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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.

Different results for the same problem solution in node.js and python

I solved the following leetCode problem with some code :
You have d dice, and each die has f faces numbered 1, 2, ..., f.
Return the number of possible ways modulo 10^9 + 7 to roll the dice so the sum of the face up numbers equals t.
I made two versions of the solution code, one in node.js using mathjs, and one in python using the math module .
In node.js
const { combinations: comb, bignumber: Big } = require("mathjs");
function dice(d, f, t) {
if (t > d * f || t < d) return 0;
var result = Big(0);
var i = 0;
var sign = 1;
var n = t - 1;
var k = t - d;
while (k >= 0 && i <= d) {
result = result.add(
comb(Big(d), Big(i))
.times(comb(Big(n), Big(k)))
.times(sign)
);
i++;
n -= f;
k -= f;
sign *= -1;
}
return result;
}
console.log(
dice(30, 30, 500).mod(
Big(10)
.pow(9)
.add(7)
)
);
In python :
import math
def dice(d, f, t):
if t > d * f or t < d:
return 0
result = 0
i = 0
sign = 1
n = t - 1
k = t - d
while k >= 0 and i <= d:
result += math.comb(d, i) * math.comb(n, k) * sign
i += 1
n -= f
k -= f
sign *= -1
return result
print(dice(30, 30, 500) % (math.pow(10, 9) + 7))
Now when i run the code with these parameters : d=30 f=30 t=500 (the last line of each version of the code), i expect the result to be 222616187 .
In the node.js version , that's exactly what i get .
But in the python version , i'm getting 811448245.0 i can't figure out why is that happening .
So why is there a difference in the results ?

The math module usesfloat, not arbitrary precision int.
math - Mathematical functions
[...]
The following functions are provided by this module. Except when
explicitly noted otherwise, all return values are floats.
Since math.pow returns a float, the leading argument to % is converted to a float as well.
The result of dice(30, 30, 500) is too large to be accurately represented as a float. Its float representation is off by -14999044413600247749080617.
The ** operator and its function version operator.pow do not force float conversion and provide an integer if all parameters are integers.
>>> print(dice(30, 30, 500) % (10 ** 9 + 7))
222616187

Solved, in a weird way. It turns out, math.pow returns a float instead of int and somehow bugged. I think int % float has a different cast operation in it and treated differently by the compiler. It can be investigated further. If you cast it to int, that would be your answer.
import math
def dice(d, f, t):
if t > d * f or t < d:
return 0
result = 0
i = 0
sign = 1
n = t - 1
k = t - d
while k >= 0 and i <= d:
result += math.comb(d, i) * math.comb(n, k) * sign
i += 1
n -= f
k -= f
sign *= -1
return result
print(dice(30, 30, 500) % int((math.pow(10, 9) + 7)))

Get reverse of an equation - JavaScript

Let's say I have this formula, for example:
function getExperience(level) {
let a = 0;
for (let x = 1; x < level; x += 1) {
a += Math.floor(x + (200 * (2 ** (x / 3))));
}
return Math.floor(a / 4);
}
for (var i = 1; i < 100; i++) {
console.log(`Level ${i}: ${getExperience(i)}`);
}
To get the experience needed for level 50, you'd do: getExperience(50).
But, how would you reverse that and get the LEVEL needed for experience? So, getLevel(20010272) would output 50.

Short answer
You can use 4.328085 * Math.log(0.00519842 * xp + 1.259921045) as a very good approximation of the corresponding level.
If you need an exact value, you could iterate over all levels until you find the desired range, as in this answer.
Long answer
Slightly modified function
I don't think it's possible to find an exact, closed-form expression for the inverse of this function. It should be possible if you modify getExperience(level) a bit, though.
First, you can notice that x grows much slower than 2 ** (x / 3).
Then, Math.floor doesn't have much influence over large numbers.
So let's remove them! Here's the slightly modified function:
function getExperienceEstimate(level) {
let a = 0;
for (let x = 1; x < level; x += 1) {
a += 200 * (2 ** (x / 3));
}
return a / 4;
}
The advantage of this method is that it's now a geometric series, so it's possible to calculate the sum directly, without any loop:
function getExperienceEstimate(level) {
let a = 50;
let r = 2 ** (1 / 3);
return a * (r**level - r) / (r - 1);
};
getExperienceEstimate(50) returns 20011971.993575357, which is only 0.0015% smaller than getExperience(50).
Inverse function
According to Wolfram Alpha, here's the inverse function of getExperienceEstimate:
function getLevelEstimate(xp){
let a = 50;
let r = 2 ** (1 / 3);
return Math.log(xp * (r - 1) / a + r) / Math.log(r);
};
With some minor precision loss, you can simplify it further:
function getLevelEstimate(xp){
return 4.328085 * Math.log(0.00519842 * xp + 1.259921045)
};
It's only an estimate, but it works pretty well and doesn't require any loop!
Test
For 20012272 XP, the approximate inverse function returns 50.00006263463371, which should be a good starting point if you want to find the exact result.
function getExperience(level) {
let a = 0;
for (let x = 1; x < level; x += 1) {
a += Math.floor(x + (200 * (2 ** (x / 3))));
}
return Math.floor(a / 4);
}
function getLevelEstimate(xp){
return 4.328085 * Math.log(0.00519842 * xp + 1.259921045)
};
for (var i = 1; i < 100; i++) {
console.log(`Level ${i} (XP = ${getExperience(i)}). Estimated level : ${getLevelEstimate(getExperience(i))}`);
}

