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0 pointsutopcell7y ago0 comments

If 'x' marks the notch of a rod of length "1", [0---------x------------1], then the implied ratio is not x/(1-x), but x/1 (so the ratio is always < 1.) Even so, your question could be "what is the information content of 1/7" (the presumed implication being that 1/7, while periodic, has an infinite decimal representation.)

But that is not the direction we are interested in. We would like, given a message, such as "l-o-v-e", or 12-15-22-05, or 0.12152205, to figure out what is the ratio that uniquely specifies it. As we can only mark one notch, we can create "only" h ~= 10^35 ratios, or represent h unique messages. We know how to distinguish between h unique elements with log(h) bits (we just enumerate them from 1 to h and write that number down in binary.)

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jtrip7y ago

Let's assume that the rod is 1 m in length, and say I wrote a book that is perfectly represented by the number 0.abcdefg...yz. If that number is perfectly represented by one of the ratios in set 'h', then have I not stored more infromation than 15 bytes?

As for x/(1-x), why not? And why limit ourselves to a 1 m rod? Why not a 22 m rod with a 7 m notch? I could then define the method of decoding the information via (Rod length)/(notch length). The I'd have 'infinite' information in the form of expression of pi.

My main issue with the parent comment is that they imply only 15 bytes of data could be stored via this method. I think that's prespoterous as the number of ratios my be only 15 bytes, the ratios themselves can have any possible size.

It becomes more a game of probability rather than that of exact numbers. Will you find the right number, from set 'h', that matches exactly what you wanted to say?

    0.abcd...xyz = [(notch length=alpha*plank length)/(rod length=beta*plank length)]

where alpha and beta are just any variables that you play with until you solve the equation.

utopcellOP7y ago

A 22m rod with a 7m notch is perfectly ok. You'd then be encoding 7/22. As others have pointed out, the rod could span the known universe and it would still not matter.

Say you found a great message in the representation of 1/7. Weird, since it is a rational so if its representation is infinite, its periodic (you can't write down 1/π or 1/e for example, as these are irrationals.)

Excited you found that message, you want to put your notch exactly at 1/7 on the rod to celebrate it.

But you can't. Your desired notch position will fall between two possible notches, spaced one planck distance apart, and you'll have to pick one of the two.

And when you do, you truncate your message to 15 bytes worth of information.

rcxdude7y ago

Not really. You have stored something which has a representation of more than 15 bytes, but you will struggle to find a way to store most things longer than 15 bytes in this way. (everything can be transformed into a representation which is infinite in size anyway, for example by using an irrational base). Information really comes down to the number of possible values you can distinguish.

If you do the ratio thing you have described you will find that alpha and beta are many, many times the size of the observable universe for something like a book. If you allow the length of the rod itself to also contribute to the information, you have added another symbol so you can store more than 15 bytes, but this doesn't even double the amount of bytes you can store.

jtrip7y ago

Total Human knowledge ~ 250 EB

Assuming all of it can be described as ascii characters, (ratio of.12152205 could be read as 12-15-22-05, or “l-o-v-e.”) then let's assume 4 numbers will be required to encode one letter. That gives us, in our limited system, 4 decimal bits to a byte (we really can't limit ourselves to just ratios that are sufficiently large and only made of 1s & 0s).

So, all human knowledge is, in our system,

    250*(1024^6)=288230376151711740000 bytes.

As we assumed that all of it can be described as ascii characters, and in the standard system 1 byte holds one character, there are now 288230376151711740000 characters. Expressing these many characters in our 4-byte decimal numbers will require a ratio with

    (288230376151711740000*4)=1152921504606846976000

numbers in it. All the ratios with 1.15 * 10^21 numbers will be the candidates which can be used to store all of humanity's contemporary knowledge.

Now, as I said earlier, the ratios may have an impressive number of numbers in them, expressed in a decimal system, and there are an infinite number of them on the number line itself, that does not mean all of those are available for use. We are limited to ratios derived from lengths which are multiples of the plank length. Assuming, for a particular rod and it's notch, there is a set 'h' that contains all the possible ratios. We will be limited to such ratios only to find our matching ratio, the ratio through chance of cosmic infinity, or not. If not, then we have to increase/decrease the design length of the rod and the notch to change the set 'h', and hope there is a number we are looking for.

How many such ratios, of the required length of 1.15 * 10^21 numbers, would exist if derived solely through the ratios, is unknown and wholly dependent on the information that has to be encoded. The longer the data, the higher the probability of not finding the right number. As you put it, there will be 15 bytes of choice, or in a one meter rod there will be 10^32 number of choices of ratios to play with. If you doubled the length of the rod, you would have twice the amount of choice, as so on. Again,

    0.abcd...xyz = [(notch length=alpha*plank length)/(rod length=beta*plank length)]

where alpha and beta are just any variables that you play with until you solve the equation.

And again, you could find the ratio you are looking for, it will be a probability game.

Please read what I wrote carefully and respond point by point.

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jtrip7y ago

It becomes more a game of probability rather than that of exact numbers. Will you find the right number, from set 'h', that matches exactly what you wanted to say?

    0.abcd...xyz = [(notch length=alpha*plank length)/(rod length=beta*plank length)]

where alpha and beta are just any variables that you play with until you solve the equation.

utopcellOP7y ago

A 22m rod with a 7m notch is perfectly ok. You'd then be encoding 7/22. As others have pointed out, the rod could span the known universe and it would still not matter.

Excited you found that message, you want to put your notch exactly at 1/7 on the rod to celebrate it.

But you can't. Your desired notch position will fall between two possible notches, spaced one planck distance apart, and you'll have to pick one of the two.

And when you do, you truncate your message to 15 bytes worth of information.

rcxdude7y ago

jtrip7y ago

Total Human knowledge ~ 250 EB

So, all human knowledge is, in our system,

    250*(1024^6)=288230376151711740000 bytes.

    (288230376151711740000*4)=1152921504606846976000

numbers in it. All the ratios with 1.15 * 10^21 numbers will be the candidates which can be used to store all of humanity's contemporary knowledge.

    0.abcd...xyz = [(notch length=alpha*plank length)/(rod length=beta*plank length)]

where alpha and beta are just any variables that you play with until you solve the equation.

And again, you could find the ratio you are looking for, it will be a probability game.

Please read what I wrote carefully and respond point by point.

1 more reply

j / k navigate · click thread line to collapse