Is there an algorithm or number such that we could basically pirate data from it by saying “start digit 9,031,643,679 with length 5,345,109 is an MP4 of Shrek”? Something that we could calculate in a day or less?
Similarly: if you write a program to randomly run through all the combinations of pixels on a decently large screen (say, 1080p) you will eventually see every important question and answer that can be expressed on a screen.
The short answer is no, and even if we could, the digit index you’d start at would have a larger binary representation than the actual data you were trying to encode.
An example I found: the string of digits 0123456789 occurs at position 17387594880. In this case, it took 11 digits to describe where to find a 10-digit number.
So I think such an algorithm would technically work, but your “start digit” would be so large it would use more data than just sending the raw file data. Not to mention the impossible amount of computing power needed.
Is there an algorithm or number such that we could basically pirate data from it by saying “start digit 9,031,643,679 with length 5,345,109 is an MP4 of Shrek”? Something that we could calculate in a day or less?
Similarly: if you write a program to randomly run through all the combinations of pixels on a decently large screen (say, 1080p) you will eventually see every important question and answer that can be expressed on a screen.
The short answer is no, and even if we could, the digit index you’d start at would have a larger binary representation than the actual data you were trying to encode.
An example I found: the string of digits 0123456789 occurs at position 17387594880. In this case, it took 11 digits to describe where to find a 10-digit number.
So I think such an algorithm would technically work, but your “start digit” would be so large it would use more data than just sending the raw file data. Not to mention the impossible amount of computing power needed.
Could we already do this by leveraging the Library of Babel?
Genuinely asking, I’m not really sure.
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