I found this snippet online along with this Stackoverflow post which converts it into a TypeScript class.
I basically copy and pasted it verbatim (because I am not qualified to modify this sort of cryptographic code), but I noticed that VS Code has a little underline in the very last function:
/**
* generates a random number on [0,1) with 53-bit resolution
*/
nextNumber53(): number {
let a = this._nextInt32() >>> 5;
let b = this._nextInt32() >>> 6;
return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
Specifically the 9007199254740992.0
VS Code says Numeric literals with absolute values equal to 2^53 or greater are too large to be represented accurately as integers.ts(80008)
I notice that if I subtract that number by one and instead make it 9007199254740991.0, then the warning goes away. But I don't necessarily want to modify the code and break it if this is indeed a significant difference.
Basically, I am unsure, because while my intuition says that having a numerical overflow is bad, my intuition also says that I shouldn't try to fix cryptographic code that was posted in several places, as it is probably correct.
But is it? Or should this number be subtracted by one?
9007199254740992 is the right value to use if you want Uniform values in [0,1), i.e. 0.0 <= x < 1.0.
This is just the automatics going awry, this value can be accurately represented by a JavaScript Number, i.e. a 64bit float. It's just 253 and binary IEEE 754 floats have no trouble with numbers of this form (it would even be represented accurately with a 32bit float).
Using 9007199254740991 would make the range [0,1], i.e. 0.0 <= x <= 1.0. Most libraries generate uniform values in [0,1) and other distributions are derived from that, but you are obviously free to do whatever is best for your application.
Note that the actual chance of getting the maximum value back is 2-53 (~1e-16) so you're unlikely not actually see it in practice.
I am working with js numbers and have lack of experience in that. So, I would like to ask few questions:
2.2932600144518896
e+160
is this float or integer number? If it's float how can I round it to two decimals (to get 2.29)? and if it's integer, I suppose it's very large number, and I have another problem than.
Thanks
Technically, as said in comments, this is a Number.
What you can do if you want the number (not its string representation):
var x = 2.2932600144518896e+160;
var magnitude = Math.floor(Math.log10(x)) + 1;
console.log(Math.round(x / Math.pow(10, magnitude - 3)) * Math.pow(10, magnitude - 3));
What's the problem with that? Floating point operation may not be precise, so some "number" different than 0 should appear.
To have this number really "rounded", you can only achieve it through string (than you can't make any operation).
JavaScript only has one Number type so is technically neither a float or an integer.
However this isn't really relevant as the value (or rather representation of it) is not specific to JavaScript and uses E-Notation which is a standard way to write very large/small numbers.
Taking this in to account 2.2932600144518896e+160 is equivalent to 2.2932600144518896 * Math.pow(10,160) and approximately 229 followed by 158 zeroes i.e. very flippin' big.
I have a problem in precision in the last digit after the comma.The javascript code generates one less Digit in compare with the C# code.
Here is the simple Node.js code
var seed = 45;
var x = Math.sin(seed) * 0.5;
console.log(x);//0.4254517622670592
Here is the simple C# code
public String pseudorandom()
{
int seed = 45;
double num = Math.Sin(seed) * (0.5);
return num.ToString("G15");//0.42545176226705922
}
How to achieve the same precision?
The JavaScript Number type is quite complex. It looks like floating point number will probably be like IEEE 754-2008 but some aspects are left to the implementation. See http://www.ecma-international.org/ecma-262/6.0/#sec-number-objects sec 12.7.
There is a note
The output of toFixed may be more precise than toString for some
values because toString only prints enough significant digits to
distinguish the number from adjacent number values. For example,
(1000000000000000128).toString() returns "1000000000000000100", while
(1000000000000000128).toFixed(0) returns "1000000000000000128".
Hence to get full digit accuracy you need something like
seed = 45;
x = Math.sin(seed) * 0.5;
x.toFixed(17);
// on my platform its "0.42545176226705922"
Also, note the specification for how the implementation of sin and cos allow for some variety in the actual algorithm. It's only guaranteed to within +/- 1 ULP.
