How Wide Do All The Possible Computed Digit Sequences Go?
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From: George Greene on 21 Jun 2010 23:13 On Jun 21, 8:21 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > How many digits wide of base 10 sequences can I finite reals > cover? You still have not defined "cover". > > The use of the word cover should be explicit in it's use above. It isn't. It IS UP TO YOU to explicitly define "cover". We're not holding our breath. > NOW what is your excuse to avoid answering the question? It's the same as your excuse for not defining "cover". And if you ONLY MEAN FINITE SEQUENCES, then you CAN SAY, "how wide a width of digit-positions does it take to cover all finite sequences?" It IS NOT the sequences that do the covering: IT'S THE NUMBER that covers the sequences, DUMBASS. That is, if you actually knew how to speak English. > If 3^2 reals are required to 'cover' > every base 3 permutation to 2 digits wide.... 3^2 reals ARE NOT required to cover every base 3 permutation to 2 digits wide, DUMBASS. Covering is by WIDTH. It goes ACROSS, NOT DOWN, DUMBASS. *TWO* digits of width are required to cover every permutation that is two digits wide, AND IT DOES NOT MATTER what the base is (as long as it's finite), DUMBASS. If you want to count DOWN (as opposed to across), then you DON'T NEED "cover" at all; you MERELY need TO COUNT: you can just ask HOW MANY two-digit sequences over a ternary digit-set are there, and, of course, the answer is 3^2. For digit-sequences of length w over a base-alphabet of d digits, the NUMBER OF all such sequences is always going to be d^w. If you are going to do reals between 0 and 1, then you can do them IN BINARY and YOU NEED to do them in binary IN ORDER TO BE ABOUT Cantor's Theorem, which is about being or not being A MEMBER OF A SUBSET (there are ONLY TWO possibilities; either the element IS or ISN'T a member, so you only get two digits). So, in the case of reals, since there are infinitely many places of width (this infinity is usually spelled with a lower-case greek omega, so here in ascii- land, I am going to spell it w, which is a fortunate coincidence -- reals are like bit- strings of width w) and 2 digits, THE NUMBER of reals is going to be 2^w. In general is set s has size |s|, then the powerset of s (the set of all of s's subsets, the set of all bit-strings telling which elements of s belong to a subset and which don't) is going to have size 2^|s| -- that's WHY THE SET IS CALLED the powerset. Cantor's Theorem is the proof that 2^|s| > |s|, FOR ALL s, and it does NOT MATTER what the size of s, PROVIDED THAT s *HAS* a size; THAT is the hard part. the number of possible sequences is 2^w
From: Graham Cooper on 22 Jun 2010 00:06 On Jun 22, 1:13 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 21, 8:21 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > > > How many digits wide of base 10 sequences can I finite reals > > cover? > > You still have not defined "cover". > > > > > The use of the word cover should be explicit in it's use above. > > It isn't. It IS UP TO YOU to explicitly define "cover". > We're not holding our breath. > > > NOW what is your excuse to avoid answering the question? > > It's the same as your excuse for not defining "cover". > And if you ONLY MEAN FINITE SEQUENCES, > then you CAN SAY, "how wide a width of digit-positions does it take > to cover all finite sequences?" It IS NOT the sequences that do the > covering: > IT'S THE NUMBER that covers the sequences, DUMBASS. > That is, if you actually knew how to speak English. > > > If 3^2 reals are required to 'cover' > > every base 3 permutation to 2 digits wide.... > > 3^2 reals ARE NOT required to cover every base 3 permutation to 2 > digits wide, > DUMBASS. > Covering is by WIDTH. It goes ACROSS, NOT DOWN, DUMBASS. > *TWO* digits of width are required to cover every permutation that is > two digits wide, > AND IT DOES NOT MATTER what the base is (as long as it's finite), > DUMBASS. > > If you want to count DOWN (as opposed to across), then you DON'T NEED > "cover" at all; you MERELY need TO COUNT: you can just ask HOW MANY > two-digit sequences over a ternary digit-set are there, and, of > course, the answer > is 3^2. For digit-sequences of length w over a base-alphabet of d > digits, > the NUMBER OF all such sequences is always going to be d^w. > > If you are going to do reals between 0 and 1, then you can do them IN > BINARY > and YOU NEED to do them in binary IN ORDER TO BE ABOUT Cantor's > Theorem, > which is about being or not being A MEMBER OF A SUBSET (there are ONLY > TWO > possibilities; either the element IS or ISN'T a member, so you only > get two digits). > > So, in the case of reals, since there are infinitely many places of > width (this infinity > is usually spelled with a lower-case greek omega, so here in ascii- > land, I am going > to spell it w, which is a fortunate coincidence -- reals are like bit- > strings of width w) > and 2 digits, THE NUMBER of reals is going to be 2^w. In general is > set s has size > |s|, then the powerset of s (the set of all of s's subsets, the set of > all bit-strings > telling which elements of s belong to a subset and which don't) is > going to have size > 2^|s| -- that's WHY THE SET IS CALLED the powerset. > > Cantor's Theorem is the proof that 2^|s| > |s|, FOR ALL s, and it > does > NOT MATTER what the size of s, PROVIDED THAT s *HAS* a size; > THAT is the hard part. > > the number of possible sequences is 2^w Ok. What if the number of possible sequences (listed computable reals) is w, regarding places of width of every permutation? Herc
From: George Greene on 22 Jun 2010 00:48 On Jun 22, 12:06 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > Ok. What if the number of possible sequences (listed computable reals) > is w, regarding places of width of every permutation? What if the sky were lime-green-and-purple-striped and the human lifespan in years were (typically, at the peak of the bell-curve) 3000 instead of 70?? This is a meaningless question. The number of possible (countably-infinitely-wide) digit-strings(over a finite alphabet) IS NOT w. It is 2^w, WHICH MUST BE BIGGER.
From: George Greene on 22 Jun 2010 00:53 On Jun 22, 12:06 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > Ok. What if the number of possible sequences (listed computable reals) > is w, regarding places of width of every permutation? This question is SO STUPID!! The PERMUTATIONS have NOTHING TO DO with the WIDTH! Every item on the list, REGARDLESS OF THE ORDER into which you permute them, STARTS AND STAYS width w! EVERY real on a list of reals IS ALWAYS countably infinitely wide, BY DEFINITION! Even the ones that LOOK FINITE ARE NOT finite -- they just end WITH AN INFINITE string of 0000's or 9999's (or of 1's, in binary, which is how you SHOULD ALWAYS be thinking of this). You ARE talking about permuting ROWS of the list, NOT COLUMNS of it, (or have I just given your ADD-addled brain another irrelevant tangent to pursue?) There are some permutations of infinity that will change the list-like character of the collection into more than 1 list; for example, if you permuted all the (formerly) even rows to come after all the (originally) odd rows, then you would wind up with TWO infinite lists, or a length (NOT a width) of w+w. But that, I repeat, IS NOT a list: THAT is TWO lists.
From: Rupert on 22 Jun 2010 01:21
On Jun 22, 6:44 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > On Jun 22, 12:08 am, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > > > > > > > On Jun 21, 10:40 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > > > > On 21/06/2010 5:03 PM, Rupert wrote: > > > > > On Jun 21, 4:28 pm, "|-|ercules"<radgray...(a)yahoo.com> wrote: > > > >> Every possible combination X wide... > > > > >> What is X? > > > > >> Now watch as 100 mathematicians fail to parse a trivial question. > > > > >> Someone MUST know what idea I'm getting at! > > > > >> This ternary set covers all possible digits sequences 2 digits wide! > > > > >> 0.00 > > > >> 0.01 > > > >> 0.02 > > > >> 0.10 > > > >> 0.11 > > > >> 0.12 > > > >> 0.20 > > > >> 0.21 > > > >> 0.22 > > > > >> HOW WIDE ARE ALL_POSSIBLE_SEQUENCES COVERED IN THE SET OF COMPUTABLE REALS? > > > > >> Herc > > > >> -- > > > >> If you ever rob someone, even to get your own stuff back, don't use the phrase > > > >> "Nobody leave the room!" ~ OJ Simpson > > > > > It would probably be a good idea for you to talk instead about the set > > > > of all computable sequences of digits base n, where n is some integer > > > > greater than one. Then the length of each sequence would be aleph- > > > > null. But not every sequence of length aleph-null would be included.. > > > > That answer looks correct. > > > > But I guarantee that Herc won't accept it. > > > > Sylvia. > > > It's truly hilarious. It's like using a Santa clause metaphor > > to explain why Santa clause is not real, > > but it will do for now. > > > Herc > > Actually on second reading I think Rupert threw a red herring > > He didn't adress the question at all. How wide are all possible > permutations of digits covered? This is different to all possible > listed sequences he just answered that numbers are inf. long! > > Herc- Hide quoted text - > > - Show quoted text - I'm afraid I don't understand the question. |