We Keep Coding

How to sum first n number of entries in an array?

I'm new to programming and learning about javascript recursion. It's my 5th day with JavaScript and following an online course.
The problem is that I need to sum up first n numbers of entries (first 3 numbers in this case) in an array.
But I'm ending up with sum of all and that also I'm not sure about.
var theArray = [1, 3, 8, 5, 7];
function sum(arr, n) {
if (n <= 0) {
return 1;
} else {
return arr[n - 1] + sum(arr, n - 2);
//assuming n is arr.length and n-1 is hence arr.length-1
}
}
console.log(sum(theArray, 3));
What am I doing wrong?
I checked most people are solving such with reduce method or with for of loop. I didn't learn those yet in the curriculum. But I know 'for loop' but that's good if the numbers are in incremental order I guess. Please explain this problem in my level.

When implementing something recursively, it's a good idea to think about the base condition. It does look like that's where you started, but your logic is flawed. n is the number of elements you would like to sum together. So let's say you want to sum the first n=3 elements of the array [2,3,4,5,6] and get 9. Here are the iterations.
return arr[n-1] + sum(arr, n-1) // arr[2] + sum(arr, 2) === 4 + sum(arr, 2) === 4 + 5
return arr[n-1] + sum(arr, n-1) // arr[1] + sum(arr, 1) === 3 + sum(arr, 2) === 3 + 2
return arr[n-1] // arr[0] === 2 (base case when n is 1)
I didn't solve the problem for you since this is obviously a school exercise, but the point is that the base case and the recursive call that you have are both slightly off.

Array of undetermined length, find the largest sum possible for any 3 consecutive numbers.
[2,0,1,100,200,10,7,2, 300,2,1 0] here 100,200,10 = 310 is the correct answer. Here only the combination of these 3 numbers produces the largest sum.
[2,1,0,20,71 = 0+20+7 = 27 is the answer that you should return.
const findMax = (numberArray) => {
let first = 0;
let second = 1;
let third = 2;
let max = 0;
for(; third < numberArray.length; first++, second++, third++) {
if(max < (numberArray[first] + numberArray[second] + numberArray[third])) {
max = numberArray[first] + numberArray[second] + numberArray[third];
}
}
return max;
}
console.log('Max sum of three consecutive numbers in array ===> ',findMax([2, 0, 100, 200, 10, 7, 2, 300, 2, 10]));

Project Euler Q#2 in "javascript"

the sum of even Fibonacci numbers below 4 mill : i am trying to do it using JavaScript, but i am getting infinity as an answer
but if i use small number such as 10, i am getting console.log() output with the result, is it possible to do that in JavaScript??
var fib = [1, 2];
for(var i =fib.length; i<4000000; i++)
{
fib[i] = fib[i-2] + fib[i-1];
}
//console.log(fib);
var arr_sum = 0;
for(var i = 0; i < fib.length; i++){
if(fib[i] % 2 === 0){
arr_sum += fib[i] ;
}
}
console.log(arr_sum);

So this is the problem:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
The correct answer is 4613732.
On the script below, the answer would be from variable "sum". I assigned it an initial value of 2 since that is the first even number on our sequence. I used 3 more variables to traverse through this sequence. The loop runs normal Fibonacci sequence and the if statement filters even numbers and adds it to the sum.
I think this is the simplest and most intuitive code implementation.
var sum = 2;
var x = 1, y = 2, fib;
while(x < 4000000) {
fib = x + y;
x = y;
y = fib;
if(fib%2===0) sum += fib;
}
console.log(sum);

Two things before I get to coding:
You need to sum the even fibonacci numbers which VALUE below 4 Million (#Alnitak)
There is an "even fibonacci series" which can be calculated in a different manner, see here.
Alright here we go:
let fib = [0, 2];
for(let i = 2; fib[i-1] < 4000000; i++) {
fib[i] = 4 * fib[i - 1] + fib[i - 2];
}
fib.pop()
let sum = fib.reduce((a, c) => a + c, 0);
console.log(sum);
Edit
Without the fib.pop() I just added, the last element of the array would be a number > 4000000. Now you should be able to get the right result.

