimensional array of doubles dist holds the airline
distance between each pair of cities: dist[i][j] is the
distance between city i and city j.
Now, for every pair of cities i and j, we'd like to consider all the flights between the two that make one stop in a third city k, and record which city k yields the shortest distance traveled in a one-stop flight between city i and city j that passes through city k. Here's the code:
const int N = some value;
assert(N > 2); // algorithm fails if N <= 2
double dist[N][N];
...
int bestMidPoint[N][N];
for (int i = 0; i < N; i++)
{
bestMidPoint[i][i] = -1; // one-stop trip to self is silly
for (int j = 0; j < N; j++)
{
if (i == j)
continue;
int minDist = maximum possible integer;
for (int k = 0; k < N; k++)
{
if (k == i || k == j)
continue;
int d = dist[i][k] + dist[k][j];
if (d < minDist)
{
minDist = d;
bestMidPoint[i][j] = k;
}
}
}
}
What is the time complexity of this algorithm, in terms of the number of basic operations (e.g., additions, assignments, comparisons) performed: Is it O(N), O(N log N), or what? Why? (Note: In this homework, whenever we ask for the time complexity, we care only about the high order term, so don't give us answers like O(N2+4N).)
The algorithm in part a doesn't take advantage of the symmetry of distances: for every pair of cities i and j, dist[i][j] == dist[j][i]. We can skip a lot of operations and compute the best midpoints more quickly with this algorithm:
const int N = some value;
assert(N > 2); // algorithm fails if N <= 2
double dist[N][N];
...
int bestMidPoint[N][N];
for (int i = 0; i < N; i++)
{
bestMidPoint[i][i] = -1; // one-stop trip to self is silly
for (int j = 0; j < i; j++) // loop limit is now i, not N
{
int minDist = maximum possible integer;
for (int k = 0; k < N; k++)
{
if (k == i || k == j)
continue;
int d = dist[i][k] + dist[k][j];
if (d < minDist)
{
minDist = d;
bestMidPoint[i][j] = k;
bestMidPoint[j][i] = k;
}
}
}
}
What is the time complexity of this algorithm? Why?
Here again is the non-member zipper function for Sequences
from Sequence.cpp:
void zipper(const Sequence& seq1, const Sequence& seq2, Sequence& result)
{
Sequence res;
int n1 = seq1.size();
int n2 = seq2.size();
int nmin = (n1 < n2 ? n1 : n2);
int resultPos = 0;
for (int k = 0; k < nmin; k++)
{
ItemType v;
seq1.get(k, v);
res.insert(resultPos, v);
resultPos++;
seq2.get(k, v);
res.insert(resultPos, v);
resultPos++;
}
const Sequence& s = (n1 > nmin ? seq1 : seq2);
int n = (n1 > nmin ? n1 : n2);
for (int k = nmin ; k < n; k++)
{
ItemType v;
s.get(k, v);
res.insert(resultPos, v);
resultPos++;
}
result.swap(res);
}
Assume that seq1, seq2, and the old value of
result each have N elements. In terms of the number of linked
list nodes visited during the execution of this function, what is its time
complexity? Why?
Here is an implementation of a related member function. The call
s3.zipper(s1, s2);
has the same effect as calling the non-member function above as
zipper(s1, s2, s3);. The implementation is
void Sequence::zipper(const Sequence& seq1, const Sequence& seq2)
{
Sequence res;
Node* p1 = seq1.m_head->m_next;
Node* p2 = seq2.m_head->m_next;
for ( ; p1 != seq1.m_head && p2 != seq2.m_head;
p1 = p1->m_next, p2 = p2->m_next)
{
res.insertBefore(res.m_head, p1->m_value);
res.insertBefore(res.m_head, p2->m_value);
}
Node* p = (p1 != seq1.m_head ? p1 : p2);
Node* pend = (p1 != seq1.m_head ? seq1 : seq2).m_head;
for ( ; p != pend; p = p->m_next)
res.insertBefore(res.m_head, p->value);
// Swap *this with res
swap(res);
// Old value of *this (now in res) is destroyed when function returns.
