CONWAY S COSMOLOGICAL THEOREM

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1 CONWAY S COSMOLOGICAL THEOREM R.A. LITHERLAND. Introduction In [C], Conway introduced an operator on strings (finite sequences) of positive integers, the audioactive (or look and say ) operator. Usually, the integers involved are single digit numbers, and a string consisting of four ones and a single three, for instance, is written as ; in general, the terms of the sequence are referred to as digits regardless of their size. To apply the operator to the above string, describe it in words as four ones, one three and then translate this to the digit string 4. More formally, suppose a string S is written as d e de den n, where de represents e consecutive digits d. Whenever this is done, it is understood that the grouping is maximal; that is, d i d i+ for i < n. Then the result of applying the operator to S is the daughter sequence α(s) = e d e d e n d n. Applying the operator repeatedly produces the descendants of S, α n (S) for n = 0,,.... A string that is equal to α n (S) for some string S and integer n is called an n-day-old string. Some strings of the form S = LR (with L and R non-empty) have the property that the descendants of L and R do not interfere with one another, in the sense that α n (S) = α n (L)α n (R) for all n. In this case, we say that S splits, and write S = L.R. A string that does not split is called an atom, or element, and every string splits uniquely into atoms, and is also called a compound. Conway proved that there is a finite set of common elements that is stable, in the sense that if S is a common element, all the elements appearing in α(s) are common, and that has the remarkable property that any sufficiently old descendant of an arbitrary string (other than and ) involves all the common elements. Astoundingly, there are precisely 9 of these elements, and (less astoundingly) Conway gave them the names Hydrogen, Helium,..., Uranium. He also found two infinite families of elements that are persistent, in the sense that each such element S appears in α n (S) for some n > 0 (in fact for n = ), and so keeps reappearing. These he called the isotopes of Neptunium and Plutonium, collectively known as the transuranic elements. His Cosmological Theorem asserts that there is some integer N such that any N-day-old string splits into common and transuranic elements. One proof of this was found by Conway and Richard Parker, and another by Mike Guy that showed that one could take N = 4. Both proofs were lost. Subsequently, Ekhad and Zeilberger [EZ] gave a proof consisting of a Maple program written by Zeilberger and executed by Ekhad, together with a short explanation of what the program does and why its output constitutes a proof. This, however, only shows that one may take N = 9. This paper discusses two proofs, both computer assisted, that indeed one may take N = 4. The C source code for the programs involved and some related programs, together with documentation, can be found at Date: April, 00. Last revised: April 4, 006. Zeilberger is a human being and Ekhad is his computer.

2 The first proof is a minor adaptation of Ekhad and Zeilberger s. Almost all the work is done by the program proof; the algorithm used, and why its execution constitutes a proof, are explained in the accompanying documentation, and this proof will not be discussed further here. The second proof is similar in spirit (I think) to Conway and Parker s. That consisted, according to Conway, of a very subtle and complicated argument, which (almost) reduced the problem to tracking a few hundred cases, which were then handled on dozens of sheets of paper (now lost). The rest of this paper gives a complicated argument reducing the problem to tracking a finite number of cases. It is clearly not subtle enough, because the number is 60. These cases are then checked by the program proof. That the number 4 cannot be improved was also shown by Guy, who found atoms of longevity 4, given on page 86 of [C]. Dropping the first two digits gives shorter elements of longevity 4, the isotopes of Thuselum, as illustrated in the documentation for my program evolve. I assume familiarity with Conway s paper; in particular, all the named theorems referred to are found there. Our notation differs in the use of α(s) rather than S for the daughter of S; subscripts on strings are just indices here. As in [C], a chunk of a string is a substring of consecutive digits. By a run of a string we mean a maximal chunk of identical digits, and by a block we mean a chunk made up of runs. A {, }-chunk or {, }-block is one containing only the digits and. The longevity of a string S [EZ] is the minimum n for which α n (S) is composed of common and transuranic elements.. Four-day-old strings Recall that a string is parsed by inserting commas to indicate its derivation from a parent string. Conway gave restrictions on the strings or parsed strings that can be chunks of - and -day-old strings; we give further restrictions for - and 4-day-old strings. Lemma.. () A -day-old string has no chunks of the form x,yx,. () A -day-old string has no chunks of the form,xy, with x >, or x. () A -day old string has no chunks of the form,, or. (4) A 4-day-old string has no chunks of the form, or. Proof. Parts () and () are Conway s One-Day and Two-Day Theorems. In part (), the first string must arise from a block, contradicting part (). By part (), the second can only be parsed as,,,, which arises from a chunk, again contrary to (). In part (4), the first string is parsed as,, by (), and so comes from a chunk xxx. One parsing contradicts (), the other (). By () and (), the other two must be parsed as,,,x, and,,,x, for some x or. These arise from xx and xx. By (), x =, and we get a contradiction to () or (). We say that a string looks 4 days old if it has a parsing satisfying the conditions of Lemma.. A chunk of a 4-day-old string need not look 4 days old. However, if S looks 4 days old, so does any block of S, as well as α(s).

