Here is the excel formula:
=E2-(J2*2*G2)-(K2*E2*2)/30+IF((L2+M2)>60,((L2+M2)-60)*H2+60*G2,(L2+M2)*G2)+(N2/$AE$1)*$AE$2+(Q2*$AE$5+P2*$AE$4+O2*$AE$3)
Here what I tried (Javascript):
var E2 = $("#sentence").val();
var J2 = $("#deduction").val();
var G2 = 55145;
var K2 = $("#absence").val();
var L2 = $("#overtime").val();
var M2 = 0;
var H2 = 50050;
var N2 = $("#transportation").val();
var sixty;
if ((L2 + M2) > 60) {
sixty = ((L2 + M2) - 60) * H2 + 60 * G2;
} else {
sixty = (L2 + M2) * G2;
};
var result = E2 - (J2 * 2 * G2) - (K2 * E2 * 2) / 30 + sixty;
I couldn't find the way to conver this part of formula:
+(N2/$AE$1)*$AE$2+(Q2*$AE$5+P2*$AE$4+O2*$AE$3)
Here I found the problem:
Even if one of the variables sets to null, then the formula does not work properly.
It looks like some pretty basic math.
let total = (N2 / AE1) * AE2 + (Q2 * AE5 + P2 * AE4 + O2 * AE3 )
This is basically impossible to translate without seeing the actual spreadsheet but that should get you started. Also, make sure to take into consideration order of operations because the computer is going to evaluate it from left to right unless there are parenthesis (where it will evaluate those first).
I'm just hoping someone could double check my implementations to make sure everything is correct on the math and JS sides. I'm wondering where my implementation went wrong b/c the formulas do not compute the same probabilities, specifically probability_three seems to have a computation error I cannot find.
Here's the JSFiddle: https://jsfiddle.net/s73Lh1fk/5/
There are 3 functions I'm hoping to get reviewed.
probability_one
probability_two
probability_three
The formulas are as follows.
P(d, n) = 1 - (d - 1 / d)^n
P(d, n, o) = 1 - (d - (d - o + 1) / d)^n
P(d, n, o, k) = nCk * (p * k) * q^(n−k)
for nCk, I implemented https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
function factorialize(num) {
// Step 1. Create a variable result to store num
var result = num;
// If num = 0 OR num = 1, the factorial will return 1
if (num === 0 || num === 1)
return 1;
// Instatiate the while loop
while (num > 1) {
// Decrement by 1
num--
// Update the result
result *= num
}
// Return
return result;
}
function nCr(n, r){
// Compute n factorial
var nFact = factorialize(n)
// Compute r factorial
var rFact = factorialize(r)
// Compute n - r factorial
var nrFact = factorialize(n - r)
// Compute nCr
var result = (nFact / (rFact * nrFact))
// Return
return result
}
If you need any more information, please don't hesitate to ask. I'll do my best to provide it.
