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;
}
}

A code wars challenge

I have been struggling with this challenge and can't seem to find where I'm failing at:
Some numbers have funny properties. For example:
89 --> 8¹ + 9² = 89 * 1
695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n. In other words:
Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k
If it is the case we will return k, if not return -1.
Note: n, p will always be given as strictly positive integers.
digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
digPow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
I'm new with javascript so there may be something off with my code but I can't find it. My whole purpose with this was learning javascript properly but now I want to find out what I'm doing wrong.I tried to convert given integer into digits by getting its modulo with 10, and dividing it with 10 using trunc to get rid of decimal parts. I tried to fill the array with these digits with their respective powers. But the test result just says I'm returning only 0.The only thing returning 0 in my code is the first part, but when I tried commenting it out, I was still returning 0.
function digPow(n, p){
// ...
var i;
var sum;
var myArray= new Array();
if(n<0)
{
return 0;
}
var holder;
holder=n;
for(i=n.length-1;i>=0;i--)
{
if(holder<10)
{
myArray[i]=holder;
break;
}
myArray[i]=holder%10;
holder=math.trunc(holder/10);
myArray[i]=math.pow(myArray[i],p+i);
sum=myArray[i]+sum;
}
if(sum%n==0)
{
return sum/n;
}
else
{
return -1;
}}

Here is the another simple solution
function digPow(n, p){
// convert the number into string
let str = String(n);
let add = 0;
// convert string into array using split()
str.split('').forEach(num=>{
add += Math.pow(Number(num) , p);
p++;
});
return (add % n) ? -1 : add/n;
}
let result = digPow(46288, 3);
console.log(result);

Mistakes
There are a few problems with your code. Here are some mistakes you've made.
number.length is invalid. The easiest way to get the length of numbers in JS is by converting it to a string, like this: n.toString().length.
Check this too: Length of Number in JavaScript
the math object should be referenced as Math, not math. (Note the capital M) So math.pow and math.trunc should be Math.pow and Math.trunc.
sum is undefined when the for loop is iterated the first time in sum=myArray[i]+sum;. Using var sum = 0; instead of var sum;.
Fixed Code
I fixed those mistakes and updated your code. Some parts have been removed--such as validating n, (the question states its strictly positive)--and other parts have been rewritten. I did some stylistic changes to make the code more readable as well.
function digPow(n, p){
var sum = 0;
var myArray = [];
var holder = n;
for (var i = n.toString().length-1; i >= 0; i--) {
myArray[i] = holder % 10;
holder = Math.trunc(holder/10);
myArray[i] = Math.pow(myArray[i],p+i);
sum += myArray[i];
}
if(sum % n == 0) {
return sum/n;
} else {
return -1;
}
}
console.log(digPow(89, 1));
console.log(digPow(92, 1));
console.log(digPow(46288, 3));
My Code
This is what I did back when I answered this question. Hope this helps.
function digPow(n, p){
var digPowSum = 0;
var temp = n;
while (temp > 0) {
digPowSum += Math.pow(temp % 10, temp.toString().length + p - 1);
temp = Math.floor(temp / 10);
}
return (digPowSum % n === 0) ? digPowSum / n : -1;
}
console.log(digPow(89, 1));
console.log(digPow(92, 1));
console.log(digPow(46288, 3));

You have multiple problems:
If n is a number it is not going to have a length property. So i is going to be undefined and your loop never runs since undefined is not greater or equal to zero
for(i=n.length-1;i>=0;i--) //could be
for(i=(""+n).length;i>=0;i--) //""+n quick way of converting to string
You never initialize sum to 0 so it is undefined and when you add the result of the power calculation to sum you will continually get NaN
var sum; //should be
var sum=0;
You have if(holder<10)...break you do not need this as the loop will end after the iteration where holder is a less than 10. Also you never do a power for it or add it to the sum. Simply remove that if all together.
Your end code would look something like:
function digPow(n, p) {
var i;
var sum=0;
var myArray = new Array();
if (n < 0) {
return 0;
}
var holder;
holder = n;
for (i = (""+n).length - 1; i >= 0; i--) {
myArray[i] = holder % 10;
holder = Math.trunc(holder / 10);
myArray[i] = Math.pow(myArray[i], p + i);
sum = myArray[i] + sum;
}
if (sum % n == 0) {
return sum / n;
} else {
return -1;
}
}
Note you could slim it down to something like
function digPow(n,p){
if( isNaN(n) || (+n)<0 || n%1!=0) return -1;
var sum = (""+n).split("").reduce( (s,num,index)=>Math.pow(num,p+index)+s,0);
return sum%n ? -1 : sum/n;
}
(""+n) simply converts to string
.split("") splits the string into an array (no need to do %10 math to get each number
.reduce( function,0) call's the array's reduce function, which calls a function for each item in the array. The function is expected to return a value each time, second argument is the starting value
(s,num,index)=>Math.pow(num,p+index+1)+s Fat Arrow function for just calling Math.pow with the right arguments and then adding it to the sum s and returning it

I have created a code that does exactly what you are looking for.The problem in your code was explained in the comment so I will not focus on that.
FIDDLE
Here is the code.
function digPow(n, p) {
var m = n;
var i, sum = 0;
var j = 0;
var l = n.toString().length;
var digits = [];
while (n >= 10) {
digits.unshift(n % 10);
n = Math.floor(n / 10);
}
digits.unshift(n);
for (i = p; i < l + p; i++) {
sum += Math.pow(digits[j], i);
j++;
}
if (sum % m == 0) {
return sum / m;
} else
return -1;
}
alert(digPow(89, 1))

