i wanted to make a turn based system like final fantasy tactics. I already created the map, which is 5x5 tiles grid and the characters which is each character places in the end of the tiles. I have 2 teams, which are named Red and Yellow.
------Red-------:
First character is at 0,0. Second character is at 0,1. Third character is at0.2, fourth character is at0.3, and the last one is at0.4`.
-----Yellow------:
First character is at 5.0. Second character is at 5.1. Third character is at 5.2, fourth character is at 5.3, and the last one is at 5.4.
I wanted Red team are moving first and make a decision (whether it is attack or wait), and after 5 characters of the Red team is already made a decision, the Yellow team is the one that make a decision (Yellow team is an AI)
But, i don't know how to move my characters into the next grid (e.g: from 0,0 to 0,1) by clicking the left mouse button and also how do i display a grid (when select a move selection) that shows how many tiles that the character able to move.
Anyone know about this? or how should i know more about this? is there any recommendations books or webs?
You have your basic data structures set up, but now you need to get some higher level code to manipulate that data.
First of all, I think you should work on selecting locations on the grid with the mouse. Once you can click and get that grid coordinate saved to a variable, you need to set up a function to move your characters. After the first click (on a character), you need to check the valid moves, and for each valid move, you need to render an image on the grid square (or highlight the square's texture).
Secondly, you need a function which iterates through all the characters in each team, according to who moves next. When you have gone through Red.length (red is an array consisting of each player), then you switch to counting through Yellow.length, and running the AI for each character. If you are trying to make a two player game, you instead ask for user input a second time for the yellow team.
I recommend that you learn about how to display your grid and set up a simple way to highlight squares on the grid. After that, you need to convert mouse coordinates into grid coordinates. Your teams should each be an array of characters. I'm not familiar with actionscript, but in the languages I know, they would look like this:
team[6] = {Character1, Character2, Character3... }
Character1.position = {x, y}
running a turn would be something like this:
while battle == not finished {
for (i = 0; i < red.length; i++) {
getInput();
move(red[i], newX, newY); //red[i].position = {newX, newY}
}
for (i = 0; i < yellow.length; i++) {
runAI();
move(yellow[i], newX, newY);
}
}
The hardest part will be the mouse selection and drawing the grid/characters. Graphics are always a nuisance. The data itself just takes a bit of thinking. Your question in particular seems to be about game programming. My advice is to make the grid, then figure out how to display the grid. Then get mouse input. Finally, worry about moving the characters and highlighting squares.
Related
I have gone through a couple of YOLO tutorials but I am finding it some what hard to figure if the Anchor boxes for each cell the image is to be divided into is predetermined. In one of the guides I went through, The image was divided into 13x13 cells and it stated each cell predicts 5 anchor boxes(bigger than it, ok here's my first problem because it also says it would first detect what object is present in the small cell before the prediction of the boxes).
How can the small cell predict anchor boxes for an object bigger than it. Also it's said that each cell classifies before predicting its anchor boxes how can the small cell classify the right object in it without querying neighbouring cells if only a small part of the object falls within the cell
E.g. say one of the 13 cells contains only the white pocket part of a man wearing a T-shirt how can that cell classify correctly that a man is present without being linked to its neighbouring cells? with a normal CNN when trying to localize a single object I know the bounding box prediction relates to the whole image so at least I can say the network has an idea of what's going on everywhere on the image before deciding where the box should be.
PS: What I currently think of how the YOLO works is basically each cell is assigned predetermined anchor boxes with a classifier at each end before the boxes with the highest scores for each class is then selected but I am sure it doesn't add up somewhere.
UPDATE: Made a mistake with this question, it should have been about how regular bounding boxes were decided rather than anchor/prior boxes. So I am marking #craq's answer as correct because that's how anchor boxes are decided according to the YOLO v2 paper
I think there are two questions here. Firstly, the one in the title, asking where the anchors come from. Secondly, how anchors are assigned to objects. I'll try to answer both.
Anchors are determined by a k-means procedure, looking at all the bounding boxes in your dataset. If you're looking at vehicles, the ones you see from the side will have an aspect ratio of about 2:1 (width = 2*height). The ones viewed from in front will be roughly square, 1:1. If your dataset includes people, the aspect ratio might be 1:3. Foreground objects will be large, background objects will be small. The k-means routine will figure out a selection of anchors that represent your dataset. k=5 for yolov3, but there are different numbers of anchors for each YOLO version.
