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My avatar is a flying toaster from after dark. I chose it because it's surreal and just plain weird. Then I put a crappy christmas hat on it a year ago for festivities, and I haven't changed it since due to laziness.
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I'm just wondering if anyone has been working on a small only run lately. Also, I can't seem to find the latest old small only WIP on this topic. If anyone could direct me to the right page, that would be great.
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The 10 pixels to the left trick was know in the previous two runs. It's actually nothing new, and can only save extra time in one stage that I am aware of, and it was cancelled out by HB luck unfortunately.
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k^2 + 3k + 1 = (k+1)(k+2) - 1
It's all so simple now!
2004's prime factors are 2^2 * 3 * 167. Therefore any term past the 167th term = 0 mod 2004, because it has factors of 4, 3, and 167, inside k!
The factorisation makes things a lot easier as well. The kth term is therefore k!((k+1)(k+2) - 1) or (k+2)! - k!. and the k+2th term = (k+4)! - (k+2)!. adding these two together gives k! + (k+4)!, and the (k+2)! term is cancelled out.
basically if we add al the odd terms together, we get 199! - 1!, because all the intermediate terms are cancelled out, and since 199! has factors of 167, 4 and 3, 1! + 199! = 2003 mod 2004. Similarly, the even terms will be 2! + 200!, which is then 2002 mod 2004. therefore, the remainder when dividing the terms by 2004 is 2001.
EDIT: my uni library has that book in stock, I might take a look at it.
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ack! The first step is similar. you prove that it's divisible by 5 and 401 by normal methods. then I get to the second step 2005 = 16 mod 17 or 58 mod 59. from then it always seems I'm left with a remainder of some sort.
I give up on this one.
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My knowledge of math ends at high school level, but you mentioned Fermat's little theorem which gives me an idea for 5)
2011 is prime, and fermat's little theorem states that a^p = a mod p. I'm thinking this is because there are only p values that a^n can be mod p, and a^n will cycle through them all as long as p is prime.
basically if 1^2011 +2^2011 + 3^2011 ... = 1+2+3... mod 2011 and 1+2+3... 2010+2011 = 2011*2012/2 mod 2011 and since 2011 = 0 mod 2011, then 1^2011 + 2^2011 + ... + 2010^2011 + 2011^2011 = 0 mod 2011.
now we know it's divisible by 2011, but we need the number to be divisible by 2011*2012/2 which is 2011*1006, so we also need the number to be divisible by 1006, or 2*503. lucky for us we can use fermat's little theorem again, but in the form a^p-1 = 1 mod p, because 2011 = 502 mod 503. so now we have 1^502 + 2^502 + ... 501^502 + 502^502 + 0^502 + 1^502 + ... + 501^502 + 502^502 mod 503. which equals 502*4 mod 503. this is where I get something wrong, because 502*4 mod 503 = 499, not 0.
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1) p is a prime greater than 4, which means it's odd. (of the form 2n + 1 for a natural n)
p^2 - 1
=(2n + 1)^2 -1
=4(n)(n+1)
(n)(n+1) is divisible by 2, since they are consecutive, so 4(n)(n+1) is divisible by 8, and so is p^2 - 1.
p is greater than 3, so it is not divisible by 3. It is either of the form 3n + 1, or 3n + 2 for a natural n. If p = 3n+1, then p^2 - 1 = 3(3n^2 + 2n), which is divisible by 3. If p = 3n + 2, then p^2 - 1 = 3(3n^2 + 4n + 1), which is also divisible by 3.
since p^2 - 1 is divisible by 8 and 3, and they are both coprime, p^2 - 1 is divisible by 24.
2)Yes we can. firstly, divide the number by 7 to get rid of it. You now have 3^2011 1's written on an enourmous blackboard. 111 is divisible by 3, but not by 9 (add the digits together). 111000 is also divisible by 3. 111000000 is also divisible by 3. basically we split the number into three digit chunks, and divide each one by 3.
we get a number m = 111/3 which is not divisible by 3, since 111 is not divisible by 9. After we divide the number by 3, the remaining number is m*1 + m*1000 + m*1000000 ect. since m is not divisible by 3, we can safely divide by m to get 100100100...001001001. but now 1001001 is divisible by 3 and not by 9, and 1001001000000000 is divisible by 3 and not by 9. we now divide by 3 and m again, but this time, m is 1001001/3. basically, every time we divide by 3, we also divide the number of 1's in the number by 3 as well. We start off with 3^2011 1's and we need to divide by 3 2011 times. eventually we get to a point where we have 3^1 ones and we need to divide by 3 once. basically the number has exactly that many factors of 3.
