Question

Write a simulation of zombies/walkers that move randomly in one dimension. Each walker begins at the origin and at each time-step it takes a step to the right or left with equal probability, so that Pright= 0.5, Pleft= 0.5. Use a lattice of spacing x =1 and discrete time-steps, t= 1. Number of steps in each walk: 20 Number of walkers: 10.000 Please make your own algorithm for this simulation. Note: You will need to use a random number generator. You can use the one included in c/c++ or make your own. Walkers Statistics Make your program to calculate and produce the following data into output files: Mean square displacement vs. time, for time from t = 0, 1, 2,3... 20. Plot this data and fit a straight line to the curve and find the slope of the curve. Can you define a physics concept that describes what this slope represents? How would you call it? Explain. Spatial distribution of walks at t = 20, normalized so the area under the curve is 1. [Note: after an even number of steps, all walks end on even-numbered sites.] In a separate function, calculate and output to a file the expected average spatial distribution at time t=20. Plot together with your simulation data. There are two ways to calculate the expected spatial distribution: Evaluating this equation PN(d)=1/2N[N d+N/2] Using the method shown below Part 2: Walkers dynamics related to Evolution Stephen Jay Gould argued that evolution is a random walk along a 1-D axis of biological complexity. Random mutations lead any organism to produce offspring that can be slightly more complex or slightly less complex, with equal probability. However, there is a restriction (lower bound) in this complexity random walk, the minimum biological complexity below which no organism can evolve. Gould argued that this lower bound is a kind of reflecting wall along the axis of complexity, so evolution produces --on average-- upward progress, even though the individual steps are random. Set up a simulation of a 1-D random walk on a lattice with Pright = Pleft = 0.5. Put a reflecting wall at x = -1, so that a walker who tries to step on x = -1 gets reflected back to x=0. Perform 10,000 walks of length 20 steps. Make a graph of <x> vs time for t=0,1,23- -20. Make another graph showing log <x> vs log t. Is it a straight line? Can this behavior represent evolution or random mutation? Following Gould's line of reasoning: what would be the result of evolution if the lower boundary represented an absorbing wall (that is, a trap) instead of a reflecting wall? Please arrange in a .pdf file: your random walk simulation code with comments, the code that calculates the mean displacements and distribution, the graphs of your data and the answers to each question in part 1 and part 2.

Answer 1

Let steps of equal length be taken along a line. Let be the probability of taking a step to the right, the probability of taking a step to the left, $n_1$ the number of steps taken to the right, and $n_2$ the number of steps taken to the left. The quantities , , $n_1$, $n_2$, and are related by

p+q=

(1)

and

$n_1+n_2=N.$

(2)

Now examine the probability of taking exactly $n_1$ steps out of to the right. There are ways of taking $n_1$ steps to the right and $n_2$ to the left, where is a binomial coefficient. The probability of taking a particular ordered sequence of $n_1$ and $n_2$ steps is $p^(n_1)q^(n_2)$. Therefore,

$P(n_1)=((n_1+n_2)!)/(n_1!n_2!)p^(n_1)q^(n_2)=(N!)/(n_1!(N-n_1)!)p^(n_1)q^(N-n_1),$

(3)

where is a factorial. But this is simply a binomial distribution, so the mean number of steps $n_1$ to the right is

$<n_1>=pN,$

(4)

and the mean number of steps to the left is

$<n_2>=N-<n_1>=N(1-p)=qN.$

(5)

Similarly, the variance is given by

$sigma_(n_1)^2=<n_1^2>-<n_1>^2=Npq,$

(6)

and the root-mean-square deviation is

$sigma_(n_1)=sqrt(Npq).$

(7)

Consider now the distribution of the distances $d_N$ traveled after a given number of steps,

$d_N=n_1-n_2=2n_1-N,$

(8)

as opposed to the number of steps in a given direction. The above plots show $d_N(p)$ for $N=200$ and three values $p=0.1$, $p=0.5$, and $p=0.9$, respectively. Clearly, weighting the steps toward one direction or the other influences the overall trend, but there is still a great deal of random scatter, as emphasized by the plot below, which shows three random walks all with $p=0.5$.

Surprisingly, the most probable number of sign changes in a walk is 0, followed by 1, then 2, etc.

