Question

为了好玩，我制作了Lua脚本，以模拟虚构物种的生长速度＆＃39; krogan＆＃39;从游戏质量效应。他们可以活1000年，一个krogan女性一年可以生产1000克罗根。基本规则是：

一群人只在1001年去世
年龄在100到300岁之间的人口中有50％是生育妇女
每年有4％的可育克罗甘女性怀孕
怀孕的krogan每年生育1000个孩子

可以找到代码（和输出）HERE。否则...

local NumYears = 1000 --How many years to estimate
local MinFertile = 100 --Age krogan start reproducing
local MaxFertile = 300 --Age krogan stop reproducing
local KroganAge = 1000 --How old a krogan can get
local Start_Population = 100000 --Number of initial krogans
local TimeSkips = 10 --Population updates displayed every # years
local Fertility = 1000 --Number of average children born by one mother each year
local PercentPregnant = 0.04 --percent of population pregnant during the year

math.randomseed(os.time())

local Krogans = {}

do --randomizes current krogan population
    local sum = 0
    for i = 1, KroganAge do
        local x = i-KroganAge/2
        x = x*5/KroganAge
        Krogans[i] = math.exp(-((x)^2)/2)/math.sqrt(2*math.pi) --bell curve
        sum = sum + Krogans[i]
    end

    --So that the ratio remains the same, and the total sum is around Start_Population
    local left = 0
    for i = 1, KroganAge do
        local a = Krogans[i]/sum*Start_Population
        left = left + (a-math.floor(a))
        local extra = math.floor(left)
        Krogans[i] = math.floor(a)+extra
        left = left - extra
    end
end


function NextYear()
    local Fertile = 0
    local Population = 0
    for i = MinFertile, MaxFertile do
        if Krogans[i] then
            Fertile = Fertile + Krogans[i]*0.5 --mult by 0.5 for fertile females (50% of population)
        end
    end
    for i = KroganAge, 2, -1 do
        if Krogans[i-1] then
            Krogans[i] = Krogans[i-1] --advance krogan by one year
            Population = Population + Krogans[i] --get population count
        end
    end
    Krogans[1] = Fertile*Fertility*PercentPregnant --Krogan born this year
    Population = Population + Krogans[1]
    return Fertile, Population
end

local words = {} --Table used for output formatting 
words[9] = "Billion"
words[12] = "Trillion"
words[15] = "Quadrillion"
words[18] = "Quintillion"
words[21] = "Sextillion"
words[24] = "Septillion"
words[27] = "Octillion"
words[30] = "Nonillion"
words[33] = "Decillion"
words[36] = "Undecillion"
words[39] = "Duodecillion"
words[42] = "Tredecillion"
words[45] = "Quattuordecillion"
words[48] = "Quindecillion"

function convert(n) --Converts/formats numbers into easy-to-read output
    if n <= 999999999 then -- credit http://r...content-available-to-author-only...n.it
        local left,num,right = string.match(n,'^([^%d]*%d)(%d*)(.-)$')
        return left..(num:reverse():gsub('(%d%d%d)','%1,'):reverse())..right
    else
        n = string.format("%18.0f",n):gsub("%s","")
        local digits = #n-1
        local choice = math.floor(digits/3)*3
        local num = (digits-choice)+1
        local str = string.sub(n,1,num).."."..string.sub(n,num+1,num+1).." "..words[choice]
        return str
    end
end

for i = 1, NumYears do
    local Fertile, Population = NextYear()
    if i%TimeSkips == 0 then
        print("Year: "..i)
        print("Total Population:\t"..convert(math.floor(Population)))
        print("Fertile Females: \t"..convert(math.floor(Fertile)))
        print("----")
    end
end

这可以预见，几百年后人口相当多。我的问题是，这是否反映了准确的克罗根人群，假设所有死亡都是从老年开始的？我尝试为人类输入一系列数字（年龄在15到44岁之间，最大年龄70岁，每个孩子出生1个孩子），但这只会导致下降的人口，而我＆＃39 ;我不确定为什么。

我猜测它与如何计算每年出生的孩子数量有关。这是非常简单的乘法：

Krogans[1] = Fertile*Fertility*PercentPregant
--[[
    Fertile: Number of Fertile Women
    Fertility: Number of children born in a pregnancy
    PercentPregnant: How many fertile women are pregnant this year (4% = 0.04)
]]

但这有什么不对吗？为什么会导致人口下降？

这种人口增长算法是否准确？

0 个答案: