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How to make the code efficient?

조회 수: 2(최근 30일)
Noor Bano
Noor Bano 2021년 6월 8일
답변: Steven Lord 2021년 6월 8일
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

답변(3개)

David Hill
David Hill 2021년 6월 8일
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)];
  댓글 수: 4
David Hill
David Hill 2021년 6월 8일
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

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Steven Lord
Steven Lord 2021년 6월 8일
Are you trying to find the points on an elliptic curve over F_n?
You could probably try a variant of the naive approach given on this Wikipedia page.

Joseph Cheng
Joseph Cheng 2021년 6월 8일
You can do things all at once cine you're not dependent on previous values.
n=4
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])
3 0 3 0 3 2 3 2
disp(['time original:' num2str(time1) 's'])
time original:0.008312s
disp(['time new:' num2str(time2) 's'])
time new:0.016845s
disp(['time delta:' num2str(time1-time2) 's'])
time delta:-0.008533s
  댓글 수: 2
David Hill
David Hill 2021년 6월 8일
Look at my code.

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