Random walk or stepping

4 ビュー (過去 30 日間)
Muzaffar Habib
Muzaffar Habib 2018 年 11 月 25 日
コメント済み: Muzaffar Habib 2018 年 11 月 27 日
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
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 )

サインインしてコメントする。

採用された回答

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

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeProgramming についてさらに検索

タグ

製品


リリース

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by