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An iPhone 7 takes pictures that have roughly 12 Megapixels. For simplicity, let's assert that the picture only encodes 256 values per red, green and blue channels such that a 1x1 pixel image has 256^3 unique representations (permutations). The number of permutations for the entire 12 Megapixel image is 256^(3 * number of pixels) = 256^(3 * 12,000,000) = 256^(36,000,000).

There are about 3 billion nucleotide bases in the human genome (which can be represented as a long string that looks like "ACATGACTTGAT..."). Since there are 4 bases by which all our DNA is expressed, the number of permutations in the human genome is 4^3,000,000,000.

For comparison, the estimate of atoms in our observable universe is 10^80

What's bigger: 4^(3,000,000,000) or 256^(36,000,000)?

I tried a couple BIG NUMBER calculators online (like this one from CASIO) but none of them seem to be able to handle numbers this big.

So if I try comparing their logarithms I get:

4^(3,000,000,000)
  =  4 ^ (3 * 10 ^ 9)
  = (3 * 10 ^ 9) * log(4)
  = (3 * 9 * log(10)) * log(4)
  = 27 * log(10) * log(4)
  ≃ 16

256^(36,000,000)
  = 256 ^ (36 * 10 ^ 6)
  = (36 * 10 ^ 6) * log(256)
  = (36 * 6 * log(10)) * log(256)
  = 216 * log(10) * log(256)
  ≃ 520

Did I do that right? If so, can it be said that there are about 33x more permutations in an iPhone picture than there are in the human genome?

Update

My math was indeed quite poor. Here's an update

4^(3,000,000,000)
  =  4 ^ (3 * 10 ^ 9)

  log (4 ^ (3 * 10 ^ 9))
    = (3 * 10 ^ 9) * log(4)
    ≃ 1806179974

256^(36,000,000)
  = 256 ^ (36 * 10 ^ 6)

  log(256 ^ (36 * 10 ^ 6))
    = (36 * 10 ^ 6) * log(256)
    ≃ 86696639

So the number of permutations in the data that describes the human genome sequence, though most of them will not yield a human, is much larger than the permutations of pixels in an iPhone image (regardless of how imperceptibly different those permuted images may be). I was tempted to try to find the ratios between these two logarithms again but realized that wouldn't be representative of the actual quotient. The actual quotient is:

(2^(6,000,000,000))/(2^(288,000,000))
  = 2^(6,000,000,000 - 288,000,000)
  = 2^(5,712,000,000)

So the data in the human genome has 2^(5,712,000,000) more permutations than the data in an iPhone picture.

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  • 1
    $\begingroup$ In fact, if you put the numbers into WolframAlpha, you find that the number of permutations of the human genome is a number that, if written down, would have about 1.8 billion digits. That's about twenty times as many digits as the number of permutations of iPhone pictures. $\endgroup$ – Rahul Apr 2 '17 at 23:30
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You've made a major arithmetic error, probably due in part to the fact you are using = to mean something more like "and then..." rather than to assert two things are equal.

For example,

$$ 256^{36,000,000} = 256^{36 \cdot 10^6}$$

You then wrote that

$$ 256^{36 \cdot 10^6} = 36 \cdot 10^6 \log(256)$$

which is clearly false; presumably, however, your intention was to say

$$ \log\left(256^{36 \cdot 10^6}\right) = 36 \cdot 10^6 \log(256)$$

so that the substitutions would let you conclude

$$ \log\left( 256^{36,000,000} \right) = 36 \cdot 10^6 \cdot \log(256)$$

Then, your big mistake was asserting

$$ 36 \cdot 10^6 \cdot \log(256) = 36 \cdot (6 \log(10)) \cdot \log(256) $$

Presumably, you were motivated by the fact that

$$ \log(10^6) = 6 \log(10) $$

But if you weren't using this weird shorthand, you would have realized that the quantity you are working with is

$$36 \cdot 10^6 \cdot \log(256) $$

which, in particular, does not have a $\log(10^6)$ to be rewritten.


To do the sort of rewriting you wanted to continue to do, you need to take logarithms again — that is, to compute

$$ \log(\log( 256^{36,000,000} ) ) = \ldots = \log( 36 \cdot 10^6 \cdot \log(256) )$$

The logarithm properties used to rewrite that last term will give something different than what you had written.