3 Lemma.. The only maximal {, }-chunks of a 4-day-old string are,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, and. Proof. Let S be any {, }-chunk of a 4-day-old string. Suppose length(s) >, and let S be the maximal chunk of S that starts and ends with commas in the parsing of S. Then S is non-empty and describes a {, }-block P of a -day-old string that contains no runs of length. We claim that P contains at most one. For two consecutive s would be a run, and non-consecutive s would give a chunk or, both of which are impossible in -day-old strings. Hence P is one of the following eleven strings: P =, P =, P =, P 4 =, P 5 =, P 6 =, P 7 =, P 8 =, P 9 =, P 0 = or P =. Therefore S is equal to α(p i ) for some i, or is obtained from it by prefixing or appending a single or, or both. Only one of or may be prefixed (the one different from the first digit of P i ), giving a maximum of six possible S for each i. Also, P 4, P 5 and P 8,..., P cannot follow a or, since it would have to be a and would give a chunk or. Further, P and P 7,...,P cannot be followed by a run of length three, since the resulting chunk would not parse. Finally, P 6 cannot be both preceded by a and followed by a run of three. This means there are at most 9 such S, and one checks that all of them, plus the 7 strings of length at most, are listed above apart from, and. The three exceptions cannot be maximal {, }-chunks, because they cannot appear at the start of a 4-day old string by part () of Lemma., nor after a by part (4), nor after x > by parts () and (). Lemma.. Let S be an atom appearing in a 4-day-old string. A maximal nonempty {, }-block of S not at the start of S is one of the following strings:,,,,,,,,,,,,, or. Proof. Such a block B appears in a chunk B] or Bx (x, ) of S, and the non-empty strings listed in Lemma. but not here would cause a splitting.b] or.bx. Let S {,} be the set of strings in Lemma. and S {,} the set of those in Lemma.. Then any atom appearing in a 4-day-old string can be written in the form (.) M = CA n B n A n B A, where: (.) C S {,} ; (.) B i S {,} for i < n; (.4) A i =, or for < i n; and (.5) A =,,, or x for some x >. (Here n = and M = CA is allowed.) We shall show that any such atom has longevity at most 0. It will be more convenient to work with strings satisfying

4 (.6) A i =, or for i n in place of (.4) and (.5). We call a string mature if it looks 4 days old and has the form (.) subject to (.), (.) and (.6). Lemma.4. If every mature string has longevity at most 0, then so does every atom appearing in a 4-day-old string. Proof. Let M be such an atom, written as in (.) subject to (.) (.5). Then M is mature unless A = or A = x >. Suppose that A =, and that M appears in α 4 (S). If M is not the last atom in this string, the next atom is. If it is, α 4 (S) ends M, so in either case M is a block of a 4-day old string, and is therefore mature. Since it splits as M., it has the same longevity as M. Finally, suppose A = x with x >. Since any occurence of x in a -day old string is neither preceded nor followed by x, in a -day-old string it is preceded by, and in a 4-day-old string by. Hence M ends x and it follows from the proof of the Ending Theorem that the last atom of α k (M) is either Np or Pu for k 8. Let M be the string obtained from M by replacing x by. Now M does not end x, since the parsings,,,x and,,x, are both impossible. Hence M is mature. Now α k (M ) is obtained from α k (M) by replacing the final x by, and since α 0 (M ) is composed of common atoms, all atoms of α 0 (M) apart from the last are common. The substrings in our decomposition of a mature string do not, of course, evolve independently. To keep track of their interactions with their neighbors, we regard each one as a state of a machine, as explained in the next section.. Machines and pipelines For our purposes, a machine M consists of a set S of states, an input alphabet I, an output alphabet O (all non-empty), and a program π: S I O S. The output function ω: S I O and the transition function τ : S I S are defined by composing π with the appropriate projections. If s is a string on I and S S, we write S s t T to indicate that M, receiving input s in state S, outputs the string t and ends in state T. (When s is a single letter, so is t, and this means that π(s, s) = (t, T). The machine in question will be clear from the context.) In the machines we actually use, the output ω(s, x) is independent of x, so we sometimes write S s t to mean that M, receiving input s in state S, outputs t (one letter longer than s), and ends in an unknown state. We also use the notation S s for infinite t sequences s and t on I and O, which we call input and output streams. Finally, parts of s that do not affect the outcome will be suppressed. Given machines M = (S, I, O, π ) and M = (S, I, O, π ) with O = I, we can form the machine M = M M, called a pipeline, by connecting the output of M to the input of M. We have M = (S, I, O, π), where S = S S, I = I, and O = O. The program for M should be intuitively clear; formally we have π(s, S, x) = (π (S, ω (S, x)), τ (S, x)). Clearly, we have a category whose arrows are machines; the order in which we write a pipeline corresponds to the functions on the left convention, and is also the natural one for our application of the idea. We now define three specific machines. The states, inputs, and outputs will all be digit strings. In fact, the inputs and 4

