I'm trying to spread 8 balls on the stage with an uneven space between them within a range. using myNum variable in this statement ball.x = 150 + i * myNum; inside for loop I was hopping that it will spread each of the 8 balls in an uneven space. However this is not what's happening, instead it position 8 balls with an even space and then about a minute later it positions another set of 8 balls with different spacing.
var minLimit: int = 25;
var maxLimit: int = 43;
var range: int = maxLimit - minLimit;
var myNum: Number = Math.ceil(Math.random() * range) + minLimit;
var balls: Array = [],
ball: bomb30a;
for (var i: int = 0; i < 8; i++) {
ball = new bomb30a();
ball.x = 150 + i * myNum;
ball.y = 242;
balls.push(ball);
addChild(ball);
}
Your code defines a random number myNum, then loops 8 times. Why would myNum change in the middle of the loop?
If you want each iteration in the loop to use a different random number, you need to move the random number code into the loop so it gets executed on each iteration.
ball.x = 150 + i * (Math.ceil(Math.random() * range) + minLimit);
Related
I'm looking for a tip for a AS3 script, have no idea how to start there
Button, if clicked the function is executed, which outputs a predefined value as the cross sum of a number string.
Example:
Cross sum should be 10
By clicking on the button, the function generates the number 55 or 82 or 37 or 523, ie numbers with the cross sum 10
An alternative way using % (modulo) instead of a string. You could write that into one line like this:
while (sum != 0) { qsum += sum % 10; sum /= 10; }
The trick is that modulo will give us only the last digit of the longer number, then we divide by 10 to trim off that last number (from longer) and we re-read a newer ending digit of the long number.
Example:
Long num = 1234, so each trim gives, 4 then 3 then 2 then 1 and we'll sum them up each time.
usage:
myInt = cross_sum(50); //makes myInt hold answer result of function (where ExpectedValue input is 50).
and the supporting function...
function cross_sum( ExpectedValue: int ) : int
{
var rand :int = Math.floor(Math.random() * 100000000000)
var sum :int = Math.abs( rand );
var qsum :int = 0;
while (sum != 0)
{
qsum += sum % 10; //get last digit of sum...
sum /= 10; //trim down sum by 1 digit...
}
if ( qsum == ExpectedValue ) { return rand; } //# stop here and give back "rand" as answer result.
else { cross_sum( expectedValue ); } //# else if wrong, try again...
}
Got it now.....
the function calculates a number, with the crosssum 50
function berechnen() {
var rand = Math.floor(Math.random() * 100000000000)
var sum = String(rand)
var qsum = 0;
for (var i = 0; i < sum.length; i++) {
qsum += Number(sum.charAt(i));
}
if (qsum == 50) {
summe.text = String(sum);
} else {
berechnen()
}
}
if I record a series of frequencies beeps into a buffer, for example:
15kHz for 50ms, 17k for 50 ms and goes on, is there any way to "go" along the time plain and to decode this freqs(with goertzel or something)?
Hey, this is an update, I've added a code that shows how I find the first delimiter in the sound buffer that I check. If I record 5 seconds of a buffer(I record into a stream buffer and not a file) The first snippet takes something like 30 seconds to analyze the index where the start delimiter starts at. I thinks it is very newbie...must find a better solution. thanks
(every delimiter is 0.2 seconds duration) and it's like that - Start delimiter = 12KHz, 1's = 13k, 0's = 14k, End delimiter = 15k
double max_power = 0;
int max_power_index = 0;
double DelimiterSamplesCount = SampleRate * DelimiterTime;
float[] samples32array = samples32.ToArray();
//Searching For Delimiter
for (int i = 0; i < (samples32array.Length); i++) //Delimiter Samples Length = SampleRate*DelimiterTimeLength,( i.e: 44100*0.2=8820 samples)
{
if ((i + (int)DelimiterSamplesCount - 1) > samples32array.Length) break;
double power = Goertzel.GoertzelFilter(samples32array, StartDelimiterFreq, i, i + (int)DelimiterSamplesCount - 1);
if(power > max_power)
{
max_power = power;
max_power_index = i;
}
}
My Goertzel is like that:
public static double GoertzelFilter(float[] samples, double freq, int start, int end)
{
double sPrev = 0.0;
double sPrev2 = 0.0;
int i;
double normalizedfreq = freq / 44100;
double coeff = 2 * Math.Cos(2 * Math.PI * normalizedfreq);
for (i = start; i < end; i++)
{
double s = samples[i] + coeff * sPrev - sPrev2;
sPrev2 = sPrev;
sPrev = s;
}
double power = sPrev2 * sPrev2 + sPrev * sPrev - coeff * sPrev * sPrev2;
return power;
}
If you know the set of frequencies and the durations, then a set of sliding Goertzel filters is a good start to building a simple demodulator. Comparing and scanning for for a peak difference between these filters is a better decision criteria than just checking for a certain magnitude output.
