How much information is stored inside a human? Not as much as you think. All you need is a mere 1.5 gigabytes to fit your entire genetic code. Veritasium did the math in his latest brain tapping video and cooked up that number using bits to understand the molecules that make up a person's genetic code.
Of course, we have a lot of cells in our body (around 40 trillion) and each of those cells contain the full 1.5 GB of our genetic code. So a real person has about 60 zettabytes (60 with 21 zeroes after) of information in total. That's huge. Veritasium says that in the year 2020, all the digital information in the world will only tally up to 40 ZB. So turns out, there's a lot of information necessary to make a person.
But! 99.9% of our genetic information is shared with everyone else on Earth. What makes us unique is much, much smaller than a ZB. In fact, it takes less than a megabyte to make a person different from the next.
So there it is. A reasonable 1.5 GB of information for our genetic code. A ridiculous 60 ZB flowing in all our bodies. And an embarrassingly tiny megabyte that makes us believe we're a unique snowflake.
(Score: 3, Insightful) by calzone on Tuesday July 08 2014, @01:38AM
The data stored in your brain is not accounted for here, which ALSO makes you different from other people. Probably accounts for a lot more of the difference than the megabyte of genetic drift.
Time to leave Soylent News [soylentnews.org]
(Score: -1, Offtopic) by Anonymous Coward on Tuesday July 08 2014, @01:53AM
A couple of white male ducks were totally gangbanging a female mallard today at the lake. It was pretty surreal, first one of the white male ducks started quacking really loudly and chasing after the mallard, but she only dashed through the water and didn't take flight -- which means she totally wanted it.
So the white duck overtakes her and forces himself on her, his tail flicking this way and that to get him keyed up with her index, and mating in nature is so savage anyway. His entire weight was on her and forced much of her body underwater as she gasped and quacked with the water level up to her neck.
But then, this other white male duck got jealous and wanted in on the fun, so at first he was jabbing her body with his neck out of anger that she was mating with another duck. He continued to jab at her, and then he felt emboldened enough to try to mounte her despite the fact that another duck had already. And the funny thing was, the aggressor male duck did not strike the other duck mating with the object of his affection, no -- he continued to strike the female duck with his beak as he too attempted to push aside his rival to mount the prize mallard.
So two white ducks jumping and flopping and flicking trying to flog the mallards' asses with their own, effectively space-docking their cloaca. A caretaker was there with two kids, and when she saw what was going on, she told the kids to look away and feed the ducks on the other side of the pier.
(Score: 2) by EvilJim on Wednesday July 09 2014, @01:06AM
I have no idea where you were going with this, but I just came here to ask the question slightly on-topic with your post... do gangbanging female pornstars hold significantly more data than the rest of us?
(Score: 3) by Nerdfest on Tuesday July 08 2014, @01:54AM
I was going to say the same. 1 MB of difference at boot time, but affected by chaos every moment afterwards.
(Score: 2) by jasassin on Tuesday July 08 2014, @05:56AM
Boot up meaning birth? How could we possibly measure the amount of information an adult mind has accumulated? I wonder if that's a question mankind could answer before it destroys itself.
jasassin@gmail.com GPG Key ID: 0xE6462C68A9A3DB5A
(Score: 2) by jasassin on Tuesday July 08 2014, @03:41AM
Very interesting. Hmmmm... Mod up.
jasassin@gmail.com GPG Key ID: 0xE6462C68A9A3DB5A
(Score: 2, Interesting) by mrchew1982 on Tuesday July 08 2014, @06:39AM
This number also ignores all of the epigenetic tagging that is semi-heritbale and may be much more important in our behavioral habits than memories. Epigenetic tags control all sorts of gene expression, even neurotransmitter levels.
(Score: 2) by frojack on Tuesday July 08 2014, @01:39AM
A gazillion cells, each with a replicate of the genome probably doesn't count as additional data. Its all the same data over and over again.
Publishing more copies than you can sell of a newspaper doesn't mean mankind is any more informed, or that the library of congress mushrooms over night.
No, you are mistaken. I've always had this sig.
(Score: 2) by egcagrac0 on Tuesday July 08 2014, @05:18AM
Right. Deduplication.
Of course, that's not entirely accurate - the cells don't replicate perfectly, every once in a while a mutation occurs. Encoding every mutation in every cell may take a bit more space, even if it's a unified diff.
As pointed out above, this doesn't encode the neural net's datastore.
You might be able to build a new set of basic hardware from the 1.5GB store, but that don't load the OS, and it sure don't get your vacation pictures back.
