# Converting a problem from Mathematica to MATLAB

2 visualizzazioni (ultimi 30 giorni)
George il 21 Feb 2011
I have this problem in mathematica and want to do it in MATLAB.
steps[m_] := Table[2 RandomInteger[] - 1, {m}]
Walk1D[n_] := FoldList[Plus, 0, steps[n]]
LastPoint1D[n_] := Fold[Plus, 0, steps[n]]
nsteps = 200; nsq = Floor[Sqrt[nsteps]];
MeanSquareDistance1D[n_Integer, m_Integer] :=
N[Sum[LastPoint1D[n]^2, {m}]/m]
r2D=MeanSquareDistance1D[100, 1000]
data = Map[({#, MeanSquareDistance1D[#,2000]})&,
Range[10, 90, 20]]
This is what i did :
steps=@ (m) randi(3,1,m)-2;
Walk1D =@ (n) cumsum(steps(n));
LastPoint1D = @ (Walk1D) (Walk1D(end));
nsteps=200;
nsq=floor(sqrt(nsteps));
MeanSquareDistance1D= @ (n,m) sum((LastPoint1D(n)).^2)./m;
r2D=MeanSquareDistance1D(100,1000)
data=zeros(5,2);
for i=10:20:90
data=[i,MeanSquareDistance1D(i,2000)]
end
The problem is 1) r2D gives me the same value as many times as i run the code,but it must change. 2) In the "data " the first column its ok (its 10:20:90) but the 2nd column gives me the same numbers as i run the code. And,also,i don't know if i defined it right(the "data")
EDITED CODE -------------------------
steps=@ (m) 2*randi([0,1],[1,m])-1;
Walk1D =@ (n) [0,cumsum(steps(n))];
findend=@ (x) x(end);
LastPoint1D=@(n) findend(Walk1D(n));
nsteps=200;
nsq=floor(sqrt(nsteps));
MeanSquareDistance1D= @ (n,m) m.*sum((LastPoint1D(n)).^2)./m;
r2D=MeanSquareDistance1D(100,1000)
data=[ ];
for i=10:20:90
data=[data; i , MeanSquareDistance1D(i,2000)]
end
EDIT-->>>
I did it like this(with help):
steps1=@ (n, m) randi([-1 1], n, m);
LastPoint_1D=@ (n, m) sum(steps1(n, m));
MeanSquareDistance1D = @(n,m) mean(LastPoint_1D(n,m).^2);
##### 2 CommentiMostra NessunoNascondi Nessuno
Oleg Komarov il 21 Feb 2011
Can you comment the code in mathematica?
George il 21 Feb 2011
Random walks in one dimens.
"step"->examines trajectories which consists from m steps
"walk1d"->defines all the intermediate positions
"lastpoint"->defines only the last position
Then ,i am trying to define the mean square distance between initial and final position of the trajectory and final
"data"->dependence of the mean square distance from the number of steps

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### Risposte (2)

Seth DeLand il 21 Feb 2011
George,
Looks like there's a problem with this line of code:
LastPoint1D = @ (Walk1D) (Walk1D(end));
I believe what you are trying to do is call the 'Walk1D' function and find the last element of the vector that is returned by 'Walk1D'. Here is one way to do that:
findend = @(x) x(end);
LastPoint1D = @ (n) findend(Walk1D(n));
The 'findend' function just takes in a vector and returns the last element of that vector. Try replacing your line of code above with the two lines I suggested and let me know if that helps.
##### 12 CommentiMostra 10 commenti meno recentiNascondi 10 commenti meno recenti
George il 23 Feb 2011
I edited my post.It's ok now!
George il 23 Feb 2011
How could i write the equivalent of "findend = @(x) x(end);"
Thank you

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Oleg Komarov il 22 Feb 2011
I found this article "One Dimensional RW", and referring to the paragraph "The root mean square distance from the origin after a random walk of n unit steps is ?n.", I derived the following function:
function Out = MeanSquareDistance1D(n,m)
% Create random walk with unit steps
Walk1D = [0 cumsum(2*randi([0,1],[1,m])-1)];
% Plot it
% plot(Walk1D)
% Means square distances from the origin for all steps
Out = Walk1D.^2./(1:m+1);
% Select just the steps at n intervals
Out = Out(n:n:m);
end
Or in handle fmt:
Walk1D = @(m) [0 cumsum(2*randi([0,1],[1,m])-1)];
MeanSquareDistance1D = @(n,m) Walk1D(m).^2./(1:m+1);
Oleg
##### 2 CommentiMostra NessunoNascondi Nessuno
George il 22 Feb 2011
First of all thanks for searching!I tried it but it gives me totally wrong results.The "data" matrix become too big.I tried "MeanSquareDistance1D = @(n,m) Walk1D(n).^2./(1:m+1);"
but it gives me :matrix dimensions must agree.
George il 23 Feb 2011
I edited my post.It's ok now!

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