Using java the printing algorithm is different. Even forcing 17 digits gives the result as 0.42545176226705920.
You can check you are getting the same bit patterns using x.toString(2) and Double.doubleToLongBits(x) in Java.
return num.ToString("G15");//0.42545176226705922
actually returns "0.425451762267059" (no significant digit + 15 decimal places in this example), and not the precision shown in the comment after.
So you would use:
return num.ToString("G16");
to get "0.4254517622670592"
(for your example - where the significant digit is always 0) G16 will be 16 decimal places.
Is it possible to get string with length of 17 by using such code?
We're saving cookie in a client's browser and then get it from this browser.
function getPartnerVisitorId() {
var pvid = cookies.get('pvid') || Math.floor(Math.random() * Math.pow(10, 15)).toString();
cookies.setOnRoot('pvid', pvid, 365 * 24 * 60 * 60);
return pvid;
};
We think that "pvid" should be maximum of 16 digits length, but sometimes (about 5% of cookies) we get 17 digits length of this cookie.
Also we've noticed that all of strange cookies are even numbers and most of them (75%) had zero as the last digit
The issue that you're dealing with has to do the nature of IEEE 754 floating point numbers - the type used to hold Number in JavaScript, double in Java and C, etc. These numbers are stored in a sort of scientific notation, in the form of
[+/-] 1.[some value] * 2^[some other value].
This results in you getting about 15 significant decimal digits that are especially adept at expressing powers of 2.
By multiplying Math.random() by Math.pow(10,15), you are encountering two problems with this scheme:
You are running out of digits, hence truncating the last few at the bottom (hence why your numbers are always even)
You are trying to represent a power of 10, which it cannot always do accurately (hence why you are sometimes getting 17-digit keys).
To fix these problems and get the constant-length random keys you want, do two things:
Multiply Math.random() by Math.pow(2,x) (where x is a number you experiment with to find the right length) - 2^anything can always be accruately expressed by floating point numbers.
In the toString(), method, insert an integer parameter that is a power of 2 (such as 8 or 16) - this will cause the number to be converted to a string with the base-8 or base-16 number system. This will prevent variable length keys and also prevent rounding.
var pvid = cookies.get('pvid') || Math.floor(Math.random() * Math.pow(8, 15)).toString(8);
I am adding client-side sub-total calculations to my order page, so that the volume discount will show as the user makes selections.
I am finding that some of the calculations are off by one cent here or there. This wouldn't be a very big deal except for the fact that the total doesn't match the final total calculated server-side (in PHP).
I know that the rounding errors are an expected result when dealing with floating point numbers. For example, 149.95 * 0.15 = 22.492499999999996 and 149.95 * 0.30 = 44.98499999999999. The former rounds as desired, the latter does not.
I've searched on this topic and found a variety of discussions, but nothing that satisfactorily addresses the problem.
My current calculation is as follows:
discount = Math.round(price * factor * 100) / 100;
A common suggestion is to work in cents rather than fractions of dollars. However, this would require me to convert my starting numbers, round them, multiply them, round the result, and then convert it back.
Essentially:
discount = Math.round(Math.round(price * 100) * Math.round(factor * 100) / 100) / 100;
I was thinking of adding 0.0001 to the number before rounding. For example:
discount = Math.round(price * factor * 100 + 0.0001) / 100;
This works for the scenarios I've tried, but I am wondering about my logic. Will adding 0.0001 always be enough, and never too much, to force the desired rounding result?
Note: For my purposes here, I am only concerned with a single calculation per price (so not compounding the errors) and will never be displaying more than two decimal places.
EDIT: For example, I want to round the result of 149.95 * 0.30 to two decimal places and get 44.99. However, I get 44.98 because the actual result is 44.98499999999999 not 44.985. The error is not being introduced by the / 100. It is happening before that.
Test:
alert(149.95 * 0.30); // yields 44.98499999999999
Thus:
alert(Math.round(149.95 * 0.30 * 100) / 100); // yields 44.98
The 44.98 is expected considering the actual result of the multiplication, but not desired since it is not what a user would expect (and differs from the PHP result).