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  • $\begingroup$ I presume you meant to write 256 everywhere you wrote 267? $\endgroup$ – zelusp Apr 2 '17 at 23:25
  • $\begingroup$ @zelusp: It is so. $\endgroup$ – user14972 Apr 2 '17 at 23:28
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Well, $4 = 2^2$ so $$4^{3000000000} = 2^{2\times3000000000} = 2^{6000000000}$$

And, $256 = 2^8$ so $$256^{36000000} = 2^{836000000} = 2^{288000000}$$

The first is a lot bigger.

Another way to look at this is how many bits would be needed to store the two data sets. One base of the genome can be encoded in $2$ bits so you need $6000000000$ bits to store the whole genome; that's 750MB which is about the capacity of a CD ROM. The image needs $36000000 \times 3 \times 8 = 864000000$ bits; that's 108MB and you can get about 7 on the same CD.

Also, I am not certain but I don't think that each of those pixels contains all three colour values; I think that they contain just one of them. So, the 36000000 in the second number should maybe be only 12000000.

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  • $\begingroup$ The first part of your answer is really helpful! The second part sorta skews the original idea of permutation. For instance there is a big difference between the range in a number like 999 (there's a thousand numbers/permutations in there) and the number of digits (which is just 3). Your second example is like trying to compare ranges by talking about digits... which may be misleading. Now I'm curious - can you show me how pixels don't contain three colour values? I always thought they had at least three $\endgroup$ – zelusp Apr 2 '17 at 23:36
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    $\begingroup$ I see a missing $\times$ in the middle term of my second line. My idea in the second section is that any computer file can be viewed as a very large binary number. On my last comment, look at this en.m.wikipedia.org/wiki/Bayer_filter, it's what I had in mind when I said that. $\endgroup$ – badjohn Apr 3 '17 at 7:29
  • $\begingroup$ Thanks for the link! From the looks of it, that article demonstrates that, at some point, RGB information would be needed to recreate the image (even if some how one of those channels wasn't overtly saved in file). After all, that information is required to display a picture to a human. R, G, or B would be called subpixels $\endgroup$ – zelusp Apr 5 '17 at 0:39
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The other answers talk about the math, so I'll talk about if you should be doing that math.

It's important to keep in mind that these are both very rough upper bounds. In the particular case of DNA, I know that there is a huge amount that is identical in virtually every human, which drastically cuts down the number of possibilities. Google tells me that $96\%$ of our DNA is shared with chimps (and therefore presumably also humans) which would reduce the number by a multiplicative factor of 0.04. In actuality, I suspect the proper factor is several orders of magnitude smaller.

On the flip side, I doubt any human can tell RGB $001100$ from RGB $001200$. I have no idea how to make the bound on this reasonably tight, but quick experimentation suggests a $8000$-fold reduction would be appropriate.

As often is the case in applied math, figuring out what math to do is a major challenge in and of itself. Drawing any kind of conclusions off of these calculations would be wildly inappropriate, especially because my (vague) estimations actually switch which number is larger. I certainly wouldn't claim that either of my estimations are particularly close to the correct value.

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  • $\begingroup$ But the DNA comment is good. The vast majority of those permutations would not represent viable human beings. I would say that way, way less than 50% would be viable. Also, even among the viable ones, there is a lot of junk which could vary with no effect. There are a lot of possible humans but much fewer than the numbers above. $\endgroup$ – badjohn Apr 2 '17 at 22:25
  • $\begingroup$ Thanks @Stella_Biderman! That's a great point! I had thought those very same things when I was writing the post but opted to ignore mentioning it since I just wanted to focus on figuring out how to compare large numbers. I commend you for keeping your head out of the theoretical clouds and encouraging others to do the same! To your point about an 8k reduction in image permutations... makes me think that one day we won't need better image sensors since anything better wouldn't yield a perceptibly better image. It would seem the day is coming sooner than later. $\endgroup$ – zelusp Apr 3 '17 at 0:56
  • $\begingroup$ ... for the average consumer that is $\endgroup$ – zelusp Apr 3 '17 at 0:58

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