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Programmatically solving Sam Loyd's Battle of Hastings puzzle - performance issues with BigInt

I'm having performance issues when trying to check whether integer n is a perfect square (sqrt is a whole number) when using BigInt. Using normal numbers below Number.MAX_SAFE_INTEGER gives reasonable performance, but attempting to use BigInt even with the same number range causes a huge performance hit.
The program solves the Battle of Hastings perfect square riddle put forth by Sam Loyd whereby my program iterates over the set of real numbers n (in this example, up to 7,000,000) to find instances where variable y is a whole number (perfect square). I'm interested in the original square root of one of the 13 perfect squares where this condition is satisfied, which is what my code generates (there's more than one).
Assuming y^2 < Number.MAX_SAFE_INTEGER which is 2^53 – 1, this can be done without BigInt and runs in ~60ms on my machine:
const limit = 7_000_000;
var a = [];
console.time('regular int');
for (let n = 1; n < limit; n++) {
if (Math.sqrt(Math.pow(n, 2) * 13 + 1) % 1 === 0)
a.push(n);
}
console.log(a.join(', '));
console.timeEnd('regular int');
Being able to use BigInt would mean I could test for numbers much higher than the inherent number variable limit 2^53 - 1, but BigInt seems inherently slower; unusably so. To test whether a BigInt is a perfect square, I have to use a third party library as Math.sqrt doesn't exist for BigInt such that I can check if the root is perfect, as all sqrt returns a floor value. I adapted functions for this from a NodeJS library, bigint-isqrt and bigint-is-perfect-square.
Thus, using BigInt with the same limit of 7,000,000 runs 35x slower:
var integerSQRT = function(value) {
if (value < 2n)
return value;
if (value < 16n)
return BigInt(Math.sqrt(Number(value)) | 0);
let x0, x1;
if (value < 4503599627370496n)
x1 = BigInt(Math.sqrt(Number(value))|0) - 3n;
else {
let vlen = value.toString().length;
if (!(vlen & 1))
x1 = 10n ** (BigInt(vlen / 2));
else
x1 = 4n * 10n ** (BigInt((vlen / 2) | 0));
}
do {
x0 = x1;
x1 = ((value / x0) + x0) >> 1n;
} while ((x0 !== x1 && x0 !== (x1 - 1n)));
return x0;
}
function perfectSquare(n) {
// Divide n by 4 while divisible
while ((n & 3n) === 0n && n !== 0n) {
n >>= 2n;
}
// So, for now n is not divisible by 2
// The only possible residual modulo 8 for such n is 1
if ((n & 7n) !== 1n)
return false;
return n === integerSQRT(n) ** 2n;
}
const limit = 7_000_000;
var a = [];
console.time('big int');
for (let n = 1n; n < limit; n++) {
if (perfectSquare(((n ** 2n) * 13n) + 1n))
a.push(n);
}
console.log(a.join(', '));
console.timeEnd('big int');
Ideally I'm interested in doing this with a much higher limit than 7 million, but I'm unsure whether I can optimise the BigInt version any further. Any suggestions?