You can use a binary search algorithm to avoid to loop over all possibilities.
Here is an example that I have adapted to your case.
You first need to create an array to map all your level => experience, this action should be done only ONCE, then you never have to do it again.
As you can see in my example, even with 1000 levels, you never have to iterate more than 9 times whatever level you are trying to find.
// You first have to create an array with all your levels.
// This has to be done only ONCE because it's an expensive one!
const list = [];
for (let i = 1; i <= 1000; i++) {
list[i] = getExperience(i);
}
function getExperience(level) {
let a = 0;
for (let x = 1; x < level; x += 1) {
a += Math.floor(x + (200 * (2 ** (x / 3))));
}
return Math.floor(a / 4);
}
function getLevel(value) {
// initial values for start, middle and end
let start = 0
let stop = list.length - 1
let middle = Math.floor((start + stop) / 2)
let iterations = 0;
// While the middle is not what we're looking for and the list does not have a single item.
while (list[middle] !== value && start < stop) {
iterations++;
if (value < list[middle]) {
stop = middle - 1
} else {
start = middle + 1
}
// Recalculate middle on every iteration.
middle = Math.floor((start + stop) / 2)
}
console.log(`${value} is Level ${middle} (Result found after ${iterations} iterations)`);
return middle;
}
// Then you can search your level according to the experience
getLevel(0);
getLevel(72);
getLevel(20010272);
getLevel(getExperience(50));
getLevel(33578608987644589722);

A brute-force (but inelegant) solution would be to just call getExperience for levels until you reach a level that requires more experience than the passed exp:
function getLevel(exp) {
if (exp === 0) return 0;
let level = 0;
let calcExp = 0;
while (exp > calcExp) {
calcExp = getExperience(level);
if (calcExp > exp) break;
level++;
}
return level - 1;
}
console.log(getLevel(20012272)); // experience required for 50 on the dot
console.log(getLevel(20012270));
console.log(getLevel(20012274));
console.log(getLevel(0));
function getExperience(level) {
let a = 0;
for (let x = 1; x < level; x += 1) {
a += Math.floor(x + (200 * (2 ** (x / 3))));
}
return Math.floor(a / 4);
}

You can use binary search to locate level value faster - in 7 steps max.
(while I doubt that gain is significant for length 100 list)

How to calculate the square root without using library and built-in methods in Javascript?

Please help me to write a function to compute the square root of positive real numbers using the formula:
x i+1 = (1/2) * (xi + (A / x1)),
where 'A' - input real number.
On the zero iteration next statements have been taken x0 = A
The error should be at least 10-6
Output
sqrt (2) = 1.414
sqrt (9) = 3
sqrt (25) = 5

You could take xi (x) and the new value of xi + 1 (x1) and check if the values are equal. Then end the series and return that value.
For starting, you need an apporopriate value like the half of the given value.
function sqrt(a) {
var x,
x1 = a / 2;
do {
x = x1;
x1 = (x + (a / x)) / 2;
} while (x !== x1);
return x;
}
console.log(sqrt (2)); // 1.414
console.log(sqrt (9)); // 3
console.log(sqrt (25)); // 5

You can also use bisection - a more general method for solving problems:
var sqrt = function(n) {
if (n<0) {
throw "are you kidding?! we are REAL here.";
}
if (n === 0) {
return 0;
}
var bisect = function(l,r) {
var avg = (l+r)/2;
if (r-l<0.00000001) {
return (l+r)/2;
}
if (avg*avg > n) {
return bisect(l, avg);
} else if (avg*avg < n) {
return bisect(avg, r);
}
}
return bisect(0, n < 1 ? 1 : n);
}

Subtracting long numbers in javascript

Why is q == 0 in the following script?
<script>
var start = 1234567890123456789;
var end = 1234567890123456799;
var q = end - start;
alert(q);
</script>
I would think the result should be 10. What is the correct way to subtract these two numbers?

Because numbers in JavaScript are floating-point. They have limited precision.
When JavaScript sees a very long number, it rounds it to the nearest number it can represent as a 64-bit float. In your script, start and end get rounded to the same value.
alert(1234567890123456789); // says: 1234567890123456800
alert(1234567890123456799); // says: 1234567890123456800
There's no built-in way to do precise arithmetic on large integers, but you can use a BigInteger library such as this one.

