Develop Reference

Rounding-off from whole numbers to whole numbers in JavaScript?

So I have some numbers x = 320232 y = 2301 z = 12020305. I want to round these numbers off using JavaScript so that they become x = 320000 y = 2300 z = 12000000.
I tried Math.round and Math.floor but turns out that they only work with decimal values like
a = 3.1; Math.round(a); // Outputs 3 and not whole numbers.
So my question is can we round of whole numbers using JavaScript and If yes then how?
Edit: I want it to the round of to the starting 3 digit places as seen in the variables above. Like If there was another variable called c = 423841 It should round off to become c = 424000.

You could work with the logarithm of ten and adjust the digits.
const
format = n => v => {
if (!v) return 0;
const l = Math.floor(Math.log10(Math.abs(v))) - n + 1;
return Math.round(v / 10 ** l) * 10 ** l;
};
console.log([0, -9876, 320232, 2301, 12020305, 123456789].map(format(3)));

The solution is to first calculate how many numbers need to be rounded away, and then use that in a round.
Math.round(1234/100)*100 would round to 1200 so we can use this to round. We then only need to determan what to replace 100 with in this example.
That is that would be a 1 followed by LENGTH - 3 zeros. That number can be calculated as it is 10 to the power of LENGTH - 3, in JS: 10 ** (length - 3).
var x = 320232;
var y = 2301;
var z = 12020305;
function my_round(number){
var org_number = number;
// calculate integer number
var count = 0;
if (number >= 1) ++count;
while (number / 10 >= 1) {
number /= 10;
++count;
}
// length - 3
count = Math.round(count) - 3;
if (count < 0){
count = 0;
}
// 10 to the power of (length - 3)
var helper = 10 ** count;
return Math.round(org_number/helper)*helper;
}
alert(my_round(x));
alert(my_round(y));
alert(my_round(z));
It is not the prettiest code, though I tried to make it explainable code.

This should work:
function roundToNthPlace(input, n) {
let powerOfTen = 10 ** n
return Math.round(input/powerOfTen) * powerOfTen;
}
console.log([320232, 2301,12020305, 423841].map(input => roundToNthPlace(input, 3)));
Output: [320000, 2000, 12020000, 424000]

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.

Calculating Pi in JavaScript using Gregory-Leibniz Series

I have to calculate value of Pi using Gregory-Leibniz series:
pi = 4 * ((1/1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + ...)
I want to write a function in JavaScript that would take the number of digits that needs to be displayed as an argument. But I'm not sure if my way of thinking is fine here.
This is what I got so far:
function pi(n) {
var pi = 0;
for (i=1; i <= n; i+2) {
pi = 4 * ((1/i) + (1/(i+2)))
}
return pi;
}
How do I write the pi calculation so it calculates values till n?

You could use an increment of 4 and multiply at the end of the function with 4.
n is not the number of digits, but the counter of the value of the series.
function pi(n) {
var v = 0;
for (i = 1; i <= n; i += 4) { // increment by 4
v += 1 / i - 1 / (i + 2); // add the value of the series
}
return 4 * v; // apply the factor at last
}
console.log(pi(1000000000));

You may also do as follows; The function will iterate 10M times and will return you PI with n significant digits after the decimal point.
function getPI(n){
var i = 1,
p = 0;
while (i < 50000000){
p += 1/i - 1/(i+2);
i += 4;
}
return +(4*p).toFixed(n);
}
var myPI = getPI(10);
console.log("myPI #n:100M:", myPI);
console.log("Math.PI :", Math.PI);
console.log("The Diff :", Math.PI-myPI);

How to convert this Pascal Code to JavaScript?

I am trying to make simple JS code to find out the amount a number from 1 to 9 occurs in a given string. I have this Pascal code that works:
Var n,i:longint;
A:array[0..9] of byte;
Begin
write('Введите число: ');readln(n);
While n>0 do
Begin
A[n mod 10]:=A[n mod 10]+1;
n:=n div 10;
End;
For i:=0 to 9 do
writeln('The number ',i,' occurs ',A[i],' amount of times');
readln;
End.
In JS I ended up with this, but that seems to have a never-ending loop:
function plosh(form) {
var list = new Array(9);
var n = form.a.value;
while (n>0) {
a = n % 10;
list[a] = list[a]+1;
n = n % 10;
}
for (var i=0; i<=9; i++)
{
alert("Цифра"+i+"встречается"+A[i]+"раз");
}
}
Would appreicate any help on where I am going wrong with this. Thanks in advance!

n = n % 10 leaves n unchanged as soon as it's lower than 10, so it will usually never reach 0, hence the endless loop.
The div operator in Pascal makes an integral division.
Change
n = n % 10
to
n = Math.floor( n / 10 );
You also have another problem : you're not properly initializing your array so you're adding 1 to undefined. Fix that like this :
function plosh(form) {
var a,
list = [],
n = form.a.value;
while (n>0) {
a = n % 10;
list[a] = (list[a]||0)+1;
n = Math.floor( n / 10 );
}
for (var i=0; i<=9; i++) {
console.log("Цифра"+i+"встречается"+A[i]+"раз"); // <- less painful than alert
}
}

n:=n div 10;
was translated as:
n = n % 10;
but should be:
n = Math.floor(n / 10);
Edit: Also, you define an array [0..9] in Pascal, which means 10 elements. When you call Array(9) you only create 9 elements.

javascript - generate a new random number

I have a variable that has a number between 1-3.
I need to randomly generate a new number between 1-3 but it must not be the same as the last one.
It happens in a loop hundreds of times.
What is the most efficient way of doing this?

May the powers of modular arithmetic help you!!
This function does what you want using the modulo operator:
/**
* generate(1) will produce 2 or 3 with probablity .5
* generate(2) will produce 1 or 3 with probablity .5
* ... you get the idea.
*/
function generate(nb) {
rnd = Math.round(Math.random())
return 1 + (nb + rnd) % 3
}
if you want to avoid a function call, you can inline the code.

Here is a jsFiddle that solves your problem : http://jsfiddle.net/AsMWG/
I've created an array containing 1,2,3 and first I select any number and swap it with the last element. Then I only pick elements from position 0 and 1, and swap them with last element.

var x = 1; // or 2 or 3
// this generates a new x out of [1,2,3] which is != x
x = (Math.floor(2*Math.random())+x) % 3 + 1;

You can randomly generate numbers with the random number generator built in to javascript. You need to use Math.random().
If you're push()-ing into an array, you can always check if the previously inserted one is the same number, thus you regenerate the number. Here is an example:
var randomArr = [];
var count = 100;
var max = 3;
var min = 1;
while (randomArr.length < count) {
var r = Math.floor(Math.random() * (max - min) + min);
if (randomArr.length == 0) {
// start condition
randomArr.push(r);
} else if (randomArr[randomArr.length-1] !== r) {
// if the previous value is not the same
// then push that value into the array
randomArr.push(r);
}
}

As Widor commented generating such a number is equivalent to generating a number with probability 0.5. So you can try something like this (not tested):
var x; /* your starting number: 1,2 or 3 */
var y = Math.round(Math.random()); /* generates 0 or 1 */
var i = 0;
var res = i+1;
while (i < y) {
res = i+1;
i++;
if (i+1 == x) i++;
}

The code is tested and it does for what you are after.
var RandomNumber = {
lastSelected: 0,
generate: function() {
var random = Math.floor(Math.random()*3)+1;
if(random == this.lastSelected) {
generateNumber();
}
else {
this.lastSelected = random;
return random;
}
}
}
RandomNumber.generate();