You may be pleased to learn that some recent improvements on V8 have sped up the BigInt version quite a bit; with a recent V8 build I'm seeing your BigInt version being about 12x slower than the Number version.
A remaining challenge is that implementations of BigInt-sqrt are typically based on Newton iteration and hence need an estimate for a starting value, which should be near the final result, so about half as wide as the input, which is given by log2(X) or bitLength(X). Until this proposal gets anywhere, that can best be done by converting the BigInt to a string and taking that string's length, which is fairly expensive.
To get faster right now, #Ouroborus' idea is great. I was curious how fast it would be, so I implemented it:
(function betterAlgorithm() {
const limit = 7_000_000n;
var a = [];
console.time('better algorithm');
let m = 1n;
let m_squared = 1n;
for (let n = 1n; n < limit; n += 1n) {
let y_squared = n * n * 13n + 1n;
while (y_squared > m_squared) {
m += 1n;
m_squared = m * m;
}
if (y_squared === m_squared) {
a.push(n);
}
}
console.log(a.join(', '));
console.timeEnd('better algorithm');
})();
As a particular short-term detail, this uses += 1n instead of ++, because as of today, V8 hasn't yet gotten around to optimizing ++ for BigInts. This difference should disappear eventually (hopefully soon).
On my machine, this version takes only about 4x as much time as your original Number-based implementation.
For larger numbers, I would expect some gains from replacing the multiplications with additions (based on the observation that the delta between consecutive square numbers grows linearly), but for small-ish upper limits that appears to be a bit slower. If you want to toy around with it, this snippet describes the idea:
let m_squared = 1n; // == 1*1
let m_squared_delta = 3n; // == 2*2 - 1*1
let y_squared = 14n; // == 1*1*13+1
let y_squared_delta = 39n; // == 2*2*13+1 - 1*1*13+1
for (let n = 1; n < limit; n++) {
while (y_squared > m_squared) {
m_squared += m_squared_delta;
m_squared_delta += 2n;
}
if (y_squared === m_squared) {
a.push(n);
}
y_squared += y_squared_delta;
y_squared_delta += 26n;
}
The earliest where this could possibly pay off is when the results exceed 2n**64n; I wouldn't be surprised if it wasn't measurable before 2n**256n or so.

Javascript Time Complexity Analysis

Hi there I have been researching and trying to learn how to check for the time complexity of certain algorithms. I've seen this video which was very helpful.
That being said I wondered off and started trying to work out the Worsts Case and an average case of certain algorithms.
1
I believe in the following snippet it is O(n) since to ind the value for sin we have to loop the entire array.
function mySin(x, iterNum) {
var mxx = -x*x;
var sin = 1;
var n = 0;
var term = 1;
for (var i = 1; i <= 2*iterNum; i++) {
n = n + 2;
term = term * mxx / ( n*(n+1) );
sin = sin + term
}
sin = x*sin;
console.log(sin + " = my function.");
console.log(Math.sin(x) + " math.sin");
}
Thanks again
2
function calculateFibonacciSum (num) {
if(cachedNumbers[num]) {
return cachedNumbers[num];
}
if(('number' === typeof num) && num <= 0) {
throw new Error ('Fibonnci series starts with 0. Please, enter any interget greater than or equal to 0');
}
else if(('number' === typeof num) && num === 0) {
return 0;
}
else if(('number' === typeof num) && (num === 1 || num === 2)) {
return 1;
}
else {
var value = calculateFibonacciSum(num-1) + calculateFibonacciSum(num-2);
cachedNumbers[num] = value;
return value;
}
}
While for this one I think it is also O(n) since in the first if/else statement the tc is O(1) since its contestant whilst the final else statement we must loop all the numbers and if the number is not calculated then call the function again (aka recurssion).
TIA

Both of these seem correct to me. Here's a bit more explanation:
1.
This is in fact O(n), as there are n iterations of the loop, the rest constant time; and n is proportional to iterNum
2.
This one is also linear time, but only since you cache the results of previous calculations. Otherwise it would be O(2n).
It is linear time since it essentially runs a loop down to the base cases (0 and 1). In fact, you could re-write this one using a loop instead of recursion.

How to get median and quartiles/percentiles of an array in JavaScript (or PHP)?

This question is turned into a Q&A, because I had struggle finding the answer, and think it can be useful for others
I have a JavaScript array of values and need to calculate in JavaScript its Q2 (50th percentile aka MEDIAN), Q1 (25th percentile) and Q3 (75th percentile) values.

