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1
Module 1: Course Overview
• Course: CSE 460
• Instructor: Dr. Eric Torng
• Grader/TA: To be determined
2
What is this course?
• Philosophy of computing course– We take a step back to think about computing
in broader terms
• Science of computing course– We study fundamental ideas/results that shape
the field of computer science
• “Applied” computing course– We learn study a broad range of material with
relevance to computing today
3
Philosophy
• Phil. of life– What is the purpose of
life?
– What are we capable of accomplishing in life?
– Are there limits to what we can do in life?
– Why do we drive on parkways and park on driveways?
• Phil. of computing– What is the purpose of
programming?
– What can we achieve through programming?
– Are there limits to what we can do with programs?
– Why don’t debuggers actually debug programs?
4
Science
• Physics– Study of fundamental
physical laws and phenomenon like gravity and electricity
• Engineering– Governed by physical
laws
• Our material– Study of fundamental
computational laws and phenomenon like undecidability and universal computers
• Programming– Governed by
computational laws
5
Applied computing
• Applications are not immediately obvious– In some cases, seeing the applicability of this
material requires advanced abstraction skills• Every year, there are people who leave this course
unable to see the applicability of the material
– Others require more material in order to completely understand their application
• for example, to understand how regular expressions and context-free grammars are applied to the design of compilers, you need to take a compilers course
6
Some applications
– Important programming languages• regular expressions (perl)
• finite state automata (used in hardware design)
• context-free grammars
– Proofs of program correctness– Subroutines
• Using them to prove problems are unsolvable
– String searching/Pattern matching– Algorithm design concepts such as recursion
7
Fundamental Theme *
• What are the capabilities and limitations of computers and computer programs?– What can we do with computers/programs?– Are there things we cannot do with
computers/programs?
8
Module 2: Fundamental Concepts
• Problems
• Programs– Programming languages
9
Problems
We view solving problems as the main application for computer
programs
10Inputs Outputs
(4,2,3,1)(3,1,2,4)
(7,5,1)
(1,2,3)
(1,2,3,4)
(1,5,7)
(1,2,3)
Definition • A problem is a mapping or function between
a set of inputs and a set of outputs
• Example Problem: Sorting
11
How to specify a problem
• Input– Describe what an input instance looks like
• Output– Describe what task should be performed on the
input– In particular, describe what output should be
produced
12
Example Problem Specifications*
– Sorting problem• Input
– Integers n1, n2, ..., nk
• Output– n1, n2, ..., nk in nondecreasing order
– Find element problem• Input
– Integers n1, n2, …, nk
– Search key S
• Output– yes if S is in n1, n2, …, nk, no otherwise
13
Programs
Programs solve problems
14
Purpose
• Why do we write programs?
• One answer– To solve problems– What does it mean to solve a problem?
• Informal answer: For every legal input, a correct output is produced.
• Formal answer: To be given later
15
Programming Language
• Definition– A programming language defines what
constitutes a legal program– Example: a pseudocode program may not be a
legal C++ program which may not be a legal C program
– A programming language is typically referred to as a “computational model” in a course like this.
16
C++
• Our programming language will be C++ with minor modifications– Main procedure will use input parameters in a
fashion similar to other procedures• no argc/argv
– Output will be returned• type specified by main function type
17
Maximum Element Problem
• Input– integer n ≥ 1
– List of n integers
• Output– The largest of the n integers
18
C++ Program which solves the Maximum Element Problem*
int main(int A[], int n) {int i, max;
if (n < 1)
return (“Illegal Input”);
max = A[0];
for (i = 1; i < n; i++)if (A[i] > max)
max = A[i];
return (max);
}
19
Fundamental Theme
Exploring capabilities and limitations of C++ programs
20
Restating the Fundamental Theme *
• We will study the capabilities and limits of C++ programs
• Specifically, we will try and identify– What problems can be solved by C++ programs– What problems cannot be solved by C++
programs
21
Question
• Is C++ general enough?
• Or is it possible that there exists some problem such that– can be solved by some program P in some
other reasonable programming language – but cannot be solved by any C++ program?
22
Church’s Thesis (modified)
• We have no proof of an answer, but it is commonly accepted that the answer is no.