For a random walk with $p=1/2$, the probability $P_N(d)$ of traveling a given distance after steps is given in the following table.

steps	5			2		0	1	2	3	4	5
0						1
1						0
2					0		0
3				0		0		0
4			0		0	16	0	16	0
5	32	0		0		0		0	32	0	32

In this table, subsequent rows are found by adding half of each cell in a given row to each of the two cells diagonally below it. In fact, it is simply Pascal's triangle padded with intervening zeros and with each row multiplied by an additional factor of 1/2. The coefficients in this triangle are given by

(9)

(Papoulis 1984, p. 291). The moments

$mu_p=sum_(d=-N,-(N-2),...,N)d^pP_N(d)$

(10)

of this distribution of signed distances are then given by

	$=$		(11)
$mu_2$	$=$		(12)
$mu_3$	$=$		(13)
$mu_4$	$=$		(14)

so the mean is $mu=0$, the skewness is $gamma_1=0$, and the kurtosis excess is

$gamma_2=-2/N.$

(15)

The expectation value of the absolute distance after steps is therefore given by

$<d_N>$	$=$	$sum_(d=-N,-(N-2),...)^(N)|d|P_N(d)$	(16)
	$=$	$1/(2^N)sum_(d=-N,-(N-2),...)^(N)(|d|N!)/(((N+d)/2)!((N-d)/2)!).$	(17)

This sum can be done symbolically by separately considering the cases even and odd. First, consider even so that $N=2J$. Then

$<d_(2J)>$	$=$		(18)
	$=$	N! 2 dl 2 dl d=-J-U-1) d= 12 ( 2	(19)
	$=$	$(N!)/(2^N)[2sum_(d=1)^(J)(2d)/((J+d)!(J-d)!)]$	(20)
	$=$	$(N!)/(2^(N-2))sum_(d=1)^(J)d/((J+d)!(J-d)!).$	(21)

But this sum can be evaluated analytically as

$sum_(d=1)^Jd/((J+d)!(J-d)!)=1/(2Gamma(J)Gamma(1+J)).$

(22)

Writing $J=N/2$, plugging back in, and simplifying gives

(23)

where is the double factorial.

Now consider odd, so $N=2J-1$. Then

	$=$	$<d_(2J-1)>$	(24)
	$=$		(25)
	$=$	$(N!)/(2^(N-1))[sum_(d=1,3,...)^(2J-1)d/(((2J-1+d)/2)!((2J-1-d)/2)!)]$	(26)
	$=$	$(N!)/(2^(N-1))[sum_(d=2,4,...)^(2J)(d-1)/(((2J-2+d)/2)!((2J-d)/2)!)]$	(27)
	$=$	$(N!)/(2^(N-1))[sum_(d=1)^(J)(2d-1)/((J+d-1)!(J-d)!)].$	(28)

But this sum can be evaluated analytically as

$sum_(d=1)^J(2d-1)/((J+d-1)!(J-d)!)=1/([Gamma(J)]^2).$

(29)

Writing , plugging back in, and simplifying gives

J=(N+1) /2

(dv odd)	$=$	$(N!)/(2^(N-1)[Gamma(1/2+1/2N)]^2)$	(30)
	$=$		(31)
	$=$		(32)

Both the even and odd solutions can be written in terms of as

$<d_J>=2/(sqrt(pi))(Gamma(J+1/2))/(Gamma(J))=((2J-1)!!)/((2J-2)!!),$

(33)

or explicitly in terms of as

$<d_N>$	$=$		(34)
	$=$		(35)

The first few values of $<d_N>$ for $N=0$, 1, ... are therefore 0, 1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, ... (OEIS A086116 and A060818; Abramowitz and Stegun 1972, Pr

Answer 2

Let steps of equal length be taken along a line. Let be the probability of taking a step to the right, the probability of taking a step to the left, $n_1$ the number of steps taken to the right, and $n_2$ the number of steps taken to the left. The quantities , , $n_1$, $n_2$, and are related by

p+q=

(1)

and

$n_1+n_2=N.$

(2)

Now examine the probability of taking exactly $n_1$ steps out of to the right. There are ways of taking $n_1$ steps to the right and $n_2$ to the left, where is a binomial coefficient. The probability of taking a particular ordered sequence of $n_1$ and $n_2$ steps is $p^(n_1)q^(n_2)$. Therefore,

$P(n_1)=((n_1+n_2)!)/(n_1!n_2!)p^(n_1)q^(n_2)=(N!)/(n_1!(N-n_1)!)p^(n_1)q^(N-n_1),$

(3)

where is a factorial. But this is simply a binomial distribution, so the mean number of steps $n_1$ to the right is

$<n_1>=pN,$

(4)

and the mean number of steps to the left is

$<n_2>=N-<n_1>=N(1-p)=qN.$

(5)

