[SOLVED] Cosc 3360 – 6310 assignment 2: encoding/decoding using walsh codes

This assignment will introduce you to interprocess
communication in UNIX using stream sockets.
Walsh Codes
A Walsh code of length n = 2k is a set of perfectly
orthogonal codewords that can be defined and
generated by the rows of a 2k × 2k Hadamard matrix.
Starting with a 1×1 matrix H1 = [0], higher-order
Hadamard matrices can be generated by the following
recursion:
�”# = %
�”#'( �”#'(
�”#'( �”#'( )))))))*
For our program, we will use H4. Given H0 = [0], we have
that
�+ = ,
0 0
0 1
0 0
0 1
0 0
0 1
1 1
1 0
/
so that the Walsh codes of length 4 represented in a
bipolar form (±1), and using positive logic (0 equal to -1,
and 1 equal to 1) are: w0 =[-1,-1,-1,-1], w1 =[-1,+1,-1,+1],
w2 =[-1,-1,+1,+1], and w3 = [-1,+1,+1,-1]. Note that these
codes are mutually orthogonal; that is, the dot product of
any pair of codes is zero.
Specifications
You must write two programs to allow communication
between three processes using the encoding/decoding
mechanism based on Walsh codes.
The sender / receiver program: the user will execute
this program using the following syntax:
./exec_filename hostname port_no < input_filename
where exec_filename is the name of your executable file,
hostname is the address where the encoder program is
located, port_no is the port number used by the encoder
program, and input_filename is the name of your input
file.
This program receives the information about the
hostname and the port number from the command line
and reads the information about the transmission
requests from a file using I/O redirection.
Input Format: the input file will always have three lines
representing the request from process 1, 2, and 3
respectively.
3 4 // Process 1 sending to process 3 the integer value 4
1 5 // Process 2 sending to process 1 the integer value 5
2 7 // Process 3 sending to process 2 the integer value 7
• The first value of each line represents the id of the
process which the current process wants to contact.
• The second value of each line represents the data
(integer value) that the current process wants to
send. The range of this value will be between zero
and seven. Based on this restriction the
encoding/decoding mechanism will always represent
the data field with three bits.
After reading the input file, the sender/receiver program
will create three child processes (representing processes
1, 2, and 3). Each child process will create a socket to:
a) send the request to the encoder program with the
destination process and data; b) receive the encoded
signal (twelve integer values), and the code (four integer
values); c) Decode the message (integer value between
zero and seven) using the received information; and d)
Print the encoded message, the code, and the decoded
message.
The encoder program: the user will execute this
program using the following syntax:
./exec_filename port_no
where exec_filename is the name of your executable file,
and port_no is the port number to create the socket.
The tasks of this program are: a) Receive the request
from each child process from the sender / receiver
program; b) Generate the encoded message using
Walsh codes; c) Send the encoded message and
corresponding code to all the child processes (waiting
one second between replies); and d) Finish its
execution.
The encoding process:
The encoder program will use Walsh codes of size four.
Each child process will have a Walsh code assigned in
the encoder program (w1 for child 1, w2 for child 2, w3
for child 3). To generate the encoded message, the
encoder program has to transform the integer value sent
by a client to its binary representation (three bits). Each
bit of the value sent by the client will be encoded
(“spread”) by the corresponding Walsh code. Based on
the input file we have:
For child process 1:
Value = 4.
Binary = +1 -1 -1 (bipolar form, positive logic)
Encoded Message 1 (EM1):
Binary * w1 =
+1 -1 -1
-1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1
EM1= -1 +1 -1 +1 +1 -1 +1 -1 +1 -1 +1 -1
For child process 2:
Value = 5.
Binary = +1 -1 +1 (bipolar form, positive logic)
Encoded Message 2 (EM2):
Binary * w2 =
+1 -1 +1
-1 -1 +1 +1 -1 -1 +1 +1 -1 -1 +1 +1
EM2= -1 -1 +1 +1 +1 +1 -1 -1 -1 -1 +1 +1
For child process 3:
Value = 7.
Binary = +1 +1 +1 (bipolar form, positive logic)
Encoded Message 3 (EM3):
Binary * w3 =
+1 +1 +1
-1 +1 +1 -1 -1 +1 +1 -1 -1 +1 +1 -1
EM3= -1 +1 +1 -1 -1 +1 +1 -1 -1 +1 +1 -1
Encoded message (EM) = EM1+ EM2 + EM3
EM = –1 +1 –1 +1 +1 –1 +1 –1 +1 –1 +1 –1 EM1
–1 –1 +1 +1 +1 +1 –1 –1 –1 –1 +1 +1 EM2
–1 +1 +1 –1 –1 +1 +1 –1 –1 +1 +1 –1 EM3
EM = –3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
The decoding process:
Each child process will receive the encoded message
and the code corresponding to the sender process. For
the proposed input file, we have:
Child process 1:
EM = –3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
w2 = –1 –1 +1 +1 (code of child process 2).
Decoded message 1 (DM1)= EM * w2
–3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
–1 –1 +1 +1 –1 –1 +1 +1 –1 –1 +1 +1
DM1 = +3 –1 +1 +1 –1 –1 +1 –3 +1 +1 +3 –1
Because each bit of the original message was spread
using codes of size four, we have that:
DM1 =
+3 –1 +1 +1 –1 –1 +1 –3 +1 +1 +3 –1
+4 –4 +4
Then dividing each element by the size of the code, we
have:
+4/4 –4/4 +4/4 =
+1 –1 +1 =
5
Child process 2:
EM = –3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
w3 = –1 +1 +1 –1 (code of child process 3).
Decoded message 1 (DM2)= EM * w3
–3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
–1 +1 +1 –1 –1 +1 +1 –1 –1 +1 +1 –1
DM2 = +3 +1 +1 –1 –1 +1 +1 +3 +1 –1 +3 +1
Because each bit of the original message was spread
using codes of size four, we have that:
DM2 =
+3 +1 +1 –1 –1 +1 +1 +3 +1 –1 +3 +1
+4 +4 +4
Then dividing each element by the size of the code, we
have:
+4/4 +4/4 +4/4 =
+1 +1 +1 =
7
Child process 3:
EM = –3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
w1 = –1 +1 –1 +1 (code of child process 1).
Decoded message 1 (DM3)= EM * w1
–3 +1 +1 +1 +1 +1 +1 –3 –1 –1 +3 –1
–1 +1 –1 +1 –1 +1 –1 +1 –1 +1 –1 +1
DM3 = +3 +1 –1 +1 –1 +1 –1 –3 +1 –1 –3 –1
Because each bit of the original message was spread
using codes of size four, we have that:
DM3 =
+3 +1 –1 +1 –1 +1 –1 –3 +1 –1 –3 –1
+4 –4 –4
Then dividing each element by the size of the code, we
have:
+4/4 –4/4 –4/4 =
+1 –1 –1 =
4
NOTES:
• Use a single-threaded server to avoid critical
sections inside the server.
• Since your server will not fork any child processes,
you should not worry about handling zombies and
can safely ignore the fireman() call in the primer.
• You can safely assume that:
a) The input files will always be in the proper format.
b) The number of lines in the input file is always
three.
c) The first value of each line in the input file
(destination process) will always be different.
d) Child 1 will always use the Walsh code w1, Child
2 will always use the Walsh code w2, and Child 3 will
always use the Walsh code w3
e) Naming convention for your source file: first
name_lastname.cpp/ first name_lastname.c
These specifications were written on Wednesday,
February 27, 2019. Please refer to the course web site
for corrections and updates.
Expected output:
Encoder program:
Here is the message from child 1: Value = 4, Destination = 3
Here is the message from child 2: Value = 5, Destination = 1
Here is the message from child 3: Value = 7, Destination = 2
Sender / Receiver program:
Child 1, sending value: 4 to child process 3
Child 2, sending value: 5 to child process 1
Child 3, sending value: 7 to child process 2
Child 1
Signal:-3 1 1 1 1 1 1 -3 -1 -1 3 -1
Code: -1 -1 1 1
Received value = 5
Child 2
Signal:-3 1 1 1 1 1 1 -3 -1 -1 3 -1
Code: -1 1 1 -1
Received value = 7
Child 3
Signal:-3 1 1 1 1 1 1 -3 -1 -1 3 -1
Code: -1 1 -1 1
Received value = 4