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)


 * I think that the point where we can call it hyper-hypercombutability is when we get to what I call

\(\omega_1,0^{CK}\), which is the supremum of all ordinal reached by using the usual \(\omega_a^{CK}\) function and recursion. Tomtom2357 (talk) 15:08, June 5, 2013 (UTC)

It could be called \(\theta_1(\theta_1(0,1))\), using my system of notating ordinals described here. Ikosarakt1 (talk ^ contribs) 12:57, June 5, 2013 (UTC)

What I meant was the first ordinal that could not be expressed with the admissible ordinal and recursion (much like \(\omega_1^{CK}\) is the first ordinal not expressible with the function a->w^a and recursion). The function you described uses recursion on the admissible ordinal function, so my ordinal is larger. Tomtom2357 (talk) 15:08, June 5, 2013 (UTC)
 * So, maybe you meant \(\theta_2(0,1)\) (which is above anything that we can express recursively using \(\theta_1(\alpha)\) or \(\theta(\alpha)\))? If not, then how your ordinal relates to my notation? Ikosarakt1 (talk ^ contribs) 19:04, June 5, 2013 (UTC)


 * Yes, \(\theta_2(0,1)\) is what I meant. Tomtom2357 (talk) 09:03, June 6, 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 must disagree. LittlePeng9 (talk) 20:23, 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? Tomtom2357 (talk) 15:11, June 5, 2013 (UTC)


 * 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)

Slight nitpick here: The bound you gave for the Xi function (7*F(n-3)), does not work for n=5, Xi(5)=6<7=7*F(2). By the way, thanks for proving the equivalence of halting and having a normal form. Tomtom2357 (talk) 04:35, June 5, 2013 (UTC)


 * \(\Xi(n)>7F_{n-2}\) for n>6. And yes, \(F_{n-2}\). If I'm mistaken (again) I'll try to improve that. LittlePeng9 (talk) 05:29, June 5, 2013 (UTC)

Now to prove some theorems to reduce the search space of the combinators. First, I will use the term optimal to refer to any length n combinator that beta reduces to a combinator of length Xi(n) (and the reduction halts there), and it does not reduce to any other combinator of length n in between. I will also use little case letters to refer to any combinator of length 1, and upper case letters to refer to any combinator expression.

Deedlit11 has already proved that any optimal combinator must begin with S, and must have the form SABC..., where C has length at least 2, so that the combinator does not beta reduce to another of the same length or less, so I will work from there.

Theorem 1: No combinator of the form SxA(yz), of length n is optimal if n>5.

Proof: SxA(yz) beta reduces to x(yz)(A(yz)). If x=I or k, then the expression will reduce to a shorter one. This has length n or less, so SIA(yz) and SKA(yz) are not optimal. If x=S or Omega, then x(yz)(A(yz)) does not beta reduce further, so SxA(yz) beta reduces to a combinator of length n+1. However, Xi(n)>n+1 for n>5, so SxA(yz) is not optimal if n>5.

Can anyone prove a stronger result than theorem 1? My basic approach is to go through all possible ways of parenthesizing Sxyz..., that fit in with LittlePeng9's proof, and use lazy evaluation to calculate (by hand) whether any combinator has more than the best record. I've been trying to write a python program that systematically goes through each possibility, and I've been able to come up with one that beta reduces any single combinator (I have to figure out the omega parts myself, because it is an uncomputable combinator), but I haven't had much success so far making it work through all combinators. Tomtom2357 (talk) 07:25, June 5, 2013 (UTC)

Actually, a similar argument shows that SwA(x(yz)) can't be optimal, or SvA(w(x(yz))), etc. Tomtom2357 (talk) 01:28, June 6, 2013 (UTC)

SSS(SI)S\(\Omega\) (according to my program, so I need to recheck this) converges to a combinator with 51 characters. Tomtom2357 (talk) 11:05, June 6, 2013 (UTC)

I have checked through all the options, and no combinator of length 7 grows to a combinator of length more than 51. I can now say that Xi(7)=51 (if I haven't made any errors). I admit it, I was wrong when I said that \(\Omega\) was redundant. So, n=7 is the point where \(\Omega\) starts being useful. I wonder when the second oracle would start being useful in the second Xi function. I am also still wondering if k has any usefulness in defining large numbers. Tomtom2357 (talk) 15:01, June 6, 2013 (UTC)


 * Maybe K doesn't have big role at the beginning, but later on it is really helpful to define recursion. In fact, I combinator is not required if we can use S and K. Trees using only S and I are sort of predicable, so either \(\Omega\) or K must take a role. LittlePeng9 (talk) 16:20, June 6, 2013 (UTC)

If at n=7, \(\Omega\) only starts to work, then maybe from n>=8 \(\Xi(n)\) shall be very large? Ikosarakt1 (talk ^ contribs) 15:13, June 6, 2013 (UTC)

The only problem is, I haven't actually proved for each apparently non-halting combinator, that it has no normal form. I have checked each of the non-halting combinators with my program, and they don't appear to halt, but that is not a proof. Would you like me to write back with a list of the non-halting combinators so we can check them (there actually aren't that many, probably about 20 or so)?

