Talk:Conway chained arrow notation

message: One of the examples is 2→3→2→2.  Along the way, we come across 2→3→8.  The example then goes to declare that equal to 2→(2→2→7)→7.  However, I think this might be a mistake.  I am not an expert in Conway's chained arrow notation, but in the case of 2→3→8, the article says we would apply the rule X→(P+1)→(Q+1) == X→(X→P→(Q+1))→Q.  In this case X=2, P=3, and Q=8.  Applying the rule, this seems to me to indicate 2→3→8 == 2→(2→2→8)→7, not 2→(2→2→7)→7. 
70.176.41.197 (talk) 06:41, 14 May 2013 (UTC)
thread_title: Examples do not seem to match description

message: As it stands, this page is gibberish. I have no doubt that the equations are "correct" but they need explanation. Are the first 3 equations alternative schemes or are they all a single scheme defining one large number. -- SGBailey 22:07, 2003 Aug 27 (UTC)
They're supposed to make up a recursive definition. They're not correct, though. They don't cover the case with more than two arrows. Matthew Woodcraft
:I understand Knuth's up-arrow notation which is explained. This "X>p>1=X>p" and "p>q=p^q" (using > for right arrow and ^ for exponent) explains nothing. I am inclined to empty the article unless someone can explain why I shouldn't. -- SGBailey 09:57, 2003 Aug 28 (UTC)
Fixed up the first part, however I don't get the chaining of 3 arrows as below. Dysprosia 10:06, 28 Aug 2003 (UTC)
thread_title: Untitled

message: "In this case the notation eventually resolves to being the leftmost number raised to some integer (usually enormous) power"
This statement is pretty silly.  I mean exponentiation just 'resolves' into repeated multiplication in the same way
as multiplication resolves into an addition and addition really resolves into repeated incrementing (by 1).  You might as well
say that the notation resolves into a really long series of increments.  Asteron 01:33, 25 January 2006 (UTC)
replies: text: I disagree.  It clarifies the structure of the result; the chained arrow notation is very hard to parse on first sight and statements like this make it easier to know what to expect.  Also, it puts the chained arrow notation in its proper place as a generalization of simple exponentiation.  For example, Knuth's up-arrow notation is also a generalization of simple exponentiation; would you say that it is silly to note that a \uparrow b is a power of a?  Obviously it would be ridiculous to describe either as "resolving" into repeated addition; the point is to relate them to the next most natural and already understood operation.  Ryan Reich 02:00, 25 January 2006 (UTC)
thread_title:  Resolve? 

message: Why did Conway invent this notation?
When did Conway invent this notation?
Where (i.e., in what book or article) did Conway specify this notation?
--Oz1cz 14:58, 3 June 2006 (UTC)
replies: text: It's specified on page 61 of The Book of Numbers, but I don't know if that's where he first specified it. Factitious 22:30, 15 September 2006 (UTC)
thread_title:  Why, when, and where? 

