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# Assignment Three: Priority Queue and its Application

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Programming Assignment Three: Priority Queue and its Application

I. Motivation
This project will give you experience in implementing priority queues using C++. You will also
empirically study the efficiency of different implementations.
II. Project Overview
In this project, you are given a rectangular grid of cells. Each cell has a number indicating its
weight, which is the cost of passing through the cell. You can assume the weights are positive
integers. The input will give you the starting coordinate and the ending coordinate. Your task
is to use priority queue to find the shortest path from the source cell to the ending cell.
III. Input

Category:

## Description

Programming Assignment Three: Priority Queue and its Application

I. Motivation
This project will give you experience in implementing priority queues using C++. You will also
empirically study the efficiency of different implementations.
II. Project Overview
In this project, you are given a rectangular grid of cells. Each cell has a number indicating its
weight, which is the cost of passing through the cell. You can assume the weights are positive
integers. The input will give you the starting coordinate and the ending coordinate. Your task
is to use priority queue to find the shortest path from the source cell to the ending cell.
III. Input
You will read from the standard input. (For the ease of testing, you can write each test case in
a file and then use Linux file redirection function “<” to read from the file.)
The format of the input is as follows:
<width m>
<height n>
<Start x> <Start y>
<End x> <End y>
<W(0,0)> <W(1,0)> … <W(m-1,0)>

<W(0,n-1)> <W(1,n-1)> … <W(m-1,n-1)>
The first and the second line give the width ? and the height ? of the grid, respectively.
They are positive integers. The third and the fourth line give the starting coordinate and the
ending coordinate, respectively. They are non-negative integers within the valid range. The
upper left corner has the coordinate (0, 0). The x-coordinate increases from left to right and
the y-coordinate increases from top to bottom. The remaining lines give the weights of the
cells in the grid. They represent a two dimensional array of ? rows and ? columns, as
shown above. ?(?,?) is the weight of the cell at coordinate (?,?) (0 ≤ ? ≤ ? − 1,0 ≤ ? ≤
? − 1). The weights are all non-negative integers.
For example, we have an input:
4
3
0 0
3 1
5 1 2 3
2 1 5 6
7 1 1 1
It specifies a grid of width 4 and height 3. We want to find the shortest path form point (0, 0),
which has weight 5, to the point (3, 1), which has weight 6.
You can assume that the testing cases are semantically and syntactically correct. Furthermore,
you can assume that the starting point and the ending point are not the same one.
IV. Algorithm
In this project, we ask you to find the shortest path between two given cells in a grid. On a
valid path, you can only either go horizontally or vertically from one cell to an adjacent cell;
you cannot go diagonally. The length of a path is defined as the sum of the weights of all the
cells on that path.
The algorithm we will use in this project is pretty similar to the Lee’s wire routing algorithm
that we previous talked about in Ve280 (the slides of that algorithm can be found in the
Progamming-Assignment-Three-Related-Files.zip). Recall that in that problem the grid cell is
of unit weight. We apply a queue to find the shortest path between two points. In order to find
the shortest path in a grid with non-uniform cell weight, we need to apply a min priority
queue. For this project, for simplicity, we assume that there are no blocked cells in the grid.
Below is the pseudo-code of the algorithm:
Let PQ be a priority queue;
start_point.pathcost = start_point.cellweight;
Mark start_point as reached;
PQ.enqueue(start_point);
while(PQ is not empty) {
C = PQ.dequeueMin(); // The key is cell’s pathcost
for each neighbor N of C that has not been reached {
N.pathcost = C.pathcost + N.cellweight;
mark cell N as reached;
mark the predecessor of N as C;
// I.e., N is reached from C.
if(end_point == N) {
trace_back_path(); // Trace and print the path
// through predecessor info
return;
}
else PQ.enqueue(N);
}
}
The primary key of the min priority queue is the path cost of a cell, which is the minimal
length from the starting point to the cell. (Updating the path cost using the above procedure
will guarantee that the pathcost entry of a cell stores the minimal path length from the
starting point to the cell.) The function trace_back_path()obtains and prints the final
path.
In order for you to get the same result as ours, we require:
1. The visit of the neighbors starts form the right neighbor and then goes in the clockwise direction, i.e., right, down, left, up. For those cells on the boundary, they may not
have a certain neighbor. Then you just skip it.
2. During a dequeueMin() operation, if multiple cells have the same smallest path
cost, choose the cell with the smallest x-coordinate. Furthermore, if there are multiple
cells with the same x-coordinate, choose the cell with the smallest y-coordinate.
3. Function trace_back_path() uses the predecessor information you have recorded during the search to recover the path. The predecessor of a cell N is the cell
from which you reached N.
V. Command Line Input
Your program should be named main. It should take the following case-sensitive command-line options:
1. -i, –implementation: a required option. It changes the priority queue implementation at runtime. An argument should immediately follow that option, being
BINARY, UNSORTED, or FIBONACCI to indicate the implementation (see Section
VII Implementations of Priority Queues).
2. -v, –verbose: an optional flag that indicates the program should print additional
outputs (see Section VI Output).
Examples of legal command lines:
 ./main –implementation BINARY < infile.txt
 ./main –verbose -i UNSORTED < infile.txt > outfile.txt
 ./main –verbose -i FIBONACCI
Note that the first two calls read the input stored in the infile.txt. The third call reads
from the standard input.
Examples of illegal command lines:
 ./main < infile.txt
No implementation is specified. Implementation is a required option.
 ./main –implementation BINARY infile.txt
You are not using input redirection “<” to read from the file infile.txt.
We require you to realize the above requirement using the function getopt_long. See its
usage and an example at
http://www.gnu.org/software/libc/manual/html_node/Getopt.html#Getopt
In testing your program, we will supply correct command-line inputs, but you are encouraged
to detect and handle errors in the command-line inputs.
VI. Output
All outputs should be printed to the standard output.
Without ‘verbose’ option, the output of main should be as follows:
The shortest path from (<start x, start y>) to (<end x, end y>)
is <length of the shortest path>.
Path:
(<start x, start y>)
(<cell_1 x, cell_1 y>)

