Cryptographically secure float

Question

How do you generate cryptographically secure floats in Javascript?

This should be a plug-in for Math.random, with range (0, 1), but cryptographically secure. Example usage

cryptoFloat.random();
0.8083966837153522

Secure random numbers in javascript? shows how to create a cryptographically secure Uint32Array. Maybe this could be converted to a float somehow?

The Mozilla Uint32Array documentation was not totally clear on how to convert from an int.
Google was not to the point, either.
Float32Array.from(someUintBuf); always gave a whole number.

Do you need only to support browsers that support `window.crypto`? — T.J. Crowder, Jan 03 '16 at 10:52
@T.J.Crowder: Actually, for my use case, Firefox would suffice. The more general, the merrier. — serv-inc, Jan 03 '16 at 11:02
Just to flag it up at a high-level: JavaScript's floats are 64-bit IEEE-754 numbers (well, you can get a 32-bit one if you want), which means they only have 53 (effective) significant binary digits. That's not enough for nearly any cryptographic purpose. — T.J. Crowder, Jan 03 '16 at 12:01
@T.J.Crowder: the warning is definitely right. Yet, for some cases (statistical distributions), it is hopefully enough. — serv-inc, Jan 03 '16 at 13:05

serv-inc · Accepted Answer · 2017-09-21T07:03:28.730

8

Since the following code is quite simple and functionally equivalent to the division method, here is the alternate method of altering the bits. (This code is copied and modified from @T.J. Crowder's very helpful answer).

// A buffer with just the right size to convert to Float64
let buffer = new ArrayBuffer(8);

// View it as an Int8Array and fill it with 8 random ints
let ints = new Int8Array(buffer);
window.crypto.getRandomValues(ints);

// Set the sign (ints[7][7]) to 0 and the
// exponent (ints[7][6]-[6][5]) to just the right size 
// (all ones except for the highest bit)
ints[7] = 63;
ints[6] |= 0xf0;

// Now view it as a Float64Array, and read the one float from it
let float = new DataView(buffer).getFloat64(0, true) - 1; 
document.body.innerHTML = "The number is " + float;

Explanation:

The format of a IEEE754 double is 1 sign bit (ints[7][7]), 11 exponent bits (ints[7][6] to ints[6][5]), and the rest as mantissa (which holds the values). The formula to compute is

To set the factor to 1, the exponent needs to be 1023. It has 11 bits, thus the highest-order bit gives 2048. This needs to be set to 0, the other bits to 1.

edited Sep 21 '17 at 07:03

answered Jan 03 '16 at 14:58

serv-inc

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9
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Cool. Did you determine that the subnormal thing wasn't an issue, or...? – T.J. Crowder Jan 03 '16 at 15:15
@T.J.Crowder: This uses the maximum number of bits. Consider that each bit of the mantissa "splits the interval" by 2: The first either adds 1/2 or 0, the second adds 1/4 or 0, the third 1/8 or 0, ... It seems as though the problem of subnormal numbers only occurs if you include the exponent. (Then the numbers are clearly logarithmically spaced). Thank you. I probably would not have made it without your help. (had not worked with JS FloatArrays etc before) – serv-inc Jan 03 '16 at 15:28
1

You're welcome, I'm glad that helped. Re the subnormals, now I've read it again a few hours later, I don't think it's a problem; your method should evenly distribute the values across the range of numbers that `Number` can represent precisely. Subnormals extend the range by compromising precision. I don't think you're losing anything from a randomness perspective by leaving them out. – T.J. Crowder Jan 03 '16 at 15:35
1

I like the solution! Thanks much! Still, while almost the whole universe is little-endian these days, it might be worth being explicit: `let float = new DataView(buffer).getFloat64(0, true) - 1;` – Tom Palmer Sep 19 '17 at 12:31

Cryptographically secure float

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