Finding the nth item in a repeating list of fixed items

I have to determine the mathematical formula to calculate a particular repeating position in a series of numbers. The list of numbers repeats ad infinitum and I need to find the number every n numbers in this list. So I want to find the *n*th item in a list of repeating y numbers.
For example, if my list has 7 digits (y=7) and I need every 5th item (n=5), how do I find that item?
The list would be like this (which I've grouped in fives for ease of viewing):
12345 67123 45671 23456 71234 56712 34567
I need to find in the first grouping number 5, then in the second grouping number 3, then 1 from the third group, then 6, then 4, then 2, then 7.
This needs to work for any number for y and n. I usually use a modulus for finding *n*th items, but only when the list keeps increasing in number and not resetting.
I'm trying to do this in Javascript or JQuery as it's a browser based problem, but I'm not very mathematical so I'm struggling to solve it.
Thanks!
Edit: I'm looking for a mathematical solution to this ideally but I'll explain a little more about the problem, but it may just add confusion. I have a list of items in a carousel arrangement. In my example there are 7 unique items (it could be any number), but the list in real terms is actually five times that size (nothing to do with the groups of 5 above) with four sets of duplicates that I create.
To give the illusion of scrolling to infinity, the list position is reset on the 'last' page (there are two pages in this example as items 1-7 span across the 5 item wide viewport). Those groups above represent pages as there are 5 items per page in my example. The duplicates provide the padding necessary to fill in any blank spaces that may occur when moving to the next page of items (page 2 for instance starts with 6 and 7 but then would be empty if it weren't for the duplicated 1,2 and 3). When the page goes past the last page (so if we try to go to page 3) then I reposition them further back in the list to page one, but offset so it looks like they are still going forwards forever.
This is why I can't use an array index and why it would be useful to have a mathematical solution. I realise there are carousels out there that do similar tasks to what I'm trying to achieve, but I have to use the one I've got!

Just loop every 5 characters, like so:
var data = "12345671234567123456712345671234567";
var results = [];
for(var i = 4; i < data.length; i += 5){
results.push(data[i]);
}
//results = [5, 3, 1, 6, 4, 2, 7]
If you want to use a variable x = 5; then your for loop would look like this:
for(var i = x - 1; i < data.length; i += x){...
There is no need to know y

If your input sequence doesn't terminate, then outputting every nth item will eventually produce its own repeating sequence. The period (length) of this repetition will be the lowest common multiple of the period of the input sequence (y) and the step size used for outputting its items (x).
If you want to output only the first repetition, then something like this should do the trick (untested):
var sequence = "1234567";
var x = 5;
var y = sequence.length;
var count = lcm(x, y);
var offset = 4;
var output = [];
for (var i = 0; i < count; i += x)
{
j = (offset + i) % y;
output.push(sequence[j]);
}
You should be able to find an algorithm for computing the LCM of two integers fairly easily.

A purely mathematical definition? Err..
T(n) = T(n-1) + K For all n > 0.
T(1) = K // If user wants the first element in the series, you return the Kth element.
T(0) = 0 // If the user want's a non-existent element, they get 0.
Where K denotes the interval.
n denotes the desired term.
T() denotes the function that generates the list.
Lets assume we want every Kth element.
T(1) = T(0) + K = K
T(2) = T(1) + K = 2K
T(3) = T(2) + K = 3K
T(n) = nk. // This looks like a promising equation. Let's prove it:
So n is any n > 1. The next step in the equation is n+1, so we need to prove that
T(n + 1) = k(n + 1).
So let's have a go.
T(n+1) = T(N+1-1) + K.
T(n+1) = T(n) + K
Assume that T(n) = nk.
T(n+1) = nk + k
T(n+1) = k(n + 1).
And there is your proof, by induction, that T(n) = nk.
That is about as mathematical as you're gonna get on SO.
Nice simple recurrence relation that describes it quite well there.

