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# COMP-200: Intro to Computer Systems
## Assignment 3
### mo khan - 3431709
Choose ONE exercise each from Chapters 6, 7, and 8
Chapter 6:
> 13. Write a complete assembly language program (including all necessary
> pseudo-ops) that reads in a series of integers, one at a time, and outputs
> the largest and smallest values. The input will consist of a list of integer
> values
```assembly
!include assignments/3/max_min.s
```
<!--
I wrote the following C code and then used the GNU C Compiler to convert the C
code into Assembly code for my machine. Below is the C code, compiler command,
and final assembly code.
main.c
```c
!include assignments/3/main.c
```
bash
```bash
$ gcc -S -fverbose-asm -02 main.c
```
main.s
```assembly
!include assignments/3/main.s
```
-->
Chapter 7:
> 16. Given the following diagram, where the numbers represent the time delays across a link:
> a. How many simple paths (those that do not repeat a node) are there from node A to G?
```plaintext
A<->B: 5
A<->D: 6
A<->F: 3
B<->C: 4
B<->E: 2
C<->D: 3
C<->E: 1
C<->G: 2
D<->E: 1
F<->G: 8
```
There are 7 simple paths
```plaintext
| Path |
| ---- |
| a,b,c,g |
| a,b,e,c,g |
| a,b,e,d,c,g |
| a,d,c,g |
| a,d,e,b,c,g |
| a,d,e,c,g |
| a,f,g |
```
> b. What is the shortest path from node A to node G? What is the overall delay?
The shortest path has a distance of 10.
The overall delay is 11 (`((11 + 10 + 13 + 11 + 15 + 10 + 11)/7) = 11`).
```plaintext
| Path | Total Distance |
| ---- | -------------- |
| a,b,c,g | 5+4+2 = 11 |
| a,b,e,c,g | 5+2+1+2 = 10 |
| a,b,e,d,c,g | 5+2+1+3+2 = 13 |
| a,d,c,g | 6+3+2 = 11 |
| a,d,e,b,c,g | 6+1+2+4+2 = 15 |
| a,d,e,c,g | 6+1+1+2 = 10 |
| a,f,g | 3+8 = 11 |
```
> c. If node E fails, does that change the shortest path? If so, what is the new shortest path?
Yes, if node E fails then there are 3 simple paths remaining with the shortest path having a distance of 11.
```plaintext
A<->B: 5
A<->D: 6
A<->F: 3
B<->C: 4
C<->D: 3
C<->G: 2
F<->G: 8
```
```plaintext
| Path | Total Distance |
| ---- | -------------- |
| a,b,c,g | 5+4+2 = 11 |
| a,d,c,g | 6+3+2 = 11 |
| a,f,g | 3+8 = 11 |
```
The following code can be used to prove these results:
```ruby
!include assignments/3/graph.rb
```
Chapter 8:
> 9. Using a Caesar cipher with s = 5, decode the received message RTAJ TZY FY IF
The following code can be used to decipher the Caesar cipher.
```ruby
!include assignments/3/caesar.rb
```
The plaintext is `MOVE OUT AT DAWN`.
```bash
$ ruby caesar.rb
# Caesar Cipher
| --------------- | --------------------------------------------------- |
| Plain Alphabet | A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z |
| Cipher Alphabet | F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A,B,C,D,E |
| --------------- | --------------------------------------------------- |
| Key | 5 |
| Ciphertext | RTAJ TZY FY IFBS |
| Plaintext | MOVE OUT AT DAWN |
```
References:
* https://blog.rchapman.org/posts/Linux_System_Call_Table_for_x86_64/
* https://cs.lmu.edu/~ray/notes/gasexamples/
* https://ftp.gnu.org/old-gnu/Manuals/gas-2.9.1/html_chapter/as_7.html
|
