The next worms title on XBLA [Archive] - Team17 Forum

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Guilty
10 Feb 2007, 17:17
I can tell the developers right now that worms will be a HUGE HUGE hit on XBLM
But for the fans not so much , We will still love it , but there is alot of stuff missing ........
Thats why i prepose this idea for the sequal ( if there is one )
better take advandtage of the space upgrade MS made for the arcade next time ( i think its like 500 mb )
Worms live party FTW
All the weapons from the past games
Custom map maker ( the one where you use that nifty paint tool )
All the game modes ( like where you have to dig bunkers to survive meteors)
And all the voices from all the countries ( Most of them atleast )
Scaleable wind effects and other options for the host (weapons , item drops)
2d GAME !!
12 player online mode! ( very important )
Feel free to add ideas:p

Guilty
10 Feb 2007, 17:19
Oh and possibly make this a pc / 360 compatable title :D
Live anywhere FTW

sPideS
10 Feb 2007, 22:38
Wierdo????????

Wrathchild
10 Feb 2007, 22:42
While I like the idea, and would love to see W:A or WWP brought to the 360, that would really be a full retail release (or conversion if they brought us W:A or WWP and not a new title) and not a Live Arcade download.
I would gladly pay $50 for it, if a conversion were made.

Plasma
11 Feb 2007, 20:18
Congratulations! You have added one more topic to that big list of "OMFG maek the ultimaet worms game!!!!!1"! Reward yourself by feeling let down by this response!
And next time, can you at least consider the fact that they had to keep the game in 50mb? T17 even wen to the trouble of making an extra pack for the stuff that couldn't fit!

quakerworm
11 Feb 2007, 21:33
12 player online mode! ( very important )
why twelve?

Etho
12 Feb 2007, 01:36
Because he likes to go outside and run some errands between his turns.

quakerworm
12 Feb 2007, 03:32
he he. what i meant is that 12 seems a bit random in the context. yes, i know that 12 has a very good set of prime factors, but if you're going that high, go for 16. first of all, you can still have 12 players, second,15 is another possibility, and that one adds another nice factor, and finally 16 is a power of two, which opens a whole lot of different possibilities, not to mention making code a bit neater.

SupSuper
12 Feb 2007, 05:58
Because he likes to go outside and run some errands between his turns.He might even be dead by the time he comes back, since the more teams, the less worms each has.

AndrewTaylor
12 Feb 2007, 10:16
he he. what i meant is that 12 seems a bit random in the context. yes, i know that 12 has a very good set of prime factors, but if you're going that high, go for 16. first of all, you can still have 12 players, second,15 is another possibility, and that one adds another nice factor, and finally 16 is a power of two, which opens a whole lot of different possibilities, not to mention making code a bit neater.
I would worry that a game requiring people to wait a quarter of an hour between turns would get boring, and that people might then start to associate Worms with boring. Then they'd be unlikely to buy future releases.
Why should this hypothetical next game be on XBLA, specifically? Why not a corss-platform release?

Haoshiro
12 Feb 2007, 17:29
I think 12-16 players could work IF... Each player only had one worm and it was FFA or team game.
But you'd still have long wait times, and half the players would be out of the game before it was even their turn... so it'd still be pretty pointless.
But hey, maybe in an enclosed level, with 200% health...

quakerworm
12 Feb 2007, 18:10
I would worry that a game requiring people to wait a quarter of an hour between turns would get boring, and that people might then start to associate Worms with boring. Then they'd be unlikely to buy future releases.
10-second turn limits would aid this significantly. or maybe even do something completely nuts. how about a relay mode, where each color group gets 2+ teams. there will be two+ teams taking a turn at the same time, one from each color group. when one team ends a turn, the turn passes to the next team in the color group. this way a color group can get more shots out if they take short turns. with 3 color groups, 4 teams in each, this can get really nuts. there would, of course, be some new net issues with such a mode of play, and the gameplay would not be exactly turn-based anymore, but since most targets will still be stationary, this will not turn liero-like until there is just one worm in each color group, which isn't even necessarily going to happen in each match.

Haoshiro
12 Feb 2007, 19:00
You could also allow 8-16 teams, but each team only gets one worm at a time.
Then when that worm is dead, you get a new worm (max of 4) that parachutes to a random location on the map ("reinforcements").
And then, yeah, keep turn time low, like 15 sec max.