In Flash AS3, how would I write a function that will:
Take in an integer (a list index, for example)
return a visually distinct hex color based on that number (and will consistently return that same color given that same number)
The purpose is to provide a visually distinct color for each item in varying-length list of items. The most I expect to support is around 200, but I don't see the count going far above 20 or so for most.
Here's my quick and dirty:
public static function GetAColor(idx:int):uint {
var baseColors:Array = [0xff0000, 0x00ff00, 0xff0080, 0x0000ff, 0xff00ff, 0x00ffff, 0xff8000];
return Math.round(baseColors[idx % baseColors.length] / (idx + 1) * 2);
}
It does OK, but it would be nice to see a more distinct set of colors that are not so visually close to one another
You could go with generator of random values that supports seed, so you will be able return same color. As for color you could build it - by randomValue * 0xFFFFFF, where randomValue between 0 and 1. And exclude values (colors) that are close.
Second option: build palette of 200 colors with step - 0xFFFFFF / 200 and shuffle palette with predefined logic, so you will have same colors.
Third option: as for really distinct colors, you could go with big jumps in every channel. Example: 0xFF * 0.2 - 5 steps in every channel.
Fourth option: go with HSV. It's easy to understand(watch image, rotate hue from 0 to 360, change saturation and value from 0 to 100) how to manipulate parameters to get distinct color:
//Very simple demo, where I'm only rotating Hue
var step:uint = 15;
var position:uint = 0;
var colors:Array = [];
for (; position < 360; position += step) {
colors.push(HSVtoRGB(position, 100, 100));
}
//Visualisation for demo
var i:uint, len:uint = colors.length, size:uint = 40, shape:Shape, posX:uint, posY:uint;
for (i; i < len; ++i) {
shape = new Shape();
shape.graphics.beginFill(colors[i]);
shape.graphics.drawRect(0, 0, size, size);
addChild(shape);
shape.x = posX;
shape.y = posY;
posX += size;
if (posX + size >= stage.stageWidth) {
posX = 0;
posY += size;
}
}
public function HSVtoRGB(h:Number, s:Number, v:Number):uint {
var r:Number = 0;
var g:Number = 0;
var b:Number = 0;
var tempS:Number = s / 100;
var tempV:Number = v / 100;
var hi:int = Math.floor(h / 60) % 6;
var f:Number = h / 60 - Math.floor(h / 60);
var p:Number = (tempV * (1 - tempS));
var q:Number = (tempV * (1 - f * tempS));
var t:Number = (tempV * (1 - (1 - f) * tempS));
switch (hi) {
case 0:
r = tempV;
g = t;
b = p;
break;
case 1:
r = q;
g = tempV;
b = p;
break;
case 2:
r = p;
g = tempV;
b = t;
break;
case 3:
r = p;
g = q;
b = tempV;
break;
case 4:
r = t;
g = p;
b = tempV;
break;
case 5:
r = tempV;
g = p;
b = q;
break;
}
return (Math.round(r * 255) << 16 | Math.round(g * 255) << 8 | Math.round(b * 255));
}
And last one, if you want go with this task like a pro, this wiki article could be helpful for you.
