## 1208: "Footnote Labyrinths"

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Istaro
Posts: 101
Joined: Mon Jan 05, 2009 6:00 pm UTC

### Re: 1208: "Footnote Labyrinths"

oryhara wrote:
Istaro wrote:If 6 is "actually a 1", that means that "(actually a 1)^(2^2)" = "(actually a 1)^4" = "(actually a 1)^3" is equal to "actually a 1"; i.e., 3 is "true".

If 3 is "true", the original text says "no^true", i.e. "no". No evidence was found in the data.

But 3=true leads to a contradiction, because it would imply 5=true^(2^(6^3))=true^(2^6)=true^(2^(1^3))=true^(2^1)=true. But then that makes 3=(not true)^(3^2)=(not true)^5=not true.

I dispute the step true^(2^6) = true^(2^(1^3)). Footnote 6 is not "1^3", it's "(actually a 1)^3", i.e. footnoted text, not a footnoted number.

lightvector
Posts: 224
Joined: Tue Jun 17, 2008 11:04 pm UTC

### Re: 1208: "Footnote Labyrinths"

Istaro wrote:Since nested footnotes are right-associative for the same reason that exponentiation is right-associative, note that footnote 6 says "actually a 1^4", i.e. "actually a 1^3". Also note that since 3 relates to truth, not numerical value, it must be applying to the entire preceding string, not just the 1: "(actually a 1)^3", not "actually a (1^3)".

Now, take a look at footnote 5: "true^(2^(6^3))".

We know that the 6^3 applied to the 2 must change said 2 somehow, because if the 2 were left as is, we'd end up having to apply it to the "true" at the bottom, and true^2 is a syntax error.

Looking at all the footnotes, there are only two conceivable ways in which 6^3 could change 2 (because there are only two footnote effects that can change numbers): increment it, or make it actually a 1.

But 6^3 can't increment 2, because 6^3 doesn't ever take us to the incrementing footnote (footnote 2). See my first line if you're unsure about that. (Specifically, 6^3 only involves the statement "actually a 1" and the stuff pertaining to truth or the lack thereof.)

So that leaves us with one option: 6^3 must equal "actually a 1"—that's the only way in which 6^3 can safely dispose of the 2 under it, preventing that 2 from illegally applying to "true".

I agree with everything so far.
Istaro wrote:If 6 is "actually a 1", that means that "(actually a 1)^(2^2)" = "(actually a 1)^4" = "(actually a 1)^3" is equal to "actually a 1"; i.e., 3 is "true".
If 3 is "true", the original text says "no^true", i.e. "no". No evidence was found in the data.

I disagree with this. I don't see how you inferred "6 = (actually a 1)", given that up until that point, you had only shown that "6^3 = (actually a 1)". In fact, the assignments "3 = true" and "6 = (actually a 1)" are inconsistent, because we also have:
3
= (Not true)^5 <definition of footnote 3>
= (Not true)^(True^(2^(6^3))) <definition of footnote 5>
= (Not true)^(True^(2^((actually a 1)^true))) <substituting assignments>
= (Not true)^(True^(2^(actually a 1))) <evaluating>
= (Not true)^(True^1) <evaluating>
= (Not true)^(True^(ignore this)) <definition of footnote 1>
= Not true <evaluating>
Which is a contradiction with our one of our assignments "3 = true".

However, if we begin with the assumption that "3 = Not true", we get the following complete assignment, which is consistent as far as I can tell. Note that footnote 5 actually has no effect.
1 = Ignore this
2 = Increment by 2 before following
3 = Not true
4 = Not true
5 = ""
6 = Not actually a 1

Verifying consistency of 3, 5, and 6:
3 = (Not true)^5 = (Not true)^("") = Not_true
5 = True^(2^(6^3)) = True^(2^((Not actually a 1)^(Not true))) = True^(2^(Actually a 1)) = True^(1) = True^(Ignore this) = ""
6 = (Actually a 1)^3 = (Actually a 1)^(Not true) = Not actually a 1

Under these interpretations (ibid meaning that 4 = 3, subfootnotes applying to the whole footnote rather than a single number or word), I don't see any other consistent assignments. So if this is the only consistent assignment, then the claim in the text is false, evidence was found.

Istaro
Posts: 101
Joined: Mon Jan 05, 2009 6:00 pm UTC

### Re: 1208: "Footnote Labyrinths"

Oops, I stand corrected. And I don't see any problems with your proposed set of assignments—nicely done!