Just for a variety you may do the same job functionally as follows without using any string operations.
function digPow(n,p){
var d = ~~Math.log10(n)+1; // number of digits
r = Array(d).fill()
.map(function(_,i){
var t = Math.pow(10,d-i);
return Math.pow(~~((n%t)*10/t),p+i);
})
.reduce((p,c) => p+c);
return r%n ? -1 : r/n;
}
var res = digPow(46288,3);
console.log(res);

algorithm connect four javascript

Hy,
I am trying to implement an Connect Four Game in javascript / jQuery. First off this is no homework or any other duty. I'm just trying to push my abilities.
My "playground" is a simple html table which has 7 rows and 6 columns.
But now I have reached my ken. I'm stuck with the main functionality of checking whether there are 4 same td's around. I am adding a class to determine which color it should represent in the game.
First I thought I could handle this with .nextAll() and .prevAll() but this does not work for me because there is no detection between.
Because I was searching for siblings, when adding a new Item and just looked up the length of siblings which were found and if they matched 4 in the end I supposed this was right, but no its not :D Is there maybe any kind of directNext() which provides all next with a css selector until something different comes up ?
I will put all of my code into this jsfiddle: http://jsfiddle.net/LcUVf/5/
Maybe somebody has ever tried the same or someone comes up with a good idea I'm not asking anybody to do or finish my code. I just want to get hints for implementing such an algorithm or examples how it could be solved !
Thanks in anyway !

DOM traversal is not particularly efficient so, when you can avoid it, I'd recommend doing so. It'd make sense for you to build this as a 2D array to store and update the state of the game. The table would only be a visual representation of the array.
I know that, normally, you would build the array with rows as the first dimension and columns as the second dimension but, for the purposes of being able to add pieces to each column's "stack," I would make the first dimension the columns and the second dimension the rows.
To do the check, take a look at this fiddle I made:
http://jsfiddle.net/Koviko/4dTyw/
There are 4 directions to check: North-South, East-West, Northeast-Southwest, and Southeast-Northwest. This can be represented as objects with the delta defined for X and Y:
directions = [
{ x: 0, y: 1 }, // North-South
{ x: 1, y: 0 }, // East-West
{ x: 1, y: 1 }, // Northeast-Southwest
{ x: 1, y: -1 } // Southeast-Northwest
];
Then, loop through that object and loop through your "table" starting at the farthest bounds that this piece can possibly contribute to a win. So, since you need 4 pieces in a row, the currently placed piece can contribute in a win for up to 3 pieces in any direction.
minX = Math.min(Math.max(placedX - (3 * directions[i].x), 0), pieces.length - 1);
minY = Math.min(Math.max(placedY - (3 * directions[i].y), 0), pieces[0].length - 1);
maxX = Math.max(Math.min(placedX + (3 * directions[i].x), pieces.length - 1), 0);
maxY = Math.max(Math.min(placedY + (3 * directions[i].y), pieces[0].length - 1), 0);
To avoid any issues with less-than and greater-than (which I ran into), calculate the number of steps before looping through your pieces instead of using the calculated bounds as your conditions.
steps = Math.max(Math.abs(maxX - minX), Math.abs(maxY - minY));
Finally, loop through the items keeping a count of consecutive pieces that match the piece that was placed last.
function isVictory(pieces, placedX, placedY) {
var i, j, x, y, maxX, maxY, steps, count = 0,
directions = [
{ x: 0, y: 1 }, // North-South
{ x: 1, y: 0 }, // East-West
{ x: 1, y: 1 }, // Northeast-Southwest
{ x: 1, y: -1 } // Southeast-Northwest
];
// Check all directions
outerloop:
for (i = 0; i < directions.length; i++, count = 0) {
// Set up bounds to go 3 pieces forward and backward
x = Math.min(Math.max(placedX - (3 * directions[i].x), 0), pieces.length - 1);
y = Math.min(Math.max(placedY - (3 * directions[i].y), 0), pieces[0].length - 1);
maxX = Math.max(Math.min(placedX + (3 * directions[i].x), pieces.length - 1), 0);
maxY = Math.max(Math.min(placedY + (3 * directions[i].y), pieces[0].length - 1), 0);
steps = Math.max(Math.abs(maxX - x), Math.abs(maxY - y));
for (j = 0; j < steps; j++, x += directions[i].x, y += directions[i].y) {
if (pieces[x][y] == pieces[placedX][placedY]) {
// Increase count
if (++count >= 4) {
break outerloop;
}
} else {
// Reset count
count = 0;
}
}
}
return count >= 4;
}

