Comments on: Translating “b is a power of 2″ into TNT

By: nomulous

nomulous — Tue, 09 Feb 2010 07:39:00 +0000

Thanks for the input, but I think you might be confused...

1) The statement is not "there exists c..." but rather "IF there exists c then...".
2) I do qualify the d, in the "if ∃c: (SSa⋅c) = b, then exists d such that", etc.

Basically, what I'm saying is that if, for any a, if a is a factor of b (that is, if there exists a c such that a * c = b), then a is a factor of 2 (there exists d such that 2 * d = a).

These things are so easily confused, but you'll find its quite alright the way it is.

By: Kirk

Kirk — Tue, 09 Feb 2010 07:26:10 +0000

∀a: <∃c: (SSa⋅c) = b ⊂ ∃d: (SS0⋅d) = SSa>

Two minor quibbles:

1/ You should not have the <> around the 'exists c' part, but rather around the sentence itself.
2/ You need to quality the d, ie. there exists d

your statement becomes
∀a: ∃c: ∃d: <(SSa⋅c) = b ⊂ ∃d: (SS0⋅d) = SSa>

But what about the case where SSa = 6? holds true ( d = 3), and <(SSa.c) = b> gives b to be any multiple of 6; i.e. 6, 12, 18, 24. Clearly these are not powers of 2!

Your logic
if all factors of b are multiples of two then b is most certainly a power of two.
is sound but you haven't followed it! Just because there 'exist' factors which are multiples of 2, does not follow that 'all' factors are multples of 2.

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By: Frank Thomas

Frank Thomas — Tue, 17 Nov 2009 04:54:22 +0000

Hi! I don't think our solutions are totally different. On the contrary, I think they are very similar. A very sloppy translation of my solution is "b has no odd factors" and I'd translate your solution to "all factors of b are even". This looks just like a double negative for me.

And by the way, if you replace "SS0.d" with "d+d" in your statement, it is 30 characters long, just as long as mine. I'm wondering if there exists a shorter solution of Hofstadter's exercise...

By: Frank Thomas

Frank Thomas — Mon, 16 Nov 2009 21:54:22 +0000

By: Ben

Ben — Mon, 16 Nov 2009 05:03:49 +0000

I'd hang my head lower in shame but I'd have to dig a hole.

Kudos.

By: nomulous

nomulous — Mon, 16 Nov 2009 03:11:22 +0000

This might make you feel like an idiot, but that's exactly what the SSa's are for. Since a + 2 can never equal 1, it works out quite nicely in the end.

But don't worry, I only know that 'cause I wrote it. I can understand why it might look like they weren't necessary at first.

By: Ben

Ben — Mon, 16 Nov 2009 03:04:58 +0000

Oh crap it's not watertight (i'm sorry i made such a claim). You need to worry about factors b and 1. Otherwise if a = 1 and c = 8 then the left hand side is true and the right is false. It shouldn't be too hard to make this amendment.

By: Ben

Ben — Mon, 16 Nov 2009 02:58:22 +0000

What you're saying is exactly right but I want to explain why your formula is not quite right.

"if there exists c such that a times c"

The SSa * c in your formula isn't what you are trying to say. This says "the successor of the successor of a times c" or (a+2) * c. You should just have it (a*c).

"d times 2 is equal to a"

The (SS0 * d) = SSa has the same issue. This says 2 * d = the successor of the successor of a (or a + 2).

Now if we take your formula and change the SSa's into a's we get:

∀a: <∃c: (a⋅c) = b ⊂ ∃d: (SS0⋅d) = a>

This is watertight. My example before about b = 12 was trying to hint that your formula wasn't quite there.

By: nomulous

nomulous — Sun, 15 Nov 2009 13:14:49 +0000

I don't understand the language in which that is expressed, but the way you translated it into English works very well. And yes, since 5 is prime you can easily use the same basic principle to express it.

In TNT, "b is a power of 5" could be expressed as the following:

∀a: <∃c: (SSa⋅c) = b ⊂ ∃d: (SSSSS0⋅d) = SSa>

Which basically says that if a is a factor of b then it is also a multiple of 5.