methinks we'll need to wait for the manual to find that all out <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->
Simple request, I don't want this thread locked and I see the potential for this to degrade to a flame war. Please excersize caution in future replies.
Some people were arguing that there should be 4 levels of upgrade for the aliens, the first being no upgrade at all. If you did argue this (as it is a viable alien status), you could not insert it into the equation along with the other upgrade. You would have too many repeats. Let's say they have no upgrades at all, there is one counted for having no Silence, one for having no Adreneline, and so on. So the combination 0 0 0 is counted 135 times. 1 0 0 is counted 9 times for each alien class, 45 times in all. Same for 2 0 0, 0 1 0, 0 0 3, et cetera.
So, we have 5 alien classes. 3 upgrade chambers, and 3 upgrades available to choose from from each. And of course the state of not having any upgrades, or not having upgrades in certain spots.
5x12x12x12 (3x4 representing every possiblity, including having no upgrade) would work, but we need some way to weed out the repeats. Thus, we could subtract 130 for the 0 0 0 (27 times counted for each alien, 5 alien classes, but it does need to be counted once for each alien, so 26 x 5), 44 for 1 0 0, and so on.
So, so far we have 5x12x12x12-130-1935 (396 being all the combinations of 1 0 0, 2 0 0, and so on, multiplied by 5 for each alien class, minus 45 that need to be counted). But we aren't yet finished, because there is always the possibilty of having two upgrades and no third, ie 1 0 3, and these repeats must also be weeded out. If my math is right, there are 27 combinations here [1 1 0, 1 0 1, 0 1 1, 2 1 0, 3 2 0, 3 0 3, and so on, ad nauseum]. Three repeats of each, so we have 405 (81 [324 {81x4} - 243 {81x3}] x 5 alien classes) taken out of there.
So now the equation is 5x12x12x12-130-1935-405. By my math that comes to 6,210 possible combinations.
Of course, two years away from calculus does strange things to a man's brain, so I could be just completely wrong. Let's just accept the fact that there are a WHOLE FRIGGIN' LOT of possibilities.
-Ryan!
"I believe the use of noise to make music will increase until we reach a music produced through the aid of electrical instruments which will make available for musical purposes any and all sounds that can be heard." -- composer John Cage, 1937
coilAmateur pirate. Professional monkey. All pance.Join Date: 2002-04-12Member: 424Members, NS1 Playtester, Contributor
Actually, the overestimate was 8,000-something. 5,000 is the correct number. You're correct that making an equation with 4 upgrade levels would result in repeats; that's where my previous page's post (and the 5,000) come from - working around that issue. Gwahir says that his equation is the correct way to express my long workaround, and I'll believe him - it's greek to me.
no stealth can be one of three things, choice of another upgrade from that chamber, using another chamber, or no upgrades at all, these cases are satisfied in the equation
the equation given is just the shorter version of the math given on the previous page
5 non-upgraded aliens (3 choose 0 = 1 and 9^0 = 1) 5*9 combinational upgrades for use of one upgrade from one chamber *3 for three chambers (3 choose 1 = 3) 5*9*9 for two chambers in use * 3 for the 3 combinations of used chambers (3 choose 2 = 3) 5*9*9*9 for all three chambers in use * 1 since they all are in use (3 choose 3 = 1). 5 + 3*5*9 + 3*5*9*9 + 1*5*9*9*9 = 5000
the lone 5 is a completely non-upgraded alien set. The 9^1 set covers all single choices within one chamber, the *3 takes care of choosing which chamber is used. Therefore all your worries are accounted for
His upgrade state can be described in a 3-digit number, where a 0 stands for no upgrade in that category, and 1, 2, 3 stands for a certain upgrade. Correct?
000 001 002 003 010 011 ... 333
There are 64 such numbers. Therefore a Lvl1 can have 64 different upgrade states.
Since there are 5 aliens, 64x5=320 different aliens altogether. Still many, but nowhere near 5000.