I released a fully working version of the game on Github.
It implements an optimised variation on the algorythm Sirko mentioned.
To avoid any unnecessary redunancy, the algorythm directly checks the DOM rather than a JS table. As that algorythm requires a minimum amount of checks, the performance overhead for accessing the DOM is neglectable.
The current player and a flag for keeping track of whether the game has ended are basicly the only statuses stored in the JS itself.
I even used the DOM to store strings. It has no external dependencies and is supported by all versions of IE from IE6 upwards as well as modern browsers.
Code is optimised for filesize and performance. The latest version also includes animation, even though the total JS code of the game is still only 1.216 bytes after minification.
The Code :
Here's the full, un-minified JS code :
(function (doc, win, onclick, gid, classname, content, showMessage) {
var
a, b, c, colorLabel, cid, players, current, finished, newgameLabel, wonLabel, laststart = 1,
cellAt = function (i, j) {
return doc[gid](cid + i + j);
},
isCurrentColor = function (i, j) {
return cellAt(i, j)[classname] === players[current];
},
start = function () {
current = laststart = (laststart + 1) % 2;
finished = 0;
colorLabel[content] = colorLabel[classname] = players[current = (current + 1) % 2];
for (a = 1; a < 7; a++)
for (b = 1; b < 8; b++)
cellAt(a, b)[classname] = '';
},
makeMove = function (i, j, s) {
s > 0 && (cellAt(s, j)[classname] = '');
cellAt(s + 1, j)[classname] = players[current];
s === i - 1 ? function (i, j) {
return function (i, j) {
for (a = j - 1; 0 < a && isCurrentColor(i, a); a--) {
}
for (b = j + 1; 8 > b && isCurrentColor(i, b); b++) {
}
return 4 < b - a;
}(i, j) || function (i, j) {
for (c = i + 1; 7 > c && isCurrentColor(c, j); c++) {
}
return 3 < c - i;
}(i, j) || function (i, j) {
for (a = i - 1, b = j - 1; 0 < a && !(1 > b) && isCurrentColor(a, b); a--)
b--;
for (c = i + 1, b = j + 1; 7 > c && !(7 < b) && isCurrentColor(c, b); c++)
b++;
return 4 < c - a
}(i, j) || function (i, j) {
for (a = i - 1, b = j + 1; 0 < a && !(7 < b) && isCurrentColor(a, b); a--)
b++;
for (c = i + 1, b = j - 1; 7 > c && !(1 > b) && isCurrentColor(c, b); c++)
b--;
return 4 < c - a;
}(i, j);
}(i, j)
? finished = 1 && win[showMessage](doc[gid](wonLabel)[content].replace("%s", players[current].toLowerCase())) && start()
: colorLabel[content] = colorLabel[classname] = players[current = (current + 1) % 2]
: setTimeout(function () {
makeMove(i, j, s + 1)
}, 20);
};
return function (n, w, c, h, p1, p2) {
cid = c;
newgameLabel = n;
wonLabel = w;
colorLabel = doc[gid](c);
players = [doc[gid](p1)[content], doc[gid](p2)[content]];
for (a = 1; a < 7; a++)
for (b = 1; b < 8; b++)
cellAt(a, b)[onclick] = function (b, a) {
return function () {
if (!finished)
for (a = 6; a > 0; a--)
if (!cellAt(a, b)[classname]) {
makeMove(a, b, 0);
break;
}
};
}(b);
;
doc[gid](h)[onclick] = function () {
win[showMessage](doc[gid](newgameLabel)[content]) && start()
};
start();
};
})(document, window, "onclick", "getElementById", "className", "innerHTML", "confirm")("newgame", "won", "color", "restart", "p1", "p2");
A screenshot :

In general a 2dimensional array would be better suited for checking for a line of 4. You could then do something like the following:
function check( lastPiece, playground, player ) {
// check length in each direction
var l = 1,
i = 1;
// top to bottom
while( (playground[ lastPiece.x ][ lastPiece.y - i ] === player) && ((lastPiece.y - i) >= 0) ) { l += 1; i += 1; };
i = 1;
while( (playground[ lastPiece.x ][ lastPiece.y + i ] === player) && ((lastPiece.y + i) <= MAX_Y) ) { l += 1; i += 1; };
if ( l >= 4 ) { return true; }
// left to right
l = 1;
while( (playground[ lastPiece.x - i][ lastPiece.y ] === player) && ((lastPiece.x - i) >= 0) ) { l += 1; i += 1; };
i = 1;
while( (playground[ lastPiece.x + i][ lastPiece.y ] === player) && ((lastPiece.x + i) <= MAX_X) ) { l += 1; i += 1; };
if ( l >= 4 ) { return true; }
// same for top left to bottom right and bottom left to top right
// . . .
// if we got no hit until here, there is no row of 4
return false;
}
EDIT: added checks for borders of the playground

Generating Fibonacci Sequence

var x = 0;
var y = 1;
var z;
fib[0] = 0;
fib[1] = 1;
for (i = 2; i <= 10; i++) {
alert(x + y);
fib[i] = x + y;
x = y;
z = y;
}
I'm trying to get to generate a simple Fibonacci Sequence but there no output.
Can anybody let me know what's wrong?

You have never declared fib to be an array. Use var fib = []; to solve this.
Also, you're never modifying the y variable, neither using it.
The code below makes more sense, plus, it doesn't create unused variables:
var i;
var fib = [0, 1]; // Initialize array!
for (i = 2; i <= 10; i++) {
// Next fibonacci number = previous + one before previous
// Translated to JavaScript:
fib[i] = fib[i - 2] + fib[i - 1];
console.log(fib[i]);
}

According to the Interview Cake question, the sequence goes 0,1,1,2,3,5,8,13,21. If this is the case, this solution works and is recursive without the use of arrays.
function fibonacci(n) {
return n < 1 ? 0
: n <= 2 ? 1
: fibonacci(n - 1) + fibonacci(n - 2)
}
console.log(fibonacci(4))
Think of it like this.
fibonacci(4) .--------> 2 + 1 = 3
| / |
'--> fibonacci(3) + fibonacci(2)
| ^
| '----------- 2 = 1 + 1 <----------.
1st step -> | ^ |
| | |
'----> fibonacci(2) -' + fibonacci(1)-'
Take note, this solution is not very efficient though.

