# Difference between revisions of "Talk:1208: Footnote Labyrinths"

Way to nerd-snipe me, Randall. Alpha (talk) 04:52, 6 May 2013 (UTC)

In the nested-footnotes interpretation, 5 has to be ignored: The 6 must be true, and the 6 says that it’s “actually a 1”, but with footnote 2+2 which says “ibid.” and thus equals footnote 3, which is true. So 6 really does mean actually a 1, which leaves 5 to be ignored. --77.186.8.191 10:47, 6 May 2013 (UTC)

The footnote for 6 is actually 1 to the 2 to the 2 Schmammel (talk) 12:36, 6 May 2013 (UTC)

Explaination is wrong : abc = a(bc) = ab^c (confer the definition of a gogol = 10^100 = 10102, and a gogolplex = 10^gogol = 10(10100), not 10^110. So since 1^2= 1, No12 really means No1. 192.54.145.66 (talk) (please sign your comments with ~~~~)

Yes, so "no1" means to ignore the "no" and the answer for the second explanation is "we found evidence for the data." By the way, it's spelled "googol." Alpha (talk) 17:51, 6 May 2013 (UTC)
Question, alternative explination

I wasn't really satisfied with the whole discarding of the infinite loop, so I worked through the problem seperately using the nested footnotes. Then, when we hit the infinite loop I split between the two possible answers (either the infinite loop ends on true or false). As I read it, they both get the same answer:

no (3)

no (not true (5))

no (not true (true (2 < 6 < 3))

no (not true (true (2 < 6 < (not true))))

no (not true (true (2 < (actually 1 < 2 < 2 (not true 3 < 2)))))

no (not true (true (2 < (actually 1 < 2 < 2 (not true (5)))))

Split!

If 6 is false (infinite loop possibility)

no (3 < 5 < 2)

no (not true (7)) - meaningless, so discard

no (not true)

If 6 is true (infinite loop possibility)

no (3 < 5 < 1 < 2 < 2)

no (3 < 5 < 1 < 4)

no (3 < 5 < 1)

no (3)

no (not true)

Both lead to the answer "... experiments to observe this and we found evidence for it in our data". -- ‎Urah (talk) (please sign your comments with ~~~~)

Yes, but at each stage you may "toggle between interpreting nested footnotes as footnotes on footnotes and interpreting them as exponents (minus one, modulo 6, plus 1)." That is, a23 may either be read as "apply note 8 (=2mod6) to text a", or as "apply note 3 to text "2", then the result to text a". 192.54.145.66 (talk) (please sign your comments with ~~~~)
There are differences in interpretation here. If we write "foo36", is it equal to "foo112" or "foo3112"? I assumed the former and you assumed the latter. My reasoning is that footnotes modify their arguments and not themselves. Alpha (talk) 17:44, 6 May 2013 (UTC)

Shouldn't 5 be true (because 6 is actually 13; therefore 5 is true2133; so the 2 is ignored regardless the truth of 3) and 3 is not true? Sebastian --178.26.118.249 18:35, 6 May 2013 (UTC)

Yet another alternative solution: Footnotes should be evaluated from top to bottom, so "no12" = "no1 + 2" = "no3". We turn to the definition of 3, which is "not true32" = "not true3 + 2" = "not true5".

Now 5 is "true263". The 6 says that the 2 footnote is really 122 = 1(4. ibid.) = 13, but the 3 tells us that the 6 is "not true5", getting us into an infinite loop. However, 263 must evaluate to 1, because otherwise we're incrementing "true" by 2, which is meaningless. This means that 3 must be equal "not true". 63 = "actually a 1"33 = "actually a 1". 5 becomes true1 which just says to ignore this footnote altogether and we can confirm that 3 is indeed not true (not true5 = not true). So the answer is that the "no" is not true, and the correct statement is "we found some evidence for it in our data." Phew. Ciamej (talk) 22:40, 6 May 2013 (UTC)

I'm not discouraging anyone from coming up with more alternate solutions, but would it be fair to say that part of the point is that there are multiple equally legit ways to run this labyrinth, and that some exit where you ignore the 'no', others exit on the other side where you don't ignore it. and then there's those who won't exit because they're busy making a map. - 70.72.16.171 23:18, 6 May 2013 (UTC)

I don't understand the proof from This means that 3 = "true". Why do you assume that footnote has to be either "true" or "false? I think it could be "ignore this", "increment by three before following", "leave the whole calculation and assume we have two pieces of evidence" etc. as well. 178.56.1.144 23:37, 6 May 2013 (UTC)

Given the footnotes' definitions I don't think it's possible to ever come up with "increment by three before following" ;)
Actually the solution I gave may be not strictly formal, but it gives some intuition why it seems to be the only valid one.
The fact that the definitions are recursive doesn't imply that the ultimate answer cannot be resolved. Ciamej (talk) 02:18, 7 May 2013 (UTC)

So what I'm hearing is this, "No means No.", yes?66.88.136.254 19:37, 8 May 2013 (UTC)

It's a real strange logic, but "No = No" means "Yes" --Dgbrt (talk) 19:44, 8 May 2013 (UTC)

Footnote logic

So... I did some footnote logic, and came up with this:

This explanation will treat footnotes as footnotes, with the order of operations from top-down, with footnotes acting on only the object they are attached to, including other footnotes.

1. no^1^2

2. no^3

3. no(not true^3^2)

4. no(not (true^5))

5. no(not (true^2^6^3))

The ^6 says the ^2 is actually a 1^2^2, but the ^3 says that the ^6 is "not true^3^2". This leads us to an infinite loop, as the "not true^3^2" in step 3 led to the addition of the additional "not true^3^2".

I assume that the loop can be reduced down to either "true" or "not true", for the purposes of following this path. I will explore both options.

if infinite loop is true:

6a. no(not true^1^2^2) (replacing 2 with 1^2^2 as per 6)

7a. no(not true^1^4)

8a. no(not true(ignore(not true^3^2))) (infinite loop again)

I guess we'll split once more.

if second loop is true:

9aa. no(not (true(ignore(not true)))) (as the second loop reduced to true, we have no more footnotes)

10aa. no(not true) (since the "ignore this exponent was not true, we can remove it)

And we finally have something simple. No is not true, so evidence was actually found.

if second loop is false:

9ab. no(not (true(ignore(not (not true))))) (again, with the second loop reduced to "not true", we have no more footnotes)

10ab. no(not (true(ignore)))

11ab. no(not)

This is a bit more confusing, as we're ignoring the true as per step 10ab, and are just left with no^not. I'm going to take this to mean true, as in, again, evidence was found.

if first loop is false:

6b. no(not (true^2)) (the ^6 which said that the ^2 was actually a 1^2^2 was negated by the ^3 (which we declared as false for this leg), therefore both the ^3 and the ^6 can be reduced to nothing.)

No idea how to proceed here, as true is not a footnote, and can't be followed or incremented.

If we just ignore the ^2, we're left with the same as 10aa. That is, evidence was again found.

Alternatively, we can say that because ^5==false led us to a nonsensical result, then ^5 must always reduce to true, meaning that the only acceptable answer is to follow the path to 10aa.

Any way you slice it, evidence was certainly found.

Kalzekdor (talk) 22:29, 25 May 2013 (UTC)