Develop Reference

Trigonometric Interpolation returns NaN

I'm a musician, who's new to programming. I use JavaScript inside Max Msp (hence the bang() and post() functions) to create a trigonometric interpolation, interpolating between given equidistant points (for testing, only values of sine from [0, 2π) and returning values from the same points). When I run the code, it returns NaN, except for x = 0, as my tau() function returns only 1 in this special case. Could it be, that it has something to do with summing Math.sin results?
var f = new Array(9);
var TWO_PI = 2*Math.PI;
bang();
function bang() {
for(var i = 0; i < f.length; i++) {
f[i] = Math.sin(i/f.length*TWO_PI);
//post("f[" + i + "]: " + Math.round(f[i]*1000)/1000 + "\n");
}
var points = new Array(f.length);
for(var i = 0; i < points.length; i++) {
var idx = i/points.length*TWO_PI;
points[i] = [i, p(idx)];
//post("p[" + points[i][0] + "]: " + Math.round(points[i][1]*1000)/1000 + "\n");
}
console.log("p(2): " + p(2/points.length*TWO_PI) + "\n");
}
function p(x) {
var result = 0;
for(var k = 0; k < f.length; k++) {
result += f[k]*tau(k, x);
}
return result;
}
function tau(k, x) {
var dividend = sinc(1/2*f.length*(x-k/f.length*TWO_PI));
var divisor = sinc(1/2*(x-k/f.length*TWO_PI));
var result = dividend/divisor;
if(f.length%2 == 0) result *= Math.cos(1/2*(x-k/f.length*TWO_PI));
if(x == 0) return 1;
return result;
}
function sinc(x) {
return Math.sin(x)/x;
}

In your tau function, if x equals k / f.length * TWO_PI (which it will since x is multiples of 1 / points.length * TWO_PI) your sinc function divides by 0, making divisor equal to NaN, which then propagates.

You have to be a bit careful in implementing sinc to avoid dividing by 0. One way is to say that if x is small enough we can replace sin(x) by the first few terms of its taylor series, and all the terms are divisible by x.
I don't know javascript but here is the function in C in case it is of use
#define SINC_EPS (1e-6)
// for small x,
// missing sinc terms start with pow(x,4)/120, and value close to 1
// so the error too small to be seen in a double
double sinc( double x)
{ if ( fabs(x) < SINC_EPS)
{ return 1.0 - x*x/6.0;
}
else
{ return sin(x)/x;
}
}

Generating alphanumerical sequence javascript

I have written a terribly slow function for generating codes that go from AA000 to ZZ999 (in sequence not random). And I have concluded that there has to be a better way to do this. Any suggestions on how to make this faster?
function generateAlphaNumeric(){
theAlphabet = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
resultArrray = [];
resultArrray2 = [];
teller = 0;
for(i in theAlphabet){
for(x in theAlphabet){
resultArrray[teller] = theAlphabet[i] + theAlphabet[x];
teller++;
}
}
teller = 0;
for(x = 0; x<10; x++){
for(y = 0; y<10; y++){
for(z = 0; z<10; z++){
resultArrray2[teller] = x.toString() + y.toString() +z.toString();
teller++;
}
}
}
teller = 0;
finalArray = [];
for(index in resultArrray){
for(i in resultArrray2){
finalArray[teller] = resultArrray[index] + resultArrray2[i];
teller++;
}
}
//console.log(resultArrray);
//console.log(resultArrray2);
console.log(finalArray);
}

This should be considerably faster:
var theAlphabet = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O',
'P','Q','R','S','T','U','V','W','X','Y','Z'];
var theDigits = ['0','1','2','3','4','5','6','7','8','9'];
var result = [];
for (var i=0 ; i<26 ; i++) {
var prefix1 = theAlphabet[i];
for (var j=0 ; j<26; j++) {
var prefix2 = prefix1 + theAlphabet[j];
for(var x = 0; x<10; x++){
var prefix3 = prefix2 + theDigits[x];
for(var y = 0; y<10; y++){
var prefix4 = prefix3 + theDigits[y];
for(var z = 0; z<10; z++){
result.push(prefix4 + theDigits[z]);
}
}
}
}
}
Key ideas:
Generate everything in one run
Reuse partial strings as much as possible
However, I don't see how such an exhaustive list is useful. There are exactly 26 * 26 * 1000 different codes. So instead of maintaining an array with all codes it could make sense to simply build a function that generates the specific code requested:
function getCode(number) {
var z = number % 10;
number -= z; number /= 10;
var y = number % 10;
number -= y; number /= 10;
var x = number % 10;
number -= x; number /= 10;
var a = number % 26;
number -= a; number /= 26;
var b = number;
return theAlphabet[a] + theAlphabet[b] + theDigits[x] + theDigits[y] + theDigits[z];
}

