Had an idea - looking for a math buff to tell me if it's possible?with today's technology.
Brett Frankenberger
rbf+nanog at panix.com
Fri May 20 19:43:45 UTC 2011
On Fri, May 20, 2011 at 09:34:59AM -1000, Paul Graydon wrote:
> On 05/20/2011 08:53 AM, Brett Frankenberger wrote:
> >On Fri, May 20, 2011 at 06:46:45PM +0000, Eu-Ming Lee wrote:
> >>To do this, you only need 2 numbers: the nth digit of pi and the number of
> >>digits.
> >>
> >>Simply convert your message into a single extremely long integer. Somewhere,
> >>in the digits of pi, you will find a matching series of digits the same as
> >>your integer!
> >>
> >>Decompressing the number is relatively easy after some sort-of recent
> >>advances in our understanding of pi.
> >>
> >>Finding out what those 2 numbers are--- well, we still have a ways to go
> >>on that.
> >Even if those problems were solved, you'd need (on average) just as
> >many bits to represent which digit of pi to start with as you'd need to
> >represent the original message.
> Not quite sure I follow that. "Start at position xyz, carry on for
> 10000 bits" shouldn't be as long as telling it all 10000 bits?
I don't know about "should", but it *will* be when "xyz" is greater
than 2^10000 (or about 10^3000). Your intuition is probably telling
you that "xyz" won't likely be a 3000 digit (or longer) number, but if
so, your intuition is wrong.
-- Brett
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