I was wondering if someone could help me with a circle meter gauage i have taken some code from a different example and i am just protypting stuff to see if i can get it to work here is a working example.
http://jsbin.com/ixuyid/28/edit
Click run with javascript
Code below
var context;
canvas = document.getElementById('myCanvas');
context = canvas.getContext('2d');
//use a reusable function
function drawCircle(num){
console.log(num);
var x = canvas.width / 2;
var y = canvas.height / 2;
var radius = 75;
var startAngle = 0 * Math.PI;
var endAngle = num * Math.PI;
var counterClockwise = false;
context.beginPath();
context.arc(x, y, radius, startAngle, endAngle, counterClockwise);
context.lineWidth = 5;
// line color
context.strokeStyle = 'black';
context.stroke();
}
drawCircle();
var num = 1;
setInterval(function(){
},1000);
+function(){
var ctx = new webkitAudioContext()
, url = '//kevincennis.com/sound/loudpipes.mp3'
, audio = new Audio(url)
// 2048 sample buffer, 1 channel in, 1 channel out
, processor = ctx.createJavaScriptNode(2048, 1, 1)
, meter = document.getElementById('meter')
, source
audio.addEventListener('canplaythrough', function(){
source = ctx.createMediaElementSource(audio)
source.connect(processor)
source.connect(ctx.destination)
processor.connect(ctx.destination)
audio.play()
}, false);
// loop through PCM data and calculate average
// volume for a given 2048 sample buffer
processor.onaudioprocess = function(evt){
var input = evt.inputBuffer.getChannelData(0)
, len = input.length
, total = i = 0
, rms
while ( i < len ) total += Math.abs( input[i++] )
rms = Math.sqrt( total / len )
meter.style.width = ( rms * 100 ) + '%';
context.clearRect(100,50,200,200);
drawCircle(rms);
}
}()
I seem to be having issue with the levels???
Any help
Change these two lines in the drawCircle function:
var startAngle = 0; //multiplying with 0 will result in 0
var endAngle = 360 * num * Math.PI / 180;
Your num seem to be a value between 0 and 1 so we need to add what we're using that with, here 360 degrees, then convert by using PI / 180.
The other problem is that the clearRect wasn't extended far enough so it left part of the arc uncleared to the right.
Tip: To make it look more realistic you can update your rms only when the new rms is higher, and if not just subtract a small value for each frame.
For example:
//global scope
var oldRMS = 0;
Inside your processor.onaudioprocess after vars:
if (rms > oldRMS) oldRMS = rms;
meter.style.width = ( oldRMS * 100 ) + '%';
context.clearRect(100,50,canvas.width,canvas.height);
drawCircle(oldRMS);
oldRMS -= 0.04; //speed of fallback
Modifcations:
http://jsbin.com/ixuyid/29/edit
If I am getting a random number, how do I increase my chances of making that random number to be even. I am not looking to make it even every time. I am just looking to generate a random number say... %70 of the time or %90 of the time.
private function randNum (high, low) {
return Math.floor(Math.random() * (1+high-low)) + low;
}
Would I pass in a greater range of numbers to this function? Or would I have to scrap this function altogether?
Thank you
private function randNum (high : Number, low : Number) : int
{
var even : Boolean = Math.random() < 0.7; //set probability of even number here
var rand : int = Math.floor(Math.random() * (1+high-low)) + low;
if (even)
while (rand % 2 != 0)
rand = Math.floor(Math.random() * (1+high-low)) + low;
else
while (rand % 2 != 1)
rand = Math.floor(Math.random() * (1+high-low)) + low;
return rand;
}
Test:
var even : int = 0;
var odd : int = 0;
for (var i : int = 0; i < 100000; i++)
{
var a : int = randNum(1, 20);
if (a % 2 == 0)
even++;
else
odd++;
}
trace(even, odd);
Output:
[trace] 69869 30131
A little too late ;) but another one with no loop and using bit masking operation :
ret & -2 will make your number even, then depending on the result of (Math.random() >= evenProbability) you set the lower bit to be 1 to give an odd number
function randomRange(low:int, high:int, evenProbability:Number = 0.5):int{
var ret:int = int( Math.random() * ( 1 + high - low ) ) + low
return ( ret & -2 ) | int( Math.random() >= evenProbability )
}
Here a live test with wonderfl : http://wonderfl.net/c/9IHx