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My test task for trainee. Javascript - quadratic equation and game theory

I've been thinking that I found solution, but tests fail (I have no access to tests).
The task:
We want to organize a chess tournament.
One day - one game.
And as entry point we have number of days - that we can afford.
Tournament rules:
We have number of teams: if number is even - we divide number of teams by 2 and that's the number of games - so they fight 1 vs 1 in pairs, who lost is excluded from the competition.
For example : 20 teams => 10 games => 10 teams left.
And if the even number of teams left - we make the same iteration: divide by 2 into pairs.
10 teams - 5 games - 5 teams left.
And now we have Odd number!
If we have odd number from start or get it in process - number of games is counted by another rule:
everyone needs to play with everyone. So the formula is = n * (n-1) / 2.
For example 5 teams = 10 games.
Or any odd number (to sum it all - if we have Odd number we always ignore first rule with pairs).
So:
We have number of days and need to find all possible team numbers for this days, it can be more than one value, for example : 3 days, number of teams that is correct is 3 and 4;
Entry point 3, function logs 3 and 4;
If that's impossible to find teams - return -1;
1 ≤ 𝑁 ≤ 1000000
Here is my solution - It ouputs correct values, but tests fail
Can someone help to find what I did wrong?
(Sorry for code mess, I had different thoughts every time)
function solve(input) {
let quadraticEquation = (number) => {
let a = 1;
let b = -1;
let c = -number*2
if(a == 0)
return 'false';
let res = {};
let D = b * b - 4 * a * c;
let tmp = [];
if(D < 0)
return 'false';
res['discriminant'] = D;
if(D == 0)
res["quadratic roots"] = (-b + Math.sqrt(D)) / (2 * a);
else if(D > 0){
tmp.push((-b + Math.sqrt(D)) / (2 * a));
tmp.push((-b - Math.sqrt(D)) / (2 * a));
res["quadratic roots"] = tmp;
}
return tmp;
}
let start = input;
let xValues = quadraticEquation(start).map(item => Math.abs(item)).filter(item => item % 2 !== 0);
let arrayToUse = []
for(i = 0; i <= xValues[xValues.length-1]; i++) {
if( i === 1) {
continue;
}
if (!(i%2)) {
continue
}
arrayToUse.push(i);
}
const answer = [];
arrayToUse.forEach(item => {
let startValue = item;
let fightsNumber = item * (item - 1) / 2;
let loop = 0;
if (fightsNumber === start) {
answer.push([startValue, loop])
} else {
do {
loop++;
fightsNumber += startValue;
if (fightsNumber === start) {
answer.push([item, loop] )
}
startValue *= 2;
} while (fightsNumber < start)
}
})
function getTeams (answer) {
const finalResult = [];
answer.forEach(item => {
if(item[1] === 0) {
finalResult.push(item[0]);
} else {
let initialValue = item[0];
for(i=0; i < item[1]; i++) {
initialValue += item[0];
item[0] *= 2;
}
finalResult.push(initialValue);
}
})
let initialValue = 2;
let fightCounter = 0;
for (i = 0; i <= start; i++) {
fightCounter += initialValue / 2
if (fightCounter === start) {
finalResult.push(initialValue);
}
initialValue *= 2;
}
return finalResult;
}
let finalString = ''
const arrayToLog = getTeams(answer).sort((a,b) => a - b);
if(arrayToLog.length !== 0) {
arrayToLog.forEach(item => {
finalString += item + '\n';
})
} else {
finalString += -1;
}
return finalString;
}
console.log(solve(325))
console.log(solve(3))
console.log(solve(15))
console.log(solve(21))
console.log(solve(10))
console.log(solve(1))
console.log(solve(5))
console.log(solve(9))

