Talk:Xi function

I think this is where we start hitting the ceiling as to where googology can go.

Next stop: infinity. FB100Z &bull; talk &bull; contribs 23:00, February 5, 2013 (UTC)

Whatsoever, infinity is not reachable, because it is not a number. There is also no largest finite number. Take something like \(\Xi(\Xi(\Sigma(100)))\) = n. To prove that isn't the largest, just double it. We know that for any n>0: \(2n > n\), so n isn't the largest. Ikosarakt1 (talk) 08:59, February 6, 2013 (UTC)


 * Yeah, but \(\Xi(\Xi(\Sigma(100)))\) is ugly. Without creating a salad ordinal, where can we go after \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\)? FB100Z &bull; talk &bull; contribs 19:52, February 6, 2013 (UTC)

Probably it is \(\omega_1\), the first uncountable ordinal. Is it limit of sequence of \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\)? Ikosarakt1 (talk) 22:02, February 6, 2013 (UTC)

I doubt it. Any countable sequence of countable ordinals has a countable limit ordinal. Just think about it. If every ordinal in a countable sequence of ordinals is countable, you can show that the limit ordinal is countable simply by diagonalizing across w^2. It's basically impossible to reach w1 via "w" just like its impossible to reach "w" via finite arithmetic. Sbiis Saibian (talk) 23:26, February 6, 2013 (UTC)


 * I don't think they're equal, but if they are, then the xi function is the end of googology. If not, what is there between \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\) and \(\omega_1\)? FB100Z &bull; talk &bull; contribs 23:24, February 6, 2013 (UTC)


 * perhaps there is no established ordinal notation after this point. That would be a perfect opportunity for googology to do something new and invent some Sbiis Saibian (talk) 23:28, February 6, 2013 (UTC)
 * So, then, we come to a Big Question:

What's between computability theory and infinity?


 * FB100Z &bull; talk &bull; contribs 23:31, February 6, 2013 (UTC)


 * I don't think that's a question that can be answered. Robert Munafo makes the point that if we went beyond formalism and computability theory, we'd by definition have nothing to define. There probably is no hard limit that we are going to bump into in googology, but ... that will still all be within a formalism that we can't escape, and that will place a certain fundamental limit on what kinds of finite numbers we can express. The catch is, we'll probably never be able to describe the size of that box without barry paradoxes or worse. So some finite numbers will probably always be inaccessible to us. We know they must exist, but that's about all we can say without being self-contradictory. If it's true that we can't escape computability theory then it proves the point I was trying to make with my site: that saying we can "continue indefinitely" is not exactly true. Not at least in the sense that there is no fundamental limit Sbiis Saibian (talk) 23:41, February 6, 2013 (UTC)

Higher-order xi functions
Any guesses about the growth rates of the order-2 xi function, etc.? FB100Z &bull; talk &bull; contribs 00:20, February 7, 2013 (UTC)

Maybe \(f_{\alpha \times 2}(n)\), where \(\alpha\) is the ordinal for the normal Xi function? It is by analogy with \(\Sigma(n)\) and \(\Sigma_2(n)\). Ikosarakt1 (talk) 20:56, February 8, 2013 (UTC)

Certainly big underestimate. Adam Goucher admited he was wrong when he first wrote about strength of \(\Sigma_2(n)\). It is actually \(\omega_2^{CK}\), well over \(\omega_1^{CK} \times 2\). My personal guess, without any analysis, is as follows: consider function \(\alpha \rightarrow \omega_\alpha^{CK}\). Strength of Xi function is smallest fixed point of that function (which is guaranteed, because it's a normal function). But there are other fixed points - actually infinitely many of them. Let \(f_\alpha \) be \(\alpha\)-th such fixed point. Then I think strength of \(\Xi_2\) function is smallest fixed point of that function. If we define higher order combinators, then strengths of corresponding Xi functions would be created according to Veblen hierarchy based on \(\alpha \rightarrow \omega_\alpha^{CK}\).

