Posts for flagitious


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Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Should the subscreen code B S S B B S S B B S S S B S B B B S S S B S B S S B S S B B S B S be allowed? It would save some time if Exim's route was used because he needs to go out of his way to get an extra potion. This game appears to accept input only once every other frame. Here are some memory locations: y = 002D x = 0031 z = 0035
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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Bag of Magic Food wrote:
Aw, now you're going to make me feel bad for recommending it to Bisqwit in the first place.
Yeah, I recommended it too a while back in an off topic topic, sorry Bisqwit :( BTW even it was some of the things that you guys mentioned, the fact still stands that they wouldn't even read his explanation, or provide details as to why
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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Thx for the map, don't know how google/me missed that one, grr. I find it amusing that the jpeg map is 2.4MB but the original game is only 128KB.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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I'm surprised to not find a topic for this game already, except the published movie page for Exim's run. There are not any maps available on the web for this game except some crude ascii ones. This game looks pretty interesting and has a great theme so I'm interested. Also of interest is an impressive 100% completion run by Lucid Faia. First order of business is to rip some maps from the game. I've never done any reverse engineering of video games or anything for that matter so it looks like a good educational experience for me. Anyone have any advice on how to rip maps from the game? I might also like to disassemble and comment some of the game if possible later. Which version of the rom should be used? Exim uses the Europe version, but I don't know why so I would probably choose the US version. Choices are: Solstice (E) [!] 72 kb Solstice (J) 74 kb Solstice (U) 72 kb Solstice (U) [a1] 72 kb Solstice (U) [o1] 72 kb
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
A company that can choose to terminate your account and not pay you without providing any proof of wrong doing is one that should be avoided. I will also advise people I know to beware of AdSense. Sorry Bisqwit.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Whoa very nice. I would have never guessed 4 seconds to be dropped and here you guys are still talking about dropping even more time!
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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Posts: 511
It sounds to me like google could be just trying to weasel their way out of paying you. If this is the case, they probably do it to lots of other people too, I'll try to find out more. --- ah couldn't find anything, all the stuff is about fraud claimed by the sites that are destination of the ads.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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Thanks for the positive comments, I'm glad you guys enjoyed it. PS: it's actually closer to 5 than 6 seconds (5.35) improvement, but seeing 6 makes me happy...
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Is it possible that fadeout is not reduced by minimizing each digit of score, but the number of white or black pixels in the score, and also more than score but lifes, coins, etc? I remember having these theories, but due to lack of consistancy I gave them up for thinking it is based on the RNG, one reason for this was because I also thought RNG was only used by monsters and thus possibly only affected by the way you interact with monsters (aka killing them, which also affects score). Now that we can look at the RAM it would be alot easier to get to the bottom of this, but I personally I think it is a waste of time to reduce this lag because it is relatively small and adds nothing to entertainment. (This is a reverse on my philosophy when Viper and I made the run, where we worked hard to reduce all fadeout). PS, should I be saying effect instead of affect?
