I'm developing a game in Java, and part of it requires that objects spawn at the top of the screen and proceed to fall down. I have three objects that can possibly spawn, and three possible x coordinates for them to spawn at, all stored in an array called xCoordinate[].
One of the objects is of a class called Enemy, which inherits a class I have called FallingThings. In the FallingThings class, I have methods to generate new objects, my enemy method is below:
public static void generateNewEnemy() {
xIndexEnemyOld = xIndexEnemy;
xIndexEnemy = new Random().nextInt(3);
if (delayTimer == 0) {
while (xIndexEnemy == xIndexEnemyOld) {
xIndexEnemy = new Random().nextInt(3);
}
}
if (xIndexEnemy != xIndexMoney && xIndexEnemy != xIndexFriend) {
Enemy enemy = new Enemy(xCoordinates[xIndexEnemy]);
enemies.add((Enemy) enemy);
} else {
generateNewEnemy();
}
}
xIndexEnemy represents the index of the xCoordinates array.
xIndexMoney and xIndexFriend are the indexes of the xCoordinates array for the two other objects (the comparisons with these values ensures that one object does not spawn directly on top of another).
The delayTimer variable represents the random delay between when new objects spawn, which was set earlier in my main class.
I store each instance of an Enemy object in an ArrayList.
Everything works except for the fact that sometimes, an object will spawn over itself (for example, the delay is 0, so two enemy objects spawn directly on top of each other, and proceed to fall down at the same speed at the same time).
I've been trying to crack this for the past two days, but I understand exactly why my code right now isn't working properly. I even tried implementing collision detection to check if another object already exists in the space, but that didn't work either.
I would be extremely grateful for any suggestions and ideas.
EDIT2
It seems that you still don't understand the problem with your function. It was addressed in the other answer but I'll try to make it more clear.
public static void generateNewEnemy() {
xIndexEnemyOld = xIndexEnemy;
This is just wrong. You can't set the Old index without having actually used a new index yet.
xIndexEnemy = new Random().nextInt(3);
if (delayTimer == 0) {
while (xIndexEnemy == xIndexEnemyOld) {
xIndexEnemy = new Random().nextInt(3);
}
}
This is actually ok. You're generating an index until you get one that is different. It may not be the most elegant of solutions but it does the job.
if (xIndexEnemy != xIndexMoney && xIndexEnemy != xIndexFriend) {
Enemy enemy = new Enemy(xCoordinates[xIndexEnemy]);
enemies.add((Enemy) enemy);
} else {
generateNewEnemy();
}
}
This is your problem (along with setting the Old index back there). Not only do you have to generate an index thats different from the Old index, it must also be different from IndexMoney and IndexFriend.
Now, what happens if, for example, IndexOld = 0, IndexMoney = 1 and IndexFriend = 2? You have to generate an index that's different from 0, so you get (again, for instance) 1. IndexMoney is 1 too, so the condition will fail and you do a recursive call. (Why do you even have a recursive call?)
OldIndex was 0, and now in the next call you're setting it to 1. So IndexOld = 1, IndexMoney = 1 and IndexFriend = 2. Do you see the problem now? The overlapped index is now wrong. And the new index can only be 0 no matter how many recursive calls it takes.
You're shooting yourself in the foot more than once. The recursive call does not result in an infinite loop (stack overflow actually) because you're changing the Old index. (Which, again is in the wrong place)
That if condition is making it so the newly generated index cannot overlap ANY of the previous indexes. From what you said before it's not what you want.
You can simplify your function like this,
public static void generateNewEnemy() {
xIndexEnemy = new Random().nextInt(3);
if (delayTimer == 0) {
while (xIndexEnemy == xIndexEnemyOld) {
xIndexEnemy = new Random().nextInt(3);
}
}
Enemy enemy = new Enemy(xCoordinates[xIndexEnemy]);
enemies.add((Enemy) enemy);
xIndexEnemyOld = xIndexEnemy;
// Now that you used the new index you can store it as the Old one
}
Will it work? It will certainly avoid overlapping when the delayTimer is 0 but I don't know the rest of your code (nor do I want to) and what do you do. It's you who should know.
About my suggestions, they were alternatives for how to generate the index you wanted. I was assuming you would know how to fit them in your code, but you're still free to try them after you've fixed the actual problem.
Original Answer
Here's one suggestion.
