Text 698, 176 rader
Skriven 2004-11-07 21:59:00 av Malcolm (1:278/230)
Ärende: Re: Tongues, curling into
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"Brett Aubrey" <brett.aubrey@shaw.ca> wrote
>
> In Richard Dawkin's The Ancestor's Tale, he mentions the 50/50 split on
> Humanity's ability to curl our tongue into a tube.  Might there have been
a
> reason for this trait, or its' absense?  Or is this just likely an example
> of a minor mutation that is advantageously neutral?  Or something else?
TIA
> Regards, Brett Aubrey.
>
When you do the sums, you will see that a trait with an adaptive advantage
too small to meaure can have a huge influence. (Try this)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

void usage(void);
void mousegrow(double advantage, double sd, int gens);
void jitter(double *x, int N, double sd);
void renormalise(double *x, int N, double sd);
double mean(double *x, int N);
double gaussrand();
double uniform(void);

int main(int argc, char **argv)
{
  if(argc != 4)
    usage();

  mousegrow( atof(argv[1]), atof(argv[2]), atoi(argv[3]) );

  return 0;
}

void usage(void)
{
  printf("Mousegrow, evolution demonstration\n");
  printf("Usage: mousegrow <advantage> <stddev> <gens>\n");
  printf("advantage - the advantage of being in the top half"
    "of the distribution (eg 0.001) \n");
  printf("jitter - the jitter factor(eg 0.01)\n");
  printf("gens - number of generations to run the simulation\n");
  printf("\nMonte Carlo simulation with an asexual population of 1000\n");

  exit(0);
}


/*
  run the simulation
*/
void mousegrow(double advantage, double sd, int gens)
{
  int i;
  int ii;
  double *pop;
  double xbar;
  int target;

  pop = malloc(1000 * sizeof(double));

  // popuate with random men 5
  for(i=0;i<1000;i++)
    pop[i] = 5.0 + gaussrand();

  // main loop
  for(i=0;i<gens;i++)
  {

 // take the mean, and add moise
 xbar = mean(pop, 1000);
    jitter(pop, 1000, sd * xbar/5.0);


 printf("Gen %d mean size %g\n", i, xbar);
 for(ii=0;ii<1000;ii++)
 {
   // if we are above average, we have a 0.5 chance of reproducing
   if(pop[ii] > xbar)
   {
     if( uniform() < 0.5 )
  {
    // just replace a random individual with a copy of yourself
    target = rand() % 1000;
    pop[target] = pop[ii];
  }
   }
   // if we are below, the chance drops to 0.5 - s
   else
   {
     if(uniform() < 0.5 - advantage)
  {
    target = rand() % 1000;
    pop[target] = pop[ii];
  }
   }
 }
  }

  free(pop);
}


/*
  add a bit of evolutionary noise to the population
  (models mutations)
*/
void jitter(double *x, int N, double sd)
{
  int i;

  for(i=0;i<N;i++)
 x[i] += gaussrand() * sd;
}


/*
  average
*/
double mean(double *x, int N)
{
  double answer = 0;
  int i;

  for(i=0;i<N;i++)
 answer += x[i];

  return answer / N;
}

/*
  random number with a normal distribution
*/
double gaussrand()
{
  static double V1, V2, S;
  static int phase = 0;
  double X;

  if(phase == 0)
  {
    do {
        double U1 = (double)rand() / RAND_MAX;
        double U2 = (double)rand() / RAND_MAX;

        V1 = 2 * U1 - 1;
        V2 = 2 * U2 - 1;
        S = V1 * V1 + V2 * V2;
 } while(S >= 1 || S == 0);
     X = V1 * sqrt(-2 * log(S) / S);
  }
  else
    X = V2 * sqrt(-2 * log(S) / S);

  phase = 1 - phase;
  return X;
}

/*
  random number on the interval 0-1
*/
double uniform(void)
{
  return rand() / (RAND_MAX + 1.0);
}
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