I recommend using two new functions: \(\Xi_-1\), \(\Xi_0\), that are lower bounds for the original (\Xi(n)\) function. \(\Xi_-1\)(n) is the largest combinator we can get using n S and I combinators, and \(\Xi_0\)(n) is the largest combinator we can get using n S, K, and I combinators. Obviously, \(\Xi_-1\)(n)<=\(\Xi_0\)(n)<=(\Xi(n)\), so we can use those functions as lower bounds. I wonder if \(\Xi_-1\)(n) is computable, and if so, what function is it? Tomtom2357 (talk) 08:24, June 10, 2013 (UTC)

What I really need a proof of is that any combinator with an (IABC...) expression in it can't be optimal. This would really simplify my analysis of the function. A proof that Xi(n)>=Xi(n-1) for n>7 would work. Tomtom2357 (talk) 09:20, June 10, 2013 (UTC)


 * I thought proof that Xi function is nondecreasing is obvious. As you wish: Let X denote n-1 symbol combinator attaining Xi(n-1). Then IX has n symbols and reduces to X, which then reaches tree of size Xi(n-1). So Xi(n-1) is lower bound for Xi(n), thus Xi(n)>=Xi(n-1). LittlePeng9 (talk) 12:48, June 10, 2013 (UTC)

Woops, should be Xi(n)>=2Xi(n-1) Tomtom2357 (talk) 00:51, June 13, 2013 (UTC)

Also, to find values of the Xi function, I am going to focus on combinators of the form SSS(SI)ABC..., which are the combinators that seem to yield the best results.

How about I focus on working out exact values and lower bounds on the Xi function for small values, and someone else can work on the long term bounds for the function, for example finding when Xi overtakes the busy beaver function, or when Xi(n)>graham's number.

Also, I have found that Xi(8)>=98 (this value being obtained by SSS(SI)S(SO)). Tomtom2357 (talk) 04:03, June 13, 2013 (UTC)

I need some help for the combinator SSS(SI)(SS)O. It has applications of O inside other applications of O.

Actually, I solved this with my new program, it diverges, so my new problem is SSS(SI)S(S(S(S(S(SO))))), Tomtom2357 (talk) 04:48, June 13, 2013 (UTC)

Generalized lower bound doesn't works at n=6, since \(F_4 = 3\) and 7*3 = 21>16. Ikosarakt1 (talk ^ contribs) 16:57, June 13, 2013 (UTC)


 * I stated n>6, not n>=6. But thanks for checking my results anyway. LittlePeng9 (talk) 18:31, June 13, 2013 (UTC)

Also, Xi(9)>=167, Xi(10)>296, Xi(11)>=513 because of SSS(SI)S(S(SO)), SSS(SI)S(S(S(SO))), SSS(SI)S(S(S(S(SO)))) respectively. I'm not sure about SSS(SI)S(S(S(S(S(SO))))), my program is still working on that. Tomtom2357 (talk) 01:20, June 14, 2013 (UTC)

We probably need to improve the bound on the Xi function, because the 7F(n-2) bound is very weak. I'll try to figure out what happens with SSS(SI)S(S(...(SO)))...), to get a better bound. Tomtom2357 (talk) 15:36, June 14, 2013 (UTC)

I tried to calculate SSS(SI)S(S(S(S(S(SO))))) with my new and improved algorithm, and I hit the recursion limit after going into 948 nested uses of the O combinator. What happens when you have infinitely many nested uses of the O function? What can we say about its beta reduction sequence? Tomtom2357 (talk) 07:02, June 15, 2013 (UTC)

Wait, you meant that Xi(12)>=948? Ikosarakt1 (talk ^ contribs) 07:59, June 15, 2013 (UTC)

No, I meant that when you are evaluating SSS(SI)S(S(S(S(S(SO))))), you get to a point where O is at the beginning (and the combinator is at least 4 expressions long, so it will beta reduce), and you have to evaluate the inner expression to see if it reduces to I. Then, while you're evaluating that, you get to another expression with O at the beginning, and you have to evaluate the inner expression of that. My program got to the 948th inner expression (okay, sometimes it was able to determine what happened to a particular combinator, but the eventual depth was 948) before it exceeded its maximum recursion depth. Tomtom2357 (talk) 10:04, June 15, 2013 (UTC)

Since \(\Xi\) function grows much faster than \(\Sigma\) function, how large n we need for \(\Xi(n)\) to beat, say, \(\Sigma(\Sigma(\cdots\Sigma(6)\cdots))\) with \(\Sigma(6)\) \(\Sigma\)'s? What about beating \(\Sigma_{1000}(10^{1000})\) (orcale machine)? &#123;hyp/^,cos&#125; (talk) 00:14, August 22, 2014 (UTC)
 * do you honestly think that these are questions that we can answer you're.so.pretty! 04:17, August 22, 2014 (UTC)
 * I think - but this is pure intuition - that the smallest \(n\) such that \(\Xi(n) > \Sigma(n)\) is around 1000, because we saw in both functions relatively fast functions with few combinators. (I think between 100 and 100,000) So I guess your first question is also around 1000.
 * For answering the second question, I use as growth rate w_w^CK. I think between 100,000 and 1,000,000,000. (Okay, that is a wide range, but I won't do any better guesses) Wythagoras (talk) 16:54, August 22, 2014 (UTC) (btw, I like your new signature)

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)

I found a shorter fixed point combinator that has 9 characters. It is SI(S(K(SI))(SII). First I used the usual fixed point combinator, then applied SI to it, because of a result (that I found in To Mock a Mockingbird) that the only fixed points of SI are fixed point combinators. Anyway, the point is that I found a shorter paradoxical statement: SI(S(K(SI)))(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)) ) which is 63 characters long.