Now riddle me this! <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->
<!--c1--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Code Sample</b> </td></tr><tr><td id="CODE"><!--ec1--> { | | | O { | | | { | | | <!--c2--></td></tr></table><span id='postcolor'><!--ec2--> that is the model for a single chamber O is a chamber | is a form of upgrade (including level) so a single chamber has 9 forms of upgrades
so having 9, we then consider them being used with other chambers 9^(number of chambers) in parrallel, think binary 1 digit 2^1 = 2 possible permutations 2 digits 2^2 = 4 3 digits 2^3 = 8 get the picture? the rest is explained earlier
Incorrect, Collateral. An alien can pick one of three upgrades for each upgrade chamber. Hence, you need to multiply this by 9. (3 * 3)
On the other hand, though, this is a slight overestimate. By this formula, level 0 carapace, level 0 regeneration, and level 0 redemption are all counted separately, although they are THE SAME THING, ie you have taken no defensive upgrade. So each type of upgrade chamber can lead to 10 possibilities. Therefore, it should be 5 * 10 * 10 * 10, or 5,000. I think.
I don't believe I just posted that. <!--emo&:p--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/tounge.gif" border="0" valign="absmiddle" alt=':p'><!--endemo-->
CollateralDamage, you are forgetting that for each upgrade chamber, there are 3 individual upgrades that could be reseached. -edit- Rhoads beat me too it with a far more detailed explanation.
lol, I think we are using different ideas of how these upgrades work...
I understand you can't have more than three upgrades on a single alien, and each has to be from a different category, of which there are three, each having four states (one of them being no update).
[edit]My head hurts, I'm going to sleep this over. Until then I still say 320 <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->[/edit]
<!--QuoteBegin--CollateralDamage+Aug. 03 2002,17:31--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Quote</b> (CollateralDamage @ Aug. 03 2002,17:31)</td></tr><tr><td id="QUOTE"><!--QuoteEBegin-->His upgrade state can be described in a 3-digit number, where a 0 stands for no upgrade in that category, and 1, 2, 3 stands for a certain upgrade. Correct?
000 001 002 ... 333
There are 64 such numbers. Therefore a Lvl1 can have 64 different upgrade states.
Since there are 5 aliens, 64x5=320 different aliens.<!--QuoteEnd--></td></tr></table><span id='postcolor'><!--QuoteEEnd--> That would be true, except that there are 9 upgrades to choose from, not just 3 upgrades that you can add on to. You have to factor that people will choose different combinations of add-ons, and then upgrade them differently.
And coil, I would say that your 5,000 is correct, but when I work out your math, I come up with 4,910 combinations. Hmmm.
Still, probably more right than mine. Your logic looks smooth. I just added the numbers together, though...I haven't looked at the equation yet.
-Ryan!
"This is my career. I have children to raise. I have to retaliate." -Mike Tyson, boxer, on why he bit off a piece of Evander Holyfield's ear
our 5000, is was a combined effort, he showed me that I failed to differentiate the different chambers. Including that and some number crunching, we came up with 5000.
to elaborate, lets hold off on the 5 aliens and work it to 1000, first you have 1 non-upgraded state, then you have 27 possible states where only one chamber is used (9 per chamber, 3 chambers). After that you have 81 possible permutations for any two chambers, one upgrade per chamber. With 3 possible combinations of two chambers, that's 243 possible upgrade combinations. If you use all 3 chambers then you have only 1 set of chambers in use, since order does not matter, and 729 combinations between all three. Take all of these 1 + 27 + 243 + 729 = 1000 possible upgrades per alien. Take the fact that there are 5 aliens and you have a total of 5000 possible combinations.
Looks like we have to clarify the whole chamber concept again.
I thought there are three different chambers: sensory, defence and movement. You seem to believe that each chamber has 9 different upgrades (i.e., 9 defence upgrades). to choose from, although coil said "From each type of chamber you can evolve one of three possible options." So how many different defence upgrades are there?
working directly with Coil on this I feel confident saying that there are 3 upgrades per chamber, only one can be chosen. But, you can build up to 3 of each of these chambers to increase the effects of these upgrades. ex. cloak: 1 sensory chamber, 25% cloak, 2 50% or 75% cloak, 3 completely invisible.
so there are 9 "forms" of upgrades per chamber type
edit: I just finished watching the movie with the same name as your handle.