Paul.Power
15 Feb 2007, 13:08
I just love the "( very important )"
He has eleven friends all eager to play a single game of Worms.

quakerworm
15 Feb 2007, 19:06
11 is an odd number of friends to have. it is also a prime number of friends to have. but the point is, it's still very random, and yes, "very important" part is what makes it all the weirder.

bonz
15 Feb 2007, 19:08
11 is an odd number of friends to have
Perhaps not friends, but the 11 most senseful of his multiple personalities.

quakerworm
15 Feb 2007, 22:31
i just pictured it. a crazy person is sitting in a room surrounded by 12 360's connected to 12 different tvs ranging from plasma screens to tiny little black and white tvs made in the 70's. all 12 screens are showing the same match. the person picks up a different controller for each turn and screams something incoherent into the head set every time he takes a bad shot.

AndrewTaylor
16 Feb 2007, 10:39
11 is an odd number
Well done, you.

Melon
16 Feb 2007, 10:45
Hahaha. You win. Trust you to notice something like that.
*Hands over award*
http://www.kieranmillar.com/atoscar.jpg

quakerworm
16 Feb 2007, 19:46
11 is an odd numberWell done, you.
yeah. all i did is applied a gcd algorithm to (11,2) pair. the gcd theorem states that gcd(11,2) = gcd(2, 11 mod 2). of course, to find the modulus, i have performed a basis conversion 10 => 2, and dropped the higher terms. so 0b1011 mod 2 gives me 0b1 or just 1. gcd(2,1) is trivially giving me, 1, telling me that the gcd(11,2) is also 1. from that i have deduced that 11 and 2 have no non-trivial factors in common, and, therefore, 11 is indeed odd.

AndrewTaylor
16 Feb 2007, 20:43
yeah. all i did is applied a gcd algorithm to (11,2) pair. the gcd theorem states that gcd(11,2) = gcd(2, 11 mod 2). of course, to find the modulus, i have performed a basis conversion 10 => 2, and dropped the higher terms. so 0b1011 mod 2 gives me 0b1 or just 1. gcd(2,1) is trivially giving me, 1, telling me that the gcd(11,2) is also 1. from that i have deduced that 11 and 2 have no non-trivial factors in common, and, therefore, 11 is indeed odd.
Thanks. Could you email me a list of all odd numbers, please?

quakerworm
16 Feb 2007, 21:46
i don't know if they'll fit in one e-mail, but i can do you one better. how about i set up a script on my server to send them to you in batches of 100 at a time every 30 seconds. what's your e-mail?

AndrewTaylor
16 Feb 2007, 22:55
Oh, I can't be bothered searching through more than one email. Just send me the first few odd numbers. Oh, and the last one, obviously.

Paul.Power
16 Feb 2007, 23:34
720720 is an interesting number.
And is it me, or does the Cyberman Oscar look like it has a really big smile on its face?

bonz
17 Feb 2007, 03:07
I want an odd perfect number.

quakerworm
17 Feb 2007, 07:53
Oh, I can't be bothered searching through more than one email. Just send me the first few odd numbers. Oh, and the last one, obviously.
no problem. i have already shown that in binary, odd numbers will end in 1. obviously, the largest odd number will consist of all 1's in binary. let us call it x for now. so we know that x=...111111. naturally, 2*x=...111110. from that follows that:
x = 2x - x = ...111111 - ...111110 = -1
so here is your list:
1, 3, 5, 7, etc, and the last one is -1.

Metal Alex
17 Feb 2007, 15:24
Thanks. Could you email me a list of all odd numbers, please?
you don't want to. A friend tryed to send me all the odd numbers discovered so far in a notepad file...
It was more than 1 Gb O_o

Paul.Power
18 Feb 2007, 17:24
you don't want to. A friend tryed to send me all the odd numbers discovered so far in a notepad file...
It was more than 1 Gb O_oYou don't discover odd numbers.
"Wow, the next highest odd number after umpty-umptillion and 1 is umpty-umptillion and 3! Who'd've thought it?"
Primes, yes. Odd numbers, no.

quakerworm
18 Feb 2007, 18:29
what we all should do is just pick a really large number, like googol, and just mod all algebra by it. do you really need numbers over 10^100? and in Z10^100 ring, you will have all the same algebra for small numbers, without hassles of infinity.