It's useful to have anchors that represent your dataset, because YOLO learns how to make small adjustments to the anchor boxes in order to create an accurate bounding box for your object. YOLO can learn small adjustments better/easier than large ones.
The assignment problem is trickier. As I understand it, part of the training process is for YOLO to learn which anchors to use for which object. So the "assignment" isn't deterministic like it might be for the Hungarian algorithm. Because of this, in general, multiple anchors will detect each object, and you need to do non-max-suppression afterwards in order to pick the "best" one (i.e. highest confidence).
There are a couple of points that I needed to understand before I came to grips with anchors:
Anchors can be any size, so they can extend beyond the boundaries of
the 13x13 grid cells. They have to be, in order to detect large
objects.
Anchors only enter in the final layers of YOLO. YOLO's neural network makes 13x13x5=845 predictions (assuming a 13x13 grid and 5 anchors). The predictions are interpreted as offsets to anchors from which to calculate a bounding box. (The predictions also include a confidence/objectness score and a class label.)
YOLO's loss function compares each object in the ground truth with one anchor. It picks the anchor (before any offsets) with highest IoU compared to the ground truth. Then the predictions are added as offsets to the anchor. All other anchors are designated as background.
If anchors which have been assigned to objects have high IoU, their loss is small. Anchors which have not been assigned to objects should predict background by setting confidence close to zero. The final loss function is a combination from all anchors. Since YOLO tries to minimise its overall loss function, the anchor closest to ground truth gets trained to recognise the object, and the other anchors get trained to ignore it.
The following pages helped my understanding of YOLO's anchors:
https://medium.com/#vivek.yadav/part-1-generating-anchor-boxes-for-yolo-like-network-for-vehicle-detection-using-kitti-dataset-b2fe033e5807
https://github.com/pjreddie/darknet/issues/568
I think that your statement about the number of predictions of the network could be misleading. Assuming a 13 x 13 grid and 5 anchor boxes the output of the network has, as I understand it, the following shape: 13 x 13 x 5 x (2+2+nbOfClasses)
13 x 13: the grid
x 5: the anchors
x (2+2+nbOfClasses): (x, y)-coordinates of the center of the bounding box (in the coordinate system of each cell), (h, w)-deviation of the bounding box (deviation to the prior anchor boxes) and a softmax activated class vector indicating a probability for each class.
If you want to have more information about the determination of the anchor priors you can take a look at the original paper in the arxiv: https://arxiv.org/pdf/1612.08242.pdf.
I am trying to make a simple arcade shooter like Galaga in TIBasic. I have successfully created some code that lets you move your character (an "X") horizontally across the screen while shooting a bullet that clears everything along the vertical path it takes. However, I'm having a problem with the "rocks" that are supposed to fall from the screen and disappear when hit. When I shoot the rock, it is cleared by the bullet, but then continues going down the screen until it hits the bottom of the screen. Here's the code for the rocks:
//outside the game loop:
1->R
//inside game loop:
If not(R=8)
Then
R+1->R
If R>1
Then
Output(E-1, 1, " " //removes last rock
End
Output(R, 1, "R" //replaces last rock with one below it (traveling towards the ground)
End
This code obviously doesn't stop the "R" from continuing to go down the screen when it is cleared (by the way, I just use Output(...," ") on wherever the bullet was to clear away anything the bullet hits). So, how would I check if the rock was cleared away on the last iteration of the game loop? Is there a way to check if something (the "R") is at a certain place on the screen in order to check if it was cleared away by a bullet on the last iteration? Or is there a better way? Thanks for the help!
It appears that the code you have doesn't test if the bullet is in the same place as the rock. To have this collision, you'd need to compare the "R" y-value of the rock with the y-value of your bullet you displayed, and compare the x-value of the rock (however you store that) with the x-value of the bullet. If both are equal to each other, then they have collided, and you can "delete" the rock by setting it to whatever value signifies it doesn't exist anymore. Assuming R is the y-value of the rock, S is the x-value of the rock, Y is the y-value of the bullet, and X is the x-value of the bullet, In the code to move the rock, I'd put this somewhere:
:If R=Y and S=X
:DelVar R
//You could also delete the bullet here because it's "stopped" by the rock
This code will test if the two are collided and set R to 0 (meaning it doesn't exist). If you wanted, you could also add a text-version of an explosion to give it some more flair. Good luck with your project, and I hoped this helped!
