Difference between revisions of "CSC231 Homework 7 Fall 2017"

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<br />
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<onlydft>
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 +
=Solutions=
 +
==Poblem 1==
 +
<br />
 +
::<source lang="asm">
 +
 +
;;; hw7a.asm
 +
;;; D. Thiebaut
 +
;;; Implements a 1-Dimensional game of life, where
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;;; the user is prompted to enter the number of generations
 +
;;; first, and then the initial cell population using a string
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;;; of spaces
 +
 +
;;; The documentation is given in the form of an original Python
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;;; program.
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;;; ===============================================================
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;;; dish = "                                      ##                                    "
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;;; dish = [ 1 if c=='#' else 0 for c in dish ]
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;;; N    = len( dish )
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;;; newDish = N * [0]
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;;; numGen  = 30
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section .data
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prompt db "> "
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dish dd 0
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N dd 0
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newDish times 256 db ' '
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numGen dd 30
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i dd 0
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j dd 0
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dishString db ' '
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section .text
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global _start
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extern _printString
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extern _printInt
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extern  _getString
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extern _getInput
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extern _println
 +
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_start:
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;;; get the number of generations from the user
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 +
mov ecx, prompt
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mov edx, 2
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call _printString
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call _getInput
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mov dword[numGen], eax
 +
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;;; get the address of the string entered by the user and save
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;;; it, as well as the number of chars in dish and N.
 +
 +
mov ecx, prompt
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mov edx, 2
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call _printString
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call _getString
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mov dword[dish], ecx
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mov dword[N], edx
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 +
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;;; if numGen == 0:
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;;; exit
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cmp dword[numGen], 0
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je theEnd
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;;; if N == 0:
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;;;    exit
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cmp dword[N], 0
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je theEnd
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;;; # iterate for many generations
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;;; for i in range( numGen ):
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;;;
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mov dword[i], 0
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forGen:
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mov eax, dword[i]
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cmp eax, dword[numGen]
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jge theEnd
 +
 +
;;;    # print current dish
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;;;    dishString = ''.join( [ '#' if c==1 else ' ' for c in dish ] )
 +
;;; print( dishString )
 +
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mov ecx, dword[dish]
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mov edx, dword[N]
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call _printString
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call _println
 +
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;;;    # generate newDish from what's in dish
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;;;    newDish = N * [0]
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;;;
 +
;;;    # loop over dish and generate newDish
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;;;    for j in range( 1, N-1 ):
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;;;        # compute fate of cell by looking at neighbors
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;;;        newDish[j] = dish[j-1] ^ dish[j+1]
 +
 +
mov dword[j], 1
 +
 +
mov esi, dword[dish]
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inc esi
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mov edi, newDish+1
 +
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forFate:mov eax, dword[j]
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mov ebx, dword[N]
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sub ebx, 1
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cmp eax, ebx
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jge forFateEnd
 +
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mov al, byte[esi-1]
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cmp al, byte[esi+1]
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je dead
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alive: mov al, '@'
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jmp setCell
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dead: mov al, ' '
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setCell:
 +
mov byte[edi],al
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inc esi
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inc edi
 +
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inc dword[j]
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jmp forFate
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forFateEnd:
 +
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;;;    # copy newDish into dish
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;;;    dish = newDish
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;;;
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mov dword[j],0
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mov esi, newDish
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mov edi, dword[dish]
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forCopy:
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mov eax, dword[j]
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cmp eax, dword[N]
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jge endForCopy
 +
 +
mov al, byte[esi]
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mov byte[edi], al
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inc esi
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inc edi
 +
 +
inc dword[j]
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jmp forCopy
 +
endForCopy:
 +
 +
;;; end of forGen loop
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inc dword[i]
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jmp forGen
 +
 +
;;; exit
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theEnd: mov eax, 1
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mov ebx, 0
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int 0x80
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 +
 +
</source>
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<br />
 +
==Problem 2==
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<br />
 +
::<source lang="C">
 +
 +
/*
 +
  hw7b.c
 +
  Program gets a number from the command line and
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  displays a Pascal triangle with that number of lines.
 +
 +
  How to compile and run
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  gcc -o hw7b hw7b.c
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  ./hw7b  10
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    1    0    0    0    0    0    0    0    0    0
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    1    1    0    0    0    0    0    0    0    0
 +
    1    2    1    0    0    0    0    0    0    0
 +
    1    3    3    1    0    0    0    0    0    0
 +
    1    4    6    4    1    0    0    0    0    0
 +
    1    5  10  10    5    1    0    0    0    0
 +
    1    6  15  20  15    6    1    0    0    0
 +
    1    7  21  35  35  21    7    1    0    0
 +
    1    8  28  56  70  56  28    8    1    0
 +
    1    9  36  84  126  126  84  36    9    1
 +
 +
*/
 +
#include <stdio.h>
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#include <stdlib.h>
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 +
 +
int main( int argc, char *argv[] ) {
 +
  if ( argc<= 1 ) {
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    printf( "Syntax %s N\n", argv[0] );
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    printf( "where N is the number of rows of Pascal triangle to be printed\n" );
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    return 0;
 +
  }
 +
  int i, j;
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  int Pascal[100];
 +
 +
  // get the number of rows
 +
  int N = atoi( argv[1] );
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  //printf( "N = %d\n", N );
 +
 +
  if ( N==0 )
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    return 0;
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  // initialize the first row with 1 and all 0s
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  Pascal[0] = 1;
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  for ( j=1; j<=N; j++ )
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    Pascal[j] = 0;
 +
 +
  // print all the rows
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  for ( i=1; i<=N; i++ ) {
 +
 +
    // print current row
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    for ( j=0; j<N; j++ ) {
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      printf( "%4d ", Pascal[j] );
 +
    }
 +
    printf( "\n" );
 +
 +
    // compute one row from the end
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    for ( j=N; j>0; j-- ) {
 +
      Pascal[j] = Pascal[j] + Pascal[j-1];
 +
    }
 +
    Pascal[0] = 1;
 +
  }
 +
  return 0;
 +
}
 +
 +
 +
 +
</source>
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</onlydft>
 
