Overcrowding Happiness Mechanics

[vinque]

I am a returning player after a long absence (6 years I believe). Anyhow, my first time around I didn't take the game too seriously. I'm trying to be "a better player" this time around and have started by trying to learn as much as I can about game mechanics. There's tons of great info in the FAQ and War Academy but I still can't find information about one mechanic - overcrowding happiness.
For example, I play a game on regent and I know (taking all variables like luxuries, wonders, etc. out of the equation) that my city will be unhappy when it reaches size 3 unless I do something about it (i.e. keep a unit in the city as MP). This leads me to believe that as my city grows each pop increase will create one unhappy person. However, this does not seem to be the case. At a certain point (somewhere around the time the town becomes a city I think) it seems each pop increase causes two (maybe 3) unhappy people. Anyhow, I'm wondering if there's information on how this "game mechanic" works?

 

[Aabraxan]

Welcome to CFC!

Hit the Civ 3 Info center. Look here: http://www.civfanatics.com/civ3/infocenter/#levels

The number of content citizens that you get automatically (without accounting for wonders, lux spending, luxuries, etc.) is a function of difficulty level At Chieftain, you get 4 before someone is born cranky. At Regent, you get 2, and at Emperor and above, only 1. Each pop increase should not be creating more than 1 unhappy person, though.

 

[vinque]

OK, well the 1 unhappiness/pop increase is good to know. Unfortunately that means I'm missing something, which is annoying since I'm at work and can't look at my current game (grrr). Are there other sources of unhappiness in the early-mid game (say around 300ad) other than WW(I'm a monarchy)?

 

[Aabraxan]

Go to the city view and click on one of the unhappy people. They'll tell you why they're unhappy. Did you pop-rush anything while you were in despotism?

 

[Spoonwood]

The number of content citizens that you get automatically (without accounting for wonders, lux spending, luxuries, etc.) is a function of difficulty level At Chieftain, you get 4 before someone is born cranky

More specifically the content citizens c function maps difficulty level to the default number of content citizens (presuming no war weariness), c:{Cheiftain, Warlord, Regent, Monarch, Emperor, Demi-God, Deity, Sid}->{1, 2, 3, 4} in the following manner

c(Cheiftain)=4
c(Warlord)=3
c(Regent)=2
c(Monarch)=2
c(Emperor)=1
c(Demi-God)=1
c(Deity)=1
c(Sid)=1.

Such a function does not have an inverse function, since supposing a function d such that
d(c(difficulty level)=d(default number of content citizens)=difficulty level, d(2) may equal Regent or Monarch level.

Are there other sources of unhappiness in the early-mid game (say around 300ad) other than WW(I'm a monarchy)?

In addition to what Aabraxan wrote, there's the effects of military police, the luxury slider, and war happiness.

 

[Lord Emsworth]

I think happyness is best understood if you conceive of every one of your population as unhappy by default. And that this default state then is altered by the various happy/content(/unhappy) faces that each city receives. As such, there is no overcrowding unhappyness; that is just the normal, unaltered state.

 

[Spoonwood]

I like that Lord_Emsworth. It reminds me of a neat problem... DIGRESSION ALERT

Suppose you've gotten assigned to conduct a chess tournament. It has 235 participants. You need to set up a bracket to determine a winner of the tournament such that there exists only one undefeated player. How many chess matches will your tournament have?

 

[CKS]

That is way too much work for me. I can't see an easy way to deal with games that end in draws. Without accounting for draws, it is, of course, much simpler.

Spoiler :

If you have a loser for each game, you need as many games as you need losers. For this tournament, you need 234 losers and one winner, so you need 234 games.

On Topic:
I think the thing that makes unhappiness hard to understand is that you can have more unhappiness than you have citizens.

 

[Nergal]

I think the solution to the unhappiness problem is simply to check each turn. The only 'unexpected' factors I know of are relating to War unhappiness/happiness. The AI seems to have a habit of asking for peace just when it would make a Wonder building city go into disorder. Changing Ages can sometimes coincide with growth, which made me think the unhappiness was related to it.( But it isnt). Just grab as many luxuries as you can and use the slider.

 

[Snarkhunter]

I like that Lord_Emsworth. It reminds me of a neat problem... DIGRESSION ALERT

Suppose you've gotten assigned to conduct a chess tournament. It has 235 participants. You need to set up a bracket to determine a winner of the tournament such that there exists only one undefeated player. How many chess matches will your tournament have?

That depends entirely on whether you are running it as a strict knockout or allowing losers to play on, without a chance of winning overall. Either is justifiable. It also depends on whether you count a bye as a match or not. and we assume you have no forfeits, or if you do, you count them as matches as well. I admit in advance, I'm feeling too lazy to do the math, or to look it up in the TD Guide I wrote years ago & work it out

kk

 

[Spoonwood]

Good point CKS. I forgot about draws, so anyone who consider such, please assume each chess match results in a victory for one player.

Snarkhunter, I meant it as a strict knockout, I guess I didn't make that clear. No byes also. All matches get played. Forfeits count as matches. There's VERY VERY little math to do in the problem. That's the catch.

 

[DWetzel]

Nittery alert: with 235 players, someone must get a bye by definition.

 

[Snarkhunter]

Good point CKS. I forgot about draws, so anyone who consider such, please assume each chess match results in a victory for one player.

