How to make the code efficient?
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I have made the following code .
x=[0:n-1];
y=[0:n-1];
k = 0;
for i=1:n
for j=1:n
if rem(((y(j))^2)-((x(i))^3)-2*(x(i))-3,n)==0
k = k+1;
xy_mtx(k,:) = [x(i) y(j)];
end
end
end
I want to make it efficient, as efficient as it can be. Is this possible
回答（3 个）
David Hill
2021-6-8
n=10;
[x,y]=meshgrid(0:n-1);
idx=mod(y.^2-x.^3-2*x-3,n)==0;
xy_mtx=[x(idx),y(idx)];
4 个评论
David Hill
2021-6-8
for n=1000, the code above is 180 times faster.
David Hill
2021-6-8
for n=10, the code above is 1.25 times faster.
David Hill
2021-6-8
How large is your n? n cannot be much larger than 2e4 or else the matrices get too large (too much memory).
David Hill
2021-6-8
编辑：David Hill
2021-6-8
You could try the help memory suggestions to try to increase memory, or you could add nested for-loop and do the meshgrids in batches of 2e4 (will be slower). Also keep in mind floating point limitations.
n=1e6;
xy_mtx=[];
for k=1:50
for j=1:50
[x,y]=meshgrid(0+(k-1)*2e4:min(2e4-1+(k-1)*2e4,n-1),0+(j-1)*2e4:min(2e4-1+(j-1)*2e4,n-1));
idx=mod(y.^2-x.^3-2*x-3,n)==0;
xy_mtx=[xy_mtx;x(idx),y(idx)];
end
end
You can do things all at once cine you're not dependent on previous values.
n=4
x=[0:n-1];
y=[0:n-1];
k = 0;
%%Orig_code
tic
for i=1:n
for j=1:n
if rem(((y(j))^2)-((x(i))^3)-2*(x(i))-3,n)==0
k = k+1;
xy_mtx(k,:) = [x(i) y(j)];
end
end
end
time1=toc;
tic
%%do everything at once
[X Y] = meshgrid(x,y); %generate combinations of x and y
REMmat = rem((Y.^2)-(X.^3)-2*(X)-3,n); %perform all rem calculations at once
[indexies]=find(REMmat==0); %find which index has a rem of 0
xy_mtxM = [X(indexies) Y(indexies)]; %put only the X and Y comb. where rem above is -
time2=toc;
disp([xy_mtx xy_mtxM])
disp(['time original:' num2str(time1) 's'])
disp(['time new:' num2str(time2) 's'])
disp(['time delta:' num2str(time1-time2) 's'])
2 个评论
Joseph Cheng
2021-6-8
hm... not sure why the time improvement above is worse but locally it appears much better
David Hill
2021-6-8
Look at my code.
Steven Lord
2021-6-8
1 个投票
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2021-6-8
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