INHERENT LIMITATIONS OF COMPUTER PROGRAMS

CSci 4011 INHERENT LIMITATIONS OF COMPUTER PROGRAMS

UNDECIDABILITY A language is a set of strings. It is a mathematical way of expressing a problem: given an input, is it in the set L? If a language is decidable, there is a computer program (TM) that can always solve the problem correctly – it terminates and has the right answer. If a language is undecidable, then no matter how smart you are, and no matter how long you give it, you cannot program a computer to always solve the problem correctly.

UNDECIDABLE PROBLEMS DTM = { 〈M〉 | M is a TM that does not accept 〈M〉 } Theorem. DTM is undecidable. Proof. Suppose machine N decides DTM. Then N accepts 〈N〉 ⇔ 〈N〉  DTM ⇔ N does not accept 〈N〉 ATM = {〈M,w〉 | M is a TM that accepts on input w } Theorem. If ATM is decidable, so is DTM. Proof. If ¬ATM is decided by the program nAccept we can decide DTM by calling nAccept(M,〈M〉). HALTTM = { 〈M,w〉 | M is a TM that halts on input w } Theorem. If HALTTM is decidable, then so is ATM.

In most cases, we will show that a language is undecidable by showing that if it is decidable, then so is ATM We reduce deciding ATM to deciding the language in question

MAPPING REDUCTIONS ƒ : Σ*  Σ* is a computable function if there is a TM that on input w, halts with ƒ(w) on its tape A m B if there is a computable ƒ such that w  A  ƒ(w)  B ƒ is called a reduction from A to B

NET = { J | J is a java applet that opens a network connection } Theorem. ATM ≤m NET. Proof. On input 〈M,w〉, output the Java applet: import cs4011.tools.TuringMachine; import java.net.*; public class OutputApplet extends Applet { public void start() { TuringMachine TM = new TuringMachine(“M”, “w”); TM.run(); if (TM.accepted()) { // open a network connection URL url = new URL(“http://is.gd/”); Object o = url.getContent(); } } }

import cs4011.tools.TuringMachine; import java.net.*; public class OutputApplet extends Applet { public void start() { TuringMachine TM = new TuringMachine(“M”, “w”); TM.run(); if (TM.accepted()) { // open a network connection URL url = new URL(“http://is.gd/”); Object o = url.getContent(); } } } If M accepts w: OutputApplet opens url. Thus OutputApplet ∈ NET. If M loops on w: OutputApplet never gets to url. If M rejects w: OutputApplet skips if {… } block. In both cases, OutputApplet ∉ NET.

What’s wrong with this reduction? On input 〈M,w〉: Run M on w. if M accepts, output: public class OutputApplet extends Applet { public void start() { URL url = new URL(“http://is.gd/”); Object o = url.getContent(); } } else output: public class OutputApplet extends Applet { public void start() { return; } }

PROBLEMS ABOUT TMs

PMw(s): NETM = { 〈M〉 | M is a TM and L(M)  } Theorem:ATM≤mNETM. Proof: Let ƒ(〈M,w〉) = PMwdefined as: if s  w: reject if s = w: run M(w) Need to show that: • If 〈M,w〉∈ATM then PMw∈NETM.  If M accepts w then L(PMw) = {w} . 2. If 〈M,w〉  ATM then PMw NETM.  If M does not accept w then L(PMw) = .

REGTM = { 〈M〉 | M is a TM and L(M) is regular} Theorem: REGTM is undecidable We show that ATM≤m REGTM. The reduction, on input 〈M,w〉 outputs the TM PMw: Proof: PMw(s): If ∃n.s= 0n1n: accept else: return M(w) If M accepts w, L(PMw) = Σ*, so PMw REGTM. If M does not accept w, then L(PMw) = { 0n1n | n ≥ 0} so PMw REGTM.

EXAMPLE: REGTM ATM REGTM ƒ L(PMw) = * 〈M,w〉 ƒ L(PMw) = 0n1n 〈M,w〉 ƒ(〈M,w〉) = PMw.

EXAMPLE SINGLETONTM = { 〈M〉 | M is a TM that accepts exactly one string} Show that ATM ≤mSINGLETONTM. ESTM = { 〈M〉 | M is a TM that accepts on input ε} Show that ATM≤mESTM.

RICE’S THEOREM Let P be a language of Turing machine encodings. IF P satisfies the following properties: For all TMs M1 and M2, where L(M1) = L(M2), 〈M1〉  P if and only if 〈M2〉  P There exist TMs 〈MIN〉  P and 〈MOUT〉  P (i.e. P is a “nontrivialproperty of the r.e. languages.”) THEN P is undecidable EXTREMELY POWERFUL

if (M(w) and MIN(s)): accept else: reject TMw(s): Theorem: Let P be anontrivialproperty of the r.e. languages.Then P is undecidable We show that either ATM ≤m P or ATM ≤m ¬P Proof: Let S be whichever of P, ¬P does not contain the TM TØ that accepts no strings, and let MIN S. On input 〈M,w〉 our reduction outputs TMw:

if (M(w) and MIN(s)): accept else: reject TMw(s): 〈M,w〉  ATM iffƒ(〈M,w〉)  S: If M accepts w, then TMw accepts whenever MIN does. Then L(TMw) = L(MIN); since S is a “language property”, TMw  S If M does not accept w, then TMw doesn’t accept any strings. So L(TMw) =  and since TØ S, neither is TMw.

Let 2STM = { 〈M〉 | M is a TM and |L(M)| = 2 }. Theorem: 2STM is undecidable. Proof: Apply Rice’s Theorem: 1. If L(M1) = L(M2) then |L(M1)| = |L(M2)| so 〈M1〉  2STM ⇔ 〈M2〉 in 2STM. 2. Let MIN(s) = “If s = 0 or s = 1 then accept. else reject.” Let MOUT(s) = accept. Then 〈MIN〉  2STM and 〈MOUT〉  2STM.

EXAMPLES Aε = { 〈M〉 | M is a TM and L(M) = {ε} } CFTM = {〈M〉 | M is a TM and L(M) is context-free} OLTM = {〈M〉 | M is a TM and accepts only odd length strings} Use Rice’s Theorem to prove that Aε , CFTM, and OLTM are undecidable.

Let EH = { 〈M〉 | M is a TM and ∃w. M(w) halts } Rice’s Theorem does not apply to EH: Let M1(s) = “reject.”, M2(s) = “loop forever.” Then L(M1) = L(M2) but 〈M1〉∈ EH and 〈M2〉  EH. EH is undecidable. Proof: HALTTM ≤m EH: ƒ(〈M,w〉) = “PMw(s): if (s  w): loop forever. else: return M(s).” WITH GREAT POWER COMES GREAT RESPONSIBILITY!

GREAT RESPONSIBILITY Prove that Rice’s Theorem does not apply to: Let ULTM = { 〈M〉 | M is a TM with a useless state } WRITE-2-1s = { 〈M〉 | M is a TM that writes exactly two ones to its tape on some input} *state q is useless if for all s, M(s) is never in state q

EXAMPLE EQTM = {〈M1,M2〉 | M1,M2 are TMs and L(M1) = L(M2)} Prove EQTM is undecidable Hint: reduce ETM to EQTM.

INHERENT LIMITATIONS OF COMPUTER PROGRAMS