3) 45. When 999...999/89 is a whole number, then the period of the digits is the number of 9's. if you have n 9's and the remainder is r, then the remainder for n+1 9's will be 10r + 9 mod 89. It just so happens that the remainder is 0 when you're on the 45th digit.
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Well, to find the highest common factor of two numbers, you just repeatedly subtract the smaller from the larger. (the Euclidean algorithm). If the HCF of 21n+4 and 14n+3 is one, then the fraction is irreducible.
Define f(a,b) as the HCF of a and b.
f(21n + 4, 14n + 3)
= f(7n + 1 , 14n + 3)
= f(7n + 1 , 7n + 2)
= f(7n + 1 , 1)
and the HCF of any number and 1 is 1, therefore the fraction is irreducible.
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You could wait a long time until most of TASvideos movies have been whittled down by frame wars until improvements are almost inconceivable; take all of the input files back into the present and begin submitting one maybe once a week, and all of the other users will hail you as a prodigy and then scratch their heads wondering what to do next.
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http://dehacked.2y.net/microstorage.php/info/1102156971/Wario%20Land%204%20TAShardgood.vbm
Took me a while, but I managed to fix the 4 problems preventing this from successfully hexing in. I was surprised that such a large amount of time could be hexed in. I think it might be because of how frequently I reset, which minimises the possibility of innacuracies carrying over.
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by (x-y)Pz I mean (x-y)!/(x-y-z)!.
You should know that P(e) = n(e)/n(s) where e is not getting an item. n(s) will then be xPz and n(e) will be (x-y)Pz, so they should actually both be P not C. But of course, 1 - ((x-y)Cz)/(xCz) also works, because the only difference is to add the z! term to both the numerator and denominator.
So yes, you're right.
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This is a fairly simple question. On the first stage you have a 3/8 chance of getting an item. On the second stage, you have a (3/8 + 3/7*5/8) chance of having an item (36/56, or 64%) so by the second stage, you are more likely to have an item than not.
A general solution to this problem is as follows: let's say you have x stages and y items. The probability of having an item by stage z is 1 - ((x-y)Pz)/(xPz). (i.e the compliment of the probability of not getting an item.) Probably can be simplified.
EDIT: Damn, too late.
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Probably not. Most versions of VBA are backwards compatable with older cores of VBA. For example, when they fixed echo RAM, old movies without the fix would still synch, because there was a flag in the movie file to ignore the respective improvement to the core. Most likely, they will do this again, if they improve the core. (basically, it will synch on a newer version, but it will be because it is actually using an older core)
The closest match I could find to the DS LCD is Sharp LQ030B7DD01. You can find a datasheet Here. Aside from RGB, the only LCD signals I needed to use were DCLK1 and GSP1 (everything is conveniently labeled on the DS PCB)
Basically, you will need to tap the signals to the LCD screens and interpret them in order to record video from a DS. To do this you will need some sort of microcontroller setup, and you will have to interface it with a PC. You might ruin a DS in the process. This sort of hack should work with pretty much any portable console.
Otherwise, there might be some software that you can run on the DS that will dump video while playing, but this will interfere with the speed of play, as the DS will need to execute more calculations than it already does.
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I'm really interested to see whether people actually play the games they TAS beforehand, or do they just start TASing straight away.
Personally, I don't. I TAS a game without playing it normally, unless, by coincidence, I've played it before in some distant past. I usually find that if I play the game first, assumptions and expectations from normal gameplay affect the way I TAS the game, which is usually not a good thing. If I haven't played the game before, I tend to be more objective.
However, I always do testruns if I am going to try something new or different, so I know that something is at least possible before I try to optimise it. This eliminates embarrasing moments when I realise I have to redo a large portion of a run, becuase I made a false assumption.
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Sorry to double post, but is there any development for a trace logger in VBA 24? I'm waiting for one for one so I can do SML2 glitched, and see if it's possible to use the pause glitch to complete the game super early.
