Can you explain this method of finding prime numbers in javascript 
It's called the Sieve of Erathestenes and it's an efficient way of finding
all prime numbers between zero and some upper limit integer.
j = i << 1
This is a bit shift operation. It shifts an integer 1 bit to the left.
Because binary is base two, this is equivalent to multiplying by two. In
decimal, the equivalent operation would be shifting, say, 1 to the left and
bringing in a zero to the right. It's base ten, so you end up with 10 (ten
times one.)
I don't believe the implementation you show is optimal. The limit to the i
value can be lower  something like sqrt(max).
You can easily find some really nice animations online of this algorithm
that show you what's going on in quite an intuitive way.

Finding number of kprime numbers; 
The algorithm you are using is know as Sieve of Eratosthenes. It is a well
known algorithm for finding prime numbers. Now to answer your questions :
1a) What is the complexity of this code
The complexity of your code is O(n log(log n)).
For and input of a and b the complexity of your code is O(b log log b). The
runtime is due to the fact that you first mark b/2 number then b/3 then b/5
and so on. So your runtime is b * (1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... +
1/prime_closest_to_b). What we have there is a prime harmonic series which
grows asymptotically as ln(ln(b+1)) (see here).
Asymptotically the upper bound is:
O(b * (1/2 + 1/3 + 1/5 + 1/7 +..)) = O(b) * O(log(log(b+1))) =
O(b*log(log(b))
1b) Can it be done in O(n)
This is tricky. I would say that for all practical purposes an O(n

MASM code for finding prime numbers not working 
Managed to fix it. Because of my poor placement of debug breakpoints I
didn't realize that the edx register is being used properly. However, two
other problems were present in my code.
Working code is below. Notice the new jmp command right before the prime:
tag, and the new reset of testNum before the return.
isPrime PROC; checks if a number is prime by dividing out values less than
it until it finds one that has a remainder of 0
;or the test value reaches the number
retest:
inc testNum
mov eax, testPrime
cmp eax, testNum
je prime
mov edx, 0
mov eax, testPrime
mov ebx, testNum
div ebx
mov eax, edx
cmp eax, ZERO
jg retest
mov eax, 1
mov testNum, eax
inc testPrime
jmp retest
prime:
mov eax, 1
mov testNum, eax
ret
isPrime ENDP

Prime numbers not printing in Javascript 
The function primes is written to expect an array, but you're passing it an
integer.
Did you mean fmt(primes(primesarray(k)))?
(That does at least print a list of numbers, but I'm afraid many of them
are not primes!)

Javascript prime number error? 
You are returning the function from inside the forloop.
So it never hits the other statements
Check Fiddle

Javascript: find first n prime numbers 
In your code, array.push[i] (with square brackets) doesn't do what you'd
like it to. It leaves the array unchanged and returns undefined.
You meant array.push(i) (with parentheses).

How to Pass JSON object to GOJS javascript file from Java/Prime Faces 
You can request data from server using an AJAX call and get the data. if
that is not an option then you can put the data in a hidden field on the
page and then access it in your load() function, like you are doing it in
your save() function using document.getElementById.

Why is my program giving me all the nonprime numbers when I ask for prime numbers? 
The problem is your loop:
while (next_prime + 1 < MAXBYTES*8 && getBit(bit_arr,
++next_prime))
{
print_prime(next_prime);
}
Your keeping printing things while the bit is set (i.e. while you know it's
not a prime). So basically, your loop is "print all the numbers I find
while looking for the next prime" rather than "look for the next prime in a
loop, then print the next prime".
I suspect you want something like:
next_prime++; // We always want to at least move on once...
while (next_prime + 1 < MAXBYTES*8 && getBit(bit_arr,
next_prime))
{
next_prime++;
}
print_prime(next_prime);
I haven't checked whether that's all that was wrong with the code, but it's
certainly an initial thing to fix.

