Develop Reference

lagrange algorithm in Javascript

I wrote a javascript version of Lagrange algorithm, but it kept going wrong when I run it, I don't know what went wrong.
I use this to calculate time.
When I pass a cSeconds as a variable, sometimes it returns a minus value which is obviously wrong...
function LagrangeForCat(cSeconds){
var y = [2592000,7776000,15552000,31104000,93312000,155520000,279936000,404352000,528768000,622080000,715392000,870912000,995328000,1119744000,1244160000,1368576000,1492992000,1617408000,1741824000,1866240000,1990656000,2115072000,2239488000,2363904000,2488320000,2612736000,2737152000,2861568000,2985984000,3110400000,3234816000,3359232000,3483648000,3608064000];
var x = [604800,1209600,1814400,2592000,5184000,7776000,15552000,23328000,31104000,46656000,62208000,93312000,124416000,155520000,186624000,217728000,248832000,279936000,311040000,342144000,373248000,404352000,435456000,466560000,497664000,528768000,559872000,590976000,622080000,653184000,684288000,715392000,746496000,777600000];
var l = 0.0;
for (var j = 0; j < 34; j++) {
var s = 1.0;
for (var i = 0; i < 34; i++) {
if (i != j)
s = s * ((cSeconds - x[i]) / (x[j] - x[i]));
}
l = l + s * y[j];
}
var result = l / (24 * 60 * 60);
var Days = Math.floor(result);
//get float seconds data
var littleData = String(result).split(".")[1];
var floatData = parseFloat("0."+littleData);
var second = floatData *60*60*24;
var hours = Math.floor(second/(60*60));
var minutes = Math.floor(second % 3600/60);
var seconds = Math.floor(second % 3600) % 60;
var returnData = {days:Days,hours: hours + ':' + minutes + ':' + seconds}
return returnData;
}

I don't believe the issue is with your code but with the data set.
I tried a few things, for instance if you have cSeconds = one of the x values, then you get the correct result (I could check that it was equal to the matching y value).
I put all the data in open office and drew the graph it was like the square root function but more extreme (the 'straight' part look very straight) then I remembered that when you interpolate you usually get a polynomial that crosses the points you want but can be very wild outside between the point.
To test my theory I modified the algorithm to control at which x/y index to start and tried for all the values:
for (let i = 0; i < 35; ++i) {
LagrangeForCat(63119321, i, 34)
}
Together with a console.log inside LagrangeForCat it gives me the interpolated y value if I use all the x/y arrays (i=0), if I ignore the first x/y point (i=1), the first two (i=2), ...
00-34 -6850462776.278063
01-34 549996977.0003194
02-34 718950902.7592317
03-34 723883771.1443908
04-34 723161627.795225
05-34 721857113.1756063
06-34 721134873.0889213
07-34 720845478.4754647
08-34 720897871.7910147
09-34 721241470.2886044
10-34 722280314.1033486
11-34 750141284.0070543
12-34 750141262.289736
13-34 750141431.2562406
14-34 750141089.6980047
15-34 750141668.8768387
16-34 750142353.3267975
17-34 750141039.138794
18-34 750141836.251831
19-34 750138039.6240234
20-34 750141696.7529297
21-34 750141120.300293
22-34 750141960.4248047
23-34 750140874.0966797
24-34 750141337.5
25-34 750141237.4694824
26-34 750141289.2150879
27-34 750141282.5408936
28-34 750141284.2094421
29-34 750141283.987999
30-34 750141284.0002298
31-34 750141284.0000689
32-34 750141283.9999985
33-34 3608064000
34-34 0
Exclude 33-34 and 34-34 (there's just not enough data to interpolate).
For the example x=63119321 you'd expect y to be between 715392000 and 870912000 you can see that if you ignore the first 2-3 values the interpolation is "believable", if you ignore more values you interpolate based off the very straight part of the curve (see how consistent the interpolation is from 11-34 onward).
I use to work on a project where interpolation was needed, to avoid those pathological cases we opted for linear interpolation trading accuracy for security (and we could generate all the x/y points we wanted). In your case I'd try to use a smaller set, for instance only two values smaller than cSeconds and two greater like this:
function LagrangeForCat(cSeconds) {
var x = [...];
var y = [...];
let begin = 0,
end = 34
for (let i = 0; i < 34; ++i) {
if (cSeconds < x[i]) {
begin = (i < 3) ? 0 : i - 2
end = (i > (x.length - 1)) ? x.length : i + 1
break
}
}
let result = 0.0;
for (let i = begin; i < end; ++i) {
let term = y[i] / (24 * 60 * 60)
for (let j = begin; j < end; ++j) {
if (i != j)
term *= (cSeconds - x[j]) / (x[i] - x[j]);
}
result += term
}
var Days = Math.floor(result);
// I didn't change the rest of the function didn't even looked at it
}
If you find this answer useful please consider marking it as answered it'd be much appreciated.

