I would like Anna to have 67% chance of being picked randomly, Bob to have a 30% change, and Tom to have a 3% chance. Is there a simpler way to do this?
This is what I have so far:
var nomes = ['Anna', 'Bob', 'Tom'];
var name = nomes[Math.ceil(Math.random() * (nomes.length - 1))];
console.log(name);
Based on this Stack Overflow I think the following code will work for you:
function randomChoice(p) {
let rnd = p.reduce((a, b) => a + b) * Math.random();
return p.findIndex(a => (rnd -= a) < 0);
}
function randomChoices(p, count) {
return Array.from(Array(count), randomChoice.bind(null, p));
}
const nomes = ['Anna', 'Bob', 'Tom'];
const selectedIndex = randomChoices([0.67, 0.3, 0.03], nomes);
console.log(nomes[selectedIndex]);
// Some testing to ensure that everything works as expected:
const odds = [0, 0, 0];
for (let i = 0; i < 100000; i++) {
const r = randomChoices([0.67, 0.3, 0.03], nomes);
odds[r] = odds[r] + 1;
}
console.log(odds.map(o => o / 1000));
Here's a short solution for you:
function pick(){
var rand = Math.random();
if(rand < .67) return "Anna";
if(rand < .97) return "Bob";
return "Tom";
}
console.log( pick() );
Related
This is inspired by a Codecadamey project, where I'm learning JavaScript.
The problem is meant to simulate a simple DNA strand. It has 15 positions in the array, and those elements are either an A, T, C, or G to represent DNA bases.
There are no limits to the amount of times a single letter (base) can show up in the array.
I create 30 arrays that are made up of at least 60% C and/or G in any of the positions, these are meant to represent strong DNA strands.
I compare the strands to each other to see what % match they are. I consider a 'match' being true when there is the same base at the same position thisDNA[i] === comparisonDNA[i]
When I test a batch of 30 of these 'strong' samples to see the best and worst match levels, I find the results very tightly grouped (I ran it 3,000 times and the highest was 87%, lowest 53%), yet it is very easy for me to concieve of two samples that will survive that are a 0% match:
const sample1 = AGACCGCGCGTGGAG
const sample2 = TCTGGCGCGCACCTC
(obviously I've cheated by building these to be a 0% match and not randomly generating them)
Here's the full code: https://gist.github.com/AidaP1/0770307979e00d4e8d3c83decc0f7771
My question is as follows: Why is the grouping of matches so tight? Why do I not see anything below a 53% match after running the test thousands of times?
Full code:
// Returns a random DNA base
const returnRandBase = () => {
const dnaBases = ['A', 'T', 'C', 'G']
return dnaBases[Math.floor(Math.random() * 4)]
}
// Returns a random single stand of DNA containing 15 bases
const mockUpStrand = () => {
const newStrand = []
for (let i = 0; i < 15; i++) {
newStrand.push(returnRandBase())
}
return newStrand
}
const pAequorFactory = (num, arr) => { //factory function for new strand specimen
return {
specimenNum: num,
dna: arr,
mutate() {
//console.log(`old dna: ${this.dna}`) //checking log
let randomBaseIndex = Math.floor(Math.random() * this.dna.length) /// chooses a location to exchange the base
let newBase = returnRandBase()
while (newBase === this.dna[randomBaseIndex]) { // Rolls a new base until newBase !== current base at that position
newBase = returnRandBase()
}
this.dna[randomBaseIndex] = newBase;
//console.log(`New dna: ${this.dna}`) //checking log
return this.dna;
},
compareDNA(pAequor) { // compare two strands and output to the console
let thisDNA = this.dna;
let compDNA = pAequor.dna;
let matchCount = 0
for (i = 0; i < this.dna.length; i++) { //cycles through each array and log position + base matches on matchCount
