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it's actually unknown. It looks like it, but it is not proven
Also is it even possible to prove it at all? My completely math inept brain thinks that it might be similar to the countable vs uncountable infinities thing, where even if you mapped every element of a countable infinity to one in the uncountable infinity, you could still generate more elements from the uncountable infinity. Would the same kind of logic apply to sequences in pi?
This is what allows pifs to work!
I'm a layman here and not a mathematician but how does it store the complete value of pi and not rounded up to a certain amount? Or do one of the libraries generate that?
You generate it when needed, using one of the known sequences that converges to π. As a simple example, the pi() recipe here shows how to compute π to arbitrary precision. For an application like pifs you can do even better and use the BBP formula which lets you directly calculate a specific hexadecimal digit of π.
I want that project continues so hard. Sounds amazing
I can't tell if this is a joke or real code... like for this sentence below.
The cat is back.
Will that repo seriously run until it finds where that is in pi? However long it might take, hours, days, years, decades, and then tell you, so you can look it up quickly?
I can’t tell if this is a joke or real code
Yes.
Will that repo seriously run until it finds where that is in pi?
Sure. It'll take a very long while though. We can estimate roughly how long - encoded as ASCII and translated to hex your sentence looks like 54686520636174206973206261636b. That's 30 hexadecimal digits. So very roughly, one of each 16^30 30-digit sequences will match this one. So on average, you'd need to look about 16^30 * 30 ≈ 4e37 digits into π to find a sequence matching this one. For comparison, something on the order of 1e15 digits of pi were ever calculated.
so you can look it up quickly?
Not very quickly, it's still n log n time. More importantly, information theory is ruthless: there exist no compression algorithms that have on average a >1 compression coefficient for arbitrary data. So if you tried to use π as compression, the offsets you get would on average be larger than the data you are compressing. For example, your data here can be written written as 30 hexadecimal digits, but the offset into pi would be on the order of 4e37, which takes ~90 hexadecimal digits to write down.
Thats very cool. It brings to mind some sort of espionage where spies are exchanging massive messages contained in 2 numbers. The index and the Metadata length. All the other spy has to do is pass it though pifs to decode. Maybe adding some salt as well to prevent someone figuring it out.
It has not been proven either way but if pi is proven to be normal then yes. https://en.m.wikipedia.org/wiki/Normal_number
no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.
for instance, write out pi in base 2, and reinterpret as base 10.
11.0010010000111111011010101000100010000101...
it is infinitely non-repeating, but nowhere will you find a 2.
i've often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven't seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.
this is correct but i think op is asking the wrong question.
at least from a mathematical perspective, the claim that pi contains any finite string is only a half-baked version of the conjecture with that implication. the property tied to this is the normality of pi which is actually about whether the digits present in pi are uniformly distributed or not.
from this angle, the given example only shows that a base 2 string contains no digits greater than 1 but the question of whether the 1s and 0s present are uniformly distributed remains unanswered. if they are uniformly distributed (which is unknown) the implication does follow that every possible finite string containing 1s and 0s is contained within, even if interpreted as a base 10 string while still base 2. base 3 pi would similarly contain every possible finite string containing the digits 0-2, even when interpreted in base 10 etc. if it is true in any one base it is true in all bases for their corresponding digits
Isnt this a stupid example though, because obviously if you remove all penguins from the zoo, you're not going to see any penguins
Its not stupid. To disprove a claim that states "All X have Y" then you only need ONE example. So, as pick a really obvious example.
it's not a good example because you've only changed the symbolic representation and not the numerical value. the op's question is identical when you convert to binary. thir is not a counterexample and does not prove anything.
The explanation is misdirecting because yes they're removing the penguins from the zoo. But they also interpreted the question as to if the zoo had infinite non-repeating exhibits whether it would NECESSARILY contain penguins. So all they had to show was that the penguins weren't necessary.
By tying the example to pi they seemed to be trying to show something about pi. I don't think that was the intention.
i just figured using pi was an easy way to acquire a known irrational number, not trying to make any special point about it.
It does contain a 2 though? Binary ‘10’ is 2, which this sequence contains?
They also say "and reinterpret in base 10". I.e. interpret the base 2 number as a base 10 number (which could theoretically contain 2,3,4,etc). So 10 in that number represents decimal 10 and not binary 10
that number is no longer pi… this is like answering the question "does the number "3548" contain 35?" by answering "no, 6925 doesnthave 35. qed"
I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”
They just proved that "X is irrational and non-repeating digits -> can find any sequence in X", as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn't be because of its irrationality or the fact that it's non-repeating, it would be some other property
Like the other commenter said its meant to be interpreted in base10.
You could also just take 0.01001100011100001111.... as an example
No, the fact that a number is infinite and non-repeating doesn't mean that and since in order to disprove something you need only one example here it is: 0.1101001000100001000001... this is a number that goes 1 and then x times 0 with x incrementing. It is infinite and non-repeating, yet doesn't contain a single 2.
This proves that an infinite, non-repeating number needn't contain any given finite numeric sequence, but it doesn't prove that an infinite, non-repeating number can't. This is not to say that Pi does contain all finite numeric sequences, just that this statement isn't sufficient to prove it can't.
you are absolutely right.
it just proves that even if Pi contains all finite sequences it's not "since it oa infinite and non-repeating"
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this post was submitted on 26 Dec 2024
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