JavaScript function "X - Y = Z" returns Y as Z value

I'm trying to apply a discount to a selection in JavaScript, but for some reason, my code is returning the total to subtract as the total price:
selectone = parseInt(selectone);
var textarea = document.getElementById('discount');
var word = '15off';
var textValue=textarea.value;
if (textValue.indexOf(word)!=-1)
{
var discval = parseFloat(selectone);
var num = parseInt(discval);
var retval = num - (num * .15);
} else {
var retval = 0
}
var total = selectone - retval;
document.getElementById("rettotal").innerHTML = "Price starts from £" + total;
}
For example, if something costs £100 and a 15% discount is applied, the total will be '£15' instead of '£100' ('retval' instead of 'total')
Is there something I've missed in this, or is something missing?
I've not done maths in JavaScript much so a bit over my head!
Many thanks

You've logic problem in math part.
You want to get amount after discount.
You're doing it:
var retval = num - (num * .15); // 100 - (100 * .15) = 85
But after You're removing discount from amount:
var total = selectone - retval; // 100 - 85 = 15
So here is the fix:
var price = parseFloat(selectone);
var discount = (textValue.indexOf('15off') != -1)?
price * .15
: 0;
var total = price - discount; // 100 - 15 = 85
or just be simple (if discount applies once):
var total = parseFloat(selectone);
if(textValue.indexOf('15off') != -1) {
total *= .85;
}
let's be flexible (applying multiple discounts to price):
var textValue = 'take this 15off and this 10off';
var price = parseFloat(1000);
var total = price;
total-= (textValue.indexOf('15off') != -1)?
price * .15
: 0;
console.log(total);
total-= (textValue.indexOf('10off') != -1)?
price * .15
: 0;
console.log(total);

Because... math.
selectone = parseInt(selectone);
...
var discval = parseFloat(selectone); // doesn't change the things, it's an int already
var num = parseInt(discval); // so num is essentially discval, which is selectone
var retval = num - (num * .15); // here you get 85% of num...
...
var total = selectone - retval; // here you get 15% back
The fix is to remove num - from retval, so as var retval = num * .15;
The code you've shown could be compressed to this:
var textarea = document.getElementById('discount');
var total = parseFloat(selectone)*(1-0.15*textarea.value.includes("15off"));
document.getElementById("rettotal").innerHTML = "Price starts from £" + total;
Or, if you have problems with includes() not being supported by your browser (in case it's IE), you could also use match():
var total = parseFloat(selectone)*(1-0.15*(textarea.value.match("15off")|0));

You have a JavaScript operator precedence and meaning problem there. That's syntax mistake on your part.
In an expression like this:
x - y = z
You are thinking that:
z = x - y //but it's not.
What you are really saying is:
y = z and x = x - z

