I'm new to JavaScript and I am trying to write a simple script that solves linear equations. So far my script solves linear equations that are plus and minus only such as "2x + 28 - 18x = 36 - 4x + 10". I want it to also be able to solve linear equations/algebra problems that contain multiplication and division such as "2x * 3x = 4 / 2x".
I kind of have an idea of what to do next but I think the script I have right now maybe overly complex and it's only going to make it more complicated to add the multiplication and division.
Below is my script. I'm hoping for a few pointers on how I could improve and simplify what I already have and what the best way to add multiplication and division?
My script on JS Bin: http://jsbin.com/ufekug/1/edit
My script:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Problem Solver</title>
<script>
window.onload = function() {
// Total Xs on each side of equation
// Example problem: 5x + 2 = 10 - 2x
var leftSideXTotal = 0; // 5
var rightSideXTotal = 0; // -2
// Total integers on each side of equation
// Example problem: 5x + 2 = 10 - 2x
var leftSideIntTotal = 0; // 2
var rightSideIntTotal = 0; // 10
// Enter a math problem to solve
var problem = "5x + 2 = 10 - 2x";
// Remove all spaces in problem
// Example problem: 5x + 2 = 10 - 2x
problem = problem.replace(/\s/g,''); // 5x+2=10-2x
// Add + signs in front of all - signs
// Example problem: 5x + 2 = 10 - 2x
problem = problem.replace(/-/gi, "+-"); // 5x+2=10+-2x
// Split problem into left and right sides
// Example problem: 5x + 2 = 10 - 2x
var problemArray = problem.split("=");
var problemLeftSide = problemArray[0]; // 5x+2
var problemRightSide = problemArray[1]; // 10+-2x
// Split values on each side into an array
var problemLeftSideValues = problemLeftSide.split("+");
var problemRightSideValues = problemRightSide.split("+");
// Go through the left side values and add them up
for (var i = 0; i < problemLeftSideValues.length; i++) {
// Current value
var currentValue = problemLeftSideValues[i];
// Length of current value
var currentValueLength = currentValue.length;
if (currentValue.charAt(currentValueLength - 1) == "x") { //Check if current value is a X value
// Remove X from end of current value
currentValue = currentValue.split("x");
// Add to total Xs on left side
leftSideXTotal = Number(leftSideXTotal) + Number(currentValue[0]);
} else {
// Add to total integers on left side
leftSideIntTotal = Number(leftSideIntTotal) + Number(problemLeftSideValues[i]);
}
}
// Go through the right side values and add them up
for (var i = 0; i < problemRightSideValues.length; i++) {
// Current value
var currentValue = problemRightSideValues[i];
// Length of current value
var currentValueLength = currentValue.length;
if (currentValue.charAt(currentValueLength - 1) == "x") { //Check if current value is a X value
// Remove X from end of current value
currentValue = currentValue.split("x");
// Add to total Xs on right side
rightSideXTotal = Number(rightSideXTotal) + Number(currentValue[0]);
} else {
// Add to total integers on right side
rightSideIntTotal = Number(rightSideIntTotal) + Number(problemRightSideValues[i]);
}
}
// Compute
var totalXs = (leftSideXTotal - rightSideXTotal)
var totalIntegers = (rightSideIntTotal - leftSideIntTotal)
var solution = (totalIntegers / totalXs)
// Display solution
document.getElementById("divSolution").innerText = solution;
}
</script>
</head>
<body>
<div id="divSolution"></div>
</body>
</html>
You need to write (or use) an operator-precedence parser.
The idea is to turn the equation into a tree, e.g.
x + 3 = 3x - 2
Is really the structure
=
/ \
+ -
/ \ / \
x 3 * 2
/ \
3 x
Where each operator describes an operation between two "branches" of the tree. Using a javascript object it shouldn't be difficult to create the structure:
function tree(lterm,op,rterm) {
t.operator = op;
t.left = lterm;
t.right = rterm;
return t;
}
expression = tree("x", "/", tree("x","+",3) ); // x / (x+3)
Then by manipulating the tree you can resolve the equation, or carry out calculations. To evaluate an expression (with no unknowns), you run through the tree starting at the terminals, and upwards from intersection to intersection. You can replace a section of the tree with a result, or annotate it with a result - add a result variable to the tree object.
Here are some useful methods to include in a tree class:
getLeft
getRight
prettyPrint
evaluate
evaluate("x",5) // x=5, now evaluate
...