But again, this is only my guess, but with some effort I think I could prove that it is lower bound LittlePeng9 (talk) 21:40, February 9, 2013 (UTC)
 * Stop it, you're making me dizzy! :) FB100Z &bull; talk &bull; contribs 00:59, February 10, 2013 (UTC)

Define the Veblen function with \(\varphi^\text{CK}_0(\alpha) = \omega^\text{CK}_\alpha\). That makes Goucher's ordinal \(\varphi^\text{CK}_1(0)\), and the ordinal you just defined is \(\varphi^\text{CK}_2(0)\). So, assuming your guess is correct, it seems natural that further xi functions would be \(\Xi_m(n) = O(f_{\varphi^\text{CK}_m(0)}(n))\). Eep. FB100Z &bull; talk &bull; contribs 17:11, February 21, 2013 (UTC)
 * Okay, challenge problem: devise a function with a growth rate of \(\Gamma^\text{CK}_0\) (the supremum of this "upper Veblen hierarchy"). FB100Z &bull; talk &bull; contribs 17:15, February 21, 2013 (UTC)
 * In every \(\Xi_m(n)\) we are allowed to use oracles with indexes up to and including m. But to define function with ordinal \(\varphi^\text{CK}_\omega(0)\) we need to alter this a bit. Going to hypercomputability makes weird things to ordinal properties. Namely, limit of admissibles isn't always admissible (e.g. \(\omega_\omega^\text{CK}\) isn't admissible), similarly there is no \(\Omega_\omega\) combinator. But it's not a problem - if we allow every possible \(\Omega_n\) combinator to appear corresponding \(\Xi_\omega(n)\) will be what we want. This is what we'll do for limit ordinals. Standard oracles will work for succesors (i.e. \(\Omega_{\omega+1}\) will do). Using this we can ad hoc create function you wished. But it isn't what you expected, is it? Don't worry, right now I'm trying to find way to make function reaching \(\Gamma^\text{CK}_0\), and even further! (Have we already reached point at which we'll call it hyper-hypercomputability?) LittlePeng9 (talk) 18:47, February 21, 2013 (UTC)

Values
We know some exact values and lower bounds for \(\Sigma(n)\), why not make it for \(\Xi(n)\)? At least, what is the value of \(\Xi(1)\)? (I believe it is not so unimaginably large). Ikosarakt1 (talk ^ contribs) 20:12, May 18, 2013 (UTC)
 * For Xi(1) we must have only one combinator. Single combinator can't beta reduce, so Xi(1)=1. If we have 2 combinators, either left one is I and tree reduces to single combinator, or we can't reduce. So Xi(2)=2. Similarily Xi(3)=3. I think also Xi(4)=4, although I didn't check all possibilities. LittlePeng9 (talk) 21:41, May 18, 2013 (UTC)
 * Good, but maybe at least \(\Xi(10)\) is already extremely large? Ikosarakt1 (talk ^ contribs) 21:45, May 18, 2013 (UTC)


 * I doubt it. I think we need more combinators to make good use of oracle, so for Xi(10) we don't have to take much care about it. LittlePeng9 (talk) 07:24, May 19, 2013 (UTC)


 * For \(\Xi(4)\), we observe that reducing I and K always reduce the length of the string, and reducing S will keep the string at the same length, unless z has length greater than 1. But for a length 4 string you can't have z greater than 1, so \(\Xi(4) = 4\). This also reduces the possibilities for \(\Xi(5)\), so perhaps we can prove that \(\Xi(5) = 6\). Deedlit11 (talk) 09:44, May 19, 2013 (UTC)


 * Now I see that \(\Xi(n)\) doesn't "explodes" so fast. Ikosarakt1 (talk ^ contribs) 09:43, May 19, 2013 (UTC)


 * Suppose leftmost combinator in n-combinator tree is I, K or \(\Omega\). For every case, after single reduction we have smaller tree which will leave same output. So, if we want to make an improvement from smaller trees, leftmost combinator must be S. But then, to make proper reduction, we need tree to have height at least 4 (including root). It makes things much easier for small trees. LittlePeng9 (talk) 10:15, May 19, 2013 (UTC)