I realized that S(KS)KA=S(KA), and that a couple of the expressions were about equal, so I used this to reduce the overall length of the combinator to 37: SI(S(K(SI)))(SII)(S(S(KS)K)(SII)(S(K(S(S(K(S(KS)K))S)(KK))))OI(SII(SII))). I will post again when I have further reducd the length of the combinator. Tomtom2357 (talk) 10:14, June 23, 2013 (UTC)

I just redid all my steps to getting the combinator and I realized that a shorter solution would be to have a P that beta reduces to OPKI. Also, more efficient search for a combinator A such that Ax=OxKI, gives that A=S(SO(KK))(KI), so using my short fixed point combinator, a 16 character contradictory statement is SI(S(K(SI)))(SII)(S(SO(KK))(KI)). I'll be surprised if anyone can beat that! Tomtom2357 (talk) 11:51, June 23, 2013 (UTC)

Actually, I can beat it, If a combinator P beta-reduces to OPPI, then if it beta reduces to I, then it beta-reduces to P, and never gets to I. If P does not beta-reduce to I, then it beta-reduces to I, contradiction. This is actually slightly easier to construct, and the combinator is: SI(S(K(SI)))(SII)(S(SOI)(KI)), which is 15 characters. Tomtom2357 (talk) 00:39, July 10, 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.

There is another problem: if we haven't enough arguments for S: e.g. SII: what to do? Ikosarakt1 (talk ^ contribs) 19:27, June 5, 2013 (UTC)

Nothing, beta reduction stops. LittlePeng9 (talk) 19:49, June 5, 2013 (UTC)

JavaScript
Can anyone made JavaScript for combinatorial expressions? Ikosarakt1 (talk ^ contribs) 11:33, June 6, 2013 (UTC)
 * i'm on it bro FB100Z &bull; talk &bull; contribs 16:57, June 6, 2013 (UTC)

Undecidable Combinators
What is the smallest combinator whose beta reduction sequence is undecidable under the current axioms of set theory? Or, alternatively, what is the smallest combinator whose beta reduction sequence is unknown and can't be easily computed? Tomtom2357 (talk) 09:34, June 10, 2013 (UTC)

Universal Combinators
I have discovered (read: found on wikipedia) that there is a combinator i that is computationally universal all by itself. It can derive S, K, and I (I=ii, K=i(i(ii)), S=iK=i(i(i(ii)))). Its definition is ix=xSK (in terms of S,K,I it is S(SI(KS))(KK)). The only problem with this combinator is that it is not a supercombinator, it refers to S and K in its definition. I was wondering, is there a combinator that can derive S and K without referring to them in the definition, or is there some sort of impossibility theorem out there preventing that sort of combinator from existing? Tomtom2357 (talk) 13:49, June 12, 2013 (UTC)

Do you mean some combinator, completely separate from S, K and I, which by itself is universal? LittlePeng9 (talk) 15:47, June 12, 2013 (UTC)

Yes, I want a combinator completely separate from S, K, and I that is universal. I realize that it would be equivalent to some SKI combinator, but I want a combinator that all its expressions can reduce without using S, K, or I. Tomtom2357 (talk) 00:56, June 13, 2013 (UTC)

Steps of Xi(6)
This are the 'steps' of Xi(6):

SSS(SI)S

S(SI)S(SI)S

SIS(S(SI)S)

I(S(SI)S)(S(S(SI)S)))

S((SI)S(S(S(SI)S))))

S(I(S(S(SI)S)))(S(S(S(SI)S)))))

I(S(S(S(SI)S))))(S(S(SI)S)))(S(S(S(SI)S))))

S(S(S(SI)S))))(S(S(SI)S)))(S(S(S(SI)S))))

Also, I think we should put the combinators for the values of XI in the article. Wythagoras (talk) 18:07, June 25, 2013 (UTC)

Some of your parentheses are incorrect. It should be:

SSS(SI)S

S(SI)(S(SI))S

SIS(S(SI)S)

I(S(SI)S)(S(S(SI)S))

S(SI)S(S(S(SI)S))

SI(S(S(SI)S))(S(S(S(SI)S)))

I(S(S(S(SI)S)))(S(S(SI)S)(S(S(S(SI)S))))

S(S(S(SI)S))(S(S(SI)S)(S(S(S(SI)S)))). Not many people will be interested in the page if we can't even balance our parentheses. Tomtom2357 (talk) 04:03, June 26, 2013 (UTC)

Also, you say, here are the combinators for 5<=n<=7, but you don't provide the combinator for Xi(7). Tomtom2357 (talk) 07:53, June 26, 2013 (UTC)

