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Get every possible combination of numbers (fastest possible method)

I am trying to get every single combination of elements into an array. Now I can use the method below, and remove the duplicates, but this way is far to slow for my use.
The code below would find every possible combination for 2 digits below 4. Now in the code I actually want to use this for, the least possible code would be 6 for loops (within each other) with the amount being 18 (rememeber this is the minimum).
The code below would execute amount^[amount of for loops], or amount^2 which in this case is 16. That means that in the code I want to use this for, it executes 18^6 times, or 34 million times. And this is the minimum, which would get much higher.
After trying to run my code (with 6 foor loops in which amount = 18), it crashed my browser... My question is: Is there any faster and more efficient (not elegant. I don't care how elegant it is) in which my browser won't crash?
Note: This question is not a duplicate question. All the other questions simply ask for a way to do this, however I already have a way. I am just trying to make it more efficient and faster so that it actually works correctly.
let combinations = [];
let amount = 4;
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
combinations.push(`${a}${b}`);
}
}
console.log(combinations);
Below is a snippet providing a possible example for how my code would work.
let possibilities = [];
let amount = 6; //Amount is set by me, so don't worry about it being incorrect
for (let a = 0; a < amount; a++) {
for (let b = 0; b < amount; b++) {
possibilities.push(a + b);
}
}
possibilities = [...new Set(possibilities)]; //Removes duplicates
possibilities.sort((a, b) => b - a); //Sorts in descending order
possibilities = possibilities.slice(0, 3); //Gets top 3 values
console.log(possibilities);

Ok, as discussed in the comments, if you need top 3 values for a particular amount, you could just do something simple like below:
let amount = 6;
let highest = amount - 1,second_highest = amount - 2,third_highest = amount - 3;
let possibilities = [
highest + highest,
highest + second_highest,
highest + third_highest
];
console.log(possibilities);

I don't know the any better solution for this, but yes there are some conditions you need to check first.
If(amount <= 0) return 'Invalid amount, Please enter a valid amount"
So if somebody enters a negative or zero value your loop will goes into infinite loop, and make the situation more worst.
if(amount === 1) return '1 possible combination'
As amount less than 1 is 0 only and combinations for 0 is 1 only, you need not to parse whole loop for 6 digits or n digits for 0 so it will get solve in complexity of 1 instead of N(no. of digits).
And for amount greater then 1 you can create manual loops, like here you created 2 loops for 2 digits, you create 6 loops for 6 digits, better create dynamic logic for this to create number of loops automatically.
You need to consider 1111, 1112 this type of combinations as well right?
Or if only 1234, 2134, 2314 this kind of scenarios are required? This can be done in very less complexity.
For duplication you can store combinations as a key value pair. and then Object.Keys will be your combinations.

Determining Powers of 2?

I am creating a simple bracket system and I need a way to check if there are a correct number of teams, OR if my program needs to compensate for bye rounds.
Right now, I am checking for "powers of two" with this function:
function validBracket(data) {
var x = data.teams.length;
return ((x != 0) && !(x & (x - 1)));
}
This works pretty well, but I am needing to know how many Bye rounds to add.
For instance, if I had 16 teams, I would not need to add anymore teams. However, if I had 12 teams, I would need the first 4 teams to get a bye round.
How can I calculate number of bye rounds to add to my bracket? And would hard-coding an array of powers of two be better?
In pseudo code, something like this is what i was thinking of:
if(validateBracket(data)) {
// Valid number of teams (power of two). Keep going.
} else {
var byeRounds = calculateByeRounds();
}
NOTE: I would rather not use an array of powers of two like below:
var powersOfTwo = [2,4,8,16,32,...];
The reasoning behind this is that I would be limiting the number of teams that could be put in the system (however, I don't think a person would have over 256 teams).

var needed = (1 << Math.ceil(Math.log2(n))) - n;
More generalized solution for extreme cases:
var needed = Math.pow(2, Math.ceil(Math.log2(n))) - n;

How to compare a number string and it's reverse to find the next highest palindrome?