When making a game in Ti-Basic, there are 2 options for which "screen" to put it on: The text screen of the calculator, using functions like Disp and Output, and the graph screen, using Text, as well as the drawing functions, stored pictures, and pxl and pt functions. I'm not sure about how to check the text out put screen for the presence of an object on the screen, but the graph function has the Pxl-Test function:
:Pxl-Test(y,x)
Someone else may have know how to check for something on the text screen, but I suggest switching to the graph screen to make use of this function.
If you are going to switch to using the graph output here are some things to note:
The text function is
:Text(row,col,output)
Row and col are switched from the Output function
Rows and cols are measured in pixels on the screen, as opposed to the space it takes to type a letter
Most letters are 3 pixels wide, spaces are 1 pixel wide, so it will take multiple spaces to clear out an unwanted character
You may need to disable the graph axis (FORMAT menu), the y-functions (VARS menu, then right to Y-VARS), and the stat-plots (STAT PLOT menu) in the begging of the program then restore them at the end.
One last not on using the Pxl-Test function is that it tests a group of pixels, while a letter. For now let's pretend you only use Rs and an X in your program (I don't know if you do or don't), so you should test a spot on that character that is relatively unique to that character.
Good luck!
I am writing now a flash game and I run into a an issue. I have a map for the game which is defined as a 2-D array, where each element represents a component of the map. The player is always in the center of the map.
The problem is when the player reaches one end of the map. Now it is empty space. I want that the player instead of seeing the empty space, to see another end of the map and in this way, the map will loo like it goes around.
So for example if the player goes to right he will eventually start seeing the the left side of the map and the world will look continuous.
Does anyone knows how to implement this functionality?
You could make the array 2 times and put the first one behind the second one again and than the second one behind the first etc etc..
It's done here with 2 pictures, just use the arrays instead:
//The speed of the scroll movement.
var scrollSpeed:uint = 2;
//This adds two instances of the movie clip onto the stage.
var s1:ScrollBg = new ScrollBg();
var s2:ScrollBg = new ScrollBg();
addChild(s1);
addChild(s2);
//This positions the second movieclip next to the first one.
s1.x = 0;
s2.x = s1.width;
//Adds an event listener to the stage.
stage.addEventListener(Event.ENTER_FRAME, moveScroll);
//This function moves both the images to left. If the first and second
//images goes pass the left stage boundary then it gets moved to
//the other side of the stage.
function moveScroll(e:Event):void{
s1.x -= scrollSpeed;
s2.x -= scrollSpeed;
if(s1.x < -s1.width){
s1.x = s1.width;
}else if(s2.x < -s2.width){
s2.x = s2.width;
}
}
You simply check if your player is about to get off the "right" or "left" edge of the map, and position him at the other edge. To draw a circular map, you can use the following technique: if you are about to draw a column of a number that exceeds the map's width, decrease that number by width and draw the column at resultant index; and if you are about to draw a column at index below zero, add width and draw the column at resultant index. If you are in troubles of making a hitcheck at continuous map's borders, you can employ the same trick to find neighbors. (The "circular array" is a pretty basic algorithmic problem, and is resolved in many ways already)
You have a few options here. You can do the pac-man style of just making your character pop up on the other side of the screen, but that would require you to abandon the cool bit of the character being in the middle at all times.
On to the real suggestions:
If you're not implementing your array as one solid object (i.e. making it draw individual collumns/rows at a time) then this is a no-brainer. Just have a function that returns the index of the next collumn/row, within certain bounds. Like, if your array is 40 elements wide, when it tries to draw element 41, subtract the size of the array, and make it draw element 1 instead.
If your array is one solid object (like if you drew it onto a stage object and are just manipulating that) and it's not very big, you could probably get away with drawing a total of four of them, and just having a new one cover up any whitespace that's about to appear. Like, as you approach the right edge of the first array, the second array moves to the right of it for a lawless transition.