Finding the javascript that changes a DOM element 
The approach I use in these situations is to examine the HTML around the
area that is being modified, note all possible ways that code could find
the appropriate DOM elements (form names, id values, class names, etc...)
and then look through the code to find where it might be querying the DOM
to find the DOM element that is being changed using one of these
identifiers. Since the identifiers can't be obscured, they should be in
the code in normal English the same as they appear in the HTML.
In addition, you can make a list of all event listeners that are being set
in the code and pay particular attention to event listeners on any objects
near the one being changed. Since it's a form submission, you can look for
the submit event or click event on a form submission button.
When you see ev

Finding the value of all H2s on a page with Javascript 
If you're using jQuery, you could run something like this
$('h2').each({
/* function */
});
Then to append to the .nav container you can run
$('h2').each(function() {
$('.nav').append($(this).text())
});

Finding properties in a javascript object 
If you restructure your JSON to nest the options/result inside the
respective parent it becomes easy to get all the possible options. You
would need to change your code to this:
$.ajax({
url: "http://www.myurl.com/jsonp.php",
type: "GET",
dataType: "jsonp",
jsonpCallback: "otmjsonp",
async: false,
success: function (JSON) {
$(".result").on("click", function () {
var currentResult = $(this).text(); //.result is the line of HTML
the user has clicked
if (JSON.hasOwnProperty(playerSelection)) {
for (var outcome in JSON[playerSelection]) {
if (JSON[playerselection].hasOwnProperty(outcome)) {
alert("This is the next outcome " +
JSON[playerSelection][outcome]);
}
}
}
})
}
});

javascript finding center of a rotated rectangle 
In the situation you describe when there is a problem, the roles of the two
dots are reversed: you HAVE the lower one and want to find the upper. Add
an if statement to detect this condition, and write a version of your code
that recognizes this reversal.

Finding the min first value of a list of pairs of numbers in JavaScript 
As Sirko said, you can do that with a simple for loop, comparing each first
value with its previous value. You can also do that with the reduce1 array
method, to achieve shorter code:
var tuples = [[1,2], [2,4], [3,5], [5,2]];
var tupleWithMinFirstValue = tuples.reduce(function(previous, current){
return current[0] < previous[0] ? current : previous;
}, [Infinity]);
console.log(tupleWithMinFirstValue);
http://jsfiddle.net/wAGMX/
1 The link also provides compatibility info, and a polyfill.

Finding inner matches using Regular Expressions in Javascript 
You can use this:
var regex = /loanId=([^&"'s]+)/;
var match = regex.exec(yourString);
console.log(match[1]);
the result is in the first capturing group. Since javascript regex doesn't
have the lookbehind feature, this is the only way.

Finding the root of a javascript function on a website 
I suggest to use Firefox Firebug. Open Console and just type "poster". It
will output function signature. You can hover on it to find out file or
click on it to navigate to Script of Firebug. Check out screenshots.
https://blog.gaurangjadia.com/?attachment_id=835
https://blog.gaurangjadia.com/?attachment_id=836
Also, you can put breakpoints and debug your scripts. It is nice and
powerful web development tool.
Example is at http://code.gaurangjadia.com/stackoverflow/18551051/

Event object in Javascript, trouble finding information 
Official W3c reccomendation
clientX and clientY are the official event property handlers you are
looking for. Although you might want to consider screenX and screenY too.
Events (API)
Events
Creating, triggering events
Event handlers
Orientation and motion data
Window Event X and Y property
Explanation
event.x and event.y, what are those?
First of all, the x and y properties are not in all events. It is relative
on the event triggered.
Here is an example:
document.body.onclick = function(){console.log(window.event.x)};
If you paste and execute that code in your browser's console, it will log
you the x position of the cursor each time you click.
Although taking a look at this example:
document.body.onkeyup = function(){console.log(window.event.x)};
The console will

Javascript  Regex finding multiple parentheses matches 
Your regex has anchors to match beginning and end of the string, so it
won't suffice to match multiple occurrences. Updated code using
String.match with the RegExp g flag (global modifier):
var userIn = 'Cu(NO3)2(CO2)3';
var inPar = userIn.match(/([^)]*)/g).map(function(s){ return s.substr(1);
});
inPar; //["NO3)", "CO2)"]
In case you need old IE support: Array.prototype.map polyfill
Or without polyfills:
var userIn = 'Cu(NO3)2(CO2)3';
var inPar = [];
userIn.replace(/(([^)]*))/g, function(s, m) { inPar.push(m); });
inPar; //["NO3)", "CO2)"]
Above matches a ( and captures a sequence of zero or more non) characters,
followed by a ) and pushes it to the inPar array.
The first regex does essentially the same, but uses the entire match
including the opening ( parenthesis (which is late

Regex for finding Java's format strings in JavaScript 
The problem lies in this group: (.[09]*)
You expect a % at first then the . in the above group is expecting one of
something (not optional). Then
(diufFeEgGxXosScCaAbBhHpn%) is also expecting
something.
So basically the only required items in your regex are:
1. %
2. the . in (.[09]*)
3. (diufFeEgGxXosScCaAbBhHpn%)
There isn't room in %s to accommodate the % and both the . and the chunk
you are wanting to pull the s.
I'm not sure exactly what you want (.[09]*) to be, but making it optional
like (.[09]*)? would atleast allow for %s to pass.