It's not just linear operations that can be "parsed" this way. Better parsers will have a list of operators that includes =*/+- but also unary operators: - ( ) sin cos...
I haven't used an operator-precedence parser in javascript, but some must exist prewritten. Surely a kind soul on this site will add a good link or two to my answer.
BTW, the tree approach has many applications. In a spreadsheet:
A2 = A1+B1
In a boolean solver:
A = not (B or C)
C = true
In XML parsing:
<main>
<part>A</part>
<part>B</part>
</main>
I have defined two functions:
getTotalX() : It will give you the count of x for any input string.
getTotalScalars() : It will give you the total of scalars (numbers).
And finally, your updated code (it still does only addition and subtraction):
<script>
window.onload = function() {
// Total Xs on each side of equation
// Example problem: 5x + 2 = 10 - 2x
var leftSideXTotal = 0; // 5
var rightSideXTotal = 0; // -2
// Total integers on each side of equation
// Example problem: 5x + 2 = 10 - 2x
var leftSideIntTotal = 0; // 2
var rightSideIntTotal = 0; // 10
// Enter a math problem to solve
var problem = "5x + 2 = 10 - 2x";
// Remove all spaces in problem
// Example problem: 5x + 2 = 10 - 2x
problem = problem.replace(/\s/g,''); // 5x+2=10-2x
// Add + signs in front of all - signs
// Example problem: 5x + 2 = 10 - 2x
problem = problem.replace(/-/gi, "+-"); // 5x+2=10+-2x
// Split problem into left and right sides
// Example problem: 5x + 2 = 10 - 2x
var problemArray = problem.split("=");
var problemLeftSide = problemArray[0]; // 5x+2
var problemRightSide = problemArray[1]; // 10+-2x
leftSideXTotal = getTotalX(problemLeftSide);
leftSideIntTotal = getTotalScalars(problemLeftSide);
rightSideXTotal = getTotalX(problemRightSide);
rightSideIntTotal = getTotalScalars(problemRightSide);
// Compute
var totalXs = (leftSideXTotal - rightSideXTotal)
var totalIntegers = (rightSideIntTotal - leftSideIntTotal)
var solution = (totalIntegers / totalXs)
// Display solution
document.getElementById("divSolution").innerText = solution;
// Find the total number of X in the string
function getTotalX(data) {
data = data.replace(/\s/g,'');
xCount = 0;
if(data.indexOf('x') != -1) {
if (data.indexOf('+') != -1) {
data = data.split('+');
for(var i = 0; i < data.length; i++) {
xCount += getTotalX(data[i]);
}
} else if (data.indexOf('-') != -1) {
data = data.split('-');
// Single negative
if(data[0] == "") {
xCount -= getTotalX(data[1]);
} else {
xCount += getTotalX(data[0]);
for(var i = 1; i < data.length; i++) {
xCount -= getTotalX(data[i]);
}
}
} else {
xCount = parseInt(data.split('x')[0]);
}
}
return xCount;
}
// Find the total of scalars
function getTotalScalars(data) {
data = data.replace(/\s/g,'');
intCount = 0;
if (data.indexOf('+') != -1) {
data = data.split('+');
for(var i = 0; i < data.length; i++) {
intCount += getTotalScalars(data[i]);
}
} else if (data.indexOf('-') != -1) {
data = data.split('-');
// Single negative
if(data[0] == "") {
intCount -= getTotalScalars(data[1]);
} else {
intCount += getTotalScalars(data[0]);
for(var i = 1; i < data.length; i++) {
intCount -= getTotalScalars(data[i]);
}
}
} else {
if(data.indexOf('x') == -1) {
intCount = parseInt(data.split('x')[0]);
} else {
intCount = 0;
}
}
return intCount;
}
}
</script>
Related
So right now, I'm trying to implement a search bar function into my d3.js plot. Right now it doesn't do anything, but that's not the issue at the moment. The problem is that when I type/delete something from the bar, there's visible lag/choppiness in the characters appearing/disappearing. I believe the issue is stemming from my plot. I have 140+ dots moving around the screen, and their position is being interpolated. So from the beginning to the end of the transition, my code has to compute 140 positions thousands of times over.
I've looked into trying to reduce the cardinality of the d3.interpolateNumber function, but it appears that there isn't a third argument to change the number of terms like in a linspace command. Right now I have an array of 1000 numbers for my function to run through, but I don't know how to pass the array to my other functions.