}
Assume that seq1, seq2, and the old value of
*this each have about N elements. In terms of the number of
linked list nodes visited during the execution of this function, what is
its time complexity? Why? Is it the same, better, or worse, than the
implementation in part a?
The file sorts.cpp contains an almost complete program that creates a randomly ordered array, sorts it in a few ways, and reports on the elapsed times. Your job is to complete it and experiment with it.
You can run the program as is to get some results for the STL sort algorithm. You won't get any result for insertion sort, because the insertion sort function right now doesn't do anything. That's one thing for you to write.
The objects in the array might not be cheap to copy (it depends on your processor), which might make a sort that does a lot of moving objects around expensive. Your other task will be to create a vector of pointers to the objects, sort the pointers using the same criterion as was used to sort the objects, and then make one pass through the vector to put the objects in the proper order.
Your two tasks are thus:
Implement the insertion_sort function. (Code is on pp.
332-333 in the Carrano book, and also here.)
Implement the compareStorePtr function and the code in
sortUsingPtrs to create and sort the array of pointers.
The places to make modifications are indicated by TODO: comments.
You should not have to make modifications anywhere else. (Our solution
doesn't.)
When your program is correct, build an optimized version of it to do some timing experiments. On cs32.seas.ucla.edu, build the executable and run it this way:
g32fast -o sorts sorts.cpp ./sorts
(You don't have to know this, but this command omits some of the runtime error checking compiler options that our g32 command supplies, and it adds the -O2 compiler option that causes the compiler to spend more time optimizing the machine language translation of your code so that it will run faster when you execute it.)
Under Xcode, select Product / Scheme / Edit Scheme.... In the left panel, select Run, then in the right panel, select the Info tab. In the Build Configuration dropdown, select Release. For Visual C++, it's a little trickier.
Try the program with about 10000 items. Depending on the speed of your processor, this number may be too large or small to get meaningful timings in a reasonable amount of time. Experiment. Once you get insertion sort working, observe the O(N2) behavior by sorting, say, 10000, 20000, and 40000 items.
To further observe the performance behavior of the STL sort algorithm, try sorting, say, 100000, 200000, and 400000 items, or even ten times as many. Since this would make the insertion sort tests take a long time, skip them by setting the TEST_INSERTION_SORT constant at the top of sorts.cpp to false. (Note for Visual C++ users: There's a default limit for the maximum heap size during execution, so you may get strange behavior for a vector of about 182,000 or more of our items (about 566MB). If you like, you can change this limit in the project's Property Pages, Configuration Properties / Linker / System, Heap Commit Size value.)
Notice that if you run the program more than once, you may not get exactly the same timings each time. This is partly because of not getting the same sequence of random numbers each time, but also because of factors like caching by the operating system.
By Monday, March 3, there will be a link on the class webpage that will enable you to turn in this homework. Turn in one zip file that contains your solutions to the homework problems. The zip file must contain eight files:
oddlist.cpp, a C++ source file with your solution to problem 1a.
oddvector.cpp, a C++ source file with your solution to problem 1b.
badlist.cpp, a C++ source file with your solution to problem 1c.
badvector.cpp, a C++ source file with your solution to problem 1d.
Sequence.h, a C++ header file with your declaration and
implementation of the class and function templates for problem 2.
list.cpp, a C++ source file with the implementation of the
two-parameter overload of the listAll function for problem 4a.
sorts.cpp, a C++ source file with your solution to problem 7.
urn in the whole file.
hw.docx or hw.txt, a Word document or a text file
with your solutions to problems 1e, 3, 4b, 5a, 5b, 6a, and 6b.
In each source file you turn in, do not comment out your implementation; you want our test scripts to see it!