I've been tasked to find the probability of getting a prime number in a Pythagorean triple for a school project, so I tried to code it, but I didn't consider multiples of Pythagorean Triples using Euclid's formula:
(a = m^2 -n^2, b = 2mn, c = m^2 + n^2.).
Ex. 3-4-5 -> 6-8-10.
var primitiveCount = 0;
var m = floor(2), n = floor(1);
var a, b, c;
var ans1, ans2, ans3;
var isPrime = function(value) {
for(var i = 2; i < value; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
};
var j=10;
for(var j = 1; j <= 150; j++){
primitiveCount = 0;
m=2;
n=1;
for(var i=0; i < j; i++) {
a = ((Math.pow(m,2))-(Math.pow(n,2)));
b = 2 * n * m;
c = ((Math.pow(m,2))+(Math.pow(n,2)));
if(a<b<c) {
if(Math.pow(a,2) + Math.pow(b,2) === Math.pow(c,2)) {
ans1 = a;
ans2 = b;
ans3 = c;
}
if(isPrime(ans1) || isPrime(ans2) || isPrime(ans3)){
println(ans1 + " " + ans2 + " " + ans3 + " ~");
primitiveCount++;
}else{
println(ans1 + " " + ans2 + " " + ans3);
}
m++;
n++;
}
}
println(primitiveCount + "/" + j);
}
My code creates unique ones, but does not give me the multiples
It is enough to use natural coefficient p
a = p * (m^2 -n^2)
b = p * 2mn
c = p * (m^2 + n^2)
With m>n, m and n coprime and with different parity this method gives all unique triplets.
Conditions together:
p = 1,2,3...
n = 1,2,3...
m = n + 1 + 2 * i, and GCD(m, n) = 1
Seems that your code does not use all values for m, so omits many triplets.
Also use m*m instead of slow pow. Also you don't need to check a^2+b^2=c^2
Related
So, for each star, i compare this one to all other stars to calculate his speed, velocity, etc.
But that didn't work, I'm not too strong in maths and I think my formula is maybe wrong? idk why that didn't work here my code :
//for each star I compare to all other stars
for(let i = 0; i < pos.length; i ++) {
for (let j = 0; j < pos.length; j ++){
if (i !== j){
// Formula part
const vector = compute_interaction(pos[i], pos[j], 1.0);
accelerations[i].x += vector.x;
accelerations[i].y += vector.y;
accelerations[i].z += vector.z;
break;
}
}
}
for (let i = 0 ; i<accelerations.length ; i++){
speedStars[i].x += accelerations[i].x * 0.001;
speedStars[i].y += accelerations[i].y * 0.001;
speedStars[i].z += accelerations[i].z * 0.001;
}
for (let i = 0 ; i<speedStars.length ; i++){
const i3 = i*3;
starsPositions[i3] += speedStars[i].x * 0.001;
starsPositions[i3 + 1] += speedStars[i].y * 0.001;
starsPositions[i3 + 2] += speedStars[i].z * 0.001;
}
function compute_interaction(currentPosition, positionOtherStar, smoothing_length)
{
const vector = new THREE.Vector3(positionOtherStar.x - currentPosition.x, positionOtherStar.y - currentPosition.y, positionOtherStar.z - currentPosition.z).normalize();
let x = vector.x / (Math.pow(positionOtherStar.x,2.0) - Math.pow(currentPosition.x,2.0)+ smoothing_length)
let y = vector.y / (Math.pow(positionOtherStar.y,2.0) - Math.pow(currentPosition.y,2.0)+ smoothing_length)
let z = vector.z / (Math.pow(positionOtherStar.z,2.0) - Math.pow(currentPosition.z,2.0)+ smoothing_length)
return new THREE.Vector3(x, y, z);
}
Here the CodePen: https://codepen.io/n0rvel/pen/ExEXbYN?editors=0010
Here is the formula/code logic I found on one OpenCL program that works:
Probably, the compute_interaction() function should be:
function compute_interaction(currentPosition, positionOtherStar, smoothing_length)
{
//const vector = new THREE.Vector3(positionOtherStar.x - currentPosition.x, positionOtherStar.y - currentPosition.y, positionOtherStar.z - currentPosition.z).normalize();
//let x = vector.x / (Math.pow(positionOtherStar.x,2.0) - Math.pow(currentPosition.x,2.0)+ smoothing_length)
//let y = vector.y / (Math.pow(positionOtherStar.y,2.0) - Math.pow(currentPosition.y,2.0)+ smoothing_length)
//let z = vector.z / (Math.pow(positionOtherStar.z,2.0) - Math.pow(currentPosition.z,2.0)+ smoothing_length)
//return new THREE.Vector3(x, y, z);
const vector = new THREE.Vector3().subVectors(positionOtherStar, currentPosition);
return vector.normalize().divideScalar(vector.lengthSq() + smoothing_length);
}
I have 3D array (height, width, depth). My global worksize is (height * width * depth). and local work size is 1. In kernel code how I can get row offset and column offset?
