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