Loader 39;s Number
Leroy Turcios
Loader's number is the output of loader.c, a C program by Ralph Loader[1] that came in first place for the Bignum Bakeoff contest, whose objective was to write a C program (in 512 characters or less) that generates the largest possible output on a theoretical machine with infinite memory.[2][3] A layman's explanation of the ideas behind loader.c can be found in the last two videos of this Youtube playlist on the Bignum Bakeoff contest.
Ralph Loader explains in the document of the program[3] that the program implements a parser, type-checker, interpreter and proof-search for the Huet-Coquand "Calculus of Constructions" (CoC), a particularly expressive lambda calculus, and David Moews stated[2] that the program diagonalizes over the CoC, but there is no actual written proof for their statements. Its output, affectionately nicknamed Loader's number, is defined as \(D^5(99)=D(D(D(D(D(99)))))\), where \(D(k)\) is an accumulation of all possible typing judgements (a statement that an expression has a some type under some context) provable within approximately \(\log(k)\) inference steps in the calculus of constructions (encoding proofs as binary numbers and expressions as power towers). The proof strength and expressiveness of the calculus of constructions makes the output of \(D(k)\) grow very large for sufficiently large k.
The final output of \(D^5(99)\) is the Loader's number, which is much larger than \(D^2(99)\) but comparison to other large numbers is not easy.[note 1] Loader's function is computable, so \(\Sigma(n) > D^5(99)\) for relatively small n, say, n = 2000, where \(\Sigma(n)\) is the Busy beaver function.
Note that this is just the number you get by removing the 0s at the end of x along with the 1 preceding the 0s, which is exactly the way the the left subtree is encoded in x. Since Z(x) stores the return value in a variable r, when calling Z(x) we get the left subtree as output and the right subtree is automatically stored in the variable r.
Loader's number is a very large number, that is kind of hard to explain (since it was itself the result of a code golfing exercise with a flexible goal). There is a definition and explanation here, but for the purposes of self-containment, I will attempt to explain it later in this post as well.
The algorithm Ralph Loader used produces one of the largest numbers of any (computable) algorithm ever written! Indeed, Loader's number is the largest "computable" number on the Googology Wiki. (By "computable" number, they mean a number defined in terms of a computation.) That means that if answer produces a number larger than Loader's number in an interesting way (i.e. not just Loader's number+1), you could go down in Googology history! That being said, programs that produce something like Loader's number+1 are definitely valid answers and contenders to this question; just don't expect any fame.
First of all, every syntactically valid program terminates. There are no infinite loops. This will be very useful, because it means that if we run an arbitrary calculus of constructions program, our program will not get stuck. The problem is that this implies the calculus of constructions is not Turing complete.
Second of all, among non-Turing complete languages, it is one of the most powerful. Essentially, if you can prove that a Turing machine will halt on every input, you can program a function in the calculus of constructions that will simulate it. (This does not make it turing complete, because there are halting turing machines that you can not prove are halting.)
In particular, loader.c defines a function called D. Approximately, D(x) iterates over all bit-strings less than x, interprets them as a coc programs, runs the syntactically valid ones, and concatenates the results (which will also be bitstrings). It returns this concatenation.
This is my first time submitting a program to code golf, but certainly not my first time implementing or inventing a large number. I probably won't go down in googology history for this. Anyway, here is my program, the explanation comes later:
I had a hard time choosing what to use. I was originally planning to implement my LINEAR(k) function, which I might do another time, but I didn't know if it would produce large enough numbers, and it would be hard to implement, so my next choice was greedy clique sequences, but those didn't work either, so I just ended up using Laver tables.
So, now that we know how Laver tables work, how does my code work? Well, s(a,b,c) denotes the operator ab where n=c, then the function f(a) uses complex while loops to find the period of the given function a.
My code then returns q^13(5), which has been proven to be greater than f_w(f_w+1(5)); note that this was for q^13(1) but q is strictly increasing so this also acts as a lower bound for q^13(5). This lower bound however is a severe understatement, since, as I mentioned earlier, q(n) cannot be proven total in ZFC while f_w(f_w+1(5)) can be by FAR.
