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 Commenti

Walter Roberson
Walter Roberson il 25 Nov 2018
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 il 25 Nov 2018
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 il 26 Nov 2018
If i hit the edge i will bounce back to the same location.
Muzaffar Habib
Muzaffar Habib il 26 Nov 2018
Thank You. Much appreciated help. I will try to code that up with randi()
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 )

Accedi per commentare.

 Risposta accettata

Eyal Ben-Hur
Eyal Ben-Hur il 26 Nov 2018
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.

Più risposte (0)

Categorie

Scopri di più su Programming in Centro assistenza e File Exchange

Prodotti

Release

R2018a

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by