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Both the avi and mp4 worked for me using vlc. I think there is probably some truth to what Mr. Boat is saying, but it might require more time investment for the encoders, which may or may not be worth it. I'm by far no expert on video encoding, but thats my 2 cents.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Here are some memory locations that are of some use (go to Cheats -> Search for new Cheats to setup watches for these): 7e0404,2u x pos 7e0406,2u y pos 7e0408,2s x speed 7e040a,2s y speed 7e0604,1s rope length 7e0602,1u rope length with stretch With these it's easy to determine game physics, but they are pretty much what you'd expect so there is no need to post 'em.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Amusing movie, although it might not catch the eye as a first person shooter would, it's a great demonstration of a single aspect of what tas are all about, great job.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
This new version is great, I love the instant display on load state. And Bisqwit+Angerfist showed the power of watching memory values, so I was pleasantly surprised to see this was already implemented in the latest improvement. There is a slight bug/misfeature of watching memory locations though: setting the data type to signed is ignored in the watch, but works in the cheat search screen. This is a bit of a pain when monitoring things like velocity, where often the value is negative. Overall though this watch stuff is really cool and I think will lead to high quality runs as well as saving everyone time, if only I found out about it before I got to the second to last level of Umihara Kawase ^^
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
If you press up+down while on a ladder you will climb downwards, and when you get to the ground, you will continue to climb though the ground. This is could be useful in theory, but I don't see anywhere to use it at least in the 2 min route.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hey spineshark, this is one of my favorite games too, I started a time attack for this game but gave up after 1 level, I hope you have more will power than me! Are you going for fastest completion or fastest all gold completion? Oh and quick tip incase you didn'k know already, R and L can switch which ability you have selected. Good luck.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hmm, these problems with c++ make it not very feasible to be super short :) The return value of cout isn't really used, you could fix it by changing it to (cout<<t/n,1). I didn't get that error, but previously I was using the ? : operator instead || and I would get that error about casting. It is interesting how many different results compilers can give, it is good evidence of two things, that a programmer should always try to write clean code and avoid risky shortcuts, and that c/c++ programmers should probably always compile with some sort of strict standards options, is -ansi sufficient? Here are some of the problems I have had in the past when trying to write real short code: Anything where a value is being modded in the same statement it is being accessed (ie x=n+n--;) Many times I was tempted to do something like this, but it would never be consistant from comp to comp. Arguement order execution (reason old sig didn't work) Array out of bounds effects (I know, very bad programming practice). This is kinda a funny story, current best pi one, before I had the array of size 9999 instead of 10001 to save a character. I knew it was not big enough, but it seemed to work fine. Then when I was changing the names of the variables so I could spell go pi. I started getting different results, I guess gcc decides where put variables in memory based somewhat on alphabetical order, but anyways it is a good thing that happened. I'm sure this stuff is all noted to be standard ambiguous stuff, but still catches me sometimes with, oh it worked, look no further. EDIT: e program improved by 3 characters, I think this will work on all comps so no need to test for me:
a[999],x,n;main(N){for(;n||(putchar(48^x),
n=998+--N);x+=10*a[--n]+!N)a[n]=x%n,x/=n;}
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
primordial soup: it prints out 11 leading 0s because fixing that would require me to make it longer, I view these 0s as not that bad, but ideally they wouldn't be there. hanzou, the second program I don't think will compile on any machine without including iostream, but I couldn't resist using it because cout<< is so much shorter than printf("%d",). The reason I don't use it in the first program is because I couldn't find a short way of doing the effect of %.4d which is to pad with 0s as to always print 4 digits. That is some very extensive testing you did, I hadn't even heard of half of those compilers. I am pretty happy with the results. What was the error on IBM XL btw? Thank you both very much for testing those out, I think it is safe for me to change my signature now :) One thing I tried recently was compiling with the -pedantric flag, and it gave me a new warning about g=g/x*p+a[p]*i+2*!o, saying it should be in parenthesis, but whatever. Someone with gcc4 told me that he couldn't get it to compile, he had to include stdio.h, and int declarations, and then would still get an error: /tmp/ccgCaxWP.o(.eh_frame+0x11): undefined reference to `__gxx_personality_v0' I have no idea what this means, but my guess was that something was weird about his gcc installation, so I had him just do a hello world test program, and he said that worked. I don't know what to make of that, does anyone have gcc4? Lastly, about the algorithm itself, I have not found much out about it yet, but here is what I have read: it can be derived from the Leibniz series (pi/4=1 - 1/3 + 1/5 - 1/7 + ...), by using Euler's Convergence Transformation. I will try to post more when I find out more. Thanks again for testing!