For example:
The result here is 9.75%
function probability_one() {
// Get number of dice
var n = document.getElementById("n1").value
// Get sides on each die
var d = document.getElementById("d1").value
// Calculate
var prob = 1 - ((d - 1) / d)**n
prob = parseFloat(prob*100).toFixed(2)+"%"
prob = `<B>${prob}</B>`
// Print the probability
var p = document.createElement('p')
p.innerHTML = prob
document.getElementById("output1").appendChild(p)
}
The result here is 9.75%
function probability_two() {
// Get number of dice
var n = document.getElementById("n2").value
// Get sides on each die
var d = document.getElementById("d2").value
// Get the specific outcome
var o = document.getElementById("o2").value
// Calculate
var prob = 1 - ((d - (d - o + 1)) / d)**n
prob = parseFloat(prob*100).toFixed(2)+"%"
prob = `<B>${prob}</B>`
// Print the probability
var p = document.createElement('p')
p.innerHTML = prob
document.getElementById("output2").appendChild(p)
}
The result here is 18.00%, not 9.75%
function probability_three(){
// USER INPUTS
// Get number of dice
var n = document.getElementById("n3").value
// Get sides on each die
var d = document.getElementById("d3").value
// Get the value that defines a success
var o = document.getElementById("o3").value
// Get the number of success needed
var k = document.getElementById("k3").value
// CALCULATIONS
// p
var p = ((d - o) + 1)/10
console.log(`p: ${p}`)
// q
var q = (1 - p)
console.log(`q: ${q}`)
// nCr
var vnCr = nCr(n, k)
console.log(`nCr: ${vnCr}`)
// p**k
var pk = p**k
console.log(`p**k: ${pk}`)
// q**n-k
var pnk = q**(n-k)
console.log(`q**n-k: ${pnk}`)
// Probability
var prob = (vnCr * pk * pnk) / (p + q)**n
console.log(`Prob.: ${prob}`)
prob = parseFloat(prob*100).toFixed(2)+"%"
prob = `<B>${prob}</B>`
// Print the probability
var p = document.createElement('p')
p.innerHTML = prob
document.getElementById("output3").appendChild(p)
}
I need to make the function so that it calculate the area when I enter function f and coefficients a, b and n on the command line.
For now I have this:
module.exports = function (f,a,b,n,next) {
parseFloat(a);
parseFloat(b);
parseInt(n);
var h = (b-a)/n;
var s = 0;
var suma = function () {
for (var i = 0; i <= n; i++) {
var ximj = a+h*(i-1);
var xi = a+h*i;
var x = (ximj + xi)/2;
s += eval(f,x);
}
return s;
};
if (n<1) {
next(new Error("Koeficijent n je broj ekvidistantnih točaka."))
}
else next(null, {povrsina : function () {return h*suma();}}
);
I think that the function suma() isn't working like it should be working.
In command line this should be look like :
f: x*x-2*x+7
a: 2
b: 4
n: 10
Povrsina ispod grafa funkcije f je 20.66000...
My prompt file looks like this:
var pov = require('./integrator.js');
var prompt = require('prompt');
prompt.get (['f','a','b','n'],function (err,koef) {
if (err) return console.log(err);
console.log("f: ",koef.f);
console.log("a: ",koef.a);
console.log("b: ",koef.b);
console.log("n: ",koef.n);
parseFloat(koef.a);
parseFloat(koef.b);
parseInt(koef.n);
pov(koef.f,koef.a,koef.b,koef.n,function(err,rj){
if (err)
console.log(err);
else
console.log("Povrsina ispod grafa funkcije f je " + rj.povrsina());
});
Thank you for help :)
First of all, if you want to calculate the area under a curve you are going to need to provide a lower and upper x limit, otherwise the answer will always be infinite. Personally I use this function that I wrote:
function findArea(lowerLimit, upperLimit, coefficients) {
let lower = 0
let upper = 0
for (let i = 0; i < coefficients.length; i++) {
lower += (coefficients[i] * Math.pow(lowerLimit, i + 1)) / (i + 1)
upper += (coefficients[i] * Math.pow(upperLimit, i + 1)) / (i + 1)
}
return Math.abs(upper - lower)
}
If you pass in two numbers as the limits and an array of coefficients in the format, a + bx + cx^2 + dx^3 ... and so on, you will get the correct answer. For example, say you have the equation y = 5 + 4x + 3x^2 + 2x^3 and you want to find the area under the curve between x = 1 and x = 3 you can call this function like so, findArea(1, 3, [5, 4, 3, 2]) and you will get the answer 92 which represents the number of units squared. Keep in mind that this works only if the area you are evaluating is either entirely above or entirely below the x axis. If your area crosses the x axis you will need to calculate the area above and below separately!
You can check the results against the Symbolab Area Under The Curve Calculator but this should work for any curve.
The eval function reads a string and evaluates it. If you want to evaluate the function f with the parameter x, then you can just write f(x) instead of calling eval(). This is because, in JavaScript, functions are first-class values and can be passed around just like numbers, strings, and so on.