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Revision as of 22:14, 9 November 2017

--D. Thiebaut (talk) 14:24, 9 November 2017 (EST)


This assignment is due on 7/20/17 at 11:55 p.m.






Problem 1: Life on a new level


For this problem you have to implement a more interesting game of life. Take a look at a demo of the solution program. You will understand right away what your program will have to do: 

cs231a@aurora ~/handout $ ./hw7a
> 0
> @@

cs231a@aurora ~/handout $ ./hw7a
> 1 
> @      @       @
@      @       @

cs231a@aurora ~/handout $ ./hw7a
> 2
> @           @            @
@           @            @
 @         @ @          @ 

cs231a@aurora ~/handout $ ./hw7a
> 10
> @                @                  @
@                @                  @
 @              @ @                @ 
  @            @   @              @  
 @ @          @ @ @ @            @ @ 
    @        @       @          @    
   @ @      @ @     @ @        @ @   
  @   @    @   @   @   @      @   @  
 @ @ @ @  @ @ @ @ @ @ @ @    @ @ @ @ 
        @@               @  @        
       @@@@             @ @@ @


Explanations


  • The program prompts for 2 pieces of information: an integer, which can be 0 or positive, and a string of spaces and '@' characters.
  • The integer represents the number of generations we want to have our dish of cells go through. 0 means nothing gets printed. 1 means the original generation entered by the user is shown. 2 or more indicates that the program goes through a loop and evolves the cells in the dish.
  • The string can be as long as what _getString allows, which is 1000 characters. However, for this assignment, we will assume that the string of live and dead cells entered by the user will never be more than 100 characters. It also means that your program will be tested only with strings of no more than 100 cells.


Requirements


  • Your program should NOT any loop instructions. Use the cmp instruction and conditional jumps to control the different loops.


Additional Information


Program Developed in Class


     getcopy  GameOfLifeClassV2.asm


Getting an int from the keyboard


Use _getInput, which is part of the 231Lib.asm library. 

        call    _getInput
        mov     dword[numGen], eax


Getting a string from the keyboard


We've used it before: _getString. It returns the address of the string in ecx, and the number of chars entered in edx. _getString has its own buffer of 1000 characters in which is saves the string. It will return the address of this buffer in ecx.

Submission


  • Save your program in a file called hw7a.asm and submit it to Moodle.
  • Document your code well!



Problem 2: Coding in C


  • Write a C program called hw7b.c that uses loops to print the first N lines of the Pascal triangle.
  • Examples with the solution program:


for n in 0 1 2 5 10 ; do echo $n ;  ./hw7b $n; echo ""; done
0

1
   1 

2
   1    0 
   1    1 

5
   1    0    0    0    0 
   1    1    0    0    0 
   1    2    1    0    0 
   1    3    3    1    0 
   1    4    6    4    1 

10
   1    0    0    0    0    0    0    0    0    0 
   1    1    0    0    0    0    0    0    0    0 
   1    2    1    0    0    0    0    0    0    0 
   1    3    3    1    0    0    0    0    0    0 
   1    4    6    4    1    0    0    0    0    0 
   1    5   10   10    5    1    0    0    0    0 
   1    6   15   20   15    6    1    0    0    0 
   1    7   21   35   35   21    7    1    0    0 
   1    8   28   56   70   56   28    8    1    0 
   1    9   36   84  126  126   84   36    9    1


Skeleton Program


  • Play with the following program to see what it does, and then build up from there.


/*                                                                                                                      
  hw7b.c                                                                                                                
  A skeleton program to get you started.                                                                                
*/
#include <stdio.h>
#include <stdlib.h>


int main( int argc, char *argv[] ) {
  if ( argc<= 1 ) {
    printf( "Syntax %s N\n", argv[0] );
    printf( "where N is the number of rows of Pascal triangle to be printed\n" );
    return 0;
  }

  int N = atoi( argv[1] );
  printf( "N = %d\n", N );

  return 0;
}








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