3 ms·
Theoretically, everything can. And there's a FUSE implementation: https://github.com/philipl/pifs https://github.com/philipl/pifs
by BuildTheRobots 12y ago
Theoretically, everything can. And there's a FUSE implementation: https://github.com/philipl/pifs https://github.com/philipl/pifs
- dannypgh 12y agoTheoretically? Citation needed. I believe this is only true if pi is normal, and while there's a lot of conjecture that pi is normal and no evidence that suggests it isn't, there is no proof.
- PeterWhittaker 12y agoNormal or not normal? If a work can be found in pi, then one would think a Finite State Gambler (FSG) would succeed on the portion of pi equivalent to the work. But FSGs cannot succeed on normal sequences, meaning this sequence would be non-normal. Of course this begs the question of whether a number normal over an infinite sequence can contain sub-sequences that are non-normal - sequences of 100 tails will occur in sufficiently large completely normal (randomized) sequences of random coin tosses, e.g.
- Fargren 12y agoA normal sequence includes all non-normal sequences, by definition of normality.
- PeterWhittaker 12y agoAs for the citation, there is this: http://en.wikipedia.org/wiki/Disjunctive_sequence http://en.wikipedia.org/wiki/Disjunctive_sequence If this is correct, then you are correct, the conjecture would hold if pi were normal. But then wouldn't an FSG be able to succeed on the sub-sequence? Of is that acceptable, provided that the FSG fails on the overall sequence?
- tzs 12y agoNormality is sufficient, but I do not think it is necessary. Normality requires that if the number is represented in any base b then each each of the b possible digits occure with density 1/b. All that is needed for every work to appear is that each digit occur at least once. They do not have to occure with equal density.
- ithinkso 12y agoNope, normality is necessary. >>All that is needed for every work to appear is that each digit occur at least once. Counter example would be a=0,123456789010011000111000011110000011111....
- millstone 12y agoNormality is not necessary. Counter-counter example is the digits of pi in groups of 1, 2, 3... with strings of zeros in between: 3.0140015900026530000.... this number is obviously nor normal since 0 occurs with greater frequency than any other digit. But if pi is normal, then this sequence contains all finite sequences too.
- tzs 12y agoEach base 10 digit occurs at least once in your number. It's not obvious to me that each base b digit occurs at least once for all bases b in your number.
- zhte415 12y agoExactly. But I've been searching Pi for the blueprints of a Klingon cloaking device for the past 5 years and have yet to turn up anything.