Yet another answer would be to use es6 generator functions.
function* fib() {
var current = a = b = 1;
yield 1;
while (true) {
current = b;
yield current;
b = a + b;
a = current;
}
}
sequence = fib();
sequence.next(); // 1
sequence.next(); // 1
sequence.next(); // 2
// ...

Here's a simple function to iterate the Fibonacci sequence into an array using arguments in the for function more than the body of the loop:
fib = function(numMax){
for(var fibArray = [0,1], i=0,j=1,k=0; k<numMax;i=j,j=x,k++ ){
x=i+j;
fibArray.push(x);
}
console.log(fibArray);
}
fib(10)
[ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ]

You should've declared the fib variable to be an array in the first place (such as var fib = [] or var fib = new Array()) and I think you're a bit confused about the algorithm.
If you use an array to store the fibonacci sequence, you do not need the other auxiliar variables (x,y,z) :
var fib = [0, 1];
for(var i=fib.length; i<10; i++) {
fib[i] = fib[i-2] + fib[i-1];
}
console.log(fib);
Click for the demo
You should consider the recursive method too (note that this is an optimised version) :
function fib(n, undefined){
if(fib.cache[n] === undefined){
fib.cache[n] = fib(n-1) + fib(n-2);
}
return fib.cache[n];
}
fib.cache = [0, 1, 1];
and then, after you call the fibonacci function, you have all the sequence in the fib.cache field :
fib(1000);
console.log(fib.cache);

The golden ration "phi" ^ n / sqrt(5) is asymptotic to the fibonacci of n, if we round that value up, we indeed get the fibonacci value.
function fib(n) {
let phi = (1 + Math.sqrt(5))/2;
let asymp = Math.pow(phi, n) / Math.sqrt(5);
return Math.round(asymp);
}
fib(1000); // 4.346655768693734e+208 in just a few milliseconds
This runs faster on large numbers compared to the recursion based solutions.

You're not assigning a value to z, so what do you expect y=z; to do? Likewise you're never actually reading from the array. It looks like you're trying a combination of two different approaches here... try getting rid of the array entirely, and just use:
// Initialization of x and y as before
for (i = 2; i <= 10; i++)
{
alert(x + y);
z = x + y;
x = y;
y = z;
}
EDIT: The OP changed the code after I'd added this answer. Originally the last line of the loop was y = z; - and that makes sense if you've initialized z as per my code.
If the array is required later, then obviously that needs to be populated still - but otherwise, the code I've given should be fine.

Another easy way to achieve this:
function fibonacciGenerator(n) {
// declare the array starting with the first 2 values of the fibonacci sequence
// starting at array index 1, and push current index + previous index to the array
for (var fibonacci = [0, 1], i = 2; i < n; i++)
fibonacci.push(fibonacci[i-1] + fibonacci[i - 2])
return fibonacci
}
console.log( fibonacciGenerator(10) )

function fib(n) {
if (n <= 1) {
return n;
} else {
return fib(n - 1) + fib(n - 2);
}
}
fib(10); // returns 55

fibonacci 1,000 ... 10,000 ... 100,000
Some answers run into issues when trying to calculate large fibonacci numbers. Others are approximating numbers using phi. This answer will show you how to calculate a precise series of large fibonacci numbers without running into limitations set by JavaScript's floating point implementation.
Below, we generate the first 1,000 fibonacci numbers in a few milliseconds. Later, we'll do 100,000!
const { fromInt, toString, add } =
Bignum
const bigfib = function* (n = 0)
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n >= 0) {
yield toString (a)
_ = a
a = b
b = add (b, _)
n = n - 1
}
}
console.time ('bigfib')
const seq = Array.from (bigfib (1000))
console.timeEnd ('bigfib')
// 25 ms
console.log (seq.length)
// 1001
console.log (seq)
// [ 0, 1, 1, 2, 3, ... 995 more elements ]
Let's see the 1,000th fibonacci number
console.log (seq [1000])
// 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
10,000
This solution scales quite nicely. We can calculate the first 10,000 fibonacci numbers in under 2 seconds. At this point in the sequence, the numbers are over 2,000 digits long – way beyond the capacity of JavaScript's floating point numbers. Still, our result includes precise values without making approximations.
console.time ('bigfib')
const seq = Array.from (bigfib (10000))
console.timeEnd ('bigfib')
// 1877 ms
console.log (seq.length)
// 10001
console.log (seq [10000] .length)
// 2090
console.log (seq [10000])
// 3364476487 ... 2070 more digits ... 9947366875
Of course all of that magic takes place in Bignum, which we will share now. To get an intuition for how we will design Bignum, recall how you added big numbers using pen and paper as a child...
1259601512351095520986368
+ 50695640938240596831104
---------------------------
?
You add each column, right to left, and when a column overflows into the double digits, remembering to carry the 1 over to the next column...
... <-001
1259601512351095520986368
+ 50695640938240596831104
---------------------------
... <-472
Above, we can see that if we had two 10-digit numbers, it would take approximately 30 simple additions (3 per column) to compute the answer. This is how we will design Bignum to work
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s, Number) .reverse ()
, toString: (b) =>
Array.from (b) .reverse () .join ('')
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
We'll run a quick test to verify our example above
const x =
fromString ('1259601512351095520986368')
const y =
fromString ('50695640938240596831104')
console.log (toString (add (x,y)))
// 1310297153289336117817472
And now a complete program demonstration. Expand it to calculate the precise 10,000th fibonacci number in your own browser! Note, the result is the same as the answer provided by wolfram alpha
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s, Number) .reverse ()
, toString: (b) =>
Array.from (b) .reverse () .join ('')
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
const { fromInt, toString, add } =
Bignum
const bigfib = function* (n = 0)
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n >= 0) {
yield toString (a)
_ = a
a = b
b = add (b, _)
n = n - 1
}
}
console.time ('bigfib')
const seq = Array.from (bigfib (10000))
console.timeEnd ('bigfib')
// 1877 ms
console.log (seq.length)
// 10001
console.log (seq [10000] .length)
// 2090
console.log (seq [10000])
// 3364476487 ... 2070 more digits ... 9947366875
100,000
I was just curious how far this little script could go. It seems like the only limitation is just time and memory. Below, we calculate the first 100,000 fibonacci numbers without approximation. Numbers at this point in the sequence are over 20,000 digits long, wow! It takes 3.18 minutes to complete but the result still matches the answer from wolfram alpha
console.time ('bigfib')
const seq = Array.from (bigfib (100000))
console.timeEnd ('bigfib')
// 191078 ms
console.log (seq .length)
// 100001
console.log (seq [100000] .length)
// 20899
console.log (seq [100000])
// 2597406934 ... 20879 more digits ... 3428746875
BigInt
JavaScript now has native support for BigInt. This allows for calculating huge integers very quickly -
function* fib (n)
{ let a = 0n
let b = 1n
let _
while (n >= 0) {
yield a.toString()
_ = a
a = b
b = b + _
n = n - 1
}
}
console.time("fib(1000)")
const result = Array.from(fib(1000))
console.timeEnd("fib(1000)")
document.body.textContent = JSON.stringify(result, null, 2)
body {
font-family: monospace;
white-space: pre;
}