I am doing the convolution operation in opencl. In C we do as follow,
// iterating through number of filters
for(c = 0; c < number_of_filters; c++)
{
for(h = 0; h < out_height; h++)
{
for(w = 0; w < out_width; w++)
{
vert_start = h * stride;
vert_end = vert_start + f_size ;
hor_start = w * stride;
hor_end = hor_start + f_size;
sum = 0;
for(c_f = 0; c_f < input_channel; c_f++)
{
for(h_f = vert_start; h_f < vert_end; h_f++)
{
for(w_f = hor_start; w_f < hor_end; w_f++)
{
// computing convolution
sum = sum +
(INPUT[(c_f * input_height * input_width) + (h * input_width) + w] *
FILTER[(c_f * filt_height* filt_width) + (h_f * filt_width) + w_f)]);
}
}
}
// storing result in output
OUTPUT[(c * out_height * out_width) + (h * out_width) + w] = sum;
}
}
}
I am not getting how to get that row offset and column offset from image for convolution in opencl?
I need to do it with recursion, but the problem is that function depends on only ONE parameter and inside function it depends on two ( k and n ), also how to find minimum value if it returns only one value?
The function is :
I've already tried to make random k, but I don't think that is really good idea.
F1(int n) {
Random random = new Random();
int k = random.Next(1,10);
if (1 <= k && k <= n){
return Math.Min(F1(k - 1) + F1(n - k) + n);
} else {
return 0;
}
}
You need to make a loop traversing all k values in range 1..n. Something like this:
F1(int n) {
if (n == 0)
return ???? what is starting value?
minn = F1(0) + F1(n - 1) + n
for (int k = 2; k <= n; k++)
minn = Math.Min(minn, F1(k - 1) + F1(n - k) + n);
return minn;
}
I am currently writing a program to show the relation between the Relaxation Factor and number of iterations it takes to achieve a solution using the Successive OverRelaxation Method
This is the concerned For loop:
for (w = 0:0.05:2)
T = -inv(D + w*L)*(w*U + (w -1) * D);
C = inv(D + w*L) * w *B ;
X = zeros(B);
for(i = 1:1:MaxIter)
X = T * X + C;
err = A * X - B;
if (abs(err) < abs(tol))
break
end
end
disp("Relaxation Factor = " + string(w) +" No. of iterations = " + string(i));
I wish to draw a plot showing the relation between Relaxation factor and no. of iterations(between w and i). How should I proceed?
Just proceed as follow
I=zeros(W)
for k = 1:size(W,"*")
w=W(k)
T = -inv(D + w*L)*(w*U + (w -1) * D);
C = inv(D + w*L) * w *B ;
X = zeros(B);
for(i = 1:1:MaxIter)
X = T * X + C;
err = A * X - B;
if (abs(err) < abs(tol))
I(k)=i
break
end
end
end
plot(W,I)
I want to define color value by using HSL(Hue, Saturation, Lightness) not RGB.
Does actionscript have a function for that?
For example, I want to write code like the following:
var color:int = hsl(10, 90, 30);
sprite.graphics.lineStyle(2, color);
No built-in way, but there are many functions available online to do the conversion. For example these two JavaScript functions (they might need to be slightly refactored to compile with AS3 but it should just be minor changes):
/**
* Converts an RGB color value to HSL. Conversion formula
* adapted from http://en.wikipedia.org/wiki/HSL_color_space.
* Assumes r, g, and b are contained in the set [0, 255] and
* returns h, s, and l in the set [0, 1].
*
* #param Number r The red color value
* #param Number g The green color value
* #param Number b The blue color value
* #return Array The HSL representation
*/
function rgbToHsl(r, g, b){
r /= 255, g /= 255, b /= 255;
var max = Math.max(r, g, b), min = Math.min(r, g, b);
var h, s, l = (max + min) / 2;
if(max == min){
h = s = 0; // achromatic
}else{
var d = max - min;
s = l > 0.5 ? d / (2 - max - min) : d / (max + min);
switch(max){
case r: h = (g - b) / d + (g < b ? 6 : 0); break;
case g: h = (b - r) / d + 2; break;
case b: h = (r - g) / d + 4; break;
}
h /= 6;
}
return [h, s, l];
}
/**
* Converts an HSL color value to RGB. Conversion formula
* adapted from http://en.wikipedia.org/wiki/HSL_color_space.
* Assumes h, s, and l are contained in the set [0, 1] and
* returns r, g, and b in the set [0, 255].
*
* #param Number h The hue
* #param Number s The saturation
* #param Number l The lightness
* #return Array The RGB representation
*/
function hslToRgb(h, s, l){
var r, g, b;
if(s == 0){
r = g = b = l; // achromatic
}else{
function hue2rgb(p, q, t){
if(t < 0) t += 1;
if(t > 1) t -= 1;
if(t < 1/6) return p + (q - p) * 6 * t;
if(t < 1/2) return q;
if(t < 2/3) return p + (q - p) * (2/3 - t) * 6;
return p;
}
var q = l < 0.5 ? l * (1 + s) : l + s - l * s;
var p = 2 * l - q;
r = hue2rgb(p, q, h + 1/3);
g = hue2rgb(p, q, h);
b = hue2rgb(p, q, h - 1/3);
}
return [r * 255, g * 255, b * 255];
}
Then use ColorTransform to get the uint RGB value:
var c:Array = hslToRgb(10, 90, 30);
var t:ColorTransform = new ColorTransform(1.0, 1.0, 1.0, 1.0, c[0], c[1], c[2]);
var color:uint = t.color;
If you use Flex4, you can use class HSBColor. It has two static methods: convertHSBtoRGB and convertRGBtoHSB.