Google Maps  Finding all the markers inside a given radius Javascript/Php 
Here is an example of how you could solve this with the geocoding API and
some simple geometry.
(Note that I have hardcoded the address and radius for brevity.)
// we assume that you have an array of markers
var markers = [];
//In order to lookup the the position of a zipcode you can use the
geocoding API:
// https://developers.google.com/maps/documentation/geocoding/
var geocode_api_base_url =
"http://maps.googleapis.com/maps/api/geocode/json?";
var params = {
adress : 05673,
components : "country:us",
sensor : false
}
// This is the result set of markers in area
var in_area = [];
//
http://maps.googleapis.com/maps/api/geocode/json?address=05673&components=country:US&sensor=false
$.getJSON( geocode_api_base_url + $.param(params), function(data) {
var loca

Regular expression for finding JavaScript arrays in html document 
A pattern like this should work:
^var w+ = new Array(.*);$
This will match the beginning of the string (or line in multiline mode),
followed a literal var, followed by one or more 'word' characters followed
by a literal = new Array( followed by zero or more of any characters,
followed by a literal ); and the end of the string (or line in multiline
mode).

Summation of nth prime 
OK, so I'm going to have a shot at this without saying too much because it
looks like it might be a homework problem.
My best guess is that a lot of people have shied away from even looking at
your code because it's a bit too messy for their level of patience. It's
not irredeemable, but you would probably get a better response if it were
considerably cleaner.
So, first, a few critical comments on your code to help you clean it up: it
is insufficiently wellcommented to make your intentions clear at any
level, including what the overall purpose of the program is; it is
inconsistently indented and spaced in an unconventional manner; and your
choice of variable names leaves something to be desired, which is
exacerbated by the absence of comments on variable declarations.
You should compile

Prime factorization  list 
You can use sieve Of Eratosthenes to generate all the primes up to (n/2) +
1 and then use a list comprehension to get all the prime factors:
def rwh_primes2(n):
#
http://stackoverflow.com/questions/2068372/fastestwaytolistallprimesbelowninpython/3035188#3035188
""" Input n>=6, Returns a list of primes, 2 <= p < n """
correction = (n%6>1)
n = {0:n,1:n1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+11
sieve[ ((k*k)/3) ::2*k]=[False]*((n/6(k*k)/61)/k+1)
sieve[(k*k+4*k2*k*(i&1))/3::2*k]=[False]*((n/6(k*k+4*k2*k*(i&1))/61)/k+1)
return [2,3] + [3*i+11 for i in xrange(1,n/3correction) if sieve[i]]
def primeFac

Loops and Prime Numbers 
Use this method to check if a given int is a prime.
public static boolean isPrime(int a)
{
if ( a == 2)
return true;
int midpoint = Math.round(a/2);
for(int i = 2; i < midpoint; i++)
{
if(a % i == 0)
return false;
}
return true;
}
Explanation:
Loop through all the numbers until midpoint and modulus until you encounter
0 or not. If you encounter 0 then return false because we know that it is
not prime, if we encounter no zero then we return true because it is prime.
We loop until midpoint because there is no need to loop further.
You can implement it in your loop via
for (int i = 2; i < n ; i++)
{
if (isPrime(i))
{
System.out.println(i);
}
}

I need Prime font for my web site 
final result:
<style type="text/css">
@fontface {
fontfamily: 'primelight';
src: url('prime_lightwebfont.eot');
src: url('prime_lightwebfont.eot?#iefix') format('embeddedopentype'),
url('prime_lightwebfont.woff') format('woff'),
url('prime_lightwebfont.ttf') format('truetype'),
url('prime_lightwebfont.svg#primelight') format('svg');
fontweight: normal;
fontstyle: normal;
}
body {
fontfamily: "primelight", Verdana, Tahoma;
}
</style>
go to http://fontfabric.com/primefreefont/
and download the font
creat a repository in your website : fonts/prime/
put the prime files inside make sure you ve something like this
:PrimeLight.otf
alternative :
download the font
convert it with all op