After your edit I make another solution;)
var n = 5, y = 7;
for (var i = 1; i<=y; i++) {
var offset = ( i*y - (i-1)*n ) % y;
var result = 0;
if (offset === n) {
result = y;
} else {
result = (n - offset) > 0 ? n - offset : offset;
}
console.log(result);
}
[5, 3, 1, 6, 4, 2, 7] in output.
JSFIDDLE: http://jsfiddle.net/mcrLQ/4/

function get(x, A, B) {
var r = (x * A) % B;
return r ? r : B;
}
var A = 5;
var B = 7;
var C = [];
for (var x = 1; x <= B; ++x) {
C.push(get(x, A, B));
}
console.log(C);
Result: [5, 3, 1, 6, 4, 2, 7]
http://jsfiddle.net/xRFTD/

var data = "12345 67123 45671 23456 71234 56712 34567";
var x = 5;
var y = 7;
var results = [];
var i = x - 1; // enumeration in string starts from zero
while ( i <= data.length){
results.push(data[i]);
i = i + x + 1;// +1 for spaces ignoring
}

Why isn't my attempted solution working?

When I run the code in the console, the browser just stops working (am assuming stack overflow).
I've come up with several different algorithms for solving this problem, but I thought this one would not cause any SOs.
The problem:
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1 3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
Failing solution:
function divisors(n){
var counter = 0;
var triangle = 3;
var triangle_add = 2;
while (counter < n){
for (var i = 1; i = triangle; i++){
if (triangle % i === 0){
counter++;
}
};
if (counter < n){
triangle_add++;
triangle = triangle + triangle_add;
counter = 0;
};
};
return triangle;
};
console.log(divisors(501));

Your solution is not working because, most probably, it is very slow. This problem can be solved much faster by the following method:
Find all the prime numbers smaller than some N (put, for example, N=100'000) using Sieve of Eratosthenes. It is quite fast.
As we know from elementary math each number can be written in the form X=p1^i1*p2^i2*...*pn^in where pj is prime number and ij is the power of corresponding prime number. The number of divisors of X is equal to (i1+1)*(i2+1)*...*(in+1) since that many different ways we can form a number which will be divisor of X. Having an array of prime numbers the number of divisors for X can be calculated quite fast (the code still has place to be optimized):
int divisorCount(long long X)
{
int c = 1;
for (int i = 0; PRIMES[i] <= X; ++i)
{
int pr = PRIMES[i];
if (X % pr == 0)
{
int p = 1;
long long r = X;
while (r % pr == 0)
{
r = r / pr;
++p;
}
c *= p;
}
}
return c;
}
Iterate through all triangle numbers and count divisor numbers for them using the above function. The i-th triangle number is i * (i + 1) / 2, so no need to keep a variable, increment it and add it each time.

Generate unique random numbers between 1 and 100

How can I generate some unique random numbers between 1 and 100 using JavaScript?

For example: To generate 8 unique random numbers and store them to an array, you can simply do this:
var arr = [];
while(arr.length < 8){
var r = Math.floor(Math.random() * 100) + 1;
if(arr.indexOf(r) === -1) arr.push(r);
}
console.log(arr);

Populate an array with the numbers 1 through 100.
Shuffle it.
Take the first 8 elements of the resulting array.

Modern JS Solution using Set (and average case O(n))
const nums = new Set();
while(nums.size !== 8) {
nums.add(Math.floor(Math.random() * 100) + 1);
}
console.log([...nums]);

Another approach is to generate an 100 items array with ascending numbers and sort it randomly. This leads actually to a really short and (in my opinion) simple snippet.
const numbers = Array(100).fill().map((_, index) => index + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers.slice(0, 8));

