## Learning Flash 9

This week I switched from Flash 8 to Flash 9.

Since last year I've
been playing
with Flash 8, using the Motion-Twin command-line
compiler `mtasc`

. I was using it to write a transportation game, and I had something running and was making progress, until Supreme Commander came out. Then I went back to playing games instead of writing them. Although I'd still like to work on that game, I've found that I also want to use Flash to build interactive demonstrations of concepts I describe on my site. For example, in my article about grids I'd like to make those diagrams interactive so that you can better see how the coordinate systems work. Diagrams are now what I'm using to learn Flash; I may go back to the game later (or maybe not).

I like `mtasc`

. However, it only supports Flash 8
(Actionscript 2), and is not going to be updated for Flash 9
(Actionscript 3). It's a dead end. Flash 9 is not only
significantly faster
(almost as fast as Java), but it also has major changes to the
libraries. Instead of `mtasc`

, you can
use HaXe, which is a new language
similar to Actionscript/Javascript/ECMAscript. HaXe looks neat
(better types, type inference), but it's a different language, not
Actionscript. Part of my goal is to publish my source code so that
others can use it, and it's less useful to publish code that isn't
usable in Actionscript. It's also less useful to publish code that
requires an expensive development environment (for example, Flex, at
$500). And it's easier to learn a language when there are lots of other users, posting tips. So I've been staying with `mtasc`

; the same code
works with both `mtasc`

and the Flash 8 development
environment.

Last week Rich Collins
pointed me to the free command-line Flash 9
compiler, `mxmlc`

. Wonderful! It's free, it's Flash 9,
it's command line—just what I was looking for. I spent a few days
learning about Flash 9, and
found this
tutorial and these tips to be most helpful. My initial thoughts:

- (
**yay**) Flash 9 has much better libraries than Flash 8. The sprites (movieclips), the event handling, and the graphics commands are all nicer. - (
**boo**) Actionscript 3 is more verbose than Actionscript 2, with types, packages,`public`

,`override`

, and other annotations. It's less of a scripting language and more like Java. This is bad. - (yay) Flash 9 is much faster than Flash 8, in part thanks to all those type annotations.
- (boo) The
`mxmlc`

compiler is significantly slower than`mtasc`

, in part because it's written in Java, which has a high startup time. - (boo) The
`mxmlc`

compiler is not open source.

I've been converting some of my code from Flash 8 to Flash 9, and despite the increased verbosity, I've been happy with it. If you want to use Flash 9 with free command-line compiler, start with this tutorial.

**Update:** [2007-07-28] [2010-03-25] You can download the Flash command-line compiler (Flex mxmlc) for free, without registration, from Adobe. Once I learned the language, the Flash 9 library reference became my #1 source of information.

Labels: flash , programming , project

## Distances on a triangular grid

As part of my article on grids, I'd like to provide basic algorithms for grids. I already know how to compute distances on square and hexagon grids, but I didn't know how to compute them on a triangle grid. I initially tried changing coordinate systems, inspired by Charles Fu's coordinate system. I normally use the coordinate system from my article on grids:

The first coordinate (which I will call `u`

) is the
position along the horizontal axis; the second (which I
call `v`

) is the position along the southwest/northeast
diagonal. The Left/Right parity introduces some compliation; I
let `R`

be 1 and `L`

be 0. This coordinate
system is nice for storing maps in an array, but it's not as clean,
because triangle grids have three axes but only two of them are
expressed in the coordinates. For the horizontal axis I
used `2u+R`

as the position; for the second axis I
used `2v+R`

. I guessed that the position along the third
axis would `u-v`

. I drew some grids on paper, wrote down
the `u,v`

values, and then computed the three positional
values. They behave properly for computing distances along the
third axis, but the system falls apart when computing distances not
on the three axes.

Frustrated, I decided to step back and approach this problem in a
more principled manner. In Charles Fu's three-axis coordinate system, taking
one step in the grid changes two of the coordinates and leaves the
third alone. For a triangle grid, the dominating feature is the
straight lines in three directions. I looked on the web for any articles
describing distances on triangular grids, but didn't find any that satisfied me. This paper was promising but unfortunately gives an iterative algorithm and not a clean formula. However, it uses an interesting coordinate system, which is useful for computing distances. If you take one step in the
grid, you will cross exactly one line. **The distance between
two locations will be the number of lines you have to
cross.** I wanted to use a coordinate system in which
crossing a line changes a coordinate:

I want to map
the `u,v,R`

coordinates I normally use into the alternate coordinate system `a,b,c`

, where
exactly one of `a,b,c`

changes when you cross a
line. Here's where I decided to use algebra. In `u,v,R`

space, the black triangle has
coordinates `0,0,0`

; the triangle to the east of it has
coordinates `0,0,1`

; to the west, `-1,0,1`

; to
the south, `0,-1,1`

. In `a,b,c`

space, the
black triangle is `0,0,0`

; the east triangle
is `0,0,1`

; the west triangle is `0,-1,0`

; and
the south triangle is `-1,0,0`

.

I want a formula that computes `a`

, and I expect it to be
linear, so I write `a = a`

. That's four unknowns, but we know the
constant will be 0, so it's really three unknowns. To get three
equations, I plug in the values for the east, west, and south
triangles, then solve for the coefficients. Let's see what happens:
_{u}*u + a_{v}*v +
a_{R}*R + constant

For each s in (a, b, c), and each triangle i:s_{i}= s_{u}*u + s_{v}*v + s_{R}*REast triangle:u,v,R = 0,0,1 s_{1}= s_{u}*0 + s_{v}*0 + s_{R}*1therefore:s_{R}= s_{1}West triangle:u,v,R = -1,0,1 s_{2}= s_{u}*-1 + s_{v}*0 + s_{R}*1therefore:s_{u}= s_{R}- s_{2}South triangle:u,v,R = 0,-1,1 s_{3}= s_{u}*0 + s_{v}*-1 + s_{R}*1therefore:s_{v}= s_{R}- s_{3}

When `s`

is axis `a`

, we look
at `a`

for triangles 1 (east), 2 (west), and 3 (south),
so `a`

is 0, _{1}`a`

is 0,
and _{2}`a`

is -1. The algebra tells us
that _{3}`a`

= 0, _{R}`a`

= 0,
and _{u}`a`

= 1. That means the formula
is _{v}`a = v`

. Pretty simple. Doing the same
for `b`

and `c`

, I get `b = u`

and `c = u + v + R`

.

These results are simple enough that if I played with the grid
enough, I would have come up with them. However the algebraic
approach works for other constraints, not only the simple ones. I
also used it to create a coordinate system with ```
2u+v+R,
u+2v+R, v-u
```

(this is 30 degrees rotated from the one above),
which may be useful for other algorithms. Algebra and calculus also
come in handy for defining movements (e.g., splines), growth rates,
equilibria in interacting systems, and other types of interesting
behaviors in games.

What is the distance between two locations in a triangular grid?
Each step involves a line crossing, and we want to count steps, so
we add up the line crossings along each axis (```
a, b,
c
```

): ```
distance = abs(Δu) + abs(Δv) +
abs(Δ(u+v+R))
```

. This looks just like the Manhattan distance formula, but for triangles instead of squares.