Program to calculate 2^128 accurately?

[Solo177]

Anyone know of a such a program that can do that calculation? Or, better yet, does anyone know how to write such a program in VB/VC++? Thanks.

-Mike

 

[Adrian Tung]

Well, the highest integer value that VC++ can go would be 2^32 - 1 using a DWORD. Supposedly you could write a program which uses five DWORD values to get 2^128, but you'll have to write your own class and methods to support the mathematical operations for this new data type.

On the other hand, you could try writing a BigInt class (which I think is a hot university programming question) which supports multiplication or indices.


🙂atwl

 

[randomlinh]

NOO.. not bigint!!! aaahhh!!!! heh.. sorry.. ap comp sci teacher trying to teach big int.. ahhahahaha...

 

[thEnEuRoMancER]

Is 3.40282366920938e+038 (according to Matlab) accurate enough for you?

 

[Becks2k]

3.4028236692093846346337460743177e+38 according to calculator hehe

 

[Becks2k]

2^106=81129638414606681695789005144064
you could just do it by hand from there

 

[thEnEuRoMancER]

Maple can calculate anything to an arbitrary number of digits.

 

[Moohooya]

340282366920938463463374607431768211456

My quick and dirty program follows

Moohoo

#include <iostream.h>

struct BigInt
{
enum { parts = 5 };
unsigned part[parts];

BigInt (void)
{
for (int p = parts -1; p >= 0; --p)
part[p] = 0;
}

unsigned Divide (unsigned rhs)
{
unsigned __int64 remainder = 0;
for (int p = parts -1; p >= 0; --p)
{
unsigned __int64 temp = remainder * 0x100000000ui64 + part[p];
remainder = temp % rhs;
part[p] = unsigned (temp / rhs);
}

return unsigned (remainder);
}

operator bool (void) const
{
for (int p = parts -1; p >= 0; --p)
if (part[p])
return true;

return false;
}
};

main ()
{
BigInt b;
int digits[50];
b.part[4] = 1;
for (int i = 0; b; ++i)
digits = b.Divide (10);

while (--i >= 0)
cout << digits;

cout << endl;
return 0;
}

 

[zzzz]

what? 2^128 is 2^128.
now solve this

x^(x^x)= (x^x)^x analytically..

 

[Argo]

3849032842038432904829304823048239429348290348932420384920384092571476206829058349205820957267283583429058492358239067258349058329058320852367582394583258342905


I just did it in head.... 😉😉😉😛😛😛😛😛😛😛

 

[thEnEuRoMancER]

zzzz:

x^(x^x)=(x^x)^x=x^(x^2) /log

log (x^(x^x))= log (x^(x^2))

x^x log x = x^2 log x

first solution: log x = 0 -> x = 1 , check: 1^(1^1) = (1^1)^1

second solution:

x^x = x^2 /log

x log x = 2 log x

assuming we already know x = 1 is one solution it's safe to divide both sides by log x and get

x = 2

So we have two solutions:

x1 = 1, x2 = 2

edit: and of course, the third solution x3 = 0 . If I remember correctly, 0^0 = 1 by definition.

 

[Installer]

Argo, what an odd number to have in your head. 😉

 

[HomerSapien]

340282366920938463463374607431768211456

is what mathematica got

 

[Argo]

My head is full with odd things 🙂

 

[Mday]

yeah argo, that is an odd number =) it ends in a 5 =)

lol

you need a lot of accuracy to calculate 2^128 correctly. one can easily calculate that with a simple program... ;-)

 

[Argo]

Oops, looks like I got an overflow...