Why this function infinity loop
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%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
回答（2 个）
John D'Errico
2021-5-7
0 个投票
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 个评论
Nguyen Huy
2021-5-7
编辑：Nguyen Huy
2021-5-7
John D'Errico
2021-5-7
What numbers are you testing this on? Because you can easily overwhelm a double. That is, if d is as large as 8 or 9, this will still fail.
By the way, using str2num and num2str repeatedly is an extremely inefficient way to do any such computation. I would point out that mod does so very well. That is, mod(X,10^d) gives you the last d digits directly.
As you can see, this code works reasonably well.
d = 1;
x = 2;
y = x;
n = 1;
flag = true;
modulus = 10^d;
while flag
n = n + 1;
y = mod(x*y,modulus);
flag = y ~= x;
end
n
Howver, if I used larger values for d and x, it could fail at some point. In that case, I might be forced to use symbolic computations.
Nguyen Huy
2021-5-8
x = randi([11 99])
d = length(num2str(x))
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
Note that x is not relatively prime with 10^d then there might not be a solution.
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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