Note: for those who didn't see it, this is a followup to the earlier question Backward recurssion.
That's a lot of code for a fairly simple problem. I don't have the inclination or the time right now to go through it carefully.
So here is how I might solve this problem, using brute force to look for all the resulting number of days for any number of players up to days + 1, and then filtering out what we want from that list:
const range = (lo, hi) =>
[... Array (hi - lo + 1)] .map ((_, i) => i + lo)
const qwe = (n) =>
n % 2 == 1
? n * (n - 1) / 2
: n / 2 + qwe (n / 2)
const solve = (days) => {
const daysByPlayers = range (1, days + 1) .map (qwe)
const res = daysByPlayers .reduce ((a, d, i) => d == days ? [...a, i + 1] : a, [])
return res .length == 0 ? [-1] : res
}
const tests = [325, 3, 15, 21, 10, 1, 5, 9]
tests .forEach (n => console .log (`${n}: ${JSON.stringify(solve(n))}`))
.as-console-wrapper {max-height: 100% !important; top: 0}
This matches your output as far as I can tell. But I don't know what tests you say are failing, so perhaps there's something wrong with this as well.
There is extra complexity in here because of your desire for a -1 output in the missing cases. If an empty array would serve, then this is cleaner:
const solve = (days) =>
rng (1, days + 1) .map (qwe)
.reduce ((a, d, i) => d == days ? [...a, i + 1] : a, [])
And there are many variants. This one returns the maximum number of players that can be accommodated in some number of days up to the given value:
const solve = (days) => {
const daysByPlayers = rng (1, days + 1) .map (qwe)
const max = Math.max (... daysByPlayers .filter (d => d <= days))
return days - daysByPlayers .reverse () .indexOf (max) + 1
}
And this one gives you all the possible number of players available to achieve that maximum:
const solve = (days) => {
const daysByPlayers = rng (1, days + 1) .map (qwe)
const max = Math.max (... daysByPlayers .filter (d => d <= days))
return daysByPlayers .reduce ((a, d, i) => d == max ? [...a, i + 1] : a, [])
}
(For instance, solve (20) would return [10, 16], which are the possible number of players to handle in exactly 15 days, the largest value found no larger than 20.)
And finally, we could list all the numbers of players we could accommodate in up to a certain number of days:
const solve = (days) => {
const daysByPlayers = rng (1, days + 1) .map (qwe)
const max = Math.max (... daysByPlayers .filter (d => d <= days))
return daysByPlayers .reduce ((a, d, i) => d <= max ? [...a, i + 1] : a, [])
}
For instance, solve (20) would return [1, 2, 3, 4, 5, 6, 8, 10, 12, 16], which are the only numbers of players that could be handled in no more than 20 days.
When I brought up the quadratic formula in the earlier question it was just to note that in certain instances, we can find an exact solution, solving for d in n * (n - 1) / 2) = d. That meant that if 8 * d + 1 is a perfect square, then we can choose n to be (1 + sqrt (8 * d + 1)) / 2 to find one solution. But that's not of any real help in the general case.

Javascript's equivalent of R's findInterval() or Python's bisect.bisect_left

I can't find how to determine to which interval an element belongs based on an Array for JavaScript. I want the behavior of bisect.bisect_left from Python. Here is some sample code:
import bisect
a = [10,20,30,40]
print(bisect.bisect_left(a,0)) #0 because 0 <= 10
print(bisect.bisect_left(a,10)) #0 because 10 <= 10
print(bisect.bisect_left(a,15)) #1 because 10 < 15 < 20
print(bisect.bisect_left(a,25)) #2 ...
print(bisect.bisect_left(a,35)) #3 ...
print(bisect.bisect_left(a,45)) #4
I know this would be easy to implement, but why re-invent the wheel?

In case anyone else lands here, here's an implementation of bisect_left that actually runs in O(log N), and should work regardless of the interval between list elements. NB that is does not sort the input list, and, as-is, will likely blow the stack if you pass it an unsorted list. It's also only set up to work with numbers, but it should be easy enough to adapt it to accept a comparison function. Take this as a starting point, not necessarily your destination. Improvements are certainly welcome!
Run it in a REPL
function bisect(sortedList, el){
if(!sortedList.length) return 0;
if(sortedList.length == 1) {
return el > sortedList[0] ? 1 : 0;
}
let lbound = 0;
let rbound = sortedList.length - 1;
return bisect(lbound, rbound);
// note that this function depends on closure over lbound and rbound
// to work correctly
function bisect(lb, rb){
if(rb - lb == 1){
if(sortedList[lb] < el && sortedList[rb] >= el){
return lb + 1;
}
if(sortedList[lb] == el){
return lb;
}
}
if(sortedList[lb] > el){
return 0;
}
if(sortedList[rb] < el){
return sortedList.length
}
let midPoint = lb + (Math.floor((rb - lb) / 2));
let midValue = sortedList[midPoint];
if(el <= midValue){
rbound = midPoint
}
else if(el > midValue){
lbound = midPoint
}
return bisect(lbound, rbound);
}
}
console.log(bisect([1,2,4,5,6], 3)) // => 2
console.log(bisect([1,2,4,5,6], 7)) // => 5
console.log(bisect([0,1,1,1,1,2], 1)) // => 1
console.log(bisect([0,1], 0)) // => 0
console.log(bisect([1,1,1,1,1], 1)) // => 0
console.log(bisect([1], 2)); // => 1
console.log(bisect([1], 1)); // => 0

