That was very interesting to read, thanks for summing things up for us. I do not fully understand what it is you are planning, but at the same time I realize that it's a personal issue.
I have a follow-up question regarding your choise of direction in life: If you were economically independent (for example, you somehow won 100 million dollars), what would you do with your time? Would you keep your current job, or start your own business or something? Would you move to some other country, or would you stay in Finland?
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On the plus side:
*I personally like this game, and find this run enjoyable to watch. So in my opinion, it would make a nice addition to the site.
On the other hand:
*It can be improved on some levels. I intend to finish the run I'm working on sometime this summer, implementing those improvements. That doesn't have to mean that this submission can't be published though, just so the encoders and publishers are aware that you'll probably see another Fire n' Ice submission soon.
*It's made with the European version of the game. The E and the U versions are very similiar, so I think this submission is in violation with this rule.
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Warp wrote:
http://www.youtube.com/watch?v=naVshTpD9l0
How can a game made in 2006 be so badly implemented? I mean, really. It's just horrible! Did the authors even play their own game? This is completely beyond comprehension.
Ugh, that really is horrible... It kinda reminds of when I bought Final Fantasy 6 for PSX. I was really annoyed that the loading times were much longer compared to the SNES version. Just entering the menu took 2-3 seconds or so... Those horrible memories...
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Gorash is indeed right - the easiest way is to move the vector c2 so that it intersects the point p1. Now you have a plane that is spanned by the vectors c1 and c2, and that includes the point p1. By moving the vector c1 so that it intersects p2 we get another plane, and the shortest distance between the lines is the distance between the two planes.
After some calculation, we find the distance, d, to be
d = |w*(c1-c2)| where w=(c1xc2)/|(c1xc2)| (the vector product of c1 and c2, with length 1)
I don't know what andymac is up to exactly, but it looks pretty cool. :P I don't know if that method works, but it would be exciting if it did.
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How well do you remember your linear algebra? Given two lines L1: p1+c1*s and L2: p2+c2*t, where p1,p2,c1 and c2 are 3-dimensional vectors and s and t are real numbers, find the shortest distance between L1 and L2.
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"floor" and "mod" aren't as complicated as they may sound. :)
*mod means the remainder after division. For example, 15mod6=3 because 15 gives the remainder 3 after division with 6.
*floor means that you round down to the nearest integer. So floor(1.2324)=1, and floor(7.8123)=7.
Instead of using these formulas, we can reason like this:
Let's take your example in this case, a movie of length 238465 frames. To find hours, we'll do as you did, and divide by 60^3, which gives us (as you've already said) 1,1040046296296 hours. To translate what 0,1040046296296 hours is in minutes, we multiply this number by 60. We now get 60*0,1040046296296 = 6,240277777776 minutes. So 6 minutes. To find out how many seconds 0,240277777776 minutes is, we multiply this number by 60 and get 60*0,240277777776 = 14,4166... seconds. So the movie length is 1 hour, 6 minutes and 14.4 seconds.
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nfq wrote:
One thing I've wondered a long time is how to for example calculate seconds to minutes or hours or frames to minutes or hours. For example if there's a TAS that is 238465 frames long, how would i calculate how long that is in hours? By dividing it thrice with 60, I get 1,1040046296296etc but how am I supposed to know how many minutes and seconds "0.1040046296296etc" hours is?
Let's say that a TAS is x frames long. To calculate its length in hours (h), minutes (m) and seconds (s), do this:
h = floor(x/216000)
m = floor((x mod 216000)/3600)
s =(x mod 3600)/60
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Thanks Aglar, this game is very edit-friendly, so there shouldn't be any problems editing it in. :)
EDIT: I edited this in now, without any problems.
EDIT 2: I have now finished world 6. Only level 6-10 have a slightly changed strategy compared to Ragowit's TAS. Here's the WIP.
EDIT 3: I have now finished world 7. There are quite a lot of new strategies on this world - levels 1,6,7,9 and 10 all have new strategies. The biggest improvements are levels 9 and 10 where I saved 284 and 114 frames respectively. Levels 1,6 and 7 all have minor improvements. Can you spot them? Here's the WIP.
EDIT 4: World 8 finsihed! Most of the levels on this world are extremely simple, their purpose is mostly to introduce the new black block and its properties. This is why I was able to complete this world to quickly. Because enemies move at a different speed in the U version compared to the E version, I have to use a slightly different strategy on 8-10. I'm not completely satisfied with how that level looks now, but I'm not so sure I can save any time over the solution i have. I tried several different strategies, but this one turned out to be the fastest. No other level in this world uses a different strategy. Here's the new WIP.
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I've completed world 5 now. There are new strategies (some minor, and some bigger) on levels 3,5,6 and 10, so quite a lot of new stuff on this world.
Overall, this world went smoothly. Here's the WIP.