Below are the pertinent functions for this issue. The commented line in tweenPatch is the original code I had that made my code run, but gave my plot computational issues. Variables arr, curr, and step were my attempt to fix the situation, but I haven't been able to figure out how to pass the array into displayPatch().
function tweenPatch() {
var patch = d3.interpolateNumber(1, 26);
var arr = [];
var curr = 1;
var step = (26 - 1) / (1000 - 1);
for (var i = 0; i < 1000; i++) {
arr.push(curr + (step * i));
}
return arr.forEach(function(d) {
console.log(arr[d]);
displayPatch(arr[d]);
});
//return function(t) { displayPatch(t); };
}
function displayPatch(patch) {
dots.data(interpolateData(patch), function(d) { return d.name; }).call(position).sort(order);
var inter = Math.floor(patch);
var seas = 8;
var patc = 1;
if (inter > 24) {
seas = 9;
patc = inter - 24;
} else {
patc = inter;
}
label.text("Patch " + seas + "." + patc);
}
function interpolateValues(values, number) {
old = Math.floor(number);
upd = Math.ceil(number);
var old_data = values.filter(function(d) {return d.internal == old;});
var new_data = values.filter(function(d) {return d.internal == upd;});
var oobj = old_data[0];
var nobj = new_data[0];
var onum = oobj[Object.keys(oobj)[4]];
var nnum = nobj[Object.keys(nobj)[4]];
var difint = number - old;
var difdis = 0;
var newnum = nnum;
if (nnum > onum) {
difdis = nnum - onum;
newnum = ((difint) * difdis) + onum;
} else if (onum > nnum) {
difdis = onum - nnum;
newnum = onum - ((difint) * difdis);
}
return newnum;
}
I believe switching my SVG to a canvas may help things, but since I have no knowledge of canvas I'd rather leave that as a last resort.
It's kind of math problem. I want to fire specific number of setTimeout (the number is based on an array length) in a specific period of time (say, 5 seconds).
The first setTimeout should start at 0 sec. and the last at 5 sec.. All timeouts between should start with an ease-in effect, so that each timeout starts faster.
There's an example which ilustrates what I want to achieve exactly.
I'm struggling around this line:
next += timePeriod/3.52/(i+1);
which works almost perfect in demo example (for any timePeriod), but obviously it doesn't work for a different letters.length as I have used static number 3.52.
How do I calculate next?
var letters = [
'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T'
];
var div = $('#container');
var timePeriod = 5000; // 5 seconds;
var perLetter = timePeriod/(letters.length-1); // it gives equal time between letters
var next = 0;
for(var i=0; i<letters.length; i++){
setTimeout(function(letter){
//div.append('<span class="letter">' + letter + '</span>');
// Used "|" instead of letter, For better redability:
div.append('<span class="letter">|</span>');
}, next, letters[i]);
// Can't find the logic here:
next += timePeriod/3.52/(i+1);
};
///////////////// FOR DEMO: ///////////////
var sec = timePeriod/1000;
var secondsInterval = setInterval(seconds, 1000);
var demoInterval = setInterval(function(){
sec >= 0 || clearInterval(demoInterval);
div.append('\'');
}, 30);
function seconds(){
sec || clearInterval(secondsInterval);
$('#sec').text(sec-- || 'DONE');
}
seconds();
.letter{
color : red;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<span id=container></span>
<span id=sec class=letter></span>
var steps = letters.length;
var target = timePeriod;
function easeOutQuad(t, b, c, d) {
t /= d;
return -c * t*(t-2) + b;
};
var arrayOfTimeouts = new Array(steps);
var n;
var prev = 0;
for(var i = 1; i <= steps; i++){
n = easeOutQuad(i, 0.0, target, steps);
arrayOfTimeouts[i-1] = n-prev;
prev = n;
}
This one should work with any input value.
fiddle
Note that the graph appears to be slightly too fast but I believe that discrepancy to be a product of timing imperfections, as the sum of my array equals the timePeriod exactly.
more on easing equations
Here's a solution based on a geometric series. It's a bit goofy but it works. It generates an array with your timeout values.
Steps = size of your array.
Target = the total time.
var steps = 50;
var target = 5000;
var fraction = 1.5 + steps / 7;
var ratio = (fraction-1) / fraction;
var n = target / fraction;
var sum = 0;
var arrayOfTimeouts = new Array(steps);
for(var i = 0; i < steps; i++){
sum += n;
arrayOfTimeouts[i] = n;
n *= ratio;
}
console.log(arrayOfTimeouts, sum);
So I've been working on re-producing the slider found here https://www.skylight.io/ ( Scroll down to find the price slider ).
So far Ive managed to create something similiar, but some numbers are hard coded, making it difficult to change and not very re-usable.
I've been researching around and I think I need to use Math.log() and Math.exp() together to achieve something like in the link above but I'm not sure.
Heres a jsfiddle of what I have so far https://jsfiddle.net/7wrvpb34/.
I feel that its the maths part of this problem that is halting me I think, so any help would be greatly appreciated.
Javascript code below:
var slider = document.getElementById("slider")
var sliderFill = document.getElementById("slider-fill")
var knob = document.getElementById("knob")
var mouseDown;
var mousePos = {x:0};
var knobPosition;
var minPrice = 20;
var price = 0;
var minRequests = 50;
var requests = 50 + ",000";
var incrementSpeed = 2;
var incrementModifier = 20;
var incrementValue = 1;
var minMillionCount = 1;
var millionCount = 1;
var previousRequestAmount = 0;
document.getElementById("price").innerHTML = price;
document.getElementById("requests").innerHTML = requests;
highlightTable(1);
document.addEventListener('mousemove', function(e) {
if(mouseDown) {
updateSlider(e);
}
})
function updateSlider(event) {
mousePos.x = event.clientX - slider.getBoundingClientRect().left;
mousePos.x -= knob.offsetWidth / 2;
console.log(mousePos.x);
if(mousePos.x < 0) {
knob.style.left = "0px";
sliderFill.style.width = "0px";
price = 0;
requests = 50 + ",000";
document.getElementById("price").innerHTML = price;
document.getElementById("requests").innerHTML = requests;
return
}
if(mousePos.x > slider.offsetWidth - 20) {
return
}
sliderFill.style.width = mousePos.x + 10 + "px";
knob.style.left = mousePos.x + "px";
//Increase requests by using X position of mouse
incrementSpeed = mousePos.x / incrementModifier;
requests = minRequests + (mousePos.x * incrementSpeed);
//Round to nearest 1
requests = Math.round(requests / incrementValue) * incrementValue;
if (requests >= 1000){
var m = requests/ 1000;
m = Math.round(m / 1) * 1;
//Problem, lower the modifier depending on requests
incrementModifier = 20 * 0.95;
document.getElementById("requests").innerHTML = m + " million";
//Adjust Prices
if(( requests >= 1000) && (requests < 10000)) {
var numOfMillions = requests / 100;
//Round to closest 10.
//10 * number of millions
var rounded = Math.round(numOfMillions / 10) * 10;
price = minPrice + rounded;
highlightTable(3);
}
//Adjust Prices
if(requests >= 10000) {
var numOfMillions = requests / 1000;
var rounded = Math.round(numOfMillions / 1) * 1;
var basePrice = minPrice * 6;
price = basePrice + rounded;
highlightTable(4);
}
} else {
incrementModifier = 20;
document.getElementById("requests").innerHTML = requests + ",000"
if(requests < 100) {
highlightTable(1);
price = 0;
} else {
highlightTable(2);
price = 20;
}
}
previousRequestAmount = requests;
document.getElementById("price").innerHTML = price;
}
knob.addEventListener('mousedown', function() {
mouseDown = true;
});
document.addEventListener('mouseup', function() {
mouseDown = false;
});
function highlightTable(rowNum) {
var table = document.getElementById("payment-table")
for(var i = 0; i < table.rows.length; ++i) {
var row = table.rows[i]
if(i == rowNum) {
row.style.background = "grey"
} else {
row.style.background = "white";
}
}
}
Thank you for your time.
If you want it to be reusable you need to create a mathematical function that assigns a result to the number of requests. I will give you a very easy example.
If you want a different result for 1,10,100,100,10000 etc
var d = Math.log10(requests);
if(d<1){
doSomething();
}else if(d<2){
doSomethingElse();
} //etc
This way if you want to change the specific values that create certain results, all you need to do is change the function.
This only works if your tiers of requests follow a math function, if they don't you need to hard code it.