I am trying to build a function that takes a price with 2 decimal points and finds the next highest palindrome. I know there are several other ways to approach this, but I am curious why my method is not working. I am new to JS so this might be a simple one. Any advice would be great.
I broke it into smaller chunks with explanations of what I want it to do below:
var ask = prompt("Enter a price");
var reverseIt = function (x) {
x = (parseFloat(x) * 100).toString();
for (var i = (x.length - 1); i >= 0; i--) {
x.substr(i, 1);
}
return
};
The reverseIt function takes an argument removes the decimal (* 100) and reverses the number.
var removeDec = function (j) {
return (parseFloat(j) * 100).toString();
}
The removeDec function takes an argument, removes the decimal point (* 100), and converts it back to a string. Is this redundant for comparing two "number" strings? Should I use the Number() and String() functions instead?
var findDrome = function (i) {
for (var i; removeDec(i) != reverseIt(i); i += (1 / 100)) {
if ((removeDec(i) + 1).toString() == reverseIt(i)) {
document.write(i + (1 / 100));
}
} return
};
findDrome(ask);
The findDrome function takes the ask prompt at the start as an argument. If the number without a decimal doesn't match the reverse without a decimal, then increments it by 0.01. Right before the loop ends, I wanted it to check if the number prior +1 (since it is * 100) is equal to the reverse and if so write the next number.
It wasn't working, so I tried adding parseFloat and toString to specify stricter/more explicit conversions. I also used the loose equality operators, but it's still not working.
My questions: Is this a conversion or syntactical problem or can you not compare the output of 2 functions? Should I instead compare 2 variables and if so how do I assign the for loop in the reverseIt function to a variable?

Your program has a number of issues. First, your reverseIt function never returns a reversed value. The variable x is passed in but it's never updated in the for loop - x.substr() creates a new string instance but it's never assigned back to x so its value never changes. As it is, your for loop in findDrome goes infinite since reverseIt returns undefined.
Another - possible - problem is that you're incrementing a floating-point number by 1/100 but floating point values have no exact representation. I don't know if this is actually affecting your code (since it currently never returns a proper value) but it's something you may have to worry about. This would likely affect parseFloat (which may return a slighly different floating-point value than the string it parses).
Using toFixed() would truncate the number to 2 decimal digits. You could then turn the number to a string and remove the decimal dot character, rather than converting the number back and forth between string and number.
You may want to read up on floating-point arithmetic (if you're not already familiar with it).
As a last comment, you should never, ever rely on Javascript terminating your statements - you should always use ; to terminate a statement like in other proper C-style languages. Leaving out ;-s (even if Javascript lets you get away with it) is considered very poor practice.

I figured it out thanks to the help above! Here is how the fixed program works:
var ask = prompt("Enter a price to find the next palindromic price");
var removeDec = function (j) {
return parseInt(j * 100);
};
var c = removeDec(ask);
This prompts a price and multiplies it by 100 to remove the decimal point and avoid floating point arithmetic. The parseInt removes any decimals smaller than the hundredths place.
var reverseIt = function (x) {
var a = ""
x = x.toString();
for (var i = (x.length - 1); i >= 0; i--) {
a = (a + String(x.substr(i, 1)));
}
return Number(a);
};
var b = reverseIt(c);
The reverseIt function takes an argument, converts it to string and adds each character in reverse to an empty string (a=""). Var a is then returned as a number. The empty string is important for storing the reverse number and is a big reason why my code wasn't working before.
var e = Math.pow(10, (String(c).length - 1));
Var e was added to take into account varying place values to left side of the decimal. Later this helps check if a number is equal to its reverse by adding a 1 to both sides of the number. Var e counts the length of var c (entered value with decimal removed) and finds the appropriate power of 10 to add later. So if you entered 14.40 * 100 then later it will check if 1440 + 1 is equal to 0441 + 1000.. or 10^3. This test is important later in order to exit the loop. This is where my code was failing before because I didn't take adding a number to the reverse into account and I was trying to add decimals which aren't as predictable.
if (c == b) {
document.write("$" + (c / 100) + "... This price is already palindrome!")
} else {
for (c; c !== b; c++) {
b = reverseIt(c);
if ((c + 1) == (b + e)) {
document.write("The next palindromic price is $" + ((Number(c) + 1) / 100));
break;
}
}
}
Here, If the original number and it's reverse are not equal then a loop begins that increments the value by 1 until the entered number + 1 is equal to the reversed number + e. So effectively the loop finds the number right before the loop ends, writes it and then breaks out of the loop. This palindrome finder seems to work smoothly with values big and small, no matter where you put the decimal point. Glad I got it working... it was a great learning experience figuring it out!

Creating a random number generator in jscript and prevent duplicates

We are trying to create a random number generator to create serial numbers for products on a virtual assembly line.
We got the random numbers to generate, however since they are serial numbers we don't want it to create duplicates.
Is there a way that it can go back and check to see if the number generated has already been generated, and then to tell it that if it is a duplicate to generate a new number, and to repeat this process until it has a "unique" number.

The point of a serial number is that they're NOT random. Serial, by definition, means that something is arranged in a series. Why not just use an incrementing number?