I updated the JavaScript translation from the first answer to use arrow functions and a bit more concise notation. The functionality remains mostly the same, except for std, which now computes the sample standard deviation (dividing by arr.length - 1 instead of just arr.length)
// sort array ascending
const asc = arr => arr.sort((a, b) => a - b);
const sum = arr => arr.reduce((a, b) => a + b, 0);
const mean = arr => sum(arr) / arr.length;
// sample standard deviation
const std = (arr) => {
const mu = mean(arr);
const diffArr = arr.map(a => (a - mu) ** 2);
return Math.sqrt(sum(diffArr) / (arr.length - 1));
};
const quantile = (arr, q) => {
const sorted = asc(arr);
const pos = (sorted.length - 1) * q;
const base = Math.floor(pos);
const rest = pos - base;
if (sorted[base + 1] !== undefined) {
return sorted[base] + rest * (sorted[base + 1] - sorted[base]);
} else {
return sorted[base];
}
};
const q25 = arr => quantile(arr, .25);
const q50 = arr => quantile(arr, .50);
const q75 = arr => quantile(arr, .75);
const median = arr => q50(arr);

After searching for a long time, finding different versions that give different results, I found this nice snippet on Bastian Pöttner's web blog, but for PHP. For the same price, we get the average and standard deviation of the data (for normal distributions)...
PHP Version
//from https://blog.poettner.de/2011/06/09/simple-statistics-with-php/
function Median($Array) {
return Quartile_50($Array);
}
function Quartile_25($Array) {
return Quartile($Array, 0.25);
}
function Quartile_50($Array) {
return Quartile($Array, 0.5);
}
function Quartile_75($Array) {
return Quartile($Array, 0.75);
}
function Quartile($Array, $Quartile) {
sort($Array);
$pos = (count($Array) - 1) * $Quartile;
$base = floor($pos);
$rest = $pos - $base;
if( isset($Array[$base+1]) ) {
return $Array[$base] + $rest * ($Array[$base+1] - $Array[$base]);
} else {
return $Array[$base];
}
}
function Average($Array) {
return array_sum($Array) / count($Array);
}
function StdDev($Array) {
if( count($Array) < 2 ) {
return;
}
$avg = Average($Array);
$sum = 0;
foreach($Array as $value) {
$sum += pow($value - $avg, 2);
}
return sqrt((1 / (count($Array) - 1)) * $sum);
}
Based on the author's comments, I simply wrote a JavaScript translation that will certainly be useful, because surprisingly, it is nearly impossible to find a JavaScript equivalent on the web, and otherwise requires additional libraries like Math.js
JavaScript Version
//adapted from https://blog.poettner.de/2011/06/09/simple-statistics-with-php/
function Median(data) {
return Quartile_50(data);
}
function Quartile_25(data) {
return Quartile(data, 0.25);
}
function Quartile_50(data) {
return Quartile(data, 0.5);
}
function Quartile_75(data) {
return Quartile(data, 0.75);
}
function Quartile(data, q) {
data=Array_Sort_Numbers(data);
var pos = ((data.length) - 1) * q;
var base = Math.floor(pos);
var rest = pos - base;
if( (data[base+1]!==undefined) ) {
return data[base] + rest * (data[base+1] - data[base]);
} else {
return data[base];
}
}
function Array_Sort_Numbers(inputarray){
return inputarray.sort(function(a, b) {
return a - b;
});
}
function Array_Sum(t){
return t.reduce(function(a, b) { return a + b; }, 0);
}
function Array_Average(data) {
return Array_Sum(data) / data.length;
}
function Array_Stdev(tab){
var i,j,total = 0, mean = 0, diffSqredArr = [];
for(i=0;i<tab.length;i+=1){
total+=tab[i];
}
mean = total/tab.length;
for(j=0;j<tab.length;j+=1){
diffSqredArr.push(Math.pow((tab[j]-mean),2));
}
return (Math.sqrt(diffSqredArr.reduce(function(firstEl, nextEl){
return firstEl + nextEl;
})/tab.length));
}