Here's what I want to know about the upgrade chambers: coil, I hope you're still following. I keep hearing that you need three chambers to have the highest form of whatever upgrade you have selected. Does that mean three of the <b>same</b> chamber? That is, you you need three Movement chambers to have 100% silent movement, or one of each chambers?
Hope that made sense.
-Ryan!
Unless hours were cups of sack, and minutes capons, and clocks the tongues of bawds, and dials the signs of leaping houses, and the blessed sun himself a fair, hot wench in flame-colored taffeta, I see no reason why thou shouldst be so superfluous to demand the time of the day. I wasted time and now doth time waste me. -- William Shakespeare
coilAmateur pirate. Professional monkey. All pance.Join Date: 2002-04-12Member: 424Members, NS1 Playtester, Contributor
Ryan: to have level-3 carapace, you need 3 defensive chambers.
People in general: trust me, the 5,000 is accurate. I'm not overestimating by counting the same non-upgraded state multiple times; I did that the first time and came up with 5x(3x4)^3 = 8,640. Gwahir and I spent way too much time on this; I'm pretty confident our math is correct. (:
take this <!--c1--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Code Sample</b> </td></tr><tr><td id="CODE"><!--ec1--> { | | | O { | | | { | | | <!--c2--></td></tr></table><span id='postcolor'><!--ec2-->
draw three of these on a sheet of paper, large enough to count on.
Now we're going to leave the different aliens till the end. Start with 1 as the non-upgraded state. Now continue counting all the | symbols. As there are 27 | symbols, you should get 28 (remember the non-upgraded state). At this point you have counted all the single upgrade states. Next put your left index finger on the first | of the first chamber and right on the first | of the second. Continue counting like this, count the | on the second chamber, and every time you reach the end, move your finger to the next | on the first and restart on the second, when you have finished on the first chamber, you are done. There are 81 such combinations between any two chambers. Do this again for the 2nd and 3rd, and 1st and 3rd chambers, or just multiply 3 * 81 and add to 28. Next do the same type thing for all three. count on the third chamber and move slowly through the second, but every time you get through the second move to the next | on the first and restart on the second and third.. In that one case you get 729 and there are no other meaningful combinations of chambers when all three are in use. So add the four situations together 1 + 27 + 243 + 729 = 1000. Take the 5 aliens into the picture, 5 * 1000 = 5000. This is the simplest this can be made, now if you need to know that 5000 is correct, take the time to count through this manually as described.
edit: freemantle: Did you learn set theory? Something likely taught in courses on discrete math or probability and stochastic processes.
O.k...I guess I don't really understand the chambers too well, but from what I can see, you are figuring the total of all possible combinations of upgrades. Is it even possible to GET all the upgrades? also, you're including in the possibility of having combinations that probably don't exist (in game not mathematically I mean) having level 3 stealth, level 3 defense, and level 3 movement, and assuming that building more chambers makes it different, but that isn't an actual upgrade from the chamber, and it's an option chosen by the player, it happens when the chamber is built, therefore you're figuring (I believe) of total upgrade options, is incorrect. If you can get 1 from each type of chamber, that's 1 motion, 1 stealth, 1 defense. It looks to me like there are really just the 3 upgrade possibilities, with a total of 9, on any given individual. The number of chambers ignored, and because we're talking about 1 alien, we'll ignore the 5 types of aliens as well. Right?
coilAmateur pirate. Professional monkey. All pance.Join Date: 2002-04-12Member: 424Members, NS1 Playtester, Contributor
Hmm... first, welcome. Second, sorry, but you don't really know what you're talking about. (: There are 3 types of chamber: defense, movement, and sensory. From each of these, you can evolve ONE of THREE different upgrades. AND there are 3 levels of upgrade - that's 28 possibilities (27, plus no upgrade) for one chamber. When you consider all possible combinations, the total number of unique aliens you can create is 5,000. (:
Comments
So, we have 5 alien classes. 3 upgrade chambers, and 3 upgrades available to choose from from each. And of course the state of not having any upgrades, or not having upgrades in certain spots.
