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ser334_unit6_exercise_sample.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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\begin_body
\begin_layout Standard
\noindent
Arizona State University
\begin_inset space \hfill{}
\end_inset
SER334: Operating Systems & System Programming
\end_layout
\begin_layout Standard
\noindent
Lecturer Acuña (4), UGTA Lisonbee (1)
\begin_inset space \hfill{}
\end_inset
Revised 3/11/2022
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\begin_layout Standard
\noindent
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LatexCommand rule
offset "0.5ex"
width "100col%"
height "1pt"
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\begin_layout Standard
\size huge
Unit 6 Sample Problems - Threads
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inverted 0
status collapsed
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\begin_inset Formula $\;$
\end_inset
(SOLN)
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\begin_inset VSpace bigskip
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $ $
\end_inset
\end_layout
\begin_layout Standard
In this exercise, we will review the ideas of Threads.
\end_layout
\begin_layout Itemize
Length: 1:15 minutes with discussion.
\end_layout
\begin_layout Itemize
Questions: Q1-Q5.
(optional: Q6)
\end_layout
\begin_layout Standard
\begin_inset VSpace 0.5cm
\end_inset
\end_layout
\begin_layout Standard
\series bold
\size larger
Multicore Programming
\end_layout
\begin_layout Enumerate
[Acuña] Consider the algorithmic task of compiling a set of source files
in a programming project.
For each source file, the compiler needs to emit an object file containing
byte-code that represents the original source code at a lower level.
Then, each file is linked together (with standard libraries) to generate
a final executable file.
\end_layout
\begin_deeper
\begin_layout Standard
Say we want to design a compiler to compile a project as quickly as possible.
Of the five issues in multi-core programming, which is the most problematic
for multi-threading this system?
\end_layout
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\series bold
Ans: [Acuña]
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\begin_layout Standard
Balance.
Source files can be of vastly different sizes (think 1 method vs 100).
Since the length of the files is proportional to how long they take to
compile, we might end up with cases where compilation of one file takes
so long that other the threads compiling other files end up idling.
(A runner up would be Data Dependency, for cases where files dependent
on one another.
Like inheritance.)
\end_layout
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña] Consider the algorithmic task of compressing a folder.
The typical threaded implementation usually divides up files in a folder
being compressed, and then sends out the work of compressing some sub-set
of them to different threads.
Each thread will produce compressed versions of each file it is given,
which are then combined into a single compression archive.
This gives a significant speedup.
\end_layout
\begin_deeper
\begin_layout Standard
Another compression scheme uses a so-called a "dictionary".
In this "dictionary", chunks of uncompressed data are mapped to indices
with compressed data.
The dictionary will end up stored in the compressed file so it can be used
to reconstruct the compressed data.
The idea is that if data repeats (e.g., same file twice), both compressed
instances of the file can share the same compressed data.
This can give a significant improvement in file size
\end_layout
\begin_layout Standard
Say we want to design a dictionary based compression tool.
Of the five issues in multi-core programming, which is the most problematic
for multi-threading this system?
\end_layout
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\begin_inset VSpace 4cm
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\end_inset
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
Ans [Acuña]:
\end_layout
\begin_layout Standard
Data Dependency.
The issue is that we are limited to a single dictionary of pieces of compressed
file.
If two threads try to compress files and decide that they need to add an
entry to the dictionary, then they will compute it in parallel, but only
one result will actually get to be stored.
Since the purpose of the dictionary is to support reuse, the slower thread
will have to discard its result (the dictionary already has an entry),
and instead reference the entry produced by the first.
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Lisonbee] Consider the following program and answer the question:
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset listings
inline false
status open
\begin_layout Plain Layout
#include <stdio.h>
\end_layout
\begin_layout Plain Layout
#include <stdlib.h>
\end_layout
\begin_layout Plain Layout
#include <pthread.h>
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
int fib(int n) {
\end_layout
\begin_layout Plain Layout
if (n == 0)
\end_layout
\begin_layout Plain Layout
return 0;
\end_layout
\begin_layout Plain Layout
else if (n == 1)
\end_layout
\begin_layout Plain Layout
return 1;
\end_layout
\begin_layout Plain Layout
return fib(n-1) + fib(n-2);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
void* runner(void* param) {
\end_layout
\begin_layout Plain Layout
int n = (int)param;
\end_layout
\begin_layout Plain Layout
printf("Fib @ %d is %d
\backslash
n", n, fib(n));
\end_layout
\begin_layout Plain Layout
pthread_exit(0);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
void threadded_fib(int* arr, int size) {
\end_layout
\begin_layout Plain Layout
pthread_t threads[size];
\end_layout
\begin_layout Plain Layout
for (int i = 0; i < size; i++) {
\end_layout
\begin_layout Plain Layout
pthread_create(&threads[i], NULL, runner, (void*)arr[i]);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
for (int i = 0; i < size; i++) {
\end_layout
\begin_layout Plain Layout
pthread_join(threads[i], NULL);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
int main() {
\end_layout
\begin_layout Plain Layout
int nums[] = {1, 6, 15, 2, 89};
\end_layout
\begin_layout Plain Layout
int size = 5;
\end_layout
\begin_layout Plain Layout
threadded_fib(nums, size);
\end_layout
\begin_layout Plain Layout
return 0;
\end_layout
\begin_layout Plain Layout
}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Of the five issues in multi-core programming, which is the most problematic
for this problem?