How to count coprime to n in a range [1,x] where x can be different from n? 
You can use InclusionExclusion Principle
Find the unique prime factors of N (they cannot be more than 1012,
Considering N and X <=10^10).
Now you can find the number of numbers <=x and divisible by 'y' just by
dividing. Try all combination of factors of n for y (you will get only 2^10
(1024) in worst case).
Use Inclusion Exclusion now to find the coprimes of n less than x.
The idea is that if a number is not coprime to n, then it will have
at least one prime factor common with n.
For our example here lets consider X=35 and N=30
First find the unique prime factor of the number. (their number must not
be greater than 1012). Unique Prime factor of N ={2,3,5}.
Find the product of each factor PAIR. {2x3, 2x5, 3x5 or 6, 10, 15}.
Find the product of each factor TRIPLET: {

Not getting value on selection of row using prime faces 
You need to set the rowKey attribute on the <p:dataTable/> to
properly enable row selection. The rowKey attribute should be set to a
unique identifier for the entity in your datatable, like the primary key
attribute. From your sample, it'll look something like:
<p:dataTable value="#{userManagedBean.dataModel}"
rowKey="#{user.userId}" var="user" pagination="true" rows="5"
selection="#{userManagedBean.selectedUser}" selectionMode="single">

Why does this prime test work? 
Let's start with the first four lines of the function's code:
def isprime(n):
if n == 2: return True
if n == 3: return True
if n % 2 == 0: return False
if n % 3 == 0: return False
The function tests to see if n is equal to 2 or 3 first. Since they are
both prime numbers, the function will return True if n is equal to either.
Next, the function tests to see if n is divisible by 2 or 3 and returning
False if either is true. This eliminates an extremely large amount of cases
because half of all numbers above two are not primes  they are divisible
by 2. The same reason applies to testing for divisibility by 3  it also
eliminates a large number of cases.
The trickier part of the function is in the next few lines:
i = 5
w = 2
while i * i <= n:
if n % i == 0:

Prime number C++ program 
i can never be equal to a  1  you're only going up to b  1. b being
a/2, that's never going to cause a match.
That means your loop ending condition that would return 1 is never true.
In the case of a prime number, you run off the end of the loop. That
causes undefined behaviour, since you don't have a return statement there.
Clang gave a warning, without any special flags:
example.cpp:22:1: warning: control may reach end of nonvoid function
[Wreturntype]
}
^
1 warning generated.
If your compiler didn't warn you, you need to turn on some more warning
flags. For example, adding Wall gives a warning when using GCC:
example.cpp: In function ‘bool primetest(int)’:
example.cpp:22: warning: control reaches end of nonvoid function
Overall, your primechecking loop is

Why does my prime generator break down? 
You need to set divided_pass back to 0 for every new i.
def find_primes(limit):
prime_holder = [2, 3, 5 ,7]
for i in range(11, 20000):
divided_pass = 0
for j in range(0, len(prime_holder)):
if i%prime_holder[j] != 0:
divided_pass += 1
if divided_pass == len(prime_holder):
prime_holder.append(i)
if len(prime_holder)1 == limit:
break
return prime_holder
>>> print find_primes(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149,
151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233]
P.S  Lists are iterable in Python, so you don't need to do for j in
range(len(...)

Can someone explain this prime list generator for me? 
looks like it's just array notation you're looking for.
[a] * 5 just becomes [a,a,a,a,a]
if sieve[i/2] is checking if the value of sieve at i/2 is True or False
and the :: defines the stride.
see this answer.

Checking for prime number  C# logic 
1 is a factor of every number, so you shouldn't check it. Start at 2. Also,
you're already looping from 2 to sqrt(num), so there's no way for i to be
equal to num.

Prime Number Simple Test 
Couple of problems in your original code:
You are doing only one iteration
You declare the array holding divisions inside the loop, causing it to
always have maximum 1 item.
Quick fix of the above would be:
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var arr2 = [];
for (var i = 0; i < divisor.length; i++) {
if(n%divisor[i] == 0) {
arr2.push(i);
}
}
return arr2.length <=1;
}
Live test case.
Optimized code that does not iterate through the whole list of divisors if
not reuired (guess that's what you were trying to achieve) is:
var primetest = function(n){
var divisor = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var arr2 = [];
for (var i = 0; i < divi

Java API for Prime Number and Factorization 
I suggest Commons Math. You can find lib here API here. The class that
would be interesting for you is Primes.

