We show that the problem of whether two Turing Machines are functionally equivalent is undecidable and explain why this is significant for the theory of repeated play and evolution

Levine, D.

Szentes, B.

English

UCL Discovery

Can A Turing Player Identify Itself? David Levine Balázs SzentesUCLA University of ChicagoAbstractWe show that the problem of whether two Turing Machines are functionally equivalent isundecidable and explain why this is signif− icant for the theory of repeated play andevolution.We thank National Science Foundation Grants SES−03−14713 and SES−05−18762 for financial support.Citation: Levine, David and Balázs Szentes, (2006) "Can A Turing Player Identify Itself?." Economics Bulletin, Vol. 1, No. 1pp. 1−6Submitted: January 29, 2006.  Accepted: March 23, 2006.URL: http://www.economicsbulletin.com/2006/volume1/EB−06A00001A.pdf1. INTRODUCTIONThe wide empirical prevalence of cooperation in prisoner's dilemma type situations where free-riding is possible has long fascinated economists, game theorists and biologists. The possibilityof punishing free-riding in repeated situations is central to the theory of repeated games, but thefolk-theorem leaves open the possibility of coordination failure and Pareto inferior equilibria. Akey question addressed in the evolutionary literature is whether cooperation is merely possible,or whether in fact there are evolutionary forces that push in the direction of cooperation. Thereare a number of afrmative answers to the question: central to evolutionary forces that lead tocooperative outcomes is the ability to identify free-riders so they can be punished.1There are three basic mechanisms through which deviant free-rider strategies can be detected:indirectly by observing behavior, through some sort of cheap talk in the form of a secret hand-shake, or directly by observing the strategies themselves. The secret handshake is a signal thatis secret from free-riders and is discussed by Robson [1990]. In the setting of play between nitestate machines secret handshakes take place through observing play as described in Rubinstein[1986], for example. The possibility of directly observing strategies is examined in Levine and Pe-sendorfer [2002] - they show a strong result that the only strategies to survive a stochastic processof evolution and imitation are strategies that cooperate when they meet themselves, and punishopponents who use different strategies.Here we take up the issue of directly observing the strategies used by opponents. In otherwords, we envisage a setting where players exchange descriptions of their strategies prior to play.There are several reasons why this is interesting. First, whether information is generated throughcheap-talk or through observation of behavior, it is not possible to have more information about anopponents intentions than through direct observation of her strategy. So direct observation givesan upper bound on how well a player can be informed about an opponent's intentions - any limita-tions when direction observation is used must also be limitations when intentions must be inferredindirectly. Second, there are interesting circumstances where direct observation of strategies ispossible. For example, in a biological setting, strategies are encoded in DNA, and direct detec-tion of DNA or its attributes is possible. A good example of this is the immune system which isbased on detecting free-riders through their chemical attributes rather than their behavior. In theeconomic setting, institutions that make it easy to observe opponents strategic rules - for exam-ple, implementation by third parties, or the ability to credibly reveal the rules that are being usedthrough institutions such as bureaucracies - are possible and indeed frequent.Technically, a strategy is simply an algorithm that responds to certain inputs. Computer sci-entists conjecture that all algorithms that can be implemented and that meet certain minimal re-quirements are Turing machines,2 so this is a natural way in which to discuss strategic algorithms.3Economists such as Binmore [1990] and Anderlini [1990] have explicitly modeled player strate-gies as Turing Machines: they have focused on showing that there does not exist a machine which1In the literature on positive assortive matching, discussed, for example in Berstrom [2002], the matching mech-anism implicitly identies free-riders by matching players based on their play. This can be interpreted as indirectlydetection through observed behavior.2This is known as Church's Thesis.3An alternative would be to consider nite state machines, also know as automata, a subclass of Turing Machineswith nite memory. This would lead to a different set of issues: how much success could we expect a machine with nstates to have in identifying the strategy of a machine with some larger number m > n states?1always gives a best-response to the opponent's strategy.4 The issues can be illustrated through theexample of matching pennies. Suppose that each player designs a Turing machine that takes asinput the Turing machine designed by the other player. An obvious strategy would be to simulatethe opponent's Turing machine, using your own Turing machine as input. The problem with thisis that when both players use such Turing machines, we get an innite regress in which I simulateyou simulating me simulating you...and there is an innite loop. In technical terms, these Turingmachine do not halt when given each other as input, and so are useless in practice.In the setting of punishing free-riders, the situationis less difcult. It is easy to design a machinethat when given opposing Turing machine