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]

Am I using the proper predicate for my lambda?

I've been working on this program for a few hours, and I finally got it to output - NaN. I dont know how this could be, I'm pushing a product of real numbers into the array... Somebody help! What did I miss? The problem is to find the largest product produced by 13 adjacent digits within the 1000 digit number assigned to _1000digits.
// what is the largest product of 13 adjacent digits within this 1000 digit number
function largestProduct() {
_1000digits = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450;
separateDigits = _1000digits.toString().split("");
products = [];
var a = 0;
var b = 1;
var c = 2;
var d = 3;
var e = 4;
var f = 5;
var g = 6;
var h = 7;
var i = 8;
var j = 9;
var k = 10;
var l = 11;
var m = 12;
while (m <= 999) {
products.push(
separateDigits[a] *
separateDigits[b] *
separateDigits[c] *
separateDigits[d] *
separateDigits[e] *
separateDigits[f] *
separateDigits[g] *
separateDigits[h] *
separateDigits[i] *
separateDigits[j] *
separateDigits[k] *
separateDigits[l] *
separateDigits[m]
);
a++;
b++;
c++;
d++;
e++;
f++;
g++;
h++;
i++;
j++;
k++;
l++;
m++;
}
products.sort((a, b) => a - b);
console.log(products.pop());
}
largestProduct();

Short:
To work with such huge numbers you'll want to use a special data structure, like BigInt.
Long:
There are a few issues with your code, the first one is trying to store such a huge number in a variable without any treatment. A JavaScript number can only store values up to 25^3 - 1, your number is a lot bigger than that.
If you run:
_1000digits = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
console.log(_1000digits)
You'll see the output is "Infinity" because that's such a huge number JavaScript doesn't know how to store it entirely.
You're also not checking if the numbers you're accessing actually exist, so if you put a smaller number in _1000digits you'll end up multiplying by undefined, which will result in NaN:
_1000digits = 700
separateDigits = _1000digits.toString().split("")
var f = 5
console.log(separateDigits[f])

Reducing a number on each iteration until it reaches exactly 0

I want to reduce n to 0 on the last iteration but my calculations are wrong. Im overlooking something. Right now im doing Math.round(n / loops). I need to find a way to disperse the substraction in such a way so that in the last iteration n becomes 0 while the range between each n´s is constant.
function check(a, b) {
let loops = a / b;
let n = 200;
for (let i = 0; i <= loops; i++) {
console.log(n);
n -= Math.round(200 / loops);
}
}
check(60, 10)
Edit: Im sorry but these downvotes are ridiculous.

The main problem with your code is that the expression n / loop is different in each iteration. You might have intended that this expression was constant, in which case the logic would have been more reasonable.
Use a separate variable that starts with the value of n and then is the subject of the subtractions, but without changing the original value of n that is used in the expression n / loop:
function check(a, b) {
let loops = a / b;
let n = 200; // don't change n after this.
let n2 = n; // use a separate variable for that.
for (let i = 0; i <= loops; i++) {
console.log(Math.round(n2)); // only round in output
n2 -= n / loops; // use n, but only change n2
}
}
check(60,10);

Generate all combinations for pair of bits set to 1?

