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Javascript print square using for loop and conditional statement only

Just started my uni course, struggling a little with javascript. I have been asked to display a square using any character, however, the solution must combine for loops and if statements.
This is what I have so far and I feel pretty close but I just can't get the second line to display. I know this can be done via two for loops, (one for iteration of the variable and another for spaces). But this is not how I have been asked to solve this problem.
Here is my code:
var size = 3;
let i;
for(i = 0; i < size; i++) {
print ("*");
if (size === i){
println ("");
}
}
For context, this is all taking place int he professors homemade learning environment.

You could use nested for loops and take a line break after each filled line.
function print(s) { document.getElementById('out').innerHTML += s; }
function println(s) { document.getElementById('out').innerHTML += s + '\n'; }
var size = 5,
i, j;
for (i = 0; i < size; i++) {
for (j = 0; j < size; j++) {
print("*");
}
println("");
}
<pre id="out"></pre>
Single loop with a check if i is unequal to zero and if the remainder is zero, then add a line break.
Using:
=== identity/strict equality operator checks the type and the value, for example if both are numbers and if the value is the same,
!== non-identity/strict inequality operator it is like above, but it checks the oposite of it,
% remainder operator, which returns a rest of a number which division returns an integer number.
&& logical AND operator, which check both sides and returns the last value if both a truthy (like any array, object, number not zero, a not empty string, true), or the first, if it is falsy (like undefined, null, 0, '' (empty string), false, the oposite of truthy).
function print(s) { document.getElementById('out').innerHTML += s; }
function println(s) { document.getElementById('out').innerHTML += s + '\n'; }
var size = 5,
i;
for (i = 0; i < size * size; i++) {
if (i !== 0 && i % size === 0) {
println("");
}
print("*");
}
<pre id="out"></pre>

Well the for loop is only iterating 3 times, printing the first line. If you want a square you'll have to print 9 stars total, right? So i'm assuming, is this is the approach you'd go for, you would need to iterate not until size, but until size * size.

I'm using console.log to 'print' the square:
var dimension = 10;
var edge = '*';
var inside = ' ';
var printLine;
for (var i = 1; i <= dimension; i++) {
if (i === 1 || i === dimension) {
printline = Array(dimension + 1).join(edge);
} else {
printline = edge + Array(dimension - 1).join(inside) + edge;
}
console.log(printline);
}
Note that in the following example, an array of length 11 gets you only 10 "a"s, since Array.join puts the argument between the array elements:
Array(11).join('a'); // create string with 10 as "aaaaaaaaaa"

You wanna make a square of * where the size is the number of * on its sides?
Let's split a task into 3 parts:
where you print top side like *****
where you print middle (left and right sides) like * *
where you print bottom (same as top)
Now let's code that, I kept the code as simple as possible, this can be done in fewer lines but I think this will be easier to understand for beginners:
var size = 5;
var i = 0;
// top
for (i = 0; i < size; i++)
console.log("*");
//middle
for (var j = 0; j < size - 2; j++){
console.log("\n"); // go to next row
// middle (2 on sides with size-2 in between)
console.log("*");
for (i = 0; i < size-2; i++)
console.log(" ");
console.log("*\n"); // goes to new row as well
}
// same as top
for (i = 0; i < size; i++)
console.log("*");
Full square is even simpler:
var size = 5;
var i = 0;
for (var i = 0; i < size; i++){ // iterates rows
for (var j = 0; j < size; j++) // iterates * in row
console.log("*");
console.log("\n") // moves to new row
}
In order to print a row, you print same sign X times. Well, to print X rows we can use just that 1 more time (only this time we are iterating over a different variable (j is * in a row, i is a number of rows).
After a row is made we go to go to next row with \n.
As for
it must contain if statement
Put this at the end:
if (youCanHandleTheTruth) console.log("It's a terrible practice to tell students their solution MUST CONTAIN CODEWORDS. If you need them to showcase something, write appropriate task that will require them to do so.");

More Efficient Way to Accomplish This?