However if say they don't follow a math function, but you know how you would like them to change based on a value then you can do this.
var changingValue = 4;
if(requests < 400*changingValue){
doSomthing();
}else if(requests <= 400*changingValue*changingValue){
doSomethingElse();
}else{// the requests is greater than any of the above
doTheOtherThing();
}
Edit:
For the second one you need to make sure that each condition if always larger than the other from top to bottom.
The description "increasingly increasing" matches an arbitrary number of functions. I assume you also want it to be continuous, since you already have a non-continuous solution.
TL;DR
Use an exponential function.
Generic approach
Assuming imin and imax are the minimal and maximal values of the slider (i for input) and omin and omax are the minimal and maximal values to be displayed, the simplest thing I can think of would be a multiplication by something based on the input value:
f(x)
{
return omin + (omax - omin) * g((x - imin) / (imax - imin));
}
This will pass 0 to g if x == imin and 1 if x == imax.
The return value r of g(y) should be
r == 0 for y == 0
r == 1 for y == 1
0 < r < y for 0 < y < 1
The simplest function that I can think of that fulfills this is an exponential function with exponent > 1.
An exponent of 1 would be a linear function.
An exponent of 2 would be make the middle of the slider display one fourth of the maximum price instead of half of it.
But you really need to find that exponent yourself, based on your needs.
I've written the following Javascript:
(function () {
var computationModule = (function foo1(stdlib, foreign, heap) {
"use asm";
var sqrt = stdlib.Math.sqrt,
heapArray = new stdlib.Int32Array(heap),
outR = 0.0,
outI = 0.0;
function computeRow(canvasWidth, canvasHeight, limit, max, rowNumber, minR, maxR, minI, maxI) {
canvasWidth = +canvasWidth;
canvasHeight = +canvasHeight;
limit = +limit;
max = max | 0;
rowNumber = +rowNumber;
minR = +minR;
maxR = +maxR;
minI = +minI;
maxI = +maxI;
var columnNumber = 0.0,
zReal = 0.0,
zImaginary = 0.0,
numberToEscape = 0;
var columnNumberInt = 0;
// Compute the imaginary part of the numbers that correspond to pixels in this row.
zImaginary = +(((maxI - minI) * +rowNumber) / +canvasHeight + minI);
// Iterate over the pixels in this row.
// Compute the number of iterations to escape for each pixel that will determine its color.
for (columnNumber = +0; + columnNumber < +canvasWidth; columnNumber = +(+columnNumber + 1.0)) {
// Compute the real part of the number for this pixel.
zReal = +(((maxR - minR) * +columnNumber) / +canvasWidth + minR);
numberToEscape = howManyToEscape(zReal, zImaginary, max, limit) | 0;
columnNumberInt = columnNumberInt + 1 | 0;
heapArray[(columnNumberInt * 4) >> 2] = numberToEscape | 0;
}
}
// Function to determine how many iterations for a point to escape.
function howManyToEscape(r, i, max, limit) {
r = +r;
i = +i;
max = max | 0;
limit = +limit;
var j = 0,
ar = 0.0,
ai = 0.0;
ar = +r;
ai = +i;
for (j = 0;
(j | 0) < (max | 0); j = (j + 1) | 0) {
iteratingFunction(ar, ai, r, i)
ar = outR;
ai = outI;
if (+(ar * ar + ai * ai) >= +(limit * limit))
return j | 0;
}
return j | 0;
}
// The iterating function defining the fractal to draw
// r and i are the real and imaginary parts of the value from the previous iteration
// r0 and i0 are the starting points
function iteratingFunction(r, i, r0, i0) {
r = +r;
i = +i;
r0 = +r0;
i0 = +i0;
computePower(r, i, 2);
// Set the output from this function to t
outR = +(r0 + outR);
outI = +(i0 + outI);
}
// Compute the result of [r,i] raised to the power n.
// Place the resulting real part in outR and the imaginary part in outI.
function computePower(r, i, n) {
// Tell asm.js that r, i are floating point and n is an integer.
r = +r;
i = +i;
n = n | 0;
// Declare and initialize variables to be numbers.
var rResult = 0.0;
var iResult = 0.0;
var j = 0;
var tr = 0.0;
var ti = 0.0;
// Declare and initialize variables that will be used only in the
// event we need to compute the reciprocal.
var abs = 0.0;
var recr = 0.0;
var reci = 0.0;
if ((n | 0) < (0 | 0)) {
// For n less than 0, compute the reciprocal and then raise it to the opposite power.
