patterncsharpMinor
Garbage collection loop in Game of Life
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loopcollectiongamelifegarbage
Problem
I'm working on a Game Of Life clone as my first project. I'm still relatively new to programming, so my code is probably not really optimised.
The biggest bottleneck is in my
My second bottleneck is somewhere in my
I've been checking with the built-in profiler of Unity3D, and this reports that the biggest bottleneck is the garbage collection.
I also tried to put half of the loop in another thread, but that didn't work out properly, because it needs to look up cells in the other half.
If you see any other ways to improve my code, and not just the performance, feel free to let me know. But please, keep it simple, as I'm still relatively new to programming.
The relevant code:
```
void Update()
{
timeElapsed += Time.deltaTime; // this will add deltaTime to timeElapsed,
if (timeElapsed >= timeBetween) // until timeElapsed exceeds timeBetween,
{ // which allows a new generation to be calculated
timeElapsed -= timeBetween;
if (gameState == GameState.Game_Playing)
{
grid = CheckAdjacent(grid); // draw new generation
UpdateInformation(); // update information variables
}
}
}
public static void DrawGeneration(bool[,] grid)
{
for (int x = 0; x < gridX; x++)
{
for (int y = 0; y < gridY; y++)
{
cubeGrid[x,y].renderer.enabled = grid[x,y]; // enables the renderer for active cells.
}
}
}
private void UpdateInformation()
{
generation++; //counts the generations
}
public static int GetAliveCells ()
{
int i = 0;
for (int x = 0; x < gridX; x++)
{
The biggest bottleneck is in my
Start() method, but that contains a lot of Unity3D specific methods which most people here probably don't know about. That's why I'm leaving that out.My second bottleneck is somewhere in my
Update() method. In here, I'm looping through my array which is currently 100 by 100 large, 10.000 in total.I've been checking with the built-in profiler of Unity3D, and this reports that the biggest bottleneck is the garbage collection.
I also tried to put half of the loop in another thread, but that didn't work out properly, because it needs to look up cells in the other half.
If you see any other ways to improve my code, and not just the performance, feel free to let me know. But please, keep it simple, as I'm still relatively new to programming.
The relevant code:
```
void Update()
{
timeElapsed += Time.deltaTime; // this will add deltaTime to timeElapsed,
if (timeElapsed >= timeBetween) // until timeElapsed exceeds timeBetween,
{ // which allows a new generation to be calculated
timeElapsed -= timeBetween;
if (gameState == GameState.Game_Playing)
{
grid = CheckAdjacent(grid); // draw new generation
UpdateInformation(); // update information variables
}
}
}
public static void DrawGeneration(bool[,] grid)
{
for (int x = 0; x < gridX; x++)
{
for (int y = 0; y < gridY; y++)
{
cubeGrid[x,y].renderer.enabled = grid[x,y]; // enables the renderer for active cells.
}
}
}
private void UpdateInformation()
{
generation++; //counts the generations
}
public static int GetAliveCells ()
{
int i = 0;
for (int x = 0; x < gridX; x++)
{
Solution
-
I would advise you to count neighbours from alive points, not for cells to be checked. Null a 100x100 int array Neighbours. For every alive cell add +1 into every neighbouring cell in Neighbours. After the pass you have number of neighbours for the every cell, but time used is about 1/4 at most, because you saved the time not counting neighbours for zero cells.
-
You are taking left neighbours for utmost left cells, for example. Out of boundaries! You are blocking this by taking a function and modulo and thus making the toroid out of plane, but thus you are counting much longer - division or modulo are slow, functions even more slow. Simply count neighbours normally for the field except boundaries, otherwards for the four sides and four corners. More to write, but less to count later.
-
If you want a shorter code, you could use a help arrays ( I have invented them for myself in 1978)
and in counting neighbours for a cell do:
-
Count the time spent for different activities, for not to waste time on optimizing part that takes 0.1% of time.
So you can throw off many objects, not to create them at all. Only hold 2 desk in 2x100x100 array - each time one will be active, the other will be old. You can even count neighbours and later make living/dead cells on the same desk. No garbage collection.
I would advise you to count neighbours from alive points, not for cells to be checked. Null a 100x100 int array Neighbours. For every alive cell add +1 into every neighbouring cell in Neighbours. After the pass you have number of neighbours for the every cell, but time used is about 1/4 at most, because you saved the time not counting neighbours for zero cells.
-
You are taking left neighbours for utmost left cells, for example. Out of boundaries! You are blocking this by taking a function and modulo and thus making the toroid out of plane, but thus you are counting much longer - division or modulo are slow, functions even more slow. Simply count neighbours normally for the field except boundaries, otherwards for the four sides and four corners. More to write, but less to count later.
-
If you want a shorter code, you could use a help arrays ( I have invented them for myself in 1978)
// neighbor-vectors in xy coordinates clockwise from algebraic x axis.
int [8] dx={1,1,0,-1,-1,-1,0,1};
int [8] dy={0,-1,-1,-1,0,1,1,1};and in counting neighbours for a cell do:
if(cell[x,y]>0)
for(i=0;i=0 & x1=0 & y1<100){
neighbours[x1,y1]++;
}
}-
Count the time spent for different activities, for not to waste time on optimizing part that takes 0.1% of time.
So you can throw off many objects, not to create them at all. Only hold 2 desk in 2x100x100 array - each time one will be active, the other will be old. You can even count neighbours and later make living/dead cells on the same desk. No garbage collection.
Code Snippets
// neighbor-vectors in xy coordinates clockwise from algebraic x axis.
int [8] dx={1,1,0,-1,-1,-1,0,1};
int [8] dy={0,-1,-1,-1,0,1,1,1};if(cell[x,y]>0)
for(i=0;i<8;i++){
x1=x+dx[i];
y1=y+dy[i];
if(x1>=0 & x1<100 & y1>=0 & y1<100){
neighbours[x1,y1]++;
}
}Context
StackExchange Code Review Q#8946, answer score: 5
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