I have code from a web site, and it looks like it should be simple, but too simple for SVG. How can I determine if this is truly SVG, and what it does? I am especially interested in what looks like nested & and dots[.], then split, map.
Snippet:
// the shape of the dragon, converted from a SVG image
'! ((&(&*$($,&.)/-.0,4%3"7$;(#/EAA<?:<9;;88573729/7,6(8&;'.split("").map(function(a,i) {
shape[i] = a.charCodeAt(0) - 32;
});
Full code:
//7 Dragons
//Rauri
// full source for entry into js1k dragons: http://js1k.com/2014-dragons/demo/1837
// thanks to simon for grunt help and sean for inspiration help
// js1k shim
var a = document.getElementsByTagName('canvas')[0];
var b = document.body;
var d = function(e){ return function(){ e.parentNode.removeChild(e); }; }(a);
// unprefix some popular vendor prefixed things (but stick to their original name)
var AudioContext =
window.AudioContext ||
window.webkitAudioContext;
var requestAnimationFrame =
window.requestAnimationFrame ||
window.mozRequestAnimationFrame ||
window.webkitRequestAnimationFrame ||
window.msRequestAnimationFrame ||
function(f){ setTimeout(f, 1000/30); };
// stretch canvas to screen size (once, wont onresize!)
a.style.width = (a.width = innerWidth - 0) + 'px';
a.style.height = (a.height = innerHeight - 0) + 'px';
var c = a.getContext('2d');
// end shim
var sw = a.width,
sh = a.height,
M = Math,
Mc = M.cos,
Ms = M.sin,
ran = M.random,
pfloat = 0,
pi = M.PI,
dragons = [],
shape = [],
loop = function() {
a.width = sw; // clear screen
for ( j = 0; j < 7; j++) {
if ( !dragons[j] ) dragons[j] = dragon(j); // create dragons initially
dragons[j]();
}
pfloat++;
requestAnimationFrame(loop);
},
dragon = function(index) {
var scale = 0.1 + index * index / 49,
gx = ran() * sw / scale,
gy = sh / scale,
lim = 300, // this gets inlined, no good!
speed = 3 + ran() * 5,
direction = pi, //0, //ran() * pi * 2, //ran(0,TAU),
direction1 = direction,
spine = [];
return function() {
// check if dragon flies off screen
if (gx < -lim || gx > sw / scale + lim || gy < -lim || gy > sh / scale + lim) {
// flip them around
var dx = sw / scale / 2 - gx,
dy = sh / scale / 2 - gy;
direction = direction1 = M.atan(dx/dy) + (dy < 0 ? pi : 0);
} else {
direction1 += ran() * .1 - .05;
direction -= (direction - direction1) * .1;
}
// move the dragon forwards
gx += Ms(direction) * speed;
gy += Mc(direction) * speed;
// calculate a spine - a chain of points
// the first point in the array follows a floating position: gx,gy
// the rest of the chain of points following each other in turn
for (i=0; i < 70; i++) {
if (i) {
if (!pfloat) spine[i] = {x: gx, y: gy}
var p = spine[i - 1],
dx = spine[i].x - p.x,
dy = spine[i].y - p.y,
d = M.sqrt(dx * dx + dy * dy),
perpendicular = M.atan(dy/dx) + pi / 2 + (dx < 0 ? pi : 0);
// make each point chase the previous, but never get too close
if (d > 4) {
var mod = .5;
} else if (d > 2){
mod = (d - 2) / 4;
} else {
mod = 0;
}
spine[i].x -= dx * mod;
spine[i].y -= dy * mod;
// perpendicular is used to map the coordinates on to the spine
spine[i].px = Mc(perpendicular);
spine[i].py = Ms(perpendicular);
if (i == 20) { // average point in the middle of the wings so the wings remain symmetrical
var wingPerpendicular = perpendicular;