message: Let's consider the function Cw1 (x, n) (Conway's first sort extended function):
Cw1 (x,n) = x->x->x->...->x with n-1 arrows according to Conway chained arrow notation.
For example, Cw1 (3, 4) = 3->3->3->3; Cw1 (10, 5) = 10->10->10->10->10 etc.
Now, let's consider another function Cw2 (x) (Conway's second sort extended function):
Cw2 (x) = Cw1 (x, x).
For example, Cw2 (4) = Cw1 (4, 4) = 4->4->4->4; Cw2 (5) = Cw1 (5, 5) = 5->5->5->5->5 etc.
Now we can construct the Extreme-Power Notation:
x~1 = x;
x~2 = Cw2 (x)= x->x->x->...->x (with x-1 arrows) (incomparably larger than Graham's number!);
x~3 = Cw2 (x~2) = x~2 -> x~2 - x~2 -> ... -> x~2 (with (x~2)-1 arrows);
x~4 = Cw2 (x~3) = x~3 -> x~3 - x~3 -> ... -> x~3 (with (x~3)-1 arrows) etc.
...
x~ ~1 = x;
x~ ~2 = x~x;
x~ ~3 = (x~ ~2)~ ~2 = (x~x)~ ~2 = (x~x)~(x~x);
x~ ~4 = (x~ ~3)~ ~3 etc.
...
x~ ~ ~1 = x;
x~ ~ ~2 = x~ ~x;
x~ ~ ~3 = (x~ ~ ~2)~ ~ ~2 = (x~ ~x)~ ~ ~2 = (x~ ~x)~ ~(x~ ~x);
x~ ~ ~4 = (x~ ~ ~3)~ ~ ~3 etc.
...
Finally, lets say that Blt (x) = x~ ~ ~...~x (with x tildes (!)) (Beloturkin's function).
For example, 1 Beloturkin = Blt (G), where G is Graham's number;
1 Beloturkinion = Blt (Beloturkin);
1 Beloturkiniard = Blt (Beloturkinion).
Beloturkin 08:23, 15 September 2006 (UTC)
replies: text: I added a remark along these lines at Large_numbers#Systematically_creating_ever_faster_increasing_sequences.--Patrick 13:06, 11 July 2007 (UTC)

text: I don't understand the point of this. Your notation doesn't construct a new paradigm. In terms of computational complexity, it is no faster than regular Chained arrow notation.
I suggest you take a gander at the "fast growing hierarchy" page. Read up on limit ordinals, the wainer hierarchy, and diagonalized functions. Then enjoy considering f_*(2) where * is the feferman-schutte ordinal. Or any large ordinal really. Even epsilon or zeta nought.
Do you really think people would be talking about Conway's chained arrow notation if all it was was a bunch of repeated and nested iterations of knuth's up arrow notation?  — Preceding unsigned comment added by 216.163.254.2 (talk) 12:22, 2 September 2011 (UTC)

text: It's not? 2601:58B:4204:B6B0:7873:3EE2:99A:88CD (talk) 21:39, 2 September 2022 (UTC)
thread_title:  New extreme-power notation proposed. 

message: The page here does not explain four-term chains; they merely give a non-explanatory example.
The way I see it, the 4th term defines a recursion. A 4-term chain such as a \rightarrow b \rightarrow c \rightarrow d is defined as a \rightarrow b \rightarrow c \rightarrow ((d-1)+1), which is defined as a \rightarrow b \rightarrow (a \rightarrow b \rightarrow (a \rightarrow b \rightarrow(...(a \rightarrow b)...)\rightarrow (d-1)) \rightarrow (d-1))
Where the number in the ellipsis is equal to c.
Am I right? What problems are there in my explanation? Is there another source I can check for information on this? ZtObOr 22:25, 9 January 2008 (UTC)
replies: text: The page defines X \to p \to (q + 1) where X stands for some chain. This includes chains of length 4 and more.--Patrick (talk) 00:55, 10 January 2008 (UTC)
replies: text: Right. That explains a lot. Thanks. ZtObOr 00:33, 11 January 2008 (UTC)
thread_title:  4-term chains? 

message: In the definition, the term "complete chain" is confusing and maybe meaningless. I had to read the definition 3 or 4 times to understand it. Apparently, all chains are complete chains. Probably it would be clearer to substitute the terms "chain" and "complete chain" by the binary relation "is subchain of" or "is contained in". 81.35.123.125 (talk) 20:09, 29 April 2008 (UTC)
thread_title:  Complete chains 

