  # Aeternus

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1. I assume your alphabet is $\{1\}$? In this case, you can just proceed by induction on the length $n$ of the string. Take a couple of base cases to make it simple. It makes things much simpler if you first of all reduce it to (111*)* = 111*, since clearly 111* has any string of length 2 or greater, which is the same as (111*)*. In the case $n = 0$, then due to the outer most Kleene Star in both cases, the empty word is in both languages. In case $n = 1$, then as we have no string in either expression of length less than 2, no amount or co
2. Why not look at things such as - http://en.wikipedia.org/wiki/Adder_(electronics) ? A nice simple example of computation being done with logic gates.
3. Note here that $n\sharp$ is not the same as $P_n\sharp$ and I imagine you should be considering $n\sharp$ if you are looking at all the primes up to $n$, which will be much better for you. However, even with this improvement, this still gives a space complexity of $O(n)$ (from the wiki), which is still worse than Sieve of Atkins. Note here, that whatever your space complexity, your time complexity must be at least that (since you need to write to that many memory locations). You still need to worry about time, since the time to calcula
4. If you look at http://en.wikipedia.org/wiki/Factorial#Rate_of_growth , you'll see that for bigger $n$, $n!$ has at least $n (\log(n) -1)$ digits, which means that for a prime with thousands of digits, you are talking of not thousands of digits but of the order of $n\log(n)$ digits where $n$ is thousands of digits itself. I.e, you are talking at least $O(n \log(n))$ space complexity as far as I can see, which is worse than that for Sieve of Atkin given at http://en.wikipedia.org/wiki/Sieve_of_Atkin#Computational_complexity ,
5. Well consider that given a number n, you need $\log_2(n)$ bits to display it, so you have $N = \log_2(n)$ $2^N = 2^{\log_2(n)}$ $2^N = n$ So you are left with $O(2^n)$. Two points to note : a) Big O and Big Theta are two different things. Big O is only an upper bound, whereas Big Theta is much tighter and must be also a lower bound. See http://en.wikipedia.org/wiki/Big_O_notation#The_family_of_Bachmann-Landau_notations . b) You only need to check for factors up to $\sqrt{n}$ rather than $\fr 6. [hide] Or the fact that there is an eldest child, (potentially) implying there aren't 2 eldest children if you only consider integer ages. [/hide] 7. Given personal experience, experiences of friends and also looking around online, I'd say that, while macs USED to be higher quality than your average PC, they are probably about on par these days. 8. Sounds right. Although you could probably get away with [math] (C \setminus A) \cup (B \setminus C)$ since $C \setminus A$ will contain the bits that you need from C.
9. I think you have the wrong idea. From what I gather, when one talks of 10 dimensions, one is not talking about different universes or anything like that (you might be thinking of http://en.wikipedia.org/wiki/Many_worlds but that suggests many more than 10 "universes") but is talking about spacial dimension, in the same way that you have 3 dimensions (i.e height, width and breadth). The idea being that you don't have just 3 spacial dimensions but in fact 10 spacial dimensions (this is very difficult to imagine, hence the video and various other resources trying to provide a way to imagine
10. The nice thing is that it's not limited to SQL backed systems, you can use LINQ to run through various datastructures that you are using or implement a few interfaces and have your own datastructure usable (through IQueryable etc I believe). Saying that, the underlying idea isn't completely original, as bascule mentioned previously, things like ActiveRecord and ADO.NET have done similar things (not specifically the query part, but bringing the handling of the data and types into the programming language) but it'll be interesting to use. Personally I'm more interested in them bringing
11. You know you can preview your post right ?
12. I don't think that's correct, you are averaging over distance, not time. I believe you can't do it. Take the speed we travel back with to be $x$. We have $\textrm{average speed} = \frac{\textrm{total distance}}{\textrm{total time}}$. We have the total distance as 60, and the time is going to be the 1 hour we have already used, added to the time it will take us to cover the remaining 30 miles, so $\frac{30}{x}$. This gives us $60 = \frac{60}{1 + \frac{30}{x}}$ $1 + \frac{30}{x} = 1$ [math]\frac{30}{x} = 0 [/
13. I think you'll find 2 + (3*4) = 14 .
14. Well, it's fairly inexpensive, so why not buy it, or get it from the local library or something. If you find it hard to read, it might make a reasonable door stop
15. Strangely it seems they are from different roots (one from greek, one from latin) http://en.wikipedia.org/wiki/Greek_and_Latin_roots_in_English
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