Fibonacci series in JavaScript - javascript

function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = (i - 1);
const b = (i - 2);
result.push(a + b);
}
return result[n];
}
console.log(fib(8));
The output of the code above is 13. I don't understand the for loop part. In very first iteration i = 2, but after second iteration i = 3 so a = 2 and b = 1 and third iteration i = 4 so a = 3, b = 2, and so on... If it's going on final sequence will be :
[0, 1, 1, 3, 5, 7, 9, 11], which is incorrect. The correct sequence will be [0, 1, 1, 2, 3, 5, 8, 13]

You were not using the previous two numbers that are already in the array to > generate the new fibonacci number to be inserted into the array.
https://www.mathsisfun.com/numbers/fibonacci-sequence.html
Here I have used the sum of result[i-2] and result[i-1] to generate the new fibonacci number and pushed it into the array.
Also to generate n number of terms you need the condition to be i < n and not i <= n.
function fib(n) {
const result = [0, 1];
for (var i = 2; i < n; i++) {
result.push(result[i-2] + result[i-1]);
}
return result; // or result[n-1] if you want to get the nth term
}
console.log(fib(8));
Return result[n-1] if you want to get the nth term.

My solution for Fibonacci series:
const fibonacci = n =>
[...Array(n)].reduce(
(acc, val, i) => acc.concat(i > 1 ? acc[i - 1] + acc[i - 2] : i),
[]
)

This function is incorrect. It cat be checked by just adding the console.log call just before the function return:
function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = (i - 1);
const b = (i - 2);
result.push(a + b);
}
console.log(result);
return result[n];
}
console.log(fib(7));
As you can see, the sequence is wrong and (for n = 7) the return value is too.
The possible change would be as following:
function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = result[i - 1];
const b = result[i - 2];
result.push(a + b);
}
console.log(result);
return result[n];
}
console.log(fib(8));
This is the "classical" Fibonacci numbers; if you really want to use the first number of 0, not 1, then you should return result[n-1], since array indexes start from zero.

One approach you could take for fibonacci sequence is recursion:
var fibonacci = {
getSequenceNumber: function(n) {
//base case to end recursive calls
if (n === 0 || n === 1) {
return this.cache[n];
}
//if we already have it in the cache, use it
if (this.cache[n]) {
return this.cache[n];
}
//calculate and store in the cache for future use
else {
//since the function calls itself it's called 'recursive'
this.cache[n] = this.getSequenceNumber(n - 2) + this.getSequenceNumber(n - 1);
}
return this.cache[n];
},
cache: {
0: 0,
1: 1
}
}
//find the 7th number in the fibbonacci function
console.log(fibonacci.getSequenceNumber(7));
//see all the values we cached (preventing extra work)
console.log(fibonacci.cache);
//if you want to output the entire sequence as an array:
console.log(Object.values(fibonacci.cache));
The code above is also an example of a dynamic programming approach. You can see that I am storing each result in a cache object the first time it is calculated by the getSequenceNumber method. This way, the second time that getSequenceNumber is asked to find a given input, it doesn't have to do any actual work - just grab the value from cache and return it! This is an optimization technique that can be applied to functions like this where you may have to find the value of a particular input multiple times.

const fib = n => {
const array = Array(n);
for (i = 0; i < array.length; i++) {
if (i > 1) {
array[i] = array[i - 1] + array[i - 2];
} else {
array[i] = 1;
}
}
return array;
}
console.log(fib(5))

What you are doing wrong is adding the iterator index (i), whereas what you need to do is add the element in the result at that index.
function fib(n) {
const result = [0, 1];
for (let i = 2; i <= n; i++) {
const a = result[(i - 1)];
const b = result[(i - 2)];
result.push(a + b);
}
console.log("Result Array: " + result);
return result[n];
}
console.log("Fibonacci Series element at 8: " + fib(8));