 * But \(\Xi(10^6)\) more than for example BB(TREE(3)) or BB(BB(..BB(1)...))? Konkhra (talk) 11:20, May 19, 2013 (UTC)
 * \(\Xi(10^6)\) is very likely to be larger than BB(TREE(3)) or BB(BB(..BB(1)...)). By the way, BB(BB(..BB(1)...)) = 1 because BB(1) = 1, BB(BB(1)) = BB(1) = 1, etc. -- I want more clouds! 11:35, May 19, 2013 (UTC)
 * Xi is the current winner. $Jiawhein$\(a\)\(l\)\(t\) 12:17, May 19, 2013 (UTC)


 * Rayo's function is way way way WAY faster growing. LittlePeng9 (talk) 12:36, May 19, 2013 (UTC)


 * Aarex's function is faster than Rayo's function. AarexTiao 16:36, June 4, 2013 (UTC)


 * I can confirm that \(\Xi(5)=6\), and there are 160 trees reaching this value - trees of form S(Sx)ab or Sxa(bc) where a,b,c are any combinators and x is either S or \(\Omega\). 6-combinator trees are smallest which can explode to infinity, so analysis there may be harder. I found \(\Xi(6)\geq 9\). LittlePeng9 (talk) 13:16, May 19, 2013 (UTC)
 * Okay, almost certainly \(\Xi(6)=11\) with this score reached by SII(SIS). I think I checked all possibilities, though I'm not sure. LittlePeng9 (talk) 18:10, May 19, 2013 (UTC)


 * Good work. A quick proof that \(\Xi(5)=6\): In order to increase past 5 symbols, the string must be S(x,y,z) with z length 2.  We get x(z,y(z)), and x must be length 1.  If x is I, K, or \(\Omega\), the string will decrease below 5, which is no good.  But if x is S, the beta reduction will stop, at length 6. Deedlit11 (talk) 02:24, May 21, 2013 (UTC)


 * This proof has a flaw - tree S(SS)SS also gives 6 combinators, though z part has one combinator. Also, if we want tree to halt, \(\Omega\) also will do. LittlePeng9 (talk) 05:33, May 21, 2013 (UTC)


 * The point is that, for any string that starts at 5 symbols and ends at more then 5 symbols, there must be a point at which it goes from 5 to more than 5 and never goes back down again. This point can only be where the string is S(x,y,zw), and after that it can only stop or get smaller, so six is the maximum.  S(SS)SS goes to SS(SS(SS)) which is of the desired form, so it fits into the proof just fine. Deedlit11 (talk) 13:05, May 25, 2013 (UTC)

It's interesting how the really fast-growing functions often have extremely slow starts. --Ixfd64 (talk) 17:46, May 19, 2013 (UTC)
 * Nonetheless, I still think that \(\Xi(10)\) can be very large, because from some argument this function can suddenly explode. Ikosarakt1 (talk ^ contribs) 18:35, May 19, 2013 (UTC)
 * The 'explosion thresholds' for some fast-growing functions are n(4), Circle(2 or 3), FuseMargin(3), Hydra(4), TREE(3), SCG(1), BH(3), Sigma(5 or 6), and Xi(around 10).
 * It seems to me that Rayo starts out extremely slow. I don't think it gets big until around Rayo(200) or so; we need a few symbols before we can take advantage of the full power of set theory. FB100Z &bull; talk &bull; contribs 18:55, May 19, 2013 (UTC)
 * For n(k) I believe it is n(3), it is already unimaginably large (in the arithmetical sense). Ikosarakt1 (talk ^ contribs) 19:02, May 19, 2013 (UTC)
 * Oops, my bad. FB100Z &bull; talk &bull; contribs 19:11, May 19, 2013 (UTC)