5x12x12x12 (3x4 representing every possiblity, including having no upgrade) would work, but we need some way to weed out the repeats. Thus, we could subtract 130 for the 0 0 0 (27 times counted for each alien, 5 alien classes, but it does need to be counted once for each alien, so 26 x 5), 44 for 1 0 0, and so on.
So, so far we have 5x12x12x12-130-1935 (396 being all the combinations of 1 0 0, 2 0 0, and so on, multiplied by 5 for each alien class, minus 45 that need to be counted). But we aren't yet finished, because there is always the possibilty of having two upgrades and no third, ie 1 0 3, and these repeats must also be weeded out. If my math is right, there are 27 combinations here [1 1 0, 1 0 1, 0 1 1, 2 1 0, 3 2 0, 3 0 3, and so on, ad nauseum]. Three repeats of each, so we have 405 (81 [324 {81x4} - 243 {81x3}] x 5 alien classes) taken out of there.
So now the equation is 5x12x12x12-130-1935-405. By my math that comes to 6,210 possible combinations.
Of course, two years away from calculus does strange things to a man's brain, so I could be just completely wrong. Let's just accept the fact that there are a WHOLE FRIGGIN' LOT of possibilities.
-Ryan!
"I believe the use of noise to make music will increase until we reach a music produced through the aid of electrical instruments which will make available for musical purposes any and all sounds that can be heard."
-- composer John Cage, 1937
Isn't "no Stealth" just the 4th kind of Stealth upgrade? Meaning there are 3 categories with 4 states each?
Please give an example for the repeats you see. Currently I see no more than 3x4x5 different aliens.
the equation given is just the shorter version of the math given on the previous page
5 non-upgraded aliens (3 choose 0 = 1 and 9^0 = 1)
5*9 combinational upgrades for use of one upgrade from one chamber *3 for three chambers (3 choose 1 = 3)
5*9*9 for two chambers in use * 3 for the 3 combinations of used chambers (3 choose 2 = 3)
5*9*9*9 for all three chambers in use * 1 since they all are in use (3 choose 3 = 1).
5 + 3*5*9 + 3*5*9*9 + 1*5*9*9*9 = 5000
the lone 5 is a completely non-upgraded alien set. The 9^1 set covers all single choices within one chamber, the *3 takes care of choosing which chamber is used. Therefore all your worries are accounted for
Let's pick one alien class, say Lvl1.
His upgrade state can be described in a 3-digit number, where a 0 stands for no upgrade in that category, and 1, 2, 3 stands for a certain upgrade. Correct?
000
001
002
003
010
011
...
333
There are 64 such numbers. Therefore a Lvl1 can have 64 different upgrade states.
Since there are 5 aliens, 64x5=320 different aliens altogether. Still many, but nowhere near 5000.
Now riddle me this! <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->
{ | | |
O { | | |
{ | | |
<!--c2--></td></tr></table><span id='postcolor'><!--ec2-->
that is the model for a single chamber
O is a chamber
| is a form of upgrade (including level)
so a single chamber has 9 forms of upgrades
so having 9, we then consider them being used with other chambers
9^(number of chambers)
in parrallel, think binary
1 digit 2^1 = 2 possible permutations
2 digits 2^2 = 4
3 digits 2^3 = 8
get the picture?
the rest is explained earlier
On the other hand, though, this is a slight overestimate. By this formula, level 0 carapace, level 0 regeneration, and level 0 redemption are all counted separately, although they are THE SAME THING, ie you have taken no defensive upgrade. So each type of upgrade chamber can lead to 10 possibilities. Therefore, it should be 5 * 10 * 10 * 10, or 5,000. I think.
I don't believe I just posted that. <!--emo&:p--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/tounge.gif" border="0" valign="absmiddle" alt=':p'><!--endemo-->
-edit- Rhoads beat me too it with a far more detailed explanation.
I understand you can't have more than three upgrades on a single alien, and each has to be from a different category, of which there are three, each having four states (one of them being no update).
[edit]My head hurts, I'm going to sleep this over. Until then I still say 320 <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->[/edit]
000
001
002
...
333
There are 64 such numbers. Therefore a Lvl1 can have 64 different upgrade states.