TL;DR
The other answers appear to have solid implementations of the "R-7" version of computing quantiles. Below is some context and another JavaScript implementation borrowed from D3 using the same R-7 method, with the bonuses that this solution is es5 compliant (no JavaScript transpilation required) and probably covers a few more edge cases.
Existing solution from D3 (ported to es5/"vanilla JS")
The "Some Background" section, below, should convince you to grab an existing implementation instead of writing your own.
One good candidate is D3's d3.array package. It has a quantile function that's essentially BSD licensed:
https://github.com/d3/d3-array/blob/master/src/quantile.js
I've quickly created a pretty straight port from es6 into vanilla JavaScript of d3's quantileSorted function (the second function defined in that file) that requires the array of elements to have already been sorted. Here it is. I've tested it against d3's own results enough to feel it's a valid port, but your experience might differ (let me know in the comments if you find a difference, though!):
Again, remember that sorting must come before the call to this function, just as in D3's quantileSorted.
//Credit D3: https://github.com/d3/d3-array/blob/master/LICENSE
function quantileSorted(values, p, fnValueFrom) {
var n = values.length;
if (!n) {
return;
}
fnValueFrom =
Object.prototype.toString.call(fnValueFrom) == "[object Function]"
? fnValueFrom
: function (x) {
return x;
};
p = +p;
if (p <= 0 || n < 2) {
return +fnValueFrom(values[0], 0, values);
}
if (p >= 1) {
return +fnValueFrom(values[n - 1], n - 1, values);
}
var i = (n - 1) * p,
i0 = Math.floor(i),
value0 = +fnValueFrom(values[i0], i0, values),
value1 = +fnValueFrom(values[i0 + 1], i0 + 1, values);
return value0 + (value1 - value0) * (i - i0);
}
Note that fnValueFrom is a way to process a complex object into a value. You can see how that might work in a list of d3 usage examples here -- search down where .quantile is used.
The quick version is if the values are tortoises and you're sorting tortoise.age in every case, your fnValueFrom might be x => x.age. More complicated versions, including ones that might require accessing the index (parameter 2) and entire collection (parameter 3) during the value calculation, are left up to the reader.
I've added a quick check here so that if nothing is given for fnValueFrom or if what's given isn't a function the logic assumes the elements in values are the actual sorted values themselves.
Logical comparison to existing answers
I'm reasonably sure this reduces to the same version in the other two answers (see "The R-7 Method", below), but if you needed to justify why you're using this to a product manager or whatever maybe the below will help.
Quick comparison:
function Quartile(data, q) {
data=Array_Sort_Numbers(data); // we're assuming it's already sorted, above, vs. the function use here. same difference.
var pos = ((data.length) - 1) * q; // i = (n - 1) * p
var base = Math.floor(pos); // i0 = Math.floor(i)
var rest = pos - base; // (i - i0);
if( (data[base+1]!==undefined) ) {
// value0 + (i - i0) * (value1 which is values[i0+1] - value0 which is values[i0])
return data[base] + rest * (data[base+1] - data[base]);
} else {
// I think this is covered by if (p <= 0 || n < 2)
return data[base];
}
}
So that's logically close/appears to be exactly the same. I think d3's version that I ported covers a few more edge/invalid conditions and includes the fnValueFrom integration, both of which could be useful.
The R-7 Method vs. "Common Sense"
As mentioned in the TL;DR, the answers here, according to d3.array's readme, all use the "R-7 method".
This particular implementation [from d3] uses the R-7 method, which is the default for the R programming language and Excel.
Since the d3.array code matches the other answers here, we can safely say they're all using R-7.
Background
After a little sleuthing on some math and stats StackExchange sites (1, 2), I found that there are "common sensical" ways of calculating each quantile, but that those don't typically mesh up with the results of the nine generally recognized ways to calculate them.
The answer at that second link from stats.stackexchange says of the common-sensical method that...
Your textbook is confused. Very few people or software define quartiles this way. (It tends to make the first quartile too small and the third quartile too large.)
The quantile function in R implements nine different ways to compute quantiles!
I thought that last bit was interesting, and here's what I dug up on those nine methods...
Wikipedia's description of those nine methods here, nicely grouped in a table
An article from the Journal of Statistics Education titled "Quartiles in Elementary Statistics"
A blog post at SAS.com called "Sample quantiles: A comparison of 9 definitions"
The differences between d3's use of "method 7" (R-7) to determine quantiles versus the common sensical approach is demonstrated nicely in the SO question "d3.quantile seems to be calculating q1 incorrectly", and the why is described in good detail in this post that can be found in philippe's original source for the php version.
Here's a bit from Google Translate (original is in German):
In our example, this value is at the (n + 1) / 4 digit = 5.25, i.e. between the 5th value (= 5) and the 6th value (= 7). The fraction (0.25) indicates that in addition to the value of 5, ¼ of the distance between 5 and 6 is added. Q1 is therefore 5 + 0.25 * 2 = 5.5.
All together, that tells me I probably shouldn't try to code something based on my understanding of what quartiles represent and should borrow someone else's solution.

