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CSc 352 (Spring 2019): Assignment 11
Due Date: 11:59PM Wed, May 1
The purpose of this assignment is to work with linked lists, memory allocation, command line
arguments, reading from a file, using free(), representing graphs, and doing more complicated
manipulation of pointers.
General Requirements
- Your C code should adhere to the coding standards for this class as listed in the
Documents section on the Resources tab for Piazza. This includes protecting against
buffer overflows whenever you read strings. - Your programs should indicate if they executed without any problems via their exit
status, i.e., the value returned by the program when it terminates:
Execution Exit Status
Normal, no problems 0
Error or problem encountered 1 - Under bash you can check the exit status of a command or program cmd by typing the
command “echo $?” immediately after the execution of cmd. A program can exit with
status n by executing “exit(n)” anywhere in the program, or by having main() execute
the statement “return(n)”. - Remember your code will be graded on lectura using a grading script. You should test
your code on lectura using the diff command to compare your output to that of the
example executable. - Your code must show no errors when run with valgind.
- You must free all your allocated memory before your program exits.
Testing
Example executables of the programs will be made available. You should copy and run these
programs on lectura to test your program’s output and to answer questions you might have about
how the program is supposed to operate. Our class has a home directory on lectura which is:
/home/cs352/spring19
You all have access to this directory. The example programs will always be in the appropriate
assignments/assg#/prob# subdirectory of this directory. They will have the same name as the
assigned program with“ex”added to the start and the capitalization changed to maintain
camelback. So, for example, if the assigned program is theBigProgram, then the example
executable will be named exTheBigProgram. You should use the appropriate UNIX
commands to copy these executables to your own directory.
Your programs will be graded by a script. This will include a timeout for all test cases. There
must be a timeout or programs that don’t terminate will cause the grading script to never finish.
This time out will never be less than 10 times the time it takes the example executable to
complete with that test input and will usually be much longer than that. If your program takes an
exceedingly long time to complete compared to the example code, you may want to think about
how to clean up your implementation.
Makefiles
You will be required to include a Makefile with each program. Running the command:
make progName
should create the executable file progName, where progName is the program name listed for the
problem. The gcc commands in your Makefile that create the object files must include the -Wall
flag. Other than that, the command may have any flags you desire.
Submission Instructions
Your solutions are to be turned in on the host lectura.cs.arizona.edu. Since the assignment will
be graded by a script, it is important you have the directory structure and the names of the files
exact. Remember that UNIX is case sensitive, so make sure the capitalization is also correct. For
all our assignments the directory structure should be as follows: The root directory will be named
assg#, where # is the number of the current assignment. Inside that directory should be a
subdirectory for each problem. These directories will be named prob# where # is the number of
the problem within the assignment. Inside these directories should be any files required by the
problem descriptions.
To submit your solutions, go to the directory containing your assg11 directory and use the
following command:
turnin cs352s19-assg11 assg11
Problems
prob1: bacon
Write a program called bacon which calculates the Bacon score
(https://en.wikipedia.org/wiki…) of various actors based on the
information given in an input file. Kevin Bacon is a movie actor and the Bacon score for any
movie actor is the number of movies it takes to connect Kevin Bacon with that actor. The
definition is as follows:
Kevin Bacon has a Bacon score of 0
Any actor who was in a movie with Kevin Bacon has a Bacon score of 1
For any actor X, if the lowest possible Bacon score of any actor X has been in a movie
with is N, the X’s Bacon score is N + 1
It may be hard to follow the outline above, but the concept is not that deep. Read through the
wiki page linked to above if you are confused.
Basically, this is a graph problem. The actors are the vertices. The edges connecting the actors
are movies. There is an edge between the vertices representing actors A and B if and only if A
and B appeared in a movie together. An actor’s Bacon score is then the smallest number of edges
connecting the actor to Kevin Bacon.
Invocation: Your program will be invoked with a required argument giving the file name
of a text file containing movie names and cast lists and an optional flag to tell the
program whether to print out just the Bacon score or to print out the score together with
the path that generated it. The invocation will look like:
bacon [-l] inFile
where inFile is the name of a file which contains the movie list from which you will
build your graph and –l (that’s a lowercase L) is an optional flag. To make our program
“UNIX like” we will allow the arguments to appear in any order and the option flag may
be specified multiple times. Here are some examples of legal invocations:
bacon myFile –l
bacon –l –l myFile
bacon myFile
The following would be illegal invocations:
bacon –l //has no file specified
bacon fileName fileName //too many arguments
bacon –ll myFile //not a legal option
If the invocation is illegal, you should print an error message to stderr indicating how the
program is used, and exit with a status of 1. If you are confused about what is legal and
what is not, experiment with the example executable.