Speaking of re-inventing the wheel, I'd like to join the conversation:
function bisectLeft(arr, value, lo=0, hi=arr.length) {
while (lo < hi) {
const mid = (lo + hi) >> 1;
if (arr[mid] < value) {
lo = mid + 1;
} else {
hi = mid;
}
}
return lo;
}
I believe that is the schoolbook implementation of bisection. Actually, you'll find something pretty much the same inside the d3-array package mentioned before.

using the D3-array npm.
const d3 = require('d3-array');
var a = [10,20,30,40];
console.log(d3.bisectLeft(a,0));
console.log(d3.bisectLeft(a,10));
console.log(d3.bisectLeft(a,15));
console.log(d3.bisectLeft(a,25));
console.log(d3.bisectLeft(a,35));
console.log(d3.bisectLeft(a,45));
output:
0
0
1
2
3
4

A faster way than the previously accepted answer that works for same size intervals is:
var array = [5, 20, 35, 50]
//Intervals:
// <5: 0
// [5-20): 1
// [20-35): 2
// [35-50): 3
// >=50: 4
var getPosition = function(array, x) {
if (array.length == 0) return;
if (array.length == 1) return (x < array[0]) ? 0 : 1;
return Math.floor((x - array[0]) / (array[1] - array[0])) + 1
}
console.log(getPosition(array, 2)); //0
console.log(getPosition(array, 5)); //1
console.log(getPosition(array, 15));//1
console.log(getPosition(array, 20));//2
console.log(getPosition(array, 48));//3
console.log(getPosition(array, 50));//4
console.log(getPosition(array, 53));//4
console.log("WHEN SIZE: 1")
array = [5];
//Intervals:
// <5: 0
// >=5: 1
console.log(getPosition(array, 3));
console.log(getPosition(array, 5));
console.log(getPosition(array, 6));

There are no built-in bisection functions in JavaScript, so you will have to roll your own. Here is my personal reinvention of the wheel:
var array = [10, 20, 30, 40]
function bisectLeft (array, x) {
for (var i = 0; i < array.length; i++) {
if (array[i] >= x) return i
}
return array.length
}
console.log(bisectLeft(array, 5))
console.log(bisectLeft(array, 15))
console.log(bisectLeft(array, 25))
console.log(bisectLeft(array, 35))
console.log(bisectLeft(array, 45))
function bisectRight (array, x) {
for (var i = 0; i < array.length; i++) {
if (array[i] > x) return i
}
return array.length
}

