Why this function infinity loop

10 visualizaciones (últimos 30 días)
Nguyen Huy
Nguyen Huy el 7 de Mayo de 2021
Comentada: Walter Roberson el 8 de Mayo de 2021
%It give an integer x which contains d digits, find the value of n (n > 1) such that
%the last d digits of x^n is equal to x. Example x=2 so n=5 because 2^5=32 and last
%digits of 32 is 2
%So why this take infinity loop
function n = bigNumRepeat(x)
d = length(num2str(x))
y=2
n=2
while 1
y = y*x
y = num2str(y)
y = str2num(y(end - d + 1:end))
if y==x
break
end
n=n+1
end
end

Respuestas (2)

John D'Errico
John D'Errico el 7 de Mayo de 2021
Because if your powers grow larger than 2^53-1 (i.e., flintmax), the number is no longer expressable exactly as an integer. All of those operations with str2num will produce garbage at some point. You CANNOT do this using doubles if the number grows too large.
  5 comentarios
Nguyen Huy
Nguyen Huy el 8 de Mayo de 2021
i tried this code,but it only true with x from 1-10,i dont understand why it fail
Walter Roberson
Walter Roberson el 8 de Mayo de 2021
x = randi([11 99])
x = 93
d = length(num2str(x))
d = 2
y = x;
n = 1;
flag = true;
modulus = 10^d;
while flag && n <= 1000
n = n + 1;
y = mod(x*y,modulus);
flag = y ~= x;
end
if flag
fprintf('not found in 1000 iterations\n')
else
n
end
n = 5
Note that x is not relatively prime with 10^d then there might not be a solution.

Iniciar sesión para comentar.


Walter Roberson
Walter Roberson el 7 de Mayo de 2021
Editada: Walter Roberson el 7 de Mayo de 2021
y=2
y = y*x
That does not give you x^n, it gives you 2 * x^n . You should initialize y = 1
  2 comentarios
Nguyen Huy
Nguyen Huy el 7 de Mayo de 2021
I tried y=1,but it only work when you know n.In this situation i dont know what is n,so if i take y=1 it will stop at the first loop because x^1=x and the condition is true,the loop is end
Walter Roberson
Walter Roberson el 7 de Mayo de 2021
then initialize y=x

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by