In suma, particularly the line var ximj = a+h*(i-1);, your for-loop multiplies h by -1, then by 0, then by 1. I suspect you meant var ximj = a+h*(i+1); since h is the width of your differential. But that would only produce small errors.
As Lends wrote, you also need to correct your Parse* calls:
a = parseFloat(a);
b = parseFloat(b);
n = parseInt(n);
You should use math.js to eval a string math expression, for example:
var math = require('mathjs');
var integral = function(f, a, b, n, next) {
a = parseFloat(a);
b = parseFloat(b);
n = parseInt(n);
var h = (b - a) / n;
var s = 0;
var suma = function() {
for (var i = 0; i <= n; i++) {
var ximj = a + h * (i - 1);
var xi = a + h * i;
var x = (ximj + xi) / 2;
s += math.eval(f, { x: x });
}
return s;
};
if (n < 1) {
next(new Error("Koeficijent n je broj ekvidistantnih točaka."))
} else {
next(null, { povrsina: function() { return h * suma(); } });
}
};
var f = 'x * x - 2 * x + 7';
var a = '2';
var b = '4';
var n = '10000';
integral(f, a, b, n, function(err, rj) {
if (err)
console.log(err);
else
console.log("Povrsina ispod grafa funkcije f je " + rj.povrsina());
});
/* output:
Povrsina ispod grafa funkcije f je 20.66806662000201
*/
You integral function seems to have an has an issue though, n must be very high to get near the expected result (like 10000). Or is it supposed to work that way? For n = 10, it gives 22 for example.
Another solution is to use an existing package or at least read the source code to get an idea.
is there a JavaScript implementation of the Inverse Error Function?
This would implement the Gauss inverse error function. Approximations are ok.
Why yes. There is.
The following code uses built-in JavaScript functions and implments Abramowitz and Stegun's algorithm as described here:
function erfinv(x){
var z;
var a = 0.147;
var the_sign_of_x;
if(0==x) {
the_sign_of_x = 0;
} else if(x>0){
the_sign_of_x = 1;
} else {
the_sign_of_x = -1;
}
if(0 != x) {
var ln_1minus_x_sqrd = Math.log(1-x*x);
var ln_1minusxx_by_a = ln_1minus_x_sqrd / a;
var ln_1minusxx_by_2 = ln_1minus_x_sqrd / 2;
var ln_etc_by2_plus2 = ln_1minusxx_by_2 + (2/(Math.PI * a));
var first_sqrt = Math.sqrt((ln_etc_by2_plus2*ln_etc_by2_plus2)-ln_1minusxx_by_a);
var second_sqrt = Math.sqrt(first_sqrt - ln_etc_by2_plus2);
z = second_sqrt * the_sign_of_x;
} else { // x is zero
z = 0;
}
return z;
}
function provided earlier in this post did not work for me... NaN result on a 33meter circle with confidence 65% represented as 65.0 ... I wrote the following based on an equation listed here https://en.wikipedia.org/wiki/Error_function#Inverse_functions and it worked fine:
var _a = ((8*(Math.PI - 3)) / ((3*Math.PI)*(4 - Math.PI)));
function erfINV( inputX )
{
var _x = parseFloat(inputX);
var signX = ((_x < 0) ? -1.0 : 1.0 );
var oneMinusXsquared = 1.0 - (_x * _x);
var LNof1minusXsqrd = Math.log( oneMinusXsquared );
var PI_times_a = Math.PI * _a ;
var firstTerm = Math.pow(((2.0 / PI_times_a) + (LNof1minusXsqrd / 2.0)), 2);
var secondTerm = (LNof1minusXsqrd / _a);
var thirdTerm = ((2 / PI_times_a) + (LNof1minusXsqrd / 2.0));
var primaryComp = Math.sqrt( Math.sqrt( firstTerm - secondTerm ) - thirdTerm );
var scaled_R = signX * primaryComp ;
return scaled_R ;
}
Here's an alternative implementation of Abramowitz and Stegun's algorithm (equivalent to ptmalcolm's answer, but more succinct and twice as fast):
function erfinv(x) {
// maximum relative error = .00013
const a = 0.147
//if (0 == x) { return 0 }
const b = 2/(Math.PI * a) + Math.log(1-x**2)/2
const sqrt1 = Math.sqrt( b**2 - Math.log(1-x**2)/a )
const sqrt2 = Math.sqrt( sqrt1 - b )
return sqrt2 * Math.sign(x)
}
You can test the speed with console.time("erfinv"); for (let i=0; i<1000000000; i++) {erfinv(i/1000000000)}; console.timeEnd("erfinv")
The if statement optimization is commented out as it doesn't seem to make a difference - presumably the interpreter recognizes that this is all one equation.
