The sad thing about this submission is that, if FractalFusion had just used the code execution to finish the game rather than promising a payload that didn't really materialise, then its reception would probably have been quite good.
^I'll be honest and admit that I somewhat share Glitcher's sentiments. This TAS seems like a victim of circumstance:
1) Coming out just a few days after another movie that is pretty much identical for 90%+ of its duration.
2) Not being nearly as impressive as its predecessor, because that movie set the bar extremely high.
3) Having to wait half an hour for a punchline which everyone who watched the first movie will know in advance.
4) And then having a worse version of that punchline.
I'd have loved for this just to be a regular TAS and for you and MrWint to take it to a frame war.
On a positive note, I'm glad you worked out the Coin Case glitch properly and I knew you would make this TAS sooner or later. I'll definitely be sticking this one in Gruefood should it be rejected.
Right, should this obsolete Mr Wint's run or not? If you consider enabling arbitrary code execution to be akin to beating the game (because it allows you to do anything), then it does so faster and should obsolete it. If you consider the game not to be beaten at all because the movie only uses code execution to muck around, then publish them separately.
The closest thing to a precedent we have for this situation is Masterjun's "Super Mario World plays SnakePong" TAS, where they were published alongside one another, but there is a difference in that case: the two movies were identical up to the point of arbitrary code execution, rather than the playaround being thirty seconds faster than the other.
Oh, I'm looking forward to an encode by the way.
Oh derp, I totally forgot SMB doesn't have a save function,so when you turn the console off presumably it doesn't remember you beat it before.
I'm having a lot of online brainmelts recently.
andrewg, why are you asking this question of all people? You are the real-time record holder for Super Mario Bros. Haven't you beaten that game 256 times yourself?
Has anyone considered a more open-ended title like, say, Civilization? Play that in democracy and it would really test the players' cohesion, as co-operatively forming a plan would be much more important, and the planning would go deeper than in an RPG.
So I found this boring because I couldn't follow what was going on at all, not helped by the fact that it was all in Japanese, nor by the fact that I feel I've seen this done in other Wizardry TASes (even if it's a different trick, it looks similar to me). Voted no. I can see I am in a minority opinion here, but for the people who voted yes I would ask them: how is this so much more entertaining than, say, the current Chrono Trigger run?
Obviously it's a great technical achievement but you could have known that without watching it just by looking at the name of the TASer.
To say that there are a lot of bugs in this game is to discredit the enormous achievement that it is to tear apart a game that had been considered very robust for over twenty years of real-time playing, and for ten years of dedicated searching with emulators and other tools designed to make finding and replicating bugs easier.
ah, you were talking about the longest subsequence of that form, not just about full sequences. Sorry I missed that. But to be fair, your description was a bit terse. ;)
Here's your star, then!
STAR GET!
As for the purpose in this thread, I don't think it's necessary to know the solution to a problem in order to ask it; however I do agree that one should at least think about the problem, if only because it might turn out to be trivially easy and not worth asking.
That does not mean that, when Warp posted some clearly incorrect reasoning, he had not thought about it; on the contrary, the fact that he had posted at all shows that he had. In fact in mathematics you can often achieve contradictory results through different lines of reasoning, in which case it is instructive to find the flaw in one of the proofs. In this case, Warp has just forgotten that, when modifying an existing sum of unique Fibonacci numbers, it is not necessarily okay to make the substitution F(n) -> F(n-2) + F(n-1) because F(n-2) or F(n-1) may already appear in the sum.
Yes, I have proved these sequences you are complaining about are never generated by the greedy algorithm: F(2n-1) + F(2n - 3) + ... + F(1) = F(2n), which can be proved recursively. I did actually note this exception.
Incidentally, not only does the Fibonacci subset that Warp mentioned fail to provide a partial sum for every integer, in fact the fraction of integers that can be represented from its partial sums is almost zero, in the technical sense that if you take the numbers high enough the fraction can become as close to zero as you like. The not-every-third-Fibonacci numbers grow at a rate of (golden ratio)^3/2 = 2.058 per term, whereas the number of possible partial sums only doubles every extra term. So eventually the size of the terms in the sequence vastly outstrips the number of possible partial sums, even if every partial sum gave a different integer.
The second part of the proof shouldn't be hard considering that we have already demonstrated that all integers are the sum of non-consecutive Fibonacci numbers.
To construct the sum of k as a sum of Fibonacci numbers disregarding F(n):
Express k as a sum of non-consecutive Fibonacci numbers. [This summation is unique with one exception: F(2k) = F(2k-1) + F(2k-3) + ... + F(1) - in this case take the first expression.] If this set does not include F(n) then we are done.
If it DOES contain F(n), then replace with F(n-2) + F(n-1). If the original sum did not contain F(n-2), then we are done. Note the sum can never contain F(n-1), by definition.
If the sequence does now contain a duplicated F(n-2), we just repeat the process again, splitting F(n-2) into F(n-4) + F(n-3), (noting that we may duplicate F(n-4), but never F(n-3)) and we can repeat this process arbitrarily many times until n reaches 2 or 1. n can never reach one because we have already prohibited precisely the sums that would not terminate before they reached n=1, two paragraphs up. If n reaches 2, we can just "split" F(2) as follows: F(2) = F(1) (=1), and the sequence terminates.
Thus we can construct an integer out of a partial sum of the set of Fibonacci numbers, minus any one element.
Can I have a gold star please?
Got it! If we remove F(n), then we must express F(n+1) - 1 as F(1) + F(2) + F(3) + ... F(n-1). Therefore we cannot maintain the property by removing F(n) and F(m) if m < n, because we have no way of expressing F(n+1) - 1. If n < m then similarly we cannot express F(m+1) - 1. Neat.
Well, there is a very trivial upper limit to the fraction of the Fibonacci numbers you can remove, because we know that the set containing the powers of 2 is the sparsest set with the property. Hence any series which grows faster than powers of 2 cannot fulfil the criterion. Since the Fibonacci numbers and the powers of 2 both grow exponentially (the Fibaonacci numbers can be defined as powers of the golden ratio, rounded to the nearest integer). Hence any set that takes away more than 1 - log(phi)/log(2) of the numbers grows too fast to form unique sums for every integer. This upper limit on the proportion of removable integers turns out to be around 0.306.
(As an aside, the way you can prove that a sequence growing faster than powers of two cannot express every integer as a sum of elements, consider that there are 2^n possible sums of the first n terms in the sequence. If your nth term is greater than 2^n you can't express every number smaller than that as the sum of previous terms because there simply aren't enough distinct possible sums to do the job.)
My little brother recently bought 2 DSes for this very reason. I told him that for the same price he could go halfsies on a Wii U with me but he didn't listen...
While we're on Fibonacci numbers, I remember my old maths teacher was very proud of being the first person to prove the following:
The only Fibonacci numbers which are also twin primes are 3, 5 and 13.
(A twin prime is a prime number that differs from another prime number by 2.)
He never showed us the proof, but it must have been relatively elementary for a proof discovered in the 80's because he never even did a master's. If anyone can find a proof, I'd be really happy indeed.
Also I guess the fact that elementary proofs keep popping up to solve age-old problems is hope for amateur mathematicians everywhere.