 * It doesn't have arithmetical sence, because imaginability is subjective, non-mathematical concept xD But I agree. Even though it is bounded by AA(5) it is still very big. LittlePeng9 (talk) 20:05, May 19, 2013 (UTC)

First general lower bound - for all \(n\geq 5\) we have \(\Xi(n)>F_{n-3}\). So past this point function grows at least exponentially. Sketch of construction: I noticed that tree SII(SIx) with x being unspecified (yet) combinator reduces to tree x(4(6)) with numbers representing subtrees of given size. Note that previous reductions are independent of x. If x is S, then we have non-reducable tree of size 11. Now say x=Sy. Then tree Sy(4(6)) reduces to (y(6)(4(6))), or y(6(10)) which is larger than y(5(10)). Now, generally, if we have y(a(b)) and y=Sz, tree reduces to z(b(a+b)), which defines Fibonacci-like sequence starting with 5 and 10. With some work we can get bound I gave. Also, I found \(\Xi(7)\geq 24\). LittlePeng9 (talk) 15:42, May 20, 2013 (UTC)

I'm sure this is wrong, but SSS(SI)S seems to expand to S(S(S(SI)S))(S(S(SI)S)(S(S(S(SI)S)))). It has 17 characters, which contradicts the result that Xi(6)=11. I have checked all the other possibilities, so if my calculation of SSS(SI)S is correct, then Xi(6)=17.

Also, can you give an example where the oracle combinator helps make the function bigger? I mean where replacing an I,K, or S with the oracle combinator makes the end result have more characters than it would otherwise.

Actually, K seems redundant too. Is there even a combinator where replacing any S by K or the oracle combinator actually helps? Tomtom2357 (talk) 01:18, May 31, 2013 (UTC)
 * Okay, I shall improve the value of \(\Xi(6)\), according to you. Ikosarakt1 (talk ^ contribs) 12:36, June 4, 2013 (UTC)

LittlePeng9, could you write the combinator of length 7 that you found that beta reduces to 24 characters?


 * I thought that combinator was SII(SI(SS)), but when I counted again it gave size 21, so I don't have combinator returning 24 at this moment. BUT I found one with output of size 40: SSS(SI)(SS) and I strengthened a general bound: \(\Xi(n)>7F_{n-3}\text{ for n}\geq 5\). LittlePeng9 (talk) 15:03, June 4, 2013 (UTC)

I am working with the idea that if a combinator does not have a normal form, then its beta reduction does not converge. Unfortunately, I don't have a proof of this. Is there a proof of this somewhere? It would allow me to rigorously prove that Xi(6)=17. Also, is there a proof of the converse statement that if a combinator does not converge it does not have a normal form (or, equivalently, that if a combinator has a normal form then it converges)? If the two statements were equivalent, it would be a whole lot easier to prove halting/non-halting of any combinator's beta reduction. Tomtom2357 (talk) 14:55, June 4, 2013 (UTC)

By definition if combinator doesn't have normal form, every form we reduce it to can be further beta-reduced. Because standard beta reduction takes place only on leftmost combinator, beta reduction is deterministic, so converse statement also is true. So both statements are equivalent, thus so are their negations (there exists normal form -> reduction converges). LittlePeng9 (talk) 15:08, June 4, 2013 (UTC)

Known non-well-founded statements
I know that, using Godel's theorem, we can construct paradoxical combinators. However, this combinator will probably be quite large. Is there a small example of a paradoxical combinator? Tomtom2357 (talk) 10:07, May 30, 2013 (UTC)
 * I'm not sure, but I suspect that the size of the smallest paradoxical combinator is related to when \(\Xi_2\) overtakes \(\Xi\). (Disclaimer: I'm not sure if we have a solid definition for \(\Xi_2\) yet.) FB100Z &bull; talk &bull; contribs 16:45, June 1, 2013 (UTC)


