How to make the code efficient?

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

Answer 1

[David Hill]

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)];

[David Hill]

for n=1000, the code above is 180 times faster.

[David Hill]

for n=10, the code above is 1.25 times faster.

[David Hill]

How large is your n? n cannot be much larger than 2e4 or else the matrices get too large (too much memory).

[David Hill]

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

Answer 2

[Joseph Cheng]

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

[Joseph Cheng]

hm... not sure why the time improvement above is worse but locally it appears much better

[David Hill]

Look at my code.

Answer 3

[Steven Lord]

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.