if (thisDNA[i] === compDNA[i]) {
matchCount += 1;
};
};
let percMatch = Math.round(matchCount / this.dna.length * 100) //multiply by 100 to make 0.25 display as 25 in console log
console.log(`specimen #${this.specimenNum} and specimen #${pAequor.specimenNum} are a ${percMatch}% DNA match.`)
return percMatch;
},
compareDNAbulk(pAequor) { //as above, but does not log to console. Used for large arrays in findMostRelated()
let thisDNA = this.dna;
let compDNA = pAequor.dna;
let matchCount = 0
for (i = 0; i < this.dna.length; i++) {
if (thisDNA[i] === compDNA[i]) {
matchCount += 1;
};
};
let percMatch = Math.round(matchCount / this.dna.length * 100) //multiply by 100 to make 0.25 display as 25 in console log
return percMatch;
},
willLikelySurvive() { // looks for >= 60% of bases as either G or C
let countCG = 0;
this.dna.forEach(base => {
if (base === 'C' || base === 'G') {
countCG += 1;
}
})
//console.log(countCG) // testing
//console.log(this.dna) // testing
return countCG / this.dna.length >= 0.6
},
complementStrand() {
return this.dna.map(base => {
switch (base) {
case 'A':
return 'T';
case 'T':
return 'A';
case 'C':
return 'G';
case 'G':
return 'C';
}
})
} //close method
} // close object
} // close factory function
function generatepAequorArray(num) { // Generatess 'num' pAequor that .willLikelySurvive() = true
let pAequorArray = []; //result array
for (i = 0; pAequorArray.length < num; i++) {
let newpAequor = pAequorFactory(i, mockUpStrand()); // runs factory until there are 30 items in the result array
if (newpAequor.willLikelySurvive() === true) {
pAequorArray.push(newpAequor)
}
}
return pAequorArray;
}
function findMostRelated(array) { // champion/challenger function to find the most related specimens
let winningScore = 0;
let winner1;
let winner2;
for (let i = 0; i < array.length; i++) {
for (let j = i; j < array.length; j++) // j = i to halve the number of loops. i = 0, j = 5 is the same as i = 5, j = 0
if (i !== j) { // Do not check specimen against itself
let currentScore = array[i].compareDNAbulk(array[j]);
if (currentScore > winningScore) { // Challenger becomes the champion if they are a closer match
winningScore = currentScore;
winner1 = array[i].specimenNum;
winner2 = array[j].specimenNum;
}
}
}
let resultArray = [winner1, winner2, winningScore] // stored together for easy return
//console.log(`The most related specimens are specimen #${winner1} and specimen #${winner2}, with a ${winningScore}% match.`)
return resultArray
}
function multiArray(loops) { //test by running finding the closest match in 30 random 'will survive' samples, repaeated 1000 times. Returns the highest and lowest match across the 1000 runs
let highScore = 0;
let lowScore = 100
for (let i = 0; i < loops; i++) {
let pAequorArray = generatepAequorArray(30);
let currentArray = findMostRelated(pAequorArray);
highScore = Math.max(highScore, currentArray[2])
lowScore = Math.min(lowScore, currentArray[2])
}
return results = {
'high score': highScore,
'low score': lowScore
}
}
console.log(multiArray(10000))
So for example I have an array with 3 waypoints:
[ [ 526, 1573, 24 ], [ 2224, 809, -1546 ], [ 6869, 96, -3074 ] ]
I also know I want to rest for lets say n times between between arriving at the first and last waypoint. So in the end I want an array of n points.
How do I go about finding those n resting-points in JS?
Thanks in advance!
EDIT: Note this is not a single object! Imagine each axis being one person. They have to stop the same amount of time and at the same time but they do not have to be at the same place obviously.
You want to use linear interpolation.