function generate() {
var str = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ',
array = [];
for (var i = 0; i < str.length; i++) {
for (var j = 0; j < str.length; j++) {
for (var k = 0; k < 10; k++) {
for (var l = 0; l < 10; l++) {
for (var m = 0; m < 10; m++) {
ar.push(str[i] + str[j] + k + l + m);
}
}
}
}
}
return array;
}
console.log(generate());
This will generate a array of all the codes .. U can save that array and parse it easily using a loop.

Try this solution:
function generate() {
var str = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ',
ar = [];
for (var index1 = 0; index1 < str.length; index1++) {
for (var index2 = 0; index2 < str.length; index2++) {
for (var index3 = 0; index3 < 1000; index3++) {
ar.push(str[index1] + str[index2] + ('000' + index3).slice(-3));
}
}
}
return ar;
}
console.log(generate());

I didn't test it, but it should do the trick
function generateAlphaNumeric()
{
var theAlphabet = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
var result = [];
// Will take a random letter inside theAlphabet
// Math.floor(Math.random() * theAlphabet.length) will generate a random number between 0 and 25
var i = 0;
while(i<2)
{
var letter = theAlphabet[Math.floor(Math.random() * theAlphabet.length)];
result.push(letter);
i++;
}
i = 0;
while(i<3)
{
// Adds a random number between 0 and 9
result.push(Math.floor(Math.random() * 10));
i++;
}
return result;
}

From a computational complexity perspective, unfortunately this is the best you can do. From a sheer number of instructions perspective, you can do a bit better (as others have pointed out), but it's still going to be the same order of complexity (remember that constants / multipliers are irrelevant in big-O complexity). You can also optimize the storage a bit.
Think about it. Your array needs to have 26 * 26 * 10 * 10 * 10 members. This means you need to at least touch that many elements.
Let N = number of elements in the alphabet
Let M = number of elements in your digit queue
Best Case Order Complexity = O(N * N * M * M * M) (if all you had to do was assign values)
Best case storage complexity = same as above (you have to store all the codes)
Right now you are using the following operations:
for(i in theAlphabet){ // *O(N)*
for(x in theAlphabet){ // *O(N)*
resultArrray[teller] = theAlphabet[i] + theAlphabet[x];// *(O(1))*
}
}
for(x = 0; x<10; x++){ // O(M)
for(y = 0; y<10; y++){ // O(M)
for(z = 0; z<10; z++){ // O(M)
resultArrray2[teller] = x.toString() + y.toString() +z.toString(); // O(1) (technically this is O(length of x + y + z)
teller++;
}
}
}
for(index in resultArrray){ // O(N * N)
for(i in resultArrray2){ // O(M * M * M(
finalArray[teller] = resultArrray[index] + resultArrray2[i]; //O(1)
teller++;
}
}
So at the end of the day your order complexity is O(N * N * M * M * M), which is the best you can do.
The bigger question is why you want to generate all the codes at all. If all you want is to create a unique code per order number or something, you can make a state machine like:
function getNextCode(previousCode) {
// in here, just increment the previous code
}
If all you want is a random identifier, consider using a hash of the timestamp + something about the request instead.
If you don't care about uniqueness, you can always just generate a random code.
All of the above are O(1).

Can a for loop increment/decrement by more than one?

Are there other ways to increment a for loop in Javascript besides i++ and ++i? For example, I want to increment by 3 instead of one.
for (var i = 0; i < myVar.length; i+3) {
//every three
}

Use the += assignment operator:
for (var i = 0; i < myVar.length; i += 3) {
Technically, you can place any expression you'd like in the final expression of the for loop, but it is typically used to update the counter variable.
For more information about each step of the for loop, check out the MDN article.