I need to know the level of a player using the amount of exp he has and the exp chart. I want to do it the most efficient way possible. This is what I got. Note: The real expChart has thousands of levels/index. All the values are in increasing order.
var expChart = [1,10,23,54,65,78,233,544,7666,22224,64654,456456,1123442];
/*
lvl 0: //0-1[ exp
lvl 1: //[1-10[ exp
lvl 2: //[10-23[ exp
*/
getLvlViaExp = function(exp){
for(var i = 0 ; i < expChart.length ; i++){
if(exp < expChart[i]) break;
}
return i;
}
This is a more efficient way to do it. Every x steps, (6 i the example, probably every hundreds with real chart), I do a quick comparation and jump to approximative index, skipping many indexes.
getLvlViaExp = function(exp){
var start = 0;
if(exp > 233) start = 6;
if(exp > 1123442) start = 12;
for(var i = start ; i < expChart.length ; i++){
if(exp < expChart[i]) break;
}
return i;
}
Is there an even better way to do this?
SOLUTION:
Array.prototype.binarySearch = function(value){
var startIndex = 0,
stopIndex = this.length - 1,
middle = Math.floor((stopIndex + startIndex)/2);
if(value < this[0]) return 0;
while(!(value >= this[middle] && value < this[middle+1]) && startIndex < stopIndex){
if (value < this[middle]){
stopIndex = middle - 1;
} else if (value > this[middle]){
startIndex = middle + 1;
}
middle = Math.floor((stopIndex + startIndex)/2);
}
return middle+1;
}

The best algorithm for searching is binary search which is O(lg n) (unless you can do it with a hashing search which is O(c).
http://www.nczonline.net/blog/2009/09/01/computer-science-in-javascript-binary-search/
Basically jump to the middle of your chart ( n / 2). Is you experience higher or lower from that number. If higher jump to the middle higher half. If lower jump to the middle of the lower half: Compare and repeat until you find what you're looking for.

var expChart = [1,10,23,54,65,78,233,544,7666,22224,64654,456456,1123442];
getLvlViaExp = function(exp){
var min=0;
var max=expChart.length-1;
var i;
while(min <=max){
i=Math.round((min+max)/2);
//document.write("<br />"+i+":"+min+":"+max+"<br />");
if(exp>=expChart[i] && exp <=expChart[i+1]) {
break;
}
if(exp>=expChart[i]){
min=i+1;
}
if(exp<=expChart[i]){
max=i-1;
}
}
return i;
}
document.write(getLvlViaExp("10"));
I have tested it and it seems to work pretty well. If you want to see how many steps it actually goes through to get to the answer, uncomment the document.write in the while loop. It was kind of fascinating watching it.

JS probability of 10 people selecting the same number between 1 - 20

I'm pretty new to to javascript so please take it easy :)
I'm trying to figure out the probability of 10 people picking the same random number (1 - 20).
When I run the code it returns the same answer every time. I think something is wrong in the 3rd for loop when comparing numbers. Some help would be much appreciated, I've been stuck on this for 3 days now :(
var counter = 0;
//Determine probability (percentage)
for (var i = 1; i <=100; i++) {
//Create array with 10 elements and assign each element with random integer (1 - 20)
for (var j = 1; j <= 10; j++) {
var rndNum = [j];
rndNum = Math.random();
rndNum = Math.floor(rndNum * 20) + 1;
}
//Increment counter if match is found
for (var p1 = 1; p1 <= 9; p1++) {
for (var p2 = p1 + 1; p2 <= 10; p2++) {
if (rndNum[p1] == rndNum[p2]) {
counter++;
}
}
}
}
document.write("The probability of a match is: " + counter + "%");

Your code to make an "array" of random numbers is part of the problem. rndNum only has one value (it's an array with only one item in it, and you're overwriting it each time). You need array.push() to add values to an array.
You want something more like this:
var rndNum = [];
for (var j = 1; j <= 10; j++) {
rndNum.push(Math.floor(Math.random()* 20) + 1);
}

You want to know the probability that twenty people will pick the same random number?
alert("The probability of a match is: " + (Math.pow(.1, 20)*100)+ "%");
Or you want to know the probability that any two of twenty people will pick the same number?
alert("The probability of a match is: " + (Math.pow(.9, 20)*100)+ "%");

Your for loop should also start at 0, not one (the first element in an array is array[0]:
for (var p1 = 0; p1 <= 8; p1++) {
for (var p2 = p1 + 1; p2 <= 9; p2++) {
if (rndNum[p1] == rndNum[p2]) {
counter++;
}
}
}
You also need to divide your results by 10 because you have 1,000 tests (if you're checking to see if two match). If you want to see if they all match you would need something like:
var ordered = rndNum.sort();
if(ordered[0] == ordered[9])
counter2++;
Here's a fiddle of the combined array declaration and match checkers.

