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import multiprocessing as mp
import numpy as np

def multiprocessing(f, data, nproc):
    chunks = np.array_split(data, nproc)
    with mp.Pool(nproc) as ex:
        res = ex.map(f, chunks)
    return np.concatenate(list(res))

def dosomething(xs):

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def smc(M, steps, prior, potential, proposal_kernel, correction_steps=1, ess_ratio=2):
    """
    smc(M, steps, prior, potential, proposal_kernel, correction_steps=1, ess_ratio=2)
    
    Creates a particle approximation of a probability distribution using the Sequential
    Monte Carlo (SMC) algorithm with Markov Chain Monte Carlo (MCMC) correction steps using
    the Metropolis-Hastings algorithm.
    
    Parameters

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brownian = @instr begin
    σ ~ Beta(2, 3)
    x = zeros(N)
    for i = 2:N
       x[i] ~ Normal(x[i-1], σ^2)
    end
end

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Function.operands = [];
Function.prototype.valueOf = function() {
  Function.operands.push(this);
  return 3;
}
function mkP(value) {
  if (value instanceof Function) return value;
  var ops = Function.operands;
  Function.operands = [];
  if (ops.length >= 2 && (value === 3 * ops.length)) {

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function rk43(f, t, x, θ, τ, k₁)
    p₁ = (θ) -> 1/6*θ*(6-9θ+4*θ^2)
    p₂ = (θ) -> θ^2-2/3*θ^3
    p₃ = (θ) -> θ^2-2/3*θ^3
    p₄ = (θ) -> 1/6*θ^2*(-3+4θ)
    
    k₂ = f(t + .5τ, x + τ*.5k₁)
    k₃ = f(t + .5τ, x + τ*.5k₂)
    k₄ = f(t + τ, x + τ*k₃)
    k₅ = f(t + τ, x + (1/6)τ*(k₁ + 2k₂ + 2k₃ + k₄))

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# Answer to: https://twitter.com/1HaskellADay/status/875063542784958465
#
# So the Initial Value Problem (IVP) you want to solve seems to be the following
# 
# dy/dx = 3*exp(-x) - 0.4*y 
# y(x_0) = y_0      with x_0 = 0 and y_0 = 5
# 
# You then ask: what is the value of y when x = 3?
#
# This can be computed using the Runge Kutta method by feeding it