No, I wrote it, supposing that Wythagoras will write the steps for Xi(7). Ikosarakt1 (talk ^ contribs) 09:16, June 26, 2013 (UTC)

I've problems with explaining the oracle, and Xi(7) has a long output and many steps, so I think we'd better don't put the steps of Xi(7) in the article. Wythagoras (talk) 09:51, June 26, 2013 (UTC)

Bounds
I've tried to improve the bounds:

XI(n) > 7*F(n-3) for n >= 6

XI(n) > 7*F(n-2) for n >= 7

XI(n) > 7*F(n-1) for n >= 8

Here seems to be a pattern, but 7*F(9) = 7*34 = 238>167. But there is still hope, XI(9) may be much larger then 167. If it is, we have a much stronger bound:

XI(n) > 7*F(2n-9).

Don't place anything of these bounds in the article since I haven't any proof for XI(n) > 7*F(n-1) for n >= 8,  XI(12) must be larger then 623 for example. Wythagoras (talk) 18:51, June 25, 2013 (UTC)

The really much stronger bound will be bound based on recursion around \(\Sigma(n)\). Ikosarakt1 (talk ^ contribs) 19:24, June 25, 2013 (UTC)

Wythagoras' bound is heuristic observation. Of course at some point it will be true (as Xi will outgrow every computable function), but we don't know where. To reach \(\Sigma\) we need to simulate Turing machine. Even for simplest ones and in lambda calculus we would need expressions using thousands of abstractions. Then conversion to SKI can require even exponential more combinators. LittlePeng9 (talk) 19:50, June 25, 2013 (UTC)

We don't know that, it could be that there is some sneaky way to simulate a Turing machine with very few combinators. Tomtom2357 (talk) 11:13, June 26, 2013 (UTC)

Or, there might be some way to obtain large numbers that has nothing to do with Turing machines. Tomtom2357 (talk) 01:20, June 27, 2013 (UTC)

I don't understand whence Xi function takes its power and how it is even possible to simulate, say, something like $$f_{\omega_{\epsilon_0}^\text{CK}}(n)$$. For me, it looks at most at level \(f_{\omega_2^\text{CK}}(n)\), more likely as fast as \(f_{\omega_1^\text{CK}*2}(n)\). Ikosarakt1 (talk ^ contribs) 15:43, June 26, 2013 (UTC)

Is it possible to get one of these starting SKI (and Omega if necessary) things that works chaotically, because if not then surely a computer program could easily have some checks put in for infinities, and be able to (eventually) calculate these? DrCeasium (talk) 16:27, June 26, 2013 (UTC)


 * Of course it is possible, but it might not be easy. As I mentioned earlier, it is hard, but possible, to construct SKI tree which works like Turing machine, so we can simulate one of unresolved 5-state machines. It depends actually on how you define "chaotically", but you can be sure that for every program there will be tree such that this program won't be able to prove that it halts nor that it doesn't. LittlePeng9 (talk) 17:05, June 26, 2013 (UTC)


 * Chaotically? Is it means that starting SKI\(\Omega\) expression will be beta-reduced so that no periods will exist in principle? Once I read that it is impossible for Turing machines, but for SKI\(\Omega\) it is probably possible. Ikosarakt1 (talk ^ contribs) 17:13, June 26, 2013 (UTC)


 * What do you mean with "no periods exist in principle"? As far as I know machine is either periodic or aperiodic, with both cases surely possible. Also, do you mean periodicity of states, tape configurations, overall configurations or something else? LittlePeng9 (talk) 17:27, June 26, 2013 (UTC)


 * I mean neither anything of it. If they're supposed to be "chaotic", then nothing will be predictable from some number of steps, otherwise we just haven't a good definition for the term "chaotic". Ikosarakt1 (talk ^ contribs) 17:31, June 26, 2013 (UTC)


 * So now comes the question of what predictability means. As TM is fully deterministic process we can  predict next step, and step after that etc., same with SKI calculus. But with oracle some sort of nondeterminism comes in - consider program \(P\rightarrow \Omega PI(SII(SII))\). If P stops it reduces to I and stops, otherwise it doesn't stop, so no matter what oracle reduces to we have no contradiction. So in that case, \(SKI\Omega\) is very likely chaotic. LittlePeng9 (talk) 19:15, June 26, 2013 (UTC)


 * You want chaotic, try SSS(SI)S(S(S(S(S(SO))))). If you find some period or somthing that prove it non-halting, or if you find where its beta reduction sequence stops, let me know. Tomtom2357 (talk) 01:26, June 27, 2013 (UTC)


 * Also, SI(S(K(SI)))(SII)(S(SO(KI))(KK)) is such an expression that LittlePeng9 described. It could beta reduce to I or K, but it is impossible to decide. Tomtom2357 (talk) 09:24, June 27, 2013 (UTC)

I too would be very interested to know why \(Xi(n)\) has growth rate \(F_{\alpha}\) where \(\alpha = \omega_{\alpha}^{\text{CK}}\). Some sort of proof that if \(Xi(n)\) reaches \(F_{\alpha}\) it reaches \(F_{\alpha+1}\) and \(F_{\omega_{\alpha}^{\text{CK}}}\) would be very appreciated. Deedlit11 (talk) 23:40, June 26, 2013 (UTC)


