{"paper_meta":{"paper_id":"arxiv:0707.4489","title":"0707.4489","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0707.4489v1 [cs.CC] 30 Jul 2007\nSmall weakly universal Turing machines\nTurlough Neary1 and Damien Woods2\n1 TASS, Department of Computer Science,\nNational University of Ireland Maynooth, Ireland. tneary@cs.may.ie\n2 Department of Computer Science,\nUniversity College Cork, Ireland. d.woods@cs.ucc.ie\nAbstract. We give small universal Turing machines with state-symbol\npairs of (6, 2), (3, 3) and (2, 4). These machines are weakly universal,\nwhich means that they have an infinitely repeated word to the left of\ntheir input and another to the right. They simulate Rule 110 and are\ncurrently the smallest known weakly universal Turing machines.\n1\nIntroduction\nShannon [22] was the first to consider the problem of finding the smallest uni-\nversal Turing machine, where size is the number of states and symbols. Here we\nsay that a Turing machine is standard if it has a single one-dimensional tape,\none tape head, and is deterministic [6]. Over the years, small universal programs\nwere given for a number of variants on the standard model. By generalising the\nmodel we often find smaller universal programs. One variation on the standard\nmodel is to allow the blank portion of the Turing machine’s tape to have an\ninfinitely repeated word to the left, and another to the right. We refer to such\nuniversal machines as weakly universal Turing machines, and they the subject\nof this work.\nBeginning in the early sixties Minsky and Watanabe engaged in a vigorous\ncompetition to see who could come up with the smallest universal Turing ma-\nchine [10,11,23,24,25]. In 1961 Watanabe [24] gave a 6-state, 5-symbol universal\nTuring machine, the first weakly universal machine. In 1962, Minsky [11] found a\nsmall 7-state, 4-symbol universal Turing machine. Not to be out-done, Watanabe\nimproved on his earlier machine to give 5-state, 4-symbol and 7-state, 3-symbol\nweakly universal machines [25,16].\nThe 7-state universal Turing machine of Minsky has received much atten-\ntion. Minsky’s machine simulates Turing machines via 2-tag systems, which\nwere proved universal by Cocke and Minsky [2]. The technique of simulating\n2-tag systems, pioneered by Minsky, was extended by Rogozhin [20] to give the\n(then) smallest known universal Turing machines for a number of state-symbol\npairs. These 2-tag simulators were subsequently reduced in size by Rogozhin [21],\nKudlek and Rogozhin [8], and Baiocchi [1]. Neary and Woods [15] gave small uni-\nversal machines that simulate Turing machines via a new variant of tag systems\ncalled bi-tag systems. Each of the smallest 2-tag or bi-tag simulators are plotted\nas circles in Figure 1. These (standard) machines induce a universal curve.\n\nb\nc : standard universal machine\nl : semi-weakly universal machine\nr\ns\n: weakly universal machine\nr\n: new weakly universal machine\n: standard universal curve\n: weakly universal curve\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\nstates\nsymbols\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nl\nl\nl\nl\nr\ns\nr\ns\nr\ns\nr\ns\nr\nr\nr\nFig. 1. State-symbol plot of small universal Turing machines. Each of our new\nweak machines is represented by a solid square. These machines induce a weakly\nuniversal curve.\nThe small weak machines of Watanabe have received little attention. In par-\nticular the 5-state and 7-state machines seem little known and are largely ignored\nin the literature. It is worth noting that while all other weak machines simulate\nTuring machine via other simple models, Watanabe’s weak machines simulate\nTuring machines directly. His machines are the most time efficient of the small\nweak machines. More precisely, let t be the running time of any deterministic\nsingle tape Turing machine M, then Watanabe’s machines are the smallest weak\nmachines that simulate M with a time overhead of O(t2).\nWe often refer to Watanabe’s machines as being semi-weak. Semi-weak ma-\nchines have an infinitely repeated word to one side of their input, and on the\nother side they have a (standard) infinitely repeated blank symbol. Recently,\nWoods and Neary [29] have given 3-state, 7-symbol and 4-state, 5-symbol semi-\nweakly universal machines that simulate cyclic tag systems. All of the smallest\nsemi-weakly universal machines are given as diamonds in Figure 1.\nCook [3] and Wolfram [26] recently gave weakly universal Turing machines,\nsmaller than Watanabe’s semi-weak machines, that simulate the universal cellu-\nlar automata Rule 110. These machines have state-symbol pairs of (7, 2), (4, 3),\n(3, 4), (2, 5) and are plotted as hollow squares in Figure 1.\n2\n\nHere we present weakly universal Turing machines with state-symbol pairs\nof (6, 2), (3, 3), (2, 4), making them the smallest known weakly universal ma-\nchines. Our machines simulate (single tape, deterministic) Turing machines in\ntime O(t4 log2 t), via Rule 110. These machines are plotted as solid squares\nin Figure 1 and induce a weakly universal curve. It is interesting to note from\nFigure 1 that the smallest universal machines, and the smallest semi-weakly uni-\nversal machines, are both symmetric about the line where states equals symbols,\nwhereas the smallest weakly universal machines are not.\nWeakness has not been the only variation on the standard model in the\nsearch for small universal Turing machines. Priese [19] gave a 2-state, 4-symbol\nmachine with a 2-dimensional tape, and a 2-state, 2-symbol machine with a\n2-dimensional tape and 2 tape heads. Margenstern and Pavlotskaya [9] gave a\n2-state, 3-symbol Turing machine that is universal when coupled with a finite\nautomaton. This machine uses only 5 instructions. Margenstern and Pavlotskaya\nalso show that the halting problem is decidable for machines of this type with 4\ninstructions. Their result implies that it is not possible to have a 4 instruction\nuniversal machine of this type, that simulates any Turing machine M and halts\nif and only if M halts. Hence they have given the smallest possible universal\nmachine of this type.\nFor results relating to the time complexity of small universal Turing machines\nsee [12,14,27,28].\n1.1\nPreliminaries\nThe Turing machines considered in this paper are deterministic and have a single\nbi-infinite tape. We let Um,n denote our weakly universal Turing machine with m\nstates and n symbols. We write c1 ⊢c2 if a configuration c2 is obtained from c1\nvia a single computation step. We let c1 ⊢s c2 denote a sequence of s computation\nsteps, and let c1 ⊢∗c2 denote zero or more computation steps.\n2\nRule 110\nRule 110 is a very simple (2 state, nearest neighbour, one dimensional) cel-\nlular automaton. It is composed of a sequence of cells . . . p−1p0p1 . . . where\neach cell has a binary state pi ∈{0, 1}. At timestep s + 1, the value pi,s+1 =\nF(pi−1,s, pi,s, pi+1,s) of the cell at position i is given by the synchronous local\nupdate function F\nF(0, 0, 0) = 0\nF(1, 0, 0) = 0\nF(0, 0, 1) = 1\nF(1, 0, 1) = 1\nF(0, 1, 0) = 1\nF(1, 1, 0) = 1\nF(0, 1, 1) = 1\nF(1, 1, 1) = 0\n(1)\nRule 110 was proven universal by Cook [3] and Wolfram [26]. Recently, Neary\nand Woods [12,13] proved that Rule 110 simulates Turing machines efficiently\n3\n\nc0\nc1\nc2\nc3...\n. . . -9 -8 -7 -6 -5 -4 -3 -2 -1 0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n. . .\nFig. 2. Seven consecutive timesteps of Rule 110. These seven timesteps are ap-\nplied to the background ether that is used in the proof [3] of universality of Rule\n110. Each black or each white square represents, a Rule 110 cell containing, state\n1 or 0 respectively. Each cell is identified by the index given above it. To the left\nof each row of cells there is a configuration label that identifies that row.\nin polynomial time O(t3 log t), an exponential improvement. Note that, in or-\nder to calculate this upper bound we substitute space bounds for time bounds\nwhenever possible in the analysis. It turns out that we can further improve the\nsimulation time to O(t2 log t) (this result is as yet unpublished). Rule 110 sim-\nulates cyclic tag systems in linear time. The weak machines in this paper, and\nin [3,26], simulate Rule 110 with a quadratic polynomial increase in time and\nhence simulate Turing machines in time O(t4 log2 t). It is worth noting that the\nprediction problem [4] for these machines is P-complete, and this is also the case\nwhen we consider only bounded initial conditions [12].\n3\nThree small weakly universal Turing machines\nThe following observation is one of the reasons for the improvement in size over\nprevious machines [3,26], and gives some insight into the simulation algorithm we\nuse. Notice from Equation (1) that the value of the update function F, with the\nexception of F(0, 1, 1) and F(1, 1, 1), may be determined using only the rightmost\ntwo states. Each of our universal Turing machines exploit this fact as follows.\nThe machines scan from right to left, and in six of the eight cases they need\nonly remember the cell immediately to the right of the current cell i in order to\ncompute the update for i. Thus for these six cases we need only store a single cell\nvalue, rather than two values. The remaining two cases are simulated as follows.\nIf two consecutive encoded states with value 1 are read, it is assumed that there\nis another encoded 1 to the left and the update F(1, 1, 1) = 0 is simulated. If\nour assumption proves false (we instead read an encoded 0), then our machine\nreturns to the wrongly updated cell and simulates the update F(0, 1, 1) = 1.\nBefore giving our three small Rule 110 simulators, we give some further back-\nground explanation. Rule 110 simulates Turing machines via cyclic tag systems.\nA Rule 110 instance that simulates a cyclic tag system computation is of the\nfollowing form (for more details see [3,26]). The input to the cyclic tag system\nis encoded in a contiguous finite number of Rule 110 cells. On the left of the\n4\n\ninput a fixed constant word (representing the ‘ossifiers’) is repeated infinitely\nmany times. On the right, another fixed constant word (representing the cyclic\ntag system program/appendants, and the ‘leaders’) is repeated infinitely many\ntimes. Both of these repeated words are independent of the input.