Develop Reference

Fill an array with distanced random integers

I need an array to be filled with random integers
Those integers should be very distinct from each other i.e. must at least be 20 units of separation between each items
This is what i have tried so far :
var all = [];
var i = 0;
randomDiff();
function randomDiff() {
var num1 = randomNumber(10, 290); //chose a first random num in the range...
all[0] = num1; //...put it in first index of array
do // until you have 12 items...
{
var temp = randomNumber(10, 290); //...you pick a temporary num
var j;
for (j = 0; j < all.length; j++) // for each item already in the array
{
if ((temp < all[i] - 10) || (temp > all[i] + 10)) // if the temporary num is different enough from others members...
{
all.push(temp); //then you can store it
i++; //increment until....
console.log(all[i]);
}
}
}
while (i < 11) // ...it is filled with 12 items in array
}
////////////Radom in int range function///////////////////////////////////////
function randomNumber(min, max) {
return Math.floor(Math.random() * (max - min) + min);
}
but always unsuccessful, including infinite loops...

Have a look on something like this:
function randomNumber(min, max) {
return Math.floor(Math.random() * (max - min) + min);
}
const LIST_SIZE = 20;
const DISTANCE = 10;
const STOP_AFTER_ATTEMPT = 2000;
const randomList = [];
let attempt = 0;
while(randomList.length < LIST_SIZE && attempt < STOP_AFTER_ATTEMPT) {
const num = randomNumber(10, 290);
const numberExistsWithSmallerDistance = randomList.some(r => Math.abs(r - num) < DISTANCE)
if (!numberExistsWithSmallerDistance) {
randomList.push(num);
}
attempt++;
}
if (randomList.length === LIST_SIZE) {
console.log(randomList);
} else {
console.log("Failed to create array with distnct values after ", attempt, " tries");
}

Here's a solution that will always work, as long as you allow enough room in the range/separation/count you choose. And it's way more efficient than a while loop. It doesn't just keep trying until it gets it right, it actually does the math to make sure it's right the first time.
This comes at the cost of tending to lean towards certain numbers more than others (like from + (i * separation)), so take note of that.
function getSeparatedRadomInts(from, through, separation, count) {
if(through < from) return getSeparatedRadomInts(through, from, separation, count);
if(count == 0) return [];
if(separation == 0) return !!console.log("Please allow enough room in the range/separation/count you choose.");
//pick values from pool of numbers evenly stepped apart by units of separation... adding 1 to from and through if from is 0 so we can divide properly
var smallFrom = Math.ceil((from || 1) / separation);
var smallThrough = Math.floor((through + (from == 0)) / separation);
var picks = randoSequence(smallFrom, smallThrough).slice(-count).sort((a, b) => a - b);
if(picks.length < count) return !!console.log("Please allow enough room in the range/separation/count you choose.");
for (var i = 0; i < picks.length; i++) picks[i] *= separation;
//go through each pick and randomize with any wiggle room between the numbers above/below it... adding 1 to from and through if from is 0
for (var i = 0; i < picks.length; i++) {
var lowerBound = picks[i - 1] + separation || from || 1;
var upperBound = picks[i + 1] - separation || (through + (from == 0));
picks[i] = rando(lowerBound, upperBound);
}
//subtract 1 from all picks in cases where from is 0 to compensate for adding 1 earlier
for (var i = 0; i < picks.length; i++) if(from == 0) picks[i] = picks[i] - 1;
return picks;
}
console.log(getSeparatedRadomInts(10, 290, 20, 12));
<script src="https://randojs.com/1.0.0.js"></script>
To be clear, from is the minimum range value, through is the maximum range value, separation is the minimum each number must be apart from each other (a separation of 20 could result in [10, 30, 50, 70], for example), and count is how many values you want to pick.
I used randojs in this code to simplify the randomness and make it easier to read, so if you want to use this code, just remember to paste this in the head of your HTML document:
<script src="https://randojs.com/1.0.0.js"></script>

How to get 2 random numbers divisible by each other?

I'm generating 2 random numbers:
1) The first number must be random from 1 to 30
2) The second number must be random from 1 to 10
I'm simply trying to have the first number divisible by the second number or vice-versa, and finally, alert the result. My question is how to get the result of the division of 2 random numbers? Can anyone point me in the right direction? Thanks a lot in advance!.
Note: the first number must be divisible by the second number.
Here's my code:
var arr = [];
for (var i = 0; i < 2; i++) {
do {
var firstRandomNumber = Math.floor(Math.random()*30) + 1;
var secondRandomNumber = Math.floor(Math.random()*10) + 1;
if(firstRandomNumber % secondRandomNumber === 0){
correctResult = result;
arr.push(correctResult);
}
} while ((firstRandomNumber % secondRandomNumber === 0));
}
console.log(arr);