5 outputs will be either the empty string or a single digit, so we can write the IO-streams without commas. By a simple string we mean a string on {,, } in which all runs have length at most. The machine A. Here S = {,, }, I = {, } and O = {,, }. The program is given by State Input Output State y, y x y x x x The machine B. Here S is the set of all non-empty simple strings that neither start nor end with a, I = {,, } and O = {, }. For S S, either α(s) S or α(s) = T with T S. The program is given by State Input Output State S, α(s) S x α(s)x S, α(s) = T x Tx The machine C. Here S is the set of all (possibly empty) simple strings that do not end with a, I = {,, } and O = {}. The program is given by State Input State S x α(s)x Consider the machine M = CA n B n A n B A for some n, where each A i or B i is a copy of A or B. If M = CA n B n A n B A is a mature string, then M has (C, A n, B n, A n,..., B, A ) as a state, and if it is started from this state with input stream, the product of the strings in the state after k inputs have been consumed will be α k (M). We say that M is run properly if the initial state corresponds to a mature string and the input is. We consider these machines in the next section. Now suppose M = M n M for some n, where each M i is a copy of A, B or C. Let M = (S, I, O, π). (I is one of {, } and {,, }, and O is one of {}, {, } and {,, }). We define a machine M = (Ŝ, I, O, π) with the same inputs and outputs as M as follows. Ŝ is the set of all non-empty simple strings whose first digit is not in O and whose last digit is in I. For S Ŝ, either α(s) Ŝ or α(s) = yt with y O and T Ŝ. The program is given by State Input Output State S, α(s) Ŝ x α(s)x S, α(s) = yt, y O x y Tx Every state (S n,...,s ) of M determines a state S n S of M. If M and M are started from corresponding states and given the same input, they will produce the same output, and at each stage their states will correspond. We call M the flattening of M. We now prove two lemmas about the machines A and B. In following the evolution of these machines from specific states, we will always know an initial chunk of the state, and usually the whole state, so we do not use Conway s [ and ] notation. Our convention is that, for instance, denotes a complete state, while denotes a state starting. We continue the convention that explicit exponents are 5

6 maximal, so that if a state is written x, it is implied that x and the state does not start xx. Lemma.. The output of A (for any initial state and input) does not contain, or. Proof. The possibilities for two consecutive steps are: ; ; ; ; ;. Lemma.. Let L = or, and let R be a string of the form x, or x or. Set ω() = and ω() =. Suppose that at some stage B has state LR. Then the subsequent output is ω(l), and at all later stages B has state LR, where R has one of the same three forms as R. Proof. Clearly the first output is ω(l) and the next state is LR where R = α(r), α(r) or α(r). If R = x, then α(r) =, and R has the same form. If R =, then α(r) = or, and R could have a different form only if α(r) were equal to, which is impossible. Finally, if R = x or then both α(r) and R have the form ( or ). The result follows. When a state S satisfying the hypotheses of this lemma arises, we indicate this by writing S as L.R. 4. Properly-run machines Throughout this section, M = CA n B n A n B A and we assume that M has an initial state corresponding to a mature string and input. For i n, we let s i be the input stream of A i, and t i its output stream. Then s i is a stream on {, }, t i is a stream on {,, }, and s =. Lemma 4.. () If appears in s i, it is followed by,, or. () If appears in s i, it is followed by. () If appears in s i, it is followed by. Proof. If i =, this is trivially true, so suppose i >. Then s i is also the output of the flattening N i of B i A i B A with input. The state transitions below are for the machine N i. () From the proof of the Starting Theorem, the state S of N i just before appears must be, or. (It cannot be just.) Suppose first S =. Because S looks 4 days old, it must have the form x or. Hence we have ( or ), so in this case is followed by. Now suppose S =. Then we have or ( or ), so in this case is followed by. Finally, suppose that S =. One possibility is, 6

