Is Wolfram and Cook's (2,5) Turing machine really universal?

Wolfram [2, p. 707] and Cook [1, p. 3] claim to prove that a (2,5) Turing machine (2 states, 5 symbols) is universal, via a universal cellular automaton known as Rule 110. The first part of this paper points out a critical gap in their argument. The second part bridges the gap, thereby giving what appears to be the first proof of universality. 1 The claim Wolfram [2, p. 707] and Cook [1, p. 3] claim to prove that the Turing Machine M with the following table is universal: State ◦ • 0 ◦ 0 ◦ 1 1 • 1 • ? Input symbol 0 0 ◦ 0 ◦ 1 1 • 1 • ? 0 ◦ 1 ◦ Here table entry “ • ? ” means “write ? and move left into state •”, entry “ 1 ◦ ” means “write 1 and move right into state ◦”, etc.1 For ease of reference, we collect together the passages from these works which together constitute the claimed proof of universality: 1. Wolfram, p. 707. . . . by using the universality of rule 110 it turns out to be possible to come up with the vastly simpler Turing machine shown below—with just 2 states and 5 possible colors. 2. Wolfram, p. 707, first caption. The rule for the simplest Turing machine currently known to be universal, based on discoveries in this book. The machine has 2 states and 5 possible colors. 3. Wolfram, p. 707, second caption. An example of how the Turing machine above manages to emulate rule 110. 4. Wolfram, p. 708. As the picture at the bottom of the previous page illustrates, this Turing machine emulates rule 110 in a quite straightforward way: its head moves systematically backwards and forwards, at each complete sweep updating all cells according to a single step of rule 110 evolution. And knowing from earlier in this chapter that rule 110 is universal, it then follows that the 2-state 5-color Turing machine must also be universal. Visiting Scholar, Computer Science Department, 353 Serra Mall, Stanford University, Stanford CA 94305, USA. Cook [1] writes 0 for 0, 1 for ? and 6= for 1. Wolfram on p. 707 [2] uses shades of grey (white, light grey, medium grey, dark grey, black for 0, 0, ?, 1, 1, resp.), enumerated 0 . . . 4 from light to dark on p. 1119. The states • ◦ are SE SO (resp.) for Cook, and •H • (resp.) for Wolfram. The reason for our notational choice becomes clear with the example in Section 3. 5. Wolfram, p. 1119 (note to p. 707). Rule 110 Turing machines. Given an initial condition for rule 110, the initial condition for the Turing machine shown here is obtained as Prepend[list, 0] with 0’s on the left and 0’s on the right.2 6. Cook, p. 3. . . . we can construct Turing machines that are universal because they can emulate the behavior of Rule 110. These machines, shown in Figure 1, are far smaller than any previously known universal Turing machines. 7. Cook, p. 4, caption to Figure 1. Some small Turing machines which are universal due to being able to emulate the behavior of Rule 110 by going back and forth over an ever wider stretch of tape, each time computing one more step of Rule 110’s activity. This list is the full extent of the Wolfram-Cook universality argument (aside from Wolfram’s depiction of an example run of the Turing machine, p. 707).3 They attempt to argue as follows: (1) The Turing machine M emulates the following cellular automaton R (“Rule 110”): 000 001 010 011 100 101 110 110 0 1 1 1 0 1 1 0 ; (2) R is universal; (3) hence M is universal. 2 The gap Unfortunately, there is a critical gap in their attempted argument. Wolfram (item 5 above) defines the emulation of R on initial configurations ⇐0 I 0⇒ where I is a word (finite sequence of 1’s and 0’s), and ⇐w (resp. w⇒) denotes the infinite repetition of a word or symbol w towards the left (resp. right). However, Wolfram and Cook demonstrated the universality of R via initial configurations ⇐X I Y⇒ for words X and Y, neither constantly 0 . In the former, there is an infinite stream of 0’s either side of the input I , hence a finite number of 1’s in total. In the latter, there are infinitely many 1’s either side of I . This breakdown in reasoning begs the question: is their (2, 5) Turing machine M really universal?4 This item is not a strictly verbatim quote: to match our notation for the Turing machine symbols, we have substituted “list” for Wolfram’s original “4 list”, and “0’s on the left” for “1’s on the left”. Wolfram uses the Turing machine tape symbol ‘4’ (depicted as a solid black square) to correspond to the cellular automaton’s 1, while we (like Cook) use the tape symbol ‘1’; where he writes the tape symbol ‘1’ (depicted as a light grey square), we write ‘0’ (Cook’s ‘0’). If there are additional details somewhere in [1, 2], they are not easy to find. In addition, after extensive web search, we were unable to find a universality proof. Note: the reader should not confuse the proof of universality of rule 110 in Cook [1], which is laid out in full, with the claimed proof of universality of the (2,5) Turing machine, the full extent of which is items 1–7 above. Naive attempts to bridge the gap from ⇐0 I 0⇒ emulations to ⇐X I Y⇒ emulations fail. For example, one could run the emulation on ⇐0X I Y n 0⇒, where W n denotes n repetitions of W , the idea being that X and Y n might contain enough gliders/particles [2, 1] to complete the computation. However, because of the halting problem, we can never predict how large n will need to be. Accordingly, we could resort to repeating the ⇐0X I Y n 0⇒ emulation again and again, with progressively larger n, since if the target computation on ⇐X I Y⇒ completes, then some n will be large enough that ⇐0X I Y n 0⇒ completes in a corresponding manner. Alas, the deus ex machina (repeatedly restarting the Turing machine) destroys any possible claim of universality.