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Warp wrote:
The classical proof that there are infinitely many primes is a proof by contradiction: Let's assume that there's a largest prime. If we multiply all the primes up to that largest prime and add 1, we get a number which is not divisible by any of the primes, and thus the assumption we made is false: There was a prime which is larger than the one we assumed was the largest.
However, consider this: 2*3*5*7*11*13 + 1 = 30031, which is a composite number.
Isn't this a contradiction to the proof? It clearly doesn't hold that the product of the first n primes plus 1 is a prime.
How to explain this apparent contradiction?
I might as well try this too. :)
Let's say that you've found all the primes from 1 to n. Construct the product of them and add 1 (call this number k). Assume that there is no prime larger than n (we want to get a contradiction with this assumption). There are two possibilities:
1: k is prime: this completes the proof, because we found a prime strictly larger than n, giving us a contradiction under our assumption.
2. k is composite (as in your question): If this is the case, then k can be factorized into primes. But k can not be divisible by any prime from 2 to n, so the smallest prime factor of k (let's call it v) satisfies v>n, which also gives us a contradiction, because now we've found a prime v strictly larger than n.
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Warp wrote:
Since this thread has been silent for a long time, here's a small problem to warm you up a bit again:
Prove that the factorization of any natural number is unique (except for a possible reordering of the factors).
I assume you mean prime factorization, otherwise it's not unique, because 12=3*4 and 12=2*2*3.
I'm gonna give this a try: Let's call the number x. Assume that x has two different prime factorizations, namely:
x = a(1)*a(2)*...*a(k) (1)
x = b(1)*b(2)*...*b(n) (2)
Now, because both of these a representations of x, we can set:
a(1)*a(2)*...*a(k) = b(1)*b(2)*...*b(n)
Because a(1) is a factor of x according to (1), we can divide by a(1), while still having having integers on both sides. a(1) disappears from the LHS, and because the LHS is an integer, the RHS has to have a(1) as a factor too, otherwise it wouldn't be an integer, giving us a contradiction. Lets call the factor in the RHS that was equal to a(1) for b(i).
a(2)*...*a(k) = b(1)*b(2)*...*b(i-1)*b(i+1)*...*b(n)
Because a(2) is a factor of the LHS, we can divide by a(2) while still having an integer in the LHS. Because the LHS is an integer, a(2) has to be a factor of the RHS. Continuing this way, after having divided with all the factors of the LHS, we have divided with all the factors of the right hand side too (because we are left with 1 in the RHS), showing that the prime factors in the RHS of eq. (1) and (2) are in fact the same.
EDIT: Here's a problem for you: Let's assume you have a sequence of numbers satisfying a(k+1) = (5*a(k)+5)mod256, with a(0)=0. As it turns out, if you look at all the numbers a(0), a(1), a(2), ..., a(255), they take every value between 0 and 255 exactly once. Why is this? If you have the sequence a(k+1) = (c*a(k) + b)mod256, a(0)=d, for what values of b,c and d does this property hold? (I don't know the answer to this, but I'd really like to know)
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Bump! I picked up this TAS again this morning and finished 4-10 and 5-1. I used the same strategy as Ragowit on these levels, but with less waiting times on level 4-10 (due to differences between the U and the E version of the game). 5-10 looks the same too, it's a very easy level.
Here's the WIP.
I'll be pretty busy with exams these upcoming weeks, but hopefully I can post a WIP here now and then.
Oh, and if anyone's interested, please give 4-10 a try, I have a hunch that it can be improved. I'll look into it some more myself, I think.
Yeah, that one is pretty much put on hold for me indefinitely. I didn't really intend to finish it when I started, I just wanted to get something started so that someone more focused and determined than me could perhaps do a complete run eventually. Though there didn't seem to be a lot of interest in that run. Anyway, although Ragowit's run is very good technicaly, it can be improved, so I personally think someone should start doing a new Fire n' Ice TAS. ;) (I personally don't have the time right now)
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I'm wondering if there's a way to adjust the position of a comment I've added to a movie-file, for example, if I want a certain comment to appear in the top right corner instead of the usual top left corner. If not, this might be a nice feature.
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That was really cool, nice job! Now I want to do something similiar myself, creating a lua-script for the sole sake of making a game more enjoyable to watch. Perhaps we should even create a thread in the forums dedicated to these kind of things. Oh, the possibilities!
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Nice run indeed, I'm fascinated by how much the SMB 3 TAS is improved over and over again. I can only agree with the others that this was an awesome run, so here's another strong yes vote.
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Thanks for the encode, nineko, I appreciate it.
This game really makes for a great TAS, and like moozooh said, it seems like there's a lot of room for optimizations. Hopefully this won't be the last TAS we see of this game. The run kept me very entertained the whole way through.
Big yes vote from me!