message: The page states that 2^...(2^16 times)..^2 is "unimaginably huge".
Someone has marked ths claim as 'needing citation':
http://en.wikipedia.org/w/index.php?title=Conway_chained_arrow_notation&diff=286008611&oldid=251908330
There are plenty of other facts on this page that could do with citations, but requiring one for this is a bit daft.  You could argue that it is understatement, or even subjective.  But it is surely not a citable fact.  —Preceding unsigned comment added by 81.101.133.255 (talk) 08:46, 18 May 2009 (UTC)
replies: text: Agreed, I even laughed out loud at seeing that. I'll remove it for now, although one might want to look into rephrasing this and other parts of the article. 83.227.162.5 (talk) 21:46, 19 May 2009 (UTC)
thread_title:  Unimaginably huge 

message: I've seen it conjectured that for any fixed p ≥ 3, the growth-rate of the function of n represented by the n-arrow chain p→p→...→p is approximately that of fω2 in the fast-growing hierarchy, and, similarly, that the growth-rate of the function of n represented by the n-arrow chain n→n→...→n is approximately that of fωω. (E.g., see Section 5 in this posting.) Can anyone cite sources for proofs adequate to allow including something along these lines in the article? — r.e.s. (talk)  16:45, 2 June 2010 (UTC)
replies: text: Although I haven't been able to find a "citable" source for this, I believe I can show fairly rigorously that the n-arrow chain n→n→...→n is less than fω2(n) for all n, disproving the mentioned conjecture re fωω. — r.e.s. (talk)  06:06, 16 June 2010 (UTC)
thread_title:  Locating ''n''-arrow chains in the fast-growing hierarchy 

message: Is anyone aware of an additional reference (besides The Book of Numbers) from any of Conway or Guy's work?  I noticed above that this was asked in 2006 and it got no reply.  Surely there is more somewhere besides a single paragraph on page 61 in The Book of Numbers.  Chasrob (talk) 13:53, 14 August 2010 (UTC)
thread_title:  Notation references 

message: Has any mathematician suggested or used an exponential notation of conway's chained arrow notation? By which I mean, as 1010 = 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10. It would display 10 → 10 → 10 → 10 → 10 → 10 → 10 → 10 → 10 → 10 as equaling 10 & it's notation to 10. Nagelfar (talk) 23:27, 14 September 2011 (UTC)
thread_title:  An exponential of the Conway chained arrow notation? 

message: Equation 4b: if q is 2 (or more), don't there need to be 2 (or more) q's at the end of the expansion, to be complete???  Chasrob (talk) 01:55, 7 August 2012 (UTC)
replies: text: Whoops, my bad, I see:).  Neato.Chasrob (talk) 01:20, 27 August 2012 (UTC)

text: Wouldn't it be logical to use the recursive steps 4a and 4b instead of step 4 in the example section?  You wouldn't have to worry about bolding, for one reason.Chasrob (talk) 00:38, 28 November 2012 (UTC)
thread_title:  Definition and overview 

message: For example A-->-->B will equal A-->A...-->A-->A with a chain length of B. If B=1, then it will equal A  — Preceding unsigned comment added by Bubby33 (talk • contribs) 21:02, 10 May 2014 (UTC)
thread_title:  Extender notation? 

message: Shouldn't this article be Conway/Guy chained arrow notation? They were co-authors of the book in which it first appeared, yes? Chasrob (talk) 18:13, 29 October 2014 (UTC)
thread_title:  Conway/Guy? 

message: We can define a {1} b as a → b, :
  \begin{matrix}
   a \{n\} b & = & \underbrace{a \{n-1\} a \{n-1\} \dots \{n-1\} a} \\
   & & b\mbox{ copies of }a
  \end{matrix} 
 
