CODEWITHSUNDEEP
matlabrandomoctaverandom-walk
Ahaan S. Rungta
Ahaan S. Rungta

Reputation: 2403

Why does the one-dimensional variant of a 2-d random walk not work?

There is a two-dimensional random walk that one can find here which works perfectly in Octave. However, when I tried to write a one-dimensional random walk program, I got an error. Here is the program: 

t=[];
x=[];
for i=1:100000
    J=rand;
    if J<0.5
        x(i+1)=x(i)+1;
        t(i+1)=t(i)+1;
    else
        x(i+1)=x(i)-1;
        t(i+1)=t(i)+1;
    end
end

plot(t,x)

Here is the error:

error: A(I): index out of bounds; value 1 out of bound 0

Thank you.

Upvotes: 0

Views: 173

Answers (4)

Luis Mendo
Luis Mendo

Reputation: 112669

No need for a loop:

N = 100000;
t = 1:N;
x = cumsum(2*(rand(1,N)<.5)-1);
plot(t,x)

enter image description here

For the 2D case you could use the same approach:

N = 100000;
%// t = 1:N; it won't be used in the plot, so not needed
x = cumsum(2*(rand(1,N)<.5)-1);
y = cumsum(2*(rand(1,N)<.5)-1);
plot(x,y)
axis square

enter image description here

Upvotes: 4

Rash
Rash

Reputation: 4336

In the first iteration, i = 1, you have x(2) = x(1) +or- 1 while x has dimension of zero. You should define the starting point for x and t, which is usually the origin, you can also change the code a little bit,

x = 0;
N = 100000;
t = 0 : N;
for i = 1 : N
    x(i+1) = x(i) + 2 * round(rand) - 1;
end
plot(t,x)

Upvotes: 1

yoh.lej
yoh.lej

Reputation: 1104

You get an error because you ask MATLAB to use x(1) in the first iteration when you actually defined x to be of length 0. So you need to either initialize x and t with the proper size:

x=zeros(1,100001);
t=zeros(1,100001);

or change your loop to add the new values at the end of the vectors:

    x(i+1)=[x(i) x(i)+1];

Upvotes: 2

Yang Zhang
Yang Zhang

Reputation: 130

Since t and x are empty, therefore, you cannot index them through x(i+1) and x(i). I believe you should intialize x and t with all zeros.

Upvotes: 1

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