\nAs in [3,26], our weakly universal machines operate by traversing a finite\namount of the tape from left to right and then from right to left. This simulates\na single timestep of Rule 110 over a finite part of the encoded infinite Rule 110\ninstance. With each simulated timestep the length of a traversal increases. So\nthat each traversal is of finite length, the left blank word l and the right blank\nword r of each of our weak machines must have a special form. These words\ncontain special subwords or symbols that terminate each traversal, causing the\ntape head to turn. When the head is turning it ‘deletes’ any symbols that caused\na turn. Thus the number of cells that are being updated increases monotonically\nover time. This technique simulates Rule 110 properly if the initial condition\nis set up so that within each repeated blank word, the subword between each\nsuccessive turn point is shifted one timestep forward in time.\nIn the sequel we describe the computation of our three machines by showing\na simulation of the update on the ether in Figure 2. In the next paragraph below,\nwe outline why this example is in fact general enough to prove universality. First,\nwe must define blank words that are suitable for this example. The left blank\nword l, on the Turing machine tape, encodes the Rule 110 sequence 0001. In\nthe initial configuration as we move left each subsequent sequence 0001 is one\ntimestep further ahead. To see this note from Figure 2 that 0001 occupies, cells\n−7 to −4 in configuration c1, cells −11 to −8 in c2, cells −15 to −12 in c3, etc.\nSimilarly, the right blank word r encodes the Rule 110 sequence 110011. Looking\nat the initial configuration, as we move right from cell 0, in the first blank word\nthe first four cells 1100 are shifted two timesteps ahead, and the next two cells\n11 are shifted a further one timestep. To see this note from Figure 2 that 1100\noccupies cells 1 to 4 in c2 and 11 occupies cells 5 and 6 in c3. In each subsequent\nsequence the first four cells 1100 are shifted only one timestep ahead and the last\ntwo cells 11 are shifted one further timestep. In each row the ether in Figure 2\nrepeats every 14 cells and if the number of timesteps s between two rows is s ≡0\nmod 7 then the two rows are identical. The periodic nature of the ether, in both\ntime and space, allows us to construct such blank words.\nIt should be noted that the machines we present here, and those in [3,26],\nrequire suitable blank words to simulate a Rule 110 instance directly. If no suit-\nable blank words can be found (i.e. if they do not contain the specific subwords\nthat we use to terminate traversals) then it may be the case that the particular\ninstance can not be simulated directly. In the sequel our machines simulate the\nbackground ether that is used in the universality proof of Rule 110 [3,26]. The\ngliders that move through this ether are periodic in time and space, and so we\ncan construct blank words where the ether includes the subwords that termi-\nnate traversals. By this reasoning, our example is sufficiently general to prove\nthat our machines simulate Turing machines via Rule 110 and we do not give a\nfull (and possibly tedious) proof of correctness. For U3,3 we explicitly simulate\n5\n\nthree updates from Figure 2, which is general enough so that an update [Equa-\ntion (1)] on each of the eight possible three state combinations is simulated.\nWe give shorter examples for the machines U2,4 and U6,2 as they use the same\nsimulation algorithm as U3,3.\nThe machines we present here do not halt. Cook [3] shows how a special\nglider may be produced during the simulation of a Turing machine by Rule 110.\nThis glider may be used to simulate halting as the encoding can be such that it\nis generated by Rule 110 if and only if the simulated machine halts. The glider\nwould be encoded on the tape of our machines as a unique, constant word.\n3.1\nU3,3\nWe begin by describing an initial configuration of U3,3. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by 0, and the\nRule 110 state 1 is encoded by either 1 or b. The word 1b0 is used to terminate\na left traversal. (Note an exception: the 1 in the subword 1b0 encodes the Rule\n110 state 0.) To the right of the tape head position, the Rule 110 state 0 is\nencoded by 1, and the Rule 110 state 1 is encoded by 0 or b. The tape symbol 0\nis used to terminate a right traversal. The left and right blank words, described\nin paragraph 4 of Section 3, are encoded as 0 0 1 b and 0 b 1 1 0 b respectively.\nu1\nu2\nu3\n0\n1Lu1 0Ru1 bLu1\n1\nbLu2 1Lu2 0Ru3\nb\nbLu3\n1Ru3\nTable 1. Table of behaviour for U3,3.\nWe give an example of U3,3 simulating the three successive Rule 110 timesteps\nc0 ⊢c1 ⊢c2 ⊢c3 given in Figure 2. In the below configurations the current\nstate of U3,3 is highlighted in bold, to the left of its tape contents. The tape\nhead position of U3,3 is given by an underline and the start state is u1. The\nconfiguration immediately below encodes c0 from Figure 2 with the tape head\nover cell index 0.\nu1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n6\n\nWhen the tape head reads the subword 1b0 the left traversal is complete and\nthe right traversal begins.\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1 0 b\n0 b 1 1 0 b . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢15 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 b b\n0 b 1 1 0 b . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2. We continue our simulation to give timestep c2 ⊢c3.\n⊢3 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢4 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢21 u1\nu1\nu1, . . . 0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0\n0 1 0 0 1 1\nb b 1 1 0 b . . .\nThe simulation of timestep c2 ⊢c3 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −15 to 6 of configuration c3 in\nFigure 2.\n3.2\nU2,4\nWe begin by describing an initial configuration of U2,4. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by either 0 or 0/\n7\n\nand the Rule 110 state 1 is encoded by either 1 or 1/ . The word 0/ 1 is used to\nterminate a left traversal. To the right of the tape head position, the Rule 110\nstate 0 is encoded by 0/ and the Rule 110 state 1 is encoded by 1/ or 0. The tape\nsymbol 0 is used to terminate a right traversal. The left and right blank words,\nfrom paragraph 4 of Section 3, are encoded as 0 0 0/ 1 and 0 1/ 0/ 0/ 0 1/ respectively.\nu1\nu2\n0\n0/ Lu1 1/ Ru1\n1\n1/ Lu2 0/ Lu2\n0/\n1/ Lu1 0Ru2\n1/\n1/ Lu1 1Ru2\nTable 2. Table of behaviour for U2,4.\nBy way of example we give U2,4 simulating the two successive Rule 110\ntimesteps c0 ⊢c1 ⊢c2 given in Figure 2. The configuration immediately below\nencodes c0 from Figure 2 with the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢6 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/1/\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nWhen the tape head reads the subword 0/ 1 the left traversal is complete and the\nright traversal begins.\n⊢6 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢2 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢2 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢4 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1/\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢15 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 1/ 1/\n0 1/ 0/ 0/ 0 1/ . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2.\n8\n\n3.3\nU6,2\nWe begin by describing an initial configuration of U6,2. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by the word\n00 and the Rule 110 state 1 is encoded by the word 11. The word 010100 is\nused to terminate a left traversal and encodes the sequence of Rule 110 states\n010. To the right of the tape head position the Rule 110 state 0 is encoded\nby the word 00 and the Rule 110 state 1 is encoded by either of the words 01\nor 10. The word 10 is used to terminate a right traversal. The left and right\nblank words, from paragraph 4 of Section 3, are encoded as 0 0 0 0 0 1 0 1 and\n1 0 0 1 0 0 0 0 1 0 0 1 respectively.\nu1\nu2\nu3\nu4\nu5\nu6\n0\n0Lu1 0Lu6 0Ru2 1Ru5 1Lu4 1Lu1\n1\n1Lu2 0Lu3 1Lu3 0Ru6 1Ru4 0Ru4\nTable 3. Table of behaviour for U6,2.\nTo illustrate the operation of U6,2 we simulate the Rule 110 timestep c0 ⊢c1\ngiven in Figure 2. The configuration immediately below encodes c0 from Figure 2\nwith the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 11\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u3\nu3\nu3, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 10 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\nWhen the tape head reads the subword 1 0 1 0 0 the left traversal is complete\nand the right traversal begins.\n9\n\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢4 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 10 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢2 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 1 0 1 0 0 0 0 1 0 0 1 . . .\nImmediately after the tape head reads a 10, during a right traversal, the simula-\ntion of timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold (recall 0 and 1 are encoded as 00 and 11 respectively) with\ncells −7 to 0 of configuration c1 in Figure 2.\n4\nDiscussion on lower bounds\nThe pursuit to find the smallest possible universal Turing machine must also\ninvolve the search for lower bounds, finding the largest set of Turing machines\nthat are in some sense non-universal. One approach is to settle the decidability\nof the halting problem, but this approach is not suitable for the machines we\nhave presented.\nIt is known that the halting problem is decidable for (standard) Turing ma-\nchines with the following state-symbol pairs (2, 2) [7,17], (3, 2) [18], (2, 3) (claimed\nby Pavlotskaya [17]), (1, n) [5] and (n, 1) (trivial), where n ⩾1. Then, these de-\ncidability results imply that a universal Turing machine, that simulates any\nTuring machine M and halts if and only if M halts, is not possible for these\nstate-symbol pairs. Hence these results give lower bounds on the size of univer-\nsal machines of this type. While it is trivial to prove that the halting problem\nis decidable for (possibly halting) weak machines with state-symbol pairs of the\nform (n, 1), it is not known whether the above decidability results generalise to\n(possibly halting) weak Turing machines.\nThe weak machines presented in this paper, and those in [3,26], do not halt.\nHence the non-universality results discussed in the previous paragraph would\nhave to be generalised to non-halting weak machines to give lower bounds that\nare relevant for our machines. This may prove difficult for two reasons. The\nfirst issue is that, intuitively speaking, weakness gives quite an advantage. For\n10\n\ninstance, the program of a universal machine may be encoded in one of the in-\nfinitely repeated blank words of the weak machine. The second issue is related\nto the problem of defining a computation. Informally, a computation could be\ndefined as a sequence of configurations that leads to a special terminal config-\nuration. For non-halting machines, there are many ways to define a terminal\nconfiguration. Given a definition of terminal configuration we may prove that\nthe terminal configuration problem (will a machine ever enter a terminal config-\nuration) is decidable for a machine or set of machines. However this result may\nnot hold as a proof of non-universality if we subsequently alter our definition of\nterminal configuration.