CODEWITHSUNDEEP
matlablargenumber
user4569486
user4569486

Reputation:

i use factor(n) in matlab but it don't work for numbers like ((10^11-1)/11),for same number what can i do?

I use factor(n) in matlab but it don't work for numbers like ((10^11-1)/11) - what can I do? My source is here.

m=input('Enter your number: ');
i=0;

while (i<m)
    if(isprime(i))
     % sum=factor((10^(i-1)-1));
        sum=factor((10^(i-1)-1)/i);
        disp(sum);
    end
    i =i+1;

end

but for large n it returns errors!!

>> FactorGen

Enter your number: 45

 3     3

 3     3    11

 3     3    11   101

 3     3     3     7    11    13    37

       3           3          11          41         271        9091

       3           3           3           7          11          13          37         101        9901

Error using factor (line 26) When n is single or double, its maximum
allowed value is FLINTMAX.

Error in FactorGen (line 7)  sum=factor((10^(i-1)-1));

I want the function factor((10^(i-1)-1)) to work for same number. How can I solve my problem?

Upvotes: 2

Views: 407

Answers (1)

brainkz
brainkz

Reputation: 1345

I think this can be partially alleviated by converting your large number into uint64 format. For R2014b maximum integer that can be handled is:

n = intmax('uint64')
n = 1.8447e+19

While the maximum double that can be handled is:

n = flintmax('double')
n = 9.0072e+15

This can be verified by simple example. Let's use factor on the number larger than flintmax. First, try double:

factor(10^16)
Error using factor (line 26)
When n is single or double, its maximum allowed value is FLINTMAX.

Now, we try uint64:

factor(uint64(10^16))
ans =  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

So, using factor(uint64(x)) increases your limit by several orders. Indeed, there's no sense in using double format for factor function.

By the way, since using factor on large numbers may freeze your system, I suggest using this function:

function f = Factorize(n)
i = 0;
while mod(n,2) == 0
    i = i + 1;
    n = n/2;
    f(i) = 2;
    disp([2,n])
end
q = round(sqrt(double(n)));
q = q + 1 - mod(q,2);
for j = 3:2:q
    while mod(n,j) == 0
        i = i + 1;
        f(i) = j;
        n = n/j;
    end
end
if n > 2;
    i = i + 1;
    f(i) = n;
end

It is much faster for large numbers and does not overload the system at large n Hope that helps

Upvotes: 1

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