Solution: I'm going to convert everything to integers to do my calculations. As the accepted answer points out, I can simplify my original conversion calculation somewhat. My idea of adding the 0.0001 is just a dirty hack. Best to use the right tool for the job.
I don't think adding a small amount will favor you, I guess there are cases where it is too much. Also it needs to be properly documented, otherwise one could see it as incorrect.
working in cents […] would require me to convert my starting numbers, round them, multiply them, round the result, and then convert it back:
discount = Math.round(Math.round(price * 100) * Math.round(factor * 100) / 100) / 100;
I think it should work as well to round afterwards only. However, you should first multiply the result so that the significant digits are the sum of the two sig digits from before, i.e. 2+2=4 decimal places in your example:
discount = Math.round(Math.round( (price * factor) * 10000) / 100) / 100;
Adding a small amount to your numbers will not be very accurate. You can try using a library to get better results: https://github.com/jtobey/javascript-bignum.
Bergi’s answer shows a solution. This answer shows a mathematical demonstration that it is correct. In the process, it also establishes some bound on how much error in the input is tolerable.
Your problem is this:
You have a floating-point number, x, which already contains rounding errors. E.g., it is intended to represent 149.95 but actually contains 149.94999999999998863131622783839702606201171875.
You want to multiply this floating-point number x by a discount value d.
You want to know the result of the multiplication to the nearest penny, performed as if ideal mathematics were used with no errors.
Suppose we add two more assumptions:
x always represents some exact number of cents. That is, it represents a number that has an exact number of hundredths, such as 149.95.
The error in x is small, less than, say, .00004.
The discount value d represents an integer percentage (that is, also an exact number of hundredths, such as .25 for 25%) and is in the interval [0%, 100%].
The error is d is tiny, always the result of correct conversion of a decimal numeral with two digits after the decimal point to double-precision (64 bit) binary floating point.
Consider the value x*d*10000. Ideally, this would be an integer, since x and d are each ideally multiples of .01, so multiplying the ideal product of x and d by 10,000 produces an integer. Since the errors in x and d are small, then rounding x*d*10000 to an integer will produce that ideal integer. E.g., instead of the ideal x and d, we have x and d plus small errors, x+e0 and d+e1, and we are computing (x+e0)•(d+e1)•10000 = (x•d+x•e1+d•e0+e0•e1)•10000. We have assumed that e1 is tiny, so the dominant error is d•e0•10000. We assumed e0, the error in x, is less than .00004, and d is at most 1 (100%), so d•e0•10000 is less than .4. This error, plus the tiny errors from e1, are not enough to change the rounding of x*d*10000 from the ideal integer to some other integer. (This is because the error must be at least .5 to change how a result that should be an integer rounds. E.g., 3 plus an error of .5 would round to 4, but 3 plus .49999 would not.)
Thus, Math.round(x*d*10000) produces the integer desired. Then Math.round(x*d*10000)/100 is an approximation of x*d*100 that is accurate to much less than one cent, so rounding it, with Math.round(Math.round(x*d*10000)/100) produces exactly the number of cents desired. Finally, dividing that by 100 (to produce a number of dollars, with hundredths, instead of a number of cents, as an integer) produces a new rounding error, but the error is so small that, when the resulting value is correctly converted to decimal with two decimal digits, the correct value is displayed. (If further arithmetic is performed with this value, that might not remain true.)
We can see from the above that, if the error in x grows to .00005, this calculation can fail. Suppose the value of an order might grow to $100,000. The floating-point error in representing a value around 100,000 is at most 100,000•2-53. If somebody ordered one hundred thousand items with this error (they could not, since the items would have smaller individual prices than $100,000, so their errors would be smaller), and the prices were individually added up, performing one hundred thousand (minus one) additions adding one hundred thousand new errors, then we have almost two hundred thousand errors of at most 100,000•2-53, so the total error is at most 2•105•105•2-53, which is about .00000222. Therefore, this solution should work for normal orders.
Note that the solution requires reconsideration if the discount is not an integer percentage. For example, if the discount is stated as “one third” instead of 33%, then x*d*10000 is not expected to be an integer.