Balázs Szentes

David K Levine

English

Research Papers in Economics

CAN A TURING PLAYER IDENTIFY ITSELF?
DAVID K. LEVINE AND AND BALÁZS SZENTES
ABSTRACT. We showthattheproblemofwhethertwo TuringMachines
are functionally equivalent is undecidable and explain why this is signif-
icant for the theory of repeated play and evolution.
Keywords: Economic Theory, Game Theory.
JEL Classication: A1; A2
1. INTRODUCTION
Consider the basic problem of cooperation in a Prisoner's dilemma type
situation. Cooperative outcomes must be enforced by punishing free-riders
- to do so it is necessary to identify free-riders so that they can be punished.
This observation is a feature of the literature on repeated play with bounded
rationality and in the theory of the evolution of cooperation. In both cases,
a good strategy is to give some sort of secret handshake to determine
if the opponent is of the same type - and so should be cooperated with -
or a different type that should be punished. In the setting of play between
nite state machines, the handshake is described in Rubinstein [1986]; in
the evolutionary setting by Robson [1990]. Typically, these secret hand-
shakes take the form of some sort of signalling during the course of play,
but any strategy must be generated by an algorithm, and in some circum-
stances it may be able to inspect the opponents algorithm prior to play -
this has obvious advantages over testing the opponent during the course of
play. An explicit theory along these lines can be found in Levine and Pe-
sendorfer [2002]. Of course an algorithm may be described in many ways
- from a strategic perspective, what is important is not details of the al-
gorithm, but rather its functionality. Is my opponent going to cheat, so I
should do likewise? Is he going to engage in a strategy that leaves him open
to exploitation?
Date: First Version: 20th January 2006, This Version: 16th March 2006.
UCLA and The University of Chicago. Corresponding Author - David K Levine: De-
partmentofEconomics,UCLA,LosAngeles,CA90095,USA.Phone/Fax: 310-825-3810.
Email: david@dklevine.com. Balász Szentes: Department of Economics, The University
of Chicago, 1126 59th Street, Chicago, IL 60637. Email: szentes@uchicago.edu.
We thank National Science Foundation Grants SES-03-14713 and SES-05-18762 for
nancial support.
1CAN A TURING PLAYER IDENTIFY ITSELF? 2
The purpose of this note is to examine the extent to which it is possible to
identifyan algorithmbased on its function, rather than on itsdescription.
1 A
natural way to describe algorithms is as a Turing Machine. Binmore [1990]
and Anderlini [1990] were among the rst to model players as Turing Ma-
chines. In their models, machines receive a description of their opponent
and the game played as inputs. The authors show that there does not exists
a machine which always gives the best response to the opponent's strategy.
2
In this note we do not require Turing Machines to predict the strategies of
their opponents. We merely want to know if the strategy of a Turing Ma-
chine can determine whether its opponent is equivalent to itself. In other
words: Are there Turing machines that can recognize whether their oppo-
nents are computationally equivalent to themselves?
2. THE QUESTION
We let t denote a Turing Machine, and t(x) denote the output of t on
input x, where t(x) = ¥ if t doesn't halt on x. By convention since there
are countably many Turing Machines, we identify their descriptions with
integers. We let hti denote the description of the Turing Machine t. In
other words t0(hti) is the output of the Turing Machine t0 when the input is
the description of the Turing Machine t. If we have two Turing Machines
t1 and t2 we say that they are equivalent if they compute the same function,
that is t1(x) = t2(x) for all x. We say that a Turing Machine is nite if it
halts on every input.
We consider the following questions, and prove that the answer is no to
all of them.
Question 1: Is it a decidable problem whether two Turing Machines are
equivalent?
Question 2: Is it a decidable problem whether two nite Turing Ma-
chines are equivalent?
Question 3: Is there some Turing Machine t for which it is a decidable
problem whether a Turing Machine is equivalent to t?
Question 4: Is there some nite Turing Machine t for which it is a de-
cidable problem whether another nite Turing Machine is equivalent to t?
Recall that for example the problem posed in Question 1 is said to be
decidable if there is a Turing Machine t such that t(hti;ht0i) = 1 if t and
t0 are equivalent and 0 if they are not.
1An alternative approach is to consider Bayesian priors as in Nachbar [1997].
2Canning [1992] shows that if the domain of possible players and games are appro-
priately restricted then Turing Machines will play Nash Equilibria. These restrictions are
fairly severe.CAN A TURING PLAYER IDENTIFY ITSELF? 3
3. THE ANSWER
Ourpointof departureisthebasicresultthatdeterminingwhetheror nota
Turing Machine halts is not a decidable problem - that is, there is no Turing
Machinewhich can take as inputthe descriptionof a TuringMachineand an