The easiest way to fix this problem is to avoid it. Use something that is monotonically increasing (like time) to form part of your serial number. To that you can prepend some fixed value that identifies the line or something.
So your serial number format could be NNNNYYYYMMDDHHMMSS, where NNNN is a 4-digit line number and YYYY is the 4 digit year, MM is a 2 digit month, ...
If you can produce multiple things per second per line, then add date components until you get to the point where only one per unit time is possible -- or simply add the count of items produced this day to the YYYYMMDD component (e.g., NNNNYYYYMMDDCCCCCC).

With a truly random number you would have to store the entire collection and review it for each number. Obviously this would mean that your generation would become slower and slower the larger the number of keys you generate (since it would have to retry more and more often and compare to a larger dataset).
This is entirely why truly random numbers just are never used for this purpose. For serial numbers the standard is always to just do a sequential number - is there any real real for them to be random?
Unique IDs are NEVER random - GUIDs and the like are based on the system time and (most often) MAC address. They're globally unique because of the algorithm used and the machine specifics - not because of the size of the value or any level of randomness.
Personally I would do everything I could to either use a sequential value (perhaps with a unique prefix if you have multiple channels) or, better, use a real GUID for your purpose.

is this what you are looking for?
var rArray;
function fillArray (range)
{
rArray = new Array ();
for(var x = 0; x < range; x++)
rArray [x] = x;
}
function randomND (range)
{
if (rArray == null || rArray.length < 1)
fillArray (range);
var pos = Math.floor(Math.random()*rArray.length);
var ran = rArray [pos];
for(var x = pos; x < rArray.length; x++)
rArray [x] = rArray [x+1];
var tempArray = new Array (rArray.length-1)
for(var x = 0; x < tempArray.length; x++)
tempArray [x] = rArray [x];
rArray = tempArray;
return ran;
}

How is randomness achieved with Math.random in javascript?

How is randomness achieved with Math.random in javascript? I've made something that picks between around 50 different options randomly. I'm wondering how comfortable I should be with using Math.random to get my randomness.

From the specifications:
random():
Returns a Number value with positive
sign, greater than or equal to 0 but
less than 1, chosen randomly or pseudo
randomly with approximately uniform
distribution over that range, using an
implementation-dependent algorithm or
strategy. This function takes no
arguments.
So the answer is that it depends on what JavaScript engine you're using.
I'm not sure if all browsers use the same strategy or what that strategy is unfortunately
It should be fine for your purposes. Only if you're doing a large amount of numbers would you begin to see a pattern

Using Math.random() is fine if you're not centrally pooling & using the results, i.e. for OAuth.
For example, our site used Math.random() to generate random "nonce" strings for use with OAuth. The original JavaScript library did this by choosing a character from a predetermined list using Math.random(): i.e.
for (var i = 0; i < length; ++i) {
var rnum = Math.floor(Math.random() * chars.length);
result += chars.substring(rnum, rnum+1);
}
The problem is, users were getting duplicate nonce strings (even using a 10 character length - theoretically ~10^18 combinations), usually within a few seconds of each other. My guess this is due to Math.random() seeding from the timestamp, as one of the other posters mentioned.

The exact implementation can of course differ somewhat depending on the browser, but they all use some kind of pseudo random number generator. Although it's not really random, it's certainly good enough for all general purposes.
You should only be worried about the randomness if you are using it for something that needs exceptionally good randomness, like encryption or simulating a game of chance in play for money, but then you would hardly use Javascript anyway.

It's 100% random enough for your purposes. It's seeded by time, so every time you run it, you'll get different results.
Paste this into your browsers address bar...
javascript:alert(Math.random() * 2 > 1);
and press [Enter] a few times... I got "true, false, false, true" - random enough :)

This is a little overkill...but, I couldn't resist doing this :)
You can execute this in your browser address bar. It generates a random number between 0 and 4, 100000 times. And outputs the number of times each number was generated and the number of times one random number followed the other.
I executed this in Firefox 3.5.2. All the numbers seem to be about equal - indicating no bias, and no obvious pattern in the way the numbers are generated.
javascript:
var max = 5;
var transitions = new Array(max);
var frequency = new Array(max);
for (var i = 0; i < max; i++)
{
transitions[i] = new Array(max);
}
var old = 0, curr = 0;
for (var i = 0; i < 100000; i++)
{
curr = Math.floor(Math.random()*max);
if (frequency[curr] === undefined)
{
frequency[curr] = -1;
}
frequency[curr] += 1;
if (transitions[old][curr] === undefined)
{
transitions[old][curr] = -1;
}
transitions[old][curr] += 1;
old = curr;
}
alert(frequency);
alert(transitions);