(Score: 1, Funny) by Anonymous Coward on Tuesday July 08 2014, @10:42AM
(does that make human body a RAID 4e13 ?~)
(Score: 2) by LoRdTAW on Tuesday July 08 2014, @01:31PM
No, it means Raid 1 with 4e13 storage nodes.
(Score: 2) by SlimmPickens on Tuesday July 08 2014, @01:39AM
I don't remember the number but I remember Kurzweil in one of his books comparing the size of the 'genetic code' being roughly comparable to the MS Word codebase of the time.
(Score: 2) by hoochiecoochieman on Tuesday July 08 2014, @09:38AM
And most likely it contains the same amount of bugs, either.
(Score: 2) by mhajicek on Tuesday July 08 2014, @03:43AM
That would include the types, masses, precise locations and inertial vectors, charges and spins etc. of every subatomic particle in the body. It basically depends on how much you weigh.
The spacelike surfaces of time foliations can have a cusp at the surface of discontinuity. - P. Hajicek
(Score: 1) by tftp on Tuesday July 08 2014, @04:04AM
That would include the types, masses, precise locations and inertial vectors, charges and spins etc. of every subatomic particle in the body
The same difference exists between two identical copies of a book. However we do not call those copies different, unless the discussion is about their subatomic physical properties, and not about the content.
(Score: 3, Interesting) by c0lo on Tuesday July 08 2014, @03:55AM
R u kidding? Do you realize how big 2^8388608=10^8388590 is? The most distant cosmic object ever observed [wikipedia.org] is a puny 13bil-light-years If you express this distance in Planck lengths, you'll need a precision of about 10^70.
If you want to express the positions all nucleons in the known universe (estimated maximally at 10^80 [wikipedia.org]) in a Cartesian ref with the Planck unit precision, you would need only 10^80*10^(70*3)=10^290.
Want to encode the evolution of the Universe with a precision of Planck time unit [wikipedia.org] and Plank length unit? Say 14b-year=7*10^54 plank-time-units=> need 10^(290+55)=10^345.
A 10^8388590 ? Mate, the entire evolution of the observable universe from BigBang to present, with maximum precision, occupies at most 360 bits from that one MB. Isn't it unique enough?
https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
(Score: 2, Interesting) by SparkyGSX on Tuesday July 08 2014, @06:39AM
That can't be right, that would imply there is only a total of 360 bits of entropy in the entire universe, otherwise you could never uniquely identify it with 360 bits, let alone actually describe it's actual state. I'm not sure about you guestimation, but I'd think 10^360 bits might not even be enough.
Even though a single megabyte might be the difference between any two humans, don't forget that a large chuck of this is probably in parts of the DNA that we think is inactive, so number of the significant bits that are actually different would be even smaller.
On the other hand, I'm glad we're so much alike; fixing sick humans is difficult enough as it is, imagine what would happen if we all were anatomically and physiologically completely different.
If you do what you did, you'll get what you got
(Score: 3, Interesting) by c0lo on Tuesday July 08 2014, @07:12AM
Mate, this is where you make the mistake: those 360 are not independent, but expressed as a number. Number on which you don't draw boundaries at "exact bit" locations (e.g u dont work in base 2, u work in base 10^90, even if you represent your numbers in binary. Look, use base 6 and represent your numbers in binary - you'll see you don't waste 110 and 111).
https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
(Score: 1) by SparkyGSX on Tuesday July 08 2014, @07:29AM
I understand these bits aren't each tied to something specific, but with 360 bits, you could only describe 2^360 different possible states of the entire universe, where every single change at quantum level would be considered a new, unique state of the universe.
It's a information theory problem; you cannot compress a system with more than 360 bits of entropy into 360 bits of space without losing data. Think of the 360 bits as a hash value of a dataset. You don't need a whole lot of datasets before you get a collision in the available 360 bits of space (think of the birthday paradox); there's a reason many cryptographic hashing algorithms use a hash size of 512 bits or even more nowadays.
The number of datasets you could possibly hash is, by definition, much smaller than the number of possible quantum states of the universe, since each dataset you hash must be represented in some kind of computer, and therefor the universe must be in a different state, as the state of the computer is part of the state of the universe.
If you do what you did, you'll get what you got
(Score: 2) by c0lo on Tuesday July 08 2014, @08:00AM
Mate it is not representing any possible state of the universe, it is about representing one specific state of the universe.
Here: if I what to represent a specific position of one billiard ball on a 1m x 1m table, with precision of 1mm, how many bits do I need? At most 20? How many for representing a specific position of all 10 balls + cue ball? 24?