 * \(\alpha\rightarrow\alpha +1\) shouldn't be that hard; if we had computable set of trees growing at rate of \(F_\alpha\), then very likely we could construct trees which iterate through them. On the other hand, \(\alpha\rightarrow\omega^\text{CK}_\alpha\) will probably be pretty hard one to formally prove. LittlePeng9 (talk) 08:58, June 27, 2013 (UTC)


 * I believe that the order is \(\alpha \rightarrow \omega_\alpha^\text{CK}\), because on the page by Goucher on the Xi function, he talked about the rank of a combinator. I believe that if there are rank m combinators, then the function is at least \(f_{\omega_m^\text{CK}}(n)\). I also think there are combinators of any rank up to (but not including) \(\omega_\alpha^\text{CK}\). Tomtom2357 (talk) 00:23, July 10, 2013 (UTC)


 * It shouldn't be hard to prove this fact. Harder thing will be to prove that combinators with required rank always exist. LittlePeng9 (talk) 14:30, July 9, 2013 (UTC)


 * I will work on finding combinators of large rank. Obviously there are combinators of any finite rank, so I will work on finding combinators of rank w. Unfortunately, this could be quite difficult, because the beta-reduction sequence of such a combinator will be infinitely long. Tomtom2357 (talk) 00:35, July 10, 2013 (UTC)


 * Uh oh, I just realized that any combinator of rank w will have to beta-reduce to something infinitely long before it beta-reduces to something finite. So this goes back to my earlier question: what do we do with infinitely long beta-reduction sequences? Tomtom2357 (talk) 02:26, July 10, 2013 (UTC)


 * If you ask about something of sort "combinator C1 tests combinator C2 which tests combinator C3 which tests..." then such combinator is not well-founded. Combinator with rank w must first check combinator of rank 0, then combinator of rank 1, then of rank 2 etc. LittlePeng9 (talk) 06:09, July 10, 2013 (UTC)


 * Ok, but what I was getting at is that if a combinator checks a combinator of length 1, then combinator of rank 2, etc, then it will take infinitely many steps to do that, and the combinator will grow to infinite length. What do we do then? Tomtom2357 (talk) 08:00, July 10, 2013 (UTC)


 * There are always combinators that refer to themselves. If P beta reduces to OPPP, then OPKI can be shown to beta-reduce to I. Tomtom2357 (talk) 08:52, July 29, 2013 (UTC)


 * But this combinator is non-well-founded, as it can be shown that if P has rank a it also has rank a+1 and if it has every rank <b it also has rank b, so no ordinal can be rank of P. Goucher's definition of Xi concerns only well-founded combinators, so your P is not competitor for Xi. LittlePeng9 (talk) 09:42, July 29, 2013 (UTC)


 * Ok then, so the function only grows as wwCK. Still cool function though! Tomtom2357 (talk) 10:04, July 29, 2013 (UTC)


 * I think we just crushed Xi function, but I think we should wait a while before editing article, to see what other people think of this. LittlePeng9 (talk) 10:39, July 29, 2013 (UTC)
 * I think we better edit the article, because it has been almost a month, and they're making a lot of changes to the site right now. Tomtom2357 (talk) 00:22, August 23, 2013 (UTC)
 * Okay, I've changed the article, lets see who notices. Tomtom2357 (talk) 13:09, August 23, 2013 (UTC)
 * 2 questions:
 * 1. why is the grow rate \(\omega_\omega^{CK}\)?
 * 2. if the grow rate is \(\omega_\omega^{CK}\), what is the grow rate of \(\Xi_2(n)\) function?
 * Wythagoras (talk) 15:43, August 23, 2013 (UTC)
 * About 1. I hope it's obvious that rank 0 trees reach at most w1CK. Rank 1 trees can use oracle these trees, so they reach w2CK. Similarily rank n trees reach w(n+1)CK. But for rank w trees problem trivialises, as they would have to use oracle on infinitely many trees, giving obvious answer "doesn't halt". So we are stuck at wwCK. I'll make a bit of analysis later for 2. question. LittlePeng9 (talk) 16:25, August 23, 2013 (UTC)
 * What are rank n trees? Wythagoras (talk) 17:06, August 23, 2013 (UTC)


 * Suppose given tree never uses oracle combinator. Then it has rank 0. If it takes under the scope of oracle only trees with rank <n for some n, then rank of tree is minimal n with this property (n can be transfinite). LittlePeng9 (talk) 17:51, August 23, 2013 (UTC)
 * I think (this is totally conjecture), that every nested application of the second oracle combinator contributes w to the rank, and therefore the rank is less than w^2, and therefore the growth is the w^2th church kleene ordinal. Tomtom2357 (talk) 02:04, August 24, 2013 (UTC)


 * Am I the only one interested in the Xi function enough to compute values? I contributed half the values and bounds that are known. Is there no one interested enough to try to find better values? Tomtom2357 (talk) 08:07, September 1, 2013 (UTC)


