# How to do this efficiently?

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S. David on 9 Aug 2014
Commented: dpb on 11 Aug 2014
Hello all,
I have this variable a[k,m]=max(k*Ts+taup,m*Ts+tauq) for k,m=0,1,...,N-1. I want to find the matrix A where [A]_{k,m}=a[k,m] efficiently. To do so, I define two matrices RowInc and ColInc as
RowInc =(0:N-1)'*ones(1,N);
ColInc =transpose(RowInc);
Then I write
for pp=1:Np
for qq=1:Np
A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));
end
end
Does this give me what I want?
Thanks
S. David on 9 Aug 2014
When you shade a part of your writing and then press "{}code" in the toolbar the shaded writing will appear in gray, right?
The (k,m)th element of A is a function of k and m for k,m=0,1,...,N-1.

dpb on 9 Aug 2014
...I want to find the matrix A where [A]{k,m}=a[k,m]..._
A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));
will end up w/ just a single value for A at the last loop of pp and qq since it overwrites the previous A each iteration. But, I don't believe it does what you want, anyway.
Should be simply
A(A==a);
if I understand the query correctly.
dpb on 9 Aug 2014
IA, I couldn't figure out what his last comment said (and little of the rest) if after "I have this variable a[k,m]..." there isn't an array a.
Guess I'll leave the field bloodied on this one...

dpb on 10 Aug 2014
OK, from the loop solution one can write
[x,y]=ndgrid(0:N-1,0:N-1);
A=max(Ts.*x+taup,Ts.*y+tauq);
trading memory for the loop. The loop solution could be simplified since there's only a dependence upon kk for the one term and mm for the other, they could be precomputed outside the loops. Preallocating also would help, of course. Not sure how the timings would come out in the end.
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dpb on 11 Aug 2014
So then what was the question???