Similarly, the variance is given by

$sigma_(n_1)^2=<n_1^2>-<n_1>^2=Npq,$

(6)

and the root-mean-square deviation is

$sigma_(n_1)=sqrt(Npq).$

(7)

Consider now the distribution of the distances $d_N$ traveled after a given number of steps,

$d_N=n_1-n_2=2n_1-N,$

(8)

as opposed to the number of steps in a given direction. The above plots show $d_N(p)$ for $N=200$ and three values $p=0.1$, $p=0.5$, and $p=0.9$, respectively. Clearly, weighting the steps toward one direction or the other influences the overall trend, but there is still a great deal of random scatter, as emphasized by the plot below, which shows three random walks all with $p=0.5$.

Surprisingly, the most probable number of sign changes in a walk is 0, followed by 1, then 2, etc.

For a random walk with $p=1/2$, the probability $P_N(d)$ of traveling a given distance after steps is given in the following table.

steps	5			2		0	1	2	3	4	5
0						1
1						0
2					0		0
3				0		0		0
4			0		0	16	0	16	0
5	32	0		0		0		0	32	0	32

In this table, subsequent rows are found by adding half of each cell in a given row to each of the two cells diagonally below it. In fact, it is simply Pascal's triangle padded with intervening zeros and with each row multiplied by an additional factor of 1/2. The coefficients in this triangle are given by

(9)

(Papoulis 1984, p. 291). The moments

$mu_p=sum_(d=-N,-(N-2),...,N)d^pP_N(d)$

(10)

of this distribution of signed distances are then given by

	$=$		(11)
$mu_2$	$=$		(12)
$mu_3$	$=$		(13)
$mu_4$	$=$		(14)

so the mean is $mu=0$, the skewness is $gamma_1=0$, and the kurtosis excess is

$gamma_2=-2/N.$

(15)

The expectation value of the absolute distance after steps is therefore given by

$<d_N>$	$=$	$sum_(d=-N,-(N-2),...)^(N)|d|P_N(d)$	(16)
	$=$	$1/(2^N)sum_(d=-N,-(N-2),...)^(N)(|d|N!)/(((N+d)/2)!((N-d)/2)!).$	(17)

This sum can be done symbolically by separately considering the cases even and odd. First, consider even so that $N=2J$. Then

$<d_(2J)>$	$=$		(18)
	$=$	N! 2 dl 2 dl d=-J-U-1) d= 12 ( 2	(19)
	$=$	$(N!)/(2^N)[2sum_(d=1)^(J)(2d)/((J+d)!(J-d)!)]$	(20)
	$=$	$(N!)/(2^(N-2))sum_(d=1)^(J)d/((J+d)!(J-d)!).$	(21)

But this sum can be evaluated analytically as

$sum_(d=1)^Jd/((J+d)!(J-d)!)=1/(2Gamma(J)Gamma(1+J)).$

(22)

Writing $J=N/2$, plugging back in, and simplifying gives

(23)

where is the double factorial.

Now consider odd, so $N=2J-1$. Then

	$=$	$<d_(2J-1)>$	(24)
	$=$		(25)
	$=$	$(N!)/(2^(N-1))[sum_(d=1,3,...)^(2J-1)d/(((2J-1+d)/2)!((2J-1-d)/2)!)]$	(26)
	$=$	$(N!)/(2^(N-1))[sum_(d=2,4,...)^(2J)(d-1)/(((2J-2+d)/2)!((2J-d)/2)!)]$	(27)
	$=$	$(N!)/(2^(N-1))[sum_(d=1)^(J)(2d-1)/((J+d-1)!(J-d)!)].$	(28)

But this sum can be evaluated analytically as

$sum_(d=1)^J(2d-1)/((J+d-1)!(J-d)!)=1/([Gamma(J)]^2).$

(29)

Writing , plugging back in, and simplifying gives

J=(N+1) /2

(dv odd)	$=$	$(N!)/(2^(N-1)[Gamma(1/2+1/2N)]^2)$	(30)
	$=$		(31)
	$=$		(32)

Both the even and odd solutions can be written in terms of as

$<d_J>=2/(sqrt(pi))(Gamma(J+1/2))/(Gamma(J))=((2J-1)!!)/((2J-2)!!),$

(33)

or explicitly in terms of as

$<d_N>$	$=$		(34)
	$=$		(35)

The first few values of $<d_N>$ for $N=0$, 1, ... are therefore 0, 1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, ... (OEIS A086116 and A060818; Abramowitz and Stegun 1972, Pr