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Cardboard wrote:
Anyway, there seem to be a few shortcuts via hidden rooms, so I'd be grateful if everyone who knows anything about this game wouldn't mind sharing their knowledge. Cheers!
It's one of my favorite games. I was planning to do tas of it, but I didn't complete previous yet. My apologies :(
I was looking into game mechanics few years ago, when I was making my hack.
You hit all secret rooms, that might be useful.
There is only three secret rooms, and one secret level.
All three rooms on mercury, two of them you visit already. One that you left behind, is not useful, because you'll return back where you enter.
I think you should test some of wierd routes, and most of time-gain you can achieve by RNG manipulating ofc :D
Most pain in ass is boss on mercury. I think you should be able to kill him from first "sequence". It goes in following way: you shoot that flying thing, and while it's stay still, he's hitting it because he's dumb asshole :D, and when it starts to fly, it finishing him off. Maybe that's not possible because of cooldowns (inv times).
I know two wierd moves. One in Mercury 2, and one in Nero 2, may be there is more of those. Ofc it would be nice to find out collision data to view it closer. As far as I know it's using same engine as Earthworm Jim 2. There is editor available somewhere nearby.
About moves, I need to make recording, because it's hard to describe by words :).
BTW my fun video:
Link to videoHere is two tricks:Link to video
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Flip wrote:
-Prove that a circle can be surrounded by exactly 6 others of its own radius. IE The above picture might actually be incomplete, there may be indistinguishable gaps in between each one; so prove it to make sure.
Easy to prove. make regular triangle, make new ones on sides of that triange, and so on... then put circles with diameter of side, and center in vertices. Voila!
Flip wrote:
-If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?
r = 1/(1/sin(pi/n)-1)*R
r - smal radius
R - big radius
n - count of small circles
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In any language you can use library for big numbers.
There are several libraries for that, or you may write one yourself.
It's not so hard, I think 1000 symbols is enough, although we don't speak about performance.
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Amaraticando wrote:
256^1000 is the number of possible strings.
Nope, it's number of different C++ programs that is possible. Assume that they all is correct, terminating, and has different output. Then you'll get 256^1000 different outputs.
If you'll allow several numbers in output of one program, then you can write such program:
Language: c
for (int i=0; i<(1<<(1<<(1<<10))); ++i)
print(i);
where you can put 1<< around 1000/4 times, and it's really very big number. I think you can make condition easier to make much bigger last number. something like *((((10)!)!)!)! (factorials) or put it into n and find n-th some rare kind of number, like perfect number, or n-th prime number (where to stop). Any way there is no sense in that big numbers.
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Task about smallest number that you cannot output.
even if you have all possible 256 characters in C++ program, you can reach maximum 256^1000 different possible outputs. And you'll get 256^1000 smallest that you can't output if and only if there is NO GAPS among all different outputs, and all programs is working, and all programs is terminating. So upper bound for answer for this task is 256^1000. and it's somewhere around 10^2408.
Summary:
lower bound is around 10^1806, and upper bound is 10^2408.
Warp wrote:
I don't really understand what you are trying to say.
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I don't see what is point to think about it.
You can output any decimal number from 0 to ~1806 digits.
Just put string in code and decode it from base64 for example.
As far as 10^1806 - is not imaginable by me, I don't see any sense.
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feos wrote:
Guidelines and organizing in the kind of art TASing is don't work as robotic as that. Who wants to come up with some solution here needs to know as much as possible about the problem, then think it over as hard as he can, and then bring it on public, to look for last fixes. This is what the system we're suggesting is based on. Abstract sophisms don't help.
Ok, you call my words robotic, and sophisms... Why? Because of your examples? It does not prove that there is no disagreeing people with your solutions. Also, I don't have any addition to my words about clear definition, so I don't care that you don't accept what I say. I think I'm right, and can't do anything with it while I don't see why I'm not right :).
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feos wrote:
Like analogy much? Alright. Please show me how many people here want "a bit less" spam, or "a bit more".
Genius! You suggest me to prove what I wasn't saying.
Let me explain your way. It looks like this:
We can't do A because we can't define it well. So
if we add "more actual gameplay" -> There will always be people that want "a bit more" gameplay included, and that want "a bit less".
In other words.
If we make A -> There will always be people who disagree.
It's redicilous, because it's true for any A. In other words, you don't need A in this sentence. So it's true anyways: "There will always be people who disagree".
So, if there always disagreeing people, if you use it like test of applicability, then you can't do anything, because "There will always be people who disagree".
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I told already, that interest != entertainment, so I agree with goldenband.
feos wrote:
There will always be people that want "a bit more" gameplay included, and that want "a bit less".
There always will be: spamers, idiots, morons, dissatisfied, disagreeing, critics. And what? Don't do anything because of that people?
This sentence normaly is used in opposite way.
