I'm trying to print all prime number between 0 and 100, but when executing this code the browser's tab just outputs nothing!!
for(var i = 2; i < 100; i++)
{
var prime = [];
for(var j = 0; j <= i; j++)
{
var p = i % j;
}
if(p != 0) prime.push(i);
else continue;
}
for(var k = 0; k < prime.length; k++)
{
document.writeln(prime[k], "<br>");
}
Try this one. I have also optimise the code (you only need to check upto sqrt(i) ).
var prime = [];
prime.push(2); //smallest prime
var flag = 0;
for(var i = 3; i < 100; i=i+2) //skip all even no
{
for(var j = 3; j*j <= i; j=j+2) //check by upto sqrt(i), skip all even no
{
if(i % j == 0) {
flag = 0;break; //not a prime, break
}
flag = 1;
}
if (flag == 1) prime.push(i); //prime, add to answer
}
for(var k = 0; k < prime.length; k++)
{
document.writeln(prime[k], "<br>");
}
Because you blank your list of primes EVERY loop cycle, move it outside the for loop
You need to make your variable prime outside of your loop
This is the code you have re-written
var prime = [];
for(var i = 2; i < 100; i++)
{
for(var j = 0; j <= i; j++)
{
var p = i % j;
}
if(p != 0) prime.push(i);
else continue;
}
for(var k = 0; k < prime.length; k++)
{
document.writeln(prime[k], "<br>");
}
I'm a fan of the sieve of Eratosthenes.
The following code should do what you wanted.
var prime = Array(101).fill(true);
for (var i = 2; i < 100; ++i){
if (prime[i]){
document.writeln(i, "<br>");
for (var j = i*i; j < 100; j += i){
prime[j] = false;
}
}
}
Or since it's only up to 100 you could just manually type the list (but, hey where's the learning if you do it that way?).
(1) Move prime outside the for loop, (2) start j at 2 and end when j < i, (3) check when p == 0 with a boolean flag and break inner loop.
var prime = []; //put prime out here so it does not reassign
for(var i = 2; i < 100; i++)
{
var isPrime = true;
for(var j = 2; j < i; j++) //start j at 2
{
var p = i % j;
if(p == 0)
{
isPrime = false;
break;
}
}
if(isPrime) prime.push(i);
}
for(var k = 0; k < prime.length; k++)
{
document.writeln(prime[k], "<br>");
}
Related
This is my code:
// Population
var Gene = function(text){
if(text){
this.text = text;
}
};
Gene.fitness = 0;
Gene.generation = 0;
var word = new Gene('Hello');
// This is where it crashes!!
// Make elements
var genArr = [];
var population = 20;
var mutation = 0;
for(var i = 0; i < population; i++){
var gene = "";
var abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
for (var j = 0; i < word.text.length; j++) {
var element = abc.charAt(Math.floor(Math.random() * abc.length));
gene += element;
}
genArr.push(gene);
}
// Divide them - fitness
// 1/20 - 0.05% each
var fitElements = [];
for (var i = 0; i < genArr.length; i++) {
var score = 0;
var curWord = Array.from(genArr[i]);
for (var j = 0; j < word.text.length; j++) {
if(genArr[j].substr(j, 1) == word.text.substr(j, 1)){
score += 1;
}
}
if(score > 0){
fitElements.push([genArr[i], (score * (1 / population)) ** 2]);
}
}
for (var i = 0; i < fitElements.length; i++) {
console.log('Element: ' + fitElements[i][0] + ', Score: ' + fitElements[i][1]);
}
My problem is that it crashes the page but doesn't gives errors. The idea is to create a simple word register in fitElements Array but I can't see what am I missing?
Thanks in advance.
In your code the nested for loop with variable j's end condition relies on i.
Take a the line:
// VVVV it relies on i instead of j
for (var j = 0; i < word.text.length; j++) {
var element = abc.charAt(Math.floor(Math.random() * abc.length));
gene += element;
}
The new code would look something like
for (var j = 0; j < word.text.length; j++)
var element = abc.charAt(Math.floor(Math.random() * abc.length));
gene += element;
}
The only difference in the entire sample is the i is changed to a j. Cheers!