Based on buboh's answer, which I have used for over a year, I have noticed some weird things for calculating the Q1 and Q3 when there are 2 numbers in the middle.
I have no clue why there is a rest value and how it is used, but by my understanding if you and up having 2 numbers in the middle you need to take the average of them to calculate the median. With that in mind I edited the function:
const asc = (arr) => arr.sort((a, b) => a - b);
const quantile = (arr, q) => {
const sorted = asc(arr);
let pos = (sorted.length - 1) * q;
if (pos % 1 === 0) {
return sorted[pos];
}
pos = Math.floor(pos);
if (sorted[pos + 1] !== undefined) {
return (sorted[pos] + sorted[pos + 1]) / 2;
}
return sorted[pos];
};

JavaScript: checking parity of alternate digits? [homework]

I have a homework question I've been having trouble with.
I have to write a function that checks if every alternate digit in a given number has the same parity. For example, both 1 2 3 3 and 2 1 3 3 are valid, but 1324 is not. I have no idea how to go about doing this, though. How do I keep track of previous digits, for one thing? Any help would be much appreciated.
Edit: My efforts so far:
Any number < 100 clearly isn't acceptable (right?) since 'every alternate digit' doesn't really make sense here. For 3-digit numbers, this should work:
function validate(n) {
var i, copy, l = [0, 0];
if (isNaN(n) || (n < 100)) {
return false;
} else {
copy = Math.round(n);
for (i = copy.toString().length; i--; n = Math.floor(n / 10)) {
l[0] = l[1];
l[1] = l[2];
l[2] = n % 10;
}
if ((l[0] % 2) == (l[2] % 2)) return true;
}
}
Edit[2]: Thanks for your help, everybody. I've managed to get an honest-to-goodness real (I think) working function based on Salix alba's first suggestion to save the parities of the first and second digits. The loops run backward over the digits.
For now, this (along with making a couple of minor edits to save the parities of the last and second-last digits instead as Salix alba said, which would make the parity = lines simpler) is my solution:
function validate(n) {
var copy, len, parity, broke = 0, i = 2;
if (!isNaN(n) || (n >= 100)) {
n = Math.round(n);
len = n.toString().length;
copy = n; // save
parity = Math.floor(n / Math.pow(10, len - 1)) % 2;
n = Math.floor(n / 10);
while (i < len) {
if (parity != ((n % 10) % 2)) {
broke++;
break;
}
i += 2;
n = Math.floor(n / 100);
}
n = copy; // restore
i = 1;
parity = (Math.floor(n / Math.pow(10, len - 2)) % 10) % 2;
while (i < len) {
if (parity != ((n % 10) % 2)) {
broke++;
break;
}
i += 2;
n = Math.floor(n / 100);
}
if (broke != 2) return true;
}
return false;
}
It's a horrible mess, of course. I would really, really appreciate any ideas for make this more efficient, easier to read, etc.
(Also going to try to write another function with jing3142's method of iterating with a flag, which might make the loops simpler.)