If you need a more accurate approximation, check out Wikipedia.
I've become interested in algorithms lately, and the fibonacci sequence grabbed my attention due to its simplicity.
I've managed to put something together in javascript that calculates the nth term in the fibonacci sequence in less than 15 milliseconds after reading lots of information on the web. It goes up to 1476...1477 is infinity and 1478 is NaN (according to javascript!)
I'm quite proud of the code itself, except it's an utter monster.
So here's my question:
A) is there a faster way to calculate the sequence?
B) is there a faster/smaller way to multiply two matrices?
Here's the code:
//Fibonacci sequence generator in JS
//Cobbled together by Salty
m = [[1,0],[0,1]];
odd = [[1,1],[1,0]];
function matrix(a,b) {
/*
Matrix multiplication
Strassen Algorithm
Only works with 2x2 matrices.
*/
c=[[0,0],[0,0]];
c[0][0]=(a[0][0]*b[0][0])+(a[0][1]*b[1][0]);
c[0][1]=(a[0][0]*b[0][1])+(a[0][1]*b[1][1]);
c[1][0]=(a[1][0]*b[0][0])+(a[1][1]*b[1][0]);
c[1][1]=(a[1][0]*b[0][1])+(a[1][1]*b[1][1]);
m1=(a[0][0]+a[1][1])*(b[0][0]+b[1][1]);
m2=(a[1][0]+a[1][1])*b[0][0];
m3=a[0][0]*(b[0][1]-b[1][1]);
m4=a[1][1]*(b[1][0]-b[0][0]);
m5=(a[0][0]+a[0][1])*b[1][1];
m6=(a[1][0]-a[0][0])*(b[0][0]+b[0][1]);
m7=(a[0][1]-a[1][1])*(b[1][0]+b[1][1]);
c[0][0]=m1+m4-m5+m7;
c[0][1]=m3+m5;
c[1][0]=m2+m4;
c[1][1]=m1-m2+m3+m6;
return c;
}
function fib(n) {
mat(n-1);
return m[0][0];
}
function mat(n) {
if(n > 1) {
mat(n/2);
m = matrix(m,m);
}
m = (n%2<1) ? m : matrix(m,odd);
}
alert(fib(1476)); //Alerts 1.3069892237633993e+308
The matrix function takes two arguments: a and b, and returns a*b where a and b are 2x2 arrays.
Oh, and on a side note, a magical thing happened...I was converting the Strassen algorithm into JS array notation and it worked on my first try! Fantastic, right? :P
Thanks in advance if you manage to find an easier way to do this.
Don't speculate, benchmark:
edit: I added my own matrix implementation using the optimized multiplication functions mentioned in my other answer. This resulted in a major speedup, but even the vanilla O(n^3) implementation of matrix multiplication with loops was faster than the Strassen algorithm.