(<end x, end y>)
For the example input shown in Section III, the output should be:
The shortest path from (0, 0) to (3, 1) is 16.
Path:
(0, 0)
(1, 0)
(1, 1)
(1, 2)
(2, 2)
(3, 2)
(3, 1)
For ‘verbose’ mode, you should print out each step in detail:
Step <x>
Choose cell (<cell_0 x, cell_0 y>) with accumulated length <l_0>.
Cell (<cell_1 x, cell_1 y>) with accumulated length <l_1> is added
into the queue.

Cell (<cell_i x, cell_i y>) with accumulated length <l_i> is added
into the queue.
Until you arrive at the ending point, print:
Step <y>
Choose cell (<cell_0 x, cell_0 y>) with accumulated length <l_0>.
Cell (<cell_1 x, cell_1 y>) with accumulated length <l_1> is added
into the queue.

Cell (<cell_i x, cell_i y>) with accumulated length <l_i> is added
into the queue.
Cell (<cell_i+1 x, cell_i+1 y>) with accumulated length <l_i+1>
is the ending point.
The shortest path from (<start x, start y>) to (<end x, end y>)
is <length of the shortest path>.
Path:
(<start x, start y>)
(<cell_1 x, cell_1 y>)

(<end x, end y>)
Note that in the last step, before you arrive at the ending point (i.e., cell_i+1), you may
first visit and enqueue some other cells (i.e., cell_1, …, cell_i). You need to treat
them in the normal way.
For the example input shown in Section III, the output should be:
Step 0
Choose cell (0, 0) with accumulated length 5.
Cell (1, 0) with accumulated length 6 is added into the queue.
Cell (0, 1) with accumulated length 7 is added into the queue.
Step 1
Choose cell (1, 0) with accumulated length 6.
Cell (2, 0) with accumulated length 8 is added into the queue.
Cell (1, 1) with accumulated length 7 is added into the queue.
Step 2
Choose cell (0, 1) with accumulated length 7.
Cell (0, 2) with accumulated length 14 is added into the queue.
Step 3
Choose cell (1, 1) with accumulated length 7.
Cell (2, 1) with accumulated length 12 is added into the queue.
Cell (1, 2) with accumulated length 8 is added into the queue.
Step 4
Choose cell (1, 2) with accumulated length 8.
Cell (2, 2) with accumulated length 9 is added into the queue.
Step 5
Choose cell (2, 0) with accumulated length 8.
Cell (3, 0) with accumulated length 11 is added into the queue.
Step 6
Choose cell (2, 2) with accumulated length 9.
Cell (3, 2) with accumulated length 10 is added into the queue.
Step 7
Choose cell (3, 2) with accumulated length 10.
Cell (3, 1) with accumulated length 16 is the ending point.
The shortest path from (0, 0) to (3, 1) is 16.
Path:
(0, 0)
(1, 0)
(1, 1)
(1, 2)
(2, 2)
(3, 2)
(3, 1)
Note that no space is attached after each line, and one space is attached after the comma
in each coordinate. The step number starts from 0.
VII. Implementations of Priority Queues
We have provided a header file, priority_queue.h, which defines a templated abstract
base class priority_queue. For this project, you are required to write three templated
implementations of priority queue as the derived class of priority_queue: binary heap,
unsorted array-based heap, and Fibonacci heap. To implement these priority queue variants,
we have given you three separate files binary_heap.h, unsorted_heap.h, and
fib_heap.h containing all the definitions for the functions declared in priority_queue.h. You will need to fill in the empty function definitions in these three header
runtime requirements and a general description of the implementation.
Fibonacci heap is an advanced data structure. You can study it by referring to the Fibonacci-Heap.pdf in Progamming-Assignment-Three-Related-Files.zip. We also show you the
pseudo-code of Fibonacci heap in Section XIII Appendix.
You are not allowed to modify priority_queue.h in any way. Nor are you allowed to
change the interface (names, parameters, and return types) of the functions that we give you
in any of the provided headers. You are allowed to add your own private helper functions and