You are working along the right lines but there are a number of issues. You code only works for three digits since your check if ((l[0] % 2) == (l[2] % 2)) return true; only operates once at the end of the loop.
So you need to set a flag, say valid and put valid = valid && ((l[0] % 2) == (l[2] % 2)) inside the loop.
The loop will now fail because the first use of l[2] in l[1] = l[2] will be undefined since it has not been defined.
If you correct these then you will need to check whether there are an even or odd number of digits else it will fail for an odd number of digits.
Also you imply in your comment that you are not to use strings and in practice you do even if it is only to find the length.
There is another way.
The clues in your working out is that if n<100 then no check can be done, and that you need to reduce n be a factor of 10 each time through n = Math.floor(n / 10) while you loop.
Rather than give you a solution as you ask for advice not a solution here is a hint.
given that n>100 to start with and if U is unit digit, T tens digit and H hundreds digit of n then while n>100 how do you calculate U and H, how do you then set the valid flag, in the loop?
EDIT Looks like you still need some help
Algorithm not code
function validate(n) {
if(n<100) return not applicable
valid=true
while(n>100) {
U=n % 10
H=Floor(n /100) % 10
valid = valid && ((U % 2) == (H % 2))
n=Floor(n/10)
}
return valid
}

Index of a really big Fibonacci Number

I need to calculate the index of a Fibonacci number with JavaScript, within the Fibonacci sequence. I need to do this without using recursion, or a loop. I found the following formula in the Math forum:
n=⌊logφ(F⋅5√+12)⌋
and coded it in JavaScript:
function fibIndex(fib)
{
fib = BigNumber(fib);
return logBasePhi(fib.times(Math.sqrt(5)).plus((1/2)));
}
function phi()
{
return (1 + Math.sqrt(5))/ 2;
}
function getBaseLog(x, y) {
return Math.log(y) / Math.log(x);
}
function logBasePhi(x)
{
return getBaseLog(phi(), x);
}
Notice the .times() and .plus() functions that are part of this BigNumber Library that has been extremely useful up to this point. This works fine, until the Fibonacci number I want to find the index for is really big.
The problem:
I need a different way to calculate the logarithm with such a big number. If I have a really big number, such as Fibonacci of 2000, I get Infinity for obvious reasons. The library itself does not have any methods to calculate the log, and I can't write this function either.
I would have never imagined that the logarithm of any number with such a small base (phi) can be bigger than the max for JavaScript integers. Can you guys point me in the right direction? Should I just leave it at obtaining the index for numbers less than Fib(1500) and call it good?

You can use BigInteger. You can see an example of how to use it here: http://reallifejs.com/the-meat/calculators/big-number-calculator/