Prime Numbers in Java  Algorithms 
What you can do is find the lowest divisor of Tn. Suppose that is p, find
the lowest divisor again for Tn/p and so on.
Now, at every step p is prime[explanation below]. So collect them and they
are the prime divisors of Tn.
For better timecomplexity, you can check for divisors up to upto
ceil(sqrt(Tn)) only, instead of Tn1.
And when you start checking for prime divisor for Tn, you can start with 2.
And once you get a prime divisor p don't start again from 2 for Tn/p.
Because, Tn/p is also a divisor of Tn and since Tn does not have divisors
less than p, Tn/p does not have it too. So start with p again[p can have
multiple power in Tn]. If p does not divide Tn, move to p+1.
Example :
Tn = 45
1. start with 2. 2 does not divides 45.
2. next test is for 3. 45 is divisible by 3. So 3 is a

Python sieve prime numbers 
for n in xrange(i,limit,i):
a[n]=False
It should be:
for n in xrange(2*i,limit,i):
a[n]=False
Or, more efficiently:
for n in xrange(i*i,limit, 2*i): #assuming you already cleared even numbers
a[n]=False
Because otherwise you end up setting i as nonprime, when it actually is a
prime.
(The sieve should mark all multiples of i as nonprimes, except i itself!)
Note that you are iterating over all the numbers, up to limit, but you can
safely stopped after reaching sqrt(limit):
import itertools as it
def primesieve(limit):
a = [True] * limit
a[0] = a[1] = False
sqrt_limit = int(limit**0.5) + 1
predicate = lambda x: x[0] <= sqrt_limit
for i, isprime in it.takewhile(predicate, enumerate(a)):
if isprime:
for n in xrange(i*i, limit

find Prime numbers up to X  complexity 
If you are using the following code, the time complexity is O(x √x).
int[] p = new int[x];
for (int i = 0; i < p.length; i++) {
p[i] = i+1;
}
for (int i = 4; i <= p.length; i++) {
for(int j = 2; j <= Math.sqrt(i) ; j += 1) {
if(i%j==0) {
p[i1] = 0;
}
}
}

Determining Prime Numbers Java 
Your problem lies with this line:
if (isPrime = true){
You made an assignment, instead of comparing to true, so the statement is
always true.
Use == to compare boolean values, or better yet, since isPrime is already a
boolean:
if (isPrime){

Compile time prime checking 
You can take a look at constexpr. It has a lot friendlier syntax than
template meta programming ( at least if you're unfamiliar with templates
like me. ) You can't use if's, or any loops. But with recursion and the
tenary opeartor you can do pretty much everything you can with template
meta programming, and it usually runs faster too.
http://cpptruths.blogspot.no/2011/07/wantspeeduseconstexprmeta.html
Here's a working example using an online compiler :
http://coliru.stackedcrooked.com/view?id=6bc10e71b8606dd2980c0c5dd982a3c06fbdb8a7476ab90c2bd2503cd4005881
Since it is executed at compile time, you can do a static assert to test
it.
static_assert(is_prime_func(x), "...");
The assert will fail if x is not a prime, meaning compilation will fail. If
x is a prime, the compilation

Prime Faces error Cannot add the same component twice 
I solved this issue by setting the transient option=true. Because I was
creating menus dynamically which resulted in "WARNING: Unable to save
dynamic action with clientId 'j_idt8:j_idt9:0:j_id3' because the
UIComponent cannot be found" and "Cannot remove the same component twice:
j_idt8:j_idt9:j_id3" problems.
public void setMenus(String type)
{
MenuItem item;
item=new MenuItem();
item.setValue("Change Password");
item.setStyle("color:black");
item.setTransient(true); /* Set this to solve the problem */
item.setUrl("/adminAccount/changepassword.xhtml");
submenus.addMenuItem(item);
}

Setting a counter using prime faces 
You don't want to have a separate table  do you mean a table in the
database? (very vague formulated question, but I think I know what you
mean). If so, add a numeric field of your choise to your Locality entity:
@Entity
public class Locality {
private String name;
private int counter;
public voidincreaseCounter() {
counter++;
}
public int getCounter() {
return this.counter;
}
}
... and when the search has happened increase the counter from that
locality:
public void searchPerformed(String searchString) {
String locality = ....// parse the search string
Locality locality = localityRepository.getLocality(localityName);
if (locality == null) {
locality = new Locality(localityName);
}
locality.increaseCounter();
System.out.println("New counter for " +