Compare to: how many bit to represent all possible positions of all 10 balls + cue ball? So, you need
* 4 bits to encode the identity of a ball
* 10^6 "slots" of 1mm x 1mm where a ball may be
=>4*10^6 bits.
https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
(Score: 1) by SparkyGSX on Tuesday July 08 2014, @08:19AM
No, you certainly don't need 4*10^6 bits, that would be the case if you would create 4 bitmaps op 1000x1000 pixels, each consisting of bit that would indicate if the ball occupied that particular pixel. It would allow for mapping 4 objects of any irregular shape and size.
Each ball would have 1000 possible positions in both directions, which is 10^6, or about 2^20, so you would need 11 * 20 bits = 220 bits to encode the position of all the balls, not 4.000.000 bits as you final conclusion states. You could even do slightly better, since the balls have a finite size, and can't overlap each other, or the edge of the field, so the total number of possible combinations is slightly less than 11*10^6.
If you can encode one entire state of the universe in 360 bits, there cannot be more than 360 bits of unpredictable information in the entire universe, that would seem like rather trivial information theory to me. If you think you actually can encode the entire state in 360 bits, you have invented the MOTHER of compression algorithms.
I think we can agree that any file we have on any computer now contains less information than the entire state of the universe, since all files are part of that state. That would mean you could compress any file, regardless of it's size, into far less than 360 bits. I'd think it's rather obvious that can't be true.
If you do what you did, you'll get what you got
(Score: 2) by c0lo on Tuesday July 08 2014, @09:05AM
https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
(Score: 1) by TwentyCharsIsNotEnou on Tuesday July 08 2014, @07:14AM
Clearly that song is not part of your observable universe. Lucky for you.
(Score: 2) by c0lo on Tuesday July 08 2014, @07:35AM
If you limit yourself in representing the position of all the nucleons in the observable universe with Planck precision, the problem become finite and a better encoding schemes can be devised. (that the difference between representing any possible configuration of the universe and a specific configuration of the universe).
https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
(Score: 2, Informative) by subs on Tuesday July 08 2014, @10:44AM
You knew just enough mathematics to confuse yourself.
The 2^8388608 is the expression of the size of the state space in 1MB of information, i.e. there are 2^8388608 possible arrangements one-megabyte words.
This is complete and utter nonsense and your calculation confuses two completely different things. x^y is the number of possible states a message of length 'y' digits in base 'x' can take. To get the number of digits needed to represent all possible states of 'z' messages, you don't multiply the size of the state space by the number of messages (z * x^y), you multiply the number of digits, since each message consists of the same number of digits (x^(y * z)). To illustrate, say you have two 8-bit messages (256 possible states). By your logic you should be able to store these two 8-bit messages in 9 bits, since 2*2^8 = 2^9. Unfortunately, state spaces do not work that way. The combined two 8-bit messages do not have 256+256 possible states to be in, but 256*256 possible states, i.e. 2^8 * 2^8 = 2^(8+8) = 2^(2 * 8).
Thus it's not 10^(70*3)*10^80 - this is multiplying two different things together (total number of particles by the possible states one particle can be in). The correct answer is 10^(70*3*10^80), because you need 70*3 decimal digits to represent the state of one particle and there are 10^80 particles, thus you need 10^80 of these digit sets. Converting to base 2 we get 2^272.
I hope you understand that 2^272 is a little larger than 360.
(Score: 0) by Anonymous Coward on Tuesday July 08 2014, @04:49PM
You'd need only 10^290 what?
Well, let's do a concrete calculation: Let's assume we want to record the position only in the observable universe.
The volume of the visible universe is, according to Wikipedia, [wikipedia.org] about 3.4*10^80 m^3. The Planck volume is about 4.2*10^(-105) m^3, according to the same source. That is, there are approximately 10^185 Planck volumes in the visible universe. To uniquely represent a single Planck volume, you therefore need 615 bits (rounded up to complete bits).
To describe the position of 10^80 nucleons you therefore need 605*10^80 bits. Compared to that, 8388608 bits is nothing.
It would indeed be very strange if the DNA (whose structure is mainly determined by the relative positions of its nucleons) could encode more information than all the nucleon positions in the visible universe together.
(Score: 2) by unitron on Tuesday July 08 2014, @10:47AM
...even if only a very subtle one, between "data" and "information"?
something something Slashcott something something Beta something something
(Score: 2) by kaszz on Wednesday July 09 2014, @02:01AM
The human genome (DNA) is like 6.4×10^9 bits and a CD-ROM is like 5.45×10^9 bits. So no 1.5 GByte. That the coding of the cells is replicated doesn't make it right to view the body as a storage device consisting of cells multiplied by the bits in each DNA.
What's perhaps more interesting is the amount of bits stored in the brain that uses replicated (nerve) cells as storage. That is the most differentiating quantity of humans in this society. And perhaps not just knowledge but the map of actual pathways.