Ger7ry
Posts: 1
Joined: Tue May 07, 2013 3:56 am UTC

### Re: 1208: "Footnote Labyrinths"

I wonder whether people are familiar with this story from Littlewood's Miscellany:

I wrote a paper for the Comptes Rendus which Prof. M. Riesz translated into French for me. At the end there were 3 footnotes. The first read (in French) 'I am greatly indebted to Prof. Riesz for translating the present paper.' The second read 'I am indebted to Prof. Riesz for translating the preceding footnote.' The third read 'I am indebted to Prof. Riesz for translating the preceding footnote', with a suggestion of reflexiveness. Actually I stop legitimately at number 3: however little French I know I am capable of copying a French sentence.

mcdigman
Posts: 77
Joined: Fri Aug 17, 2012 6:32 pm UTC

### Re: 1208: "Footnote Labyrinths"

rogerhere wrote:Reminds me about "CLOG" - The Computer Linograph, an Ada package for automatic footnote generation.
Communications of the ACM, April 1984. You don't happen to remember if it was released on any specific date in April, do you?

StClair
Posts: 409
Joined: Fri Feb 29, 2008 8:07 am UTC

### Re: 1208: "Footnote Labyrinths"

If you choose to stand and fight the Wumpus, turn to12.

orthogon
Posts: 3103
Joined: Thu May 17, 2012 7:52 am UTC
Location: The Airy 1830 ellipsoid

### Re: 1208: "Footnote Labyrinths"

Klear wrote:But if it is a House of Leaves reference, than it makes it a bit pitiful, since footnotes are emplayed a lot better there. I think I'd enjoy this comic way more if I hadn't read it.

I was disappointed to find that I'm not one of today's lucky 10,000: sadly "emplay" isn't a word after all. But it should be! And we can make it so!

emplay v. [portmanteau of employ + play]: To put into play; to employ playfully.
Klear, 2013: "... footnotes are emplayed a lot better there"

Edit: part of speech fail.
Edit2: It might even be a malamanteau.
xtifr wrote:... and orthogon merely sounds undecided.

Blake
Posts: 13
Joined: Thu Jan 31, 2013 11:58 pm UTC

### Re: 1208: "Footnote Labyrinths"

Klear wrote:But if it is a House of Leaves reference, than it makes it a bit pitiful, since footnotes are emplayed a lot better there. I think I'd enjoy this comic way more if I hadn't read it.

Well, we are talking about a thrice weekly, often humorous webcomic, not a several hundred page densely written metafictional descent into madness.

Klear
Posts: 1965
Joined: Sun Jun 13, 2010 8:43 am UTC
Location: Prague

### Re: 1208: "Footnote Labyrinths"

Blake wrote:
Klear wrote:But if it is a House of Leaves reference, than it makes it a bit pitiful, since footnotes are emplayed a lot better there. I think I'd enjoy this comic way more if I hadn't read it.

Well, we are talking about a thrice weekly, often humorous webcomic, not a several hundred page densely written metafictional descent into madness.

Doesn't matter. Both xkcd and House of Leaves is filed under "best thing ever" in my brain. As such, both are subject to high expectations and apologies such as yours are way less forceful than they reasonably should be.

Trickster
Posts: 89
Joined: Fri Sep 05, 2008 5:56 pm UTC

### Re: 1208: "Footnote Labyrinths"

Argh. I miss a week of XKCD and then this happens.

Now I have to relearn Prolog so I can confirm the logic for this thing. crusnik02
Posts: 1
Joined: Fri May 24, 2013 9:27 pm UTC

### Re: 1208: "Footnote Labyrinths"

Both interpretations of footnotes imply that evidence was found.

Recursive footnotes:
Spoiler:
(no)^1^2
(no)^1^(increment by 2 before following)
(no)^3
(no)^(not true)^3^2
(no)^(not true)^3^(increment by 2 before following)
(no)^(not true)^5
(no)^(not true)^(true)^2^6^3

Therefore, 3 = (not true)^(true)^2^6^3 is recursively defined.
We now find the fixed points of 3.

Case 3 = true:
true = (not true)^(true)^2^6^(true)
= (not true)^(true)^2^6
= (not true)^(true)^2^(actually a 1)^2^2
= (not true)^(true)^2^(actually a 1)^2^(increment by 2 before following)
= (not true)^(true)^2^(actually a 1)^4
= (not true)^(true)^2^(actually a 1)^(ibid = 3 = true)
= (not true)^(true)^2^(actually a 1)
= (not true)^(true)^1
= (not true)^(true)^(ignore this)
true = not true
CONTRADICTION

Case 3 = not true:
not true = (not true)^(true)^2^6^(not true)
= (not true)^(true)^2^((actually a 1)^2^2)^(not true)
= (not true)^(true)^2^(not (actually a 1)^2^2)
= (not true)^(true)^2^(not (actually a 1)^2^(increment by 2 before following))
= (not true)^(true)^2^(not (actually a 1)^4))
= (not true)^(true)^2^(not (actually a 1)^(ibid = 3 = not true))
= (not true)^(true)^2^(not not (actually a 1))
= (not true)^(true)^2^(actually a 1)
= (not true)^(true)^1
= (not true)^(true)^(ignore this)
not true = not true
TAUTOLOGY

Therefore, 3 = not true is the fixed point of 3

(no)^3
(no)^(not true)
not no
some

"We found some evidence"

Exponents:
Spoiler:
(no)^1^2
(no)^1
(no)^(ignore this)

"We found evidence"

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