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
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回答 (2 件)
John D'Errico
2021 年 5 月 7 日
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 件のコメント
Walter Roberson
2021 年 5 月 8 日
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 年 5 月 7 日
編集済み: Walter Roberson
2021 年 5 月 7 日
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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