Generate permutation of 100 numbers and then choose serially.
Use Knuth Shuffle(aka the Fisher-Yates shuffle) Algorithm.
JavaScript:
function fisherYates ( myArray,stop_count ) {
var i = myArray.length;
if ( i == 0 ) return false;
int c = 0;
while ( --i ) {
var j = Math.floor( Math.random() * ( i + 1 ) );
var tempi = myArray[i];
var tempj = myArray[j];
myArray[i] = tempj;
myArray[j] = tempi;
// Edited thanks to Frerich Raabe
c++;
if(c == stop_count)return;
}
}
CODE COPIED FROM LINK.
EDIT:
Improved code:
function fisherYates(myArray,nb_picks)
{
for (i = myArray.length-1; i > 1 ; i--)
{
var r = Math.floor(Math.random()*i);
var t = myArray[i];
myArray[i] = myArray[r];
myArray[r] = t;
}
return myArray.slice(0,nb_picks);
}
Potential problem:
Suppose we have array of 100 numbers {e.g. [1,2,3...100]} and we stop swapping after 8 swaps;
then most of the times array will look like {1,2,3,76,5,6,7,8,...numbers here will be shuffled ...10}.
Because every number will be swapped with probability 1/100 so
prob. of swapping first 8 numbers is 8/100 whereas prob. of swapping other 92 is 92/100.
But if we run algorithm for full array then we are sure (almost)every entry is swapped.
Otherwise we face a question : which 8 numbers to choose?

The above techniques are good if you want to avoid a library, but depending if you would be alright with a library, I would suggest checking out Chance for generating random stuff in JavaScript.
Specifically to solve your question, using Chance it's as easy as:
// One line!
var uniques = chance.unique(chance.natural, 8, {min: 1, max: 100});
// Print it out to the document for this snippet so we can see it in action
document.write(JSON.stringify(uniques));
<script src="http://chancejs.com/chance.min.js"></script>
Disclaimer, as the author of Chance, I am a bit biased ;)

To avoid any long and unreliable shuffles, I'd do the following...
Generate an array that contains the number between 1 and 100, in order.
Generate a random number between 1 and 100
Look up the number at this index in the array and store in your results
Remove the elemnt from the array, making it one shorter
Repeat from step 2, but use 99 as the upper limit of the random number
Repeat from step 2, but use 98 as the upper limit of the random number
Repeat from step 2, but use 97 as the upper limit of the random number
Repeat from step 2, but use 96 as the upper limit of the random number
Repeat from step 2, but use 95 as the upper limit of the random number
Repeat from step 2, but use 94 as the upper limit of the random number
Repeat from step 2, but use 93 as the upper limit of the random number
Voila - no repeated numbers.
I may post some actual code later, if anybody is interested.
Edit: It's probably the competitive streak in me but, having seen the post by #Alsciende, I couldn't resist posting the code that I promised.
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<title>8 unique random number between 1 and 100</title>
<script type="text/javascript" language="Javascript">
function pick(n, min, max){
var values = [], i = max;
while(i >= min) values.push(i--);
var results = [];
var maxIndex = max;
for(i=1; i <= n; i++){
maxIndex--;
var index = Math.floor(maxIndex * Math.random());
results.push(values[index]);
values[index] = values[maxIndex];
}
return results;
}
function go(){
var running = true;
do{
if(!confirm(pick(8, 1, 100).sort(function(a,b){return a - b;}))){
running = false;
}
}while(running)
}
</script>
</head>
<body>
<h1>8 unique random number between 1 and 100</h1>
<p><button onclick="go()">Click me</button> to start generating numbers.</p>
<p>When the numbers appear, click OK to generate another set, or Cancel to stop.</p>
</body>

I would do this:
function randomInt(min, max) {
return Math.round(min + Math.random()*(max-min));
}
var index = {}, numbers = [];
for (var i=0; i<8; ++i) {
var number;
do {
number = randomInt(1, 100);
} while (index.hasOwnProperty("_"+number));
index["_"+number] = true;
numbers.push(number);
}
delete index;