Snarkhunter, I meant it as a strict knockout, I guess I didn't make that clear. No byes also. All matches get played. Forfeits count as matches. There's VERY VERY little math to do in the problem. That's the catch.

Yeah, but I was too lazy to do even that--it's been that kind of day since my surgery Monday

You can dispense with the draw problem by making it a Go tournament; conventionally, komi always includes a half point in the value, so there are no draws.

There will be multiple byes, incidentally, for 235 players. This doesn't seem to change the observation that there must be one game per loser, but does change the number of matches if byes are considered as a match. For pairing purposes in any system I can think of off the top of my head, they would be. For example a 16 player tourney requires 15 games; a 15 player tourney requires 14--and 1 bye: 15 matches. I suppose that means that each bye increases the effective number of players by 1, then the easy calculation can be made.

kk

 

[splunge the 2nd]

235 players:
First round 118 matches, second round 59 matches, 3rd 30, 4th 15, 6th 8...4...2..1
Sum is 237 unless my math is off. If there were a power of 2 number of players it would have come out even.

 

[Spoonwood]

235 players:
First round 118 matches, second round 59 matches, 3rd 30, 4th 15, 6th 8...4...2..1
Sum is 237 unless my math is off. If there were a power of 2 number of players it would have come out even.

Yep, it's off. Using that method we have floor[235/2]=234/2=117 matches for the first round, 117+1(the bye player now plays)=118, 118/2=59 matches for the second round, then
floor[59/2]=58/2=29 for the next round, then 30/2=15 for the next round (another bye player pops back in), then floor[15/2]=14/2=7 matches, then 8/2=4, then 4/2=2, then 2/1=1 for 117+59+29+15+7+4+2+1=234 as we should get. This method has the advantage in that it helps to reveal the number of necessary byes. We just count the number of times we had an odd number in the numerator of the fractions above, and it's 3 here. CKS's solution (which I originally had in mind) doesn't have this advantage, even though it solves the intended problem faster.

For example a 16 player tourney requires 15 games; a 15 player tourney requires 14--and 1 bye: 15 matches. I suppose that means that each bye increases the effective number of players by 1, then the easy calculation can be made.

Quite a quick conjecture from such a small sample size in my opinion. I don't quite understand you really. Can you prove or disprove what you mean?

 

[CKS]

What Snarkhunter means is that if you count byes as matches, they are matches that have no losers. If you add in imaginary people to lose these matches, you will need to add in one per bye. Then each match has a loser, and you can count them up.

While one might think of byes as matches for pairing purposes, I don't think they are. They don't need a table to play at, they won't have spectators, and no one will be interested in the outcome.

 

[Snarkhunter]

Quite a quick conjecture from such a small sample size in my opinion. I don't quite understand you really. Can you prove or disprove what you mean?

Not rigorously, I don't think. (And why the heck would I want to disprove what I mean???) But the example speaks for itself: if there are 15 players there are 15 matches, not 14; a bye has to count as a match in tournament pairing schemes. (Sorry, CKS--that's just the way it goes. And as a competitor, I assure you that I am definitely interested in whether my opponent caught a bye at any point during the proceedings!) Equally, if there are 16 players, there are 15 matches. So at least in this case, a bye acts as if it were a player for the purpose of the quick calculation. My experience as a TD suggests to me that it always does in this scheme.

There are 2 other ways to do the pairings for a KO with a non-2^some-power entry field: 1) bye the top N players where N is a number such that N + (Field-N)/2 = 2^some power. In this case, we would want 21 byes & 107 first round matches, to give us 128 second round players. this allows the TD to use a traditional tree to write/record the results for all to see, in a straightforward fashion. If so, you then have 128 + 127 matches overall. 2) Treat the 107 matches as a "play-in" round to the tournament proper. If so, then technically the 21 seeds might not be considered byed & the 107 matches not part of the tournament, as the tournament proper has not yet begun. If so, then there are only 127 matches

The proper answer, then, depends on how you define what you are looking for & how much you hew to standard directing practices and which ones you decide to follow. I admit that 1) above would be unusual procedure, based on the theory of minimizing the number of byes, but not impossible procedure, as it allows for definite seeds. But 2) is fairly common--if you don't force the field into a 2^N number from the outset, based on invitations and suchlike.

Actually, I'd treat the whole thing as a Swiss-McMahon; KO's are such a bore for most of the field

kk

 

[vinque]

Well I see I've been completely threadjacked, but for anyone besides myself who cares, my problem was a simple math miscalculation. All my unhappiness was due to overcrowding. With 3MP, 2 "free" content (Regent level), and 4 luxuries I figured, with a marketplace, my bigger cities could grow to size 12 without rioting instead of size 11. Luckily, after correcting my math error, I "acquired" another luxury from the Arabs and the problem resolved itself.

 

[TheOverseer714]

Funny how much an extra luxury helps, isn't it?

 

[Spoonwood]

And why the heck would I want to disprove what I mean???

You might that figure out if what you said actually worked or not. That said, good points in general Snarkhunter. I hadn't considered the problem and byes from the perspective of actual players and what the tournament director actually has to do before when setting up such a bracket. Or how to keep the tournament interesting for players. I guess thinking of tournaments in terms of professional sports doesn't work so well, since more and more teams end up sitting the bench as spectators as the tournament goes further and further on.

Sorry about the threadjacking vinque... I tried to warn everyone this might happen.