However, there are certain games with unlockable modes, second quests, or other things of interest that can only be accessed if a save file (or an otherwise "dirty" SRAM) is present. If you really wish to submit a movie made on such a mode, you will need a verification movie made and provided alongside it.
[...]
In any case, ensure a run of your unlocked mode provides meaningful content over a fresh game.
I don't know the game but it apparently "adds a ton of bonus levels" and makes the game more interesting. If you think that this is enough meaningful content over a fresh game then go ahead and do it!
(hint: entertaining movies are usually received very well)
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Alyosha wrote:
If you write RNG1=A7,...,A0 and RNG2=B7,...,B0 then after 1 of the 11 cycles you get
RNG2=B6,...,B0,A7
RNG1=A6,...,A0,NOT(XOR(XOR(A0,B7),A1)) (if I did all my math right)
But simplifying the loop, I think not, since after 11 cycles, RNG2 will look like:
A3,A2,A1,A0,NOT(XOR(XOR(A0,B7),A1),.... and RNG1 will be a mess.
Not bad.
I decided I'd try it on my own and managed to get to this final result. The way you read this is, the next two bytes of rng are in the form abcdefgh ijklmnop and let's say you want to figure out what value n will have. You look at the column for n and see that you have to XOR 1, g, h and i (from the given rng) together to get the value for n. There, it's easy to see why 0x00 0x00 will have 0x92 0x04 as the next value.
I agree that it is unlikely to be able reduce it even further.
Too bad.
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ASL: left shift where the new LSB (least significant bit) becomes 0 and the old MSB (most significant bit) goes into carry.
LSR: right shift where the new MSB becomes 0 and the old LSB goes into carry.
ROL: left shift where the new LSB becomes the old carry and the old MSB goes into carry.
ROR: right shift where the new MSB becomes the old carry and the old LSB goes into carry.
I think XOR and AND are pretty basic.
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Ever since I wrote this script for predicting RNG in this run I wondered if the formula could be reduced.
Currently you call the function nRNG (line 44) with two bytes to let the function advance them one step further. As you can see, the bytes run through a bunch of binary operations like XOR and ROL which loops 11 times.
As an example, calling the function nRNG with bytes 0x00 and 0x00 will give you 0x92 and 0x04. With only two bytes there are only a limited number of solutions (0x10000 or simply 65536).
After some testing I found out that when you start at, for example, 0x00 0x00, you will go through every possibility once until you get back to 0x00 0x00... except for the combination 0x55 0x55 and 0xAA 0xAA. These are not occurring which makes sense because calling the nRNG function with 0x55 0x55 will give you 0xAA 0xAA and vice versa.
So my question is: is it possible to reduce the formula or function to a simpler one?
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Derakon wrote:
Does exposure (author of submission 4667) deserve co-author status on this movie? It seems a little cruel to have someone submit a movie, then go "Hey, nice submission, I see what you did there but I can do it better, no publication for you."
It's Spikestuff, get used to it.
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Orange Claw Hammer wrote:
why do the acquisition of the orb item and the cloud item differ sufficiently to warrant them being separated on a categorical level?
Because getting a goal orb doesn't execute open bus, it always does the same.
Getting a cloud executes open bus and thus has the potential to crash the game, finish the game or miraculously return to the game, depending on the RAM state.
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If so, then I have to tell you that it is not possible.
A .fm2 file is a movie file that contains the input which the emulator uses to play back your movie with the correct button presses. Without an emulator you just have a bunch of buttons flying around in a file without any meaning whatsoever.
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NordicAnomaly wrote:
Btw, I just remembered a question that my teacher was unable to answer a few years back... maybe someone here wants to give it a shot? Here it is:
"How many digits does the number 200^2010 have?"
At this point I hadn't been introduced to logarithms, so it should be possible to solve without them.
Without logarithms, huh? All I can come up with is a good approximation :D
200^2010
= (100*2)^2010
= 100^2010 * 2^2010
= 100^2010 * (2^10)^201
since 2^10=1024 is around 1000 we approximate
≈ 100^2010 * 1000^201
= 10^4020 * 10^603
= 10^4623
which has 4624 digits, which is close to the correct result 4626 (2010*log(200) rounded down and +1).
(Other approximations could be 2^103≈10^31, 2^196≈10^59, 2^681≈10^205, 2^1166≈10^351, etc. but they are not useful here)
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The_Trust wrote:
Thank you Masterjun for not watching my video because I use internet explorer.
Using Internet Explorer often implies a rather limited knowledge about the usage of software in any kind.
Your so called "new users" are just new to TASing, however, they might not be new to other areas. I know a lot of people which think that the people that use Internet Explorer only use it, because they don't know better (which is usually the case).
And that is a bad attitude to have towards a guy trying to teach you stuff.
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Once I see a person use Internet Explorer in a video, I stop watching immediately.
To be worth a sticky thread, the video has to be much better than this.
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Or I just don't understand the problem at hand...
I mean you can even show signed values in the search list in the settings...