7 and in this case is followed by, by the Starting Theorem. Otherwise we have and in this case is followed by. ( ) () From the proof of the Starting Theorem, the state of N i just before appears must be x, and the next step is x. That this is followed by the output 4 is part of the proof of part (). () From the proof of the Starting Theorem, the state of N i just before appears must be x, or x, and the result follows. Lemma 4.. () If appears in t i, it is followed by,, or. () If appears in t i, it is followed by. () If 4 appears in t i, it is followed by. (4) If appears in t i, it is followed by,, or. (5) If appears in t i, it is followed by. (6) If 4 appears in t i, it is followed by. Proof. In cases () () (resp. (4) (6)), the state of A i just before the given sequence appears in its output stream must be (resp. ), and the part of the input stream s i about to be read must start, or. In each case the possibilities for the continuation of s i are given by the previous lemma, and we just need to compute the corresponding outputs. The following computations show the output starting with the given sequence. () ; ;. (). (4) ; ;. (5). (,6) ;. 7

8 Recall that the initial state of A i (resp. B i ) is the string A i (resp. B i ) from the decomposition (.) of our mature string M. Lemma 4.. If B i is, or, then t i does not start with. If B i is or, t i does not start with. Proof. We need to show that in the first three cases, A i is not, and in the others it is not. This is so because M looks 4 days old. In Lemmas , we determine the possible output streams s i+ of B i for the allowable initial states B i S {,}. In each case, the proof consists of listing the evolutions of the state for all possible inputs until it becomes periodic. That we have considered all possible inputs t i to B i can be seen from Lemmas 4.,. and 4.. In most cases, that the output becomes periodic follows from Lemma., or from a case already dealt with. We shall therefore not give these justifications explicitly. Lemma 4.4. If B i is one of the eight strings in the left column below, the output stream s i+ of B i is as indicated. Proof. We have B i s i+,,, 5. ; or. ;. ; or or ; or. ;. Let s be a stream on {, } and t a finite or infinite sequence on {,, }. In Lemmas , the notation t s means that, for the given value of B i, if t i starts with t, then s i+ = s. Lemma 4.5. If B i =, we have 6 and. Proof. We have 5 and. Lemma 4.6. If B i =, we have or, and 4. 8

9 Proof. The first two cases are given by and the third by. and., or.. Lemma 4.7. If B i =, we have or ; or ; and 4. Proof. We have. ;. ; ; 4. ; Lemma 4.8. If B i =, we have 7, 4, a a+ for 0 a, and. Proof. First, Second, noting that, and. 6 a a Lemma 4.9. If B i =, we have, 7, 4,, a a+ for a,, and. 7, 4,,, and (). Proof. The first case comes from.. Otherwise, the evolution starts along some path in the following diagram. (On the vertical arrows, inputs are on the left and outputs on the right.).. 9 ()

10 Considering the possible continuations from the occurences of the state completes the proof. Lemma 4.0. If B i =, we have, and a a+ () for 0 a ; a a+ 7 and a a+ 4 for a = 0 or ; a b a+ b+ for 0 a and b ; 4 and a a+ for 0 a ; a a+ for 0 a ; and a a+ for a = 0 or ; Proof. The first case is given by.. Noting that, the second is clear, and all the rest come from an evolution starting a for 0 a a. Lemma 4.. For i < n and any B i S {,}, s i+ is one of the following streams. a () for 0 a ; a for 0 a ; a for a ; a for 0 a 7; a b for 0 a and b =,, 4, 5 or 7; a for a =,, 4 or 5; a for a =,, 4 or 7; a 4 for 0 a ; a for a = or. Proof. This is just a matter of checking that all the values of s i+ from Lemmas appear in the above list. In fact, not all the above streams can occur. Lemma 4.. For i < n and any B i S {,}, s i+ is not equal to a 5 for 0 a, 4, or. Proof. From Lemmas , these streams could only arise as s i+ if B i and an initial portion t of t i were as follows. s i+ B i t 5 a 5, a a 4 0

