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# How can I fix the error in the given below program?

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Manoj el 8 de Feb. de 2018
clc;
clear all;
close all;
disp('***** y^2 = x^3 + ax + b mod p *****');
n=input('insert prime p: ');
a=input('insert coefficient a: ');
b=input('insert coefficient b: ');
x=[0:n-1];
y=[0:n-1];
figure
for i=1:n
for j=1:n
if rem((vpi(y(j))^2)-(vpi(x(i))^3)-a*(vpi(x(i)))-b,n)==0
plot(x(i),y(j),'r o');
hold on;
end
end
end
grid;
hold off;
Maximum variable size allowed by the program is exceeded.
Error in sir (line 9)
x=[0:n-1];
##### 3 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Manoj el 8 de Feb. de 2018
n=730750818665451459101842416358141509827966272589 a=361221832160941494419407229753461201862246374272 b=346937986907100912443917613080678747571038706015
Star Strider el 8 de Feb. de 2018
Using double-precision representation, ‘x’ would then require 5.8460e+048 bytes.

### Respuestas (1)

Jan el 8 de Feb. de 2018
Editada: Jan el 8 de Feb. de 2018
input replies a double. According to the IEEE754 standard, a double has about 15 significant digits. Therefore you cannot input "730750818665451459101842416358141509827966272589" and expect it to be stored exactly. Even using the command vpi afterwards does not reconstruct the rounded digits magically. You can insert directly:
n = 7.30750818665451e47
etc. But then the shown algorithm cannot produce a meaningful output.
In addition creating a vector 0:7.3e47 needs 7.3e47*8 Bytes. There is no computer on the earth with 5.8*10^41 GigaByte RAM. And you want to create 2 of these vectors and process them in two nested loops over all elements?!
The universe is too small for this computation. Then most likely Matlab is, too. The message
Maximum variable size allowed by the program is exceeded.
is pure understatement.

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