One thing you could do is to have these enemies "borrow" elements from the array. Say you have an array,
ArrayList< Float > coordinates = new ArrayList< Float >();
// Add the coordinates you want ...
You can select one of the indexes as you're doing, but use the maximum size of the array instead and then remove the element that you choose. By doing that you are removing one of the index options.
int nextIndex = new Random().nextInt( coordinates.size() );
float xCoordinate = coordinates.get( nextIndex );
coordinates.remove( nextIndex ); // Remove the coordinate
Later, when you're done with the value (say, when enough time has passed, or the enemy dies) you can put it back into the array.
coordinates.add( xCoordinate );
Now the value is available again and you don't have to bother with checking indexes.
Well, this is the general idea for my suggestion. You will have to adapt it to make it work the way you need, specifically when you place the value back into the array as I don't know where in your code you can do that.
EDIT:
Another alternative is, you keep the array that you previously had. No need to remove values from it or anything.
When you want to get a new coordinate create an extra array with only the values that are available, that is the values that won't overlap other objects.
...
if (delayTimer == 0) {
ArrayList< Integer > availableIndexes = new ArrayList< Integer >();
for ( int i = 0; i < 3; ++i ) {
if ( i != xIndexEnemyOld ) {
availableIndexes.add( i );
}
}
int selectedIndex = new Random().nextInt( availableIndexes.size() );
xIndexEnemy = availableIndexes.get( selectedIndex );
}
// Else no need to use the array
else {
xIndexEnemy = new Random().nextInt( 3 );
}
...
And now you're sure that the index you're getting should be different, so no need to check if it overlaps.
The downside is that you have to create this extra array, but it makes your conditions simpler.
(I'm keeping the "new Random()" from your code but other answers/comments refer that you should use a single instance, remember that)
As I see, if delay == 0 all is good, but if not, you have a chance to generate new enemy with the same index. Maybe you want to call return; if delayTimer != 0?
UPDATED
Look what you have in such case:
OldEnemyIndex = 1
NewEnemyIndex = random(3) -> 1
DelayTimer = 2
Then you do not pass to your if statement, then in the next if all is ok, if your enemy has no the same index with money or something else, so you create new enemy with the same index as previous
Related
I'm creating a game within Java and i'm currently implementing the amount of lives the player has. At the start they have 5 (this is shown as a frog on the screen, each frog = 1 life, lose a life = one frog is removed from the screen).
I've created a class called Life which takes in the image for the life and its x and y positions. If the player dies, a frog is supposed to be removed from the screen.
I've managed to do this by creating 5 new Life objects and giving them different names i.e. life1, life2 etc. and then removing them using their given names during the object instantiation, but it's very inefficient as i'm just chaining if statements together.
I was thinking of giving each new frog an ID within the constructor from 1 to 5, and then having a while loop which will remove the lives if the player dies (which would be the condition) and a counter decrementing from 5 to 1, however, for this to work, I would somehow need to get the separate Life objects, so something like this (in pseudocode):
numOfLives = 5;
while (i < numOfLives)
{
life = Life.getLifeFromID(numOfLives);
background.removeLife(life);
numOfLives--;
}
So is it possible to create a method such as this? (written in pseudocode, i'm not entirely sure how you would specify this in actual code)
public Life getLifeFromID(int id)
{
return specific Life object based on ID from argument
}
So if numOfLives = 4 , the line life = Life.getLifeFromID(numOfLives); would be assigned the Life object of 4, which can then be removed in the next line.
Thanks!
I think you can put your Life instances into a List, then when you lose a life, you just need to remove from List with specific index.
I am working on a game too that uses a system to reference objects based on an ID.
You can start by giving 'Life' a variable called ID or whatever else you want to call it and add it to the constructor.
here is an example where I use entities in a game world.
for (int i = 0; i < spawnLimit; i++) {
int x = (int) (Math.random() * 200) + 10;
int y = (int) (Math.random() * 200) + 10;
entities.add(new Entity(x, y, this.width, this.height, world_scale, , entityCount));
entityCount += 1;
}
and then when you want to get an object based on the ID:
return entities.get(id);
So I'm writing a Rush Hour solver in Java, which is meant to be able to solve the configurations here. However, even the simplest puzzle from that page results in the solver running infinitely and eventually running out of memory. I'm using a breadth first search to work my way through all possible moves arising from each board state (using a HashSet to ensure I'm not repeating myself), and mapping each state to the move that got it there so I can backtrack through them later.