• Church’s Thesis (three identical statements)– C++ is a general model of computation– Any algorithm can be expressed as a C++
program– If some algorithm cannot be expressed by a C+
+ program, it cannot be expressed in any reasonable programming language
23
Summary *• Problems
– When we talk about what programs can or cannot “DO”, we mean what PROBLEMS can or cannot be solved
24
Module 3: Classifying Problems
• One of the main themes of this course will be to classify problems in various ways– By solvability
• Solvable, “half-solvable”, unsolvable
• We will focus our study on decision problems– function (one correct answer for every input)
– finite range (yes or no is the correct output)
25
Classification Process
• Take some set of problems and partition it into two or more subsets of problems where membership in a subset is based on some shared problem characteristic
Set of Problems
Subset 1 Subset 2 Subset 3
26
Classify by Solvability
• Criteria used is whether or not the problem is solvable– that is, does there exist a C++ program which
solves the problem?Set of All Problems
Solvable Problems Unsolvable Problems
27
Function Problems
• We will focus on problems where the mapping from input to output is a function
Set of All ProblemsNon-Function Problems Function Problems
28
General (Relation) Problem
• the mapping is a relation– that is, more than one output is possible for a
given input
Inputs Outputs
29
Criteria for Function Problems
• mapping is a function– unique output for each input
Inputs Outputs
30
Example Non-Function Problem
• Divisor Problem– Input: Positive integer n– Output: A positive integral divisor of n
Inputs Outputs
9 13
9
31
Example Function Problems
• Sorting
• Multiplication Problem– Input: 2 integers x and y– Output: xy
Inputs Outputs
2,5 10
32
Another Example *
• Maximum divisor problem– Input: Positive integer n– Output: size of maximum divisor of n smaller
than n
Inputs Outputs
9 13
9
33
Decision Problems
• We will focus on function problems where the correct answer is always yes or no
Set of Function ProblemsNon-Decision Problems Decision Problems
34
Criteria for Decision Problems• Output is yes or no
– range = {Yes, No}
• Note, problem must be a function problem– only one of Yes/No is correct
Inputs Outputs
Yes
No
35
Example
• Decision sorting– Input: list of integers– Yes/No question: Is the list in nondecreasing
order?
Inputs Outputs
Yes
No
(1,3,2,4)
(1,2,3,4)
36
Another Example
• Decision multiplication– Input: Three integers x, y, z– Yes/No question: Is xy = z?
Inputs Outputs
Yes
No
(3,5,14)
(3,5,15)
37
A Third Example *
• Decision Divisor Problem– Input: Two integers x and y– Yes/No question: Is y a divisor of x?
Inputs Outputs
Yes
No
(14,5)
(14,7)
38
Focus on Decision ProblemsSet of All Problems
Solvable Problems Unsolvable Problems
DecisionProblems
OtherProbs
• When studying solvability, we are going to focus specifically on decision problems– There is no loss of generality, but we will not explore
that here
39
Finite Domain Problems
• These problems have only a finite number of inputs
Set of All ProblemsFinite Domain Problems Infinite Domain Problems
40
Lack of GeneralitySet of All Problems
Solvable Problems Unsolvable Problems
• All finite domain problems can be solved using “table lookup” idea
FiniteDomain
InfiniteDomain
Empty
41
Table Lookup Program
int main(string x) {
switch x {
case “Bill”: return(3);
case “Judy”: return(25);
case “Tom”: return(30);
default: cerr << “Illegal input\n”;
}
42
Key Concepts
• Classification Theme
• Decision Problems– Important subset of problems– We can focus our attention on decision
problems without loss of generality– Same is not true for finite domain problems
• Table lookup
43
Module 4: Formal Definition of Solvability
• Analysis of decision problems– Two types of inputs:yes inputs and no inputs
– Language recognition problem
• Analysis of programs which solve decision problems– Four types of inputs: yes, no, crash, loop inputs
– Solving and not solving decision problems
• Classifying Decision Problems– Formal definition of solvable and unsolvable decision
problems
44
Analyzing Decision Problems
Can be defined by two sets
45
Decision Problems and Sets• Decision problems consist of 3 sets
– The set of legal input instances (or universe of input instances)
– The set of “yes” input instances – The set of “no” input instances
Yes Inputs No InputsSet of All Legal Inputs
46
Redundancy *
• Only two of these sets are needed; the third is redundant– Given
• The set of legal input instances (or universe of input instances)
– This is given by the description of a typical input instance
• The set of “yes” input instances – This is given by the yes/no question
– We can compute• The set of “no” input instances
47
Typical Input Universes
• *: The set of all finite length strings over finite alphabet – Examples
• {a}*: {/\, a, aa, aaa, aaaa, aaaaa, … }• {a,b}*: {/\, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, … }• {0,1}*: {/\, 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, … }
• The set of all integers• If the input universe is understood, a decision
problem can be specified by just giving the set of yes input instances
48
Language Recognition Problem
• Input Universe– * for some finite alphabet
• Yes input instances– Some set L subset of *
• No input instances– * - L
• When is understood, a language recognition problem can be specified by just stating what L is.