I would suggest a more functional approach: create a function that creates two random numbers, and returns them if one is divisible by the other. Then, just call that function until you get a truthy result:
function tryGetDivisible() {
var firstRandomNumber = Math.floor(Math.random() * 30) + 1;
var secondRandomNumber = Math.floor(Math.random() * 10) + 1;
if (firstRandomNumber % secondRandomNumber === 0) {
console.log(firstRandomNumber, secondRandomNumber);
return firstRandomNumber / secondRandomNumber;
}
}
let result;
while (!result) result = tryGetDivisible();
const arr = [result];
console.log(arr);

Few things:
Your while loop should be looping until firstRandomNumber % secondRandomNumber === 0, so you want to just keep looping while it's not true.
result isn't set anywhere, so I added the result in the array
The if(firstRandomNumber % secondRandomNumber === 0){ is redundant. When the do/while loop completes, it will have the do numbers that matched. Simply move the arr.push() outside that loop.
var arr = [];
for (var i = 0; i < 2; i++) {
do {
var firstRandomNumber = Math.floor(Math.random()*30) + 1;
var secondRandomNumber = Math.floor(Math.random()*10) + 1;
} while ((firstRandomNumber % secondRandomNumber !== 0));
console.log('first', firstRandomNumber, 'second', secondRandomNumber);
arr.push(firstRandomNumber / secondRandomNumber);
}
console.log(arr);

A much simpler approach is to get the first random number, and then try getting the second random number until they are divisible. So here will be the code:
var firstRandomNumber = Math.floor(Math.random()*30) + 1;
while (firstRandomNumber % secondRandomNumber !== 0) {
var secondRandomNumber = Math.floor(Math.random()*10) + 1;
}
console.log(firstRandomNumber + "," + secondRandomNumber);

Since the first must be divisible by the second, my approach would be:
Generate the second number.
Determine the maximum multiple of the second number that is no more than 30 (e.g., Math.floor(30/firstNumber)).
Select a multiple at random and use that as the first number. You simply need to select a random number between 1 and the largest allowed multiplier (inclusive).
This way, there's no need to do a generate-and-test loop, which could go on an unbounded number of times before a successful pair is generated.

If you want to avoid the while loop, you can pick the first number, then assemble the possible second numbers in an array. Then randomly pick one of these:
let first = Math.floor(Math.random() * 10) + 1
// all possible divisible numbers
let factors = Array.from({length: 30}, (_, i) => i + 1)
.filter(i => first % i === 0 || i % first === 0)
//pick one
let second = factors[Math.floor(Math.random() * factors.length)]
console.log(first, second)

Replace numbers in a string with asterisks, and reveal them again

I have a problem with the following task. I want to achieve this output for n = 5:
* 2 3 4 5
* * 3 4 5
* * * 4 5
* * * * 5
* * * * *
* * * * *
* * * * 5
* * * 4 5
* * 3 4 5
* 2 3 4 5
I'm stuck in the second part of the exercise. My code for now:
var n = 5;
var numbers = '';
for (var i = 1; i <= n; i++) {
numbers += i;
}
for (var i = 0; i < n; i++) {
numbers = numbers.replace(numbers[i], '*');
console.log(numbers);
}
So far I have this result:
*2345
**345
***45
****5
*****
So now I need to add the spaces between numbers/stars, and make a reverse loop. I have no idea how to do it.
In addition, there is probably a faster solution to this task than I did.

You can save each of the numbers you generate on a stack (an array), and then pop them from the stack in reverse order:
var n = 5;
var numbers = '';
var stack = []; // <--- add this
for (var i = 1; i <= n; i++) {
numbers += i + ' '; // add a space here
}
for (var i = 0; i < n; i++) {
numbers = numbers.replace(i, '*'); // find/replace the digit
console.log(numbers);
stack.push(numbers); // <--- push on stack
}
while (stack.length > 0) {
numbers = stack.pop(); // <--- pull in reverse order
console.log(numbers); // <--- and print
}
.as-console-wrapper { max-height: 100% !important; top: 0; }
A similar way, without the use of a stack, delays the output, and gathers all the strings in two longer strings which each will have multiple lines of output:
var n = 5;
var numbers = '';
var stack = [];
var output1 = ''; // <-- add this
var output2 = ''; //
for (var i = 1; i <= n; i++) {
numbers += i + ' ';
}
numbers += '\n'; // <-- add a newline character
for (var i = 0; i < n; i++) {
numbers = numbers.replace(i, '*');
output1 += numbers;
output2 = numbers + output2; // <-- add reversed
}
console.log(output1 + output2); // <-- output both
.as-console-wrapper { max-height: 100% !important; top: 0; }

Sticking with something similar to your approach:
var n = 5;
var numbers = '';
for (var i = 1; i <= n; i++) {
numbers += i + ' ';
}
for (var i = 0; i < n; i++) {
numbers = numbers.substr (0, i * 2) + '*' + numbers.substr (i * 2 + 1);
console.log(numbers);
};
for (var i = n - 1; i >= 0; i--) {
console.log(numbers);
numbers = numbers.substr (0, i * 2) + (i + 1) + numbers.substr (i * 2 + 1);
};
The disadvantage of this approach is that it only works for 0-9 because the string positions break when the numbers aren't single digits.