as input checks to see if that machine is the same ordifferent.5 The strategy is then to cooperate if the opposing machine is the same and punish if itis different. This is the basis for the Pesendorfer-Levine [2002] result, for example. However, thesame strategy may be described by many different Turing machines. For example, one machinemight engage in some irrelevant preliminary calculations that are later discarded, while anotherdoes not. This raises the question whether it is possible to identify an opponent not by the detailsof how their strategy is described, but according to the strategy or function that their algorithmimplements. That is the question addressed in this note. It is important for three reasons. First, theinteresting question about opponent play is how they will behave, not how their code is written.The question we would like to know from examining an opponent's strategy are of the type: Is myopponent going to cheat, so I should do likewise? Is he going to engage in a strategy that leaveshim open to exploitation? Second, there is an obvious social disadvantage if machines that differmerely in external appearance engage in costly punishment against one another. Third, if strategiescan only be determined through observing behavior, then it is evident that only the function of amachine can be identied.The questions we will examine, then, is the extent to which it is possible to identify an algorithmbased on its function, rather than on its description.6 We are not going to require Turing Machinesto predict the strategies of their opponents, but focus on the narrower question at issue in sus-taining cooperation: can Turing Machine recognize whether their opponents are computationallyequivalent to themselves?2. THE QUESTIONWe let t denote a Turing Machine, and t(x) denote the output of t on input the integer x,7where t(x) is an integer, or t(x) = ¥ if t does not halt on input x. By convention, since there arecountably many Turing Machines, we identify their descriptions with integers. We let hti denotethe description of the Turing Machine t. In other words t0(hti) is the output of the Turing Machinet0 when the input is the description of the Turing Machine t. If we have two Turing Machines t1and t2 we say that they are equivalent if they compute the same function, that is t1(x) = t2(x) forall x. We say that a Turing Machine is nite if it halts on every input. A problem with a yes/no4Canning [1992] shows that if the domain of possible players and games are appropriately restricted then TuringMachines will play Nash Equilibria. These restrictions are fairly severe.5This is obvious for Turing Machines. Since a program that checks an input to see if it is identical to itself requiresonly enough memory to store itself and the input, it is also true for nite state machines.6An alternative approach is to consider Bayesian priors as in Nachbar [1997].7Occasionally we refer to Turing machines that take as input pairs of integers; since there is a computable 1-1 mapfrom pairs of integers to integers, such a machine can be understood as taking as input the single integer generated bya xed map of this type.2answer is decidable if there is a Turing machine that provides the output zero if the answer is noand the output one if the answer is yes.We consider a series of questions about the decidability of the functional equivalence of twoTuring machines. In each case, we ask if there is a Turing machine that can determine whether twoTuring machines are computationally equivalent. The answer to each question is no: the problemis not decidable, or in other words, no Turing machine can determine equivalence in any of thesecases.Question 1: Is it a decidable problem whether two Turing Machines are equivalent?Question 2: Is it a decidable problem whether two nite Turing Machines are equivalent?Question 3: Is there some Turing Machine t for which it is a decidable problem whether aTuring Machine is equivalent to t?Question 4: Is there some nite Turing Machine t for which it is a decidable problem whetheranother nite Turing Machine is equivalent to t?We should emphasize that the problem of functional equivalence of Turing machines has notescaped the notice of computer scientists. For example, Question 1 is essentially a restatement ofRice's Theorem.8 Since the proofs are short, and the terminology of index sets used in Rice'sTheorem may not be familiar, we givesimple direct proofs that the answers to all of these questionsis no.3. THE ANSWEROur point of departure is the famous Halting Lemma that determining whether or not a TuringMachine halts is not a decidable problem - that is, there is no Turing Machine which can take asinput the description of a Turing Machine and an integer n and compute 1 if it halts on n and 0 ifit does not. In other words, whether t(hti) halts or not is not decidable. For expositional purposeswe provide a proof for this result from which the answers to Questions 1-4 will follow.Halting Lemma. Whether a Turing Machine halts on itself is undecidable.Proof. Suppose by contradiction, that there exists a Turing Machine t such that t(hti) = 1 ift(hti) halts and zero otherwise. Construct the Turing Machine t0 as follows.t0(hti) =0 if t(hti) = 0¥ if t(hti) = 1:Then t0(ht0i) = 0 , t(ht0i) = 0 by the construction of t0. However, by the denition of t,t(ht0i)= 0 is true if and only if t0 does not halt, that is, t0(ht0i)= ¥. So t0(ht0i)=0,t0(ht0i)=¥ an obvious contradiction. We now answer each question in the negative by showing that if it had a positive answer thenwe could nd a Turing Machine that determines whether t(hti) halts for any Turing Machine