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I think I see what you're talking about adelikat, but it would be nice if you cold post a .fcm so I don't misunderstand anything.
I changed tactics in the seconds cave, this time I don't get hit by the wolfman plus I kill two enemies on the way out. This cost me a total of 30 frames, which isn't that much, and on the upside, I now have 20 exp. and a full life bar. I'll submit an updated WIP later today.
Right now, I am 30 seconds and 10 exp ahead of Arc's run.
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Could you explain the jump + fairy spell glitch please? I'm not familiar with that...
Here are some problems I've run into so far:
1. I'm experiencing some lag in the first cave (around frame 1400). This probably can't be avoided, but it still bugs me.
2. I have to wait for the black blob to pass me on the world map (around frame 2320). This costs me around 15 frames. So far I know how to manipulate luck to avoid this, however, that luck manipulation costs me more than 15 frames to perform, therefor I wait those 15 frames. I'd really like to know how to manipulate luck in this game without it costing me any time.
3. I get hit by the "wolfman" (or whatever he is) in the second cave when I'm holding up the trophee (around frame 3300). I don't see how this can be avoided though.
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Hehe, so if 4*(4↑↑↑↑↑↑↑↑↑↑↑44) is that huge, I'm guessing that 4^(4↑↑↑↑↑↑↑↑↑↑↑44) is so huge that our brains would explode if we found out the magnitude of that number. :)
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The first part of this is quite easy to do, so here's the updated WIP
I'm having some trouble manipulating luck in this game, such as what kind of enemies appear on the world map. Perhaps someone could help me with this?
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I know that there is a simple formula for calculating the average of 1dS with a damage reduction of R, I could post it here if you'd like, I have written it down on some paper that should be somewhere in my room i think.
Thanks for taking the time to solve this, xebra. I think I'll let someone who's good at programming write a program for this though :)
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Yeah, I'm getting kind of tired of watching runs with loads of glitches, this run is meant to be a "standard" Zelda 2 run. I don't think a full run of Zelda 2 using the left+right glitch would be interesting to watch anyway, since the viewer would have a hard folowing Link.
And yes, I agree that it would be nice to have a Zelda 2 run in .fcm format instead of .fmv. Plus, the .avi-quality will be a little better too! :)
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xebra: Okay, sorry for misunderstanding the hyperpower function. And I am aware of the fact that a number divided by zero is not actually a defined number, I mostly took that example up to make sure i'd win if you accepted that as a number :)
Maybe 4*(4↑↑↑↑↑↑↑↑↑↑↑44) is higher than 4 ^ ( 4 ^ ( 4 ^ ( 4!!!!!!!!! ))), I don't know.
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I searhed the forum looking for a thread named "Zelda II - The adventure of Link" and forgot the japanese name. Sorry about starting a new topic for the same game...
However, since this is a run of the english version, which name actually is "Zelda II - The adventure of Link", perhaps this thread-name fits better.
If you'd like, Bisqwit, I could of course continue this discussion on the thread "Zelda no Densetsu 2: Link no Bouken" if you think that would be better.
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I'm trying out a Zelda II TAS at the moment, and so far it's going really good. A lot of the time I shaved off comes from the fact that there is no loading points in the NES version compared to the Famicom version. Anyhow, here is the first WIP, if you're interested I could keep working on this to make it a full run.
Oh, and as you'll notice when watching the movie, I don't use the left+right trick for this run, since I'd like to make this run look as "real" as possible. Although if you'd like I could redo it all and use the left+right trick.
Here's the WIP
Emulator used: FCEU Ultra 0.98.12-blip
ROM name: Zelda II - The Adventure of Link (U).nes
Please comment if you find anything that could be improved or changed.
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Hey, just watched your movie, and I can say that you've done a great job. This movie gets an obvious yes-vote from me. I do have some comments on the movie though:
You wobble (turn left, right, left right really fast) a little too much in my opinion. This is not really that annoying, it was just a detail that I thought of.