\nAcknowledgements\nTurlough Neary is funded by the Irish Research Council for Science, Engineering\nand Technology. Damien Woods is funded by Science Foundation Ireland grant\nnumber 04/IN3/1524.\nReferences\n1. Claudio Baiocchi. Three small universal Turing machines. In Maurice Margenstern\nand Yurii Rogozhin, editors, Machines, Computations, and Universality (MCU),\nvolume 2055 of LNCS, pages 1–10, Chi ̧sin ̆au, Moldova, May 2001. Springer.\n2. John Cocke and Marvin Minsky. Universality of tag systems with P = 2. Journal\nof the ACM, 11(1):15–20, January 1964.\n3. Matthew Cook. Universality in elementary cellular automata. Complex Systems,\n15(1):1–40, 2004.\n4. Raymond Greenlaw, H. James Hoover, and Walter L. Ruzzo. Limits to parallel\ncomputation: P -completeness theory. Oxford university Press, Oxford, 1995.\n5. Gabor Hermann. The uniform halting problem for generalized one state Turing\nmachines. In Proceedings, Ninth Annual Symposium on Switching and Automata\nTheory (FOCS), pages 368–372, Schenectady, New York, October 1968. IEEE Com-\nputer Society Press.\n6. John E. Hopcroft and Jeffrey D. Ullman. Introduction to automata theory, lan-\nguages, and computation. Addison-Wesley Series in Computer Science. Addison-\nWesley, Reading, Mass., 1979.\n7. Manfred Kudlek.\nSmall deterministic Turing machines.\nTheoretical Computer\nScience, 168(2):241–255, November 1996.\n8. Manfred Kudlek and Yurii Rogozhin. A universal Turing machine with 3 states\nand 9 symbols. In Werner Kuich, Grzegorz Rozenberg, and Arto Salomaa, editors,\nDevelopments in Language Theory (DLT) 2001, volume 2295 of LNCS, pages 311–\n318, Vienna, May 2002. Springer.\n9. Maurice Margenstern and Liudmila Pavlotskaya. On the optimal number of in-\nstructions for universality of Turing machines connected with a finite automaton.\nInternational Journal of Algebra and Computation, 13(2):133–202, April 2003.\n10. Marvin Minsky. A 6-symbol 7-state universal Turing machines. Technical Report\n54-G-027, MIT, August 1960.\n11\n\n11. Marvin Minsky. Size and structure of universal Turing machines using tag systems.\nIn Recursive Function Theory, Symp. in Pure Math., volume 5, pages 229–238.\nAMS, 1962.\n12. Turlough Neary and Damien Woods. P-completeness of cellular automaton Rule\n110. In M. Bugliesi et al., editor, International Colloquium on Automata Languages\nand Programing 2006, (ICALP) Part I, volume 4051 of LNCS, pages 132–143,\nVenice, July 2006. Springer.\n13. Turlough Neary and Damien Woods. The P-completeness of cellular automaton\nRule 110. Technical Report 04/2006, Boole Centre for Research in Informatics,\nUniversity College Cork, Ireland, 2006.\n14. Turlough Neary and Damien Woods. Small fast universal Turing machines. The-\noretical Computer Science, 362(1–3):171–195, November 2006.\n15. Turlough Neary and Damien Woods. Four small universal Turing machines. In\nJ ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computations,\nand Universality (MCU), volume 4664 of LNCS, pages 242–254, Or ́elans, France,\nSeptember 2007. Springer.\n16. A. Nozaki. On the notion of universality of Turing machine. Kybernetika Academia\nPraha, 5(1):29–43, 1969.\n17. Liudmila Pavlotskaya.\nSolvability of the halting problem for certain classes of\nTuring machines. Mathematical Notes (Springer), 13(6):537–541, June 1973.\n18. Liudmila Pavlotskaya.\nDostatochnye uslovija razreshimosti problemy ostanovki\ndlja mashin T’juring. Problemi kibernetiki, pages 91–118, 1978. (Sufficient condi-\ntions for the halting problem decidability of Turing machines.In Russian).\n19. Lutz Priese. Towards a precise characterization of the complexity of universal and\nnon-universal Turing machines. Siam journal of Computing, 8(4):508–523, 1979.\n20. Yurii Rogozhin. Sem’ universal’nykh mashin T’juringa. Systems and theoretical\nprogramming, Mat. Issled, 69:76–90, 1982. (Seven universal Turing machines. In\nRussian).\n21. Yurii Rogozhin. Small universal Turing machines. Theoretical Computer Science,\n168(2):215–240, November 1996.\n22. Claude Elwood Shannon. A universal Turing machine with two internal states.\nAutomata Studies, Annals of Mathematics Studies, 34:157–165, 1956.\n23. Shigeru Watanabe. On a minimal universal Turing machines. Technical report,\nMCB Report, Tokyo, August 1960.\n24. Shigeru Watanabe. 5-symbol 8-state and 5-symbol 6-state universal Turing ma-\nchines. Journal of ACM, 8(4):476–483, October 1961.\n25. Shigeru Watanabe. 4-symbol 5-state universal Turing machines. Information Pro-\ncessing Society of Japan Magazine, 13(9):588–592, 1972.\n26. Stephen Wolfram. A new kind of science. Wolfram Media, 2002.\n27. Damien Woods and Turlough Neary. On the time complexity of 2-tag systems and\nsmall universal Turing machines. In 47th Annual IEEE Symposium on Foundations\nof Computer Science (FOCS), pages 132–143, Berkeley, California, October 2006.\nIEEE.\n28. Damien Woods and Turlough Neary. The complexity of small universal Turing\nmachines. In S. Barry Cooper, Benedikt L ̈owe, and Andrea Sorbi, editors, Com-\nputability in Europe 2007, volume 4497 of LNCS, pages 791–798, Siena, Italy, June\n2007. Springer.\n29. Damien Woods and Turlough Neary. Small semi-weakly universal Turing machines.\nIn J ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computa-\ntions, and Universality (MCU), volume 4664 of LNCS, pages 306–323, Or ́elans,\nFrance, September 2007. Springer.\n12","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0707.4489v1 [cs.CC] 30 Jul 2007\nSmall weakly universal Turing machines\nTurlough Neary1 and Damien Woods2\n1 TASS, Department of Computer Science,\nNational University of Ireland Maynooth, Ireland. tneary@cs.may.ie\n2 Department of Computer Science,\nUniversity College Cork, Ireland. d.woods@cs.ucc.ie\nAbstract. We give small universal Turing machines with state-symbol\npairs of (6, 2), (3, 3) and (2, 4). These machines are weakly universal,\nwhich means that they have an infinitely repeated word to the left of\ntheir input and another to the right. They simulate Rule 110 and are\ncurrently the smallest known weakly universal Turing machines.\n1\nIntroduction\nShannon [22] was the first to consider the problem of finding the smallest uni-\nversal Turing machine, where size is the number of states and symbols. Here we\nsay that a Turing machine is standard if it has a single one-dimensional tape,\none tape head, and is deterministic [6]. Over the years, small universal programs\nwere given for a number of variants on the standard model. By generalising the\nmodel we often find smaller universal programs. One variation on the standard\nmodel is to allow the blank portion of the Turing machine’s tape to have an\ninfinitely repeated word to the left, and another to the right. We refer to such\nuniversal machines as weakly universal Turing machines, and they the subject\nof this work.\nBeginning in the early sixties Minsky and Watanabe engaged in a vigorous\ncompetition to see who could come up with the smallest universal Turing ma-\nchine [10,11,23,24,25]. In 1961 Watanabe [24] gave a 6-state, 5-symbol universal\nTuring machine, the first weakly universal machine. In 1962, Minsky [11] found a\nsmall 7-state, 4-symbol universal Turing machine. Not to be out-done, Watanabe\nimproved on his earlier machine to give 5-state, 4-symbol and 7-state, 3-symbol\nweakly universal machines [25,16].\nThe 7-state universal Turing machine of Minsky has received much atten-\ntion. Minsky’s machine simulates Turing machines via 2-tag systems, which\nwere proved universal by Cocke and Minsky [2]. The technique of simulating\n2-tag systems, pioneered by Minsky, was extended by Rogozhin [20] to give the\n(then) smallest known universal Turing machines for a number of state-symbol\npairs. These 2-tag simulators were subsequently reduced in size by Rogozhin [21],\nKudlek and Rogozhin [8], and Baiocchi [1]. Neary and Woods [15] gave small uni-\nversal machines that simulate Turing machines via a new variant of tag systems\ncalled bi-tag systems. Each of the smallest 2-tag or bi-tag simulators are plotted\nas circles in Figure 1. These (standard) machines induce a universal curve."},{"paragraph_id":"p2","order":2,"text":"b\nc : standard universal machine\nl : semi-weakly universal machine\nr\ns\n: weakly universal machine\nr\n: new weakly universal machine\n: standard universal curve\n: weakly universal curve\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\nstates\nsymbols\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nl\nl\nl\nl\nr\ns\nr\ns\nr\ns\nr\ns\nr\nr\nr\nFig. 1. State-symbol plot of small universal Turing machines. Each of our new\nweak machines is represented by a solid square. These machines induce a weakly\nuniversal curve.\nThe small weak machines of Watanabe have received little attention. In par-\nticular the 5-state and 7-state machines seem little known and are largely ignored\nin the literature. It is worth noting that while all other weak machines simulate\nTuring machine via other simple models, Watanabe’s weak machines simulate\nTuring machines directly. His machines are the most time efficient of the small\nweak machines. More precisely, let t be the running time of any deterministic\nsingle tape Turing machine M, then Watanabe’s machines are the smallest weak\nmachines that simulate M with a time overhead of O(t2).\nWe often refer to Watanabe’s machines as being semi-weak. Semi-weak ma-\nchines have an infinitely repeated word to one side of their input, and on the\nother side they have a (standard) infinitely repeated blank symbol. Recently,\nWoods and Neary [29] have given 3-state, 7-symbol and 4-state, 5-symbol semi-\nweakly universal machines that simulate cyclic tag systems. All of the smallest\nsemi-weakly universal machines are given as diamonds in Figure 1.\nCook [3] and Wolfram [26] recently gave weakly universal Turing machines,\nsmaller than Watanabe’s semi-weak machines, that simulate the universal cellu-\nlar automata Rule 110. These machines have state-symbol pairs of (7, 2), (4, 3),\n(3, 4), (2, 5) and are plotted as hollow squares in Figure 1.\n2"},{"paragraph_id":"p3","order":3,"text":"Here we present weakly universal Turing machines with state-symbol pairs\nof (6, 2), (3, 3), (2, 4), making them the smallest known weakly universal ma-\nchines. Our machines simulate (single tape, deterministic) Turing machines in\ntime O(t4 log2 t), via Rule 110. These machines are plotted as solid squares\nin Figure 1 and induce a weakly universal curve. It is interesting to note from\nFigure 1 that the smallest universal machines, and the smallest semi-weakly uni-\nversal machines, are both symmetric about the line where states equals symbols,\nwhereas the smallest weakly universal machines are not.