That test run, and this 110-part walkthrough (clear time about 24 hours, but many of the battles and the level grinding near the end to reach the secret Rainbow and Final Boss ranks are cut from the video), make clear just how slowly this game runs on my old laptop, whether in DeSmuME or in NO$GBA: http://www.youtube.com/watch?v=k--9_Z4-pW0funnyhair wrote:
Technically, isn't the first game Super Mario RPG?
The first game in the Mario & Luigi RPG series is Superstar Saga for GBA; if he wanted to gain experience with an earlier game the only good option is Partners in Time, the other Mario RPG on the DS...but seriously IMO he should just stick to this game, it's not like the others are any easier to optimize, and Superstar Saga IIRC still has emulation problems that explain why nobody has submitted a run for it yet.
So far I have not found one that forces this... but the best one I used is the TeaTimer that comes with Spybot S&D. No matter what browser you are using you should really use Spybot S&D, because the immunization function is one of the greatest innovation I ever encountered.
I ended up disabling TeaTimer because it hogged system resources and kept bothering me even about legitimate Registry changes...even as I was about to reboot because of a software update, giving me almost no time to confirm or deny.
However I do make sure to immunize on a regular basis (it's a bit outdated though, like it doesn't work with the new Opera urlfilter.ini location), and also to use SpywareBlaster, which only partially immunizes, and then only works on IE and browsers based on Firefox.
The answer to my previous math question is the set E is a modified version of the union of the removed intervals of this: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set
instead of summing (1/4 + 1/8 + 1/16 +...), we need the sum to be (c/2 + c/4 + c/8 + c/16 +...) which is easy to do.
Does my solution also work?
Here it is in more concise form:
Let a sequence of subsets of [0,1] Cn(m,r), where n is in W, m is in N, and 0<r<1, be defined recursively by setting C0(m,r)=[0,1] and constructing, for 0<n, Cn(m,r) by removing from each closed sub-interval of Cn-1(m,r) the centered open interval of r times its length, and additionally, if m|n, by adding into each open interval between two closed intervals in the set the centered closed interval of r times its length. Let C be the set that Cn(m,r) approaches in the limit as n approaches infinity, and let A=[0,1]\C; then it can be proven that r can be chosen so that the measure of A lies between 0 and m/(1+m), so m merely needs to be chosen large enough so that c<m/(1+m), with the warning that m must be finite, for otherwise the measure of A would be 1/2 regardless of r.
the post in which I typed out some of the work involved got a bit mangled because I forgot to check "Disable HTML in this post" :-(
I won't be as silly as arflech, but buy as much RAM as you can afford. 32-bit Windows OSes can handle up to 3GB.
They can handle up to 4GB (2^32 B=2^2*2^30 B=4*1GB=4GB), it's just that some of that is reserved by the kernel for use by your hardware, so your OS will report less (my desktop with 4GB installed says 3.5GB, while my laptop with 2GB, which is the max. for its motherboard, says 1.96GB)
ITT we do someone's graduate real analysis homework
Cmuben wrote:
Randil wrote:
Perhaps it involves the cantor set, or some strange set including only irrational numbers? I'll give this some more thought...
cantor set is a good hint
I originally thought about a process involving translations of portions of the Cantor set to "fill up" the holes that had been removed, but then the measure would still be 0
Another idea may be to start with a similar infinite process in which, at each step, the middle third of the hole left behind is filled in again with a closed interval, so the second step of the construction would be
[0,1/9]U[2/9,1/3]U[4/9,5/9]U[2/3,7/9]U[8/9,1]
and the measures of the sets at each step, starting from step 0, would be 1, 2/3, 5/9, 14/27, ..., and would go to 1/2 in the limit, and so would its complement within [0,1], which IIRC would be an open set dense in [0,1].
More generally, consider this type of process with a more general factor r (above was the special case r=1/3); at each step n the measure of the set is An and the measure of its complement is 1-An, and then An+1=(1-r)An+(1-An)r=r+(1-2r)An, with A0=1.
Then A0=1, A1=1-r, A2=1-2r+2r^2, and more generally...
An=r+(1-2r)An-1=r+(1-2r)(r+(1-2r)An-2)=...=r+(1-2r)r+(1-2r)^2*r+(1-2r)^3*r+...+(1-2r)^(n-1)*r+(1-2r)^n=sum((1-2r)^j,j,0,n-1)r+(1-2r)^n=(1-(1-2r)^n)r/(2r)+(1-2r)^n=(1+(1-2r)^n)/2.
...so I guess this one doesn't work either, because regardless of the factor chosen (of course 0<r<1>infinity, and because as I said, 0<r<1, this means -2<-2r<0, so 1<3-2r<3, so 1/3<3-2r<1>infinity is 1/(2r^2-5r+4), and because 0<r<1>infinity of r/(1-(1-2r)(1-r)^(m-1)), and indeed, as r->1 (if m is finite) this expression approaches 1, and as r->0 this expression approaches 1/(1+m), and approaching the case of a more Cantor-like set, as m->infinity regardless of r, it approaches 0.
Then the complement of the set so constructed is an open dense set with area between 0 and m/(1+m), so pick a value of m, an integer greater than 1, so that 1/(1+m)<1-c, and then pick r, a real number between 0 and 1, so that 1/(1+m)<1-r<1-c and so that An=1-c (I realize now that maybe I should triply-index A by n, m, and r but whatev).
The perfect closeout to a story on CNN about a 103-year-old black woman who still drives...
...is a clip of an old Coolio song including the word "nigga": http://www.youtube.com/watch?v=8oF1pf5V-SQ
Kinda sounds like the issue that McAfee had with svchost.exe. Be thankful that it's not rebooting your computer constantly.
It also sounds like the issue that Norton had with plugin-container.exe, which Firefox 3.6.4 uses to host the out-of-process plugins; that's the main thing pushing the release date back almost a month, from 4 May to 1 June.
