awesty

Had a go at making some metaballs… because they’re pretty.

Click the image to see them in real time.

Was bored, so I made this. It’s just simple 2D cellular automation.

The default settings match those of “Conway’s Game of Life”. [link]

Instructions:

• The seed determines which cells start off alive.
• Min. Neighbours is the minimum number of live neighbours each live cells can have to continue living.
• Max. Neighbours is the maximum number of live neighbours each live cells can have to continue living.
• “Resurrection Range” is the range of the number of neighbours each dead cell needs to come back to life.
• Birth survival chance is simply the probability that each cell has at initiation of being alive. (1 in X chance).

Click restart to apply the settings.

If nothing is happening change the seed and click restart until something starts happening.

I’ve started working on a new game. I got a small prototype up and running yesterday.

I’m not going to give any details about the game out yet. I still don’t have a story planned, but I have an idea about how I want the gameplay. Expect to see more updates on this soon (hopefully!).

David Danielsson sent in a tutorial. It’s so you can make something shoot left and right, like in a sidescrolling shooter. You can add a bit of code to it to make it shoot in 4 directions easily though. Here it is:

Problem 2 was another easy one.

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

Find the sum of all the even-valued terms in the sequence which do not exceed four million.

I brute forced it again, with a bit of optimisation.

We only need to add the even Fibonacci numbers, and if you look closely at the example above you might notice a pattern.

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

An even number shows up every 3 numbers. So instead of calculating every number in the sequence below 4,000,000 and then test if it’s even, we can calculate every third number and add it straight onto the total. Here is what I came up with.

sum = 0
prev = 1
current = 2
while current < 4000000:
    sum = sum + current
    n = current
    current += 2 * (current + prev)
    prev += 2 * n
    print current
print sum

For fun I also did it in 4 lines. This way is very inefficient, but it is small.

a = [1, 2]
while a[len(a) - 1] < 4000000:
    a.append(a[len(a) - 1] + a[len(a) - 2])
print sum(filter(lambda x: x % 2 == 0, a))

Can anyone think of a shorter and/or more efficient way?

Problem 1 is extremely simple:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

I chose to just brute force it. I looped through numbers 1 – 999, checked if it was a multiple of 3 or 5 and if it was added it onto the total.

sum = 0
for i in range(1, 1000):
    if i % 3 == 0 or i % 5 == 0:
        sum = sum + i
print sum

After a bit of editing I was able to get it onto 1 line:

print sum(filter(lambda x: x % 3 == 0 or x % 5 == 0, range(1,1000)))

Not the most efficient solution, but it works.

Lately I have been spending a lot of my time on Project Euler. Project Euler is a site with a heap of math problems (209 to be exact) ranging in difficulty. You pick your favourite programming language and try to solve them.

They start of pretty easy:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

And get very, very hard:

A best approximation to a real number x for the denominator bound d is a rational number r/s (in reduced form) with s  d, so that any rational number p/q which is closer to x than r/s has q d.

Usually the best approximation to a real number is uniquely determined for all denominator bounds. However, there are some exceptions, e.g. 9/40 has the two best approximations 1/4 and 1/5 for the denominator bound 6. We shall call a real number x ambiguous, if there is at least one denominator bound for which x possesses two best approximations. Clearly, an ambiguous number is necessarily rational.

How many ambiguous numbers x = p/q, 0  x  1/100, are there whose denominator q does not exceed 10⁸?

I’ve done 15 so far. I’ve found that it’s great if your trying to learn a new language. I’m doing mine in python, which I’ve never really used before.

I’m going to start making a few posts about trying to solve them.

First tutorial in 9 months, so bare with me.

We are going to make a script that will make circles collide with each other. This would be very hard to do with hitTest()’s so we are going to use a small amount of trigonometry. Don’t let the big word or its meaning scare you. All we are going to need is 13 lines of code. Here is what it should look like:

Read the rest of this entry »

No, I’m not dead. If you were a fan of my tutorials you would’ve probably noticed that I hadn’t written one for a long time. Actually, I haven’t even updated my old site since January, and it wouldn’t be much less since I’ve replied to a comment.

I’m starting fresh from awestyproductions.com. I think awesty.com sounds and looks better without the extra eleven letters hanging off the end.

But awesty, what’s going to happen to all the tutorials!? Don’t worry Timmy, I’m not going to take down the old site. Those tutorials will be there for a long time to come (well, at least until around this time next year), so don’t panic. But I’m also not going to import all the posts from the old site over to here. I feel like starting fresh. I know alot more now than I did when I started awestyproductions.com (man, thats a lot of letters, lets just call it ap) and a lot of the tutorials have mediocre code, because I really didn’t know very much. I’m not going to focus this blog around tutorials like I did with ap, but I will still occasionally write them. I’m almost exclusively using Linux these days so I don’t use flash anymore so you might see some more variety.

What is this blog going to be for then? Anything I feel like. This time it is actually going to be a blog, instead of a tutorial site.