I would like align row in function of the Name (column B & D), for exemple on the picture (1) the 2 red rectangles should be on the same line like for the 2 green rectangles but I don't know why, it doesn't work, here the code :
function onOpen(event){
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheetByName("test1");
var data = sheet.getDataRange().getValues();
for(var j = 1; j < data.length; j++) {
if(data[j][1].trim() != '') {
for (var i = 1; i < data.length; i++) {
if(data[i][3].trim() != '') {
if(data[j][1] == data[i][3]) {
if(j != i) {
var j1 = j + 1;
var i1 = i + 1;
sheet.getRange('D'+j1+':E'+j1).moveTo(sheet.getRange('F1:G1'));
sheet.getRange('D'+i1+':E'+i1).moveTo(sheet.getRange('D'+j1+':E'+j1));
sheet.getRange('F1:G1').moveTo(sheet.getRange('D'+i1+':E'+i1));
}
break;
}
}
}
}
}
}
I'm pretty sure this code should work
Oddly it works if I'm running the code step by step like this, firstly :
for(var j = 1; j < 2; j++)
after
for(var j = 2; j < 3; j++)
after
...
until
for(var j = 6; j < 7; j++)
Well I found a workaround :)
function onOpen(event){
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheetByName("test1");
var data = sheet.getDataRange().getValues();
var blank = 0;
// I count the blank case so I don't run the code effortlessly on blank case
for(var j = 1; j < data.length; j++) {
if(data[j][1].trim() == '')
blank++;
}
var length = data.length - blank;
// I don't make a real switch in fact, I'm using a tempory space where I order the data
for(var j = 1; j < length; j++) {
for (var i = 1; i < length; i++) {
if(data[j][1] === data[i][3]) {
var j1 = j + 1;
var i1 = i + 1;
sheet.getRange('D'+i1+':E'+i1).moveTo(sheet.getRange('F'+j1+':G'+j1)); // column F & G is the tempory space where I put the ordered data
break;
}
}
}
// I move back the tempory space ordered now to his initial place
sheet.getRange('F2:G'+length).moveTo(sheet.getRange('D2:E'+length));
}
The algorithm is taken from LeetCode: https://leetcode.com/problems/maximum-product-of-word-lengths/description/
Here is the jsperf I created (I have some local tests which gives the same result): https://jsperf.com/maximum-product-of-word-lengths
Here is the first "slow" implementation:
function maxProduct (words) {
if (!words || !words.length) return 0;
let len = words.length;
let values = [];
// console.log(values)
for (let i = 0; i < len; ++i) {
let tmp = words[i];
let num = 0, len = tmp.length;
for (let j = 0; j < len; ++j) {
num |= 1 << (tmp.charCodeAt(j) - 'a'.charCodeAt(0));
}
values[i] = {
num: num,
len: tmp.length
};
}
let maxProduct = 0;
for (let i = 0; i < len; ++i) {
for (let j = 0; j < len; ++j) {
if ((values[i].num & values[j].num) == 0) {
maxProduct = Math.max(maxProduct, values[i].len * values[j].len);
}
}
}
return maxProduct;
};
Here is the "fast" implementation:
function maxProductFast (words) {
var temp = [];
for(var i = 0; i < words.length; i++){
var tempObj = {};
tempObj.item = words[i];
var num = 0;
for(var j = 0; j < words[i].length; j++){
num |= 1 << (words[i].charCodeAt(j) - 97);
}
tempObj.num = num;
temp.push(tempObj);
}
var res = 0;
for(var i = 0; i < temp.length; i++){
for(var j = i + 1; j < temp.length; j++){
var item1 = temp[i];
var item2 = temp[j];
if((item1.num & item2.num) == 0) {
res = Math.max(res, item1.item.length * item2.item.length);
}
}
}
return res;
}
They're not the same. The second algorithm has a loop with a complexity of (n*(n+1))/2 where each progressive step is from i+1 to the length of temp. the first algorithm has a two nested for loops each with a cost of n^2. the complexity of both will reduce to O(n^2). I believe that both of these will have a similar performance with a significantly large enough set.
The reason you would do n+1 for each sub iteration is because you are trying to find the max of any pair of items. if you place your elements in a grid you will notice that any diagonal pair a_3 * a_2 = a_2 * a_3 produces the same value. you can basically halve the collection and save a few cycles.
This is a bad coded solution to a problem in "advent of code": link to problem
I don't know the reason because my code is not working properly, I had an error related to regular expressions cause I didn't reset the pointer of the regexp object, now that error is fixed I think, but something is escaping to my knowledge in what I've done bad.
The problem is that the solution that my code displays is not correct, you can submit a solution on the link I've provided and get feedback of your solutions.