This is a very generic function I have written to generate random unique/non-unique integers for an array. Assume the last parameter to be true in this scenario for this answer.
/* Creates an array of random integers between the range specified
len = length of the array you want to generate
min = min value you require
max = max value you require
unique = whether you want unique or not (assume 'true' for this answer)
*/
function _arrayRandom(len, min, max, unique) {
var len = (len) ? len : 10,
min = (min !== undefined) ? min : 1,
max = (max !== undefined) ? max : 100,
unique = (unique) ? unique : false,
toReturn = [], tempObj = {}, i = 0;
if(unique === true) {
for(; i < len; i++) {
var randomInt = Math.floor(Math.random() * ((max - min) + min));
if(tempObj['key_'+ randomInt] === undefined) {
tempObj['key_'+ randomInt] = randomInt;
toReturn.push(randomInt);
} else {
i--;
}
}
} else {
for(; i < len; i++) {
toReturn.push(Math.floor(Math.random() * ((max - min) + min)));
}
}
return toReturn;
}
Here the 'tempObj' is a very useful obj since every random number generated will directly check in this tempObj if that key already exists, if not, then we reduce the i by one since we need 1 extra run since the current random number already exists.
In your case, run the following
_arrayRandom(8, 1, 100, true);
That's all.

Shuffling the numbers from 1 to 100 is the right basic strategy, but if you need only 8 shuffled numbers, there's no need to shuffle all 100 numbers.
I don't know Javascript very well, but I believe it's easy to create an array of 100 nulls quickly. Then, for 8 rounds, you swap the n'th element of the array (n starting at 0) with a randomly selected element from n+1 through 99. Of course, any elements not populated yet mean that the element would really have been the original index plus 1, so that's trivial to factor in. When you're done with the 8 rounds, the first 8 elements of your array will have your 8 shuffled numbers.

var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
if(arr.indexOf(randomnumber) === -1){arr.push(randomnumber)}
}
document.write(arr);
shorter than other answers I've seen

Implementing this as a generator makes it pretty nice to work with. Note, this implementation differs from ones that require the entire input array to be shuffled first.
This sample function works lazily, giving you 1 random item per iteration up to N items you ask for. This is nice because if you just want 3 items from a list of 1000, you don't have to touch all 1000 items first.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let ys = xs.slice(0);
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield ys.splice(i,1)[0];
n--; len--;
}
}
// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
// get 3 random items
for (let i of sample(3) (items))
console.log(i); // f g c
// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
console.log(i); // 3 8 7
// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]
I chose to implement sample in a way that does not mutate the input array, but you could easily argue that a mutating implementation is favourable.
For example, the shuffle function might wish to mutate the original input array. Or you might wish to sample from the same input at various times, updating the input each time.
// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
let len = xs.length;
while (n > 0 && len > 0) {
let i = (Math.random() * len) >> 0;
yield xs.splice(i,1)[0];
n--; len--;
}
}
// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));
// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');
// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))
console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]
// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]
sample is no longer a pure function because of the array input mutation, but in certain circumstances (demonstrated above) it might make more sense.
Another reason I chose a generator instead of a function that just returns an array is because you may want to continue sampling until some specific condition.
Perhaps I want the first prime number from a list of 1,000,000 random numbers.
"How many should I sample?" – you don't have to specify
"Do I have to find all the primes first and then select a random prime?" – Nope.
Because we're working with a generator, this task is trivial
const randomPrimeNumber = listOfNumbers => {
for (let x of sample(Infinity) (listOfNumbers)) {
if (isPrime(x))
return x;
}
return NaN;
}
This will continuously sample 1 random number at a time, x, check if it's prime, then return x if it is. If the list of numbers is exhausted before a prime is found, NaN is returned.
Note:
This answer was originally shared on another question that was closed as a duplicate of this one. Because it's very different from the other solutions provided here, I've decided to share it here as well

var numbers = [];
for (let i = 0; i < 8; i++) {
let a = true,
n;
while(a) {
n = Math.floor(Math.random() * 100) + 1;
a = numbers.includes(n);
}
numbers.push(n);
}
console.log(numbers);