<pre><script>
var fib = {};
(function() {
var sqrt_5 = Math.sqrt(5),
phi = (1 + sqrt_5) / 2;
fib.round = function(n) {
return Math.floor(Math.pow(phi, n) / sqrt_5 + 0.5);
};
})();
(function() {
fib.loop = function(n) {
var i = 0,
j = 1;
while(n--) {
var tmp = i;
i = j;
j += tmp;
}
return i;
};
})();
(function () {
var cache = [0, 1];
fib.loop_cached = function(n) {
if(n >= cache.length) {
for(var i = cache.length; i <= n; ++i)
cache[i] = cache[i - 1] + cache[i - 2];
}
return cache[n];
};
})();
(function() {
//Fibonacci sequence generator in JS
//Cobbled together by Salty
var m;
var odd = [[1,1],[1,0]];
function matrix(a,b) {
/*
Matrix multiplication
Strassen Algorithm
Only works with 2x2 matrices.
*/
var c=[[0,0],[0,0]];
var m1=(a[0][0]+a[1][1])*(b[0][0]+b[1][1]);
var m2=(a[1][0]+a[1][1])*b[0][0];
var m3=a[0][0]*(b[0][1]-b[1][1]);
var m4=a[1][1]*(b[1][0]-b[0][0]);
var m5=(a[0][0]+a[0][1])*b[1][1];
var m6=(a[1][0]-a[0][0])*(b[0][0]+b[0][1]);
var m7=(a[0][1]-a[1][1])*(b[1][0]+b[1][1]);
c[0][0]=m1+m4-m5+m7;
c[0][1]=m3+m5;
c[1][0]=m2+m4;
c[1][1]=m1-m2+m3+m6;
return c;
}
function mat(n) {
if(n > 1) {
mat(n/2);
m = matrix(m,m);
}
m = (n%2<1) ? m : matrix(m,odd);
}
fib.matrix = function(n) {
m = [[1,0],[0,1]];
mat(n-1);
return m[0][0];
};
})();
(function() {
var a;
function square() {
var a00 = a[0][0],
a01 = a[0][1],
a10 = a[1][0],
a11 = a[1][1];
var a10_x_a01 = a10 * a01,
a00_p_a11 = a00 + a11;
a[0][0] = a10_x_a01 + a00 * a00;
a[0][1] = a00_p_a11 * a01;
a[1][0] = a00_p_a11 * a10;
a[1][1] = a10_x_a01 + a11 * a11;
}
function powPlusPlus() {
var a01 = a[0][1],
a11 = a[1][1];
a[0][1] = a[0][0];
a[1][1] = a[1][0];
a[0][0] += a01;
a[1][0] += a11;
}
function compute(n) {
if(n > 1) {
compute(n >> 1);
square();
if(n & 1)
powPlusPlus();
}
}
fib.matrix_optimised = function(n) {
if(n == 0)
return 0;
a = [[1, 1], [1, 0]];
compute(n - 1);
return a[0][0];
};
})();
(function() {
var cache = {};
cache[0] = [[1, 0], [0, 1]];
cache[1] = [[1, 1], [1, 0]];
function mult(a, b) {
return [
[a[0][0] * b[0][0] + a[0][1] * b[1][0],
a[0][0] * b[0][1] + a[0][1] * b[1][1]],
[a[1][0] * b[0][0] + a[1][1] * b[1][0],
a[1][0] * b[0][1] + a[1][1] * b[1][1]]
];
}
function compute(n) {
if(!cache[n]) {
var n_2 = n >> 1;
compute(n_2);
cache[n] = mult(cache[n_2], cache[n_2]);
if(n & 1)
cache[n] = mult(cache[1], cache[n]);
}
}
fib.matrix_cached = function(n) {
if(n == 0)
return 0;
compute(--n);
return cache[n][0][0];
};
})();
function test(name, func, n, count) {
var value;
var start = Number(new Date);
while(count--)
value = func(n);
var end = Number(new Date);
return 'fib.' + name + '(' + n + ') = ' + value + ' [' +
(end - start) + 'ms]';
}
for(var func in fib)
document.writeln(test(func, fib[func], 1450, 10000));
</script></pre>
yields
fib.round(1450) = 4.8149675025003456e+302 [20ms]
fib.loop(1450) = 4.81496750250011e+302 [4035ms]
fib.loop_cached(1450) = 4.81496750250011e+302 [8ms]
fib.matrix(1450) = 4.814967502500118e+302 [2201ms]
fib.matrix_optimised(1450) = 4.814967502500113e+302 [585ms]
fib.matrix_cached(1450) = 4.814967502500113e+302 [12ms]
Your algorithm is nearly as bad as uncached looping. Caching is your best bet, closely followed by the rounding algorithm - which yields incorrect results for big n (as does your matrix algorithm).
