I am working on a matrix on Gate Level Delay Computing-Commom Subexpression Elimination (GLDC-CSE)

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Rutika Titre 2015-11-5

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评论： Rutika Titre 2015-11-22

my matrix is

mat=[0 0 0 0 0;0 0 -1 0 0;0 0 -1 -1 -1;0 0 -1 0 -1;-1 -1 0 -1 -1]

I have written code for CSE, I have to apply GLDC now. It consist of an algorithm in which suppose the row is ar=[0 0 -1 0 0]

I have to sort the row after sorting [-1 0 0 0 0]
then have to take two positive delay values here it is d1=0 and d2=0
The initial delay will be di=-1 and finaldelay=max(d1,d2)+1.
then repeat repeat steps 2 to 4 till u get one positive value.

Explation: after sorting [-1 0 0 0 0]; d1=0 and d2=0; so di=-1 and finaldelay=max(0,0)+1=1.row[-1 -1 1 0 0]

Again sorting [-1 -1 0 0 1]; d1=0 and d2=0; di=-1 finaldelay=max(0,0)+1=1.so[-1 -1 -1 1 0]

again sorting [-1 -1 -1 0 1];d1= 0 and d2=1; di=-1 finaldelay=max(0,1)+1=2 so [-1 -1 -1 -1 2].

I am not able to extract two positive elements and replace it with -1.

Any kind of help will help me.

Thanks!!!

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Walter Roberson 2015-11-12

在 MATLAB Online 中打开

"2. then have to take two positive delay values here it is d1=0 and d2=0"

But 0 is never a positive delay value.

Rutika Titre 2015-11-22

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Hello sir, I have written this code...it is working partially.I just want your help how should I go back to the starting of the loop. I have read continue and while loop these are the two things i can use.where to use continue to start the loop again or while loop.Help plz

x=[0 0 -1 0 0]; s=sort(x); T=zeros(size(s)); for ii=1:numel(s) if s(ii)==-1 T(ii)=-1; continue; end % ii=ii+1; if s(ii)==0 d1=0; T(ii)=-1; end ii=ii+1; if s(ii)==0 d2=0; difi=max(d1,d2)+1; T(ii)=difi; break end

    if s(ii)>0
        d3=s(ii);
        fidi=max(d2,d3)+1;
        T(ii)=fidi;
        break
    end
end
R=sort(T);
ii=ii+1;

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Answer 1

Walter Roberson 2015-11-12

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在 MATLAB Online 中打开

Your code has

d1=ar(ar(j:j+1)>=0);

You are examining two positions, out of which 0, 1, or 2 values will be >= 0. You are selecting those elements into d1.

You take d3=max(d1) . If the vector is length 1 or 2 then that is okay, you would get a scalar that contained the maximum. But if the vector was of length 0, then max() of the empty vector will be empty. And then you do difi=d3+1 . If none of the d1 values were >= 0 then d3 came out empty, and empty + 1 is empty. You then try to go around storing that emptiness into real locations...