A quick example:
const POINTS = [ [ 526, 1573, 24 ], [ 2224, 809, -1546 ], [ 6869, 96, -3074 ] ];
const N = 10;
function getDistance(point1, point2) {
// speed in 3d space is mutated according only to the X distance,
// to keep speed constant in X dimension
return Math.abs(point1[0] - point2[0]);
}
function go(points, n) {
const pointDistances = points.slice(1).map((point, index) => getDistance(points[index], point));
const fullDistance = pointDistances.reduce((sum, distance) => sum + distance, 0);
const distancePerSection = fullDistance / n;
return points.slice(1)
.reduce((last, point, index) => {
const thisDistance = pointDistances[index];
const numRestPoints = Math.max(0, Math.floor(thisDistance / distancePerSection) - 1);
if (!numRestPoints) {
return last.concat([point]);
}
const thisYVector = point[1] - points[index][1];
const thisZVector = point[2] - points[index][2];
return last.concat(new Array(numRestPoints).fill(0)
.reduce((section, item, restIndex) => {
return section.concat([[
points[index][0] + (restIndex + 1) * distancePerSection,
points[index][1] + (restIndex + 1) * thisYVector * distancePerSection / thisDistance,
points[index][2] + (restIndex + 1) * thisZVector * distancePerSection / thisDistance
]]);
}, [])
.concat([point])
);
}, points.slice(0, 1));
}
function test() {
const result = go(POINTS, N);
if (result.length !== N) {
throw new Error('Must be N length');
}
if (!result[0].every((value, index) => value === POINTS[0][index])) {
throw new Error('Doesn\'t start at the first point');
}
if (!result[N - 1].every((value, index) => value === POINTS[POINTS.length - 1][index])) {
throw new Error('Doesn\'t end at the last point');
}
if (!POINTS.slice(1, N - 1).every(point =>
result.some(resultPoint => resultPoint.every((value, index) => value === point[index]))
)) {
throw new Error('Doesn\'t go through every provided point');
}
console.log(result.slice(1).map((point, index) => getDistance(point, result[index])));
console.log('The result passed the tests!');
console.log(JSON.stringify(result, null, 2));
}
test();
I'm basically going through the list of points, and determining if there should exist any rest points between them, inserting them if so.
Please comment if you want further clarification!
I also solved this problem now with linear interpolation:
My solution:
var waypoints = [[526,1573,24],[2224,809,-1546],[6869,96,-3074]];
var pauses = 20;
generateWaypopints();
function generateWaypopints(){
var newWaypoints = [];
var progressAtMainPoints = 1 / (waypoints.length - 1)
var pausesBetweenWaypoints = pauses * progressAtMainPoints;
var progressAtPauses = 1 / pausesBetweenWaypoints;
newWaypoints.push(waypoints[0]);
var sector = 0;
var pausesInSector = 0;
for(var i = 0; i < pauses; i++){
var progress = progressAtPauses * (pausesInSector + 1)
var x = Math.round(waypoints[sector][0] + (waypoints[sector + 1][0] - waypoints[sector][0]) * progress);
var y = Math.round(waypoints[sector][1] + (waypoints[sector + 1][1] - waypoints[sector][1]) * progress);
var z = Math.round(waypoints[sector][2] + (waypoints[sector + 1][2] - waypoints[sector][2]) * progress);
if(progress >= 1){
sector++;
pausesInSector = 0;
}else
pausesInSector++;
newWaypoints.push([x,y,z]);
}
console.log(newWaypoints);
return newWaypoints;
}
So I am looking to create look up tables. However I am running into a problem with integer ranges instead of just 1, 2, 3, etc. Here is what I have:
var ancient = 1;
var legendary = 19;
var epic = 251;
var rare = 1000;
var uncommon = 25000;
var common = 74629;
var poolTotal = ancient + legendary + epic + rare + uncommon + common;
var pool = general.rand(1, poolTotal);
var lootPool = {
1: function () {
return console.log("Ancient");
},
2-19: function () {
}
};
Of course I know 2-19 isn't going to work, but I've tried other things like [2-19] etc etc.
Okay, so more information:
When I call: lootPool[pool](); It will select a integer between 1 and poolTotal Depending on if it is 1 it will log it in the console as ancient. If it hits in the range of 2 through 19 it would be legendary. So on and so forth following my numbers.
EDIT: I am well aware I can easily do this with a switch, but I would like to try it this way.