AndrewTaylor
18 Feb 2007, 19:36
what we all should do is just pick a really large number, like googol, and just mod all algebra by it. do you really need numbers over 10^100? and in Z10^100 ring, you will have all the same algebra for small numbers
Well yeah, until you want to divide anything. I'm pretty sure you can't meaningfully divide in ring algebra.

quakerworm
18 Feb 2007, 22:42
Well yeah, until you want to divide anything. I'm pretty sure you can't meaningfully divide in ring algebra.
you can, as long as your ring is a field. the only condition is that for every a=/=0 in ring there exists a unique b such that ab=1. (note that 1 and 0 are defined as multiplicative and additive identities) then, b is 1/a, and you have your division. simplest example is a ring of all rational numbers. a slightly more abstract example is any modulo prime ring. consider Z5. 1*1 = 1, 2*3 = 6 mod 5 = 1, 4*4 = 16 mod 5 = 1. so 1/1=1, 1/2=3, 1/3=2 and 1/4=4. now, if you want, for example, to compute 4/3 = 4*(1/3) = 4*2 = 8 mod 5 = 3. there is a simple theorem that proves that this works for any prime.

AndrewTaylor
19 Feb 2007, 10:34
1/1=1, 1/2=3, 1/3=2 and 1/4=4.
That sounds like it has just bucketloads of practical applications.
Edit: How can you have a ring of all rational numbers? There are infinite of them and there isn't a highest or lowest one.

Paul.Power
19 Feb 2007, 12:26
That sounds like it has just bucketloads of practical applications.
Edit: How can you have a ring of all rational numbers? There are infinite of them and there isn't a highest or lowest one.Who said rings had to be finite?

quakerworm
19 Feb 2007, 12:31
it's still called a ring, regardless of whether it is finite or not. a ring is any set with two operations, addition and multiplication, defined for it. a ring must be closed under these two operations. (for every a,b in ring R, a+b and a*b are also in R) in addition, a ring must have an additive identity, every element must have an additive inverse, addition must commute, and associative and distributive laws must hold for addition and multiplication.
as for practical applications of division in integer rings, if you take an integer ring modulated over a large enough prime, you will not feel a difference from working with rational numbers. Z5 looks odd because it is so small. with a prime large enough, you will never actually work with inverses of numbers. for example, in Z1021, 1/2 = 511, 1/3 = 681, 1/5 = 817. as you can see, the gap grows quite large. for a very big prime, you don't need to worry what 1/7 is. you can call it 1/7, knowing that it is somewhere in your ring. so essentially, all algebra remains the same as algebra of rational numbers.

AndrewTaylor
19 Feb 2007, 14:30
as for practical applications of division in integer rings, if you take an integer ring modulated over a large enough prime, you will not feel a difference from working with rational numbers. Z5 looks odd because it is so small. with a prime large enough, you will never actually work with inverses of numbers. for example, in Z1021, 1/2 = 511, 1/3 = 681, 1/5 = 817. as you can see, the gap grows quite large. for a very big prime, you don't need to worry what 1/7 is. you can call it 1/7, knowing that it is somewhere in your ring. so essentially, all algebra remains the same as algebra of rational numbers.
Righto... so a "ring" needn't actually be in any way cyclic or in actual plain English, a ring. Got you. But I still don't see how any system that says 1/2 is over five hundred is actually useful, aside from as an interesting mathematical diversion.
Who said rings had to be finite?
Everybody but mathematicians and Tolkein, apparently.

Paul.Power
19 Feb 2007, 15:46
Everybody but mathematicians and Tolkein, apparently.This is my personal favourite ring: http://en.wikipedia.org/wiki/Eisenstein_integers
EDIT: Oh, and you forgot Wagner :p.

quakerworm
19 Feb 2007, 22:42
But I still don't see how any system that says 1/2 is over five hundred is actually useful, aside from as an interesting mathematical diversion.
the fact that it is over 500 isn't useful at all. the fact that is useful is that 1/2 exists somewhere in your ring. that's all you really need to know. after that, you can work with quotient rules, and know that you will end up somewhere in the ring. so if you want to compute 3/4 * 2/5, you simply follow rules of algebra of rational numbers, and get 3/10. you also know that this corresponds to some integer, but odds are, you only ever going to use it as 3/10, and not as an integer that it corresponds to.
in other words, what i mean is that, you really don't care how it works. the thing you care about is that rules of division still work, and you are working in some finite ring. the fact that it is finite means that there is the smallest non-zero fraction, and that there is no such thing as infinity in your algebra.
oh, and the best part, of course, is that it becomes very easy to name all odd integers. there are exactly none of these, since a/2 will always give you an integer result, just like division by any other number.

Luther
21 Feb 2007, 16:34
the thing you care about is that rules of division still work, and you are working in some finite ring.
Wanna bet?

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