I like the fact that there are so many ways to create a fibonacci sequence in JS. I will try to reproduce a few of them. The goal is to output a sequence to console (like {n: 6, fiboNum: 8})
Good ol' closure
// The IIFE form is purposefully omitted. See below.
const fiboGenClosure = () => {
let [a, b] = [0, 1];
let n = 0;
return (fiboNum = a) => {
[a, b] = [b, a + b];
return {
n: n++,
fiboNum: fiboNum
};
};
}
// Gets the sequence until given nth number. Always returns a new copy of the main function, so it is possible to generate multiple independent sequences.
const generateFiboClosure = n => {
const newSequence = fiboGenClosure();
for (let i = 0; i <= n; i++) {
console.log(newSequence());
}
}
generateFiboClosure(21);
Fancy ES6 generator
Similar to the closure pattern above, using the advantages of generator function and for..of loop.
// The 'n' argument is a substitute for index.
function* fiboGen(n = 0) {
let [a, b] = [0, 1];
while (true) {
yield [a, n++];
[a, b] = [b, a + b];
}
}
// Also gives a new sequence every time is invoked.
const generateFibonacci = n => {
const iterator = fiboGen();
for (let [value, index] of iterator) {
console.log({
n: index,
fiboNum: value
});
if (index >= n) break;
}
}
generateFibonacci(21);
Tail call recursion
This one is a little tricky, because, now in late 2018, TC optimization is still an issue. But honestly – if you don't use any smart tricks to allow the default JS engine to use a really big numbers, it will get dizzy and claims that the next fibonacci number is "Infinity" by iteration 1477. The stack would probably overflow somewhere around iteration 10 000 (vastly depends on browser, memory etc…). Could be probably padded by try… catch block or check if "Infinity" was reached.
const fibonacciRTC = (n, i = 0, a = 0, b = 1) => {
console.log({
n: i,
fibonacci: a
});
if (n === 0) return;
return fibonacciRTC(--n, ++i, b, a + b);
}
fibonacciRTC(21)
It can be written as a one-liner, if we throe away the console.log thing and simply return a number:
const fibonacciRTC2 = (n, a = 0, b = 1) => n === 0 ? a : fibonacciRTC2(n - 1, b, a + b);
console.log(fibonacciRTC2(21))
Important note!
As I found out reading this mathIsFun article, the fibonacci sequence is valid for negative numbers as well! I tried to implement that in the recursive tail call form above like that:
const fibonacciRTC3 = (n, a = 0, b = 1, sign = n >= 0 ? 1 : -1) => {
if (n === 0) return a * sign;
return fibonacciRTC3(n - sign, b, a + b, sign);
}
console.log(fibonacciRTC3(8)); // 21
console.log(fibonacciRTC3(-8)); // -21

There is also a generalization of Binet's formula for negative integers:
static float phi = (1.0f + sqrt(5.0f)) / 2.0f;
int generalized_binet_fib(int n) {
return round( (pow(phi, n) - cos(n * M_PI) * pow(phi, -n)) / sqrt(5.0f) );
}
...
for(int i = -10; i < 10; ++i)
printf("%i ", generalized_binet_fib(i));