A for loop:
for(INIT; TEST; ADVANCE) {
BODY
}
Means the following:
INIT;
while (true) {
if (!TEST)
break;
BODY;
ADVANCE;
}
You can write almost any expression for INIT, TEST, ADVANCE, and BODY.
Do note that the ++ operators and variants are operators with side-effects (one should try to avoid them if you are not using them like i+=1 and the like):
++i means i+=1; return i
i++ means oldI=i; i+=1; return oldI
Example:
> i=0
> [i++, i, ++i, i, i--, i, --i, i]
[0, 1, 2, 2, 2, 1, 0, 0]

for (var i = 0; i < 10; i = i + 2) {
// code here
}​

Andrew Whitaker's answer is true, but you can use any expression for any part.
Just remember the second (middle) expression should evaluate so it can be compared to a boolean true or false.
When I use a for loop, I think of it as
for (var i = 0; i < 10; ++i) {
/* expression */
}
as being
var i = 0;
while( i < 10 ) {
/* expression */
++i;
}

for (var i = 0; i < myVar.length; i+=3) {
//every three
}
additional
Operator Example Same As
++ X ++ x = x + 1
-- X -- x = x - 1
+= x += y x = x + y
-= x -= y x = x - y
*= x *= y x = x * y
/= x /= y x = x / y
%= x %= y x = x % y

You certainly can. Others have pointed out correctly that you need to do i += 3. You can't do what you have posted because all you are doing here is adding i + 3 but never assigning the result back to i. i++ is just a shorthand for i = i + 1, similarly i +=3 is a shorthand for i = i + 3.

For those who are looking to increment pair of numbers (like 1-2 to 3-4):
Solution one:
//initial values
var n_left = 1;
var n_right = 2;
for (i = 1; i <= 5; i++) {
console.log(n_left + "-" + n_right);
n_left =+ n_left+2;
n_right =+ n_right+2;
}
//result: 1-2 3-4 5-6 7-8 9-10
Solution two:
for (x = 0; x <= 9; x+=2) {
console.log((x+1) + "-" + (x+2));
}
//result: 1-2 3-4 5-6 7-8 9-10

The last part of the ternary operator allows you to specify the increment step size. For instance, i++ means increment by 1. i+=2 is same as i=i+2,... etc.
Example:
let val= [];
for (let i = 0; i < 9; i+=2) {
val = val + i+",";
}
console.log(val);
Expected results: "2,4,6,8"
'i' can be any floating point or whole number depending on the desired step size.

There is an operator just for this. For example, if I wanted to change a variable i by 3 then:
var someValue = 9;
var Increment = 3;
for(var i=0;i<someValue;i+=Increment){
//do whatever
}
to decrease, you use -=
var someValue = 3;
var Increment = 3;
for(var i=9;i>someValue;i+=Increment){
//do whatever
}

Javascript loop with dynamic start and end variables

This seems pretty basic, but I can't find the best method to do this... I'm trying to set up a function that loops between a user selected start and end variables. This is what I ended up with but I'm sure there is a better way to do it (demo).
Note: the x & y variables are indexed to one, not zero.
getWidths1 = function(x, y) {
var start = (x < y) ? x : y,
end = (x < y) ? y : x,
total = 0;
for (; start < end; start++) {
total += values[start - 1] || 0;
}
return total;
};
I tried this function, but the results are one result off when y > x:
getWidths2 = function(x, y) {
var total = 0,
diff = (x < y) ? 1 : -1;
while (x !== y) {
total += values[x - 1] || 0;
x += diff;
}
return w;
};
So, is the first function the best, or does someone have a better method?

The first isn't bad. I think this is slightly more traditional:
for (var i = start; i < end; i++){
}
Only real difference is that it doesn't affect start and end.

I'd make a few changes:
Use Math.min and Math.max - much more readable.
Don't subtract one from start if the first value you want is values[start].
var getWidths1 = function(x, y) {
var start = Math.min(x,y), end = Math.max(x,y);
var total = 0;
for (; start < end; start++) {
total += values[start] || 0;
}
return(total);
}

I agree with #kingjiv with the added caveat that if you want to include the item at y then you need:
for (var i = start; i <= end; i++){
...
}
As it is your code (both versions) will total the values from x inclusive to y exclusive.

Reverse decimal digits in javascript

How do I reverse the digits of a number using bitwise?
input:
x = 123;
output:
x = 321;
How Do this?

That's not inverting bits; that's reversing the order of decimal digits, which is completely different. Here's one way:
var x = 123;
var y = 0;
for(; x; x = Math.floor(x / 10)) {
y *= 10;
y += x % 10;
}
x = y;
If you actually want to invert bits, it's:
x = ~x;
As a function:
function reverse(n) {
for(var r = 0; n; n = Math.floor(n / 10)) {
r *= 10;
r += n % 10;
}
return r;
}

If you wanted to make a simple reversal:
var x = 123;
var y = x.toString();
var z = y.split("").reverse().join("");
var aa = Number(z);
document.write(aa);
http://jsfiddle.net/jasongennaro/gV39e/