There always will be: spamers, idiots, morons, dissatisfied, disagreeing, critics. So listen them, but do what you think is right. Ignoring them (except spamers) - is bad choice.
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Noone is perfect. I just knew that this TAS already exists, but it took time for me to find it out, where I've seen it. So probably you didn't find it, so I post it.
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Dimon12321 wrote:
I know TAS route without the help of it though it and also mission 3 and final mission shows that my route is the same. Also I'm ~3,5 secs ahead for now!
Who knows what you know except you? What if you didn't see this TAS and your resulting TAS would be slower? There would be laugh and rejection, because bad route/optimization/not used bug.
I just want to be sure that you seen existed work! Note, this TAS isn't in TASVideos database! So who knows, maybe you look into tasvideos submissions only.
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Check this out: http://www.youtube.com/watch?v=fh7X-zAh_tQ and further
btw I love this game, and I complete it on real genesis during 1997+ years when I had only genesis :)
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hegyak wrote:
Mono not working?
It's better solution to make Qt new input editor instead of tug Mono with your application (where you're using code of your editor), and deal with copyrights.
creaothceann wrote:
WST wrote:
C# again? :(
Why not choose Qt, so Linux TASers can use it too?
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feos.
I knew about tasvideos for a long time, but at some point feos sent me link to this thread, and suggest to help with info about game. I still need to complete this run but...
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Also, there is people (or bots?) on twitch who are streaming TAS as they own gameplay. Hilarious! Including normal speedruns of other people. Of course they can't stream if speerun with commentary.
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feos wrote:
r57shell wrote:
And that's why I laugh so much about it, because you want to solve problem with simple deletion of part of system that has issue. It's like cut off hand instead of its treatment.
I'll remind my personal reason to participate in this discussion: I don't want to feel "uneasy" or "excusing" for accepting to "wrong" tiers.
Captain obvious comming! :D
It's topic about "Vault Tier", and main question is: "Should we keep the Vault as it is?"
You're talking about borderline cases. It's offtopic :D. So I can simple ignore you. lol
But I'll reply. And forgive me again. I'm repeating again....
Instead of cutting off hand, I offer you treatment: remove borderlines at all.
Choose some X, and Y values, and automatically set icons in following way:
nothing < X <= moons < Y <= stars
according to raiting.
If in rules stated: moons if rating >= X, then noone can dispute that rating is < X. Rating of TAS is some Z, it can be >= X or < X nothing else. You can't argue.
Your main argument will be: there would be always flames about right selection of X, and Y. Everyone can flood about it. But honestly, who cares about exact values of X and Y? It can be selected approximately.
Why? Because X is absolute, and Y is absolute.
If your TAS in < X list: all in this category with rating < X. You can say that you best among them and what's a point? But what if you telling us that you must be in >= X list? You simply wrong, because ALL of those TAS with rating >= X is actually has rating >= X but your TAS is not >= so you can't stay among them. Same with borderline from moons to stars. TAS with < Y it's obviously lower by rating of any rating from >= Y list (stars).
I actually don't even care about not perfect rating votes system, it's still can be measured. Let say X = 0.7 (moons border) and Y=0.9 (stars border). Then moons will show that raiting >= 0.7, and stars will show that rating >=0.9. And anyone who tells you that his movie is moons, and TAS rating is < 0.7, he is definitely wrong, because all TAS in moons has rating >= 7.
Also, if you so hate this kind of X, Y, you can make icon "TOP" and with number from right side. Again, noone can say that it's higher in top, than it's showed. Also, you have to choose again from what position to show this icon. For example: show TOP icon for TOP 100 only. And again, there will be people who start flaming about it.
I hope again, that this is my last post in this topic.
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pff you don't need to proove that complicated thing.
It's much easier.
You must know well known property:
(a*b) mod c = (a mod c)*(b mod c)
proof:
a = u*c + (a mod c) (select right u)
b = v*c + (b mod c) (select right v)
then
(a*b) = (u*c + (a mod c))*(v*c + (b mod c))
rearrange
a*b = u*v*c*c + ((a mod c)+(a mod c))*v*c + (a mod c)*(a mod c)
remove all what is multiplied by c and you'll get (a*b) mod c.
so you can always use (a mod c) instead of a.
a^x - 1 = 0 (mod b)
a^x = 1 (mod b)
(a mod b)^x = 1 (mod b)
your question is: what b I need to pick, so it would work for any x?
It must work for x = 1, so you need to pick any such b:
(a mod b) = 1 (mod b)
in other words
a = 1 (mod b)
also if it's so, then
(a^x) = 1 (mod b)
because
((a mod b)^x) = (1^x) = 1 (mod b)
how many of such b?
a = 1 (mod b)
a - 1 = 0 (mod b)
in other words, a - 1 must be devisible by b.
a^x - 1 divisible by b for any x if and only if (a-1) divisible by b.
So, factorize (a-1) and make all combinations of factors: you'll get all such b.