A quick way to get ~75
ty #geeves for the catch, I replaced Math.floor for Math.round which seems to get it up to 76 where floating point issues come into play :/ ...
either way, I wouldn't want to be using recursion up and until that point.
/**
* Binet Fibonacci number formula for determining
* sequence values
* #param {int} pos - the position in sequence to lookup
* #returns {int} the Fibonacci value of sequence #pos
*/
var test = [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025,20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041,1548008755920,2504730781961,4052739537881,6557470319842,10610209857723,17167680177565,27777890035288,44945570212853,72723460248141,117669030460994,190392490709135,308061521170129,498454011879264,806515533049393,1304969544928657,2111485077978050,3416454622906707,5527939700884757,8944394323791464,14472334024676221,23416728348467685,37889062373143906,61305790721611591,99194853094755497,160500643816367088,259695496911122585,420196140727489673,679891637638612258,1100087778366101931,1779979416004714189,2880067194370816120,4660046610375530309,7540113804746346429,12200160415121876738,19740274219868223167,31940434634990099905,51680708854858323072,83621143489848422977,135301852344706746049,218922995834555169026];
var fib = function (pos) {
return Math.round((Math.pow( 1 + Math.sqrt(5), pos)
- Math.pow( 1 - Math.sqrt(5), pos))
/ (Math.pow(2, pos) * Math.sqrt(5)));
};
/* This is only for the test */
var max = test.length,
i = 0,
frag = document.createDocumentFragment(),
_div = document.createElement('div'),
_text = document.createTextNode(''),
div,
text,
err,
num;
for ( ; i < max; i++) {
div = _div.cloneNode();
text = _text.cloneNode();
num = fib(i);
if (num !== test[i]) {
err = i + ' == ' + test[i] + '; got ' + num;
div.style.color = 'red';
}
text.nodeValue = i + ': ' + num;
div.appendChild(text);
frag.appendChild(div);
}
document.body.appendChild(frag);

You can get some cache to speedup the algorithm...
var tools = {
fibonacci : function(n) {
var cache = {};
// optional seed cache
cache[2] = 1;
cache[3] = 2;
cache[4] = 3;
cache[5] = 5;
cache[6] = 8;
return execute(n);
function execute(n) {
// special cases 0 or 1
if (n < 2) return n;
var a = n - 1;
var b = n - 2;
if(!cache[a]) cache[a] = execute(a);
if(!cache[b]) cache[b] = execute(b);
return cache[a] + cache[b];
}
}
};

If using ES2015
const fib = (n, prev = 0, current = 1) => n
? fib(--n, current, prev + current)
: prev + current
console.log( fib(10) )

If you need to build a list of fibonacci numbers easily you can use array destructuring assignment to ease your pain:
function fibonacci(n) {
let fibList = [];
let [a, b] = [0, 1]; // array destructuring to ease your pain
while (a < n) {
fibList.push(a);
[a, b] = [b, a + b]; // less pain, more gain
}
return fibList;
}
console.log(fibonacci(10)); // prints [0, 1, 1, 2, 3, 5, 8]

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>fibonacci series</title>
<script type="text/javascript">
function generateseries(){
var fno = document.getElementById("firstno").value;
var sno = document.getElementById("secondno").value;
var a = parseInt(fno);
var result = new Array();
result[0] = a;
var b = ++fno;
var c = b;
while (b <= sno) {
result.push(c);
document.getElementById("maindiv").innerHTML = "Fibonacci Series between "+fno+ " and " +sno+ " is " +result;
c = a + b;
a = b;
b = c;
}
}
function numeric(evt){
var theEvent = evt || window.event;
var key = theEvent.keyCode || theEvent.which;
key = String.fromCharCode(key);
var regex = /[0-9]|\./;
if (!regex.test(key)) {
theEvent.returnValue = false;
if (theEvent.preventDefault)
theEvent.preventDefault();
}
}
</script>
<h1 align="center">Fibonacci Series</h1>
</head>
<body>
<div id="resultdiv" align="center">
<input type="text" name="firstno" id="firstno" onkeypress="numeric(event)"><br>
<input type="text" name="secondno" id="secondno" onkeypress="numeric(event)"><br>
<input type="button" id="result" value="Result" onclick="generateseries();">
<div id="maindiv"></div>
</div>
</body>
</html>

I know this is a bit of an old question, but I realized that many of the answers here are utilizing for loops rather than while loops.
Sometimes, while loops are faster than for loops, so I figured I'd contribute some code that runs the Fibonacci sequence in a while loop as well! Use whatever you find suitable to your needs.
function fib(length) {
var fibArr = [],
i = 0,
j = 1;
fibArr.push(i);
fibArr.push(j);
while (fibArr.length <= length) {
fibArr.push(fibArr[j] + fibArr[i]);
j++;
i++;
}
return fibArr;
};
fib(15);

sparkida, found an issue with your method. If you check position 10, it returns 54 and causes all subsequent values to be incorrect. You can see this appearing here: http://jsfiddle.net/createanaccount/cdrgyzdz/5/
(function() {
function fib(n) {
var root5 = Math.sqrt(5);
var val1 = (1 + root5) / 2;
var val2 = 1 - val1;
var value = (Math.pow(val1, n) - Math.pow(val2, n)) / root5;
return Math.floor(value + 0.5);
}
for (var i = 0; i < 100; i++) {
document.getElementById("sequence").innerHTML += (0 < i ? ", " : "") + fib(i);
}
}());
<div id="sequence">
</div>