}
} else {
// i is 0 - first point in spine
spine[i] = {x: gx, y: gy, px: 0, py: 0};
}
}
// map the dragon to the spine
// the x co-ordinates of each point of the dragon shape are honoured
// the y co-ordinates of each point of the dragon are mapped to the spine
c.moveTo(spine[0].x,spine[0].y)
for (i=0; i < 154; i+=2) { // shape.length * 2 - it's symmetrical, so draw up one side and back down the other
if (i < 77 ) { // shape.length
// draw the one half from nose to tail
var index = i; // even index is x, odd (index + 1) is y of each coordinate
var L = 1;
} else {
// draw the other half from tail back to nose
index = 152 - i;
L = -1;
}
var x = shape[index];
var spineNode = spine[shape[index+1]]; // get the equivalent spine position from the dragon shape
if (index >= 56) { // draw tail
var wobbleIndex = 56 - index; // table wobbles more towards the end
var wobble = Ms(wobbleIndex / 3 + pfloat * 0.1) * wobbleIndex * L;
x = 20 - index / 4 + wobble;
// override the node for the correct tail position
spineNode = spine[ index * 2 - 83 ];
} else if (index > 13) { // draw "flappy wings"
// 4 is hinge point
x = 4 + (x-4) * (Ms(( -x / 2 + pfloat) / 25 * speed / 4) + 2) * 2; // feed x into sin to make wings "bend"
// override the perpindicular lines for the wings
spineNode.px = Mc(wingPerpendicular);
spineNode.py = Ms(wingPerpendicular);
}
c.lineTo(
(spineNode.x + x * L * spineNode.px) * scale,
(spineNode.y + x * L * spineNode.py) * scale
);
}
c.fill();
}
}
// the shape of the dragon, converted from a SVG image
'! ((&(&*$($,&.)/-.0,4%3"7$;(#/EAA<?:<9;;88573729/7,6(8&;'.split("").map(function(a,i) {
shape[i] = a.charCodeAt(0) - 32;
});
loop();
While the context this is used in is <canvas>, the origin may well be a SVG <polyline>.
In a first step, the letters are mapped to numbers. A bit of obscuration, but nothing too serious: get the number representing the letter and write it to an array.
const shape = [];
'! ((&(&*$($,&.)/-.0,4%3"7$;(#/EAA<?:<9;;88573729/7,6(8&;'.split("").map(function(a,i) {
shape[i] = a.charCodeAt(0) - 32;
});
results in an array
[1,0,8,8,6,8,6,10,4,8,4,12,6,14,9,15,13,14,16,12,20,5,19,2,23,4,27,8,32,15,37,33,33,28,31,26,28,25,27,27,24,24,21,23,19,23,18,25,15,23,12,22,8,24,6,27]
Now just write this array to a points attribute of a polyline, joining the numbers with a space character:
const outline = document.querySelector('#outline');
const shape = [];
'! ((&(&*$($,&.)/-.0,4%3"7$;(#/EAA<?:<9;;88573729/7,6(8&;'.split("").map(function(a,i) {
shape[i] = a.charCodeAt(0) - 32;
});
outline.setAttribute('points', shape.join(' '))
#outline {
stroke: black;
stroke-width: 0.5;
fill:none;
}
<svg viewBox="0 0 77 77" width="300" height="300">
<polyline id="outline" />
</svg>
and you get the basic outline of (half) a dragon. The rest is repetition and transformation to make things a bit more complex.
I'm using simple moving average in Math.Net, but now that I also need to calculate EMA (exponential moving average) or any kind of weighted moving average, I don't find it in the library.
I looked over all methods under MathNet.Numerics.Statistics and beyond, but didn't find anything similar.
Is it missing in library or I need to reference some additional package?