For anyone else looking for this function, here it is using this BigInteger library:
function fibIndex(fib)
{
fib = BigInteger(fib);
var x = fib.multiply(Math.sqrt(5)).add((1/2));
return Math.round(x.log() / Math.log(phi()));
}
function phi()
{
return (1 + Math.sqrt(5))/ 2;
}
I still use the same equation explained in my question above, and it returns the index of Fibonacci's of any size.
fibIndex("35522938794321715091272953991088875073660950670711879743399900326436254083421380378927750257524675311447286915610820861302904371152466182968261111009824391725637150862745505342130220586979511719255023895335108709522075314248260664483166479670588221585277069887873168196490963561219694518077864879100421788205515385380434545975662001723555342440392621808579760295648531141638822913590607533540054087452041707826153271185259107394199852367649618298517093117009455894918353503525076230125819543123779319167440820789626564459764725339684808290073756385496248142195843240135064507885354877296572024804408624272941753639812538039913142028651903030529871116793317789757893606550341466951324756526825899737667945813833853722262630433262780974915425005732866591818868174705546087022106127052877310847951571707582794820376128579789767721485109492542287764348615868723395725124814856415577763540656765591673162724146048330852788081439178288706881889502843933839383437965572895385440792960391702268984769357859686271266574632871727046024303184663919395401465801528726015901456333025573481247959101652204602988035141532490361245742139050819433077833707742246312835208439293469725777437940254819086871672146128972238328251414589544434435170261367824782155103657578194196270111748570034449297964612564456266891635499257186520205662004190179581465184858273590634696557067719668344569716772604494379268256417559005989196664062339943367426392267549671696091620704483335705235401024668972377058959013548701899237423163317609813480075906438821567501678027453981255872940165896765562906948275888682233026018398591561683968279253311810352982216449990605138841279476258998291555393112171672512247299540528273985304628059093340049555047981741901553118436996372565143437092164040501385121979087826864836002852441013290435451405818936965791830088594057993174801701555239838033825491101182302604693483923297155552732646664230339899386949247469662146601783799159535265663192798622519600080199294778264021930327804674406845390858689361183645138036024622999759181149374409868339056190354930762438018253181839721998646473911299168577029520666199783681191268719288387969624745653240780708319950931159323616116725759084631179863296728766212415593748082930558151101350076376704295363472805637813559350925898715117938481138744212886965977892516525139040863376874438253015614485120277306681922196720541898193702570355885540352668267759850827312025869672621201575016416207760471674541668295376322809412095582968275396449970226064500618788480102243996614437085271546164050332641040829307354667753670012241015315160013952802535500838629086649248253271677865717482331893600871123634025348607623548331397239596180750809096946397974233223417735790158178612741331748855629088340732705900553246041710742016160018303725512211509204034880759596775427996675371964469431717567054234107252511625358715489171574578479304777517899774723598872665991091538945488811618222438651723224465992160327444696552759313881273021480919406887970238509074105071808066821703115066838126027585207922256205186141921352880657758551963602504587265334948468963725795943612659061581738118921217900480358979991209140061985794462152498458564473369295078153567296201818251720281822962062936831573631653751528074225190111823253702351177610664803940345503699699208037095784870495785646943997234474258262569998608041243140247674073513323374199936066218946098722092264140092669753657824017634461763981521997119226219136508584203375683292957948563073632975937642581947768042371117470198355444599517718979158713784804470849343173517943800269775988799065209263727016757620989100649249277790139290067789688270481157537247255077854388424596882337360026236416764073030845206282134294301701605369452555608749844089881840152464688185413471165531348083450071936767808847178828505212544681448346850412422392584457798023562617507156540143036586660455094150665003248778589943633804236061867787254092465761897721097498706848304838082130151573830146281598905862080528755999259386441406295844212093958532689277331339576745477613093048842162872506248493879631984787279577095875465635013803960469019743694441996546910736934590725390122252181310568062868015617422771375425422209106466232597689466636780861666245204430735375550444974466951762888881642801337840709202391876433786768647147807880162471534334272202034098411011768290957484345424507121327462388443014612800575348423810123382495642833743034606424879522789397956839996920113680951463518836156462019057063161795046895734075593501902084338246542048532779483281408634769806186279989881229648075555962086774926497206070780542404761166097604241890965888018873735027199940548827053350115337885438800728312460914286268127990478092896975620706029422142841447344514680046143167682001640750053397540223424322177217456434741847020047763710403144096996427837529811812126999093061373016438435440619803496909856986800826405322182728111872725881192065183612822832173197471616932926246556344247662468294551754101114527143077792003917544284111176961413199426663155326179333146951914261328112918116870606040456416800180399145364936151721824514256765308265696290759951243242140057433018143404698921069198350343599629915865217541917472815612277351716569260985624821969133328022587501");
will return 25,001, which is the index of the above fib.

Instead use this formula:
(Fn) = (Fn-1) + (Fn-2)
n is a subindex, for understanding I say...
So let's code :D
function fibonacci(n) {
var f = new Array();
f[0] = 1;
f[1] = 1;
if(n == 1 && n == 2) {
return 1;
}
for(var i = 2; i < n; i++) {
f[i] = f[i - 1] + f[i - 2];
}
return f[n - 1];
}