By the way, I'm a bit confused. In the userfile page you wrote: "also known as "Pocket Monsters Aka (Japan) (Rev A)" ", but the (Rev A) version is actually a different ROM.
Great job. Why didn't you submitted it?
The first half or so of the TAS is probably the weakest in terms of entertainment; it gave the impression that the TAS was going to be mostly running through empty fields that all look the same. It is only after numerous forced encounters later in the run that the game really shows what you can do. Then the final fight against Fudo Myoou is completely different from everything else I've seen so far, and it was great.
Some things about the game itself caught my eye:
- The text at some point says "Act 2: Hurry to Edo, Musashi Province to recover the Soul of Momohime stolen by Rankai." You mean, the soul of the character you are playing as right now? XD
- So the Ippondatara sequence is really weird. There's a giant foot, then a giant monster trying to hit you with a giant hand, then a giant pig, then a bath scene with the same giant pig. I mean, what?
- Big Oni eats you. The first time it happened, it caught me off guard. It's kind of funny now. Also, that voice.
- Raijin is really annoying. Then again, with a name like "Raijin", I expected an awesome boss, not whatever this boss turned out to be.
Ranking in full detail. (you get it, spoilers)
20) Mega Man Legends (JP title: Rockman DASH Hagane no Bōkenshin (ロックマンDASH 鋼の冒険心)) [Hetfield90's TAS] (mislabeled as being done on the N64 version instead of the PSX version)
Habreno, as I said in my original post, I am aware that I am shouting until I am blue in the face unto the void. That is not going to stop me from trying to correct peoples' use of terminology, particularly on a forum where technical precision is required.
I can't really find an "authoritative" source on this matter, mostly because this is probably a matter of convention, and terminology is often very inconsistent.
The footnote goes on to say: [3] Decreasing an integer beyond its minimum value is often referred to as an integer underflow, although technically this term refers to a floating point condition.
For those who don't want to register at nicovideo: just exchange nicovideo.jp with nicozon.net and it will work. So in this case it would be: http://www.nicozon.net/watch/sm29962607. In my opinion, it should always be nicozon.
>partyboy1a I see. I add link for no nico users to this submission page.
"If the positive whole number M divides the positive whole number N, then all the factors of M are also factors of N."
wasn't there a trick to enter a major dungeon earlier,using the generic ruin entrance nearby?
1 Nxb4#
(take back)
1 ... hxg2#
(take back)
1 Rxa7#
(take back)
1 ... Qxg2#
(take back)
1 Bxb7#
(setup board)
1 ... gxf6
2 Nxg6 fxe5
3 Bexd5 exd4
4 Bxc4 Qxg6
5 Nxg5 Qxf5
6 Bxb3 Qxe4
7 Bxa2 Nxa2
8 Rxg1 Qxd3+
9 Kxd3 dxc3
10 Rxf1 Kxc7
11 Rxe1 Kxc6
12 Rxd1 Nxd1
13 Kxc2 Kxb5
1 3 4 6 8 9 11 12 14 16 17 19 21...
2 5 7 10 13 15 18 20 23 26 28 31 34...
Note that, since α is irrational, the line does not pass through an intersection point of the grid other than (0,0). The gray squares (the ones which contain any part of this line) form a cell path from the bottom-left up and to the right. The kth step of this path is a rightward step if k is of the form n+[n*(α-1)]=[n*α]. By reflecting the diagram about y=x, it follows that the kth step is an upward step if k is of the form n+[n/(α-1)]=[n*(α/(α-1))]=[n*β]. Since there are only upward and rightward steps, [n*α] and [n*β] partition the set {1,2,... }.
(There is a way to allow α to be a rational number, but one of the floor functions has to be replaced with a modified floor function, the smallest integer strictly less than x.)
For example, the image above uses a line of slope φ-1, and the kth step is a rightward step if k is any one of the values an (1,3,4,6,8,9,... ), and an upward step if k is any one of the bn (2,5,7,10,13,15,... ).
Letting α=φ gives (after some calculation) β=α/(α-1)=φ+1. (Note: Solving x+1=x/(x-1), x>1 gives φ as the only solution.) Therefore an=[n*φ], bn=[n*(φ+1)]=n+[n*φ], and bn-an=n.
Note that the rows can also be generated by recursively letting an be the smallest number not yet listed, then bn=an+n. (There is only one way to place all natural numbers exactly once in two ascending rows {an} and {bn} where bn-an=n.) This ties into, for example, the winning positions of Wythoff's game; the winning positions are (0,0), and (an,bn) and (bn,an) for all n.
Similarly for the Zebra/Giraffe, they can move an odd number of squares, so no immediate upper bound emerges, but I have no idea how to show whether their respective tours can be generated.
Riddler Classic First: Just a Knight's tour? It's been known that it can go around the entire board for a long time now. Second: A Camel which moves 3+1, IE an even number of squares, will always land on its own colour, thus placing a boundary of at most 32 squares. Whether it can visit them all is another matter. Similarly for the Zebra/Giraffe, they can move an odd number of squares, so no immediate upper bound emerges, but I have no idea how to show whether their respective tours can be generated.
FractalFusion, the way I understood it, you had to make the seven bets ahead of time, and if you label them from A to G...
This wasn't much of a challenge, but I was disappointed to see no Extra Credit (maybe something like "what if the World Series had an arbitrary odd number of games?"): http://fivethirtyeight.com/features/cubs-world-series-puzzles-for-fun-and-profit/
A 0 1 2 3 4
B
0 100
1 100
2 100
3 100
4 -100 -100 -100 -100
A 0 1 2 3 4
B
0 0 31.25 62.5 87.5 100
1 -31.25 0 37.5 75 100
2 -62.5 -37.5 0 50 100
3 -87.5 -75 -50 0 100
4 -100 -100 -100 -100
A 0 1 2 3 4
B
0 31.25 31.25 25 12.5
1 31.25 37.5 37.5 25
2 25 37.5 50 50
3 12.5 25 50 100
4
Suppose we have a membrane in two dimensions that is both homogeneous and isotropic.