Creating a slider between two numbers

So I've been working on re-producing the slider found here https://www.skylight.io/ ( Scroll down to find the price slider ).
So far Ive managed to create something similiar, but some numbers are hard coded, making it difficult to change and not very re-usable.
I've been researching around and I think I need to use Math.log() and Math.exp() together to achieve something like in the link above but I'm not sure.
Heres a jsfiddle of what I have so far https://jsfiddle.net/7wrvpb34/.
I feel that its the maths part of this problem that is halting me I think, so any help would be greatly appreciated.
Javascript code below:
var slider = document.getElementById("slider")
var sliderFill = document.getElementById("slider-fill")
var knob = document.getElementById("knob")
var mouseDown;
var mousePos = {x:0};
var knobPosition;
var minPrice = 20;
var price = 0;
var minRequests = 50;
var requests = 50 + ",000";
var incrementSpeed = 2;
var incrementModifier = 20;
var incrementValue = 1;
var minMillionCount = 1;
var millionCount = 1;
var previousRequestAmount = 0;
document.getElementById("price").innerHTML = price;
document.getElementById("requests").innerHTML = requests;
highlightTable(1);
document.addEventListener('mousemove', function(e) {
if(mouseDown) {
updateSlider(e);
}
})
function updateSlider(event) {
mousePos.x = event.clientX - slider.getBoundingClientRect().left;
mousePos.x -= knob.offsetWidth / 2;
console.log(mousePos.x);
if(mousePos.x < 0) {
knob.style.left = "0px";
sliderFill.style.width = "0px";
price = 0;
requests = 50 + ",000";
document.getElementById("price").innerHTML = price;
document.getElementById("requests").innerHTML = requests;
return
}
if(mousePos.x > slider.offsetWidth - 20) {
return
}
sliderFill.style.width = mousePos.x + 10 + "px";
knob.style.left = mousePos.x + "px";
//Increase requests by using X position of mouse
incrementSpeed = mousePos.x / incrementModifier;
requests = minRequests + (mousePos.x * incrementSpeed);
//Round to nearest 1
requests = Math.round(requests / incrementValue) * incrementValue;
if (requests >= 1000){
var m = requests/ 1000;
m = Math.round(m / 1) * 1;
//Problem, lower the modifier depending on requests
incrementModifier = 20 * 0.95;
document.getElementById("requests").innerHTML = m + " million";
//Adjust Prices
if(( requests >= 1000) && (requests < 10000)) {
var numOfMillions = requests / 100;
//Round to closest 10.
//10 * number of millions
var rounded = Math.round(numOfMillions / 10) * 10;
price = minPrice + rounded;
highlightTable(3);
}
//Adjust Prices
if(requests >= 10000) {
var numOfMillions = requests / 1000;
var rounded = Math.round(numOfMillions / 1) * 1;
var basePrice = minPrice * 6;
price = basePrice + rounded;
highlightTable(4);
}
} else {
incrementModifier = 20;
document.getElementById("requests").innerHTML = requests + ",000"
if(requests < 100) {
highlightTable(1);
price = 0;
} else {
highlightTable(2);
price = 20;
}
}
previousRequestAmount = requests;
document.getElementById("price").innerHTML = price;
}
knob.addEventListener('mousedown', function() {
mouseDown = true;
});
document.addEventListener('mouseup', function() {
mouseDown = false;
});
function highlightTable(rowNum) {
var table = document.getElementById("payment-table")
for(var i = 0; i < table.rows.length; ++i) {
var row = table.rows[i]
if(i == rowNum) {
row.style.background = "grey"
} else {
row.style.background = "white";
}
}
}
Thank you for your time.

If you want it to be reusable you need to create a mathematical function that assigns a result to the number of requests. I will give you a very easy example.
If you want a different result for 1,10,100,100,10000 etc
var d = Math.log10(requests);
if(d<1){
doSomething();
}else if(d<2){
doSomethingElse();
} //etc
This way if you want to change the specific values that create certain results, all you need to do is change the function.
This only works if your tiers of requests follow a math function, if they don't you need to hard code it.
However if say they don't follow a math function, but you know how you would like them to change based on a value then you can do this.
var changingValue = 4;
if(requests < 400*changingValue){
doSomthing();
}else if(requests <= 400*changingValue*changingValue){
doSomethingElse();
}else{// the requests is greater than any of the above
doTheOtherThing();
}
Edit:
For the second one you need to make sure that each condition if always larger than the other from top to bottom.

The description "increasingly increasing" matches an arbitrary number of functions. I assume you also want it to be continuous, since you already have a non-continuous solution.
TL;DR
Use an exponential function.
Generic approach
Assuming imin and imax are the minimal and maximal values of the slider (i for input) and omin and omax are the minimal and maximal values to be displayed, the simplest thing I can think of would be a multiplication by something based on the input value:
f(x)
{
return omin + (omax - omin) * g((x - imin) / (imax - imin));
}
This will pass 0 to g if x == imin and 1 if x == imax.
The return value r of g(y) should be
r == 0 for y == 0
r == 1 for y == 1
0 < r < y for 0 < y < 1
The simplest function that I can think of that fulfills this is an exponential function with exponent > 1.
An exponent of 1 would be a linear function.
An exponent of 2 would be make the middle of the slider display one fourth of the maximum price instead of half of it.
But you really need to find that exponent yourself, based on your needs.

is there a JavaScript implementation of the Inverse Error Function, akin to MATLAB erfinv()?

is there a JavaScript implementation of the Inverse Error Function?
This would implement the Gauss inverse error function. Approximations are ok.