49
Language Recognition Problem *
• Traditional Formulation– Input
• A string x over some finite alphabet
– Task• Is x in some language L
subset of ?
• 3 set formulation• Input Universe
– * for a finite alphabet • Yes input instances
– Some set L subset of *
• No input instances– * - L
• When is understood, a language recognition problem can be specified by just stating what L is.
50
Equivalence of Decision Problems and Languages
• All decision problems can be formulated as language recognition problems– Simply develop an encoding scheme for
representing all inputs of the decision problem as strings over some fixed alphabet
– The corresponding language is just the set of strings encoding yes input instances
• In what follows, we will often use decision problems and languages interchangeably
51
Visualization *
Yes Inputs
Original DecisionProblem
No Inputs
EncodingScheme overalphabet
Language L
* - L
CorrespondingLanguage RecognitionProblem
52
Analyzing Programs which Solve Decision Problems
Four possible outcomes
53
Program Declaration
• Suppose a program P is designed to solve some decision problem . What does P’s declaration look like?
• What should P return on a yes input instance?
• What should P return on a no input instance?
54
Program Declaration II
• Suppose a program P is designed to solve a language recognition problem . What does P’s declaration look like?– bool main(string x) {
• We will assume that the string declaration is correctly defined for the input alphabet
– If = {a,b}, then string will define variables consisting of only a’s and b’s
– If = {a, b, …, z, A, …, Z}, then string will define variables consisting of any string of alphabet characters
55
Programs and Inputs
• Notation– P denotes a program
– x denotes an input for program P
• 4 possible outcomes of running P on x– P halts and says yes: P accepts input x
– P halts and says no: P rejects input x
– P halts without saying yes or no: P crashes on input x• We typically ignore this case as it can be combined with rejects
– P never halts: P infinite loops on input x
56
Programs and the Set of Legal Inputs
• Based on the 4 possible outcomes of running P on x, P partitions the set of legal inputs into 4 groups– Y(P): The set of inputs P accepts
• When the problem is a language recognition problem, Y(P) is often represented as L(P)
– N(P): The set of inputs P rejects
– C(P): The set of inputs P crashes on
– I(P): The set of inputs P infinite loops on• Because L(P) is often used in place of Y(P) as described above, we
use notation I(P) to represent this set
57
Illustration
All Inputs
Y(P) N(P) C(P) I(P)
58
Analyzing Programs and Decision Problems
Distinguish the two carefully
59
Program solving a decision problem
• Formal Definition:– A program P solves decision problem if and only if
• The set of legal inputs for P is identical to the set of input instances of
• Y(P) is the same as the set of yes input instances for • N(P) is the same as the set of no input instances for
– Otherwise, program P does not solve problem • Note C(P) and I(P) must be empty in order for P to solve
problem
Y(P) N(P) C(P) I(P)
60
Solvable Problem
• A decision problem is solvable if and only if there exists some C++ program P which solves – When the decision problem is a language
recognition problem for language L, we often say that L is solvable or L is decidable
• A decision problem is unsolvable if and only if all C++ programs P do not solve – Similar comment as above
61
Illustration of Solvability
Inputs of Program P
Y(P) N(P) C(P) I(P)C(P) I(P)
Inputs of Problem
Yes Inputs No Inputs
62
Program half-solving a problem
• Formal Definition:– A program P half-solves problem if and only if
• The set of legal inputs for P is identical to the set of input instances of
• Y(P) is the same as the set of yes input instances for • N(P) union C(P) union I(P) is the same as the set of no
input instances for
– Otherwise, program P does not half-solve problem • Note C(P) and I(P) need not be empty
Y(P) N(P) C(P) I(P)
63
Half-solvable Problem
• A decision problem is half-solvable if and only if there exists some C++ program P which half-solves – When the decision problem is a language
recognition problem for language L, we often say that L is half-solvable
• A decision problem is not half-solvable if and only if all C++ programs P do not half-solve
64
Illustration of Half-Solvability *
Inputs of Program P
Y(P) N(P) C(P) I(P)
Inputs of Problem
Yes Inputs No Inputs
65
Hierarchy of Decision Problems
Solvable
The set of half-solvable decision problems is a proper subset of the set of all decision problems
The set of solvable decision problems is a proper subset of the set of half-solvable decision problems.
