1

The human genome consists of sequences of BASE Pairs A G C T Convert the number PI to base 4. Does my unique human genome exist in the sequence of digits?

user316076
  • 21
  • 3
    Yes, if $\pi$ is a normal number. That's probable but not known. – Ethan Bolker Feb 20 '16 at 00:58
  • @EthanBolker -- Presumably you are using the word "normal" in some technical sense, not in the usual English sense? – bubba Feb 20 '16 at 01:03
  • 1
    Do you know your unique human genome sequence? – SS_C4 Feb 20 '16 at 01:07
  • It's possible but there is no guarantee unless you find it. – Carser Feb 20 '16 at 01:07
  • 1
    "Normal" in mathematical sense means a number whose decimal expansion (in any base) will have normal distribution of digits. If so every string will eventually appear. We don't know if pi is normal but it probably is. – fleablood Feb 20 '16 at 01:24
  • Related : https://en.wikipedia.org/wiki/Infinite_monkey_theorem – SS_C4 Feb 20 '16 at 01:25
  • 1
    Rational numbers are not normal the digits are periodic or finite for all but 0 or 9. A number whose decimal expansion contains only 5s and 9s is not normal. No number in the cantor set is normal. The number .1212312341234512345612345678123456789123456789101234567891011... is not normal. Etc. Surprisingly proving a number is normal is ... practically impossible. – fleablood Feb 20 '16 at 01:30
  • 1
    @fleablood: Your example number may not be normal, but it does contain user316076's genome. – TonyK Feb 20 '16 at 13:05
  • Wha... oh. Yes, yes it does. Doesn't it? And unlike pi we know where, don't we? I wasn't really thinking about that. The number (base 10) .101001000100001... is not normal and probably doesn't contain the op's DNA either unless the op's DNA is very unnormallike. – fleablood Feb 20 '16 at 16:34
  • You can search for five digit (base 10) strings in $\pi$ here and at other places on the net: http://www.facade.com/legacy/amiinpi/ – Ethan Bolker Feb 21 '16 at 13:56

2 Answers2

1

A heuristic argument would go as follows: Assume your genome $G$ is a string of $n\gg1$ digits over $\{0,1,2,3\}$. Denote by $x_k$ the $k^{\rm th}$ digit of $\pi$ in base $4$. For each $r\geq0$ the probability that $$(x_{rn+1},x_{rn+2},\ldots,x_{rn+n-1}, x_{(r+1)n})\ne G$$ amounts to $(4^n-1)/ 4^n<1$. Therefore the probability that for no $r\geq0$ we have a coincidence is $$\lim_{N\to\infty}\left({4^n-1\over 4^n}\right)^N=0\ .$$

Christian Blatter
  • 226,825
  • This assumes that pi has a normal distribution of decimal digits in base 4, does it not? So this probability is only correct if pi is a normal number (still an open problem). – Mark B Feb 20 '16 at 12:54
  • @MarkB: Of course. This has been addressed in the comments to the question. But I'm not so confident about the OP's sophistication. Maybe you have overseen the word "heuristic" in my answer. – Christian Blatter Feb 20 '16 at 13:22
0

The facts are these:

  • nobody knows the answer for sure;
  • every mathematician in the world expects that the answer is Yes.

But we don't have a proof one way or the other, in spite of much effort invested in the problem.

TonyK
  • 64,559