 * No, your're not the only one. I just found \(\Xi(12)\) ≥ 846, without using program. I found out that the oracle expression don't reduce to I because it will go on forever. I also think \(Xi(13)\) ≥ 1,008 using bound \(\Xi(n) > 7F_{n-1}\) for n≥8. Wythagoras (talk) 10:45, September 1, 2013 (UTC)


 * I like all things which are related to extremely large numbers, but I must learn more programming first, because without aid of good computer program I probably can't improve bounds. Ikosarakt1 (talk ^ contribs) 10:05, September 1, 2013 (UTC)


 * I hope to one day finish my SKI Turing machine, it might be some help with looking at expressions. Of course, it will be only some help, but I can't do better, as I'm better at programming abstract machines that real programming languages :P LittlePeng9 (talk) 11:07, September 1, 2013 (UTC)


 * It actually didn't require much programming on my part. I did make a python program that will betareduce an arbitrary combinator, but that was for the special cases, and for the values for 8, 9, 10, and 11. I didn't have to use it in many cases, it took a little bit of guesswork to find the pattern, but one I found it I was able to test only those combinators for the n=8 case.


 * For the n=7 case, it just took a few hours  days of going through every possible parenthetisation of the combinator, there were about 20, and every possible combinator for that parenthetisation. I used lazy evaluation, which made it a lot faster than a naive search. I only had to look through about 10-30 combinators (on average) for each parenthetisation, as opposed to 4^7=16384. Also, the theorems I proved further up this page really helped to narrow it down. That is why I was really hoping for some improved theorems that will reduce the possibilities even further.


 * Unfortunately, for n=8, the are about 100 possible ways to parenthesize the combinators. I gave up about the 5th one. It may require some more programming to do. I never got my program to be able to check all possible ways, it had too many bugs in it. I can post my python program that reduces each combinator if you want (it may require some explanation though). Tomtom2357 (talk) 13:56, September 1, 2013 (UTC)

When we are using only brute force search, we shall not obtain good bounds for quite large n with reasonable time. The good idea is making a bijection between Xi structures and ordinals in FGH. Ikosarakt1 (talk ^ contribs) 12:43, September 1, 2013 (UTC)


 * P.S: Turing machine?! I'm not sure that is going to work. If you can program it with any less than 1000 states I will be very impressed. It could be possible, but I don't think we will find the program that does that any time soon. Tomtom2357 (talk) 13:56, September 1, 2013 (UTC)


 * If you published your code, that would be awesome. My Turing machine doesn't really do what your program does, it just reduces SKI combinator and halting when O is met. That's why I said it's only some help. LittlePeng9 (talk) 14:11, September 1, 2013 (UTC)


 * I meant not constructing TM which makes that bijection, but estabilish it by hand. For example, as Wythagoras made it in his blog post. We must be able to compare each Xi structure to some ordinal up to and excluding \(\omega_\omega^\text{CK}\). Ikosarakt1 (talk ^ contribs) 14:43, September 1, 2013 (UTC)


 * Oh, okay, that works. Tomtom2357 (talk) 14:52, September 1, 2013 (UTC)


 * Every such bijection must be non-recursive. This is for similar reason Kleene's O is not injective - for every matching of ordinal to tree we can't know if given two trees have same ordinal. LittlePeng9 (talk) 19:25, September 1, 2013 (UTC)


 * Sorry, it took me a while before I noticed this debate.


 * It is possible, with S, K, I and Omega, to construct an oracle capable of deciding any statement in first-order Peano arithmetic. In particular, you can make an 'existential quantifier' which successively tests an expression with each natural number, and reduces to the identity if one of those expressions evaluates as 'true'; otherwise, it would continue forever. Now apply an Omega combinator to this machine. Hence, just by emulating first-order logic we attain the level of \(\omega_\omega^\text{CK}\). But you can then build a machine that systematically checks each statement in first-order Peano arithmetic of a particular form, and use the Omega combinator to obtain an oracle. Then, the Busy Beaver function over *those* machines is \(\omega_{\omega+1}^\text{CK}\).


 * More generally, if we can construct all machines (trees) corresponding to an ordinal \(\beta\), then we can also get \(\beta+1\) by constructing an oracle for those machines. Also, we can take a limit of a sequence of ordinals, provided we have a computable (by S,K,I,Omega) way of systematically constructing each of those machines. This would give \(\omega_{\omega_1^\text{CK}}^\text{CK}\) at the very minimum, by using a Turing machine to systematically construct well-founded machines of increasing complexity. But then machines of this form could be used to construct trees, giving \(\omega_{\omega_{\omega_1^\text{CK}}^\text{CK}}^\text{CK}\), and so on. I don't think it's possible for a single tree to perform these successive complexity jumps (but haven't proved that), which would mean that the lower bound \(\alpha = \omega_{\alpha}^\text{CK}\) is tight.


 * -- Adam P. Goucher, 2013-09-09, 14:17


 * I'm not sure about that. If you could give us some examples of machines that check every statement in first order peano arithmetic, that would help. Also, an example of a rank w combinator would be nice.


