[_m8fo]

AFAIK a single character is 1 byte.

An A4 sheet of paper is 62370 sqmm. A good printer can print a 1 sq mm character. Therefore, at a minimum you have 62370 bytes * 2, so basically 0.12 of a megabyte, which is pretty bad.

However, say you store something on the piece of paper that uses the second from midnight as a translator, such that the contents mean something else, depending on which second in the day you read it. Let's also say this translation is not using a hash type function, but is completely arithmetic and the formula to do this can be stored on the back side of the sheet in its entirety. This would take the 0.12MB /2 (you're not using both sides now) and multiply that by every second of the day, until the next, so 86400 seconds * 0.06MB = about 5 gigabytes.

Honestly I think you could get into the zetabytes. There are other factors you could use that I haven't considered:

1. Smell of the paper

2. Electrical charge

3. Feel of the paper (who said we're not printing in three dimensions)?

4. Taste (you could use a single piece of paper formed from different types, which have unique tastes).

I think the main limitation is as you add more factors to increase the compression you increase the complexity and time in which it takes to decompress, or get the information back. I'm sure there's some sort of law on this. Let's add in the orientation, in degrees of the piece of paper as well. Let's say you can reliably use all 360 degrees to permute the existing formula. Now you have 5 gigabytes * 360 = 1800 gigabytes. Let's just call this 2 terabytes.

[odonnellryan]

Even with such a function, aren't there limits to how much sensible data can be stored? Something something Claude Shannon something something. I forget.

Otherwise you'd have infinite storage capacity, on any medium, which is very obviously impossible.

Why infinite? Well, there are infinite numbers. If your algorithm was just "keep applying f(i,j) by an increasing integer to get the next page of the data" then yeah you've actually just discovered infinite storage.

This is literally a compression problem, isn't it? :)

Edit: http://mathforum.org/library/drmath/view/65726.html

Something I came across a long time ago.

[_m8fo]

Indeed. However, the hard limit here is the number of "discernible" and distinct characters that can fit on the page. The whole second thing is really just a way of permuting the existing data. So I'd say there would be infinite data if the amount of characters that could fit on the page were also infinite. Applying the function really was just a way to represent the mechanism that gave you each combination. The whole seconds from midnight thing may have been needlessly convoluted.

Though, for that portion of my comment, it would've been easier to simply say the information is approximately equal to the amount of characters that fit on a page and all combinations of its arrangement.

Of course, if the actual "processing" of the information lies outside of the paper, I feel like that's kind of cheating. What do you think?

[odonnellryan]

I don't think it'd be cheating, just like using programs to compress files isn't cheating.

However, there's two ways to look at this.

1) designing "some kind of data" specifically for this problem.

2) a general-purpose solution, where you could could put any data on the paper.

You could hit some obscenely-high number for 1), using some tricks or whatever.

But 2 probably has some sensible solution on the order of MB.

Example: if our function is as simple as "raise each number in the sequence to the next" we'd get some obscenely-large number, and we can put that function right on the page.

But, finding an obscenely large number representing some kind of data that actually means something, then coming up with a rule like that to reduce it?

Anyway, my argument would be: no, having an external compression algorithm isn't cheating, but formulating your data to fit the problem is.

Anyway, there exists no general-purpose compression algorithm, so compression would largely be out of #2, unless we're taking a subset of the problem: "how many English words can fit..." which of course we can come up with a good compression algorithm for (I think!) which would make it work.