\end_layout
\begin_layout Standard
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inverted 0
status open
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\end_inset
\begin_inset Branch soln
inverted 0
status open
\begin_layout Standard
\series bold
Ans: [Lisonbee]
\end_layout
\begin_layout Standard
Balance.
Each thread computes a single fibonacci number, but the placement of the
number in the sequence will determine how much work that thread does.
In the example sequence, the first thread will get the value '1' and return
'1' without recursively calling fib at all.
The fifth thread gets '89' and returns the result of fib(89) (a very large
value; will overflow countless times), but the computation necessary to
calculate this value is many times greater than the first thread.
That being said, this thread will take much longer to finish it's work
than the first thread, leading to imbalance between the amount of work
each thread has done.
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Newpage pagebreak
\end_inset
\end_layout
\begin_layout Standard
\series bold
\size larger
Thread Libraries
\end_layout
\end_deeper
\begin_layout Enumerate
[Silberschatz 4.15 edited] Consider the following code fragment and answer
the following questions:
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset listings
inline false
status open
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pid_t pid;
\end_layout
\begin_layout Plain Layout
temp = fork();
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
if (temp == 0) {
\end_layout
\begin_layout Plain Layout
fork();
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
thread_create(...);
\end_layout
\end_inset
\end_layout
\begin_layout Enumerate
Using
\begin_inset Quotes eld
\end_inset
lifeline notation
\begin_inset Quotes erd
\end_inset
, draw the creation of processes and threads during execution.
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Branch blank
inverted 0
status collapsed
\begin_layout Standard
\begin_inset VSpace 7cm
\end_inset
\end_layout
\end_inset
\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
\begin_inset Graphics
filename unit06/exe/ser334_unit5_exercise_sample_soln_trace.png
scale 40
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
How many unique processes are created? (Do not include the initial process.)
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Branch blank
inverted 0
status collapsed
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
2
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
How many unique threads are created? (Hint: processes don't count!)
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Branch blank
inverted 0
status collapsed
\begin_layout Standard
\begin_inset VSpace 2cm
\end_inset
\end_layout
\end_inset
\begin_inset Branch soln
inverted 0
status collapsed
\begin_layout Standard
3
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Newpage pagebreak
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
[Acuña] Consider the following program written with the pthreads library:
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset listings
inline false
status open
\begin_layout Plain Layout
#include <pthread.h>
\end_layout
\begin_layout Plain Layout
#include <stdio.h>
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
int t1, t2, input;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
void* runner(void *param) {
\end_layout
\begin_layout Plain Layout
int result, upper = (int)param;
\end_layout
\begin_layout Plain Layout
int a = 1, b = 1, c = 1;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
for (int i = 2; i < input-upper; i++) {
\end_layout
\begin_layout Plain Layout
c = a + b;
\end_layout
\begin_layout Plain Layout
a = b;
\end_layout
\begin_layout Plain Layout
b = c;
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
if(upper == 1)
\end_layout
\begin_layout Plain Layout
t1 = c;
\end_layout
\begin_layout Plain Layout
else
\end_layout
\begin_layout Plain Layout
t2 = c;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_exit(0);
\end_layout
\begin_layout Plain Layout
}
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
int main(int argc, char *argv[]) {
\end_layout
\begin_layout Plain Layout
pthread_t tid1, tid2;
\end_layout
\begin_layout Plain Layout
pthread_attr_t attr;
\end_layout
\begin_layout Plain Layout
input = 10;
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_attr_init(&attr);
\end_layout
\begin_layout Plain Layout
pthread_create(&tid1, &attr, runner, 1);
\end_layout
\begin_layout Plain Layout
pthread_create(&tid2, &attr, runner, 2);
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
pthread_join(tid1, NULL);
\end_layout
\begin_layout Plain Layout
pthread_join(tid2, NULL);
\end_layout
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printf("result = %d
\backslash
n", t1+t2);
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}
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What is this programming trying to compute and how does it uses threads?
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Ans [Acuña]:
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This program is computing the 10th Fibonacci number.
It uses one thread to compute fib(n-1) and store it in t1, and another
to compute fib(n-2) and store it into t2.
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[Acuña] Consider the problem of finding the maximum element in an array
of values, which is solved by the following program:
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