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