Secondly, the autoscrolling part on the last stage was a little boring. It would have been nicer if you had finished off those purple ninjas instead of running past them.
Also, there seem to be a lot of lag on the first part of stage 4_2, maybe this could be avoided by killing the enemies that are currently on the screen, if this doesn't waste too much time.
Other than that, it was really fun watching a ninja that is high on coffein (judging by the amazingly fast way he moved) kicking some asses.
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Here's a bigger one:
4^(4^(4^[4↑]!!!!!!!))
Or, if division with 0 is allowed: (4+4)/(4-4)
That number goes towards infinity, IF division with 0 is allowed here. If it is, than there is no number that could possibly be bigger than this.
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Gigafrost wrote:
I'm not sure I can do more than post the code to my program, though. It'd be the code that simply brute-forces the solution (in other words, it'll be fine as long as N doesn't get very big.)
It'll be just fine if you'd post your programcode, preferebly in C++ code, that's the language I'm most familiar with. And I would really appreciate it if you came up with a simple mathematic formula, that would be awesome.
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Gorash wrote:
But the greatest value will probably be:
4 ^ ( 4 ^ ( 4 ^ ( 4!!!!!!!!! )))
That number is... huge.
Yup, that number is really, really huge. I thought about the exact same thing when I wrote my previous post, that the highest value would probably be when you use all of your ! in your highest exponent. But perhaps there is a number that is higher than 4 ^ ( 4 ^ ( 4 ^ ( 4!!!!!!!!! ))), but since that number i amazingly huge, I'm guessing you need tremendous computer power to compare number of that size.
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Let's say that you are allowed to use 12 operators at most, and you still have to use your four fours.
An operator is something that changes the value, such as +, - or !. You may still use () and [] freely.
If it's okay with you, you can use the operators that xebra talked about in his post, such as hyperpower. See his post for more info on this.
This is just a suggestion for the rules for this, if you don't like something about them it's okay if you change them.
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Just watched your twitchy lolo run. I don't really know much about this game, but I didn't find any mistakes or anything, it looked really good. Keep up the good work Bag of Magic Food!
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itstoearly wrote:
shouldnt it just be (S+1-R)/2?
No, the (S+1)/2 is the average of 1dS and the R is the damage reduction, and the equation means "average damage" - "damage reduction", which is (S+1)/2 - R.
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Here's an idea: Why not see who can find out the highest integer that be be represented by 4 fours, using programs. Although that will turn all this to who can code the best program, I, for one, would find it quite interesting. What do you think?
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Gigafrost wrote:
But, you wanted a program and not just a formula, right? Can you compile/run Java programs?
Yes, I'd like a program that could calculate this, be it a Java- or C++ program or whatever, as long as it can do it. And no, I can't compile Java programs. The only programming skills I have are basic C++ knowledge, nothing advanced. But if I got an easy equation I could probably compile a program that does the calculations.
Oh, and sorry about misunderstanding your previous post, Gigafrost. By the way, have you played a lot of D&D before? Because normally when I talk to people about this they're mostly "what the crap is going on in that guy's head???" you know. Well, I'm just glad you know what I'm talking about. :)
And just for fun, a friend of mine has done a lot of dice-calculations for roleplaying games such as D&D, like what the average is of 4d6 if you take away the dice with the lowest outcome. If anyone's interested I could post these here sometime.
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Actually, the formula N*(S+1)/2-R isn't correct. The reason for this is that you have to subtract R from every possible outcome (the result of the outcome) of NdS. To try to make things clearer, let's calculate the average of 1d8 with a damage reduction of 5:
The average of 1d8 is: (1+2+3+4+5+6+7+8)/8 = 4,5
The sum 1+2+3+4+5+6+7+8 represent all the possible outcomes (the results of the outcomes) of 1d8. With a damage reduction of 5 we will have to subtract 5 from each and every term. But since you can't inflict negative damage, all negative terms are changed to 0. This gives us:
(0+0+0+0+0+1+2+3)/8 = 6/8 = 0,75
As you can see, every term has been subtracted from every term because the damage reduction affect every possible outcome. There is still 8 possible outcomes, that's why we still divide the sum with 8.