Half-solvable
All decision problems
66
Why study half-solvable problems?
• A correct program must halt on all inputs
• Why then do we define and study half-solvable problems?
• One Answer: the set of half-solvable problems is the natural class of problems associated with general computational models like C++ – Every program half-solves some decision problem
– Some programs do not solve any decision problem
• In particular, programs which do not halt do not solve their corresponding decision problems
67
Key Concepts
• Four possible outcomes of running a program on an input
• The four subsets every program divides its set of legal inputs into
• Formal definition of– a program solving (half-solving) a decision problem– a problem being solvable (half-solvable)
• Be precise: with the above two statements!
68
Module 5
• Topics– Proof of the existence of unsolvable problems
• Proof Technique– There are more problems/languages than there are
programs/algorithms
– Countable and uncountable infinities
69
Overview
• We will show that there are more problems than programs– Actually more problems than programs in any
computational model (programming language)
• Implication– Some problems are not solvable
70
Preliminaries
Define set of problems
Observation about programs
71
Define set of problems
• We will restrict the set of problems to be the set of language recognition problems over the alphabet {a}.
• That is– Universe: {a}*– Yes Inputs: Some language L subset of {a}*– No Inputs: {a}* - L
72
Set of Problems *• The number of distinct problems is given by the
number of languages L subset of {a}*– 2{a}* is our shorthand for this set of subset languages
• Examples of languages L subset of {a}*– 0 elements: { }
– 1 element: {/\}, {a}, {aa}, {aaa}, {aaaa}, …
– 2 elements: {/\, a}, {/\, aa}, {a, aa}, …
– Infinite # of elements: {an | n is even}, {an | n is prime}, {an | n is a perfect square}
73
Infinity and {a}*
• All strings in {a}* have finite length
• The number of strings in {a}* is infinite
• The number of languages L in 2{a}* is infinite
• The number of strings in a language L in 2{a}* may be finite or infinite
74
Define set of programs
• The set of programs we will consider are the set of legal C++ programs as defined in earlier lectures
• Key Observation– Each C++ program can be thought of as a finite
length string over alphabet P
• P = {a, …, z, A, …, Z, 0, …, 9, white space, punctuation}
75
Example *
int main(int A[], int n){ {26 characters including newline}
int i, max; {13 characters including initial tab}
{1 character: newline}
if (n < 1) {12 characters}
return (“Illegal Input”); {28 characters including 2 tabs}
max = A[0]; {13 characters}
for (i = 1; i < n; i++) {25 characters}
if (A[i] > max) {18 characters}max = A[i]; {15 characters}
return (max); {15 characters}
} {2 characters including newline}
76
Number of programs
• The set of legal C++ programs is clearly infinite
• It is also no more than |P*|
– P = {a, …, z, A, …, Z, 0, …, 9, white space, punctuation}
77
Goal
• Show that the number of languages L in 2{a}* is greater than the number of strings in P*
– P = {a, …, z, A, …, Z, 0, …, 9, white space, punctuation}
• Problem– Both are infinite
78
How do we compare the relative sizes of infinite sets?
Bijection (yes)
Proper subset (no)
79
Bijections
• Two sets have EQUAL size if there exists a bijection between them– bijection is a 1-1 and onto function between
two sets
• Examples– Set {1, 2, 3} and Set {A, B, C}– Positive even numbers and positive integers
80
Bijection Example
• Positive Integers Positive Even Integers 1 2
2 4
3 6
... ...
i 2i
… ...