I don't see any EMA in MathNet.Numerics, however it's trivial to program. The routine below is based on the definition at Investopedia.
public double[] EMA(double[] x, int N)
{
// x is the input series
// N is the notional age of the data used
// k is the smoothing constant
double k = 2.0 / (N + 1);
double[] y = new double[x.Length];
y[0] = x[0];
for (int i = 1; i < x.Length; i++) y[i] = k * x[i] + (1 - k) * y[i - 1];
return y;
}
Occasionally I found this package: https://daveskender.github.io/Stock.Indicators/docs/INDICATORS.html It targets to the latest .NET framework and has very detailed documents.
Try this:
public IEnumerable<double> EMA(IEnumerable<double> items, int notationalAge)
{
double k = 2.0d / (notationalAge + 1), prev = 0.0d;
var e = items.GetEnumerator();
if (!e.MoveNext()) yield break;
yield return prev = e.Current;
while(e.MoveNext())
{
yield return prev = (k * e.Current) + (1 - k) * prev;
}
}
It will still work with arrays, but also List, Queue, Stack, IReadOnlyCollection, etc.
Although it's not explicitly stated I also get the sense this is working with money, in which case it really ought to use decimal instead of double.
my result , but the shadow is so hard.
for (SceneLight* light : scene->lights)
{
Vector3D dir_to_light;
float dist_to_light;
float pdf;
int num_light_samples = light->is_delta_light() ? 1 : ns_area_light;
double scale = 1.0 / num_light_samples;
for (int i = 0; i < num_light_samples; i++) {
Spectrum light_L = light->sample_L(hit_p, &dir_to_light, &dist_to_light, &pdf);
Vector3D w_in = w2o * dir_to_light;
double cos_theta = std::max(0.0, w_in[2]);
Spectrum f = isect.bsdf->f(w_out, w_in);
Ray shadow_ray(hit_p + EPS_D * dir_to_light, dir_to_light, dist_to_light - (EPS_D * dir_to_light).norm(), 0);
if (!bvh->intersect(shadow_ray))
{
L_out += (f * light_L * (cos_theta * scale / pdf));
}
}
}
**
abolve is my some code and render result. The shadow looks so hard.If i want to make the shadow softer, What can I do? I am writing path tracing.
Thanks.**
Here's the link to the question..
http://www.codechef.com/problems/J7
I figured out that 2 edges have to be equal in order to give the maximum volume, and then used x, x, a*x as the lengths of the three edges to write the equations -
4*x + 4*x + 4*a*x = P (perimeter) and,
2*x^2 + 4*(a*x *x) = S (total area of the box)
so from the first equation I got x in terms of P and a, and then substituted it in the second equation and then got a quadratic equation with the unknown being a. and then I used the greater root of a and got x.
But this method seems to be giving the wrong answer! :|
I know that there isn't any logical error in this. Maybe some formatting error?
Here's the main code that I've written :
{
public static void main(String[] args)
{
TheBestBox box = new TheBestBox();
reader = box.new InputReader(System.in);
writer = box.new OutputWriter(System.out);
getAttributes();
writer.flush();
reader.close();
writer.close();
}
public static void getAttributes()
{
t = reader.nextInt(); // t is the number of test cases in the question
for (int i = 0; i < t; i++)
{
p = reader.nextInt(); // p is the perimeter given as input
area = reader.nextInt(); // area of the whole sheet, given as input
a = findRoot(); // the fraction by which the third side differs by the first two
side = (double) p / (4 * (2 + a)); // length of the first and the second sides (equal)
height = a * side; // assuming that the base is a square, the height has to be the side which differs from the other two
// writer.println(side * side * height);
// System.out.printf("%.2f\n", (side * side * height));
writer.println(String.format("%.2f", (side * side * height))); // just printing out the final answer
}
}
public static double findRoot() // the method to find the 2 possible fractions by which the height can differ from the other two sides and return the bigger one of them
{
double a32, b, discriminant, root1, root2;
a32 = 32 * area - p * p;
b = 32 * area - 2 * p * p;
discriminant = Math.sqrt(b * b - 4 * 8 * area * a32);
double temp;
temp = 2 * 8 * area;
root1 = (- b + discriminant) / temp;
root2 = (- b - discriminant) / temp;
return Math.max(root1, root2);
}
}
could someone please help me out with this? Thank You. :)
I also got stuck in this question and realized that can be done by making equation of V(volume) in terms of one side say 'l' and using differentiation to find maximum volume in terms of any one side 'l'.
