Difference between revisions of "CSC231 Homework 8 2010"
(→Game of Life in 1D) |
(→Details) |
||
| Line 49: | Line 49: | ||
** If a cell is dead in one generation, but has only one alive neighbor, it will be alive in the next generation. This correspond to a positive birth rate under preferred population size. | ** If a cell is dead in one generation, but has only one alive neighbor, it will be alive in the next generation. This correspond to a positive birth rate under preferred population size. | ||
* Your program will print each generation one above the next, one line at a time. | * Your program will print each generation one above the next, one line at a time. | ||
| − | * | + | * Your program should print at least 20 generations, but no more than 60, so as to limit the size of the printout I'll hand back to you. |
| + | * Below is an example generated by the applet on Garret Wilson's [http://www.garretwilson.com/education/institutions/soas/dissertation/lifecellauto1d.html page], where the initial population was of size 6, distributed as follows: ..XXX.XXX....... where dots are dead, and Xs are alive. | ||
<br /> | <br /> | ||
| Line 58: | Line 59: | ||
<br /> | <br /> | ||
<br /> | <br /> | ||
| + | |||
| + | =Tricks= | ||
| + | |||
| + | * At some point you will have to compute the number of neighbors for each cell. A simple way to do this is to maintain your array of cells as a collection of integers that are either 0 or 1, 0 for dead, one for alive. Then when you want to compute the number of neighbors of cell Cell[i], you add together Cell[i-1] to Cell[i+1]. You will get 0, 1, or 2. You can create another array containing the rules of the game, so that you know how what the fate of the cell is by looking it up in the array. | ||
| + | |||
| + | fate db 0, 1, 0 | ||
| + | |||
| + | if the number of neighbors is stored in ebx, for example, then mov al,[fate+ebx] will put the fate of the cell in al. This is the type of trick you can use when you can't compare ebx to 0, 1 or 2. | ||
| + | |||
| + | =Submission= | ||
| + | |||
| + | submit hw8 hw8.asm | ||
Revision as of 08:55, 11 November 2010
--D. Thiebaut 13:36, 11 November 2010 (UTC)
Page under construction!
This homework is due Wednesday evening, 11/17/10, at midnight. You can work on this assignment in pairs.
Game of Life in 1D
- Assignment 1
Watch Conway talk about the Game of Life on YouTube (Parts 1 and 2).
- Assignment 2
Read the page on Conway's Game of Life on Wikipedia. Great introduction.
- Assignment 3
Your assignment is simple, and you have to write a program that you may have already written in one or two different languages before: [Conway's Game of Life]. The challenge this time, is that it must be done in assembly, using functions where parameters are passed through the stack and not via registers, and you have to write it without the compare instruction that would normally be used for this purpose (I'll show you a way around it).
Details
- Your program should be called hw8.asm
- It should maintain the cells in a 1-dimensional array of size 70
- Each cell can be either dead or alive. Dead cells are presented as spaces, life cells with a character of your choice.
- You are free to choose the initial population in the array
- The rules are simple:
- If a cell is alive in one generation, it will be dead the next. The lifespan is finite.
- If a cell is dead in one generation, but has only one alive neighbor, it will be alive in the next generation. This correspond to a positive birth rate under preferred population size.
- Your program will print each generation one above the next, one line at a time.
- Your program should print at least 20 generations, but no more than 60, so as to limit the size of the printout I'll hand back to you.
- Below is an example generated by the applet on Garret Wilson's page, where the initial population was of size 6, distributed as follows: ..XXX.XXX....... where dots are dead, and Xs are alive.
Tricks
- At some point you will have to compute the number of neighbors for each cell. A simple way to do this is to maintain your array of cells as a collection of integers that are either 0 or 1, 0 for dead, one for alive. Then when you want to compute the number of neighbors of cell Cell[i], you add together Cell[i-1] to Cell[i+1]. You will get 0, 1, or 2. You can create another array containing the rules of the game, so that you know how what the fate of the cell is by looking it up in the array.
fate db 0, 1, 0
if the number of neighbors is stored in ebx, for example, then mov al,[fate+ebx] will put the fate of the cell in al. This is the type of trick you can use when you can't compare ebx to 0, 1 or 2.
Submission
submit hw8 hw8.asm