There are two issues with the logic:
Variables a and b currently refer to i - 1 and i - 2. Instead they should refer to the elements of result array, i.e. result[i - 1] and result[i - 2].
If you need 8th element of the array, you need to call result[7]. So the returned value should be result[n - 1] instead of result[n].
function fib(n) {
const result = [0, 1];
for (var i = 2; i < n; i++) {
const a = result[i - 1];
const b = result[i - 2];
result.push(a + b);
}
console.log(result);
return result[n - 1];
}
console.log(fib(8));

simple solution for Fibonacci series:
function fib(n){
var arr = [];
for(var i = 0; i <n; i++ ){
if(i == 0 || i == 1){
arr.push(i);
} else {
var a = arr[i - 1];
var b = arr[i - 2];
arr.push(a + b);
}
}
return arr
}
console.log(fib(8))

This is certainly one of those "more than one way to clean chicken" type situations, this JavaScript method below works for me.
function fibCalc(n) {
var myArr = [];
for (var i = 0; i < n; i++) {
if(i < 2) {
myArr.push(i);
} else {
myArr.push(myArr[i-2] + myArr[i-1]);
}
}
return myArr;
}
fibCalc(8);
When called as above, this produces [0,1,1,2,3,5,8,13]
It allows me to have a sequence of fib numbers based on n.

function fib(n) {
const result = [0];
if (n > 1) {
result.push(1);
for (var i = 2; i < n; i++) {
const a = result[result.length - 1]
const b = result[result.length - 2];
result.push(a + b);
}
}
console.log(result);
}

i came up with this solution to get the n index fibonacci value.
function findFac(n){
if (n===1)
{
return [0, 1];
}
else
{
var s = findFac(n - 1);
s.push(s[s.length - 1] + s[s.length - 2]);
return s;
}
}
function findFac0(n){
var vv1 = findFac(n);
return vv1[n-1];
}
console.log(findFac0(10));

Here, you have it, with few argument check, without using exception handling
function fibonacci(limit){
if(typeof limit != "number"){return "Please enter a natural number";}
if(limit <=0){
return "limit should be at least 1";
}
else if(limit == 1){
return [0];
}
else{
var series = [0, 1];
for(var num=1; num<=limit-2; num++){
series.push(series[series.length-1]+series[series.length-2]);
}
return series;
}
}

I came up with this solution.
function fibonacci(n) {
if (n == 0) {
return [0];
}
if ( n == 1) {
return [0, 1];
} else {
let fibo = fibonacci(n-1);
let nextElement = fibo [n-1] + fibo [n-2];
fibo.push(nextElement);
return fibo;
}
}
console.log(fibonacci(10));

function fibonacciGenerator (n) {
var output = [];
if(n===1){
output=[0];
}else if(n===2){
output=[0,1];
}else{
output=[0,1];
for(var i=2; i<n; i++){
output.push(output[output.length-2] + output[output.length-1]);
}
}
return output;
}
output = fibonacciGenerator();
console.log(output);

function fibonacci(end) {
if (isNaN(end) === false && typeof (end) === "number") {
var one = 0, res, two = 1;
for (var i = 0; i < end; ++i) {
res = one + two;
one = two;
two = res;
console.log(res);
}
} else {
console.error("One of the parameters is not correct!")
}
}
fibonacci(5);

var input = parseInt(prompt(""));
var a =0;
var b=1;
var x;
for(i=0;i<=input;i++){
document.write(a+"<br>")
x = a+b;
a =b;
b= x;
}

