# How to create the contour (closed surface) utilizing matlab functions or loopings using the given Matrix under certain conditions

조회 수: 1 (최근 30일)
M.S. Khan 2019년 8월 14일
댓글: M.S. Khan 2019년 8월 20일
##### 댓글 수: 6이전 댓글 4개 표시이전 댓글 4개 숨기기
Guillaume 2019년 8월 15일
Ok, that makes it a bit clearer, I still don't understand why you say that in:
[ 0 0 0 0 0 3 0]
you say that the row has a closing 3 but not starting 3. Why couldn't you say it has a starting 3 but no closing 3?
So, it looks like you want each row and column to have either no 3 at all, or at least two 3s. So for the rows/columns with just one 3, any preference where the missing 3 should be added?
M.S. Khan 2019년 8월 15일
i want to generate a closed surface just like a circular surface. i want to create boundary for that surface by filling 3s on the boundary.
let me make it more clear. From the below matrix, there is an ouline which is not closed, i want to close it by fillings 3s.M
[ 0 0 0 3 3 0 0 0
0 0 3 0 0 0 3 0
0 3 0 0 0 0 3 0
3 0 0 0 0 0 0 3
3 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0
0 3 0 0 0 0 3 0
0 0 0 0 0 0 3 0
0 0 0 0 0 0 0 0
0 0 3 3 0 0 0 0 ]
if you see it, its some circular closed shape. i want to maintain the shape by filling the missing values of 3s in rows and columns so that i can get the interior of the shape.
Thanks and regards for advance for all help and cooperation.

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### 답변 (1개)

Dheeraj Singh 2019년 8월 19일
I understand that you want to fill the matrix such that the boundaries are filled circularly.
And for starting 3, it should be before the middle row or column and vice-versa for the closing.
So, for say matrix:
[ 0 0 0 3 3 0 0 0
0 0 3 0 0 0 3 0
0 3 0 0 0 0 3 0
3 0 0 0 0 0 0 3
3 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0
0 3 0 0 0 0 3 0
0 0 0 0 0 0 3 0
0 0 0 0 0 0 0 0
0 0 3 3 0 0 0 0 ]
You would like the output to be something like this:
[ 0 0 0 3 3 0 0 0
0 0 3 0 0 3 3 0
0 3 0 0 0 0 3 0
3 0 0 0 0 0 0 3
3 0 0 0 0 0 0 3
3 0 0 0 0 0 0 3
0 3 0 0 0 0 3 0
0 3 0 0 0 0 3 0
0 3 0 0 0 0 3 0
0 0 3 3 3 3 0 0]
You can do it by dividing the whole matrix into 4 quadrants similar to the circle.
The following code below implements it for when the number of columns is greater than number of rows.
a=zeros(8,10);
N=a(1,:);
M=a(:,1);
m1=N/2;
e1=m1+1;
m2=M/2;
e2=m2+1;
j=1;
for i=m2:-1:1
if a(j,i)~=3
a(j,i)=3;
end
j=j+1;
j=min([j,m1]);
end
j=e1;
for i=1:m2
if a(j,i)~=3
a(j,i)=3;
end
j=j+1;
j=min([j,N]);
end
j=N;
for i=e2:M
if a(j,i)~=3
a(j,i)=3;
end
j=j-1;
j=max([j,e1]);
end
j=m1;
for i=M:-1:e2
if a(j,i)~=3
a(j,i)=3;
end
j=j-1;
j=max([j,1]);
end
You can the modify the above code for the case where no of rows are more than columns.
##### 댓글 수: 1이전 댓글 -1개 표시이전 댓글 -1개 숨기기
M.S. Khan 2019년 8월 20일
Thanks Dheeraj Singh. i am tryiing to apply it on different matrices of different shapes.

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