Javascript generate random unique number every time

Ok so i need to create four randomly generated numbers between 1-10 and they cannot be the same. so my thought is to add each number to an array but how can I check to see if the number is in the array, and if it is, re-generate the number and if it isnt add the new number to the array?
so basically it will go,
1.create new number and add to array
2.create second new number, check to see if it exist already, if it doesn't exist, add to array. If it does exist, re-create new number, check again etc...
3.same as above and so on.

You want what is called a 'random grab bag'. Consider you have a 'bag' of numbers, each number is only represented once in this bag. You take the numbers out, at random, for as many as you need.
The problem with some of the other solutions presented here is that they randomly generate the number, and check to see if it was already used. This will take longer and longer to complete (theoretically up to an infinite amount of time) because you are waiting for the random() function to return a value you don't already have (and it doesn't have to do that, it could give you 1-9 forever, but never return 10).
There are a lot of ways to implement a grab-bag type solution, each with varying degrees of cost (though, if done correctly, won't ever be infinite).
The most basic solution to your problem would be the following:
var grabBag = [1,2,3,4,5,6,7,8,9,10];
// randomize order of elements with a sort function that randomly returns -1/0/1
grabBag.sort(function(xx,yy){ return Math.floor(Math.random() * 3) - 1; })
function getNextRandom(){
return grabBag.shift();
};
var originalLength = grabBag.length;
for(var i = 0; i < originalLength; i++){
console.log(getNextRandom());
}
This is of course destructive to the original grabBag array. And I'm not sure how 'truly random' that sort is, but for many applications it could be 'good enough'.
An slightly different approach would be to store all the unused elements in an array, randomly select an index, and then remove the element at that index. The cost here is how frequently you are creating/destroying arrays each time you remove an element.

Here are a couple versions using Matt's grabBag technique:
function getRandoms(numPicks) {
var nums = [1,2,3,4,5,6,7,8,9,10];
var selections = [];
// randomly pick one from the array
for (var i = 0; i < numPicks; i++) {
var index = Math.floor(Math.random() * nums.length);
selections.push(nums[index]);
nums.splice(index, 1);
}
return(selections);
}
You can see it work here: http://jsfiddle.net/jfriend00/b3MF3/.
And, here's a version that lets you pass in the range you want to cover:
function getRandoms(numPicks, low, high) {
var len = high - low + 1;
var nums = new Array(len);
var selections = [], i;
// initialize the array
for (i = 0; i < len; i++) {
nums[i] = i + low;
}
// randomly pick one from the array
for (var i = 0; i < numPicks; i++) {
var index = Math.floor(Math.random() * nums.length);
selections.push(nums[index]);
nums.splice(index, 1);
}
return(selections);
}
And a fiddle for that one: http://jsfiddle.net/jfriend00/UXnGB/

Use an array to see if the number has already been generated.
var randomArr = [], trackingArr = [],
targetCount = 4, currentCount = 0,
min = 1, max = 10,
rnd;
while (currentCount < targetCount) {
rnd = Math.floor(Math.random() * (max - min + 1)) + min;
if (!trackingArr[rnd]) {
trackingArr[rnd] = rnd;
randomArr[currentCount] = rnd;
currentCount += 1;
}
}
alert(randomArr); // Will contain four unique, random numbers between 1 and 10.
Working example: http://jsfiddle.net/FishBasketGordo/J4Ly7/

var a = [];
for (var i = 0; i < 5; i++) {
var r = Math.floor(Math.random()*10) + 1;
if(!(r in a))
a.push(r);
else
i--;
}
That'll do it for you. But be careful. If you make the number of random numbers generated greater than the may number (10) you'll hit an infinite loop.

I'm using a recursive function. The test function pick 6 unique value between 1 and 9.
//test(1, 9, 6);
function test(min, max, nbValue){
var result = recursValue(min, max, nbValue, []);
alert(result);
}
function recursValue(min, max, nbValue, result){
var randomNum = Math.random() * (max-min);
randomNum = Math.round(randomNum) + min;
if(!in_array(randomNum, result)){
result.push(randomNum);
nbValue--;
}
if(nbValue>0){
recursValue(min, max, nbValue, result);
}
return result;
}
function in_array(value, my_array){
for(var i=0;i< my_array.length; i++){
if(my_array[i] == value){
console.log(my_array+" val "+value);
return true;
}
}
return false;
}