So if n = googolplex, than n {n} n is very large, much larger than n [n] n in hyperoperation.  — Preceding unsigned comment added by 140.115.138.5 (talk) 06:02, 5 November 2014 (UTC)
replies: text: It's not well-defined, as the operations are not associative.  — Arthur Rubin  (talk) 19:20, 6 November 2014 (UTC)
thread_title:  Another notation 

message: Just for a thought - can the numbers in the chain be somehow resolved to some operation which can be applied? Especially, it seems that the last member of a chain stands for the number of applications of a certain operation on the remaining members: in a chain of length 1, it is the number of times zero has to be incremented; in a chain of length 2, it is the number of times the first number has to be multiplied with itself; in a chain of length 3, it is the number of Knuth arrows which has to be put between the first and the second member. Can a chain of length 4 be written as some fourth-member-fold operation on the first three members? And, generally, can a chain of length n be written as an n-th-member-fold operation (depending on n) on the first n-1 members?  And if the answer is yes, is there any algorithm to get the operation in question (that is, to say which operation is meant) without actually calculating out the chain?--85.177.249.99 (talk) 12:17, 15 March 2015 (UTC)
replies: text: The background of my question, by the way, is another question: can this notation be sensibly extended to non-natural numbers as chain members? Zero? Negatives? Rationals? Or even complex numbers?--85.177.112.83 (talk) 22:55, 17 March 2015 (UTC)
thread_title:  Connections between the chains and any mathematical operations? 

message: the linked Peter Hurford (musician) is not the one who did the extension of the Conway chains.  — Preceding unsigned comment added by 193.83.106.80 (talk) 21:22, 11 March 2018 (UTC)
thread_title:  wrong link

message: Why the chained arrow notation a \to b \to c is defined as a[c+2]b instead of a[c]b? Why this c is increased by 2? Where a[c]b is the hyperoperation, i.e. a[1]b = a+b, a[2]b = ab, a[3]b = ab, ...  — Preceding unsigned comment added by Xayahrainie43 (talk • contribs) 22:00, 26 August 2018 (UTC)
thread_title:  Definition of this notation 

message: In the current version of Definition and overview section, rule 3 states that p \rightarrow q \rightarrow r is equivalent to p [\uparrow ^ r] q. However, this seems to already follow from other rules.
Rule 4 for the special case of the chain of length 3 can be stated as:
p \to q \to (r + 1) is equivalent to p \to ( p \to ( \cdots (p \to p  \to r)\cdots ) \to r ) \to r (with q copies of p, q-1 copies of r, and q-2 pairs of parentheses; applies for r > 0).
Let f_{p,r}(q) = p \to q \to r. Then, using functional powers, the rule can be more concisely restated as: 
:p \to q \to (r + 1) = f_{p,r+1}(q) = f^{q-1}_{p,r}(p), 
where f^{n+1}_{a,b}(c) = f_{a,b}(f^n_{a,b}(c)) and f^1_{a,b}(c) = f_{a,b}(c). Also note that, using rules 2 and 5, it follows: 
:f_{p,1}(q) = p \to q \to 1 = p \to q = p^q
Starting from this and using the recursion above, it turns out f_{p,2}(q) = f_{p,1}^{q-1}(p) = {}^q p = p \uparrow\uparrow q and, more generally, f_{p,r}(q) = p [\uparrow ^ r] q. 
This implies p\to q \to r = p [\uparrow ^ r] q, which is the statement of the rule 3, which therefore seems redundant since it seems to follow from rules 2, 4 and 5. 193.198.162.14 (talk) 09:41, 22 September 2021 (UTC)
thread_title:  Is rule 3 redundant? 

message: What's the difference between X → (X) and X → X? 2601:58B:4204:B6B0:7873:3EE2:99A:88CD (talk) 21:41, 2 September 2022 (UTC)
thread_title:  Why doesn't this page explain anything clearly? 

message: The 5 step rules seemed adequate to me. So were the examples using the right arrows instead of another notation (up arrows). Chasrob (talk) 02:41, 29 December 2022 (UTC)
thread_title:  5 to 6 rules? Why the examples using another notation (Knuth's)? 

message: ВСЕМ Пламенное Здарова!  — Preceding unsigned comment added by 95.153.161.86 (talk) 11:41, 20 June 2023 (UTC)
thread_title:  Vsem Privetik, Drugany