Related

Calculate and push fibonacci numbers in an array

let result = [];
function recurseFivonaci(n){
if(n < 2){
return n;
}
return recurseFivonaci(n - 2) + recurseFivonaci(n - 1);
}
console.log(recurseFivonaci(10).map(r => result.push(r)))
TypeError: recurseFivonaci(...).map is not a function
at <anonymous>:16:33
at dn (<anonymous>:16:5449)
How can I push in the result of fibonacci function into an array?
I expect this to work but doesn't:
result = [0,1,1,2,...];
Here's a possible solution
let result=[];
function fibonacci_series(n)
{
if(n === 0){
return [0];
}
if (n===1)
{
return [0, 1];
}
else
{
result = fibonacci_series(n - 1);
result.push(result[result.length - 1] + result[result.length - 2]);
return result;
}
};
console.log(fibonacci_series(10))
Map() is a function available on an Array type in JavaScript and since you're returning a number from your function, it is throwing this error. What you need instead is something like:
let result=[];
function recurseFibonaci (n) {
if(n < 2){
return n;
}
return recurseFibonaci(n-2) + recurseFibonaci(n-1);
}
result.push(recurseFibonaci(10));
console.log(result)
There may be some neato means to achieve it but a basic iteration seems fine?
const arr = [0,1]
for (let i = 0; i < 10; i++) {
arr.push(arr[i] + arr[i + 1]);
}
console.log(arr);

Javascript/ variable place/ what is wrong with the code

I want to write a Generator for Fibonacci numbers in Javascript;
0,1,1,2,5,7,12..... (to make the sequence you have to add the last two numbers)
But I have this problem when I assign the the output.length to a variable the code is not working, if I write it down straight instead of "newNumber" the code down is however working, but I don't understand what is wrong with the first one. Is it something wrong with the place of the variables?
function fibonacciGenerator(n) {
var output = [];
var lastNumber = output[output.length - 1];
var nPrev = output[output.length - 2];
var newNumber = lastNumber + nPrev;
if (n === 1) {
output = [0];
} else if (n === 2) {
output = [0, 1];
} else {
output = [0, 1];
for (var i = 2; i < n; i++) {
output.push(newNumber);
}
}
return output
}
console.log(fibonacciGenerator(5));
function fibonacciGen(n) {
const output = [0, 1];
// Return an empty array if n is less than 1
if (n < 1) {
return [];
}
// If n is 1 or 2, we can return the array now
if (n <2) {
return output;
}
// Loop through the remaining numbers in the sequence
for (let i = 2; i < n; i++) {
// Calculate the next number in the sequence
let lastNumber = output[output.length - 1];
let nPrev = output[output.length - 2];
let newNumber = lastNumber + nPrev;
// Add the new number to the output array
output.push(newNumber);
}
// Return the output array
return output;
}
console.log(fibonacciGen(5));
you have to declare your logic of next term in loop because first time length of output is zero
function fibonacciGenerator (n) {
var output =[];
if (n < 1) {
return [];
}
if (n===1){
output=[0];
}
else if (n===2){
output=[0,1];
}
else{
output=[0,1];
for( var i = 2; i < n; i++){
var lastNumber=output[output.length-1];
var nPrev=output[output.length-2];
var newNumber=lastNumber+nPrev;
output.push(newNumber);
}
}
return output
}

How to write a function generating Fibonacci series smaller than a given number (javascript)?

