Roundation In Binary Problem

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Turgut
Turgut 2012 年 11 月 6 日
Hello everyone, I need a programme help. I think it will be obvious if I will examine it with an example.
Consider we have a matrix (6x6) in binary;
0 0 1 0 1 0
0 1 1 0 1 1
y= 1 0 0 0 1 0
0 1 0 1 1 0
0 0 0 1 0 1
0 0 0 0 0 1
I would like to make roundation. Means it will look the rows' one by one, if the last number is 1, it will add 1 to the number (in binary) and in the end it will return matrix 5x5 (for this example).
For this example:
0 0 1 0 1
0 1 1 1 0
y= 1 0 0 0 1
0 1 0 1 1
0 0 0 1 1
0 0 0 0 1
And it should do it for any matrix which is (nxn)
Thanks from now ;)
  2 件のコメント
Walter Roberson
Walter Roberson 2012 年 11 月 6 日
What code have you tried?
Image Analyst
Image Analyst 2012 年 11 月 6 日
I don't understand. I've never heard of "roundation". What does "Add 1 to the number in binary" mean? If it's binary (logical) adding 1 will still leave it at 1 - it will get clipped. And how does doing that chop off the last column?

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Matt J
Matt J 2012 年 11 月 6 日
編集済み: Matt J 2012 年 11 月 6 日
out = dec2bin( y(:,1:end-1)*2.^((n-2:-1:0).') + y(:,end) )-'0';
  1 件のコメント
Turgut
Turgut 2012 年 11 月 6 日
thank you so much it works perfect ;)

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