 * First of all, this can be derived without Godel's theorems. Trick is to see that system at least as strong as Turing machines can make any computable operation on input string. Specifically, there is program P which reduces to \(\Omega P (SII(SII)) I\). Basically, if P reduces to I, \(\Omega P (SII(SII)) I\) becomes SII(SII) which grows infinitely, contradiction. But if P doesn't reduce to I, \(\Omega P (SII(SII)) I\) reduces to I, contradiction. Such program P can be derieved from existence of . LittlePeng9 (talk) 18:13, June 1, 2013 (UTC)


 * In the above example, the smallest paradoxical statement I could come up with is SII(LA), where Lxy=x(yy), and Ax=Omega x(SII(SII))I. However, this is about 66 characters long Tomtom2357 (talk) 08:11, June 2, 2013 (UTC)


 * If you could explictly construct such statement you would make step into combinators which really use \(\Omega\). At such size, Xi function may break out beyond Church-Turing thesis. LittlePeng9 (talk) 09:20, June 2, 2013 (UTC)

P=SII( S(S(KS)K)(K(SII) )(( S(KS)K (S(KS)K( S(S(K(S(KS)K))S)(KK) )) (S(KS)K(S(S(K(S(KS)K))S)(KK))(S(KS)K(S(S(K(S(KS)K))S)(KK)))) ) OI(SII(SII)) ))

is a paradoxical statement (O is the oracle combinator). This isn't quite the P you suggested, because I shortened it a little using SII, so it would beta-reduce into the combinator that LittlePeng9 suggested. It is 66 characters long, but I'm sure there is a shorter statement, because with mine, the site I used to find the combinators used just S,K, rather than S,K,I.

Also, I have another question: The first oracle combinator decides whether any combinator beta reduces to I, the second combinator decides whether any combinator is contradictory. In the article it can only decide whether a SKI(Omega) combinator is contradictory, but I see no reason why it can't just decide whether any combinator is contradictory (yes, that may allow for a contradiction, but then, the 2nd combinator can work from that, so that isn't a problem). However, what does the third oracle combinator do? Tomtom2357 (talk) 03:25, June 4, 2013 (UTC)

About another question - I guess this is minor mistake in article, \(\Omega_2\) should be allowed to refer to own statements. Third combinator checks if it is "super-contradictory". Such statement gives paradox even if asked whatever it gives paradox. Example - suppose program \(P_2 \rightarrow_\beta \Omega_2 P_2 P I\). I'll use convention \(\Omega_2xyz\) returns y if x is well behaved and z otherwise. If P_2 is well behaved, it will reduce to P, which isn't well behaved, contradiction. If P_2 isn't well behaved, it'll reduce to I, which is obviously well behaved, contradiction. \(\Omega_3\) seeks for such programs exactly. LittlePeng9 (talk) 08:55, June 4, 2013 (UTC)

I suggest a better version of the oracle combinator: Oxyz=y if (and only if) x beta reduces to y and assigning y does not lead to a contradiction, and z otherwise. Then my P above beta reduces to Q, which beta reduces to OQ(SII(SII))I, would then beta reduce to I, but this is no contradiction, because the conjunction of "x beta reduces to I" and "assigning y to the formula does not lead to a contradiction" is false, because Oxyz=y is a contradiction because it implies that x beta reduces to I and that it does not beta reduce to I. I don't think that there is any contradictory combinator with this new oracle. This does not change the output of the Oxyz at any other time than when it would be a contradiction under the original definition. Tomtom2357 (talk) 10:07, June 4, 2013 (UTC)

3-argument I
How to handle I(x,y,z)? Ikosarakt1 (talk ^ contribs) 14:23, June 4, 2013 (UTC)

When we have any combinator \(xyz...\) we think of it as about \(((...(xy)z)...\) so Ixyz is actually (Ix)yz=xyz. LittlePeng9 (talk) 14:35, June 4, 2013 (UTC)

If I have IxyzKxyzSxyzIxyz, I must evaluate I's first, then K, then S? Ikosarakt1 (talk ^ contribs) 14:49, June 4, 2013 (UTC)

Your combinator is equivalent to ((((Ix)y)z)K)... so first we evaluate I, then output of Ix (which is x), then output of xy etc.