81
Proper subset
• Finite sets– S1 proper subset of S2 implies S2 is strictly
bigger than S1• Example
– women proper subset of people
– number of women less than number of people
• Infinite sets– Counterexample
• even numbers and integers
82
Two sizes of infinity
Countable
Uncountable
83
Countably infinite set S *
• Definition 1– S is equal in size (bijection) to N
• N is the set of natural numbers {1, 2, 3, …}
• Definition 2 (Key property)– There exists a way to list all the elements of set
S (enumerate S) such that the following is true• Every element appears at a finite position in the
infinite list
84
Uncountable infinity *
• Any set which is not countably infinite
• Examples– Set of real numbers– 2{a}*, the set of all languages L which are a
subset of {a}*
• Further gradations within this set, but we ignore them
85
Proof
86
(1) The set of all legal C++ programs is countably infinite
• Every C++ program is a finite string
• Thus, the set of all legal C++ programs is a language LC
• This language LC is a subset of P*
87
For any alphabet , * is countably infinite
• Enumeration ordering– All length 0 strings
• ||0 = 1 string:
– All length 1 strings• || strings
– All length 2 strings• ||2 strings
– …
• Thus, P* is countably infinite
88
Example with alphabet {a,b} *
• Length 0 strings– 0 and
• Length 1 strings– 1 and a, 2 and b
• Length 2 strings– 3 and aa, 4 and ab, 5 and ba, 6 and bb, ...
• Question– write a program that takes a number as input and
computes the corresponding string as output
89
(2) The set of languages in 2{a}* is uncountably infinite
• Diagonalization proof technique – “Algorithmic” proof– Typically presented as a proof by contradiction
90
Algorithm Overview *
• To prove this set is uncountably infinite, we construct an algorithm D that behaves as follows:– Input
• A countably infinite list of languages L[] subset of {a}*
– Output• A language D(L[]) which is a subset of {a}* that is not
on list L[]
91
Visualizing D
List L[]
L[0]
L[1]
L[2]
L[3]
...
LanguageD(L[]) not inlist L[]
Algorithm D
92
Why existence of D implies result
• If the number of languages in 2{a}* is countably infinite, there exists a list L[] s.t.– L[] is complete
• it contains every language in 2{a}*
– L[] is countably infinite
• The existence of algorithm D implies that no list of languages in 2{a}* is both complete and countably infinite– Specifically, the existence of D shows that any
countably infinite list of languages is not complete
93
Visualizing One Possible L[ ] *
L[0]
L[1]
L[2]
L[3]
L[4]...
a aa aaa aaaa ...
IN ININININ
OUTIN IN INOUT
OUT OUT OUT OUT OUT
IN INOUT OUT OUT
ININOUT OUTOUT
•#Rows is countably infinite
•Given
•#Cols is countably infinite
• {a}* is countably infinite
• Consider each string to be a feature– A set contains or does not contain each string
94
Constructing D(L[ ]) *• We construct D(L[]) by using a unique feature
(string) to differentiate D(L[]) from L[i]– Typically use ith string for language L[i]– Thus the name diagonalization
L[0]
L[1]
L[2]
L[3]
L[4]...
a aa aaa aaaa ...
IN ININININ
OUTIN IN INOUT
OUT OUT OUT OUT OUT
IN INOUT OUT OUT
ININOUT OUTOUT
OUT
IN
IN
IN
OUT
D(L[])
95
D(L[]) cannot be any language L[i]• D(L[]) cannot be language L[0] because D(L[])
differs from L[0] with respect to string a0 = /\.– If L[0] contains /\, then D(L[]) does not, and vice versa.
• D(L[]) cannot be language L[1] because D(L[]) differs from L[1] with respect to string a1 = a.– If L[1] contains a, then D(L[]) does not, and vice versa.
• D(L[]) cannot be language L[2] because D(L[]) differs from L[2] with respect to string a2 = aa.– If L[2] contains aa, then D(L[]) does not, and vice versa.
• …
96
Questions
• Do we need to use the diagonal?– Every other column and every row?– Every other row and every column?
• What properties are needed to construct D(L[])?
L[0]
L[1]
L[2]
L[3]
L[4]...
a aa aaa aaaa ...
IN ININININ
OUTIN IN INOUT
OUT OUT OUT OUT OUT
IN INOUT OUT OUT
ININOUT OUTOUT
97
Visualization
Solvable Problems
The set of solvable problems is a proper subset of the set of all problems.
All problems
98
Summary
• Equal size infinite sets: bijections– Countable and uncountable infinities
• More languages than algorithms– Number of algorithms countably infinite– Number of languages uncountably infinite– Diagonalization technique
• Construct D(L[]) using infinite set of features
• The set of solvable problems is a proper subset of the set of all problems
99
Module 6• Topics
– Program behavior problems• Input of problem is a program/algorithm
• Definition of type program
– Program correctness• Testing versus Proving
100
Number Theory Problems
• These are problems where we investigate properties of numbers– Primality
• Input: Positive integer n
• Yes/No Question: Is n a prime number?