Movie File: The movie file whose name is given as a command line argument will
contain a list of movies and their cast lists. The format of the file is as follows:
Move: <name of movie>
<actor 1>
<actor 2>
<actor n>
Movie: <name of movie>
<actor 1>
. . .
where the actors listed (one per line) after the movie name are the actors in that movie.
An example input file is in the subdirectory for this project of the class home directory on
lectura. The file may contain blank lines (lines containing only white space) and these
should be ignored. A single space will follow the keyword “Movie:”. The name of the
movie is considered to be the string starting at the character beyond that space and
continuing until the end of the line but NOT including the ‘\n’. The actor’s name is
everything on the line except for the possible newline (‘\n’) at the end. To simplify the
program, do not worry about trimming white space from the ends of the lines (other than
the ‘\n’) or capitalization. In other words the actor’s “John Wayne”, “John Wayne “, “john
wayne”, and ” John Wayne” can all be considered to be different by your program.
Behavior: After reading the file of cast lists and creating your graph, your program will
do the following: - read in a line from stdin containing an actor’s name
- compute the Bacon score for that actor (using a breadth first search)
- print the Bacon score (and the connecting actors and movies if –l option invoked)
to stdout according to the format specified below.
until no further input can be read.
? Assumptions: You can assume the following:
o The movie file, if it exists, is in correct format. In other words you may assume
the first nonblank line contains a movie name. You may also assume that if a line
starts with “Movie: “, then the line continues with some name. Note that since
blank lines are legal, and empty file is in fact a legal movie file.
Output: Output should be printed to stdout. To print out the score, use the format
printf(“Score: %d\n”, score)
where score is the calculated Bacon score. If there is no path between the actor and Kevin
Bacon, your program should print “Score: No Bacon!” on a single line.
If the actor queried is not in the graph, print a message to stderr and continue reading
queries. As always, whenever you print to stderr, when your program finally exits it
should exit with a status of 1.
The –l option
If the –l option is specified when the program is invoked, your program should print the
list of movies and actors connecting the queried actor to Kevin Bacon. The format for this
output should be
<Queried Actor>
was in <Movie Name> with
<Actor>
was in <Movie Name> with
<Actor>
. . .
was in <Movie Name> with
Kevin Bacon
Keeping track of this list is more difficult and only 10% of the test cases will involve this
option and they will be for extra credit. In other words you can still get 90/90 (100%) on
the assignment even if you fail to implement the –l option (you must still recognize an
invocation with the –l option as legal though), or 100/90 if you correctly implement the –l
option. Also note that while the Bacon score is unique, the list printed out may not be.
This means that if you are testing this option you may end up with a different output than
the example executable and still be correct. Don’t worry about getting the same list as the
example as long as your list is correct. When I test this part of the program I will use test
cases that have only one path to generate the Bacon score. You may want to create such
test cases yourself. (The example movie file should do.)
? Data Structures:
Once again you will use linked lists to represent the graph. The vertices of the graph will
be the actors. The edges will be the movies. If you are implementing the –l option the
edges will need to contain the movie name.
There are certainly variations on this. I used a linked list of actors. Each actor contains a
linked list of movies (nodes which point to a movie). Each movie contains a linked list of
actors (nodes pointing to an actor).
The Algorithm:
A depth first search will not work here. A depth first search finds if there is a path
between two vertices. We want to find the shortest path between two vertices. To
accomplish this, we will use a breadth first search. A depth first search recursively
searches all the children of each vertex. A breadth first search puts all the children on a
queue. This way all the children are search before the grandchildren, etc. Here is the
algorithm for a breadth first search:
BFS(Start, Target)
mark Start as queued
set level of Start to 0
add Start to the queue
while queue is not empty do
Take A from top of queue
For all children C of A do
if C == Target
return level of A + 1 //found, score A.level + 1
if C not queued
mark C as queued
set level of C to level of A + 1
add C to queue
return Target not found //If you get through loop without
//finding Target, there is no
//path from Start to Target
Note that in C you will have to implement this queue with a linked list. Don’t forget to
free it again after you’re done using it.
Here’s a hint for implementing the –l option. It requires you be able to trace the path from
Start to Target. If some actor A is put on the queue when looking at the children of some
actor B, the path to A comes from B. If your structure for nodes of the queue contains a
link to this “parent” node, then you can following these links from the queue node for the
Target back to the queue node for the Start. (i.e. It records your path.)
Error Conditions:
o Fatal errors: Bad invocation of program; input file cannot be opened for reading.
o Non-fatal errors: Queried actor is not in the graph.
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