Heller M, Woodin W.H. (eds.) Infinity. New research frontiers (CUP, 2011)(ISBN 1107003873)(O)(327s)_MAml_

the turing machine and the busy beaver 71A0 → (A1,R), A1 → (H ), and an input tape. The machine starts in state A, printing1 as long as it reads 0, and will keep moving to the right until it reads 1, when it willstop having printed a string of consecutive 1 ′ s to its left; it will not stop otherwise.Is it possible to understand whether such a Turing machine will stop or not? Thesimplest way to approach the problem is to consider a blank tape of 0 as the initial tapeand machines with N states only and two symbols 0 and 1. There are only finitely manypossible tables of instructions. Consider now only the tables I for which the Turingmachine stops. Is it possible to determine the maximum number of ones that may beprinted on the tape, the maximum length needed for the tape, and the maximum numberof steps before the Turing machine halts? These problems are all of a similar nature,and mathematicians call the first one the “busy beaver problem.” The beaver (i.e., aTuring machine with N states, two symbols, and a table of instructions) starts workingin a river (the blank tape) carrying sticks (the bits equal to 1), to build the biggest dam(the largest number of 1 when the machine halts). If the beaver never stops working,call it a “foolish beaver.” Otherwise, there will be a “champion beaver” (possibly morethan one) finishing with the biggest dam. Call it a “busy beaver” and let (N) bethenumber of sticks (i.e., 1’s on the tape) at the end of the work of the busy beaver.For small values of N, the value of (N) can be computed by a clever analysisof all cases, so it is known that (1) = 1, (2) = 4, (3) = 6, (4) = 13. However,the number of Turing machines to analyze grows very quickly, because thereare (4n + 4) 2n Turing machines with n states and two symbols. There are already63, 403, 380, 965, 376 Turing machines with five states and two symbols, so a caseby-caseanalysis is difficult, to say the least. However, by exploiting symmetries andfinding conditions necessary for the Turing machine to stop, mathematicians haveshown that for determining (5) there remain 164 Turing machines for which thehalting problem has not been solved yet (there is no a priori reason why the haltingproblem should be decidable for all of them). Current lower bounds for the nextvalues of the busy beaver function are (5) ≥ 4098 (by Marxen and Buntrock in1989), (6) > 2.50 ∗ 10 881 (by Ligocki and Ligocki in 2007). If S(n) is the maximumnumber of steps that an n-state, two-symbol Turing machine as shown above canperform before halting, it is known that S(1) = 1, S(2) = 6, S(3) = 21, S(4) = 107,S(5) > 47, 176, 870, and S(6) > 8.929 ∗ 10 1762 .One should not be misled into thinking that the working of Turing machines withonly a few possible states and symbols must be simple. Any description of a Turingmachine (i.e., states, symbols, rules, initial tape) can be encoded in a tape of symbols.Turing showed that a universal machine U exists that emulates any Turing machine M.In his words, “It is possible to invent a single machine which can be used to computeany computable sequence. If this machine U is supplied with a tape on the beginningof which is written the S.D 6 of some computing machine M, then U will compute thesame sequence as M” (see Davis 1965, p. 127).The existence of U was, at the time, a great surprise, and it may be consideredas the theoretical equivalent of the modern computer. Note, however, that, unlike thecomputer, the Turing machine has no fast access to its memory (the tape), so the Turing6 The S.D is the “standard description” given by Turing for a machine.