 * Also, since you are the function's inventor, could you clarify a few things? If, within the oracle, the machine runs indefinitely, but we know that it never reaches I, can we still use that combinator? We have also encountered problems with possibly infinite nesting of oracles. Would you allow the combinator to be used if we could somehow prove that one of those nested combinators never reached I? Tomtom2357 (talk) 14:12, September 9, 2013 (UTC)

The reason that the argument doesn't work is that it is analogous to saying that an oracle machine capable of solving the regular halting problem only needs to check Turing machines that do not run forever. However, such an 'oracle' would only be as powerful as an ordinary Turing machine -- it's the fact that it can handle the non-halting machines that makes a difference.
 * Thank you for your comment in this debate, would you be able to point out why argument I used doesn't work? I understand your argument, but it seems to simply contradict mine.
 * Tomtom, if a combinator checked in order OIII, O(OIII)II, O(O(OIII)II)II... it would check combinators of every finite rank, thus having rank w. I'm too lazy to actually make such job, but I think you see why such combinator exists. LittlePeng9 (talk) 14:55, September 9, 2013 (UTC)
 * I think the strength is more than \(\omega_\omega^\text{CK}\) beacause of \(Xi(12)\). \(Xi(12)\) has already over 948 nested oracle operations. And for \(Xi(>12)\) there can be a lot of nested oracle operations. Wythagoras (talk) 16:35, September 9, 2013 (UTC)

It's also best to think not of explicit closed combinators, but rather 'function' combinators that accept a natural number* parameter: no individual case of the halting problem is undecidable, whereas the general halting problem is undecidable. It's then possible to make an automaton (also a combinator tree) that tests a particular function over any recursive (or indeed arithmetic) set of inputs.


 * encoded as a Church numeral, naturally.

-- Adam P. Goucher, 2013-09-09, 18:52

Number of possible expressions
How many ways there are to parenthesis a string with n terms long? I guess that (n-1)!, and found that it holds true for 1 <= n <= 5. If it is indeed true, then, as we have 4^n strings for Xi(n) (4 combinators, n of them), then the resulting number of expressions for Xi(n) will be 4^n*(n-1)! Can anyone verify it? Ikosarakt1 (talk ^ contribs) 17:56, June 29, 2013 (UTC)


 * While it is true that the number of ways to order the operations of a string with n terms is (n-1)!, this is different from the number of ways to parenthesize a string with n terms. You can see this by looking at (ab)(cd); there are two ways to perform the operations: you can either multiply a and b and then c and d, or c and d and then a and b. But the order which you do this doesn't matter; you get the same answer in the end. So the number of ways to parenthesize a sequence of n terms will be less than (n-1)!. The answer is C_{n-1}, where C_n = C(2n,n)/(n+1) = (2n)!/(n!(n+1)!). C_n is known as the Catalan sequence, and comes up in a bunch of different places, for example C_n is the number of simple walks from (0,0) to (2n,0) that stays above the y-axis. Anyway, the number of expressions for Xi(n) will be 4^n * C_{n-1}. Deedlit11 (talk) 18:21, June 29, 2013 (UTC)

New bound for XI(n)
I found

\(\Xi(12) \geq 846\),

\(\Xi(13) \geq 1464\) and

\(\Xi(14) \geq 2526\). And I found a new general bound:

\(\Xi(n) \geq (23n+48)F_{n-8}+(14n+29)F_{n-9}+1\) for \(n>11\)

Wythagoras (talk) 11:50, September 1, 2013 (UTC)

Can I see the combinators for those? Tomtom2357 (talk) 13:55, September 1, 2013 (UTC)

The combinators are in the form of SSS(SI)S(S(...(S(SΩ))...)).

I found the general bound when I looked to the halting combinators for Xi(12), Xi(13) and Xi(14): they are Ω(a)((b)a), Ω((b)a)(a((b)a)) and Ω(a((b)a))(((b)a)(a((b)a))). I noticed that they will grow like the Fibonnacci sequence, combinator a will grow with 23 and combinator b with 14. Wythagoras (talk) 16:01, September 1, 2013 (UTC)

Did you actually evaluate these combinators? I tried, but my python program choked at a depth of 948 nested O applications.

I've evaluated the combinators for xi(12), xi(13) and xi(14). You can stop with trying to evaluate infinite nested Ω applications. It won't reduce to anything, so it doesn't reduce to I, so the output is z. after that I got a new S(S(...(S(SΩ))...)), but when you've evaluated all the S, you can't beta-reduce because there are only 2 arguments, in case of xi(12) they are (a) and ((b)a) with b a string of length 197 and a a string of length 324. Wythagoras (talk) 05:15, September 2, 2013 (UTC)

Ok, try it with xi(8), xi(9), xi(10), xi(11), and see if it still works. Then I will agree with it. Tomtom2357 (talk) 05:44, September 2, 2013 (UTC)

If tree has infinite nested oracles, it is necessarily not well-founded, thus its value can't be bound for Xi(n). LittlePeng9 (talk) 06:36, September 2, 2013 (UTC)

This is the point where we make a decision: Do we stick with the original definition that doesn't have a very good growth rate, or do we allow combinators that might have an infinte nesting of the O combinator. I vote for the second, it makes it much easier to use. You can do things like prove that the nested combinator will never reduce to I, so avoid the computation altogether. It also mat allow for slightly faster growth.