variables as needed in binary_heap.h, unsorted_heap.h and fib_heap.h, so
long as you still follow the requirements outlined in both the specifications and the comments
in the provided files. You should not construct your program such that one priority queue im-
file.
Note: We may compile your priority queue implementations with our own code to ensure
that you have correctly and fully implemented them. To ensure that this is possible (and that
you do not lose credit for these test cases), do not define your main function in one of the
For the templated version of priority queue, it takes two template parameters TYPE and
COMP, where COMP is a functor. An example of functor that can be used with the priority
queue in this project is given to you in the file test_heap.cpp (which can be found in the
Priority-Queue-Test folder within Progamming-Assignment-Three-Related-Files.zip).
VIII. Implementation Requirements and Restrictions
Your code for solving the shortest path problem should call the priority queue implementations you write. Except some constraints on priority_queue.h, binary_heap.h, unsorted_heap.h, and fib_heap.h you are free to write your
own .cpp and .h files to solve the shortest path problem.
You are allowed to use std::vector, std::list, and std::deque.You are not
allowed to use other STL containers. Specifically, this means that the use of std::stack,
std::queue, std::priority_queue, std::set, std::map,
std::unordered_set, std::unordered_map, and the ‘multi’ variants of the
aforementioned containers are forbidden.
You are not allowed to use std::partition, std::partition_copy,
std::stable_partition, std::make_heap, std::push_heap,
std::pop_heap, std::sort_heap, or std::qsort.
Furthermore, you may not use any STL component that trivializes the implementation of
IX. Testing
We have provided you a test_heap.cpp to test your binary heap implementation. It can be
found in the Priority-Queue-Test folder within Progamming-Assignment-Three-Related-Files.zip. To test, copy the original prority_queue.h
and your modified binary_heap.h into the Priority-Queue-Test folder and type
make to compile a program test_heap. The output of the program should be exactly like
what is in the file test_heap.out. (Note: The Makefile in that folder also shows you
how to compile codes involving templates.) You can modify test_heap.cpp to test your
other implementations.
In the folder Shortest-Path-Test within Progamming-Assignment-Three-Related-Files.zip, we include two small test cases for the shortest
path problem. They are test-1.in and test-2.in. You can use them as inputs by applying the Linux input redirection function “<”. The outputs with and without the –v option
are *-verbose.out and *-brief.out, respectively. Use diff to compare your output
with these outputs.
These are the minimal amount of tests you should run to check your program. Those programs
that do not pass these tests are not likely to receive much credit. You should also write other
different test cases yourself to test your program extensively.
You should also check whether there is any memory leak using valgrind as we discussed
before. The Makefile in the Priority-Queue-Test folder shows you a way to put
valgrind command into a Makefile. To check whether test_heap program has
memory leak, type make memcheck.
X. Performance Comparison
We also ask you to compare the performance of the three priority queue implementations. To
do this, you should first generate random inputs with different grid sizes. Then you should
apply your implementations to these inputs. Finally, you should plot a figure showing the
runtime of each implementation versus the grid size. For comparison purpose, you should plot
three curves corresponding to the three implementations in the same figure. (You do not need to