If your array is a solid object and is very big, perhaps you could make eight buffer objects (one per edge and one per corner) that hold approximately half a screen's worth of the array. That way as you approach the right edge, you see the left edge, but then when you cross into the buffer zone, you could teleport the player to the corresponding position on the left of the array, which has the buffer for the right size. To the player, nothing has changed, but now they're on the other side of the world.
I'm working on a game where there are 4 characters on screen, each with several different states and corresponding MovieClips so that my library looks a little something like this:
Character 1:
IdleAnimation
SleepAnimation
..
Character 2:
IdleAnimation
SleepAnimation
..
Each animation (provided by a 3rd party) has a different anchor point, so when character A is at position 100, 100 on screen and moves from the idle to sleep animation he suddenly jumps 20 pixels to the right due to the anchor point.
Usually I would store all the different states in one MovieClip in separate frames and manually adjust the position of each one til they matched up, however I am also trying to port this to Android and therefore must keep the number of children on screen at once to a minimum.
Is there anything else I can do other than store offset x and y values for each character and each animation, and simple set the x/y via code whenever their state changes?
To keep the numChildren down I guess you have no choice
but to keep track of x- and y-offsets of the animations of each character
stored perhaps in an Object variable of form { state: registration_point }.
Then when changing states you can set the registration point
through this method by Emanuel Feronato:
http://www.emanueleferonato.com/2010/08/04/changing-a-movieclip-registration-point-on-the-fly-with-as3/
An idea to try though is to simply place the registration point to the center of each animation state:
var reg_x:Number = mc.width / 2;
var reg_y:Number = mc.height / 2;
I'm trying to position an image on top of another image based upon the make-up of the smaller image. The smaller image is a cut-out of a larger image and I need it to be positioned exactly on the larger image to make it look like a single image, but allow for separate filters and alphas to be applied. As the images are not simple rectangles or circles, but complex satellite images, I cannot simply redraw them in code. I have quite a few images and therefore do not feel like manually finding the position of each image every and hard setting them manually in actionscript. Is there any way for me to sample a small 5-10 sq. pixel area against the larger image and set the x and y values of the smaller image if a perfect match is found? All the images are in an array and iterating through them has already been set, I just need a way to sample and match pixels. My first guess was to loop the images pixel by pixel right and down, covering the whole bitmap and moving to the next child in the array once a match was found, leaving the matched child where it was when the perfect match was found.
I hope I understood your question correctly.
There may be an option that uses copypixels to achieve what you want. You can use the bitmapdata.rect value to determine the size of the sample you want, and loop through the bigger bitmap using thet rectangle and a moving point. Let's see if I can code this out...
function findBitmapInBitmap(tinyimg:BitmapData, largeimg:BitmapData):Point {
var rect:Rectangle = tinyimg.rect;
var xbound:uint = largeimg.rect.width;
var ybound:uint = largeimg.rect.height;
var imgtest:BitmapData = new BitmapData(tinyimg.rect.width, tinyimg.rect.height);
for (var ypos:uint = 0, y <= ybound, y++) {
for (var xpos:uint = 0, x <= xbound, x++) {
imgtest.copyPixels(largeimg, rect, new Point(xpos, ypos);
if (imgtest.compare(tinyimg) == 0) return new Point(xpos, ypos);
}
}
return new Point(-1,-1); // Dummy value, indicating no match.
}
Something along those lines should work - I'm sure there's room for code elegance and possible optimization. However, it seems like something like this method would be very slow, since you'd have to check each pixel for a match.
There is a better way. Split your big image into layers, and use the blitting technique to composite them at runtime. In your case, you could create a ground texture without satellites, and then create the satellites separately, and use the copyPixels method to place them whereever you want. Google "blitting in as3" to find some good tutorials. I'm currently working on a game project that uses this technique and it's a very good method.
Good luck!
Edit: Forgot to code in a default return statement. Using this method, you'd have to return an invalid point (like (-1,-1)) and check for it outside the function. Alternatively, you could just copy your small bitmap to the big one within the function, which would be much more logical, but I don't know your requirements.
You need to find pixel sequence in the big image. BitmapData.getPixel gives you pixel value. So get first pixel from small image, find it in big image, then continue comparing until you find full match. If you have trouble to code that, feel free to ask.
For the actual comparison, there's BitmapData.compare which returns the number 0 if the BitmapData objects are equivalent.