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Much ado about (almost) nothing

Today is Pi day, and indeed a typical way to impress people with the power of Mathematica is to display Pi to an obscene number of decimal places.

It is the purpose of this note, however, to exhibit a more refined display of power in which Mathematica creates elaborate structures purely from that venerable vessel of vacuity, the empty set, which we specify by

empty = {  };

Mathematica is so powerful it can do geat things even when running on empty, so to speak. Consider the list

nonempty = NestList[Subsets, empty, 4]

This is better appreciated in the form

Grid[Transpose@{nonempty}, Frame -> All]

This can, of course be continued, and the quantity

nonempty = NestList[Subsets, empty, 5];

becomes quite large.

Short of an explicit display, nonempty does have some amusing properties:

Length /@ nonempty
{0, 1, 2, 4, 16, 65536}
Length[ToCharacterCode[ToString[#]]] & /@ nonempty
{2, 4, 10, 30, 274, 9109506}`
empty == Flatten@nonempty
True`

It should finally be noted that, as a gesture of mercy to a finite universe, Mathematica refuses to show the result for

NestList[Subsets, empty, 6];
POSTED BY: David Vasholz
6 Replies

I think the problem is that the empty set has those pesky borders, one on each side. So we have to deal with power sets of braces, and that leads to a brace of power sets. How many to a brace? I always wonder about such things (not that I expect I'll ever be hunting pigeons in the countryside of whatever parts of Olde England have people hunting them by the brace). In this case, I guess because we have two borders that preceed and follow one power set, it follows that the result would be a set of two to the power of its predecessor. So that gets big fast.

Possibly I should have waited until March 32 (aka "two-to-the-fifth day" since March is the only month with that many days) to post..

POSTED BY: Daniel Lichtblau

Hi,

the point regarding the finite universe is well made indeed! I assume that the number of elements will follow this sequence:

NestList[2^# &, 1, 5]

A function which Mathematica evaluates easily. As you say the last number is quite large. If we display the logarithm to base 10 of these numbers we get:

Log[10, #] & /@ NestList[2^# &, 1, 5] // 
(*{0., 0.30103, 0.60206, 1.20412, 4.81648, 19728.3}*)

The last element has 19729 digits. Given that WolframAlpha says

== Number of atoms in the Universe

that there are only about $10^{80}$ atoms in the universe it might be difficult with the little matter in our Universe to actually print out these about $10^{19728}$ elements. Also, even if we wanted to display them, say each for a millisecond, it might -depending on your favourite model of the universe - take quite a bit longer than the Universe, as we know it, will exist. Well, there is a theory that says that if we wait a bit longer, $10^{10^56}}$ years or so, there might be a new big bang and we can start all over again. Even though the authors themselves seem to doubt that number.

Best wishes,

Marco

PS: If I ask Mathematica to calculate:

Log[10, #] & /@ NestList[2^# &, 1, 6] 

it bails out on me.

POSTED BY: Marco Thiel

That last would be approximately 6.655582364766226*10^19728. Just saying.

POSTED BY: Daniel Lichtblau

Hi Daniel,

I think that the $6.655582364766226\cdot10^{19728}$ was for the fifth iteration as I said in my post. I cannot get the corresponding number for the 6th iteration....

Cheers,

M.

POSTED BY: Marco Thiel

What I meant is that it is the log base 10 of the result of the last value in the NestList to 6 iterations. Agreed, computing the NestList itself that far is out of the question.

POSTED BY: Daniel Lichtblau
Posted 9 years ago

As always, Dave, your comments are thought provoking! I didn't know that Nothing was so interesting. Hope you've been well.

POSTED BY: Bruce Colletti
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