– Divisor• Input: Integers m,n
• Yes/No question: Is m a divisor of n?
101
Graph Theory Problems
• These are problems where we investigate properties of graphs– Connected
• Input: Graph G
• Yes/No Question: Is G a connected graph?
– Subgraph• Input: Graphs G1 and G2
• Yes/No question: Is G1 a subgraph of G2?
102
Program Behavior Problems
• These are problems where we investigate properties of programs and how they behave
• Give an example problem with one input program P
• Give an example problem with two input programs P1 and P2
103
Program Representation• Program variables
– Abstractly, we define the type “program”• graph G, program P
– More concretely, we define type program to be a string over the program alphabet P = {a, …, z, A, …, Z, 0, …, 9, punctuation, white space}
• Note, many strings over P are not legal programs
• We consider them to be programs that always crash
• Possible declaration of main procedure– bool main(program P)
104
Program correctness
• How do we determine whether or not a program P we have written is correct?
• What are some weaknesses of this approach?
• What might be a better approach?
105
Testing versus Analyzing
ProgramP
Test Inputsx1
x2
x3
...
OutputsP(x1)P(x2)P(x3)
...
AnalyzerProgram P Analysis ofProgram P
106
2 Program Behavior Problems *• Correctness
– Input• Program P
– Yes/No Question• Does P correctly solve the primality problem?
• Functional Equivalence– Input
• Programs P1, P2
– Yes/No Question• Is program P1 functionally equivalent to program P2
107
Module 7
• Halting Problem– Fundamental program behavior problem– A specific unsolvable problem– Diagonalization technique revisited
• Proof more complex
108
Definition• Input
– Program P • Assume the input to program P is a single nonnegative integer
– This assumption is not necessary, but it simplifies the following unsolvability proof
– To see the full generality of the halting problem, remove this assumption
– Nonnegative integer x, an input for program P
• Yes/No Question– Does P halt when run on x?
• Notation– Use H as shorthand for halting problem when space is a
constraint
109
Example Input *• Program with one input of type unsigned
bool main(unsigned Q) {
int i=2;
if ((Q = = 0) || (Q= = 1)) return false;
while (i<Q) {
if (Q%i = = 0) return (false);
i++;
}
return (true);
}
• Input x4
110
Three key definitions
111
Definition of list L *
• P* is countably infinite where P = {characters, digits, white
space, punctuation}• Type program will be type string with P as the alphabet• Define L to be the strings in P
* listed in enumeration order– length 0 strings first– length 1 strings next– …
• Every program is a string in P – For simplicity, consider only programs that have
• one input• the type of this input is an unsigned
• Consider strings in P* that are not legal programs to be
programs that always crash (and thus halt on all inputs)
112
Definition of PH *
• If H is solvable, some program must solve H
• Let PH be a procedure which solves H
– We declare it as a procedure because we will use PH as a subroutine
• Declaration of PH
– bool PH(program P, unsigned x)• In general, the type of x should be the type of the input to P
• Comments– We do not know how PH works
– However, if H is solvable, we can build programs which call PH as a subroutine
113
bool main(unsigned y) /* main for program D */{ program P = generate(y);
if (PH(P,y)) while (1>0); else return (yes); }
/* generate the yth string in P* in enumeration order */
program generate(unsigned y) /* code for program of slide 21 from module 5 did this for
{a,b}* */
bool PH(program P, unsigned x)/* how PH solves H is unknown */
Definition of program D
114
Generating Py from y *• We won’t go into this in detail here
– This was the basis of the question at the bottom of slide 21 of lecture 5 (alphabet for that problem was {a,b} instead of P).