Same permutation algorithm as The Machine Charmer, but with a prototyped implementation. Better suited to large number of picks. Uses js 1.7 destructuring assignment if available.
// swaps elements at index i and j in array this
// swapping is easy on js 1.7 (feature detection)
Array.prototype.swap = (function () {
var i=0, j=1;
try { [i,j]=[j,i]; }
catch (e) {}
if(i) {
return function(i,j) {
[this[i],this[j]] = [this[j],this[i]];
return this;
}
} else {
return function(i,j) {
var temp = this[i];
this[i] = this[j];
this[j] = temp;
return this;
}
}
})();
// shuffles array this
Array.prototype.shuffle = function() {
for(var i=this.length; i>1; i--) {
this.swap(i-1, Math.floor(i*Math.random()));
}
return this;
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.shuffle().slice(0,n);
}
pick(8,1,100);
Edit:
An other proposition, better suited to small number of picks, based on belugabob's answer. To guarantee uniqueness, we remove the picked numbers from the array.
// removes n random elements from array this
// and returns them
Array.prototype.pick = function(n) {
if(!n || !this.length) return [];
var i = Math.floor(this.length*Math.random());
return this.splice(i,1).concat(this.pick(n-1));
}
// returns n unique random numbers between min and max
function pick(n, min, max) {
var a = [], i = max;
while(i >= min) a.push(i--);
return a.pick(n);
}
pick(8,1,100);

for arrays with holes like this [,2,,4,,6,7,,]
because my problem was to fill these holes. So I modified it as per my need :)
the following modified solution worked for me :)
var arr = [,2,,4,,6,7,,]; //example
while(arr.length < 9){
var randomnumber=Math.floor(Math.random()*9+1);
var found=false;
for(var i=0;i<arr.length;i++){
if(arr[i]==randomnumber){found=true;break;}
}
if(!found)
for(k=0;k<9;k++)
{if(!arr[k]) //if it's empty !!MODIFICATION
{arr[k]=randomnumber; break;}}
}
alert(arr); //outputs on the screen

The best earlier answer is the answer by sje397. You will get as good random numbers as you can get, as quick as possible.
My solution is very similar to his solution. However, sometimes you want the random numbers in random order, and that is why I decided to post an answer. In addition, I provide a general function.
function selectKOutOfN(k, n) {
if (k>n) throw "k>n";
var selection = [];
var sorted = [];
for (var i = 0; i < k; i++) {
var rand = Math.floor(Math.random()*(n - i));
for (var j = 0; j < i; j++) {
if (sorted[j]<=rand)
rand++;
else
break;
}
selection.push(rand);
sorted.splice(j, 0, rand);
}
return selection;
}
alert(selectKOutOfN(8, 100));

Here is my ES6 version I cobbled together. I'm sure it can be a little more consolidated.
function randomArray(i, min, max) {
min = Math.ceil(min);
max = Math.floor(max);
let arr = Array.from({length: i}, () => Math.floor(Math.random()* (max - min)) + min);
return arr.sort();
}
let uniqueItems = [...new Set(randomArray(8, 0, 100))]
console.log(uniqueItems);

How about using object properties as a hash table? This way your best scenario is to only randomize 8 times. It would only be effective if you want a small part of the range of numbers. It's also much less memory intensive than Fisher-Yates because you don't have to allocate space for an array.
var ht={}, i=rands=8;
while ( i>0 || keys(ht).length<rands) ht[Math.ceil(Math.random()*100)]=i--;
alert(keys(ht));
I then found out that Object.keys(obj) is an ECMAScript 5 feature so the above is pretty much useless on the internets right now. Fear not, because I made it ECMAScript 3 compatible by adding a keys function like this.
if (typeof keys == "undefined")
{
var keys = function(obj)
{
props=[];
for (k in ht) if (ht.hasOwnProperty(k)) props.push(k);
return props;
}
}

var bombout=0;
var checkArr=[];
var arr=[];
while(arr.length < 8 && bombout<100){
bombout++;
var randomNumber=Math.ceil(Math.random()*100);
if(typeof checkArr[randomNumber] == "undefined"){
checkArr[randomNumber]=1;
arr.push(randomNumber);
}
}​
// untested - hence bombout

