Using julia software with extremely large numbers

[enzocreti]

I am trying to factor 2^200 with Julia software. So I tried to write factor(2^200) but the result is 0 which is clearly wrong. How does it work BigInt in Julia for handling extremely large numbers? Have I to install some extra-package?

[Batalov]

Originally Posted by enzocreti

I am trying to factor 2^200 ...

Good luck with that!

That's an extremely large number. It has like more than sixty digits in it. It's beyond capabilities of modern computers.

[paulunderwood]

Originally Posted by enzocreti

I am trying to factor 2^200 with Julia software. So I tried to write factor(2^200) but the result is 0 which is clearly wrong. How does it work BigInt in Julia for handling extremely large numbers? Have I to install some extra-package?

According to the documentation, it looks like you have to cast it to BigInt thusly:

factor(BigInt(2^200))

There might be another way to factor the number

[ET_]

enzo, 2^200 = 2*2*2*2*2*2*.......*2*2*2
That is, 200 times.
So 2^200 factors like 200 times 2.

[henryzz]

Originally Posted by paulunderwood

According to the documentation, it looks like you have to cast it to BigInt thusly:

factor(BigInt(2^200))

There might be another way to factor the number

Would that work or would it try to calculate the 2^200 before converting to BigInt? This wouldn't work in a lot of languages.

[danaj]

Henry is right and I have tested. big(2^200) gives 0. big(2)^200 will give you what you want. Hence factor(big(2)^200).

With Julia v0.5 and later, factor is in the Prime package, you need to run Pkg.add("Primes") if it isn't already installed. Then

julia> using Primes

julia> factor(big(2)^2000 * big(3)^1000 * 57 * 25)
2^2000 ⋅ 3^1001 ⋅ 5^2 ⋅ 19

julia> factor(big(2^2000))
0

[enzocreti]

Originally Posted by danaj

Henry is right and I have tested. big(2^200) gives 0. big(2)^200 will give you what you want. Hence factor(big(2)^200).

With Julia v0.5 and later, factor is in the Prime package, you need to run Pkg.add("Primes") if it isn't already installed. Then

julia> using Primes

julia> factor(big(2)^2000 * big(3)^1000 * 57 * 25)
2^2000 ⋅ 3^1001 ⋅ 5^2 ⋅ 19

julia> factor(big(2^2000))
0

Ok thank you, I and Peter already managed to use it, but we realized that the software is too much slow for our purposes. If you want, you can always help us to find new ec-primes. Thanks!

[enzocreti]

Originally Posted by danaj

Henry is right and I have tested. big(2^200) gives 0. big(2)^200 will give you what you want. Hence factor(big(2)^200).

With Julia v0.5 and later, factor is in the Prime package, you need to run Pkg.add("Primes") if it isn't already installed. Then

julia> using Primes

julia> factor(big(2)^2000 * big(3)^1000 * 57 * 25)
2^2000 ⋅ 3^1001 ⋅ 5^2 ⋅ 19

julia> factor(big(2^2000))
0

Do you think that with Amazon ec2 virtual machine I can reach exponent 1 million in the search of ec-primes? Primes of the form (2^k-1)*10^d+2^(k-1)-1?

[Batalov]

yes.
just go on and rent one.
the proof of the pudding is in the eating.