– This is the main place where our assumption about the input type for program P is important
• for other input types, how to do this would vary
• Specification– 0 maps to program – 1 maps to program a– 2 maps to program b– 3 maps to program c– …– 26 maps to program z– 27 maps to program A– …
115
Proof that H is not solvable
116
Argument Overview *
H is solvable
D is NOT on list L
PH exists
Definition ofSolvability
D exists
D’s code
D is on list L
L is list ofall programs
D does NOT exist
PH does NOT exist
H is NOT solvable
p → q is logically equivalent to (not q) → (not p)
117
Proving D is not on list L
• Use list L to specify a program behavior B that is distinct from all real program behaviors (for programs with one input of type unsigned)– Diagonalization argument similar to the one for proving the
number of languages over {a} is uncountably infinite
– No program P exists that exhibits program behavior B
• Argue that D exhibits program behavior B– Thus D cannot exist and thus is not on list L
118
Non-existent program behavior B
119
Visualizing List L *
P0
P1
P2
P3
P4...
1 2 3 4 ...
H HHHH
NHH H HNH
NH NH NH NH NH
H H H H H
HHNH NHNH
•#Rows is countably infinite
• p* is countably infinite
•#Cols is countably infinite
• Set of nnnegative integers is countably infinite
• Consider each number to be a feature– A program halts or doesn’t halt on each integer– We have a fixed L this time
120
Diagonalization to specify B *
P0
P1
P2
P3
P4...
1 2 3 4 ...
H HHHH
NHH H HNH
NH NH NH NH NH
H H H H H
HHNH NHNH
•We specify a non-existent program behavior B by using a unique feature (number) to differentiate B from Pi
NH
H
H
H
NH
B
121
Arguing D exhibits program behavior B
122
bool main(unsigned y) /* main for program D */{ program P = generate(y);
if (PH(P,y)) while (1>0); else return (yes); }
/* generate the yth string in P* in enumeration order */
program generate(unsigned y) /* code for extra credit program of slide 21 from lecture 5 did
this for {a,b}* */
bool PH(program P, unsigned x)/* how PH solves H is unknown */
Code for D
123
Visualization of D in action on input y
• Program D with input y– (type for y: unsigned)
Given input y, generate the program (string) Py
Run PH on Py and y
• Guaranteed to halt since PH solves H
IF (PH(Py,y)) while (1>0); else return (yes);
P0
P1
P2
Py...
1 2 ...
HH
H H
NH NH
H
NH
NH
D ... y
...
HNH
124
D cannot be any program on list L• D cannot be program P0 because D behaves
differently on 0 than P0 does on 0.– If P0 halts on 0, then D does not, and vice versa.
• D cannot be program P1 because D behaves differently on 1 than P1 does on 1.– If P1 halts on 1, then D does not, and vice versa.
• D cannot be program P2 because D behaves differently on 2 than P2 does on 2.– If P2 halts on 2, then D does not, and vice versa.
• …
125
Alternate Proof
126
Alternate Proof Overview
• For every program Py, there is a number y that we associate with it
• The number we use to distinguish program Py from D is this number y
• Using this idea, we can arrive at a contradiction without explicitly using the table L– The diagonalization is hidden
127
H is not solvable, proof II
• Assume H is solvable– Let PH be the program which solves H
– Use PH to construct program D which cannot exist
• Contradiction– This means program PH cannot exist.
– This implies H is not solvable
• D is the same as before
128
Arguing D cannot exist
• If D is a program, it must have an associated number y
• What does D do on this number y?
• 2 cases– D halts on y
• This means PH(D,y) = NO
– Definition of D
• This means D does not halt on y
– PH solves H
• Contradiction
• This case is not possible
129
Continued
– D does not halt on this number y
• This means PH(D,y) = YES
– Definition of D
• This means D halts on y– PH solves H
• Contradiction
• This case is not possible
– Both cases are not possible, but one must be for D to exist
– Thus D cannot exist
130
Placing the Halting Problem
Solvable
ll Problems
H
131
Implications *
• The Halting Problem is one of the simplest problems we can formulate about program behavior
• We can use the fact that it is unsolvable to show that other problems about program behavior are also unsolvable
• This has important implications restricting what we can do in the field of software engineering– In particular, “perfect” debuggers/testers do not exist
– We are forced to “test” programs for correctness even though this approach has many flaws
132
Summary
• Halting Problem definition– Basic problem about program behavior
• Halting Problem is unsolvable– We have identified a specific unsolvable
problem– Diagonalization technique
• Proof more complicated because we actually need to construct D, not just give a specification B
133
Module 8
• Halting Problem revisited– Universal Turing machine half-solves halting
problem– A universal Turing machine is an operating
system/general purpose computer
134
Half-solving the Halting Problem
• State the Halting Problem
• Give an input instance of the Halting Problem
• We saw last time that the Halting Problem is not solvable. How might we half-solve the Halting Problem?