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Oh I didn't come up with it. It is a common formula for computing pi compactly. I have no idea why or how it works, but it is on my todo list to try to prove this formula. I will post more information about it when I start doing that.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Post subject: Pi Program
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
If the digits of pi were as interesting as a video game, and the characters of the program were frames, then this is exactly like a time attack. But that's a bit of a stretch so this better go in the Off Topic. I have been working on a new pi program, and I think I have come up with a program that I am satisfied with. Now I want to make sure that it runs properly on all machines, because I do some things that aren't exactly good programming practices to save space. My signature right now, suffers from this problem, it won't actually work on many machines/compilers.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o:
53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
It should calculate pi to 2880 decimal places. I would like to know if there are any issues you have getting it to compile. There are some problems I already know about but am not willing to add the characters to fix: -not giving type for g,o,p,i,a,main,x. To fix just add int everywhere -printf not declared, include stdio.h If these errors occur, I would still like to know to get an idea of what percent of people have these problems. I have made slightly shorter versions but I am pretty sure that even if they compile, they won't give the same output on all machines. All help is greatly appreciated. In case you are curious the algorithm used is: 2 + 1/3*(2 + 2/5*(2 + 3/7*(2 + 4/9*(...)))) If anyone is interested, here are some other tiny codes I have written: pi to 140 digits in 102 chars
b=1e6,a[27],t,n=995,i;main(){for(;i>24?i=t=!n--,
n--:n^3||cout<<t/n;t=n/2*a[i]+t%n*b,a[i++]=t/n)*a+=8;}
e to 992 digits in 87 chars
N,a[999],x;main(n){for(;n;--n?a[n]=x%n:printf(
"%d",x,n=998+--N))x=x/n+10*(a[n-1]+!~N);}
The algorithm is just like pi, but simpler: 2 + 1/2*(1 + 1/3*(1 + 1/4*(1 + 1/5*(...)))) (same as 1/0!+1/1!+1/2!+1/3!+...)
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hey, still haven't started the run, but while not even looking for it, out of sheer coincidence I found the .scm files of all single players mission files. http://www.staredit.net/index.php?download=1857 I am looking at them now, this is interesting, perhaps useless, but it is a good way to make sure there aren't any shortcuts for missions that we could miss.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
I am only in the second world, it is hard for me to play with keyboard, and the usb controller I have is really bad, I should buy a new one. Thanks for explaining that fly out of pipe thing.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hmm when I started to remake it along time ago using the hopping glitch I found it fastest in castle 1 to do hoping glitch from the start up until that long jump right before half way, the jump is still too long so you have to get your speed up again, but it was still a little faster. In SW4 the reason we started with such a high fly was to get high enough to grab a shell, I don't remember it that well, so maybe it can be improved after all. Also, was there some trick where you can start flying asap after some door transfers or something, how did that work? Just so you know, I am interested in all the new tricks you guys found in SDW, the reason I am not up to speed in following your guys runs is because that game looks very cool and most of the challenge and fun is finding the secrets exits and stuff for yourself. If I watched your guys run then the game would be spoiled for me when I play it. But first thing I will do after I get 120* or close to it is watch all your guys runs.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hey, glad to hear someone is incorperating the hopping trick. While it sounds like you are saving less time that you expected, I think you will save more time than you expected toward the very last levels. This is partly because towards the end I independently discovered the hopping trick and so I knew the run would not be perfect so I just wanted to get it over with. In the last star world level, if you use left + right, I think you can save time getting the key in the hole (accelerate while stuck to the ceiling). And in bowsers castle in the dark room you can save a bunch of time with the hopping trick. I am unfamiliar with the new tricks that have been discovered in SDW, are there any that can apply? Anyways best of luck and looking forward to your run.
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
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Posts: 511
Very nice. I especially liked the parts where you control the bird or climb ropes fast. My only suggestion would be to get hit by some of those poison clouds that do things like reverse your controls and stuff, in the final fight. I watched the avi version and using vlc I had some problems watching it, I think my comp was just not fast enough to decompress on some levels, specifically the ones in brambles or honey combs, which is unfortunate because these are some of the nicest levels. Is there a way to use less cpu intensive compression? How much bigger would the filesize be?
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
Experienced Forum User, Published Author, Player (201)
Joined: 7/6/2004
Posts: 511
Hmm not feeling motivated yet, but I will some day...
g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}
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