The thing is, I've tried it with more trivial puzzles that I've come up with myself, and it's able to solve them (albeit slowly).
Is there anything blatantly wrong with how I'm approaching this problem? I can put up some code from the relevant classes as well if I need to, but I've tested them pretty thoroughly and I'm pretty sure the problem lies somewhere in the code below. My gut says it's something to do with the HashSet and making sure I'm not repeating myself (since the Queue's size regularly reaches the hundred thousands).
Any suggestions?
// Start at the original configuration
queue.add(originalBoard);
// We add this to our map, but getting here did not require a move, so we use
// a dummy move as a placeholder move
previous.put(originalBoard, new Move(-1, -1, "up"));
// Breadth first search through all possible configurations
while(!queue.isEmpty()) {
// Dequeue next board and make sure it is unique
Board currentBoard = queue.poll();
if (currentBoard == null) continue;
if (seen.contains(currentBoard)) continue;
seen.add(currentBoard);
// Check if we've won
if (currentBoard.hasWon()) {
System.out.println("We won!");
currentBoard.renderBoard();
return solved(currentBoard);
}
// Get a list of all possible moves for the current board state
ArrayList<Move> possibleMoves = currentBoard.allPossibleMoves();
// Check if one of these moves is the winning move
for (Move move : possibleMoves) {
Board newBoard = move.execute(currentBoard);
// We don't need to enqueue boards we've already seen
if (seen.contains(newBoard)) continue;
queue.add(newBoard);
// Map this board to the move that got it there
previous.put(newBoard, move);
}
}
As requested, here are my initialisations of the HashSet (they're class level variables):
private static HashSet<Board> seen = new HashSet<>();
And my Board.equals() method:
#Override
public boolean equals (Object b) {
Board otherBoard = (Board) b;
boolean equal = false;
if (this.M == otherBoard.getM() && this.N == otherBoard.getN()) {
equal = true;
// Each board has an ArrayList of Car objects, and boards are only
// considered equal if they contain the exact same cars
for (Car car : this.cars) {
if (otherBoard.getCar(car.getPosition()) == null) {
equal = false;
}
}
}
System.out.println(equal);
return equal;
}
You must implement Board.hashCode() to override the default Object-based version, in such a way that, per its contract, any two equal Board objects have the same hash code. If you do not, then your seen set does not in fact accomplish anything at all for you.
On another issue, I suspect that the way you're checking the boards' cars is not fully correct. If it works the way I think it does, it would consider these two boards to be equal:
. = empty space
* = part of a car
......
.**.*.
....*.
.*....
.*.**.
......
......
.*..**
.*....
......
.**.*.
....*.
I'm creating a game (Java) with scene2d.
I wrote function for collision detection but I think it's bad function. It looks bad.
How can I optimize it? Make faster and more beautiful.
private void deleteEnemies()
{
for(int i = 0; i < getActors().size - 1; i++)
{
if(getActors().get(i) != null && getActors().get(i) instanceof Enemy)
{
////////////////
for (int j = 0; j < getActors().size - 1; j++)
{
if(getActors().get(j) != null && getActors().get(j) instanceof Ball)
{
if (actorsIntersecting(getActors().get(i), getActors().get(j)))
{
getActors().get(i).remove();
getActors().get(j).remove();
}
}
}
//////////////
}
}
}
Put getActors().get(i) in a variable, dont call it twice in the outer if
Same for getActors().get(j) in the inner if
use these variable in the most inner if's condition and body
save the size in a variable because now the .size function is being called on every iteration when the for condition is checked
You shouldn't use a size that can dynamically change during the loop for the loop condition (because you are removing items as you go) which brings us back to #4.
Other than that its pretty much ok coding style perspective and I doubt you can make it more efficient than with what I told you (Other than using threads)
Since you will do this frequently, consider storing the Enemies and Balls in their own structures (List or Set or whatever works). That prevents you from looping through actors you don't need, and avoids the instanceof checks.
Well, my first idea was to check only "nearest" enemies and not all of them. Somehow try to decrease size of that list.
2. Second one - please check your and conditions in and one by one - now you are checking 2 conditions always. Try to put "heavier" if later, for example:
from:
if(getActors().get(i) != null && getActors().get(i) instanceof Enemy)
to:
if(getActors().get(i) != null) {
if(getActors().get(i) instanceof Enemy) {
.....