Since there are 5 aliens, 64x5=320 different aliens.<!--QuoteEnd--></td></tr></table><span id='postcolor'><!--QuoteEEnd-->
That would be true, except that there are 9 upgrades to choose from, not just 3 upgrades that you can add on to. You have to factor that people will choose different combinations of add-ons, and then upgrade them differently.
And coil, I would say that your 5,000 is correct, but when I work out your math, I come up with 4,910 combinations. Hmmm.
Still, probably more right than mine. Your logic looks smooth. I just added the numbers together, though...I haven't looked at the equation yet.
-Ryan!
"This is my career. I have children to raise. I have to retaliate."
-Mike Tyson, boxer, on why he bit off a piece of Evander Holyfield's ear
to elaborate, lets hold off on the 5 aliens and work it to 1000, first you have 1 non-upgraded state, then you have 27 possible states where only one chamber is used (9 per chamber, 3 chambers). After that you have 81 possible permutations for any two chambers, one upgrade per chamber. With 3 possible combinations of two chambers, that's 243 possible upgrade combinations. If you use all 3 chambers then you have only 1 set of chambers in use, since order does not matter, and 729 combinations between all three. Take all of these 1 + 27 + 243 + 729 = 1000 possible upgrades per alien. Take the fact that there are 5 aliens and you have a total of 5000 possible combinations.
I thought there are three different chambers: sensory, defence and movement. You seem to believe that each chamber has 9 different upgrades (i.e., 9 defence upgrades). to choose from, although coil said "From each type of chamber you can evolve one of three possible options." So how many different defence upgrades are there?
ex. cloak: 1 sensory chamber, 25% cloak, 2 50% or 75% cloak, 3 completely invisible.
so there are 9 "forms" of upgrades per chamber type
edit: I just finished watching the movie with the same name as your handle.
5 Aliens: 5
3 Chambers: 5x3
3 Upgrades Per Chamber: 5x3x3
3 States for every upgrade: 5x3x3x3
3 Empty upgrade spaces: 5x3x3x3x3
Total: 405
Hope that made sense.
-Ryan!
Unless hours were cups of sack, and minutes capons, and clocks the tongues
of bawds, and dials the signs of leaping houses, and the blessed sun himself
a fair, hot wench in flame-colored taffeta, I see no reason why thou shouldst
be so superfluous to demand the time of the day. I wasted time and now doth
time waste me.
-- William Shakespeare
People in general: trust me, the 5,000 is accurate. I'm not overestimating by counting the same non-upgraded state multiple times; I did that the first time and came up with 5x(3x4)^3 = 8,640. Gwahir and I spent way too much time on this; I'm pretty confident our math is correct. (:
take this
<!--c1--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Code Sample</b> </td></tr><tr><td id="CODE"><!--ec1-->
{ | | |
O { | | |
{ | | |
<!--c2--></td></tr></table><span id='postcolor'><!--ec2-->
draw three of these on a sheet of paper, large enough to count on.
Now we're going to leave the different aliens till the end. Start with 1 as the non-upgraded state. Now continue counting all the | symbols. As there are 27 | symbols, you should get 28 (remember the non-upgraded state). At this point you have counted all the single upgrade states. Next put your left index finger on the first | of the first chamber and right on the first | of the second. Continue counting like this, count the | on the second chamber, and every time you reach the end, move your finger to the next | on the first and restart on the second, when you have finished on the first chamber, you are done. There are 81 such combinations between any two chambers. Do this again for the 2nd and 3rd, and 1st and 3rd chambers, or just multiply 3 * 81 and add to 28. Next do the same type thing for all three. count on the third chamber and move slowly through the second, but every time you get through the second move to the next | on the first and restart on the second and third.. In that one case you get 729 and there are no other meaningful combinations of chambers when all three are in use. So add the four situations together 1 + 27 + 243 + 729 = 1000. Take the 5 aliens into the picture, 5 * 1000 = 5000. This is the simplest this can be made, now if you need to know that 5000 is correct, take the time to count through this manually as described.
edit: freemantle: Did you learn set theory? Something likely taught in courses on discrete math or probability and stochastic processes.