Here is a recursive function what are you looking for.
"howMany" parameter is count of how many unique numbers you want to generate.
"randomize" parameter is biggest number that function can generate.
for example : rand(4,8) function returns an array that has 4 number in it, and the numbers are between 0 and 7 ( because as you know, Math.random() function generates numbers starting from zero to [given number - 1])
var array = [];
var isMatch= false;
function rand(howMany, randomize){
if( array.length < howMany){
var r = Math.floor( Math.random() * randomize );
for( var i = 0; i < howMany; i++ ){
if( array[i] !== r ){
isMatch= false;
continue;
} else {
isMatch= true;
break;
}
}
if( isMatch == false ){
array.push(r);
ran(howMany, randomize);
}
ran(howMany, randomize);
return array;
}
}

In your answer earlier, you do have a small bug. Instead of
var originalLength = grabBag.length;
for(var i = 0; i < originalLength .length; i++){
console.log(getNextRandom());
}
I believe you meant:
var originalLength = grabBag.length;
for(var i = 0; i < originalLength; i++){
console.log(getNextRandom());
}
Thanks.

Sorting function?

I need to organize an array of strings of random length into the least number of new strings with a max size. Is there a function or something in javascript, or something that can be translated to javascript, that will do this?
For example, the new strings might have max lengths of 1000 characters. The array might have strings of lengths 100, 48, 29, etc. I would want to combine those strings into as few new strings as possible.
edit: Sorry if this doesn't make sense, I tried my best.

No standard method in Javascript, but plenty of theoretical work has been done on this (i.e. the bin packing problem).
http://en.wikipedia.org/wiki/Bin_packing_problem
Some sample pseudo code in the link - should be trivial to translate to javascript.
The algorithm shown isn't going to be optimal in every case. To find the optimal solution to your example you'll just need to iterate over every possibility which might not be that bad depending on how many strings you have.

For my own entertainment, I wrote a simple bin packing algorithm. I picked a simple algorithm which is to sort the input strings by length. Create a new bin. Put the first (longest remaining) string into the bin and then keep filling it up with the longest strings that will fit until no more strings will fit. Create a new bin, repeat. To test it, I allocate an array of strings of random lengths and use that as input. You can see the output visually here: http://jsfiddle.net/jfriend00/FqPKe/.
Running it a bunch of times, it gets a fill percentage of between 91-98%, usually around 96%. Obviously the fill percentage is higher if there are more short strings to fill with.
Here's the code:
function generateRandomLengthStringArrays(num, maxLen) {
var sourceChars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXY1234567890";
var sourceIndex = 0;
var result = [];
var len, temp, fill;
function getNextSourceChar() {
var ch = sourceChars.charAt(sourceIndex++);
if (sourceIndex >= sourceChars.length) {
sourceIndex = 0;
}
return(ch);
}
for (var i = 0; i < num; i++) {
len = Math.floor(Math.random() * maxLen);
temp = new String();
fill = getNextSourceChar();
// create string
for (var j = 0; j < len; j++) {
temp += fill;
}
result.push(temp);
}
return(result);
}
function packIntoFewestBins(input, maxLen) {
// we assume that none of the strings in input are longer than maxLen (they wouldn't fit in any bin)
var result = [];
// algorithm here is to put the longest string into a bin and
// then find the next longest one that will fit into that bin with it
// repeat until nothing more fits in the bin, put next longest into a new bin
// rinse, lather, repeat
var bin, i, tryAgain, binLen;
// sort the input strings by length (longest first)
input.sort(function(a, b) {return(b.length - a.length)});
while (input.length > 0) {
bin = new String(); // create new bin
bin += input.shift(); // put first one in (longest we have left) and remove it
tryAgain = true;
while (bin.length < maxLen && tryAgain) {
tryAgain = false; // if we don't find any more that fit, we'll stop after this iteration
binLen = bin.length; // save locally for speed/convenience
// find longest string left that will fit in the bin
for (i = 0; i < input.length; i++) {
if (input[i].length + binLen <= maxLen) {
bin += input[i];
input.splice(i, 1); // remove this item from the array
tryAgain = true; // try one more time
break; // break out of for loop
}
}
}
result.push(bin);
}
return(result);
}
var binLength = 60;
var numStrings = 100;
var list = generateRandomLengthStringArrays(numStrings, binLength);
var result = packIntoFewestBins(list, binLength);
var capacity = result.length * binLength;
var fillage = 0;
for (var i = 0; i < result.length; i++) {
fillage += result[i].length;
$("#result").append(result[i] + "<br>")
}
$("#summary").html(
"Fill percentage: " + ((fillage/capacity) * 100).toFixed(1) + "%<br>" +
"Number of Input Strings: " + numStrings + "<br>" +
"Number of Output Bins: " + result.length + "<br>" +
"Bin Legnth: " + binLength + "<br>"
);