Here are examples how to write fibonacci using recursion, generator and reduce.
'use strict'
//------------- using recursion ------------
function fibonacciRecursion(n) {
return (n < 2) ? n : fibonacciRecursion(n - 2) + fibonacciRecursion(n - 1)
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciRecursion(i))
}
//-------------- using generator -----------------
function* fibonacciGenerator() {
let a = 1,
b = 0
while (true) {
yield b;
[a, b] = [b, a + b]
}
}
// usage
const gen = fibonacciGenerator()
for (let i = 0; i < 10; i++) {
console.log(gen.next().value)
}
//------------- using reduce ---------------------
function fibonacciReduce(n) {
return new Array(n).fill(0)
.reduce((prev, curr) => ([prev[0], prev[1]] = [prev[1], prev[0] + prev[1]], prev), [0, 1])[0]
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciReduce(i))
}

I just would like to contribute with a tail call optimized version by ES6. It's quite simple;
var fibonacci = (n, f = 0, s = 1) => n === 0 ? f : fibonacci(--n, s, f + s);
console.log(fibonacci(12));

There is no need for slow loops, generators or recursive functions (with or without caching). Here is a fast one-liner using Array and reduce.
ECMAScript 6:
var fibonacci=(n)=>Array(n).fill().reduce((a,b,c)=>a.concat(c<2?c:a[c-1]+a[c-2]),[])
ECMAScript 5:
function fibonacci(n){
return Array.apply(null,{length:n}).reduce(function(a,b,c){return a.concat((c<2)?c:a[c-1]+a[c-2]);},[]);
}
Tested in Chrome 59 (Windows 10):
fibonacci(10); // 0 ms -> (10) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
JavaScript can handle numbers up to 1476 before reaching Infinity.
fibonacci(1476); // 11ms -> (1476) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...]

Another implementation, while recursive is very fast and uses single inline function. It hits the javascript 64-bit number precision limit, starting 80th sequence (as do all other algorithms):
For example if you want the 78th term (78 goes in the last parenthesis):
(function (n,i,p,r){p=(p||0)+r||1;i=i?i+1:1;return i<=n?arguments.callee(n,i,r,p):r}(78));
will return: 8944394323791464
This is backwards compatible all the way to ECMASCRIPT4 - I tested it with IE7 and it works!

This script will take a number as parameter, that you want your Fibonacci sequence to go.
function calculateFib(num) {
var fibArray = [];
var counter = 0;
if (fibArray.length == 0) {
fibArray.push(
counter
);
counter++
};
fibArray.push(fibArray[fibArray.length - 1] + counter);
do {
var lastIndex = fibArray[fibArray.length - 1];
var snLastIndex = fibArray[fibArray.length - 2];
if (lastIndex + snLastIndex < num) {
fibArray.push(lastIndex + snLastIndex);
}
} while (lastIndex + snLastIndex < num);
return fibArray;
};

This is what I came up with
//fibonacci numbers
//0,1,1,2,3,5,8,13,21,34,55,89
//print out the first ten fibonacci numbers
'use strict';
function printFobonacciNumbers(n) {
var firstNumber = 0,
secondNumber = 1,
fibNumbers = [];
if (n <= 0) {
return fibNumbers;
}
if (n === 1) {
return fibNumbers.push(firstNumber);
}
//if we are here,we should have at least two numbers in the array
fibNumbers[0] = firstNumber;
fibNumbers[1] = secondNumber;
for (var i = 2; i <= n; i++) {
fibNumbers[i] = fibNumbers[(i - 1)] + fibNumbers[(i - 2)];
}
return fibNumbers;
}
var result = printFobonacciNumbers(10);
if (result) {
for (var i = 0; i < result.length; i++) {
console.log(result[i]);
}
}

Beginner, not too elegant, but shows the basic steps and deductions in JavaScript
/* Array Four Million Numbers */
var j = [];
var x = [1,2];
var even = [];
for (var i = 1;i<4000001;i++){
j.push(i);
}
// Array Even Million
i = 1;
while (i<4000001){
var k = j[i] + j[i-1];
j[i + 1] = k;
if (k < 4000001){
x.push(k);
}
i++;
}
var total = 0;
for (w in x){
if (x[w] %2 === 0){
even.push(x[w]);
}
}
for (num in even){
total += even[num];
}
console.log(x);
console.log(even);
console.log(total);

My 2 cents:
function fibonacci(num) {
return Array.apply(null, Array(num)).reduce(function(acc, curr, idx) {
return idx > 2 ? acc.concat(acc[idx-1] + acc[idx-2]) : acc;
}, [0, 1, 1]);
}
console.log(fibonacci(10));

I would like to add some more code as an answer :), Its never too late to code :P
function fibonacciRecursive(a, b, counter, len) {
if (counter <= len) {
console.log(a);
fibonacciRecursive(b, a + b, counter + 1, len);
}
}
fibonacciRecursive(0, 1, 1, 20);
Result
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181

function fibo(count) {
//when count is 0, just return
if (!count) return;
//Push 0 as the first element into an array
var fibArr = [0];
//when count is 1, just print and return
if (count === 1) {
console.log(fibArr);
return;
}
//Now push 1 as the next element to the same array
fibArr.push(1);
//Start the iteration from 2 to the count
for(var i = 2, len = count; i < len; i++) {
//Addition of previous and one before previous
fibArr.push(fibArr[i-1] + fibArr[i-2]);
}
//outputs the final fibonacci series
console.log(fibArr);
}
Whatever count we need, we can give it to above fibo method and get the fibonacci series upto the count.
fibo(20); //output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