\nWeakness has not been the only variation on the standard model in the\nsearch for small universal Turing machines. Priese [19] gave a 2-state, 4-symbol\nmachine with a 2-dimensional tape, and a 2-state, 2-symbol machine with a\n2-dimensional tape and 2 tape heads. Margenstern and Pavlotskaya [9] gave a\n2-state, 3-symbol Turing machine that is universal when coupled with a finite\nautomaton. This machine uses only 5 instructions. Margenstern and Pavlotskaya\nalso show that the halting problem is decidable for machines of this type with 4\ninstructions. Their result implies that it is not possible to have a 4 instruction\nuniversal machine of this type, that simulates any Turing machine M and halts\nif and only if M halts. Hence they have given the smallest possible universal\nmachine of this type.\nFor results relating to the time complexity of small universal Turing machines\nsee [12,14,27,28].\n1.1\nPreliminaries\nThe Turing machines considered in this paper are deterministic and have a single\nbi-infinite tape. We let Um,n denote our weakly universal Turing machine with m\nstates and n symbols. We write c1 ⊢c2 if a configuration c2 is obtained from c1\nvia a single computation step. We let c1 ⊢s c2 denote a sequence of s computation\nsteps, and let c1 ⊢∗c2 denote zero or more computation steps.\n2\nRule 110\nRule 110 is a very simple (2 state, nearest neighbour, one dimensional) cel-\nlular automaton. It is composed of a sequence of cells . . . p−1p0p1 . . . where\neach cell has a binary state pi ∈{0, 1}. At timestep s + 1, the value pi,s+1 =\nF(pi−1,s, pi,s, pi+1,s) of the cell at position i is given by the synchronous local\nupdate function F\nF(0, 0, 0) = 0\nF(1, 0, 0) = 0\nF(0, 0, 1) = 1\nF(1, 0, 1) = 1\nF(0, 1, 0) = 1\nF(1, 1, 0) = 1\nF(0, 1, 1) = 1\nF(1, 1, 1) = 0\n(1)\nRule 110 was proven universal by Cook [3] and Wolfram [26]. Recently, Neary\nand Woods [12,13] proved that Rule 110 simulates Turing machines efficiently\n3"},{"paragraph_id":"p4","order":4,"text":"c0\nc1\nc2\nc3...\n. . . -9 -8 -7 -6 -5 -4 -3 -2 -1 0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n. . .\nFig. 2. Seven consecutive timesteps of Rule 110. These seven timesteps are ap-\nplied to the background ether that is used in the proof [3] of universality of Rule\n110. Each black or each white square represents, a Rule 110 cell containing, state\n1 or 0 respectively. Each cell is identified by the index given above it. To the left\nof each row of cells there is a configuration label that identifies that row.\nin polynomial time O(t3 log t), an exponential improvement. Note that, in or-\nder to calculate this upper bound we substitute space bounds for time bounds\nwhenever possible in the analysis. It turns out that we can further improve the\nsimulation time to O(t2 log t) (this result is as yet unpublished). Rule 110 sim-\nulates cyclic tag systems in linear time. The weak machines in this paper, and\nin [3,26], simulate Rule 110 with a quadratic polynomial increase in time and\nhence simulate Turing machines in time O(t4 log2 t). It is worth noting that the\nprediction problem [4] for these machines is P-complete, and this is also the case\nwhen we consider only bounded initial conditions [12].\n3\nThree small weakly universal Turing machines\nThe following observation is one of the reasons for the improvement in size over\nprevious machines [3,26], and gives some insight into the simulation algorithm we\nuse. Notice from Equation (1) that the value of the update function F, with the\nexception of F(0, 1, 1) and F(1, 1, 1), may be determined using only the rightmost\ntwo states. Each of our universal Turing machines exploit this fact as follows.\nThe machines scan from right to left, and in six of the eight cases they need\nonly remember the cell immediately to the right of the current cell i in order to\ncompute the update for i. Thus for these six cases we need only store a single cell\nvalue, rather than two values. The remaining two cases are simulated as follows.\nIf two consecutive encoded states with value 1 are read, it is assumed that there\nis another encoded 1 to the left and the update F(1, 1, 1) = 0 is simulated. If\nour assumption proves false (we instead read an encoded 0), then our machine\nreturns to the wrongly updated cell and simulates the update F(0, 1, 1) = 1.\nBefore giving our three small Rule 110 simulators, we give some further back-\nground explanation. Rule 110 simulates Turing machines via cyclic tag systems.\nA Rule 110 instance that simulates a cyclic tag system computation is of the\nfollowing form (for more details see [3,26]). The input to the cyclic tag system\nis encoded in a contiguous finite number of Rule 110 cells. On the left of the\n4"},{"paragraph_id":"p5","order":5,"text":"input a fixed constant word (representing the ‘ossifiers’) is repeated infinitely\nmany times. On the right, another fixed constant word (representing the cyclic\ntag system program/appendants, and the ‘leaders’) is repeated infinitely many\ntimes. Both of these repeated words are independent of the input.\nAs in [3,26], our weakly universal machines operate by traversing a finite\namount of the tape from left to right and then from right to left. This simulates\na single timestep of Rule 110 over a finite part of the encoded infinite Rule 110\ninstance. With each simulated timestep the length of a traversal increases. So\nthat each traversal is of finite length, the left blank word l and the right blank\nword r of each of our weak machines must have a special form. These words\ncontain special subwords or symbols that terminate each traversal, causing the\ntape head to turn. When the head is turning it ‘deletes’ any symbols that caused\na turn. Thus the number of cells that are being updated increases monotonically\nover time. This technique simulates Rule 110 properly if the initial condition\nis set up so that within each repeated blank word, the subword between each\nsuccessive turn point is shifted one timestep forward in time.\nIn the sequel we describe the computation of our three machines by showing\na simulation of the update on the ether in Figure 2. In the next paragraph below,\nwe outline why this example is in fact general enough to prove universality. First,\nwe must define blank words that are suitable for this example. The left blank\nword l, on the Turing machine tape, encodes the Rule 110 sequence 0001. In\nthe initial configuration as we move left each subsequent sequence 0001 is one\ntimestep further ahead. To see this note from Figure 2 that 0001 occupies, cells\n−7 to −4 in configuration c1, cells −11 to −8 in c2, cells −15 to −12 in c3, etc.\nSimilarly, the right blank word r encodes the Rule 110 sequence 110011. Looking\nat the initial configuration, as we move right from cell 0, in the first blank word\nthe first four cells 1100 are shifted two timesteps ahead, and the next two cells\n11 are shifted a further one timestep. To see this note from Figure 2 that 1100\noccupies cells 1 to 4 in c2 and 11 occupies cells 5 and 6 in c3. In each subsequent\nsequence the first four cells 1100 are shifted only one timestep ahead and the last\ntwo cells 11 are shifted one further timestep. In each row the ether in Figure 2\nrepeats every 14 cells and if the number of timesteps s between two rows is s ≡0\nmod 7 then the two rows are identical. The periodic nature of the ether, in both\ntime and space, allows us to construct such blank words.\nIt should be noted that the machines we present here, and those in [3,26],\nrequire suitable blank words to simulate a Rule 110 instance directly. If no suit-\nable blank words can be found (i.e. if they do not contain the specific subwords\nthat we use to terminate traversals) then it may be the case that the particular\ninstance can not be simulated directly. In the sequel our machines simulate the\nbackground ether that is used in the universality proof of Rule 110 [3,26]. The\ngliders that move through this ether are periodic in time and space, and so we\ncan construct blank words where the ether includes the subwords that termi-\nnate traversals. By this reasoning, our example is sufficiently general to prove\nthat our machines simulate Turing machines via Rule 110 and we do not give a\nfull (and possibly tedious) proof of correctness. For U3,3 we explicitly simulate\n5"},{"paragraph_id":"p6","order":6,"text":"three updates from Figure 2, which is general enough so that an update [Equa-\ntion (1)] on each of the eight possible three state combinations is simulated.\nWe give shorter examples for the machines U2,4 and U6,2 as they use the same\nsimulation algorithm as U3,3.\nThe machines we present here do not halt. Cook [3] shows how a special\nglider may be produced during the simulation of a Turing machine by Rule 110.\nThis glider may be used to simulate halting as the encoding can be such that it\nis generated by Rule 110 if and only if the simulated machine halts. The glider\nwould be encoded on the tape of our machines as a unique, constant word.\n3.1\nU3,3\nWe begin by describing an initial configuration of U3,3. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by 0, and the\nRule 110 state 1 is encoded by either 1 or b. The word 1b0 is used to terminate\na left traversal. (Note an exception: the 1 in the subword 1b0 encodes the Rule\n110 state 0.) To the right of the tape head position, the Rule 110 state 0 is\nencoded by 1, and the Rule 110 state 1 is encoded by 0 or b. The tape symbol 0\nis used to terminate a right traversal. The left and right blank words, described\nin paragraph 4 of Section 3, are encoded as 0 0 1 b and 0 b 1 1 0 b respectively.\nu1\nu2\nu3\n0\n1Lu1 0Ru1 bLu1\n1\nbLu2 1Lu2 0Ru3\nb\nbLu3\n1Ru3\nTable 1. Table of behaviour for U3,3.\nWe give an example of U3,3 simulating the three successive Rule 110 timesteps\nc0 ⊢c1 ⊢c2 ⊢c3 given in Figure 2. In the below configurations the current\nstate of U3,3 is highlighted in bold, to the left of its tape contents. The tape\nhead position of U3,3 is given by an underline and the start state is u1. The\nconfiguration immediately below encodes c0 from Figure 2 with the tape head\nover cell index 0.\nu1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n6"},{"paragraph_id":"p7","order":7,"text":"When the tape head reads the subword 1b0 the left traversal is complete and\nthe right traversal begins.\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1 0 b\n0 b 1 1 0 b . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢15 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 b b\n0 b 1 1 0 b . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2. We continue our simulation to give timestep c2 ⊢c3.\n⊢3 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢4 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢21 u1\nu1\nu1, . . . 0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0\n0 1 0 0 1 1\nb b 1 1 0 b . . .\nThe simulation of timestep c2 ⊢c3 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −15 to 6 of configuration c3 in\nFigure 2.\n3.2\nU2,4\nWe begin by describing an initial configuration of U2,4. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by either 0 or 0/\n7"},{"paragraph_id":"p8","order":8,"text":"and the Rule 110 state 1 is encoded by either 1 or 1/ . The word 0/ 1 is used to\nterminate a left traversal. To the right of the tape head position, the Rule 110\nstate 0 is encoded by 0/ and the Rule 110 state 1 is encoded by 1/ or 0. The tape\nsymbol 0 is used to terminate a right traversal. The left and right blank words,\nfrom paragraph 4 of Section 3, are encoded as 0 0 0/ 1 and 0 1/ 0/ 0/ 0 1/ respectively.\nu1\nu2\n0\n0/ Lu1 1/ Ru1\n1\n1/ Lu2 0/ Lu2\n0/\n1/ Lu1 0Ru2\n1/\n1/ Lu1 1Ru2\nTable 2. Table of behaviour for U2,4.\nBy way of example we give U2,4 simulating the two successive Rule 110\ntimesteps c0 ⊢c1 ⊢c2 given in Figure 2. The configuration immediately below\nencodes c0 from Figure 2 with the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢6 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/1/\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nWhen the tape head reads the subword 0/ 1 the left traversal is complete and the\nright traversal begins.\n⊢6 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢2 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢2 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢4 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1/\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢15 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 1/ 1/\n0 1/ 0/ 0/ 0 1/ . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2.\n8"},{"paragraph_id":"p9","order":9,"text":"3.3\nU6,2\nWe begin by describing an initial configuration of U6,2. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by the word\n00 and the Rule 110 state 1 is encoded by the word 11. The word 010100 is\nused to terminate a left traversal and encodes the sequence of Rule 110 states\n010. To the right of the tape head position the Rule 110 state 0 is encoded\nby the word 00 and the Rule 110 state 1 is encoded by either of the words 01\nor 10. The word 10 is used to terminate a right traversal. The left and right\nblank words, from paragraph 4 of Section 3, are encoded as 0 0 0 0 0 1 0 1 and\n1 0 0 1 0 0 0 0 1 0 0 1 respectively.\nu1\nu2\nu3\nu4\nu5\nu6\n0\n0Lu1 0Lu6 0Ru2 1Ru5 1Lu4 1Lu1\n1\n1Lu2 0Lu3 1Lu3 0Ru6 1Ru4 0Ru4\nTable 3. Table of behaviour for U6,2.\nTo illustrate the operation of U6,2 we simulate the Rule 110 timestep c0 ⊢c1\ngiven in Figure 2. The configuration immediately below encodes c0 from Figure 2\nwith the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 11\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u3\nu3\nu3, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 10 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\nWhen the tape head reads the subword 1 0 1 0 0 the left traversal is complete\nand the right traversal begins.\n9"},{"paragraph_id":"p10","order":10,"text":"⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢4 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 10 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢2 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 1 0 1 0 0 0 0 1 0 0 1 . . .\nImmediately after the tape head reads a 10, during a right traversal, the simula-\ntion of timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold (recall 0 and 1 are encoded as 00 and 11 respectively) with\ncells −7 to 0 of configuration c1 in Figure 2.\n4\nDiscussion on lower bounds\nThe pursuit to find the smallest possible universal Turing machine must also\ninvolve the search for lower bounds, finding the largest set of Turing machines\nthat are in some sense non-universal. One approach is to settle the decidability\nof the halting problem, but this approach is not suitable for the machines we\nhave presented.\nIt is known that the halting problem is decidable for (standard) Turing ma-\nchines with the following state-symbol pairs (2, 2) [7,17], (3, 2) [18], (2, 3) (claimed\nby Pavlotskaya [17]), (1, n) [5] and (n, 1) (trivial), where n ⩾1. Then, these de-\ncidability results imply that a universal Turing machine, that simulates any\nTuring machine M and halts if and only if M halts, is not possible for these\nstate-symbol pairs. Hence these results give lower bounds on the size of univer-\nsal machines of this type. While it is trivial to prove that the halting problem\nis decidable for (possibly halting) weak machines with state-symbol pairs of the\nform (n, 1), it is not known whether the above decidability results generalise to\n(possibly halting) weak Turing machines.\nThe weak machines presented in this paper, and those in [3,26], do not halt.\nHence the non-universality results discussed in the previous paragraph would\nhave to be generalised to non-halting weak machines to give lower bounds that\nare relevant for our machines. This may prove difficult for two reasons. The\nfirst issue is that, intuitively speaking, weakness gives quite an advantage. For\n10"},{"paragraph_id":"p11","order":11,"text":"instance, the program of a universal machine may be encoded in one of the in-\nfinitely repeated blank words of the weak machine. The second issue is related\nto the problem of defining a computation. Informally, a computation could be\ndefined as a sequence of configurations that leads to a special terminal config-\nuration. For non-halting machines, there are many ways to define a terminal\nconfiguration. Given a definition of terminal configuration we may prove that\nthe terminal configuration problem (will a machine ever enter a terminal config-\nuration) is decidable for a machine or set of machines. However this result may\nnot hold as a proof of non-universality if we subsequently alter our definition of\nterminal configuration.\nAcknowledgements\nTurlough Neary is funded by the Irish Research Council for Science, Engineering\nand Technology. Damien Woods is funded by Science Foundation Ireland grant\nnumber 04/IN3/1524.\nReferences\n1. Claudio Baiocchi. Three small universal Turing machines. In Maurice Margenstern\nand Yurii Rogozhin, editors, Machines, Computations, and Universality (MCU),\nvolume 2055 of LNCS, pages 1–10, Chi ̧sin ̆au, Moldova, May 2001. Springer.\n2. John Cocke and Marvin Minsky. Universality of tag systems with P = 2. Journal\nof the ACM, 11(1):15–20, January 1964.\n3. Matthew Cook. Universality in elementary cellular automata. Complex Systems,\n15(1):1–40, 2004.\n4. Raymond Greenlaw, H. James Hoover, and Walter L. Ruzzo. Limits to parallel\ncomputation: P -completeness theory. Oxford university Press, Oxford, 1995.\n5. Gabor Hermann. The uniform halting problem for generalized one state Turing\nmachines. In Proceedings, Ninth Annual Symposium on Switching and Automata\nTheory (FOCS), pages 368–372, Schenectady, New York, October 1968. IEEE Com-\nputer Society Press.\n6. John E. Hopcroft and Jeffrey D. Ullman. Introduction to automata theory, lan-\nguages, and computation. Addison-Wesley Series in Computer Science. Addison-\nWesley, Reading, Mass., 1979.\n7. Manfred Kudlek.\nSmall deterministic Turing machines.\nTheoretical Computer\nScience, 168(2):241–255, November 1996.\n8. Manfred Kudlek and Yurii Rogozhin. A universal Turing machine with 3 states\nand 9 symbols. In Werner Kuich, Grzegorz Rozenberg, and Arto Salomaa, editors,\nDevelopments in Language Theory (DLT) 2001, volume 2295 of LNCS, pages 311–\n318, Vienna, May 2002. Springer.\n9. Maurice Margenstern and Liudmila Pavlotskaya. On the optimal number of in-\nstructions for universality of Turing machines connected with a finite automaton.\nInternational Journal of Algebra and Computation, 13(2):133–202, April 2003.\n10. Marvin Minsky. A 6-symbol 7-state universal Turing machines. Technical Report\n54-G-027, MIT, August 1960.\n11"},{"paragraph_id":"p12","order":12,"text":"11. Marvin Minsky. Size and structure of universal Turing machines using tag systems.\nIn Recursive Function Theory, Symp. in Pure Math., volume 5, pages 229–238.\nAMS, 1962.\n12. Turlough Neary and Damien Woods. P-completeness of cellular automaton Rule\n110. In M. Bugliesi et al., editor, International Colloquium on Automata Languages\nand Programing 2006, (ICALP) Part I, volume 4051 of LNCS, pages 132–143,\nVenice, July 2006. Springer.\n13. Turlough Neary and Damien Woods. The P-completeness of cellular automaton\nRule 110. Technical Report 04/2006, Boole Centre for Research in Informatics,\nUniversity College Cork, Ireland, 2006.\n14. Turlough Neary and Damien Woods. Small fast universal Turing machines. The-\noretical Computer Science, 362(1–3):171–195, November 2006.\n15. Turlough Neary and Damien Woods. Four small universal Turing machines. In\nJ ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computations,\nand Universality (MCU), volume 4664 of LNCS, pages 242–254, Or ́elans, France,\nSeptember 2007. Springer.\n16. A. Nozaki. On the notion of universality of Turing machine. Kybernetika Academia\nPraha, 5(1):29–43, 1969.\n17. Liudmila Pavlotskaya.\nSolvability of the halting problem for certain classes of\nTuring machines. Mathematical Notes (Springer), 13(6):537–541, June 1973.\n18. Liudmila Pavlotskaya.\nDostatochnye uslovija razreshimosti problemy ostanovki\ndlja mashin T’juring. Problemi kibernetiki, pages 91–118, 1978. (Sufficient condi-\ntions for the halting problem decidability of Turing machines.In Russian).\n19. Lutz Priese. Towards a precise characterization of the complexity of universal and\nnon-universal Turing machines. Siam journal of Computing, 8(4):508–523, 1979.\n20. Yurii Rogozhin. Sem’ universal’nykh mashin T’juringa. Systems and theoretical\nprogramming, Mat. Issled, 69:76–90, 1982. (Seven universal Turing machines. In\nRussian).\n21. Yurii Rogozhin. Small universal Turing machines. Theoretical Computer Science,\n168(2):215–240, November 1996.\n22. Claude Elwood Shannon. A universal Turing machine with two internal states.\nAutomata Studies, Annals of Mathematics Studies, 34:157–165, 1956.\n23. Shigeru Watanabe. On a minimal universal Turing machines. Technical report,\nMCB Report, Tokyo, August 1960.\n24. Shigeru Watanabe. 