The way I might approach the problem is by having a variable that keeps track of up to which number needs to be an asterisk, doing the first half, then using a whole new for loop to do the second half.
For instance,
String result = '';
String line = '';
int counter = 1;
for (int line = 1; line =< 5; line++) {
for (int i = 1; i =< 5; i++) { // note that we start at 1, since the numbers do
if (i <= counter) {
line += '*'; // use asterisk for positions less than or equal to counter
else {
line += i; // otherwise use the number itself
}
line += ' '; // a space always needs to be added
}
result += line + '\n'; // add the newline character after each line
counter++; // move the counter over after each line
}
Then you can do the same loop, but make the counter go backwards. To do that, set counter to 5 before you begin the loop (since Strings are zero-indexed) and do counter-- after each line.
Alternatively if you don't want to write two loops, you can increase the outer for loop's limit to 10 and have an if statement check if you should be subtracting from counter instead of adding, based on the value of line

Decided to use this as an excuse to get more practice with immutable mapping and reducing. I used an array to hold all the rows, and reduce them at the end to a string. Each row starts as an array holding 1 to n, and each column number is then mapped to an asterisk based on the case:
if rowIndex <= number:
rowIndex.
else:
rowIndex - (2 * (rowIndex - number) - 1)
Essentially, [n + 1, n * 2] maps to (1, 3, 5, ..., n - 3, n - 1), which subtracted from the original range becomes [n, 1]. For the row, check if the currently selected column is less than or equal to its row's translated index, and return an asterisk or the number.
// expansion number (n by 2n)
const maxNum = 5;
// make an array to size to hold all the rows
const result = Array(maxNum * 2)
// Fill each row with an array of maxNum elements
.fill(Array(maxNum).fill())
// iterate over each row
.map((row, rowIndex) =>
// iterate over each column
row.map((v, column) => (
// check if the column is less than the translated rowIndex number (as mentioned above)
column < ((rowIndex <= maxNum) ?
rowIndex + 1 :
2 * maxNum - rowIndex
// if it is, replace it with an asterisk
)) ? "*" : column + 1)
// combine the row into a string with each column separated by a space
.reduce((rowAsString, col) => rowAsString + " " + col)
// combine all rows so they're on new lines
).reduce((rowAccum, row) => rowAccum + "\n" + row);
console.log(result);

JavaScript: Random number in Loop

I was wondering how I can make a JavaScript loop that have to guess a correct number, 1-500.
Every loop there should be a new unique number that hasn't been guessed before. However it should guess in a random order.
Yes:
351, 201, 97 ...
No:
1, 2, 3, 4, 5, 5 ...
My code so far:
var number;
setInterval(function(){
number = Math.floor(Math.random() * 500) + 1;
console.log(number);
if (number == 350) {
console.log("Correct Number!")
}
}, 1000)

Here is my attempt at solving your problem using a fisher yates shuffle on an array with all numbers between 500.
var number = 121;
var numbArr = [];
for(var x = 1; x <= 500; x++){
numbArr.push(x);
}
function shuffle(array) {
var m = array.length, t, i;
// While there remain elements to shuffle…
while (m) {
// Pick a remaining element…
i = Math.floor(Math.random() * m--);
// And swap it with the current element.
t = array[m];
array[m] = array[i];
array[i] = t;
}
return array;
}
shuffle(numbArr);
for(var x = 0; x < numbArr.length; x++){
if(number == numbArr[x]){
console.log("Found number at entry " + x);
}
}
The guessing with a random number comes by using the shuffle of all the possible numbers which means that it shall not be guessed again.