Why yes. There is.
The following code uses built-in JavaScript functions and implments Abramowitz and Stegun's algorithm as described here:
function erfinv(x){
var z;
var a = 0.147;
var the_sign_of_x;
if(0==x) {
the_sign_of_x = 0;
} else if(x>0){
the_sign_of_x = 1;
} else {
the_sign_of_x = -1;
}
if(0 != x) {
var ln_1minus_x_sqrd = Math.log(1-x*x);
var ln_1minusxx_by_a = ln_1minus_x_sqrd / a;
var ln_1minusxx_by_2 = ln_1minus_x_sqrd / 2;
var ln_etc_by2_plus2 = ln_1minusxx_by_2 + (2/(Math.PI * a));
var first_sqrt = Math.sqrt((ln_etc_by2_plus2*ln_etc_by2_plus2)-ln_1minusxx_by_a);
var second_sqrt = Math.sqrt(first_sqrt - ln_etc_by2_plus2);
z = second_sqrt * the_sign_of_x;
} else { // x is zero
z = 0;
}
return z;
}

function provided earlier in this post did not work for me... NaN result on a 33meter circle with confidence 65% represented as 65.0 ... I wrote the following based on an equation listed here https://en.wikipedia.org/wiki/Error_function#Inverse_functions and it worked fine:
var _a = ((8*(Math.PI - 3)) / ((3*Math.PI)*(4 - Math.PI)));
function erfINV( inputX )
{
var _x = parseFloat(inputX);
var signX = ((_x < 0) ? -1.0 : 1.0 );
var oneMinusXsquared = 1.0 - (_x * _x);
var LNof1minusXsqrd = Math.log( oneMinusXsquared );
var PI_times_a = Math.PI * _a ;
var firstTerm = Math.pow(((2.0 / PI_times_a) + (LNof1minusXsqrd / 2.0)), 2);
var secondTerm = (LNof1minusXsqrd / _a);
var thirdTerm = ((2 / PI_times_a) + (LNof1minusXsqrd / 2.0));
var primaryComp = Math.sqrt( Math.sqrt( firstTerm - secondTerm ) - thirdTerm );
var scaled_R = signX * primaryComp ;
return scaled_R ;
}

Here's an alternative implementation of Abramowitz and Stegun's algorithm (equivalent to ptmalcolm's answer, but more succinct and twice as fast):
function erfinv(x) {
// maximum relative error = .00013
const a = 0.147
//if (0 == x) { return 0 }
const b = 2/(Math.PI * a) + Math.log(1-x**2)/2
const sqrt1 = Math.sqrt( b**2 - Math.log(1-x**2)/a )
const sqrt2 = Math.sqrt( sqrt1 - b )
return sqrt2 * Math.sign(x)
}
You can test the speed with console.time("erfinv"); for (let i=0; i<1000000000; i++) {erfinv(i/1000000000)}; console.timeEnd("erfinv")
The if statement optimization is commented out as it doesn't seem to make a difference - presumably the interpreter recognizes that this is all one equation.
If you need a more accurate approximation, check out Wikipedia.

Create Javascript Array With 2 Unique Numbers

I am trying to do following:
Create 2 unique numbers that are both within a certain range and they are at least n bigger/smaller.
In example:
Range is 0-600
Minimum "difference" is 150
So the generated numbers could be: [2,400],[120,310],[82,530]
But Not [900,400] or [200,220].
Thats what I have so far:
var posYArray = [];
for(i=0; i < 2; i++){
var posY = (Math.random() * 200).toFixed();
if(i < 1){
posYArray.push(posY);
}else{
for(i=0; i < posYArray.length; i++){
if(posY < posYArray[i]+100){
posYArray.push(posY);
}else{
//Restart loop??
}
}
}
}
But this randomly crashes the browser and also I didnt know a good way to restart the loop when the numbers are too close...

You could do this in two steps.
Generate your first random number.
Reduce the pool of random numbers which to the only possible valids.
Select your next random number in the reduced pool.
var upperBound = 200,
minDelta = 90,
firstRandom = Math.floor(Math.random() * upperBound);
var validPool = [];
for (var i = 0; i < upperBound; i++) {
if (i < firstRandom - minDelta || i > firstRandom + minDelta) {
validPool.push(i);
}
}
var secondRandom = validPool[Math.floor(Math.random() * validPool.length)];
jsFiddle.
It might be slower than randomly choosing and comparing, but at least it has a guaranteed running time :)

You can shift the gap in which not to choose a number, i.e. pick a random number x between 0 and range - gap, than pick first one between 0 and x and the second between x + gap and range. That would be somewhat more efficient.
var range = 600, gap = 150;
var x = Math.floor(Math.random() * (range - gap));
var posX = (Math.random() * (x)).toFixed();
var posY = (Math.random() * (range - x - gap) + x + gap).toFixed();
Works in O(1).

What about simply trying them? Very easy and when the difference is a lot smaller then the max size (example 150 and 600) you have a good pair in 1/2 of the possibilities.
For example:
var posY = (Math.random() * 600).toFixed();
var posX = (Math.random() * 600).toFixed();
while(abs(poX-posY) < 150){
posX = (Math.random() * 600).toFixed()
}
Not that efficient, but when you only have 2 numbers to generate that wont matter !