Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
459 views
in Technique[技术] by (71.8m points)

javascript - 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.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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(`

Max safe number to a precision of ${digits} digits after the decimal point: ${n}((MAX + 1) / ${(Number.MAX_SAFE_INTEGER + 1) / (n + 1)} - 1)

`);
        return n;
      } else {
        lastSafe = n;
      }
    }
    n = Math.round((lastSafe + lastUnsafe) / 2);
  }
}

console.log(findMaxSafeFloat(1));

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...