<p>Well, this is not exact.
Clearly there exists an irrational number that contains all patterns - simply concatenate all finite patterns (e.g. in lexicographical order, from shortest and up)</p>

<p>However, not every number contains all these patterns. Indeed, you can construct an irrational number by concatenating the binary sequences "000" and "111", which will give you a very limited set of patterns (albeit infinite).</p>

<p>Also, your question seems to assume that if a number contains all possible patterns, then it is also computable where each pattern starts - this is most likely untrue, unless it is a very specific number (e.g. the one described above).</p>

<p>A somewhat related idea is <a href="http://en.wikipedia.org/wiki/Chaitin%27s_constant" rel="nofollow">Chaitin's Constant</a>, perhaps you will find it relevant.</p>

<p>Also, observe that the term "quickly calculate the decimal sequence" is not well-defined. You cannot calculate in the sense of writing it down, as it is infinite and might not have a finite representation. What you can ask is for an algorithm that gives the $i$-th digit, in time polynomial in $i$ (or indeed, just that the $i$-th digit is computable for every $i$). The link above demonstrates a number for which it is provably impossible.</p>

CC BY-SA 3.0
2013-03-21T10:01:59.590Z
2013-03-21T10:08:01.930Z
2013-03-21T10:08:01.930Z
Shaull
<p>I guess the question is </p>

<blockquote>
  <p>"does an 'infinite' number of patterns imply 'every' number of
  patterns?"</p>
</blockquote>

<p>For instance, if you could quickly calculate the decimal sequence of π, could you not (in theory of course) come up with an algorithm to search that sequence for some pre-determined sequence?</p>

<p>Then you could do this:</p>

<pre><code>start = findInPi(sequence)
</code></pre>

<p>So "sequence" in theory could be a decimal representation of the movie "The Life of Pi".  The implication is that all digital knowledge (past, present and future) is bound up in irrational numbers (not just the group of irrational numbers, but each irrational number), and we just need to know the index to pull data out.</p>

<p>Once you know the index and length of data, then  you could simply pass this long.  </p>

<pre><code>playMovie(piSequence(start, length))
</code></pre>

<p>From an encryption standpoint, you could pass the start, length pair around, and the irrational number would be known only by the private key holder.</p>

<p>Am I off base here?</p>

CC BY-SA 3.0
2013-03-21T11:27:52.727Z
I think this belongs more on Mathematics.SE.
Niel de Beaudrap
CC BY-SA 3.0
2013-03-21T13:35:19.490Z
Thanks.. I'll try that.  When I looked for the right spot for some reason I found mathmatica but not "math" - I see it now.
ChronoFish
CC BY-SA 3.0
2013-03-21T22:05:46.263Z
The number 0.01001000100001000001... is irrational yet most patterns never come up in its expansion.
Mahdi Cheraghchi
<ds.algorithms><soft-question><homomorphic-encryption>
Do irrational numbers contain an infinite number of (or all possible) patterns of sequences?
ChronoFish
# Do irrational numbers contain an infinite number of (or all possible) patterns of sequences?

(Asked by: username_0 on 2013-03-21)

I guess the question is 

"does an 'infinite' number of patterns imply 'every' number of
  patterns?"

For instance, if you could quickly calculate the decimal sequence of π, could you not (in theory of course) come up with an algorithm to search that sequence for some pre-determined sequence?
Then you could do this:
start = findInPi(sequence)

So "sequence" in theory could be a decimal representation of the movie "The Life of Pi".  The implication is that all digital knowledge (past, present and future) is bound up in irrational numbers (not just the group of irrational numbers, but each irrational number), and we just need to know the index to pull data out.
Once you know the index and length of data, then  you could simply pass this long.  
playMovie(piSequence(start, length))

From an encryption standpoint, you could pass the start, length pair around, and the irrational number would be known only by the private key holder.
Am I off base here?

**Tags:** ds.algorithms, soft-question, homomorphic-encryption
> **Comments:**
> **username_1**: I think this belongs more on Mathematics.SE.  
> **username_0**: Thanks.. I'll try that.  When I looked for the right spot for some reason I found mathmatica but not "math" - I see it now.  
> **username_2**: The number 0.01001000100001000001... is irrational yet most patterns never come up in its expansion.  

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## 1 answers

### Answer by username_3 on 2013-03-21. Score: 1

Well, this is not exact.
Clearly there exists an irrational number that contains all patterns - simply concatenate all finite patterns (e.g. in lexicographical order, from shortest and up)

However, not every number contains all these patterns. Indeed, you can construct an irrational number by concatenating the binary sequences "000" and "111", which will give you a very limited set of patterns (albeit infinite).

Also, your question seems to assume that if a number contains all possible patterns, then it is also computable where each pattern starts - this is most likely untrue, unless it is a very specific number (e.g. the one described above).

A somewhat related idea is [Chaitin's Constant](http://en.wikipedia.org/wiki/Chaitin%27s_constant), perhaps you will find it relevant.

Also, observe that the term "quickly calculate the decimal sequence" is not well-defined. You cannot calculate in the sense of writing it down, as it is infinite and might not have a finite representation. What you can ask is for an algorithm that gives the $i$-th digit, in time polynomial in $i$ (or indeed, just that the $i$-th digit is computable for every $i$). The link above demonstrates a number for which it is provably impossible.


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