Here's my code:
function fibs(num) {
//generate Fibonacci numbers:
let arr = [1,1]
let i = 2;
function fibsRange(i) {
arr[i] = arr[i-1] + arr[i-2]
if (arr[i] < num) {
fibsRange(i+1);//call function one more time;
}
return arr
}
return fibsRange();
}
console.log(fibs(5));
Assume that given number (num) is bigger than 2. Where do I get wrong?
Note: I edited my code.
first :as mentionned by Phil your function (the first one) needs to return something.
second: you don't need 2 functions, one will do the job.
here's a working version of your code (I also modified some unnecessary code.
Edit: I added the case when you need to parse just one parameter
function fibsRange(i,arr,num) {
if(arr[i-1]+arr[i-2] > num) return arr
else{
arr[i] = arr[i - 1] + arr[i - 2]
return fibsRange(i + 1,arr,num);
}
}
//in case you need just one parameter
function fib(num){
let arr = [1, 1];
let i = 2;
return fibsRange(i,arr,num);
}
//those(declaration of arr and i) are unnecessary in case you use fib
let arr = [1, 1]
let i = 2;
console.log(fibsRange(i,arr,5));
console.log(fib(5));
You never called your fibsRange() function and never returned anything from fibs().
The following version works.
function fibs(num) {
//generate Fibonacci numbers:
let arr = [1, 1]
if (num>1) fibsRange(2);
function fibsRange(i) {
arr[i] = arr[i - 1] + arr[i - 2]
if (arr[i] < num) {
fibsRange(i + 1); //call function one more time;
}
}
return arr;
}
console.log(fibs(5)); //undified;
Your code is checking if arr[i] < num, then invoke the fibsRange(i+1). This is a wrong logic, you will always get the fibbonacci number just higher than the num because you are already setting it at arr[i] and then checking the condition. However, you should first check if arr[i] < num, and then push it into arr.
function fibs(num) {
//generate Fibonacci numbers:
let arr = [1, 1];
let i = 2;
function fibsRange(i) {
const lastFibNumber = arr[i - 1] + arr[i - 2];
if (lastFibNumber < num) {
arr.push(lastFibNumber);
fibsRange(i + 1); //call function one more time;
}
return arr;
}
return fibsRange(i);
}
console.log(fibs(5));
function fibs(num) {
//generate Fibonacci numbers:
let arr = [1, 1];
while(arr[arr.length-1] <= num){
let y = arr[arr.length-1] + arr[arr.length-2];
if(y > num){
break; // break if new y is greater than num
}
arr.push(y); // add y to the serie
}
return arr; //// return the serie
}
console.log(fibs(50));

How to Make Consecutive Numbers from an Array

I have an array like this
[-2,4,5,6,7,8,10,11,15,16,17,18,21]
Is anyone know, how to make the output from that array become integer like this
-2,4-8,10-11,15-18,21
The output will take the consecutive number become one number
This things is new for me, any help would be appreciated, thanks
Below I created function for replacing a sequence in an array with a string containing its range. There are three functions.
getConsectiveCount will take array and index as arguments and will get the count of consecutive numbers after that.
replaceFirstConsective will take array and will replace only first sequence in the array.
replaceAllConsectives will replace all the sequences in an array.
const arr = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
const getConsectiveCount = (arr, index) => {
let count = 0;
