Question

在Solidity中工作，以太坊EVM和小数不存在。有没有办法我仍然可以在数学上使用整数创建泊松分布？不一定是完美的，即四舍五入或丢失一些数字是可以接受的。

Answer 1

让我作为序言，指出接下来的内容将不会（直接）对您带来以太/固体的帮助。但是，它会生成可能用于工作的概率表。

我最终对将Poisson概率表示为有理有多精确的问题深感兴趣，因此我在Ruby中将以下脚本组合在一起进行尝试：

def rational_poisson(lmbda)
  Hash.new.tap do |h|   # create a hash and pass it to this block as 'h'.
    # Make all components of the calculations rational to allow
    # cancellations to occur wherever possible when dividing
    e_to_minus_lambda = Math.exp(-lmbda).to_r
    factorial = 1r
    lmbda = lmbda.to_r
    power = 1r
    (0...).each do |x|
      unless x == 0
        power *= lmbda
        factorial *= x
      end
      value = (e_to_minus_lambda / factorial) * power
      # the following double inversion/conversion bounds the result
      # by the significant bits in the mantissa of a float
      approx = Rational(1, (1 / value).to_f)
      h[x] = approx
      break if x > lmbda && approx.numerator <= 1
    end
  end
end

if __FILE__ == $PROGRAM_NAME
  lmbda = (ARGV.shift || 2.0).to_f  # read in a lambda (defaults to 2.0)
  pmf = rational_poisson(lmbda)     # create the pmf for a Poisson with that lambda
  pmf.each { |key, value| puts "p(#{key}) = #{value} = #{value.to_f}" }
  puts "cumulative error = #{1.0 - pmf.values.inject(&:+)}"  # does it sum to 1?
end

浏览代码时要了解的内容。将.to_r附加到值或表达式可将其转换为有理数，即两个整数之比；后缀为r的值是有理常数； (0...).each是一个开放式迭代器，它将循环运行，直到满足break条件为止。

那个小脚本会产生如下结果：

localhost:pjs$ ruby poisson_rational.rb 1.0
p(0) = 2251799813685248/6121026514868073 = 0.36787944117144233
p(1) = 2251799813685248/6121026514868073 = 0.36787944117144233
p(2) = 1125899906842624/6121026514868073 = 0.18393972058572117
p(3) = 281474976710656/4590769886151055 = 0.061313240195240384
p(4) = 70368744177664/4590769886151055 = 0.015328310048810096
p(5) = 17592186044416/5738462357688819 = 0.003065662009762019
p(6) = 1099511627776/2151923384133307 = 0.0005109436682936699
p(7) = 274877906944/3765865922233287 = 7.299195261338141e-05
p(8) = 34359738368/3765865922233287 = 9.123994076672677e-06
p(9) = 67108864/66196861914257 = 1.0137771196302974e-06
p(10) = 33554432/330984309571285 = 1.0137771196302975e-07
p(11) = 33554432/3640827405284135 = 9.216155633002704e-09
p(12) = 4194304/5461241107926203 = 7.68012969416892e-10
p(13) = 524288/8874516800380079 = 5.907792072437631e-11
p(14) = 32768/7765202200332569 = 4.2198514803125934e-12
p(15) = 256/909984632851473 = 2.8132343202083955e-13
p(16) = 16/909984632851473 = 1.7582714501302472e-14
p(17) = 1/966858672404690 = 1.0342773236060278e-15
cumulative error = 0.0

是否可以使用基于整数的概率进行泊松分布？

1 个答案: