hello, i have a 5x5 matrix and i am standing at 2,2 location, now from here i have to take one step and jump to new value. i can take jump in all 4 directions i,e i-1,i+1,j-1,j+1 with equal probability of 0.25, and also i have to update my location for making a counter loop for continous movement inside matrix.how can i code it using randperm ? or any other command available ?

5 件のコメント

Walter Roberson
Walter Roberson 2018 年 11 月 25 日
What should happen if you reach an edge and randomly select to go outside?
randperm() is not going to help you with this. Consider randi()
Image Analyst
Image Analyst 2018 年 11 月 25 日
Is this homework? Just compute the candidate next location. If it's outside of your boundaries, then just get another value, which will most likely be in a different direction and be inside the box.
Muzaffar Habib
Muzaffar Habib 2018 年 11 月 26 日
If i hit the edge i will bounce back to the same location.
Muzaffar Habib
Muzaffar Habib 2018 年 11 月 26 日
Thank You. Much appreciated help. I will try to code that up with randi()
Walter Roberson
Walter Roberson 2018 年 11 月 26 日
To "bounce back to the same location" you can use techniques such as
new_x_position = max( min(new_x_position, maximum_x_position), minimum_x_position )

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Eyal Ben-Hur
Eyal Ben-Hur 2018 年 11 月 26 日

0 投票

Here is a suggestion:
A = zeros(5); % the matrix
p ={2,2}; % initial location
A(p{:}) = 1; % set initial location
N = 20; % no. of steps
for k = 1:N
s = 1-(rand<0.5)*2; % random +/-
vh = randi(2); % random vertical or horizontal
p{vh} = p{vh}+s; % move
p{vh} = abs(p{vh}-5*(p{vh}==6 || p{vh}==0)); % if outbound, return from the other side
A = zeros(5); % remove last location
A(p{:}) = 1; % set new location
end
Since it looks like homwork I leave it to you to understand the details.

1 件のコメント

Muzaffar Habib
Muzaffar Habib 2018 年 11 月 27 日
Thank U. it did work with some ammendments.

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