}
}
3. call your getActors().get(i) one time - save to variable.
4. I'm thinking why is it necessary to check if an actor is null, maybe just remove nulls from list or keep uninitialized actors on another list. Also try this with Balls and Enemies, please don't keep every actor on a single list.
I would rewrite the models a bit, so they can test the intersection itself and then do the delete like that (probably it can still be improved)
private void deleteEnemies () {
List<Actor> actors = getActors();
List<Actor> toRemove = new ArrayList<Actor>();
int actorsSize = actors.size();
Actor first = null, second = null;
for(int i = 0; i < actorsSize; ++i) {
first = actors.get(i);
for(int j = 0; j < actorsSize; ++j) {
if(i == j) continue;
second = actors.get(j);
if(first.intersects(second)) {
toRemove.add(first);
toRemove.add(second);
}
}
}
actors.removeAll(toRemove);
}
Don't use size(), define a variable
Try not to cast. Try not to uae instanceof.
Maybe, sort lists by zsort or the like so u can, sometimes, start and or stop the loops sooner??
Adding to the (very good) suggestions of the other participant: cache the enemies and projectiles in separate structures, so you don't have to check what they are at all.
Use the time vs space trade-off as much as you can: the standard approach, as hinted by Tomek, in this kind of situations is to reduce the number of checks (=iterations) by pruning the enemies and projectiles that cannot possibly collide within the current frame (they are way to far).
Anyway, a word of advice: go on with the game, complete as much as you can so that it will run correctly (if slowly), and only then go for the optimization.
That because
by optimizing preemptively in this way you will never finish it
you don't know how the final game really will be, perhaps: maybe after finishing 90% of it, you will see some easy chances for optimization.
As others have said, the real improvement to speed would be two collections, one with balls and the other with enemies. As for making it look nicer, you could something like this:
for (Actor enemy : getActors()) {
if (enemy != null && enemy instanceof Enemy) {
for (Actor ball : getActors()) {
if (ball != null && ball instanceof Ball && actorsIntersecting(enemy, ball)) {
ball.remove();
enemy.remove();
}
}
}
}
It's my first working on a quite big project, and I've been asked to obtain the best performances.
So I've thouhgt to replace my for loops with a ListIterator, because I've got around 180 loops which call list.get(i) on lists with about 5000 elements.
So I've got two questions.
1) Are those 2 snippets equal? I mean, do them produce the same output? If no, how can I correct the ListIterator thing?
ListIterator<Corsa> ridesIterator = rides.listIterator();
while (ridesIterator.hasNext()) {
ridesIterator.next();
Corsa previous = ridesIterator.previous(); //rides.get(i-1)
Corsa current = ridesIterator.next(); //rides.get(i)
if (current.getOP() < d.getFFP() && previous.getOA() > d.getIP() && current.wait(previous) > DP) {
doSomething();
break;
}
}
__
for (int i = 1; i < rides.size(); i++) {
if (rides.get(i).getOP() < d.getFP() && rides.get(i - 1).getOA() > d.getIP() && rides.get(i).getOP() - rides.get(i - 1).getOA() > DP) {
doSomething();
break;
}
}
2) How will it be the first snippet if I've got something like this? (changed i and its exit condition)
for (int i = 0; i < rides.size() - 1; i++) {
if (rides.get(i).getOP() < d.getFP() && rides.get(i + 1).getOA() > d.getIP() && rides.get(i).getOP() - rides.get(i + 1).getOA() > DP) {
doSomething();
break;
}
}
I'm asking because it's the first time that I'm using a ListIterator and I can't try it now!
EDIT: I'm not using an ArrayList, it's a custom List based on a LinkedList
EDIT 2 : I'm adding some more infos.
I can't use a caching system because my data is changing on evry iteration and managing the cache would be hard as I'd have to deal with inconsistent data.
I can't even merge some of this loops into one big loop, as I've got them on different methods because they need to do a lot of different things.
So, sticking on this particular case, what do you think is the best pratice?
Is ListIterator the best way to deal with my case? And how can I use the ListIterator if my for loop works between 0 and size-1 ?
If you know the maximum size, you will get the best performance if you resign from collections such as ArrayList replacing them with simple arrays.