for(let i = index; i < arr.length; i++){
if(arr[i + 1] === arr[index] + (i - index) + 1){
count++;
}
}
return count;
}
console.log(getConsectiveCount(arr, 1));
const replaceFirstConsective = (arr) => {
for(let i = 0; i < arr.length; i++){
let count = getConsectiveCount(arr,i);
if(count){
return [...arr.slice(0, i), `${arr[i]}-${arr[i + count]}`, ...arr.slice(i + count + 1)]
}
}
return arr;
}
const replaceAllConsectives = (arr) => {
for(let i = 0; i < arr.length;i++){
arr = replaceFirstConsective(arr)
}
return arr;
}
console.log(JSON.stringify(replaceAllConsectives(arr)))
const inp = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
let res = [];
for(let i=0;i<inp.length;i++){
let b = inp[i];
let j = i+1;
while(j<inp.length){
if(b+1 == inp[j]){
b = inp[j++];
continue;
}
break;
}
if(i == j-1){
res.push(inp[i]);
}
else{
res.push(inp[i]+"-"+inp[j-1]);
i=j-1;
}
}
console.log(res);
Check this if it helps.
I have done that :
const arr1 = [-2,4,5,6,7,8,10,11,15,16,17,18,21]
const arr2 = arr1.reduce((a,c,i,{[i+1]:nxt})=>
{
if (!a.s1) a.s1 = c.toString(10)
if ( (c+1) !== nxt )
{
a.s1 += a.s2 ? `_${a.s2}` : ''
a.r.push(a.s1)
a.s1 = a.s2 = ''
}
else a.s2 = nxt.toString(10)
return (nxt===undefined) ? a.r : a
},{r:[],s1:'',s2:''})
console.log(JSON.stringify( arr2 ))
.as-console-wrapper { max-height: 100% !important; top: 0; }
It's an interesting problem. The way I solved it was to iterate over the array and then find the index of the last consecutive number from the current one. We can either write the single number into the result array or write the range string in there and continue from the next number after the range.
function lastConsecutive(arr, start)
{
let ind = start;
while(ind < arr.length && (arr[ind] + 1) == arr[ind + 1])
{
ind++;
}
return ind;
}
function consecCollapse(nums)
{
let i = 0;
const result = [];
while (i < nums.length)
{
let n = lastConsecutive(nums, i);
result.push((n == i) ? nums[n]+"" : nums[i]+"-"+nums[n]);
i = n + 1;
}
return result;
}
console.log(consecCollapse([-2,4,5,6,7,8,10,11,15,16,17,18,21]));
const arr = [-2,4,5,6,7,8,10,11,15,16,17,18,21];
const _newArray = [];
let start = arr[0];
let end = start;
for(let i=1; i<=arr.length; i++) {
let elem = arr[i];
if (elem === end+1) {
end = elem; // update the end value (range)
}else {
if (end !== start) {
_newArray.push(`${start}-${end}`);
} else {
_newArray.push(start);
}
start = elem;
end = start;
}
}
console.log(_newArray.join(','))
If a sorted array with unique numbers, this is how I find the range of numbers with dash:
function findRange(arr) {
const results = []
for (let i = 0; i < arr.length; i++) {
// only more than 2 consecutive numbers can be form a range
if (arr[i + 1] === arr[i] + 1 && arr[i + 2] === arr[i] + 2) {
// store the first number of a range
results.push(arr[i])
// loop until meet the next one is not consecutive
while (arr[i] + 1 === arr[i + 1]) {
i++
}
// store the last number of a range with '-' in between
results[results.length - 1] = results[results.length - 1] + '-' + arr[i]
} else {
// if only 2 consecutive number or not consecutive at all
results.push(arr[i])
}
}
return results
}
console.log(findRange([1, 2, 3, 4, 6, 7, 8, 9]))
console.log(findRange([1, 2, 4, 6, 7, 8, 9]))
console.log(findRange([-2,4,5,6,7,8,10,11,15,16,17,18,21]))