Fibonacci (one-liner)
function fibonacci(n) {
return (n <= 1) ? n : fibonacci(n - 1) + fibonacci(n - 2);
}
Fibonacci (recursive)
function fibonacci(number) {
// n <= 1
if (number <= 0) {
return n;
} else {
// f(n) = f(n-1) + f(n-2)
return fibonacci(number - 1) + fibonacci(number - 2);
}
};
console.log('f(14) = ' + fibonacci(14)); // 377
Fibonacci (iterative)
function fibonacci(number) {
// n < 2
if (number <= 0) {
return number ;
} else {
var n = 2; // n = 2
var fn_1 = 0; // f(n-2), if n=2
var fn_2 = 1; // f(n-1), if n=2
// n >= 2
while (n <= number) {
var aa = fn_2; // f(n-1)
var fn = fn_1 + fn_2; // f(n)
// Preparation for next loop
fn_1 = aa;
fn_2 = fn;
n++;
}
return fn_2;
}
};
console.log('f(14) = ' + fibonacci(14)); // 377
Fibonacci (with Tail Call Optimization)
function fibonacci(number) {
if (number <= 1) {
return number;
}
function recursion(length, originalLength, previous, next) {
if (length === originalLength)
return previous + next;
return recursion(length + 1, originalLength, next, previous + next);
}
return recursion(1, number - 1, 0, 1);
}
console.log(`f(14) = ${fibonacci(14)}`); // 377

Math.ceil to nearest five at position 1

Okay....
I have a lot of uncontrolled numbers i want to round:
51255 -> 55000
25 -> 25
9214 -> 9500
13135 -> 15000
25123 -> 30000
I have tried modifying the numbers as string and counting length....
But is there a simple way using some Math function maybe?

Here's my late answer. Uses no Math methods.
function toN5( x ) {
var i = 5;
while( x >= 100 ) {x/=10; i*=10;}
return ((~~(x/5))+(x%5?1:0)) * i;
}
DEMO: http://jsbin.com/ujamoj/edit#javascript,live
[51255, 24, 25, 26, 9214, 13135, 25123, 1, 9, 0].map( toN5 );
// [55000, 25, 25, 30, 9500, 15000, 30000, 5, 10, 0]
Or this is perhaps a bit cleaner:
function toN5( x ) {
var i = 1;
while( x >= 100 ) {x/=10; i*=10;}
return (x + (5-((x%5)||5))) * i;
}
DEMO: http://jsbin.com/idowan/edit#javascript,live
To break it down:
function toN5( x ) {
// v---we're going to reduce x to the tens place, and for each place
// v reduction, we'll multiply i * 10 to restore x later.
var i = 1;
// as long as x >= 100, divide x by 10, and multiply i by 10.
while( x >= 100 ) {x/=10; i*=10;}
// Now round up to the next 5 by adding to x the difference between 5 and
// the remainder of x/5 (or if the remainder was 0, we substitute 5
// for the remainder, so it is (x + (5 - 5)), which of course equals x).
// So then since we are now in either the tens or ones place, and we've
// rounded to the next 5 (or stayed the same), we multiply by i to restore
// x to its original place.
return (x + (5-((x%5)||5))) * i;
}
Or to avoid logical operators, and just use arithmetic operators, we could do:
return (x + ((5-(x%5))%5)) * i;
And to spread it out a bit:
function toN5( x ) {
var i = 1;
while( x >= 100 ) {
x/=10;
i*=10;
}
var remainder = x % 5;
var distance_to_5 = (5 - remainder) % 5;
return (x + distance_to_5) * i;
}

var numbers = [51255, 25, 9214, 13135, 25123, 3, 6];
function weird_round(a) {
var len = a.toString().length;
var div = len == 1 ? 1 : Math.pow(10, len - 2);
return Math.ceil(a / 5 / div) * div * 5;
}
alert(numbers.map(weird_round));
Also updated for numbers below 10. Won't work properly for negative numbers either, just mention if you need this.
DEMO

I'm not sure why, but I thought it would be fun with regular expressions:
var result = +(number.toString().replace(/([1-9])([0-9])(.+)/, function() {
return Math.ceil(+(arguments[1] + '.' + arguments[2])) * 10 - (+arguments[2] < 5?5:0) + arguments[3].replace(/./g, '0');
}));
Working Demo

with(Math) {
var exp = floor(log(number)/log(10)) - 1;
exp = max(exp,0);
var n = number/pow(10,exp);
var n2 = ceil(n/5) * 5;
var result = n2 * pow(10,exp);
}
http://jsfiddle.net/NvvGf/4/
Caveat: only works for the natural numbers.

function round(number) {
var numberStr = number + "",
max,
i;
if (numberStr[1] > '4') {
numberStr[0] = parseInt(numberStr[0]) + 1;
numberStr[1] = '0';
} else {
numberStr[1] = '5';
}
for (i = 2; max = numberStr.length; i < max; i += 1) {
numberStr += '0';
}
return parseInt(numberStr);
}

Strange coincidence, I wrote something really similar not so long ago!
function iSuckAtNames(n) {
var n = n.toString(), len = n.length, res;
//Check the second number. if it's less than a 5, round down,
//If it's more/equal, round up
//Either way, we'll need to use this:
var res = parseFloat(n[0]) * Math.pow(10, len - 1); //e.g. 5 * 10^4 = 50000
if (n[1] <= 5) {
//we need to add a 5 right before the end!
res += 5 * Math.pow(10, len - 2);
}
else {
//We need another number of that size
res += Math.pow(10, len - 1);
}
return res;
}