## Archive for March, 2008

### Demon

Wednesday, March 26th, 2008

I’ve been meaning to make a “pretty” implementation of an Xscreensaver hack for ages. It’s called “demon”, and it features a random grid of colored cells that end up in swirling color patterns. I finally got around to trying it out, and used it as an opportunity to play with Mac OS X’s graphics APIs.

### Code!

Without any particular context, here is how I iterate. Note that I make the edges wrap around. If the syntax is slightly wacky-looking that’s because it’s Objective C.

#### Iteration

The logic itself is fairly simple; at each iteration, one particular color of cell gets to “eat” any neighbors of the previous color. To wit: First the red pixels gets to eat their purple neighbors, then the orange pixels get to eat their red neighbors, then the yellow the orange, and so on, until red comes around again. Stable arrangements come up when you have cyclical patterns where the colors can chase one another forever without getting consumed or any one color is entirely eliminated, at which point a cycle can never emerge.

```- iterate{ static int c = 0; char **newgrid; int x, y;   if(grid == grid1){ newgrid = grid2; }else{ newgrid = grid1; }   int C = (c + 1) % ncolors; for(x = 0; x < width; x++){ for(y = 0; y < height; y++){ if((grid[x][y] == c) && ( (x > 0 && grid[x-1][y] == C) || (x == 0 && grid[width-1][y] == C) || (x < width - 1 && grid[x+1][y] == C) || (x == width - 1 && grid[0][y] == C) || (y > 0 && grid[x][y-1] == C) || (y == 0 && grid[x][height - 1] == C) || (y < height - 1 && grid[x][y+1] == C) || (y == height - 1 && grid[x][0] == C))){ newgrid[x][y] = C; }else{ newgrid[x][y] = grid[x][y]; } } } c = (c + 1) % ncolors;   grid = newgrid;   [self redraw];   return 0; }```

#### Drawing

Unfortunately, the means of drawing I used is rather slow. This is actually my second attempt, as the first was even slower. For reference, the first attempt drew using `NSRectFill` and associates.

```- (void)drawRect:(NSRect)rect{ if(grid){ int x, y; CGRect pixel; CGContextRef context = [[NSGraphicsContext currentContext] graphicsPort];   pixel.size.height = size; pixel.size.width = size;   for(x = 0; x < width; x++){ pixel.origin.x = x * size; for(y = 0; y < height; y++){ pixel.origin.y = y * size; Color c = colors[grid[x][y]]; CGContextSetRGBFillColor(context, c.r, c.g, c.b, 1); CGContextFillRect(context, pixel); } } }else{ [[NSColor blackColor] set]; NSRectFill ([self bounds]); } }```

### Future

I will next try and draw to a bitmap and see if that’s any faster. I also suspect that I may want to iterate with a second thread rather than having a callback timer go off every quarter-second. When I’m satisfied with the performance, I’ll post an app on here.

### Demon Preview

Tuesday, March 18th, 2008

In order to get Dion off my back, I am posting this screenshot of the next mini-project I’ve been working on. It’s an implementation of the cellular automaton from the xscreensaver module “demon”.

A full writeup and code will follow.

### Dice Rolling 2

Saturday, March 1st, 2008

In the last piece I wrote, I was describing how I was trying to make a tool to efficiently calculate the probability of rolling or exceeding a particular sum on a number of dice. I had gotten as far as a recursive solution which allowed me to roll any number of `s`-sided dice, but which rapidly slowed down as more dice were added.

I spent a good amount of time at this point tearing my hair out looking for patterns in the numbers that would allow me to directly calculate the total number of ways to roll a particular number on `n` `s`-sided dice. I never came up with one. What I did end up doing was mentioning my problem to a mathematically inclined officemate, who found a paper on an interesting technique for me

### The Technique

The general gist of this method is to, for each face of your current die, add it to the various sums you made on your previous dice.

Let’s give it a try, because that explanation was nearly worthless. I’ll roll two six-sided dice (by hand) to keep things simple, and verify my results against what my prior calculator generated. Then I’ll write a program that will use this method for any `n` `s`-sided dice, as before.

For the first die, I can roll any number 1 through 6 exactly one way.

pipsways
11
21
31
41
51
61

Now, when I roll the second die, I can get a 1 through 6, and add it to each of the pips/ways pairs above. I’ll show what I mean. In the case that my second die is a 1, the table updates like this:

total pipsold pipsnew pips
211
321
431
541
651
761

OK, so that chart isn’t completely enlightening, at least not yet. Please do note that total pips is old pips + new pips. Now I’m going to see what happens if my die comes up a 2.

total pipsold pipsnew pips
312
422
532
642
752
862

Then I add the charts from rolling 1 and 2 and count the number of ways I can get each number.

total pipstotal ways
10
21
32
42
52
62
72
81

I’ll do the whole thing again for a three. Possible outcomes:

total pipsold pipsnew pips
413
523
633
743
853
963

And add it to the running count of new possibilities:

total pipstotal ways
10
21
32
43
53
63
73
82
91

If you recall from last time what the chart of how many ways you can roll a given number on two dice looked like, you’ll see that we’re well on our way to generating it. I’m just going to jump ahead and show the results for applying this from 1 all the way to 6 for the second die.

total pipstotal ways
10
21
32
43
54
65
76
85
94
103
112
121

So, let’s have some code.

### The Code

```#!/usr/bin/python   def roll(n, s): # roll n s-sided dice numresults = s ** n # s^n results   # make two buffers, with 1 for each value on 1 die # neededd because we only do a proper calculation for 2+ dice buff = [0] + [1] * s + [0] * ((n - 1) * s) oldbuff = [0] + [1] * s + [0] * ((n - 1) * s)   for d in range(2, n + 1): # for dice 2 through n # clear buffer buff = [0] * (s * n + 1)   for i in range(1, s + 1): # for each face, sum with old outcomes for j in range(d - 1, (d - 1) * s + 1): # We are doing die 'd' now # prior results are thus in the range of (d-1, (d-1)*s) # We found one way * oldbuff[j] ways = oldbuff[j] ways to get here buff[i + j] += oldbuff[j]   oldbuff = buff   return (buff, numresults)   def printres(results, numresults): remaining = numresults print "N\t#N\t%N\t#N+\t%N+" for i in range(len(results)): chance = 100.0 * results[i] / numresults pluschance = 100.0 * remaining / numresults print "%d\t%d\t%0.02f\t%d\t%0.02f" % \ (i, results[i], chance, remaining, pluschance) remaining -= results[i]   (results, numresults) = roll(2, 6) printres(results, numresults)```

This produces the same results in only a fraction of the time. It actually goes about the summing in a slightly less efficient manner than the paper that I was shown, but makes up for it by being less mysterious.

### The Results

How much faster is it?

When I mentioned calculating 10 10-sided dice last time I actually told a little lie. I didn’t actually bother running the test to the end. During the course of writing this entry I started it running and although I’m not using the fastest machine out there it took more than 2 hours to complete. This version can calculate those results in less than 0.02 seconds.

To be a bit more mathematical, the way the recursive solution is nesting `for` loops is O(s^n). This solution, however, is O(s*n).