Javascript least common multiple algorithm

I'm trying to script a function that takes two numbers and returns the smallest common multiple that is also divisible by all the numbers between those numbers, what I've got only works for 1,1 through 1,12, but for some reason stops working at 1,13. Other set like 12,14 work but I can't figure out why or what the pattern is.
function smallestCommons(arr) {
arr.sort(function(a, b) {
return a-b;
});
var arr1 = [];
var arr2 = [];
for (var k = arr[0]; k<=arr[1]; k++) {
arr1.push(k);
}
function remainder(val1, val2) {
return val1%val2;
}
var b = arr1.reduce(function(a, b) {
return a*b;
});
var i = arr1[arr1.length-1]*arr1[arr1.length-2];
while (i<=b) {
for (var m = 0; m<arr1.length; m++) {
var a = remainder(i, arr1[m]);
arr2.push(a);
}
var answer = arr2.reduce(function(c, d) {
return c+d;
});
if (answer === 0) {
return i;
} else {
arr2 = [];
i++;
}
}
}
I guess you can do as follows in JavaScript; It can calculate the common LCM up to an 216 item array, such as [1,2,3,...,216] in less than 0.25 ms.
function gcd(a,b){
var t = 0;
a < b && (t = b, b = a, a = t); // swap them if a < b
t = a%b;
return t ? gcd(b,t) : b;
}
function lcm(a,b){
return a/gcd(a,b)*b;
}
var arr = [1,2,3,4,5,6,7,8,9,10,11,12,13],
brr = Array(216).fill().map((_,i) => i+1), // limit before Infinity
result = arr.reduce(lcm);
console.log(result);
console.time("limit");
result = brr.reduce(lcm);
console.timeEnd("limit");
console.log(result);
A way is to keep multiplying the largest number in your range with an increasing number and check if all the others are divisible by that. If yes, return that or continue the loop.
Here is my solution in typescript...
function findLowestCommonMultipleBetween(start: number, end: number): number {
let numbers: number[] = [];
for (let i = start; i <= end; i++) {
numbers.push(i);
}
for (let i = 1; true; i++) {
let divisor = end * i;
if (numbers.every((number) => divisor % number == 0)) {
return divisor;
}
}
}
...but for larger ranges, this is a more efficient answer :)
As far as I can tell your algorithm is giving you a correct answer.
I am far from being a professional programmer so anyone who wants please give options to improve my code or its style :)
If you want to be able to check for the answer yourself you can check this fiddle:
https://jsfiddle.net/cowCrazy/Ld8khrx7/
function multiplyDict(arr) {
arr.sort(function (a, b) {
return a - b;
});
if (arr[0] === 1) {
arr[0] = 2;
}
var currentArr = [];
for (var i = arr[0]; i <= arr[1]; i++) {
currentArr.push(i);
}
var primeDivs = allPrimes(arr[1]);
var divsDict = {};
for (var j = currentArr[0]; j <= currentArr[currentArr.length -1]; j++){
divsDict[j] = [];
if (primeDivs.indexOf(j) > -1) {
divsDict[j].push(j);
} else {
var x = j;
for (var n = 2; n <= Math.floor(j / 2); n++) {
if (x % n === 0) {
divsDict[j].push(n);
x = x / n;
n--;
continue;
}
}
}
}
return divsDict;
}
function allPrimes(num) {
var primeArr = [];
var smallestDiv = 2;
loopi:
for (var i = 2; i <= num; i++) {
loopj:
for (var j = smallestDiv; j <= largestDiv(i); j++) {
if (i % j === 0) {
continue loopi;
}
}
primeArr.push(i);
}
return primeArr;
}
function largestDiv (a) {
return Math.floor(Math.sqrt(a));
}
multiplyDict([1,13]);
it gives a dictionary of the requested array and the divisors of each element.
from there you can go on your own to check that your algorithm is doing the right job or you can check it here:
https://jsfiddle.net/cowCrazy/kr04mas7/
I hope it helps
It is true! The result of [1, 13] is 360360. and after this we have [1, 14].
14 = 2 * 7 and we now 360360 is dividable to 2 and 7 so the answer is 360360 again.
[1, 15]: 15 = 3 * 5 and result is same.
[1, 16]: result is 720720.
[1, 17]: result is: 12252240
[1, 18]: 18 = 2 * 9 and result is 12252240 same as 17
[1, 19]: for my computer this process is so heavy and can not do this. But in a strong machine it will work. I promise. But your code is not good in performance.
To find the LCM in N numbers.
It is Compatible with ES6, and consider that is there is no control for boundaries in case that we need to find for large numbers.
var a = [10, 40, 50, 7];
console.log(GetMinMultiple(a));
function GetMinMultiple(data) {
var maxOf = data.reduce((max, p) => p > max ? p : max, 0);
var incremental = maxOf;
var found = false;
do {
for (var j = 0; j < data.length; j++) {
if (maxOf % data[j] !== 0) {
maxOf += incremental;
break;
}
else {
if (j === data.length - 1) {
found = true;
break;
}
}
}
} while (!found);
return maxOf;
}
https://jsfiddle.net/djp30gfz/
Here is my solution in Typescript
function greatestCommonDivider(x: number, y: number): number {
if (y === 0) {
return x;
}
return greatestCommonDivider(y, x % y);
}
function singleLowestCommonMultiply(x: number, y: number): number {
return (x * y) / greatestCommonDivider(x, y);
}
function lowestCommonMultiply(...numbers: number[]): number {
/**
* For each number, get it's lowest common multiply with next number.
*
* Then using new number, compute new lowest common multiply
*/
return numbers.reduce((a, b) => {
return singleLowestCommonMultiply(a, b);
});
}
lowestCommonMultiply(2, 3); // Outputs 6
lowestCommonMultiply(2, 3, 5); // Outputs 30
Playground - click here

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