So instead creating ArrayList<Corsa> with 5000 elements, do Corsa[] rides = new Corsa[5000]. Instead of hard-coding 5000 use it as final static int MAX_RIDES = 5000 for example, to avoid magic number in the code. Then iterate with normal for, referring to rides[i].
Generally if you look for performance, you should code in Java, as if it was C/C++ (of course where you can). The code is not so object-oriented and beautiful, but it's fast. Remember to do optimization always in the end, when you are sure, you have found a bottleneck. Otherwise, your efforts are futile, only making the code less readable and maintainable. Also use a profiler, to make sure your changes are in fact upgrades, not downgrades.
Another downside of using ListIterator is that it internally allocates memory. So GC (Garbage Collector) will awake more often, which also can have impact on the overall performance.
No they do not do the same.
while (ridesIterator.hasNext()) {
ridesIterator.next();
Corsa previous = ridesIterator.previous(); //rides.get(i-1)
Corsa current = ridesIterator.next(); //rides.get(i)
The variables previous and current would contain the same "Corsa" value, see the ListIterator documentation for details (iterators are "in between" positions).
The correct code would look as follows:
while (ridesIterator.hasNext()) {
Corsa previous = ridesIterator.next(); //rides.get(i-1)
if(!ridesIterator.hasNext())
break; // We are already at the last element
Corsa current = ridesIterator.next(); //rides.get(i)
ridesIterator.previous(); // going back 1, to start correctly next time
The code would actually look exactly the same, only the interpretation (as shown in the comments) would be different:
while (ridesIterator.hasNext()) {
Corsa previous = ridesIterator.next(); //rides.get(i)
if(!ridesIterator.hasNext())
break; // We are already at the last element
Corsa current = ridesIterator.next(); //rides.get(i+1)
ridesIterator.previous(); // going back 1, to start correctly next time
From a (premature?) optimization viewpoint the ListIterator implementation is better.
LinkedList is a doubly-linked list which means that each element links to both its predecessor (previous) as well as its successor (next). So it does 3 referals per loop. => 3*N
Each get(i) needs to go through all previous elements to get to the i index position. So on average N/4 referals per loop. (You'd think N/2, but LinkedList starts from the beginning or the end of the list.) => 2 * N * N/4 == N^2 /2
Here are some suggestions, hopefully one or two will be applicable to your situation.
Try to do only one rides.get(x) per loop.
Cache method results in local variables as appropriate for your code.
In some cases the compiler can optimize multiple calls to the same thing doing it just once instead, but not always for many subtle reasons. As a programmer, if you know for a fact that these should deliver the same values, then cache them in local variables.
For example,
int sz = rides.size ();
float dFP = d.getFP (); // wasn't sure of the type, so just called if float..
float dIP = d.getIP ();
Corsa lastRide = rides.get ( 0 );
for ( int i = 1; i < sz; i++ ) {
Corsa = rides.get ( i );
float rOP = r.getOP ();
if ( rOP < dFP ) {
float lastRideOA = lastRide.getOA (); // only get OA if rOP < dFP
if ( lastRideOA > dIP && rOP - lastRideOA > DP ) {
doSomething ();
// maybe break;
}
}
lastRide = r;
}
These are optimizations that may not work in all cases. For example, if your doSomething expands the list, then you need to recompute sz, or maybe go back to doing rides.size() each iteration. These optimizations also assumes that the list is stable in that the elements don't change during the get..()'s. If doSomething makes changes to the list, then you'd need to cache less. Hopefully you get the idea. You can apply some of these techniques to the iterator form of the loop as well.
I am writing a snake game, specifically, is a centipede game. It needs me to draw a snake and that snake will automatically move one line by one line.
I did draw a snake, and it can move from left side to right side. However, the problem is:
I can't make the snake changes line, if it finish the first line, I need it changes to the second line and which starts from the right side.