5-symbol 8-state and 5-symbol 6-state universal Turing ma-\nchines. Journal of ACM, 8(4):476–483, October 1961.\n25. Shigeru Watanabe. 4-symbol 5-state universal Turing machines. Information Pro-\ncessing Society of Japan Magazine, 13(9):588–592, 1972.\n26. Stephen Wolfram. A new kind of science. Wolfram Media, 2002.\n27. Damien Woods and Turlough Neary. On the time complexity of 2-tag systems and\nsmall universal Turing machines. In 47th Annual IEEE Symposium on Foundations\nof Computer Science (FOCS), pages 132–143, Berkeley, California, October 2006.\nIEEE.\n28. Damien Woods and Turlough Neary. The complexity of small universal Turing\nmachines. In S. Barry Cooper, Benedikt L ̈owe, and Andrea Sorbi, editors, Com-\nputability in Europe 2007, volume 4497 of LNCS, pages 791–798, Siena, Italy, June\n2007. Springer.\n29. Damien Woods and Turlough Neary. Small semi-weakly universal Turing machines.\nIn J ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computa-\ntions, and Universality (MCU), volume 4664 of LNCS, pages 306–323, Or ́elans,\nFrance, September 2007. Springer.\n12"}],"pages":[{"page":1,"text":"arXiv:0707.4489v1 [cs.CC] 30 Jul 2007\nSmall weakly universal Turing machines\nTurlough Neary1 and Damien Woods2\n1 TASS, Department of Computer Science,\nNational University of Ireland Maynooth, Ireland. tneary@cs.may.ie\n2 Department of Computer Science,\nUniversity College Cork, Ireland. d.woods@cs.ucc.ie\nAbstract. We give small universal Turing machines with state-symbol\npairs of (6, 2), (3, 3) and (2, 4). These machines are weakly universal,\nwhich means that they have an infinitely repeated word to the left of\ntheir input and another to the right. They simulate Rule 110 and are\ncurrently the smallest known weakly universal Turing machines.\n1\nIntroduction\nShannon [22] was the first to consider the problem of finding the smallest uni-\nversal Turing machine, where size is the number of states and symbols. Here we\nsay that a Turing machine is standard if it has a single one-dimensional tape,\none tape head, and is deterministic [6]. Over the years, small universal programs\nwere given for a number of variants on the standard model. By generalising the\nmodel we often find smaller universal programs. One variation on the standard\nmodel is to allow the blank portion of the Turing machine’s tape to have an\ninfinitely repeated word to the left, and another to the right. We refer to such\nuniversal machines as weakly universal Turing machines, and they the subject\nof this work.\nBeginning in the early sixties Minsky and Watanabe engaged in a vigorous\ncompetition to see who could come up with the smallest universal Turing ma-\nchine [10,11,23,24,25]. In 1961 Watanabe [24] gave a 6-state, 5-symbol universal\nTuring machine, the first weakly universal machine. In 1962, Minsky [11] found a\nsmall 7-state, 4-symbol universal Turing machine. Not to be out-done, Watanabe\nimproved on his earlier machine to give 5-state, 4-symbol and 7-state, 3-symbol\nweakly universal machines [25,16].\nThe 7-state universal Turing machine of Minsky has received much atten-\ntion. Minsky’s machine simulates Turing machines via 2-tag systems, which\nwere proved universal by Cocke and Minsky [2]. The technique of simulating\n2-tag systems, pioneered by Minsky, was extended by Rogozhin [20] to give the\n(then) smallest known universal Turing machines for a number of state-symbol\npairs. These 2-tag simulators were subsequently reduced in size by Rogozhin [21],\nKudlek and Rogozhin [8], and Baiocchi [1]. Neary and Woods [15] gave small uni-\nversal machines that simulate Turing machines via a new variant of tag systems\ncalled bi-tag systems. Each of the smallest 2-tag or bi-tag simulators are plotted\nas circles in Figure 1. These (standard) machines induce a universal curve."},{"page":2,"text":"b\nc : standard universal machine\nl : semi-weakly universal machine\nr\ns\n: weakly universal machine\nr\n: new weakly universal machine\n: standard universal curve\n: weakly universal curve\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\nstates\nsymbols\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nb\nc\nl\nl\nl\nl\nr\ns\nr\ns\nr\ns\nr\ns\nr\nr\nr\nFig. 1. State-symbol plot of small universal Turing machines. Each of our new\nweak machines is represented by a solid square. These machines induce a weakly\nuniversal curve.\nThe small weak machines of Watanabe have received little attention. In par-\nticular the 5-state and 7-state machines seem little known and are largely ignored\nin the literature. It is worth noting that while all other weak machines simulate\nTuring machine via other simple models, Watanabe’s weak machines simulate\nTuring machines directly. His machines are the most time efficient of the small\nweak machines. More precisely, let t be the running time of any deterministic\nsingle tape Turing machine M, then Watanabe’s machines are the smallest weak\nmachines that simulate M with a time overhead of O(t2).\nWe often refer to Watanabe’s machines as being semi-weak. Semi-weak ma-\nchines have an infinitely repeated word to one side of their input, and on the\nother side they have a (standard) infinitely repeated blank symbol. Recently,\nWoods and Neary [29] have given 3-state, 7-symbol and 4-state, 5-symbol semi-\nweakly universal machines that simulate cyclic tag systems. All of the smallest\nsemi-weakly universal machines are given as diamonds in Figure 1.\nCook [3] and Wolfram [26] recently gave weakly universal Turing machines,\nsmaller than Watanabe’s semi-weak machines, that simulate the universal cellu-\nlar automata Rule 110. These machines have state-symbol pairs of (7, 2), (4, 3),\n(3, 4), (2, 5) and are plotted as hollow squares in Figure 1.\n2"},{"page":3,"text":"Here we present weakly universal Turing machines with state-symbol pairs\nof (6, 2), (3, 3), (2, 4), making them the smallest known weakly universal ma-\nchines. Our machines simulate (single tape, deterministic) Turing machines in\ntime O(t4 log2 t), via Rule 110. These machines are plotted as solid squares\nin Figure 1 and induce a weakly universal curve. It is interesting to note from\nFigure 1 that the smallest universal machines, and the smallest semi-weakly uni-\nversal machines, are both symmetric about the line where states equals symbols,\nwhereas the smallest weakly universal machines are not.\nWeakness has not been the only variation on the standard model in the\nsearch for small universal Turing machines. Priese [19] gave a 2-state, 4-symbol\nmachine with a 2-dimensional tape, and a 2-state, 2-symbol machine with a\n2-dimensional tape and 2 tape heads. Margenstern and Pavlotskaya [9] gave a\n2-state, 3-symbol Turing machine that is universal when coupled with a finite\nautomaton. This machine uses only 5 instructions. Margenstern and Pavlotskaya\nalso show that the halting problem is decidable for machines of this type with 4\ninstructions. Their result implies that it is not possible to have a 4 instruction\nuniversal machine of this type, that simulates any Turing machine M and halts\nif and only if M halts. Hence they have given the smallest possible universal\nmachine of this type.\nFor results relating to the time complexity of small universal Turing machines\nsee [12,14,27,28].\n1.1\nPreliminaries\nThe Turing machines considered in this paper are deterministic and have a single\nbi-infinite tape. We let Um,n denote our weakly universal Turing machine with m\nstates and n symbols. We write c1 ⊢c2 if a configuration c2 is obtained from c1\nvia a single computation step. We let c1 ⊢s c2 denote a sequence of s computation\nsteps, and let c1 ⊢∗c2 denote zero or more computation steps.\n2\nRule 110\nRule 110 is a very simple (2 state, nearest neighbour, one dimensional) cel-\nlular automaton. It is composed of a sequence of cells . . . p−1p0p1 . . . where\neach cell has a binary state pi ∈{0, 1}. At timestep s + 1, the value pi,s+1 =\nF(pi−1,s, pi,s, pi+1,s) of the cell at position i is given by the synchronous local\nupdate function F\nF(0, 0, 0) = 0\nF(1, 0, 0) = 0\nF(0, 0, 1) = 1\nF(1, 0, 1) = 1\nF(0, 1, 0) = 1\nF(1, 1, 0) = 1\nF(0, 1, 1) = 1\nF(1, 1, 1) = 0\n(1)\nRule 110 was proven universal by Cook [3] and Wolfram [26]. Recently, Neary\nand Woods [12,13] proved that Rule 110 simulates Turing machines efficiently\n3"},{"page":4,"text":"c0\nc1\nc2\nc3...\n. . . -9 -8 -7 -6 -5 -4 -3 -2 -1 0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n. . .\nFig. 2. Seven consecutive timesteps of Rule 110. These seven timesteps are ap-\nplied to the background ether that is used in the proof [3] of universality of Rule\n110. Each black or each white square represents, a Rule 110 cell containing, state\n1 or 0 respectively. Each cell is identified by the index given above it. To the left\nof each row of cells there is a configuration label that identifies that row.\nin polynomial time O(t3 log t), an exponential improvement. Note that, in or-\nder to calculate this upper bound we substitute space bounds for time bounds\nwhenever possible in the analysis. It turns out that we can further improve the\nsimulation time to O(t2 log t) (this result is as yet unpublished). Rule 110 sim-\nulates cyclic tag systems in linear time. The weak machines in this paper, and\nin [3,26], simulate Rule 110 with a quadratic polynomial increase in time and\nhence simulate Turing machines in time O(t4 log2 t). It is worth noting that the\nprediction problem [4] for these machines is P-complete, and this is also the case\nwhen we consider only bounded initial conditions [12].\n3\nThree small weakly universal Turing machines\nThe following observation is one of the reasons for the improvement in size over\nprevious machines [3,26], and gives some insight into the simulation algorithm we\nuse. Notice from Equation (1) that the value of the update function F, with the\nexception of F(0, 1, 1) and F(1, 1, 1), may be determined using only the rightmost\ntwo states. Each of our universal Turing machines exploit this fact as follows.\nThe machines scan from right to left, and in six of the eight cases they need\nonly remember the cell immediately to the right of the current cell i in order to\ncompute the update for i. Thus for these six cases we need only store a single cell\nvalue, rather than two values. The remaining two cases are simulated as follows.\nIf two consecutive encoded states with value 1 are read, it is assumed that there\nis another encoded 1 to the left and the update F(1, 1, 1) = 0 is simulated. If\nour assumption proves false (we instead read an encoded 0), then our machine\nreturns to the wrongly updated cell and simulates the update F(0, 1, 1) = 1.\nBefore giving our three small Rule 110 simulators, we give some further back-\nground explanation. Rule 110 simulates Turing machines via cyclic tag systems.