*only* be 2^128 possible texts

only? Only!? Do you have any idea how big 2^128 is?

2^128 = 340,282,366,920,938,463,463,374,607,431,768,211,456
> 3*10^38

Which is bigger than the number of cups of water in all the oceans (6*10^21)

Bigger than the distance from one end of the universe to the other in inches...(2*10^28)

Bigger than the volume of the sun in cubic inches...(8*10^31)

Bigger than the area of the galaxy in square miles...(3*10^35)

Approaching the number of atoms in our atmosphere.....(2*10^44)

(from bignum)

Perhaps a secondary check is in order, but I'd hardly use 'only' when talking about 2^128 hash buckets.

**Update:**

Ok, lets play with the numbers some more:

Let's assume perlmonks has 300,000 nodes (3*10^{5} ) and has 3*10^{38} buckets in its hashing algorithm. The ratio of
nodes/buckets is 3*10^{5} : 3*10^{38} or **1 : 10**^{33}.

Now, consider this lottery where you pick six different numbers from 1-49.
Get all six right and you win the jackpot. As the page
above notes, the chances of winning with one ticket are:

1 : 13,983,816 ( (49*48*47*46*45*44)/(6*5*4*3*2*1) ) or about:

1 : 10^{7}

Lets buy one ticket a week for four weeks... odds of winning *all* four lotteries with our four tickets are:
1 : (10^{7})^{4} or 1 : 10^{28} .

That *still* doesn't get you there... after winning your four lotteries, we'll take you to one of the new huge NFL stadiums being built, and you have to gamble all your winnings on picking a specific, randomly-chosen seat (1 : 10^{5})

So the chances of my next post colliding with a node already in the database (1:10^{33} ) are about the same as you winning four lotteries on four tickets, then picking the single correct seat out of a gigantic stadium (1 : 10^{28}*10^{5})

-Blake