I am trying to create a hangman game in html,css,js.
So i am now to give it a randomisation on based how the question will be displayed -
Create the Random number - Math.floor((Math.random() * 10) + 1)
First of all check if the array(this array contains which questions are already done) has the random number
if not
if yes
then allow the below statement to happen
then let the function run again and agiin until random number is not there in th array
First question based on the random number and second olso and so on
But actually i cannot figure out how to check in the array has the random number. I tried:-
this works
function NextQuestion() {
let done = ['1','2'];
let random = Math.floor((Math.random() * 10) + 1);
let include = done.includes('1');
if (include == true) {
alert('true');// this works
}
}
this doesn't
function NextQuestion() {
let done = ['1','2'];
let random = Math.floor((Math.random() * 10) + 1);
let include = done.includes(random);
if (include == true) {
alert('true');// this does not works
}
}
The difference is that in the first one it has includes('1'); but in second it has includes(random);
Do you know why this happens if you do please let me know and please give the solution to this problem
Thanks in advance.
This will explain why
let random = Math.floor((Math.random() * 10) + 1);
console.log(random)
console.log(
["1","2","3","4","5","6","7","8","9","10"].includes(random)
);
console.log(
[1,2,3,4,5,6,7,8,9,10].includes(random)
);
So your hangman will work if you store the random as they are created
#mplungian correctly answered the question you raised in your title. But I believe you are going the wrong way about it if want to ask a given number of questions in a random sequence. For this you should simply step through the questions array with a for loop. But before you do that the array should be shuffled:
function shfl(a){
// Durstenfeld shuffle:
for(let i=a.length;i>1;){
j=Math.floor(Math.random()*i--);
if (i!=j) [a[i],a[j]]=[a[j],a[i]]
}
return a
}
const q=[1,2,3,4,5,6,7,8,9,10];
shfl(q);
q.forEach((n,i)=>console.log(i+1+". Q"+n));
On my pricing page, I've got a pricing estimator; input an order value and a return rate (%) and price estimate will print into a div block below. The formula should look like this: ((orderValue * 0.05, 2) + 4) + ((returnRate / 100) * 4)... Can anyone see why this isn't working?
var elDeliveryPrice = document.getElementById('deliveryPrice');
var elOrderValue = document.getElementById('orderValue');
var elReturnRate = document.getElementById('returnRate');
var formatter = new Intl.NumberFormat('gb-GB', { style: 'currency', currency: 'GBP' });
elOrderValue.addEventListener('keyup', _ => {
let curVal = elOrderValue.value;
let curValFloat = parseFloat(curVal);
if (isNaN(curValFloat)) {
elDeliveryPrice.innerHTML = '';
return;
}
elDeliveryPrice.innerHTML = formatter.format(Math.max(curValFloat * 0.05,2) + 4) + (parseInt("elReturnRate") / 100)) * 4;
});
There are a few things that cause your issues. For starters you should also listen to 'keyup' on the returnRate input box. Secondly you should handle NaN of the returnRate value.
For example:
var elDeliveryPrice = document.getElementById('deliveryPrice');
var elOrderValue = document.getElementById('orderValue');
var elReturnRate = document.getElementById('returnRate');
var formatter = new Intl.NumberFormat('gb-GB', { style: 'currency', currency: 'GBP' });
function calc() {
let curVal = elOrderValue.value;
let curValFloat = parseFloat(curVal);
let curRate = parseFloat(elReturnRate.value);
if (isNaN(curValFloat) || isNaN(curRate)) {
elDeliveryPrice.innerHTML = '£ _';
return;
}
elDeliveryPrice.innerHTML = formatter.format(Math.max(curValFloat * 0.05,2) + 4 + (curRate / 100) * 4);
}
elOrderValue.addEventListener('keyup', calc);
elReturnRate.addEventListener('keyup', calc);
You can divide a percent number by 100 to get the correct number to include in your calculation. If the return rate input is given in text, you can convert to a number using parseInt(returnRate) e.g.
Math.max(curValFloat * 0.05,2) + 4 + (parseInt(returnRate) / 100) * 4
Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.
The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/
You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.
Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.
I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/
I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.
It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}
Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.
There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.
What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best
You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>
Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>
Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}
var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}
The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.
Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}
You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}
If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()
The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.
This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.
I have some code to randomly highlight one name from a list (this works - see this fiddle):
function pickRandom() {
var random = Math.floor(Math.random() * 6);
$('.stname').css('background','none').eq(random).css('background','yellow');
}
But I'd like to make sure that the same names don't come up over and over. So I intend to remember the last 3 chosen indexes as a blacklist:
var recentlyAsked = new Array();
function pickRandom() {
var random;
do {
random = Math.floor(Math.random() * 6);
} while ($.inArray(random,recentlyAsked));
recentlyAsked.push(random);
if (recentlyAsked.length >= 4) recentlyAsked.shift();
$('.stname').css('background','none').eq(random).css('background','yellow');
}
This is not working; see this fiddle. Warning: it causes the browser to hang.