For smaller n, your algorithm performs even worse than everything else:
fib.round(100) = 354224848179263100000 [20ms]
fib.loop(100) = 354224848179262000000 [248ms]
fib.loop_cached(100) = 354224848179262000000 [6ms]
fib.matrix(100) = 354224848179261900000 [1911ms]
fib.matrix_optimised(100) = 354224848179261900000 [380ms]
fib.matrix_cached(100) = 354224848179261900000 [12ms]
There is a closed form (no loops) solution for the nth Fibonacci number.
See Wikipedia.
There may well be a faster way to calculate the values but I don't believe it's necessary.
Calculate them once and, in your program, output the results as the fibdata line below:
fibdata = [1,1,2,3,5,8,13, ... , 1.3069892237633993e+308]; // 1476 entries.
function fib(n) {
if ((n < 0) || (n > 1476)) {
** Do something exception-like or return INF;
}
return fibdata[n];
}
Then, that's the code you ship to your clients. That's an O(1) solution for you.
People often overlook the 'caching' solution. I once had to write trigonometry routines for an embedded system and, rather than using infinite series to calculate them on the fly, I just had a few lookup tables, 360 entries in each for each of the degrees of input.
Needless to say, it screamed along, at the cost of only about 1K of RAM. The values were stored as 1-byte entries, [actual value (0-1) * 16] so we could just do a lookup, multiply and bit shift to get the desired value.
My previous answer got a bit crowded, so I'll post a new one:
You can speed up your algorithm by using vanilla 2x2 matrix multiplication - ie replace your matrix() function with this:
function matrix(a, b) {
return [
[a[0][0] * b[0][0] + a[0][1] * b[1][0],
a[0][0] * b[0][1] + a[0][1] * b[1][1]],
[a[1][0] * b[0][0] + a[1][1] * b[1][0],
a[1][0] * b[0][1] + a[1][1] * b[1][1]]
];
}
If you care for accuracy and speed, use the caching solution. If accuracy isn't a concern, but memory consumption is, use the rounding solution. The matrix solution only makes sense if you want results for big n fast, don't care for accuracy and don't want to call the function repeatedly.
edit: You can even further speed up the computation if you use specialised multiplication functions, eliminate common subexpressions and replace the values in the existing array instead of creating a new array:
function square() {
var a00 = a[0][0],
a01 = a[0][1],
a10 = a[1][0],
a11 = a[1][1];
var a10_x_a01 = a10 * a01,
a00_p_a11 = a00 + a11;
a[0][0] = a10_x_a01 + a00 * a00;
a[0][1] = a00_p_a11 * a01;
a[1][0] = a00_p_a11 * a10;
a[1][1] = a10_x_a01 + a11 * a11;
}
function powPlusPlus() {
var a01 = a[0][1],
a11 = a[1][1];
a[0][1] = a[0][0];
a[1][1] = a[1][0];
a[0][0] += a01;
a[1][0] += a11;
}
Note: a is the name of the global matrix variable.
Closed form solution in JavaScript: O(1), accurate up for n=75
function fib(n){
var sqrt5 = Math.sqrt(5);
var a = (1 + sqrt5)/2;
var b = (1 - sqrt5)/2;
var ans = Math.round((Math.pow(a, n) - Math.pow(b, n))/sqrt5);
return ans;
}
Granted, even multiplication starts to take its expense when dealing with huge numbers, but this will give you the answer. As far as I know, because of JavaScript rounding the values, it's only accurate up to n = 75. Past that, you'll get a good estimate, but it won't be totally accurate unless you want to do something tricky like store the values as a string then parse those as BigIntegers.
