CODEWITHSUNDEEP
javascriptnode.jsfloating-point
WoLfPwNeR
WoLfPwNeR

Reputation: 1324

Node.js Maximum Safe Floating-point Number

In Node.js, is there a maximum safe floating-point number like Number.MAX_SAFE_INTEGER?

I had a little experiment to find out the (approximate) number I can use for subtracting 0.13 from it:

console.log(Math.floor(Number.MAX_SAFE_INTEGER));  // 9007199254740991
console.log(Math.floor(Number.MAX_SAFE_INTEGER)-0.13);  // 9007199254740991

console.log(Math.floor(Number.MAX_SAFE_INTEGER/2));  // 4503599627370495
console.log(Math.floor(Number.MAX_SAFE_INTEGER/2)-0.13);  // 4503599627370495

console.log(Math.floor(Number.MAX_SAFE_INTEGER/4));  // 2251799813685247
console.log(Math.floor(Number.MAX_SAFE_INTEGER/4)-0.13);  // 2251799813685246.8

console.log(Math.floor(Number.MAX_SAFE_INTEGER/64));  // 140737488355327
console.log(Math.floor(Number.MAX_SAFE_INTEGER/64)-0.13);  // 140737488355326.88

console.log(Math.floor(Number.MAX_SAFE_INTEGER/128));  // 70368744177663
console.log(Math.floor(Number.MAX_SAFE_INTEGER/128)-0.13);  // 70368744177662.87

My guess is that as the target precision increases, the maximum value decreases.

Upvotes: 18

Views: 22208

Answers (4)

giladbarnea
giladbarnea

Reputation: 691

To a precision of 1 decimal digits, the maximum number you can work with is 562949953421311.

To a precision of 2 decimal digits, it's 70368744177663. Interestingly, the first number is equal to:

(Number.MAX_SAFE_INTEGER + 1) / 16 - 1

And the second number is equal to:

(Number.MAX_SAFE_INTEGER + 1) / 128 - 1

What we're looking for, is the maximum safe number to support a precision of d digits after the decimal point. By "support" I mean "can reliably do basic arithmetic".

For example, we know that Number.MAX_SAFE_INTEGER (aka 2**53-1) is not safe, because basic arithmetic is broken:

Number.MAX_SAFE_INTEGER - 0.1 === Number.MAX_SAFE_INTEGER
>>> true // unsafe

And we know that 0 is safe, since:

0 + 0.1 === 0
>>> false // safe

BTW, 0 is reliable as far as 1e-323 (including):

0 + 1e-323 === 0
>>> false // safe

0 + 1e-324 === 0
>>> true // unsafe

I binary-searched between 0 and Number.MAX_SAFE_INTEGER for the biggest number that answers that definition, and came up with these numbers.

Here's the code (pass any other number to findMaxSafeFloat() at the end of snippet)

/**Returns whether basic arithmetic breaks between n and n+1, to a precision of `digits` after the decimal point*/
function isUnsafe(n, digits) {
  // digits = 1 loops 10 times with 0.1 increases.
  // digits = 2 means 100 steps of 0.01, and so on.
  let prev = n;
  for (let i = 10 ** -digits; i < 1; i += 10 ** -digits) {
    if (n + i === prev) { // eg 10.2 === 10.1
      return true;
    }
    prev = n + i;
  }
  return false;


}

/**Binary search between 0 and Number.MAX_SAFE_INTEGER (2**53 - 1) for the biggest number that is safe to the `digits` level of precision.
 * digits=9 took ~30s, I wouldn't pass anything bigger.*/
function findMaxSafeFloat(digits, log = false) {
  let n = Number.MAX_SAFE_INTEGER;
  let lastSafe = 0;
  let lastUnsafe = undefined;
  while (true) {
    if (log) {
      console.table({
        '': {
          n,
          'Relative to Number.MAX_SAFE_INTEGER': `(MAX + 1) / ${(Number.MAX_SAFE_INTEGER + 1) / (n + 1)} - 1`,
          lastSafe,
          lastUnsafe,
          'lastUnsafe - lastSafe': lastUnsafe - lastSafe
        }
      });
    }
    if (isUnsafe(n, digits)) {
      lastUnsafe = n;
    } else { // safe
      if (lastSafe + 1 === n) { // Closed in as far as possible
        console.log(`\n\nMax safe number to a precision of ${digits} digits after the decimal point: ${n}\t((MAX + 1) / ${(Number.MAX_SAFE_INTEGER + 1) / (n + 1)} - 1)\n\n`);
        return n;
      } else {
        lastSafe = n;
      }
    }
    n = Math.round((lastSafe + lastUnsafe) / 2);
  }
}