So, equations are like this :-
P = 4(l+b+h);
S = 2(l*b+b*h+l*h);
V = l*b*h;
so equation in l for V = (l^3) - (l^2)P/4 + lS/2 -------equ(1)
After differentiation we get:-
d(V)/d(l) = 3*(l^2) - l*P/2 + S/2;
to get max V we need to equate above equation to zero(0) and get the value of l.
So, solutions to a quadratic equation will be:-
l = ( P + sqrt((P^2)-24S) ) / 24;
so substitute this l in equation(1) to get max volume.
Ok, so I have a histogram (represented by an array of ints), and I'm looking for the best way to find local maxima and minima. Each histogram should have 3 peaks, one of them (the first one) probably much higher than the others.
I want to do several things:
Find the first "valley" following the first peak (in order to get rid of the first peak altogether in the picture)
Find the optimum "valley" value in between the remaining two peaks to separate the picture
I already know how to do step 2 by implementing a variant of Otsu.
But I'm struggling with step 1
In case the valley in between the two remaining peaks is not low enough, I'd like to give a warning.
Also, the image is quite clean with little noise to account for
What would be the brute-force algorithms to do steps 1 and 3? I could find a way to implement Otsu, but the brute-force is escaping me, math-wise. As it turns out, there is more documentation on doing methods like otsu, and less on simply finding peaks and valleys. I am not looking for anything more than whatever gets the job done (i.e. it's a temporary solution, just has to be implementable in a reasonable timeframe, until I can spend more time on it)
I am doing all this in c#
Any help on which steps to take would be appreciated!
Thank you so much!
EDIT: some more data:
most histogram are likely to be like the first one, with the first peak representing background.
Use peakiness-test. It's a method to find all the possible peak between two local minima, and measure the peakiness based on a formula. If the peakiness higher than a threshold, the peak is accepted.
Source: UCF CV CAP5415 lecture 9 slides
Below is my code:
public static List<int> PeakinessTest(int[] histogram, double peakinessThres)
{
int j=0;
List<int> valleys = new List<int> ();
//The start of the valley
int vA = histogram[j];
int P = vA;
//The end of the valley
int vB = 0;
//The width of the valley, default width is 1
int W = 1;
//The sum of the pixels between vA and vB
int N = 0;
//The measure of the peaks peakiness
double peakiness=0.0;
int peak=0;
bool l = false;
try
{
while (j < 254)
{
l = false;
vA = histogram[j];
P = vA;
W = 1;
N = vA;
int i = j + 1;
//To find the peak
while (P < histogram[i])
{
P = histogram[i];
W++;
N += histogram[i];
i++;
}
//To find the border of the valley other side
peak = i - 1;
vB = histogram[i];
N += histogram[i];
i++;
W++;
l = true;
while (vB >= histogram[i])
{
vB = histogram[i];
W++;
N += histogram[i];
i++;
}
//Calculate peakiness
peakiness = (1 - (double)((vA + vB) / (2.0 * P))) * (1 - ((double)N / (double)(W * P)));
if (peakiness > peakinessThres & !valleys.Contains(j))
{
//peaks.Add(peak);
valleys.Add(j);
valleys.Add(i - 1);
}
j = i - 1;
}
}
catch (Exception)
{
if (l)
{
vB = histogram[255];
peakiness = (1 - (double)((vA + vB) / (2.0 * P))) * (1 - ((double)N / (double)(W * P)));
if (peakiness > peakinessThres)
valleys.Add(255);
//peaks.Add(255);
return valleys;
}
}
//if(!valleys.Contains(255))
// valleys.Add(255);
return valleys;
}