abs = +sqrt(r * r + i * i);
recr = r / abs;
reci = -i / abs;
r = recr;
i = reci;
n = -n | 0;
}
rResult = r;
iResult = i;
for (j = 1;
(j | 0) < (n | 0); j = (j + 1) | 0) {
tr = rResult * r - iResult * i;
ti = rResult * i + iResult * r;
rResult = tr;
iResult = ti;
}
outR = rResult;
outI = iResult;
} // end computePower
return {
computeRow: computeRow
};
})(self, foreign, heap);
// Return computationModule that we just defined.
return computationModule;
})();
There's nothing particularly unusual about this Javascript, except that I want to make my web application display the Javascript in an ACE text editor (http://ace.c9.io/) so that the user can modify the code at runtime.
I load the Javascript using jQuery AJAX and then set the contents of the ACE Editor to the Javascript code. After the user modifies the code, he can click a button to run the code. (This uses eval)
The problem I'm having is that ACE is displaying strange characters instead of spaces.
Oddly enough, if I try to copy text from the ACE editor, the strange characters disappear and the spaces are normal. Furthermore, the code runs fine even with these strange characters being displayed.
I also noticed that the problem does not appear when using Firefox, but it appears for Chrome and IE 11.
Finally, the problem only occurs when I put the code on a real web server. It doesn't reproduce in my development environment.
Looking at this some more, I can see that it's not just the text I'm loading via AJAX. Even when I type new spaces, more text characters appear!
What could be going wrong so that the text doesn't display properly?
Here's a link to the application: http://danielsadventure.info/html5fractal/
Use charset="utf-8" in the script tag where you include ace.
<script src="path/to/ace.js" charset="utf-8">
This may have something to do with this:
When no explicit charset parameter is provided by the sender, media
subtypes of the "text" type are defined to have a default charset
value of "ISO-8859-1" when received via HTTP. Data in character sets
other than "ISO-8859-1" or its subsets MUST be labeled with an
appropriate charset value. See section 3.4.1 for compatibility
problems.
http://www.w3.org/Protocols/rfc2616/rfc2616-sec3.html#sec3.7.1
So the string passed to the script are in an encoding different than what ACE (or JS in general) expects, which is UTF-8.
I have a variable that has a number between 1-3.
I need to randomly generate a new number between 1-3 but it must not be the same as the last one.
It happens in a loop hundreds of times.
What is the most efficient way of doing this?
May the powers of modular arithmetic help you!!
This function does what you want using the modulo operator:
/**
* generate(1) will produce 2 or 3 with probablity .5
* generate(2) will produce 1 or 3 with probablity .5
* ... you get the idea.
*/
function generate(nb) {
rnd = Math.round(Math.random())
return 1 + (nb + rnd) % 3
}
if you want to avoid a function call, you can inline the code.
Here is a jsFiddle that solves your problem : http://jsfiddle.net/AsMWG/
I've created an array containing 1,2,3 and first I select any number and swap it with the last element. Then I only pick elements from position 0 and 1, and swap them with last element.
var x = 1; // or 2 or 3
// this generates a new x out of [1,2,3] which is != x
x = (Math.floor(2*Math.random())+x) % 3 + 1;
You can randomly generate numbers with the random number generator built in to javascript. You need to use Math.random().
If you're push()-ing into an array, you can always check if the previously inserted one is the same number, thus you regenerate the number. Here is an example:
var randomArr = [];
var count = 100;
var max = 3;
var min = 1;
while (randomArr.length < count) {
var r = Math.floor(Math.random() * (max - min) + min);
if (randomArr.length == 0) {
// start condition
randomArr.push(r);
} else if (randomArr[randomArr.length-1] !== r) {
// if the previous value is not the same
// then push that value into the array
randomArr.push(r);
}
}
As Widor commented generating such a number is equivalent to generating a number with probability 0.5. So you can try something like this (not tested):
var x; /* your starting number: 1,2 or 3 */
var y = Math.round(Math.random()); /* generates 0 or 1 */
var i = 0;
var res = i+1;
while (i < y) {
res = i+1;
i++;
if (i+1 == x) i++;
}
The code is tested and it does for what you are after.
var RandomNumber = {
lastSelected: 0,
generate: function() {
var random = Math.floor(Math.random()*3)+1;
if(random == this.lastSelected) {
generateNumber();
}
else {
this.lastSelected = random;
return random;
}
}
}
RandomNumber.generate();