With NdS you would have to add together every possible outcome, and then divide it with the number of possible outcomes.
I know this is quite confusing, sometimes I even confuse myself with this, but I hope you understand what I mean.
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Yaaaaay! :)
Thanks a bunch flagitious, this problem has been bugging me for quite a while. And the equation was a lot simpler than I had thought, and that's good. Even I could code a program now that calculates this now.
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Yeah, here I calculated with pen and paper a lot of these equations, and here they go coding programs that do that stuff for them! Mathematic judges you! :)
Just kidding, but seriously, I agree with you, Mr. Kelly R. Flewin, that this is quite sad... I was prepared to go to at least 500 with this, but hey, making programs that does this for you is also kind of cool, although the topic should change name to perhaps The Four Fours challenge (programming) or something.
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Dacicus wrote:
Randil wrote:
The average is NOT N*(S+1/2) - R.
Isn't the second parenthesis supposed to be before the division?
Yes, you are right, it should be "The average is NOT N*(S+1)/2 - R." Thanks for pointing that out.
Dacicus wrote:
Randil wrote:
What you must do is subtract R from every possible outcome of N*(S+1/2)
I don't quite understand this. Do you mean to subtract R from every possible average of NdS (what you wrote, I think) or from every possible result of NdS? I'm inclined to think that you meant every possible result, but please correct me if I'm wrong. In any case, there are S^N possible results when you roll N dice with S sides, so that could get quite complicated to program. Well, maybe not with loops.
What I meant was every possible result of NdS. For example, here is the average of 1d6:
(1+2+3+4+5+6)/6 = 21/6 = 3,5
And here is the average of 1d6 with a damage reduction of 4:
(0+0+0+0+1+2)/6 = 3/6 = 0,5
You subtract the damage reduction (in this case 4) from every possible result of 1d6. That's why 1,2,3 and 4 deals 0 damage, as you can see in the equation. Only when you get 5 or 6 on the dice you will deal damage, if you roll 5 the damage will be 1 (5-4), and if you roll 6 it will be 2 (6-4). The same thing goes for NdS, you must subtract the damage reduction from every possible result of the outcome and divide ot with the possible number of outcomes. I hope this cleared things up a little.
And it would be really awesome if someone could come up with a program that calculates this stuff, but don't feel any pressure on you, it's not like I force you to do it or anything.
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There is a math problem that has been troubling me for awhile now... Here's the thing. Those of you who have played roleplaying games such as Dungeons and Dragons will recognize a lot of terms in this thread:
1 dice with 6 sides is hereby called 1d6, just as one dice with 10 sides is hereby called 1d10. So one dice with S sides is called 1dS.
If you roll N dice with S sides each, this is hereby called ANdS. Example:
If you roll 7 dice with 12 sides each (N = 7, S = 12) this is called 7d12.
The average of 1dS is not that hard to calculate, it's just (S+1)/2. The average for 1d6 is (6+1/)2 = 3,5. The average of NdS is not very hard to calculate either, it's just N*(S+1)/2, the average of 5d4 is 5*(4+1)/2.
Here is the problem: Let's say that on a roleplaying game, like Neverwinter Nights, or whatever game that is dice-based, you attack an enemy with a weapon that does a total of NdA damage. The average damage would be N*(S+1)/2. However, this enemy has a certain damage reduction, R. My question is, what is the average of the outcome when you attack an enemy, who has damage reduction R, with a weapon that does NdS damage?
The average is NOT N*(S+1/2) - R. What you must do is subtract R from every possible outcome of N*(S+1/2) and divide it with the number of possible outcomes. If an outcome becomes negative then change that outcome to 0, since you can't possibly do negative damage on an enemy.
If someone could perhaps code a program that calculates the average of N*(S+1/2) - R, where you can input values for N, S and R, that would be great.
I hope that you understood what I talked about here, just ask me if you want me to be clearer on something.