Rather than making a huge lookup table (which is quite possible, but very inelegant), I'd suggest making a (small) object, choosing a random number, and then finding the first entry in the object whose value is greater than the random number:
// baseLootWeight: weights are proportional to each other
const baseLootWeight = {
ancient: 1,
legendary: 19,
epic: 251,
rare: 1000,
uncommon: 25000,
common: 74629,
};
let totalWeightSoFar = 0;
// lootWeight: weights are proportional to the total weight
const lootWeight = Object.entries(baseLootWeight).map(([rarity, weight]) => {
totalWeightSoFar += weight;
return { rarity, weight: totalWeightSoFar };
});
console.log(lootWeight);
const randomType = () => {
const rand = Math.floor(Math.random() * totalWeightSoFar);
return lootWeight
.find(({ rarity, weight }) => weight >= rand)
.rarity;
};
for (let i = 0; i < 10; i++) console.log(randomType());
Its not a lookup, but this might help you.
let loots = {
"Ancient": 1,
"Epic": 251,
"Legendary": 19
};
//We need loots sorted by value of lootType
function prepareSteps(loots) {
let steps = Object.entries(loots).map((val) => {return {"lootType": val[0], "lootVal": val[1]}});
steps.sort((a, b) => a.lootVal > b.lootVal);
return steps;
}
function getMyLoot(steps, val) {
let myLootRange;
for (var i = 0; i < steps.length; i++) {
if((i === 0 && val < steps[0].lootVal) || val === steps[i].lootVal) {
myLootRange = steps[i];
break;
}
else if( i + 1 < steps.length && val > steps[i].lootVal && val < steps[i + 1].lootVal) {
myLootRange = steps[i + 1];
break;
}
}
myLootRange && myLootRange['lootType'] ? console.log(myLootRange['lootType']) : console.log('Off Upper Limit!');
}
let steps = prepareSteps(loots);
let pool = 0;
getMyLoot(steps, pool);
I need help with a function which can calculate mortgage annuity amortization table with output in cols and rows of payment amount, principal payment, Interest payment, etc. sorted by each month like in link below:
Output example
The output file format should be .cvs. At the moment I am stuck with this witch is still far away from result:
var i = 5/100;
var loanAmount = 15000;
var m = 12;
var monthlyPayment = loanAmount*(i/12)*Math.pow((1+i/12), m) / (Math.pow((1+i/12), m)-1)
var currentBalance = loanAmount;
var paymentCounter = 1;
var totalInterest = 0;
monthlyPayment = monthlyPayment;
while(currentBalance > 0) {
//this calculates the portion of your monthly payment that goes towards interest
towardsInterest = (i/12)*currentBalance;
if (monthlyPayment > currentBalance){
monthlyPayment = currentBalance + towardsInterest;
}
towardsBalance = monthlyPayment - towardsInterest;
totalInterest = totalInterest + towardsInterest;
currentBalance = currentBalance - towardsBalance;
}
Would really appreciate any help on this.
Try something like this:
const annuity = (C, i, n) => C * (i/(1-(1+i)**(-n)));
const balance_t = (C, i, P) => {
const period_movements = {
base: C
}
period_movements.interest = C*i;
period_movements.amortization = P - (C*i);
period_movements.annuity = P;
period_movements.final_value = Math.round((C - period_movements.amortization)*100)/100;
return period_movements;
}
const display_mortgage = (C, i, n) => {
const payements = annuity(C, i, n);
let movements = balance_t(C, i, payements);
while (movements.final_value>-.01){
console.log(movements);
movements = balance_t(movements.final_value, i, payements);
}
}
display_mortgage(20000, 0.05, 4);
Output:
{ base: 20000,
interest: 1000,
amortization: 4640.236652069255,
annuity: 5640.236652069255,
final_value: 15360 }
{ base: 15360,
interest: 768,
amortization: 4872.236652069255,
annuity: 5640.236652069255,
final_value: 10488 }
{ base: 10488,
interest: 524.4,
amortization: 5115.836652069255,
annuity: 5640.236652069255,
final_value: 5372 }
{ base: 5372,
interest: 268.6,
amortization: 5371.6366520692545,
annuity: 5640.236652069255,
final_value: 0 }
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