Hello,
I have this Matrix
2 4
4 11
5 7
5 10
7 12
8 10
11 12
2 8
I want the output to be something like this
2 4 11 12 7 5 10 8 2
So basically start with the first element 2 and build the connection. since 2 is connected to 4, and 4 is connected to 11, 11 is connected to 12, 12 is connected to 7 and so on. I am not sure how to program this in a simple way.
Another example to make it clear
1 2
2 3
3 6
6 1
to be changed to
1 2 3 6 1
Hope the question is clear.

3 件のコメント

Image Analyst
Image Analyst 2015 年 7 月 26 日
I don't understand your algorithm. OK, 11 is "connected" or adjacent to 12, but an 11 is also next to an 8, 10, 2, 5, and 7. Why are you picking only the 12 to come next in your row vector output?
Image Analyst
Image Analyst 2015 年 7 月 26 日
編集済み: Image Analyst 2015 年 7 月 26 日
OK, it looks like you're considering only numbers on the same row, then when a row is "used" you look for the number in a different row. What if the next number shows up in 5 rows? Which of the 5 rows do you pick next? What if there is a row that is not connected to anything prior? For example, what if row two had 991 and 999 in it? You'd never get to that row. There could be many such rows that are "unattached".
Hesham Ismail
Hesham Ismail 2015 年 7 月 26 日
@image analyst Thanks these condition will be consider as a new row in the final output

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 採用された回答

Azzi Abdelmalek
Azzi Abdelmalek 2015 年 7 月 26 日
編集済み: Azzi Abdelmalek 2015 年 7 月 26 日

0 投票

a= [2 4
4 11
5 7
5 10
7 12
8 10
11 12
2 8];
test=0;
out=a(1,: );
d=a(2:end,:);
while test==0
[ii,jj]=find(d==out(end),1);
if ~isempty(ii)
out(end+1)=d(ii,setdiff(1:2,jj));
d(ii,:)=[];
else
test=1;
end
end
out

2 件のコメント

Hesham Ismail
Hesham Ismail 2015 年 7 月 26 日
Thanks it works, is there a way to preallocate out variable in your code?
Azzi Abdelmalek
Azzi Abdelmalek 2015 年 7 月 26 日
It's possible to pre-allocate the variable out to size(a,1), then take what we have to take. I tried it, but it doesn't improve the speed.

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