Correct solution: 543903
Given solution: 418954
// day 6 of advent of code
var input = "removed, take from problem";
function processInput(input, matrix) {
var linesOfInput = input.split("\n");
var matches;
var turnOnRE = /turn on (\d+),(\d+).*?(\d+),(\d+)/g;
var turnOffRE = /turn off (\d+),(\d+).*?(\d+),(\d+)/g;
var toggleRE = /toggle (\d+),(\d+).*?(\d+),(\d+)/g;
// regular expression objects lastIndex property must be 'reseted' in order to work well
for (var i = 0 ; i < linesOfInput.length; i++) {
turnOnRE.lastIndex = 0;
turnOffRE.lastIndex = 0;
toggleRE.lastIndex = 0;
matches = turnOnRE.exec(linesOfInput[i]);
if (matches != null) {
manipulateLights(matrix, matches[1], matches[2], matches[3], matches[4], true);
continue;
}
matches = turnOffRE.exec(linesOfInput[i]);
if (matches != null) {
manipulateLights(matrix, matches[1], matches[2], matches[3], matches[4], false);
continue;
}
matches = toggleRE.exec(linesOfInput[i]);
manipulateLights(matrix, matches[1], matches[2], matches[3], matches[4]);
}
}
function manipulateLights(matrix, startI, startJ, endI, endJ, newValue) {
if (newValue == undefined) { // toogle
for (var i = startI ; i <= endI; i++) {
for (var j = startJ ; j <= endJ; j++) {
matrix[i][j] = !matrix[i][j];
}
}
console.log(startI, startJ, endI, endJ, newValue);
} else {
for (var i = startI ; i <= endI; i++) {
for (var j = startJ ; j <= endJ; j++) {
matrix[i][j] = newValue;
}
}
console.log(startI, startJ, endI, endJ, newValue);
}
console.log(countTurnedOnLights(matrix));
}
function countTurnedOnLights(matrix) {
var turnedOn = 0;
for (var i = 0 ; i < matrixWidth; i++) {
for (var j = 0 ; j < matrixHeigth; j++) {
if (matrix[i][j] == true) {
turnedOn++;
}
}
}
return turnedOn;
}
var matrixHeigth = 1000;
var matrixWidth = 1000;
// define a bidimensional array, is almost like in C++
var lightMatrix = new Array(matrixWidth);
for (var i = 0 ; i < matrixWidth; i++) {
lightMatrix[i] = new Array(matrixHeigth);
}
// turn off all lights
for (var i = 0 ; i < matrixWidth; i++) {
for (var j = 0 ; j < matrixHeigth; j++) {
lightMatrix[i][j] = false;
}
}
processInput(input, lightMatrix);
console.log(countTurnedOnLights(lightMatrix));
OK I figured out the error - your regular expression matches are being treated as strings when you first create your for loops.
for (var i = startI ; i <= endI; i++) {
for (var j = startJ ; j <= endJ; j++) {
When you hit a combo like 756,53 through 923,339, it thinks 53 > 339 and it exits the loop immediately. Either wrap each "start" variable with Number() in your for loops, or do so when passing the parameters.
The code is supposed to check whether the first few triangular numbers are prime (they are not), but it does not run.
<!DOCTYPE HTML>
<html>
<head><title>C1E9P1</title>
<script>
for(i = 3; i < 13; i++){
prime = true;
n = (i*(i+1))/2;
for(i = 2; i < Math.sqrt(n)+1; i++){
if(n%i == 0){
prime = false;
}
}
if(prime){
document.write(n+" is prime.");
document.write("<br>");
} else {
document.write(n+" is composite.");
document.write("<br>");
}
}
</script>
</head>
</html>
You used same variable for your two for loops that are nested!
for(i = 3; i < 13; i++){ <-- i
for(i = 2; i < Math.sqrt(n)+1; i++){ <-- i
Try this and look at your JavaScript console :
for(var i = 3; i < 13; i++){
var prime = true;
var n = (i*(i+1))/2;
for(var j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
if(prime){
console.log(n+" is prime.");
console.log("<br>");
} else {
console.log(n+" is composite.");
console.log("<br>");
}
}
I've modified script, but don't known if algorithm is correct.
for(var i = 3; i < 13; i++){
var prime = true;
var n = (i*(i+1))/2;
for(var j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
if(prime){
document.write(n+" is prime.");
document.write("<br>");
} else {
document.write(n+" is composite.");
document.write("<br>");
}
}
You should change the inner for-loop's variable name to something else other than i.
for(i = 3; i < 13; i++){
prime = true;
n = (i*(i+1))/2;
for(j = 2; j < Math.sqrt(n)+1; j++){
if(n%j == 0){
prime = false;
}
}
You should be all good after that.