I'm trying to generate all possible combinations for pair of 1's within given bit width.
Let's say the bit width is 6, i.e. number 32. This is what I would like to generate:
000000
000011
000110
001100
001111
011000
011011
011110
110000
110011
110110
111100
111111
If I have variables:
var a = 1,
b = 2;
num = a | b;
and create a loop that I'll loop over width - 1 times, and where I shift both a << 1 and b << 1, I'll get all combinations for one pair. After that, I'm pretty much stuck.
Could someone , please, provide some help.
Update: working example
Based on Barmar's mathematical approach, this is what I managed to implement
var arr = [],
arrBits = [];
function getCombs(pairs, startIdx) {
var i, j, val = 0, tmpVal, idx;
if (startIdx + 2 < pairs) {
startIdx = arr.length - 1;
pairs -= 1;
}
if (pairs < 2) {
return;
}
for (i = 0; i < pairs-1; i++) {
idx = startIdx - (i * 2);
val += arr[idx];
}
for (j = 0; j < idx - 1; j++) {
arrBits.push((val + arr[j]).toString(2));
}
getCombs(pairs, startIdx-1);
}
(function initArr(bits) {
var i, val, pairs, startIdx;
for (i = 1; i < bits; i++) {
val = i == 1 ? 3 : val * 2;
arr.push(val);
arrBits.push(val.toString(2));
}
pairs = Math.floor(bits / 2);
startIdx = arr.length - 1;
getCombs(pairs, startIdx);
console.log(arrBits);
}(9));
Working example on JSFiddle
http://jsfiddle.net/zywc5/

The numbers with exactly one pair of 1's are the sequence 3, 6, 12, 24, 48, ...; they start with 3 and just double each time.
The numbers with two pairs of 1's are 12+3, 24+3, 24+6, 48+3, 48+6, 48+12, ...; these are the above sequence starting at 12 + the original sequence up to n/4.
The numbers with three pairs of 1's are 48+12+3, 96+12+3, 96+24+3, 96+24+6, ...
The relationship between each of these suggests a recursive algorithm making use of the original doubling sequence. I don't have time right now to write it, but I think this should get you going.

if the bit width isn't that big then you'll be way better off creating bit representations for all numbers from 0 to 31 in a loop and simply ignore the ones that have an odd number of "ones" in the bit representation.

Maybe start counting normally in binary and replace all 1's with 11's like this:
n = 5
n = n.toString(2) //= "101"
n = n.replace(/1/g, "11") //= "11011"
n = parseInt(n, 2) //= 27
So you'll get:
0 -> 0
1 -> 11
10 -> 110
11 -> 1111
100 -> 1100
101 -> 11011
110 -> 11110
111 -> 111111
And so on. You'll have to count up to 31 or so on the left side, and reject ones longer than 6 bits on the right side.

See http://jsfiddle.net/SBH6R/
var len=6,
arr=[''];
for(var i=0;i<len;i++){
for(var j=0;j<arr.length;j++){
var k=j;
if(getNum1(arr[j])%2===1){
arr[j]+=1;
}else{
if(i<len-1){
arr.splice(j+1,0,arr[j]+1);
j++;
}
arr[k]+=0;
}
}
}
function getNum1(str){
var n=0;
for(var i=str.length-1;i>=0;i--){
if(str.substr(i,1)==='1'){n++;}
else{break;}
}
return n;
}
document.write(arr.join('<br />'));
Or maybe you will prefer http://jsfiddle.net/SBH6R/1/. It's simpler, but then you will have to sort() the array:
var len=6,
arr=[''];
for(var i=0;i<len;i++){
for(var k=0,l=arr.length;k<l;k++){
if(getNum1(arr[k])%2===1){
arr[k]+=1;
}else{
if(i<len-1){
arr.push(arr[k]+1);
}
arr[k]+=0;
}
}
}
function getNum1(str){
var n=0;
for(var i=str.length-1;i>=0;i--){
if(str.substr(i,1)==='1'){n++;}
else{break;}
}
return n;
}
document.write(arr.sort().join('<br />'));
See http://jsperf.com/generate-all-combinations-for-pair-of-bits-set-to-1 if you want to compare the performance. It seems that the fastest code is the first one on Chrome but the second one on Firefox.

You can also do this with bit twiddling. If the lowest two bits are zero, we need to set them, which is equivalent to adding 3. Otherwise, we need to replace the lowest block of ones by its top bit and a 1-bit to the left of it. This can be done as follows, where x is the current combination:
x3 = x + 3;
return (((x ^ x3) - 2) >> 2) + x3;

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();