Similarly, this code should return a number greater than Loader's number since, although sadly there is no proof, Loader's function is commonly believed to be provably total in ZFC; I do note however that just because a function which is provably total in a stronger theory doesn't mean that it will eventually dominate the one provably total in the weaker theory. It is usually the case though. So, all said, I WIN!!!
See Binary198's answer for the definition of q and an argument that q15(2) is large enough. If the argument is correct, then this is an impressively concise way to beat Loader's number. Unfortunately their implementation is broken.
s(b,a) computes row a, column b+1 of the Laver table of size c, using 0...c-1 as the representatives of Z/cZ instead of the usual (for these tables) 1...c. The period-finding code uses the fact that the period can only be a power of 2. Other than that, there are only standard golfing tricks here.
I deliberately computed the same value as Binary198's answer, though it could be made larger at no cost, because I'm not trying to compete on size; I just wanted a working implementation of this interesting function (for some definition of "working").
This function simulates all Turing machines with n states and n colours for a number of steps (initially 9), then sums the number of steps that it took for any machine that halted to halt, storing that value as the number of steps to take on the next iteration. It keeps doing this until it reaches a steady state.
So far, I've just described a function which is polynomial at best. Notably, it grows slower than the busy beaver function on these Turing machines. Thus, we can be confident that there is at least one machine of size n which halts but we have not yet found. So we search for that one (this function is undefined on inputs which are too small).
Once we find it, we continue on as we were. We'll reach another steady state. This time, we can't be confident that there is another halting Turing machine of size n, but as BB(n+1)>F(BB(n)) for any reasonable polynomial function F (which this is), we can be confident that there is a halting Turing machine of size n+1 which we have not found. If we then find a halting machine of size n, we know that we can perform this step on n+1 again.
We repeat this process until we've attempted this on every size of machine we're looking at (from n to n+V[0]), and thus can no longer be confident that there is a machine that halts of those sizes that we have not seen.
While this number is certainly less than BB(n), I'm not sure how one would go about determining just how big it is. The guarantee that it halts is basically "BB(n) grows much faster than this does". I'm fairly confident that's a safe assumption for values of R(n) greater than, say, R(5).
Actually, it occurs to me that there's no reason why (in principle) we can know that R(n) < BB(n) (when BB(n) is defined to be on Turing machines with n colours and states). Determining exactly how big R(n) is is related to a limited version of the halting problem, as it isn't guaranteed that $$BB(n)*2^n*2^3n^2
while we can know that R(n) halts on big enough inputs, it's certainly possible that the result is in every case larger than BB(n). Proving such (if true) would be equivalent to the halting problem, of course.
Loader's number is the output of loader.c, a C program by Ralph Loader that came in first place for the Bignum Bakeoff contest, whose objective was to write a C program (in 512 characters or less) that generates the largest possible output on a theoretical machine with infinite memory.[1][2] It is among the largest numbers with computable definitions ever devised. A layman's explanation of the ideas behind loader.c can be found in the last two videos of this Youtube playlist on the Bignum Bakeoff contest.
The program diagonalizes over the Huet-Coquand calculus of constructions, a particularly expressive lambda calculus. Its output, affectionately nicknamed Loader's number, is defined as \(D^5(99)=D(D(D(D(D(99)))))\), where \(D(k)\) is an accumulation of all possible expressions provable within approximately \(log(k)\) inference steps in the calculus of constructions (encoding proofs as binary numbers and expressions as power towers). The proof strength and expressiveness of the calculus of constructions makes the output of \(D(k)\) grow very large for sufficiently large k.
The final output of \(D^5(99)\) is much larger than TREE(3), SCG(13), and, say, BH(100). It is overpowered by finite promise games and greedy clique sequences[citation needed]. Loader's function is computable, so \(\Sigma(n) > D^5(99)\) for relatively small n, say, n = 2000.
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