\nA Rule 110 instance that simulates a cyclic tag system computation is of the\nfollowing form (for more details see [3,26]). The input to the cyclic tag system\nis encoded in a contiguous finite number of Rule 110 cells. On the left of the\n4"},{"page":5,"text":"input a fixed constant word (representing the ‘ossifiers’) is repeated infinitely\nmany times. On the right, another fixed constant word (representing the cyclic\ntag system program/appendants, and the ‘leaders’) is repeated infinitely many\ntimes. Both of these repeated words are independent of the input.\nAs in [3,26], our weakly universal machines operate by traversing a finite\namount of the tape from left to right and then from right to left. This simulates\na single timestep of Rule 110 over a finite part of the encoded infinite Rule 110\ninstance. With each simulated timestep the length of a traversal increases. So\nthat each traversal is of finite length, the left blank word l and the right blank\nword r of each of our weak machines must have a special form. These words\ncontain special subwords or symbols that terminate each traversal, causing the\ntape head to turn. When the head is turning it ‘deletes’ any symbols that caused\na turn. Thus the number of cells that are being updated increases monotonically\nover time. This technique simulates Rule 110 properly if the initial condition\nis set up so that within each repeated blank word, the subword between each\nsuccessive turn point is shifted one timestep forward in time.\nIn the sequel we describe the computation of our three machines by showing\na simulation of the update on the ether in Figure 2. In the next paragraph below,\nwe outline why this example is in fact general enough to prove universality. First,\nwe must define blank words that are suitable for this example. The left blank\nword l, on the Turing machine tape, encodes the Rule 110 sequence 0001. In\nthe initial configuration as we move left each subsequent sequence 0001 is one\ntimestep further ahead. To see this note from Figure 2 that 0001 occupies, cells\n−7 to −4 in configuration c1, cells −11 to −8 in c2, cells −15 to −12 in c3, etc.\nSimilarly, the right blank word r encodes the Rule 110 sequence 110011. Looking\nat the initial configuration, as we move right from cell 0, in the first blank word\nthe first four cells 1100 are shifted two timesteps ahead, and the next two cells\n11 are shifted a further one timestep. To see this note from Figure 2 that 1100\noccupies cells 1 to 4 in c2 and 11 occupies cells 5 and 6 in c3. In each subsequent\nsequence the first four cells 1100 are shifted only one timestep ahead and the last\ntwo cells 11 are shifted one further timestep. In each row the ether in Figure 2\nrepeats every 14 cells and if the number of timesteps s between two rows is s ≡0\nmod 7 then the two rows are identical. The periodic nature of the ether, in both\ntime and space, allows us to construct such blank words.\nIt should be noted that the machines we present here, and those in [3,26],\nrequire suitable blank words to simulate a Rule 110 instance directly. If no suit-\nable blank words can be found (i.e. if they do not contain the specific subwords\nthat we use to terminate traversals) then it may be the case that the particular\ninstance can not be simulated directly. In the sequel our machines simulate the\nbackground ether that is used in the universality proof of Rule 110 [3,26]. The\ngliders that move through this ether are periodic in time and space, and so we\ncan construct blank words where the ether includes the subwords that termi-\nnate traversals. By this reasoning, our example is sufficiently general to prove\nthat our machines simulate Turing machines via Rule 110 and we do not give a\nfull (and possibly tedious) proof of correctness. For U3,3 we explicitly simulate\n5"},{"page":6,"text":"three updates from Figure 2, which is general enough so that an update [Equa-\ntion (1)] on each of the eight possible three state combinations is simulated.\nWe give shorter examples for the machines U2,4 and U6,2 as they use the same\nsimulation algorithm as U3,3.\nThe machines we present here do not halt. Cook [3] shows how a special\nglider may be produced during the simulation of a Turing machine by Rule 110.\nThis glider may be used to simulate halting as the encoding can be such that it\nis generated by Rule 110 if and only if the simulated machine halts. The glider\nwould be encoded on the tape of our machines as a unique, constant word.\n3.1\nU3,3\nWe begin by describing an initial configuration of U3,3. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by 0, and the\nRule 110 state 1 is encoded by either 1 or b. The word 1b0 is used to terminate\na left traversal. (Note an exception: the 1 in the subword 1b0 encodes the Rule\n110 state 0.) To the right of the tape head position, the Rule 110 state 0 is\nencoded by 1, and the Rule 110 state 1 is encoded by 0 or b. The tape symbol 0\nis used to terminate a right traversal. The left and right blank words, described\nin paragraph 4 of Section 3, are encoded as 0 0 1 b and 0 b 1 1 0 b respectively.\nu1\nu2\nu3\n0\n1Lu1 0Ru1 bLu1\n1\nbLu2 1Lu2 0Ru3\nb\nbLu3\n1Ru3\nTable 1. Table of behaviour for U3,3.\nWe give an example of U3,3 simulating the three successive Rule 110 timesteps\nc0 ⊢c1 ⊢c2 ⊢c3 given in Figure 2. In the below configurations the current\nstate of U3,3 is highlighted in bold, to the left of its tape contents. The tape\nhead position of U3,3 is given by an underline and the start state is u1. The\nconfiguration immediately below encodes c0 from Figure 2 with the tape head\nover cell index 0.\nu1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 0 b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n0 0 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 1 b\n1 1 b b\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n6"},{"page":7,"text":"When the tape head reads the subword 1b0 the left traversal is complete and\nthe right traversal begins.\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\n0 b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1 0 b\n0 b 1 1 0 b . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 0 b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u1\nu1\nu1, . . . 0 0 1 b\n0 0 1 b\n0 0 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢3 u3\nu3\nu3, . . . 0 0 1 b\n0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1 0 b\n0 b 1 1 0 b . . .\n⊢15 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 b b\n0 b 1 1 0 b . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2. We continue our simulation to give timestep c2 ⊢c3.\n⊢3 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢4 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 0 1 1\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 1\n0 0 1 b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢5 u1\nu1\nu1, . . . 0 0 1 b\n0 0 0 1\n0 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢2 u2\nu2\nu2, . . . 0 0 1 b\n0 0 0 b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢6 u3\nu3\nu3, . . . 0 0 1 b\n1 1 b b\n1 b b b\nb b 1 1\n1 b 1 1 b b\n0 b 1 1 0 b . . .\n⊢21 u1\nu1\nu1, . . . 0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0\n0 1 0 0 1 1\nb b 1 1 0 b . . .\nThe simulation of timestep c2 ⊢c3 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −15 to 6 of configuration c3 in\nFigure 2.\n3.2\nU2,4\nWe begin by describing an initial configuration of U2,4. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by either 0 or 0/\n7"},{"page":8,"text":"and the Rule 110 state 1 is encoded by either 1 or 1/ . The word 0/ 1 is used to\nterminate a left traversal. To the right of the tape head position, the Rule 110\nstate 0 is encoded by 0/ and the Rule 110 state 1 is encoded by 1/ or 0. The tape\nsymbol 0 is used to terminate a right traversal. The left and right blank words,\nfrom paragraph 4 of Section 3, are encoded as 0 0 0/ 1 and 0 1/ 0/ 0/ 0 1/ respectively.\nu1\nu2\n0\n0/ Lu1 1/ Ru1\n1\n1/ Lu2 0/ Lu2\n0/\n1/ Lu1 0Ru2\n1/\n1/ Lu1 1Ru2\nTable 2. Table of behaviour for U2,4.\nBy way of example we give U2,4 simulating the two successive Rule 110\ntimesteps c0 ⊢c1 ⊢c2 given in Figure 2. The configuration immediately below\nencodes c0 from Figure 2 with the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢6 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0/1/\n0/ 0/ 1/ 1/\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nWhen the tape head reads the subword 0/ 1 the left traversal is complete and the\nright traversal begins.\n⊢6 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\nImmediately after the tape head reads a 0, during a right traversal, the simulation\nof timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold with cells −7 to 0 of configuration c1 in Figure 2. We\ncontinue our simulation to give timestep c1 ⊢c2.\n⊢2 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 1 1\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢2 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 0 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1\n0 1/ 0/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢4 u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1\n0 0 0 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0/ 1\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢u2\nu2\nu2, . . . 0 0 0/ 1\n0 0 0/ 1/\n0/ 0/ 1/ 1/\n0/ 1/ 1/ 1/\n1/ 1/ 0/ 0/ 0 1/\n0 1/ 0/ 0/ 0 1/ . . .\n⊢15 u1\nu1\nu1, . . . 0 0 0/ 1\n0 0 0 1\n0 0 1 1\n0 1 1 1\n1 1 0 0 1/ 1/\n0 1/ 0/ 0/ 0 1/ . . .\nThe simulation of timestep c1 ⊢c2 is complete. To see this, compare the part\nof the Turing machine tape in bold with cells −11 to 4 of configuration c2 in\nFigure 2.\n8"},{"page":9,"text":"3.3\nU6,2\nWe begin by describing an initial configuration of U6,2. To the left of, and in-\ncluding, the tape head position, the Rule 110 state 0 is encoded by the word\n00 and the Rule 110 state 1 is encoded by the word 11. The word 010100 is\nused to terminate a left traversal and encodes the sequence of Rule 110 states\n010. To the right of the tape head position the Rule 110 state 0 is encoded\nby the word 00 and the Rule 110 state 1 is encoded by either of the words 01\nor 10. The word 10 is used to terminate a right traversal. The left and right\nblank words, from paragraph 4 of Section 3, are encoded as 0 0 0 0 0 1 0 1 and\n1 0 0 1 0 0 0 0 1 0 0 1 respectively.\nu1\nu2\nu3\nu4\nu5\nu6\n0\n0Lu1 0Lu6 0Ru2 1Ru5 1Lu4 1Lu1\n1\n1Lu2 0Lu3 1Lu3 0Ru6 1Ru4 0Ru4\nTable 3. Table of behaviour for U6,2.\nTo illustrate the operation of U6,2 we simulate the Rule 110 timestep c0 ⊢c1\ngiven in Figure 2. The configuration immediately below encodes c0 from Figure 2\nwith the tape head over cell index 0.