Why this function infinity loop

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Nguyen Huy
Nguyen Huy 2021-5-7
评论： Walter Roberson 2021-5-8
%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

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回答（2 个）

John D'Errico
John D'Errico 2021-5-7
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.
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Nguyen Huy
Nguyen Huy 2021-5-8
i tried this code,but it only true with x from 1-10,i dont understand why it fail
Walter Roberson
Walter Roberson 2021-5-8
Ran in:
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.

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Walter Roberson
Walter Roberson 2021-5-7
编辑：Walter Roberson 2021-5-7
y=2
y = y*x
That does not give you x^n, it gives you 2 * x^n . You should initialize y = 1
See also https://www.mathworks.com/matlabcentral/fileexchange/932-big-x-y-modulo-function and see the symbolic powermod() function
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Nguyen Huy
Nguyen Huy 2021-5-7
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

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