How about memoizing the results that where already calculated, like such:
var IterMemoFib = function() {
var cache = [1, 1];
var fib = function(n) {
if (n >= cache.length) {
for (var i = cache.length; i <= n; i++) {
cache[i] = cache[i - 2] + cache[i - 1];
}
}
return cache[n];
}
return fib;
}();
Or if you want a more generic memoization function, extend the Function prototype:
Function.prototype.memoize = function() {
var pad = {};
var self = this;
var obj = arguments.length > 0 ? arguments[i] : null;
var memoizedFn = function() {
// Copy the arguments object into an array: allows it to be used as
// a cache key.
var args = [];
for (var i = 0; i < arguments.length; i++) {
args[i] = arguments[i];
}
// Evaluate the memoized function if it hasn't been evaluated with
// these arguments before.
if (!(args in pad)) {
pad[args] = self.apply(obj, arguments);
}
return pad[args];
}
memoizedFn.unmemoize = function() {
return self;
}
return memoizedFn;
}
//Now, you can apply the memoized function to a normal fibonacci function like such:
Fib = fib.memoize();
One note to add is that due to technical (browser security) constraints, the arguments for memoized functions can only be arrays or scalar values. No objects.
Reference: http://talideon.com/weblog/2005/07/javascript-memoization.cfm
To expand a bit on Dreas's answer:
1) cache should start as [0, 1]
2) what do you do with IterMemoFib(5.5)? (cache[5.5] == undefined)
fibonacci = (function () {
var FIB = [0, 1];
return function (x) {
if ((typeof(x) !== 'number') || (x < 0)) return;
x = Math.floor(x);
if (x >= FIB.length)
for (var i = FIB.length; i <= x; i += 1)
FIB[i] = FIB[i-1] + FIB[i-2];
return FIB[x];
}
})();
alert(fibonacci(17)); // 1597 (FIB => [0, 1, ..., 1597]) (length = 17)
alert(fibonacci(400)); // 1.760236806450138e+83 (finds 18 to 400)
alert(fibonacci(1476)); // 1.3069892237633987e+308 (length = 1476)
If you don't like silent errors:
// replace...
if ((typeof(x) !== 'number') || (x < 0)) return;
// with...
if (typeof(x) !== 'number') throw new TypeError('Not a Number.');
if (x < 0) throw new RangeError('Not a possible fibonacci index. (' + x + ')');
Here is a very fast solution of calculating the fibonacci sequence
function fib(n){
var start = Number(new Date);
var field = new Array();
field[0] = 0;
field[1] = 1;
for(var i=2; i<=n; i++)
field[i] = field[i-2] + field[i-1]
var end = Number(new Date);
return 'fib' + '(' + n + ') = ' + field[n] + ' [' +
(end - start) + 'ms]';
}
var f = fib(1450)
console.log(f)
I've just written my own little implementation using an Object to store already computed results. I've written it in Node.JS, which needed 2ms (according to my timer) to calculate the fibonacci for 1476.
Here's the code stripped down to pure Javascript:
var nums = {}; // Object that stores already computed fibonacci results
function fib(n) { //Function
var ret; //Variable that holds the return Value
if (n < 3) return 1; //Fib of 1 and 2 equal 1 => filtered here
else if (nums.hasOwnProperty(n)) ret = nums[n]; /*if the requested number is
already in the object nums, return it from the object, instead of computing */
else ret = fib( n - 2 ) + fib( n - 1 ); /* if requested number has not
yet been calculated, do so here */
nums[n] = ret; // add calculated number to nums objecti
return ret; //return the value
}
//and finally the function call:
fib(1476)
EDIT: I did not try running this in a Browser!
EDIT again: now I did. try the jsfiddle: jsfiddle fibonacci Time varies between 0 and 2ms
Much faster algorithm:
const fib = n => fib[n] || (fib[n-1] = fib(n-1)) + fib[n-2];
fib[0] = 0; // Any number you like
fib[1] = 1; // Any number you like