if you need more unique you must generate a array(1..100).
var arr=[];
function generateRandoms(){
for(var i=1;i<=100;i++) arr.push(i);
}
function extractUniqueRandom()
{
if (arr.length==0) generateRandoms();
var randIndex=Math.floor(arr.length*Math.random());
var result=arr[randIndex];
arr.splice(randIndex,1);
return result;
}
function extractUniqueRandomArray(n)
{
var resultArr=[];
for(var i=0;i<n;i++) resultArr.push(extractUniqueRandom());
return resultArr;
}
above code is faster:
extractUniqueRandomArray(50)=>
[2, 79, 38, 59, 63, 42, 52, 22, 78, 50, 39, 77, 1, 88, 40, 23, 48, 84, 91, 49, 4, 54, 93, 36, 100, 82, 62, 41, 89, 12, 24, 31, 86, 92, 64, 75, 70, 61, 67, 98, 76, 80, 56, 90, 83, 44, 43, 47, 7, 53]

Adding another better version of same code (accepted answer) with JavaScript 1.6 indexOf function. Do not need to loop thru whole array every time you are checking the duplicate.
var arr = []
while(arr.length < 8){
var randomnumber=Math.ceil(Math.random()*100)
var found=false;
if(arr.indexOf(randomnumber) > -1){found=true;}
if(!found)arr[arr.length]=randomnumber;
}
Older version of Javascript can still use the version at top
PS: Tried suggesting an update to the wiki but it was rejected. I still think it may be useful for others.

This is my personal solution :
<script>
var i, k;
var numbers = new Array();
k = Math.floor((Math.random()*8));
numbers[0]=k;
for (var j=1;j<8;j++){
k = Math.floor((Math.random()*8));
i=0;
while (i < numbers.length){
if (numbers[i] == k){
k = Math.floor((Math.random()*8));
i=0;
}else {i++;}
}
numbers[j]=k;
}
for (var j=0;j<8;j++){
alert (numbers[j]);
}
</script>
It randomly generates 8 unique array values (between 0 and 7), then displays them using an alert box.

function getUniqueRandomNos() {
var indexedArrayOfRandomNo = [];
for (var i = 0; i < 100; i++) {
var randNo = Math.random();
indexedArrayOfRandomNo.push([i, randNo]);
}
indexedArrayOfRandomNo.sort(function (arr1, arr2) {
return arr1[1] - arr2[1]
});
var uniqueRandNoArray = [];
for (i = 0; i < 8; i++) {
uniqueRandNoArray.push(indexedArrayOfRandomNo[i][0]);
}
return uniqueRandNoArray;
}
I think this method is different from methods given in most of the answers, so I thought I might add an answer here (though the question was asked 4 years ago).
We generate 100 random numbers, and tag each of them with numbers from 1 to 100. Then we sort these tagged random numbers, and the tags get shuffled randomly. Alternatively, as needed in this question, one could do away with just finding top 8 of the tagged random numbers. Finding top 8 items is cheaper than sorting the whole array.
One must note here, that the sorting algorithm influences this algorithm. If the sorting algorithm used is stable, there is slight bias in favor of smaller numbers. Ideally, we would want the sorting algorithm to be unstable and not even biased towards stability (or instability) to produce an answer with perfectly uniform probability distribution.

This can handle generating upto 20 digit UNIQUE random number
JS
var generatedNumbers = [];
function generateRandomNumber(precision) { // input --> number precision in integer
if (precision <= 20) {
var randomNum = Math.round(Math.random().toFixed(precision) * Math.pow(10, precision));
if (generatedNumbers.indexOf(randomNum) > -1) {
if (generatedNumbers.length == Math.pow(10, precision))
return "Generated all values with this precision";
return generateRandomNumber(precision);
} else {
generatedNumbers.push(randomNum);
return randomNum;
}
} else
return "Number Precision shoould not exceed 20";
}
generateRandomNumber(1);
jsFiddle