As of January 2020, BigInt datatype is going to be added to Javascript. The proposal is currently in Stage 4. It will enable precise calculation for number which are more than 2^53-1 (Number.MAX_SAFE_INTEGER).
BigInt has been shipped in Chrome, Node, Firefox, and is underway in Safari. Read more here.
var start = BigInt('1234567890123456789');
var end = BigInt('1234567890123456799');
var q = end - start;
alert(q)
A BigInt is created by appending n to the end of an integer literal — 10n — or by calling the function BigInt(). It is also different from Number so 1 + 1n will fail.
You can read more about it here from MDN pages

Jason already posted the why. For a solution, you can get a Javascript BigInt library at http://www-cs-students.stanford.edu/~tjw/jsbn/

const subtract = (a, b) => [a, b].map(n => [...n].reverse()).reduce((a, b) => a.reduce((r, d, i) => {
let s = d - (b[i] || 0)
if (s < 0) {
s += 10
a[i + 1]--
}
return '' + s + r
}, '').replace(/^0+/, ''))
Better use big-integer library for these things so as to handle all different test cases.This is just for the a general case you can use....

It is explained in the JavaScript documentation:
According to the ECMAScript standard, there is only one number type: the double-precision 64-bit binary format IEEE 754 value (numbers between -(253-1) and 253-1). There is no specific type for integers.
Wikipedia page about double precision floating point format explains:
Between 252= 4,503,599,627,370,496 and 253= 9,007,199,254,740,992 the representable numbers are exactly the integers. For the next range, from 253 to 254, everything is multiplied by 2, so the representable numbers are the even ones, etc.
(All integer numbers smaller than 252 are represented exactly.)
1234567890123456789 and 1234567890123456799 are larger than 260= 1152921504606846976. At this magnitude only about 1% of the integer numbers are stored exactly using the double-precision floating point format.
These two cannot be stored exactly. They both are rounded to 1234567890123456800.
The JavaScript documentation also explains how to tell if a an integer number is stored exactly:
[...] and starting with ECMAScript 6, you are also able to check if a number is in the double-precision floating-point number range using Number.isSafeInteger() as well as Number.MAX_SAFE_INTEGER and Number.MIN_SAFE_INTEGER. Beyond this range, integers in JavaScript are not safe anymore and will be a double-precision floating point approximation of the value.

function add(x, y) {
//*********************************************************************//
// This function adds or subtracts two extremely large decimal numbers //
// Inputs x and y should be numbers, i.e. commas are removed already //
// Use this function to remove commas and convert to number: //
// x = parseFloat(strNumber.replaceAll(",","").trim()); //
// Inputs x and y can be both positive, or both negative, //
// or a combination (i.e. one positive and one negative in any //
// position whether as x or as y) which means subtraction //
//*********************************************************************//
var temp, borrow=false, bothNeg=false, oneNeg=false, neg=false;
if (x < 0 && y < 0) { bothNeg = true; x = -x; y = -y; }
else if (x < 0 || y < 0) {
oneNeg = true;
if (Math.abs(x) == Math.abs(y)) { x = 0; y = 0; }
else if (x < 0 && Math.abs(x) > Math.abs(y)) { neg = true; x = -x; y = -y; }
else if (x < 0 && Math.abs(x) < Math.abs(y)) { temp = y; y = x; x = temp; }
else if (y < 0 && Math.abs(x) < Math.abs(y)) { neg = true; temp = y; y = -x; x = -temp; }
}
x = parseInt(x*1000000000/10).toString();
y = parseInt(y*1000000000/10).toString();
var lenx=x.length, leny=y.length, len=(lenx>leny)?lenx:leny, sum="", div=0, x1, y1, rem;
for (var i = 0; i < len; i++) {
x1 = (i >= lenx) ? 0 : parseInt(x[lenx-i-1]);
y1 = (i >= leny) ? 0 : parseInt(y[leny-i-1]);
y1 = (isNaN(y1)) ? 0 : y1;
if (oneNeg) y1 = -y1;
if (borrow) x1 = x1 - 1;
if (y < 0 && x1 > 0 && Math.abs(x1) >= Math.abs(y1)) { borrow=false; div=0; }
if (y < 0 && y1 <= 0 && (x1 < 0 || Math.abs(x1) < Math.abs(y1))) { borrow=true; rem=(x1+y1+div+10)%10; div=10; }
else { rem=(x1+y1+div)%10; div=Math.floor((x1+y1+div)/10); }
sum = Math.abs(rem).toString() + sum;
}
if (div > 0) sum = div.toString() + sum;
sum = parseFloat(sum*10/1000000000);
if (bothNeg || neg) sum = -sum;
return sum;
}