\nu1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 11\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u3\nu3\nu3, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 00 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢5 u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u2\nu2\nu2, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 1 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 10 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 01\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\nWhen the tape head reads the subword 1 0 1 0 0 the left traversal is complete\nand the right traversal begins.\n9"},{"page":10,"text":"⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n1 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 1 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 0 1 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢4 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 01 0 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u5\nu5\nu5, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 10 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 01\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢2 u4\nu4\nu4, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n1 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u6\nu6\nu6, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 0 0 1 0 0 0 0 1 0 0 1 . . .\n⊢u1\nu1\nu1, . . . 0 0 0 0 0 1 0 1\n0 0 0 0 0 0 1 1\n0 0 0 0 1 1 1 1\n0 1 0 1 0 0 0 0 1 0 0 1 . . .\nImmediately after the tape head reads a 10, during a right traversal, the simula-\ntion of timestep c0 ⊢c1 is complete. To see this, compare the part of the Turing\nmachine tape in bold (recall 0 and 1 are encoded as 00 and 11 respectively) with\ncells −7 to 0 of configuration c1 in Figure 2.\n4\nDiscussion on lower bounds\nThe pursuit to find the smallest possible universal Turing machine must also\ninvolve the search for lower bounds, finding the largest set of Turing machines\nthat are in some sense non-universal. One approach is to settle the decidability\nof the halting problem, but this approach is not suitable for the machines we\nhave presented.\nIt is known that the halting problem is decidable for (standard) Turing ma-\nchines with the following state-symbol pairs (2, 2) [7,17], (3, 2) [18], (2, 3) (claimed\nby Pavlotskaya [17]), (1, n) [5] and (n, 1) (trivial), where n ⩾1. Then, these de-\ncidability results imply that a universal Turing machine, that simulates any\nTuring machine M and halts if and only if M halts, is not possible for these\nstate-symbol pairs. Hence these results give lower bounds on the size of univer-\nsal machines of this type. While it is trivial to prove that the halting problem\nis decidable for (possibly halting) weak machines with state-symbol pairs of the\nform (n, 1), it is not known whether the above decidability results generalise to\n(possibly halting) weak Turing machines.\nThe weak machines presented in this paper, and those in [3,26], do not halt.\nHence the non-universality results discussed in the previous paragraph would\nhave to be generalised to non-halting weak machines to give lower bounds that\nare relevant for our machines. This may prove difficult for two reasons. The\nfirst issue is that, intuitively speaking, weakness gives quite an advantage. For\n10"},{"page":11,"text":"instance, the program of a universal machine may be encoded in one of the in-\nfinitely repeated blank words of the weak machine. The second issue is related\nto the problem of defining a computation. Informally, a computation could be\ndefined as a sequence of configurations that leads to a special terminal config-\nuration. For non-halting machines, there are many ways to define a terminal\nconfiguration. Given a definition of terminal configuration we may prove that\nthe terminal configuration problem (will a machine ever enter a terminal config-\nuration) is decidable for a machine or set of machines. However this result may\nnot hold as a proof of non-universality if we subsequently alter our definition of\nterminal configuration.\nAcknowledgements\nTurlough Neary is funded by the Irish Research Council for Science, Engineering\nand Technology. Damien Woods is funded by Science Foundation Ireland grant\nnumber 04/IN3/1524.\nReferences\n1. Claudio Baiocchi. Three small universal Turing machines. In Maurice Margenstern\nand Yurii Rogozhin, editors, Machines, Computations, and Universality (MCU),\nvolume 2055 of LNCS, pages 1–10, Chi ̧sin ̆au, Moldova, May 2001. Springer.\n2. John Cocke and Marvin Minsky. Universality of tag systems with P = 2. Journal\nof the ACM, 11(1):15–20, January 1964.\n3. Matthew Cook. Universality in elementary cellular automata. Complex Systems,\n15(1):1–40, 2004.\n4. Raymond Greenlaw, H. James Hoover, and Walter L. Ruzzo. Limits to parallel\ncomputation: P -completeness theory. Oxford university Press, Oxford, 1995.\n5. Gabor Hermann. The uniform halting problem for generalized one state Turing\nmachines. In Proceedings, Ninth Annual Symposium on Switching and Automata\nTheory (FOCS), pages 368–372, Schenectady, New York, October 1968. IEEE Com-\nputer Society Press.\n6. John E. Hopcroft and Jeffrey D. Ullman. Introduction to automata theory, lan-\nguages, and computation. Addison-Wesley Series in Computer Science. Addison-\nWesley, Reading, Mass., 1979.\n7. Manfred Kudlek.\nSmall deterministic Turing machines.\nTheoretical Computer\nScience, 168(2):241–255, November 1996.\n8. Manfred Kudlek and Yurii Rogozhin. A universal Turing machine with 3 states\nand 9 symbols. In Werner Kuich, Grzegorz Rozenberg, and Arto Salomaa, editors,\nDevelopments in Language Theory (DLT) 2001, volume 2295 of LNCS, pages 311–\n318, Vienna, May 2002. Springer.\n9. Maurice Margenstern and Liudmila Pavlotskaya. On the optimal number of in-\nstructions for universality of Turing machines connected with a finite automaton.\nInternational Journal of Algebra and Computation, 13(2):133–202, April 2003.\n10. Marvin Minsky. A 6-symbol 7-state universal Turing machines. Technical Report\n54-G-027, MIT, August 1960.\n11"},{"page":12,"text":"11. Marvin Minsky. Size and structure of universal Turing machines using tag systems.\nIn Recursive Function Theory, Symp. in Pure Math., volume 5, pages 229–238.\nAMS, 1962.\n12. Turlough Neary and Damien Woods. P-completeness of cellular automaton Rule\n110. In M. Bugliesi et al., editor, International Colloquium on Automata Languages\nand Programing 2006, (ICALP) Part I, volume 4051 of LNCS, pages 132–143,\nVenice, July 2006. Springer.\n13. Turlough Neary and Damien Woods. The P-completeness of cellular automaton\nRule 110. Technical Report 04/2006, Boole Centre for Research in Informatics,\nUniversity College Cork, Ireland, 2006.\n14. Turlough Neary and Damien Woods. Small fast universal Turing machines. The-\noretical Computer Science, 362(1–3):171–195, November 2006.\n15. Turlough Neary and Damien Woods. Four small universal Turing machines. In\nJ ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computations,\nand Universality (MCU), volume 4664 of LNCS, pages 242–254, Or ́elans, France,\nSeptember 2007. Springer.\n16. A. Nozaki. On the notion of universality of Turing machine. Kybernetika Academia\nPraha, 5(1):29–43, 1969.\n17. Liudmila Pavlotskaya.\nSolvability of the halting problem for certain classes of\nTuring machines. Mathematical Notes (Springer), 13(6):537–541, June 1973.\n18. Liudmila Pavlotskaya.\nDostatochnye uslovija razreshimosti problemy ostanovki\ndlja mashin T’juring. Problemi kibernetiki, pages 91–118, 1978. (Sufficient condi-\ntions for the halting problem decidability of Turing machines.In Russian).\n19. Lutz Priese. Towards a precise characterization of the complexity of universal and\nnon-universal Turing machines. Siam journal of Computing, 8(4):508–523, 1979.\n20. Yurii Rogozhin. Sem’ universal’nykh mashin T’juringa. Systems and theoretical\nprogramming, Mat. Issled, 69:76–90, 1982. (Seven universal Turing machines. In\nRussian).\n21. Yurii Rogozhin. Small universal Turing machines. Theoretical Computer Science,\n168(2):215–240, November 1996.\n22. Claude Elwood Shannon. A universal Turing machine with two internal states.\nAutomata Studies, Annals of Mathematics Studies, 34:157–165, 1956.\n23. Shigeru Watanabe. On a minimal universal Turing machines. Technical report,\nMCB Report, Tokyo, August 1960.\n24. Shigeru Watanabe. 5-symbol 8-state and 5-symbol 6-state universal Turing ma-\nchines. Journal of ACM, 8(4):476–483, October 1961.\n25. Shigeru Watanabe. 4-symbol 5-state universal Turing machines. Information Pro-\ncessing Society of Japan Magazine, 13(9):588–592, 1972.\n26. Stephen Wolfram. A new kind of science. Wolfram Media, 2002.\n27. Damien Woods and Turlough Neary. On the time complexity of 2-tag systems and\nsmall universal Turing machines. In 47th Annual IEEE Symposium on Foundations\nof Computer Science (FOCS), pages 132–143, Berkeley, California, October 2006.\nIEEE.\n28. Damien Woods and Turlough Neary. The complexity of small universal Turing\nmachines. In S. Barry Cooper, Benedikt L ̈owe, and Andrea Sorbi, editors, Com-\nputability in Europe 2007, volume 4497 of LNCS, pages 791–798, Siena, Italy, June\n2007. Springer.\n29. Damien Woods and Turlough Neary. Small semi-weakly universal Turing machines.\nIn J ́erˆome Durand-Lose and Maurice Margenstern, editors, Machines, Computa-\ntions, and Universality (MCU), volume 4664 of LNCS, pages 306–323, Or ́elans,\nFrance, September 2007. Springer.\n12"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"each cell has a binary state pi ∈{0, 1}. At timestep s + 1, the value pi,s+1 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"F(0, 0, 0) = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"F(1, 0, 0) = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"F(0, 0, 1) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"F(1, 0, 1) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"F(0, 1, 0) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"F(1, 1, 0) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"F(0, 1, 1) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"F(1, 1, 1) = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"is another encoded 1 to the left and the update F(1, 1, 1) = 0 is simulated. If","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"returns to the wrongly updated cell and simulates the update F(0, 1, 1) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"2. John Cocke and Marvin Minsky. Universality of tag systems with P = 2. Journal","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":31623,"parse_confidence":0.5,"equation_parse_rate_proxy":0.6,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}