This solution uses the hash which is much more performant O(1) than checking if the resides in the array. It has extra safe checks too. Hope it helps.
function uniqueArray(minRange, maxRange, arrayLength) {
var arrayLength = (arrayLength) ? arrayLength : 10
var minRange = (minRange !== undefined) ? minRange : 1
var maxRange = (maxRange !== undefined) ? maxRange : 100
var numberOfItemsInArray = 0
var hash = {}
var array = []
if ( arrayLength > (maxRange - minRange) ) throw new Error('Cannot generate unique array: Array length too high')
while(numberOfItemsInArray < arrayLength){
// var randomNumber = Math.floor(Math.random() * (maxRange - minRange + 1) + minRange)
// following line used for performance benefits
var randomNumber = (Math.random() * (maxRange - minRange + 1) + minRange) << 0
if (!hash[randomNumber]) {
hash[randomNumber] = true
array.push(randomNumber)
numberOfItemsInArray++
}
}
return array
}
document.write(uniqueArray(1, 100, 8))

You can also do it with a one liner like this:
[...((add, set) => add(set, add))((set, add) => set.size < 8 ? add(set.add(Math.floor(Math.random()*100) + 1), add) : set, new Set())]

getRandom (min, max) {
return Math.floor(Math.random() * (max - min)) + min
}
getNRandom (min, max, n) {
const numbers = []
if (min > max) {
return new Error('Max is gt min')
}
if (min === max) {
return [min]
}
if ((max - min) >= n) {
while (numbers.length < n) {
let rand = this.getRandom(min, max + 1)
if (numbers.indexOf(rand) === -1) {
numbers.push(rand)
}
}
}
if ((max - min) < n) {
for (let i = min; i <= max; i++) {
numbers.push(i)
}
}
return numbers
}

Using a Set is your fastest option. Here is a generic function for getting a unique random that uses a callback generator. Now it's fast and reusable.
// Get a unique 'anything'
let unique = new Set()
function getUnique(generator) {
let number = generator()
while (!unique.add(number)) {
number = generator()
}
return number;
}
// The generator. Return anything, not just numbers.
const between_1_100 = () => 1 + Math.floor(Math.random() * 100)
// Test it
for (var i = 0; i < 8; i++) {
const aNumber = getUnique(between_1_100)
}
// Dump the 'stored numbers'
console.log(Array.from(unique))

This is a implementation of Fisher Yates/Durstenfeld Shuffle, but without actual creation of a array thus reducing space complexity or memory needed, when the pick size is small compared to the number of elements available.
To pick 8 numbers from 100, it is not necessary to create a array of 100 elements.
Assuming a array is created,
From the end of array(100), get random number(rnd) from 1 to 100
Swap 100 and the random number rnd
Repeat step 1 with array(99)
If a array is not created, A hashMap may be used to remember the actual swapped positions. When the second random number generated is equal to the one of the previously generated numbers, the map provides the current value in that position rather than the actual value.
const getRandom_ = (start, end) => {
return Math.floor(Math.random() * (end - start + 1)) + start;
};
const getRealValue_ = (map, rnd) => {
if (map.has(rnd)) {
return getRealValue_(map, map.get(rnd));
} else {
return rnd;
}
};
const getRandomNumbers = (n, start, end) => {
const out = new Map();
while (n--) {
const rnd = getRandom_(start, end--);
out.set(getRealValue_(out, rnd), end + 1);
}
return [...out.keys()];
};
console.info(getRandomNumbers(8, 1, 100));
console.info(getRandomNumbers(8, 1, Math.pow(10, 12)));
console.info(getRandomNumbers(800000, 1, Math.pow(10, 15)));

Here is an example of random 5 numbers taken from a range of 0 to 100 (both 0 and 100 included) with no duplication.
let finals = [];
const count = 5; // Considering 5 numbers
const max = 100;
for(let i = 0; i < max; i++){
const rand = Math.round(Math.random() * max);
!finals.includes(rand) && finals.push(rand)
}
finals = finals.slice(0, count)