My code is like this:
private void move()
{
myCentipedes[0] =
new Centipede(Settings.centipedeStartSize, Settings.RIGHT,
Settings.DOWN);
myCentipedes[0].segments = new Point[Settings.centipedeStartSize];
myCentipedes[0].segments[0] = new Point(0, 0);
boolean dr = true;
if (dr == true) {
if (myCentipedes[0].segments[0].x < 30) {
System.out.println(myCentipedes[0].segments[0].x +
" " +
myCentipedes[0].segments[0].y);
myCentipedes[0].segments[0] = new Point(x, 0);
for (int i = 1; i < 10; i++) {
myCentipedes[0].segments[i] =
new Point(myCentipedes[0].segments[i - 1].x - 1,
myCentipedes[0].segments[i - 1].y);
}
x++;
}
}
if (myCentipedes[0].segments[0].x == 29) {
x = 29;
dr = false;
}
if (dr == false) {
if (myCentipedes[0].segments[0].x > 0) {
myCentipedes[0].segments[0] = new Point(x, 1);
for (int i = 1; i < 10; i++) {
myCentipedes[0].segments[i] =
new Point(myCentipedes[0].segments[i - 1].x + 1, 1);
}
x--;
}
}
}
It appears to me that you re-create your entire centipede on every single move:
private void move()
{
myCentipedes[0] =
new Centipede(Settings.centipedeStartSize, Settings.RIGHT,
Settings.DOWN);
Is re-creating the centipede every move() intentional? Or should move() run the centipede entirely down the board, from start to finish? (If so, you'll need to add some looping to this method.)
I assume the myCentipedes[0] is simply a placeholder for future extensions, involving two or more centipedes on the board simultaneously. This sort of over-generic programming can sometimes make the code more difficult to read and write while initially programming, and almost certainly doesn't help matters. You can always re-factor a move() method that works on one centipede to a move(int centipede) method that works on a specific centipede and a move() method that calls move(int) for every centipede on the board. Or maybe you'll find it easier to place the movement code into the Centipede class, and need to remove the array indexes then and use class member storage instead.
boolean dr = true;
if (dr == true) {
dr will always equal true at this point. You might as well remove the variable and the test.
for (int i = 1; i < 10; i++) {
myCentipedes[0].segments[i] =
new Point(myCentipedes[0].segments[i - 1].x - 1,
myCentipedes[0].segments[i - 1].y);
}
Since you're counting up, you'll actually copy the value from segment[0] through to all elements in the array, one element at a time. Can't you just assign the Point objects new array indexes? Starting from i=centipede.segments.length and counting down, it'll look more like this:
for (int i=myCentipede[0].segments.length; i > 0; i--) {
myCentipede[0].segments[i] = myCentipede[0].segments[i-1];
}
myCentipede[0].segments[0] = new Point(...,...);
Some of your tests can be simplified:
if (myCentipedes[0].segments[0].x == 29) {
x = 29;
dr = false;
}
if (dr == false) {
if (myCentipedes[0].segments[0].x > 0) {
If dr == false at this point, you might as well have written it like this instead:
if (myCentipedes[0].segments[0].x == 29) {
x = 29;
if (myCentipedes[0].segments[0].x > 0) {
But then the second if is obviously not needed -- after all, 29 > 0.
While you're here, clean up all those hard-coded 10 with either a constant (Settings.centipedeStartSize) or find the actual length of the centipede (myCentipedes[0].segments.length).
Now that I've critiqued your current approach, I'd like to suggest a different tack:
Take a step back and break your problem down into smaller methods.
You've embedded two for loops that move the centipede one segment at a time by assigning to segment[i] the values from segment[i-1]. Instead of duplicating the code, write a new method with the body of the for loop to move the centipede forward. Make it take a Point object for the new first element each trip through the function. (Don't forget to make it count down rather than up.)
Once you've broken apart the for loops, I think it will be easier to make whatever changes are necessary for traveling left-to-right and right-to-left. You will probably want to write it with nested for loops -- one to control the vertical dimension, and within it, perhaps one or two new for loops to control the horizontal dimension. Make these loops work with a simple Centipede c, rather than the complicated expression you've currently got.
Breaking apart the larger function into smaller function will give you a better opportunity to test your functions in isolation -- test movement manually, with simple test methods like this:
move_forward(Centipede c, Point p) {
/* code to move forward one space to occupy `p` */
}
test_right() {
Centipede c = new Centipede(/* ... */);
move_forward(c, new Point(0,0));
move_forward(c, new Point(1,0));
move_forward(c, new Point(2,0));
move_forward(c, new Point(3,0));
move_forward(c, new Point(4,0));
move_forward(c, new Point(5,0));
/* ... */
}
Take it slow, test every method as you write them, and I think you'll find this is an easier problem than it currently looks.