Although, I do believe that the combinators above do not have infinite nested oracles, it is just nested so deep that it is really hard to get to the bottom of it. I think that there is a better combinator than the ones we have found. Tomtom2357 (talk) 07:43, September 2, 2013 (UTC)

Using the combinator A=S(S(KS)K)I (Axy=x(xy)), the combinator SSS(SI)(AAAASO) goes to a combinator of length >F(65536) (F is fibonacci). Extending this by adding more A's: Xi(6n+7)>F(2^^n). 182.52.92.46 00:06, December 9, 2013 (UTC)


 * Are you sure your parentheses are good? Also, what do you mean with Axy=x(xy)? But, when your bound is correct it will be a great bound! Wythagoras (talk) 16:47, December 9, 2013 (UTC)

Yet another question
Infinite time SKIO calculus, anyone? FB100Z &bull; talk &bull; contribs 20:53, October 7, 2013 (UTC)

I think someone should first think of IT SKI, then try to add oracle. LittlePeng9 (talk) 21:14, October 7, 2013 (UTC)

Why Xi is uncomputable
What's the proof that SKI calculus is even Turing complete? Can anyone give an example of non-halting expression? Ikosarakt1 (talk ^ contribs) 10:03, June 21, 2014 (UTC)


 * SKI is complete by reduction from lambda calculus. Reduction is simple, proof of correctness - not so much.
 * SII(SII) has no normal (irreducible) form. LittlePeng9 (talk) 10:22, June 21, 2014 (UTC)

New Bounds?
I've had an attempt at making some new bounds (found here). If they're correct, then Xi(50) > Graham's number. DrCeasium (talk) 17:29, August 19, 2014 (UTC)

How does it work?
There is NO definition on the article about strings of S, K and I, neither for S having less than 3 entries and K having less than 2 entries.

Are new googologists gonna understand that?

190.96.244.228 18:29, July 12, 2015 (UTC)

Woops, didn't log in.

-- From the googol and beyond -- 18:29, July 12, 2015 (UTC)


 * I did, but I'm not sure about the others. Examples of how the strings evaluate should be enough to clarify that S, K, and I don't evaluate when they don't have enough entries. King2218 (talk) 18:50, July 12, 2015 (UTC)

And what would S, K, I, and Ω (the symbol form) mean?

-- From the googol and beyond -- 19:01, July 12, 2015 (UTC)


 * it's there King2218 (talk) 19:05, July 12, 2015 (UTC)


 * If the leftmost combinator doesn't have enough arguments, beta-reduction terminates. I have clarified this in the article. -- vel! 19:09, July 12, 2015 (UTC)

I mean, it doesn't define if it has no entries, it's just the symbol itself.

And if it's already defined, what does a string of symbols which is not on a mean?

-- From the googol and beyond -- 19:48, July 12, 2015 (UTC)


 * I can't make heads or tails of this comment. -- vel! 21:42, July 12, 2015 (UTC)

Then how does SKIΩ Works?

190.96.244.228 23:22, July 12, 2015 (UTC)

Forgot to log in again.

-- From the googol and beyond -- 23:23, July 12, 2015 (UTC)
 * SKIΩ is described in the article. If you have difficulties understanding the definition, then please try to phrase your questions more comprehensibly. I'm sorry, but I don't understand "it doesn't define if it has no entries, it's just the symbol itself."
 * I can try to take a stab at what's causing the misunderstanding. The combinator notation K(I,S,I) looks like a function call, but combinators are not functions. Think of S(I,K,I) as a meaningless tree of letters, and beta-reduction is an operation that we apply to it that transforms it into K(I,I(I)) and then I. I will revise the definition so it no longer uses =, to make it more clear that we manipulate abstract symbols and not standard mathematical notation.
 * Overall, I find our current introduction to SKI to be poor and lacking in clarity. I'll keep revising it. -- vel! 23:55, July 12, 2015 (UTC)

They don't define #@($)# or #$$$...$$$# where # is the part of the expression before #, @ is any valid expression except S, K, I or Ω, and $ is any valid expression.

-- From the googol and beyond -- 01:52, July 13, 2015 (UTC)

Why does nowiki have no effect in LaTeX?

-- From the googol and beyond -- 01:55, July 13, 2015 (UTC)


 * I've made some revisions to the definition. Let me know if it's easier to understand now. -- vel! 06:08, July 14, 2015 (UTC)

Sorry, but it isn't any better.

1. It doesn't define what things inside mean.

2. It still doesn't define what S or K or I (as entries) mean.

3. It doesn't define the order of entries.

190.96.244.228 20:30, July 14, 2015 (UTC)

Sorry, but it isn't any better.

1. It doesn't define what things inside mean.

2. It still doesn't define what S or K or I (as entries) mean.

3. It doesn't define the order of entries.

-- From the googol and beyond -- 20:31, July 14, 2015 (UTC)


 * The misunderstandings are because you haven't quite shaken off the old definition yet. Try looking up what a "binary tree" is and don't think of the combinators as "meaning" anything. -- vel! 20:43, July 14, 2015 (UTC)

Oh! But it doesn't define in what order to resolve the tree.

-- From the googol and beyond -- 02:16, July 16, 2015 (UTC)