Here is another way...
var reversed = num.toString().split('').reverse().join('');
jsFiddle.
If you wanted it again as a Number, use parseInt(reversed, 10). Keep in mind though, leading 0s are not significant in a decimal number, and you will lose them if you convert to Number.

you also use this function
function myfunction(a){
var x=a.toString();
var y= x.split("");
var z=y.reverse();
var result=z.join("");
return result;
}
myfunction(123);

Simple and quick solution: Let's assume that you want to reverse a number 4546. You will take the reminder from each division by 10 and append it to the result until the number is > 0. And simultaneously updating the num variable by dividing it by 10.
var x = '';
var num = 4546;
while(num>0){
x = x + (num%10);
num = parseInt(num/10);
}
console.log(x);

Reversing The Positive/ Negative Integer Number
function reverseInt(n) {
return parseInt(n.toString().split('').reverse().join()) * Math.sign(n)
}
If n is -5, then Math.sign(n)==> will return -1
If n is 5, then Math.sign(n)==> will return 1

Here are reversible array functions in JavaScript that handle integers or strings:
function reverse(array)
{
var left = null;
var right = null;
var length = array.length;
for (left = 0, right = length - 1; left < right; left += 1, right -= 1)
{
var temporary = array[left];
array[left] = array[right];
array[right] = temporary;
}
return array;
}
function toDigitsArrayFromInteger(integer, isReverse)
{
var digits = [];
if (integer > 0)
{
var floor = window.Math.floor;
while (integer > 0)
{
digits.push(floor(integer % 10));
integer = floor(integer / 10);
}
// Array is populated in reverse order. Un-reverse it to make it normal.
if (!isReverse)
{
digits = reverse(digits);
}
}
else if (integer < 0)
{
digits = toDigitsArrayFromInteger(-integer, isReverse);
}
else if (integer === 0)
{
digits.push(0);
}
return digits;
}
function toDigitsArrayFromString(string, isReverse)
{
var digits = [];
string += ""; // Coerce to string.
var i = null;
var length = string.length;
for (i = 0; i < length; i += 1)
{
var integer = parseInt(string.charAt(i), 10);
if (isFinite(integer))
{
digits.push(integer);
}
}
if (isReverse)
{
digits = reverse(digits);
}
return digits;
}
Once you have the digits as an array, you can reverse the array easily to get the digits starting from the left or from the right.
The string function is more versatile because it can find any digit in a string, whereas the integer function is limited to integers.
Benchmarks:
http://jsperf.com/todigitsarray
The benchmarks between the two functions show that in Firefox 10 and Chrome 12, the string function is 30% to 60% faster than the integer function. In Opera 12, the integer function is slightly faster by about 10%.

//reverse integer
const revInt = (num)=>{
//turn into string
if(Math.sign(num)===1)
return parseInt(num.toString().split('').reverse().join(''));
else return -1*parseInt(num.toString().split('').reverse().join(''));
}
console.log(revInt(-501));

<html>
<script>
function reverseInt(n){
var r=0;
while(n!=0){
r*=10;
r+=n%10;
n=Math.floor(n/10);
}
return r;
}
</script>
</html>

try this
var n = 352;
function loop(n, r){
if(!n) return r;
r = (r ? r * 10 : 0) + n % 10;
return loop(Math.floor( n / 10), r);
}
console.log(loop(n));

OK, how about using and chaining these popular tricks in JavaScript in one-line function as below...
const reverseNum = num => +("" + ~~num.split("").reverse().join(""));
And call it like these:
reverseNum(123); //321
reverseNum(423.09); //324
reverseNum(23305.1); //50332
reverseNum(89112); //21198
reverseNum(568434.2389); //434865

This takes Number x as a parameter and returns the reversed number.
const reverse = (x) => Number(x.toString().split("").reverse().join(""));

Memory Usage: 35.3 MB, less than 100.00% of JavaScript online submissions for Reverse Integer on leetcode.com.
Runtime: 80 ms, faster than 61.48% of JavaScript online submissions for Reverse Integer.
Time complexity is O(log10(n)).
function reverse(x) {
let rev = 0;
const isNegative = Math.sign(x) === -1;
const isOverflow = n => n > 2**31;
x = Math.abs(x);
while (x) {
let pop = x % 10;
x = Math.floor(x / 10);
rev = rev * 10 + pop;
if (isOverflow(rev)) {
return 0;
}
}
return isNegative ? rev * -1 : rev;
}

The code block below should do the trick
<script type = "text/javascript">
var input;
input=window.prompt ("Please enter a number to be reversed.");
x=input.length;
while(x > 0)
{
x=x-1;
document.write(input[x]);
}
</script>