You will need to store the numbers you have already randomed.
If you encountered a number you have already randomed, random again.
var usedNumbers = [];
var interval = setInterval(function(){
var number = Math.floor(Math.random() * 500) + 1;
while(usedNumbers.includes(number)){
number = Math.floor(Math.random() * 500) + 1;
}
usedNumbers.push(number);
if (number == 350){
console.log("Correct Number!");
clearInterval(interval);
return;
}
console.log(number);
}, 500)

JavaScript - Improving algorithm for finding square roots of perfect squares without Math.sqrt

I'm trying to learn algorithms and coding stuff by scratch. I wrote a function that will find square roots of square numbers only, but I need to know how to improve its performance and possibly return square roots of non square numbers
function squareroot(number) {
var number;
for (var i = number; i >= 1; i--) {
if (i * i === number) {
number = i;
break;
}
}
return number;
}
alert(squareroot(64))
Will return 8
Most importantly I need to know how to improve this performance. I don't really care about its limited functionality yet

Here is a small improvement I can suggest. First - start iterating from 0. Second - exit loop when the square of root candidate exceeds the number.
function squareroot(number) {
for (var i = 0; i * i <= number; i++) {
if (i * i === number)
return i;
}
return number; // don't know if you should have this line in case nothing found
}
This algo will work in O(√number) time comparing to initial O(n) which is indeed performance improvement that you asked.
Edit #1
Just even more efficient solution would be to binary search the answer as #Spektre suggested. It is known that x2 is increasing function.
function squareroot(number) {
var lo = 0, hi = number;
while(lo <= hi) {
var mid = Math.floor((lo + hi) / 2);
if(mid * mid > number) hi = mid - 1;
else lo = mid + 1;
}
return hi;
}
This algo has O(log(number)) running time complexity.

The stuff that you try to do is called numerical methods. The most rudimentary/easy numerical method for equation solving (yes, you solve an equation x^2 = a here) is a Newtons method.
All you do is iterate this equation:
In your case f(x) = x^2 - a and therefore f'(x) = 2x.
This will allow you to find a square root of any number with any precision. It is not hard to add a step which approximate the solution to an integer and verifies whether sol^2 == a

function squareRoot(n){
var avg=(a,b)=>(a+b)/2,c=5,b;
for(let i=0;i<20;i++){
b=n/c;
c=avg(b,c);
}
return c;
}
This will return the square root by repeatedly finding the average.
var result1 = squareRoot(25) //5
var result2 = squareRoot(100) //10
var result3 = squareRoot(15) //3.872983346207417
JSFiddle: https://jsfiddle.net/L5bytmoz/12/

Here is the solution using newton's iterative method -
/**
* #param {number} x
* #return {number}
*/
// newstons method
var mySqrt = function(x) {
if(x==0 || x == 1) return x;
let ans, absX = Math.abs(x);
let tolerance = 0.00001;
while(true){
ans = (x+absX/x)/2;
if(Math.abs(x-ans) < tolerance) break;
x = ans;
}
return ans;
};

Separates Newton's method from the function to approximate. Can be used to find other roots.
function newton(f, fPrime, tolerance) {
var x, first;
return function iterate(n) {
if (!first) { x = n; first = 1; }
var fn = f(x);
var deltaX = fn(n) / fPrime(n);
if (deltaX > tolerance) {
return iterate(n - deltaX)
}
first = 0;
return n;
}
}
function f(n) {
return function(x) {
if(n < 0) throw n + ' is outside the domain of sqrt()';
return x*x - n;
};
}
function fPrime(x) {
return 2*x;
}
var sqrt = newton(f, fPrime, .00000001)
console.log(sqrt(2))
console.log(sqrt(9))
console.log(sqrt(64))

Binary search will work best.
let number = 29;
let res = 0;
console.log((square_root_binary(number)));
function square_root_binary(number){
if (number == 0 || number == 1)
return number;
let start = 0;
let end = number;
while(start <= end){
let mid = ( start + end ) / 2;
mid = Math.floor(mid);
if(mid * mid == number){
return mid;
}
if(mid * mid < number){
start = mid + 1;
res = mid;
}
else{
end = mid - 1;
}
}
return res;
}

If you analyze all natural numbers with their squares you might spot a pattern...
Numbers Squares Additives
1 1 3
2 4 5
3 9 7
4 16 9
5 25 11
6 36 13
7 49 15
Look at the first row in the squares column (i.e 1) and add it with the first row in the additives column (ie. 3). You will get four which is in the second row of the squares column.
If you keep repeating this you'll see that this applies to all squares of natural numbers. Now if you look at the additives column, all the numbers below are actually odd.
To find the square root of a perfect square you should keep on subtracting it with consecutive odd numbers (starting from one) until it is zero. The number of times it could be subtracted is the square root of that number.
This is my solution in typescript...
function findSquareRoot(number: number): number {
for (let i = 1, count = 0; true; number -= i, i += 2, count++) {
if (number <= 0) {
return number === 0 ? count : -1; // -1 if number is not a perfect square
}
}
}
Hopefully this has better time complexity :)

I see this solution on Github which is the much better and easiest approach to take a square root of a number without using any external library
function TakingPerfectSquare(Num) {
for (var i = 0; i <= Num; i++) {
var element = i;
if ((element == element) && (element*element == Num)) {
return true;
}
}
return false;
}
console.log(TakingPerfectSquare(25));