Any suggestions, please?
do {
random = Math.floor(Math.random() * 6);
} while ($.inArray(random,recentlyAsked));
Runs forever because inArray returns -1 when an item is not found in the array, which is a truthy value. 0 is the only number that is a falsy value. Your array is initially empty so nothing is found.
Fix it with :
do {
random = Math.floor(Math.random() * 6);
} while ($.inArray(random,recentlyAsked) > -1);
This will stop when it returns -1(not found)
var ids=['a','b','c'];
var old=['d','e','f'];//At the beginning this will need populated with 3 random values
var ran=Math.floor(Math.random() * ids.length);
var ele=ids.splice(ran,1);
old.push(ele);
ids.push(old.shift());
highlight(ele);
Here is a slightly alternative way to do what you want. The idea is just remove chosen elements and then add it back in to the original array.
Just thought I would throw my code out there:
var randomArray = new Array(0, 1, 2, 3, 4, 5);
var pastArray = new Array();
function pickRandom() {
var random = Math.floor(Math.random() * randomArray.length);
$('.stname').css('background', 'none').eq(randomArray[random]).css('background', 'yellow');
if (pastArray.length < 3) {
pastArray.unshift(randomArray[random]);
randomArray.splice(random, 1);
} else {
pastArray.unshift(randomArray[random]);
randomArray.splice(random, 1, pastArray.pop());
}
console.log("possible values: [" + randomArray + "]");
console.log("past values: [" + pastArray + "]");
}
Values are moved back and forth from the current and past values. There is no need to prepopulate values as 'past', so it starts out truly random.
I found i little snipet on internet, about PMT calculate.
function PMT(i, n, p) {
return i * p * Math.pow((1 + i), n) / (1 - Math.pow((1 + i), n));
}
function CalculatePMTFromForm(idLoanAmount, idAnnualInterestRate, idMonths, idResult) {
var i = jQuery('#' + idAnnualInterestRate).val() / 1200;
var n = jQuery('#' + idMonths).val();
var p = jQuery('#' + idLoanAmount).val();
var pmt = PMT(i, n, -p);
jQuery('#' + idResult).val(pmt.toFixed(2));
}
function performCalc() {
CalculatePMTFromForm('LoanAmount', 'InterestRate', 'Months', 'Payment');
}
jQuery(document).ready(function() { performCalc(); jQuery('.calc').keyup(performCalc); });
When the page is load, in the result input box I see "NaN" , and when i tpye some irrelevant number then "-Infinity" msg appear.
I search to "NaN" in files and i found in jquery.js, but after I modify, nothing change. And I can't find Infinity
How can I change this messages?
Edit
Calling code:-
function performCalc() {
CalculatePMTFromForm('LoanAmount', 'InterestRate', 'Months', 'Payment');
}
jQuery(document).ready(function() {
performCalc(); jQuery('.calc').keyup(performCalc);
});
This is worked for me:
if(pmt>0 && pmt<Number.MAX_VALUE) {jQuery('#' + idResult).val(pmt.toFixed(2));}
This question's been dead for over a year, but I recently needed to do the same thing. Here's what I came up with:
function pmt(rate_per_period, number_of_payments, present_value, future_value, type){
if(rate_per_period != 0.0){
// Interest rate exists
var q = Math.pow(1 + rate_per_period, number_of_payments);
return -(rate_per_period * (future_value + (q * present_value))) / ((-1 + q) * (1 + rate_per_period * (type)));
} else if(number_of_payments != 0.0){
// No interest rate, but number of payments exists
return -(future_value + present_value) / number_of_payments;
}
return 0;
}
type needs to be 1 or 0, same as Excel's. The rate_per_period needs to be a decimal (eg: 0.25, not 25%).
An example:
/* Example: */
var interest = 0.07, // Annual interest
years = 5, // Lifetime of loan (in years)
present = 10000, // Present value of loan
future = 20000, // Future value of loan
beginning = 1; // Calculated at start of each period
var payment = -pmt(interest / 12, // Annual interest into months
years * 12, // Total months for life of loan
present,
future,
beginning);
And the payment for the example period (month) is ~$474.60.
Note the negation of the result, as the amount is a dedection - ie: costs you $474 - the result is a negative value. Were the result to be a credit, the result would be a positive. Generally you'll want to keep it as a negative/positive, but if you were displaying it in a format like Total Debt: $XYZ, you'd want to convert it to a positive.
NaN means "Not A Number".
Make sure you check for each input value whether it is numeric. Throw an error message if one is not, or set it to 0, depending on whether it's essential to your calculation or not.
A good collection of the best ways to check a value for whether it's numeric is this SO question:
Validate numbers in JavaScript - IsNumeric()
Try it like this:-
function CalculatePMTFromForm(idLoanAmount, idAnnualInterestRate, idMonths, idResult) {
var i = parseFloat($('#' + idAnnualInterestRate).val()) / 1200;
var n = parseFloat($('#' + idMonths).val());
var p = parseFloat($('#' + idLoanAmount).val());
var pmt = PMT(i, n, -p);
$('#' + idResult).val(pmt.toFixed(2));
}
The .val() is likely returning a string type not a number type.