integer n and compute 1 if it halts on n and 0 if it does not. More precisely,
from the proof of the Halting Lemma whether t(hti) halts or not is not
decidable. For expositional purposes we provide a proof for this result.
HaltingLemma. Whether a TuringMachine haltson itself is undecidable.3
Proof. Suppose by contradiction, that there existsa Turing Machine t such
that t(hti) = 1 if t(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 then we could nd a Turing Machine that determines
whether t(hti) halts for any Turing Machine t, contradicting the fact that
this problem is undecidable.
Answer to Question 1: Fix a Turing Machine t. Dene t0 as follows.
If x 6= hti then t0(x) = t(x). If x = hti then t0(hti) = ¥. To construct t0
is easy. It is almost identical to t, except that rst it checks whether the
input is hti or not. If it is, it runs forever, otherwise it simulates t. Notice,
that in order to dene t0 one does not need to know whether t(hti) halts or
not. Observe, that t and t0 are equivalent if and only if t(hti) does not halt.
Hence, if there was a Turing Machine which determines whether t and t0
are equivalent, it could also determine whether t(hti) halts or not.
Answer to Question 2: Fix a Turing Machine t. Dene the Turing Ma-
chine 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 all n. No-
tice that b t and e t are nite, and in addition they are equivalent if and only
if t(hti) does not halt. Hence a machine that could determine if two nite
machines are equivalent could determine if t(hti) halts.
3An alternative line of proof is to use Rice's Theorem. Like the Halting Lemma, this
is a basic result in computability proven in any text on the subject. In fact Question 1 is
essentially a restatement of Rice's Theorem. However for the benet reader unfamiliar
with the terminology of index sets used in Rice's Theorem we give simple direct proofs
of all of the results.CAN A TURING PLAYER IDENTIFY ITSELF? 4
Answer to Question 3: Before we start to answer the question, note that
there is a Turing Machine ts such that if t halts on at least one input, then
ts(hti) is an inputonwhich t halts. The way to constructts is the following.
First ts simulates the rst step of t on 1. Then if ts simulated the nth step
of t on k, for n 6= 1, it will simulate the n 1th step of t on k+1, while for
n = 1 it will simulate the (k+1)th step on input 1. In other words
step input
1 1
2 1
1 2
3 1
2 2
1 3
4 1
and so forth. Once ts nds a step where t halts, it stops and its output is the
input of t on which it halted.
Let t be some arbitrary Turing Machine, and let t¥ denote a Turing Ma-
chine which does not halt on any input. Suppose by contradiction that there
exists a Turing Machineb t such that it is decidable whether or notb t is equiv-
alent toanother TuringMachine. This wouldmean that there existsa Turing
Machine t such that t(ht0i)= 1 if t0 is equivalenttob t and zero otherwise.
We now construct a Turing Machine e t as follows. The construction de-
pends on which of the following two cases holds.
Case 1: t(ht¥i) = 1. Then dene e t as follows. If x 6= 1;e t(x) =b t(x). If
x = 1, thene t simulates t 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 halts for some input. Hence ts(hb ti) is well-dened. This
enables us to dene e t(x) =b t(x) whenever x 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.
Soe t(ts(hb ti)) halts if and only if e t is functionally equivalent to b t.
Weconcludethatt(hti)haltsifandonlyif(t(t¥);t(e t))2f(1;0);(0;1)g.
So t can determine if t(hti) halts.
Answer to Question 4: Suppose bycontradictionthat there existsa nite
Turing Machine b t such that it is decidable whether or not b t is equivalent to
another nite Turing Machine. Fix a Turing Machine t and construct a
Turing Machine e t as follows. On input n; e t simulates the rst n steps of
t on hti. Then e t computes b t(n): If t halts before the nth step on hti then
e t(n) 6=b t(n); otherwisee t(n) =b t(n). Obviously,e t is nite becauseb t is alsoCAN A TURING PLAYER IDENTIFY ITSELF? 5
nite. Furthermore, e t and b t are equivalent if and only if t(hti) does not
halt.
REFERENCES
[1] Anderlini, L. [1990], Some Notes on Chruch's Thesis and the Theory of Games,
Theory and Decision 29, 19-52.
[2] Binmore, K. [1990], Essays on the Foundation of Game Theory. Oxford: Basil Black-
well.
[3] Canning, D. [1992], Rationality, Computability, and Nash Equilibrium, Economet-
rica 60, 877-888.
[4] Levine, D. K. and W. Pesendorfer [2002], Evolution of Cooperation Through Imita-
tion, UCLA.
[5] Nachbar, J. [1997], Prediction, Optimization, and Learning in Repeated Games,
Econometrica 65, 275-309.
[6] Robson, A. J. [1990], Efciency in evolutionary games: Darwin, Nash and the secret
handshake, Journal of Theoretical Biology 144, 379-96.
[7] RubinsteinA. [1986],FiniteAutomataPlay theRepeatedPrisoner'sDilemma,Jour-
nal of Economic Theory 39, 83-96.

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

Essays on the Foundation of Game Theory.

Can A Turing Player Identify Itself?

http://www.dklevine.com/papers/turing10.pdf

Can A Turing Player Identify Itself?

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