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AlotOfReading a day ago

Another procedure based on a similar problem I worked on with a friend: you both pick positive integers a and b, then add them together to create c. Either sqrt(c) or sqrt(c+1) is irrational and the fractional digits provide your random numbers. If you need a new sequence, you take some digits from the current expansion and sqrt() them again.

Might not be unbiased, but good luck proving it.

pama a day ago | parent | next [-]

Even without a formal proof you could test it empirically if you generate a large number of samples and run regular tests for pseudorandom number generators. [Edit: a quick test on a million samples and relatively simple RNG tests suggests that this is indeed good enough; maybe worth working out a proof if this hasn't been done already. Edit2: I guess the main problem you'd hit in practice with short sequences of digits would be to avoid accidental recurrences with too short a period, but it should be possible to make it statistically unlikely in practice with enough compute/digits.]

AlotOfReading a day ago | parent [-]

Passes dieharder and PractRand, so it's pretty dang good.

pama a day ago | parent [-]

Yeah I like the simplicity and power of it. You might want to tackle the math and write it up.

crdrost a day ago | parent | prev | next [-]

I'm not entirely sure what algebraic property you would prove with this, but you probably could prove something about it. The issue is that they have repeating continued fraction representations, and large numbers in the continued fraction correspond to very good rational approximations, and so you'd find that a bunch of these chosen at random have pretty good rational approximations, which assuming the denominator is co-prime to 10, probably means that it explores the space of digits too uniformly? Something like that.

AlotOfReading a day ago | parent [-]

The approach I was thinking of is that you'd prove normality or the lack thereof, a notoriously open problem for virtually all irrational roots. Continued fractions might be fruitful, but I suspect you'd eventually run one of the many other open problems in that space instead.

gowld 7 hours ago | parent | prev | next [-]

You need to go mod 10^(-n) to avoid the bias in the first digits.

I'd prefer that one person choose a non-square integer, and the other choose (very large) n.

echoangle a day ago | parent | prev [-]

Can you calculate the digits in your head without help?

a day ago | parent [-]
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