upload the source codes for this comparison program.)
Hint:
1. You may want to write another program that calls different implementations to do this
study.
2. You can use mrand48() to generate integers uniformly distributed in the range
[-2
31, 231
-1].
3. You can use clock() function to get the runtime. See
http://www.cplusplus.com/reference/ctime/clock/
4. For simplicity, you can test on grid of which the width and the height are the same. With
this, the x-axis in the plotted figure is the width. Choose the starting point to be the upper
left corner of the grid and the ending point to be the lower right corner.
5. The runtime of your algorithm also depends on the detailed layout of the grid, not just its
width. Thus, for each width you pick, you should generate a number of grids with that
width and obtain the average runtime over all these grids. Also, for fair comparison, the
same set of grids should be applied to all the three implementations.
6. You should try at least 5 representative widths.
XI. Submitting and Due Date
You should submit all the source code files, a Makefile, and a report of the runtime comparison. The Makefile compiles a program named main, which solves the short path
problem. The report should be in pdf format. It should include a description of the experiment
setup. At the end of your report, please attach your source code for the program that solves the
short path problem. See announcement from the TAs for details about how to submit these files.
The submission deadline is 11:59 pm on Nov. 7
th
, 2017.
1. Functional correctness
2. Implementation constraints
3. General style
4. Performance
5. Report on the performance study
Functional correctness is determined by running a variety of test cases against your program,
checking your solution using our automatic testing program. We will grade Implementation
constraints to see if you have met all of the implementation requirements and restrictions.
General style refers to the ease with which TAs can read and understand your program, and the
cleanliness and elegance of your code. Part of your grade will also be determined by the performance of your algorithm. We will test your program with some large test cases. If your
program is not able to finish within a reasonable amount of time, you will lose the performance
score for those test cases. Finally, we will also read your report and grade it based on the quality
XIII. Appendix
Pseudo-code for Fibonacci heap:
Make_Fibonacci_Heap() {
H.n = 0; // n stores the number of elements in the heap
H.min = NULL; // min refers to the minimal element in the heap
return H;
}
Fibonacci_Heap_Get_Min(H) {
return H.min;
}
Fibonacci_Heap_Enqueue(H, key) {
create a new node x;
x.degree = 0;
x.parent = NULL;
x.child = NULL;
x.key = key;
// x.mark = FALSE; // This statement is commented out, because you
// do not need it in this project.
if(H.min == NULL) {
create a root list for H only containing x;
H.min = x;
}
else {
insert x into the root list of H;
if(x.key < H.min.key)
H.min = x;
}
H.n = H.n + 1;
}
Fibonacci_Heap_Dequeue_Min(H) {
z = H.min;
if(z != NULL) {
for(each child x of z) {
add x to the root list of H;
x.parent = NULL;
}
remove z from the root list of H;
H.n = H.n – 1;
if(H.n == 0) H.min = NULL;
else Consolidate(H);
}
return z;
}
Consolidate(H) {
let A[0..D(H.n)] be a new array; // D(N) is the maximum degree of
// any node in a Fibonacci heap of N nodes
for(i = 0 to D(H.n))
A[i] = NULL;
for(each node w in the root list of H) {
x = w;
d = x.degree;
while(A[d] != NULL) {
y = A[d];
if(x.key > y.key) exchange x with y;
A[d] = NULL;
d = d + 1;
}
A[d] = x;
}
H.min = NULL;
for(i = 0 to D(H.n)) {
if(A[i] != NULL) {
if(H.min == NULL) {
create a root list for H only containing A[i];
H.min = A[i];
}
else {
add A[i] to the root list of H;
if(A[i].key < H.min.key)
H.min = A[i];
}
}
}
}