console.log(findMaxSafeFloat(1));

An interesting thing I've found by lining up the safe numbers, is that the exponents don't step up in a consistent manner. Look at the table below; once in a while, the exponent increases (or decreases) by 4, and not 3. Not sure why.

| Precision | First UNsafe                | 2^53/x                   |
|-----------|-----------------------------|--------------------------|
| 1         | 562,949,953,421,312 = 2^49  | x = 16 = 2^4             |
| 2         |  70,368,744,177,664 = 2^46  | x = 128 = 2^7            |
| 3         |   8,796,093,022,208 = 2^43  | x = 1,024 = 2^10         |
| 4         |     549,755,813,888 = 2^39  | x = 16,384 = 2^14        |
| 5         |      68,719,476,736 = 2^36  | x = 131,072 = 2^17       |
| 6         |       8,589,934,592 = 2^33  | x = 1,048,576 = 2^20     |
| 7         |         536,870,912 = 2^29  | x = 16,777,216 = 2^24    |
| 8         |          67,108,864 = 2^26  | x = 134,217,728 = 2^27   |
| 9         |           8,388,608 = 2^23  | x = 1,073,741,824 = 2^30 |

Upvotes: 59

Lcf.vs
Lcf.vs

Reputation: 1842

The exponent decreases in a consistent manner, because it depends on the decimal part size (binary).

  • For a precision 1, the max decimal part size is 9 -> 1001 -> size = 4
  • For a precision 2, the max decimal part size is 99 -> 1100011 -> size = 7
  • For a precision 3, the max decimal part size is 999 -> 1111100111 -> size = 10
  • ...

As you can see, we have the same numbers than @giladbarnea found.

Based on that observation, we can easily write a simpliest solution to find all precisions that can be related to any integer.

const precisions = [...new Array(16).keys()]
  .reverse()
  .map(value => {
    const int = 53 - (value && BigInt('9'.repeat(value)).toString(2).length)

    return int > 0
      ? 2 ** int - 1
      : 0
  })

function getSafePrecision (value) {
  const abs = Math.abs(value)

  return 15 - precisions.findIndex(precision => precision >= abs)
}

Example:

getSafePrecision(2 ** 43 - 1) // 3
getSafePrecision(2 ** 43) // 2

Upvotes: 0

shaochuancs
shaochuancs

Reputation: 16246

Update: My understanding about this question is: Is there a maximum floating number, between 0 and that, all floating number operation can be safely delivered.

If that is the question, short answer is: No

Actually, there is no MAX_SAFE_FLOAT in all programming language (will be very glad if there is one). Number in programming language is stored by 0 or 1 bits. As long as there is a limit for the storage (32bits, 64bits etc), numbers that can be represented is finite. However, the number of floating-number is infinite.

Consider floating-number between 0 and 0.000000001, how many numbers need to be represented? Infinite. It's impossible to let computer store infinite possibility accurately. That's why there would never be MAX_SAFE_FLOAT.

p.s. In JavaScript, all numbers are 64bit double-precision floating-number. There is no floating-number v.s. interger-number in JavaScript.

Upvotes: 5

Pointy
Pointy

Reputation: 413828

You're looking for Number.MAX_VALUE and Number.MIN_VALUE.

Number.MAX_VALUE is 1.7976931348623157e+308 and Number.MIN_VALUE is 5e-324.

Upvotes: -3

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