11 In all but one case, t i contains or, which implies that s i contains ; but this is impossible by the previous lemma. In the remaining case, t =, which implies that s i starts with, which is also impossible. We let S {,} be the set of streams appearing in Lemma 4. but not in Lemma 4.. Theorem. Every mature string has longevity at most 0. From Lemma.4, this implies Guy s version of the Cosmological Theorem. Corollary. Every digit string has longevity at most 4. Proof of Theorem. Let the state of M after k inputs (all ) have been consumed be ( C (k), A (k) n, B (k) n, A(k) n,..., B(k), ) A(k), so that α k (M) = C (k) A (k) n B (k) n A(k) n B(k) A(k). Let N i be the flattening of B i A i for i < n, and of CA n for i = n. Give N n N the initial state corresonding to M and input, and let its state after k inputs be ( (k) N n,..., N (k) ), so that also α k (M) = N n (k) N (k). N i has input s i for i n and output s i+ for i < n. Let i < n. Since s i+ S {,}, s i+ = u i v i, where length(u i ) = 0 and v i =,, or. If v i =, α 0 (M) splits as N n (0) N (0) i+.n(0) i N (0). Streams ending in () only arise in Lemmas 4.9 and 4.0, and it follows that if v i = or then B (0) i = or and t i contains only after its first 0 terms. This implies that A (0) i = and s i contains only after its first 0 terms, and it follows that α 0 (M) splits as N (0) n N (0) i+ ( or )..N(0) i N(0). Now suppose that v i =. Examining the proofs of Lemmas , we see that this can arise in three ways. First, we may have B (0) i = R with R as in Lemma.. In this case, α 0 (M) splits as N n (0) N (0) i+.ra(0) i N (0) i N(0). Second, we may have B (0) i = and t i containing only after its first 0 terms. This implies that A (0) i = and s i contains only after its first 0 terms, and it follows that α 0 (M) splits as N n (0) N (0) i+..n(0) i N(0). Finally, we may have B i = and t i =. This implies that A i = and s i =, and it follows that α 0 (M) splits as N n (0) N (0) i+.n(0) i N(0). In all cases, we may write N (0) i = L i R i, where L i =, or if v i =, or, respectively, and L i = or if v i =, and α 0 (M) splits as N (0) n N (0) i+ L i.r i N (0) i N(0).

12 Setting L 0 =, we have the splitting N n (0) L n.r n L n..r L 0 of α 0 (M), so it suffices to show that all the factors in this splitting have longevity at most 0. There are only finitely many possibilities for these factors and they may be obtained as follows. For j = or, let X j be the set of triples (S, s, L), where S = (S, S ) S j {,} {,, }, S S looks 4 days old, s S {,}, and, writing s = uv with length(u) = 0, L =, or if v =, or, respectively, and L = or if v =. Let Y X consist of those triples with L. For (S, s, L) X, let N(S, s) be the state of CA, starting from S, after the first L n = N(S, s)l for some (S, s, L). Similarly,for (S, s, L) X, let S be the state of BA, starting from S, after the first 0 inputs have been taken from s. If the first atom of S is,, or, let R(S, s) be the remainder of S, and otherwise let R(S, s) = S. Then, for i < n, R i L i = R(S, s)l for some (S, s, L). For (S, s, L) X, N(S, s) and R(S, s) are both defined, and N(S, s)l = T.R(S, s)l for some string T, so it is enough to show that N(S, s)l has longevity at most 0 for all (S, s, L) X. If s ends in, the longevity of N(S, s) is at most that of N(S, s), since for any common atom ending, replacing the final by gives a common atom. Hence we need only consider (S, s, L) Y. Unfortunately, there are still many cases. Of the 7 strings in S {,}, 9 cannot be followed by in a string that looks 4 days old, and 8 others not by, so there are 84 choices for S. Since there are 40 streams in S {,}, we have Y = = 60. The corresponding strings N(S, s)l are computed, and 0 inputs have been taken from s. Then N (0) n their longevities checked, by the program proof referred to earlier, completing the proof. References [C] J.H. Conway, The weird and wonderful chemistry of audioactive decay, in: Open Problems in Communication and Computation, T.M. Cover and B. Gopinath, eds., Springer, 987, pp [EZ] Shalosh B. Ekhad and Doron Zeilberger, Proof of Conway s Lost Cosmological Theorem, Electron. Res. Announc. Amer. Math. Soc. (997), 78 8; available at Department of Mathematics, Louisiana State University, Baton Rouge, LA address: lither@math.lsu.edu