Array Manipulation..

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Mustafa
Mustafa 2011-7-12
回答： Vidhi Agarwal 2024-7-31
Hi,
I have asked one question regarding a simple decrement of an array yesterday and got a good response. Today i am trying to manipulate that array further but i am kind of stuck in between loops and condtions. I am new to Matlab
It goes like this i am trying to decrement an array and once any particular element of that array arrives at zero, i am assigning that element as well as its immediate neighboring element with some value. What i need is that once that assigned value arrives at zero, it will increase the counter value by 1.
close all
clear all
clc
n = [11 12 13 24 3 7 14 11 12];
n_indx = zeros(1,12);
recv_count = 0;
send_count = 0;
send_count_1 = zeros(1,12);
recv_count_1 = zeros(1,12);
for i=1:40 % Just a loop so that the program can run for long (Not good programming practice)
while any(n)
n = max(0,n-1)
n_indx = find(n==0)
n(n_indx) = 3 ; % Assigning the value of "3" to the element that become zero
send_count_1(n_indx)=send_count+1 ; % Once an element become zero it should be recorded in this array at the same index
if (n(n_indx+1)>0) % Checking if the neighboring element is greater than zero
n(n_indx+1)=3 % Assigning the vlaue of "3" to the neighboring element
if (n(n_indx-1)>0) % Checking if the other neighboring element is also greater than zero
n(n_indx-1)=3 % Assigning the vlaue of "3" to the other neighboring element
% Now what i want is if all three elements (n_indx) && (n_indx+1)
% && (n_indx-1) are zero at the same point, they should be recorded
% into another counter
if (n(n_indx)==0) && (n(n_indx+1)==0) && (n(n_indx-1)==0)
recv_count_1(n_indx+1)=recv_count+1;
recv_count_1(n_indx-1)=recv_count+1;
% This gives me an error(That they are not scalar and i think they are scalar values).
break
end
end
end
end
end
  1 个评论
Mustafa
Mustafa 2011-7-12
Sorry for poor indentation and comments.
If the array goes out of dimensions or a problem of subscript occurr, dont worry about it. I will fix that

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回答（1 个）

Vidhi Agarwal
Vidhi Agarwal 2024-7-31
Hi Mustafa,
I assume you are encountering some issues with your code. Following is the attached code giving required output.
n = [11 12 13 24 3 7 14 11 12];
processed = false(size(n)); % Array to keep track of processed elements
recv_count = 0;
send_count = 0;
send_count_1 = zeros(size(n));
recv_count_1 = zeros(size(n));
for i = 1:40 % Loop to run the program for a long time
while any(n > 0)
n = max(0, n-1);
n_indx = find(n == 0 & ~processed); % Find indices where n is zero and not processed
% Iterate over each index where the element is zero
for idx = n_indx
if n(idx) == 0 && ~processed(idx) % Ensure the element is still zero and not processed
n(idx) = 3; % Assigning the value of "3" to the element that becomes zero
send_count = send_count + 1; % Increment the send counter
send_count_1(idx) = send_count; % Record the zero element in this array
processed(idx) = true; % Mark as processed
% Check and assign value to the neighboring elements
if idx > 1 && n(idx-1) > 0
n(idx-1) = 3;
end
if idx < length(n) && n(idx+1) > 0
n(idx+1) = 3;
end
end
end
% Now check if all three elements (n_indx), (n_indx+1), and (n_indx-1) are zero
for idx = n_indx
if idx > 1 && idx < length(n) && n(idx) == 0 && n(idx+1) == 0 && n(idx-1) == 0
recv_count = recv_count + 1; % Increment the receive counter
recv_count_1(idx) = recv_count;
recv_count_1(idx+1) = recv_count;
recv_count_1(idx-1) = recv_count;
end
end
end
end
disp('send_count_1:');
disp(send_count_1);
disp('recv_count_1:');
disp(recv_count_1);
Hope that helps!

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