t,contradicting the fact that this problem is undecidable.In addition to the Halting Lemma, the other key ingredient in our proofs will be the idea of oneTuring machine simulating another. That is, if we are given as input the description of a Turingmachine hti and an input x we can as part of our design simply execute t(x) to see what the inputTuring machine t would do on the input x. Of course, t(x) might not halt, in which case themachine we are designing would not halt either.8See Corollary 6.1.5 in Cutland [1980].3Answer to Question 1: Is it a decidable problem whether two Turing Machines are equivalent?No. Fix a Turing Machine t. Dene t0 as follows. If x 6= hti then t0(x) = t(x). If x = hti thent0(hti) = ¥. To construct t0 is easy. It is almost identical to t, except that rst it checks whetherthe input is hti or not. If it is, it runs forever, otherwise it simulates t. Notice, that in order to denet0 one does not need to know whether t(hti) halts or not. Observe, that t and t0 are equivalent ifand only if t(hti) does not halt. Hence, if there was a Turing Machine which determines whethert and t0 are equivalent, it could also determine whether t(hti) halts or not.Answer to Question 2: Is it a decidable problem whether two nite Turing Machines are equiv-alent? No. Fix a Turing Machine t. Dene the Turing Machine b t as follows: b t(n) = 1 if t(hti)did not halt after n steps, and zero otherwise. Let e t be a Turing Machine such that e t(n) = 1 for alln. Notice that b t and e t are nite, and in addition they are equivalent if and only if t(hti) does nothalt. Hence a machine that could determine if two nite machines are equivalent could determineif t(hti) halts.Answer to Question 3: Is there some Turing Machine t for which it is a decidable problemwhether a Turing Machine is equivalent to t? No. Before we start to answer the question, notethat there is a Turing Machine ts such that if t halts on at least one input, then ts(hti) is an inputon which t halts. The way to construct ts is the following. First ts simulates the rst step of t on1. Then if ts simulated the nth step of t on k, for n 6= 1, it will simulate the n 1th step of t onk+1, while for n = 1 it will simulate the (k+1)th step on input 1. In other wordsstep input1 12 11 23 12 21 34 1and so forth. Once ts nds a step where t halts, it stops and its output is the input of t on which ithalted.Let t be some arbitrary Turing Machine, and let t¥ denote a Turing Machine which does nothalt on any input. Suppose by contradiction that there exists a Turing Machine b t such that it isdecidable whether or not b t is equivalent to another Turing Machine. This would mean that thereexists a Turing Machine t such that t(ht0i) = 1 if t0 is equivalent tob t and zero otherwise.We now construct a Turing Machine e t as follows. The construction depends on which of thefollowing two cases holds.Case 1: t(ht¥i) = 1. Then denee t as follows. If x 6= 1;e t(x) =b t(x). If x = 1, thene t simulatest on hti. Thuse t(1) halts if and only if t(hti) does.Case 2: t(ht¥i) = 0. Since this means that b t is not equivalent to t¥, it implies that b t haltsfor some input. Hence ts(hb ti) is well-dened. This enables us to dene e t(x) = b t(x) wheneverx 6= ts(hb ti). If x = ts(hb ti), then First: e t simulates t on hti. If it halts then Second: e t simulatesb t on x = ts(hb ti) and provides the same output as b t.Soe t(ts(hb ti)) halts if and only if e t is functionally equivalent to b t.We conclude that t(hti) halts if and only if (t(t¥);t(e t)) 2 f(1;0);(0;1)g. So t can deter-mine if t(hti) halts.4Answer to Question 4: Is there some nite TuringMachinet for whichit is a decidableproblemwhether another nite Turing Machine is equivalent to t? No. Suppose by contradiction that thereexists a nite Turing Machine b t such that it is decidable whether or not b t is equivalent to anothernite Turing Machine. Fix a Turing Machine t and construct a Turing Machine e t as follows. Oninput n; e t simulates the rst n steps of t on hti. Then e t computes b t(n): If t halts before the nthstep on hti thene t(n) 6=b t(n); otherwisee t(n) =b t(n). Obviously,e t is nite becauseb t is also nite.Furthermore,e t andb t are equivalent if and only if t(hti) does not halt.REFERENCES[1] Anderlini,L.[1990],SomeNotesonChurch'sThesisandtheTheoryofGames, TheoryandDecision29, 19-52.[2] Bergstrom, T. [2002], Evolution of Social Behavior: Individual and Group Selection, The Journal of EconomicPerspectives 16, 67-88.[3] Binmore, K. [1990], Essays on the Foundation of Game Theory. Oxford: Basil Blackwell.[4] Canning, D. [1992], Rationality, Computability, and Nash Equilibrium, Econometrica 60, 877-888.[5] Cutland, N. J. [1980],Computability: An introductionto Recursive FunctionTheory, CambridgeUniversityPress.[6] Levine, D. K. and W. Pesendorfer [2002], Evolution of Cooperation Through Imitation, UCLA.[7] Nachbar, J. [1997], Prediction, Optimization, and Learning in Repeated Games, Econometrica 65, 275-309.[8] Robson, A. J. [1990], Efciency in evolutionary games: Darwin, Nash and the secret handshake, Journal ofTheoretical Biology 144, 379-96.[9] Rubinstein A. [1986], Finite Automata Play the Repeated Prisoner's Dilemma, Journal of EconomicTheory 39,83-96.5

[1980],Computability: An introductionto Recursive FunctionTheory,

Efciency in evolutionary games: Darwin, Nash and the secret handshake,

Essays on the Foundation of Game Theory.

Evolution of Social Behavior: Individual and Group Selection,

Finite Automata Play the Repeated Prisoner's Dilemma,

Can a Turing Player identify itself?

Can a Turing Player identify itself?

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