Write a JS function to find the first prime pair with a given step value

** Here is the question from Codewars.
The prime numbers are not regularly spaced. For example from 2 to 3 the step is 1. From 3 to 5 the step is 2. From 7 to 11 it is 4. Between 2 and 50 we have the following pairs of 2-steps primes:
3, 5 - 5, 7, - 11, 13, - 17, 19, - 29, 31, - 41, 43
We will write a function step with parameters:
g (integer >= 2) which indicates the step we are looking for,
m (integer >= 2) which gives the start of the search (m inclusive),
n (integer >= m) which gives the end of the search (n inclusive)
In the example above step(2, 2, 50) will return [3, 5] which is the first pair between 2 and 50 with a 2-steps.
So this function should return the first pair of the two prime numbers spaced with a step of g between the limits m, n if these g-steps prime numbers exist otherwise nil or null or None or Nothing (depending on the language).
Examples:
step(2, 5, 7) --> [5, 7] or (5, 7) or {5, 7}
step(2, 5, 5) --> nil or null
step(4, 130, 200) --> [163, 167] or (163, 167) or {163, 167}
Here is my solution- it is taking too long to pass the tests- how can I make it more efficient?
function step(g, m, n) {
var arr = [];
function isPrime(num){
for (var k=2; k<num; k++){
if(num%k ===0){
return false;
}
}return true;
}
for (var i= m; i < n; i++){
if(isPrime(i)=== true){
arr.push(i);
}
}
var endArr=[];
for(var l=0;l<arr.length;l++)
for(var p=1;p<arr.length;p++){
if(arr[l]-arr[l-p]=== g){
endArr.push(arr[l])
endArr.push(arr[l]-g)
}
}
return endArr.slice(0,2).sort(function(a,b){
return a-b;
})
}
Thanks!

There are several points of interest here.
First, your implementation of isPrime is inefficient. The fastest way would be using the precached set (it's easy to prepare one), but even without it you should only check a candidate by dividing it by primes, dropping out when the next prime divisor becomes greater than sqrt(cand).
For example, if you already know that 2, 3, 5, and 7 are primes, you can detect that 11 is a prime by dividing it only by 3 (as 5 > Math.sqrt(11)) Oh, and did I mention that you'd never check even numbers greater than 2?
Second, your implementation attempts to find all the primes in the given range, and after that tries to find all the primes that satisfy the "difference condition". But remember, your task actually is finding the first pair of the two prime numbers with the given step.
With that in mind, why don't you just check a number (n), and if it's a prime, attempt to check n + g's primeness? You can cache the result of that check, so that you skip it afterwards.
Here's one possible approach:
function findPrimesByStep(g, m, n) {
let primes = new Set([2]);
function isPrimeNumber(candidate) {
if (primes.has(candidate)) {
return true;
}
let lim = Math.sqrt(candidate);
for (let prime of primes) {
if (prime > lim) {
break;
}
if (candidate % prime === 0) {
return false;
}
}
primes.add(candidate);
return true;
}
n = n - g;
for (let i = 3; i <= n; i += 2) {
let isPrime = isPrimeNumber(i);
if (isPrime && i >= m && isPrimeNumber(i + g)) {
return [i, i + g];
}
}
return null;
}
This is both incomplete and unoptimized (it doesn't check against trivial cases, it skips 2, one doesn't need primes greater than sqrt(n)), but shows how it can be done.

How can I recursively generate an Array of the Fibonacci sequence?

I've seen several posts on generating a given fibonacci sequence, such as this one. However, I can't figure out how to generate the sequence (return an array) of fibonnaci numbers for a given n using recursion. What I have clearly doesn't work, but I really can't figure out how to do it otherwise.
var fibArray = function(n) {
var f = [];
n < 2 ? f.push(n) : f.push(fibArray(n-1) + fibArray(n-2));
return f;
};

A slightly modified version from the previous answer:
function fib(n) {
if (n == 0) return [0]
if (n == 1) return [0, 1]
const arr = fib(n - 1)
return [...arr, arr[n-1] + arr[n-2]]
}
console.log(fib(15))

Notice that you start each function call with an empty array and then only add 1 member to it. That won't work.
You have to add the new element to the array that was returned from the previous fib(n - 1) step. Like so:
function fib (n) {
if (n < 2) {
return [1];
}
if (n < 3) {
return [1, 1];
}
var a = fib(n - 1);
a.push(a[n - 2] + a[n - 3]);
return a;
};
The nth number appears on the position n - 1 on the array. That justifies the n - 2 = n - 1 - 1 and n - 3 = n - 2 - 1.

This is an option without spread operator and with an option to start the sequence when you want:
function fibonacciRecursion(amountNumbers = 4, sequence = [0, 1]) {
if (amountNumbers > 0) {
sequence.push(sequence[sequence.length - 1] + sequence[sequence.length - 2]);
return fibonacciRecursion(amountNumbers - 1, sequence);
}
return sequence
}
console.log(fibonacciRecursion(10, [3,5]))

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