Had an idea - looking for a math buff to tell me if it's possible with today's technology.

mikea mikea at mikea.ath.cx
Fri May 20 14:48:18 CDT 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.
> >
> >      -- Brett
> 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?

This depends strongly on the size of the number expressing "position xyz".
Pi is infinitely long, so there is no guarantee that for some random string
which can be found starting at "position xyz" in, say, the binary, decimal,
or hexadecimal expansion of pi, xyz can be expressed in fewer than 10000
(or indeed any fixed number N) bits.

-- 
Mike Andrews, W5EGO
mikea at mikea.ath.cx
Tired old sysadmin 




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