135
Example InputProgram P
bool main(unsigned Q) {
int i=2;
if ((Q = = 0) || (Q= = 1)) return false;
while (i<Q) {
if (Q%i = = 0) return (false);
i++;
}
return (true);
}
Nonnegative integer
4
136
OrganizationUniversal Turing machine’s Memory
Program P
Program P’s Memory
Program P
bool main(unsigned Q) {
int i=2;
if ((Q = = 0) || (Q= = 1)) return false;
while (i<Q) {
if (Q%i = = 0) return (false);
i++;
}
return (true);
}
Program Counter
int i,Q;4
Line 1
137
Description of universal Turing machine *
• Basic Loop– Find current line of program P– Execute current line of program P
• Update program P’s memory
– Update program counter– Return to Top of Loop
Program P
Program P’s Memory
Program Counter
138
Past, Present, Future• Turing came up with the concept of a universal Turing machine in
the 1930’s– This is well before the invention of the general purpose computer
– People were still thinking of computing devices as special-purpose devices (calculators, etc.)
– Turing helped move people beyond this narrow perspective
• Turing/Von Neumann perspective– Computers are general purpose/universal algorithms
• Focused on computation• Stand-alone
• Today, we are moving beyond this view– Internet, world-wide web
– However, results in Turing perspective still relevant
139
Debuggers
• How are debugger’s like gdb or ddd related to universal Turing machines?
• How do debuggers simplify the debugging process?
140
Placing the Halting Problem
Solvable
Half-solvable
ll Problems
H
141
Summary
• Universal Turing machines– 1930’s, Turing– Introduces general purpose computing concept– Not a super intelligent program, merely a
precise follower of instructions
• Halting Problem half-solvable but not solvable
142
Module 9
• Closure Properties– Definition– Language class definition
• set of languages
– Closure properties and first-order logic statements
• For all, there exists
143
Closure Properties
• A set is closed under an operation if applying the operation to elements of the set produces another element of the set
• Example/Counterexample– set of integers and addition– set of integers and division
144
Integers and Addition
Integers
2
5
7
145
Integers and Division
Integers
2
5
.4
146
Language Classes
– We will be interested in closure properties of language classes
• A language class is a set of languages
• Thus, the elements of a language class (set of languages) are languages which are sets themselves
– Crucial Observation• When we say that a language class is closed under
some set operation, we apply the set operation to the languages (elements of the language classes) rather than the language classes themselves
147
Example Language Classes *
• In all these examples, we do not explicitly state what the underlying alphabet is
• Solvable languages (REC)– Languages whose language recognition problem can be solved
• Half-solvable languages (RE)– Languages whose language recognition problems can be half-
solved
• Finite languages– Languages with a finite number of strings
• CARD-3– Languages with at most 3 strings
148
Finite Sets and Set Union *
Finite Languages
{a,b,aa}
{a,b,bb}
{a,b,aa,bb}
All Languages
149
All Languages
CARD-3 and Set Union
CARD-3
{a,b,aa}
{a,b,bb}
{a,b,aa,bb}
150
All Languages
Finite Languages and Set Complement
Finite Languages
{a,b,ab}
{/\,aa,ba,bb,aaa,...}
151
Infinite Number of Facts
• A closure property often represents an infinite number of facts
• Example: The set of finite languages is closed under the set union operation– {} union {} is a finite language– {} union {} is a finite language– {} union {0} is a finite language– ...– {} union {} is a finite language– ...
152
First-order logic and closure properties *
• A way to formally write (not prove) a closure property– L1, ...,Lk in LC, op (L1, ... Lk) in LC
– Only one expression is needed because of the for all quantifier
• Number of languages k is determined by arity of the operation op
153
Example F-O logic statements *
• L1,L2 in FINITE, L1 union L2 in FINITE
• L1,L2 in CARD-3, L1 union L2 in CARD-3
• L in FINITE, Lc in FINITE
• L in CARD-3, Lc in CARD-3
154
Stating a closure property is false
• What is true if a set is not closed under some k-ary operator?– There exist k elements of that set which, when
combined together under the given operator, produce an element not in the set
– L1, ...,Lk in LC, op (L1, …, Lk) not in LC
• Example– Finite sets and set complement
155
Complementing a F-O logic statement
• Complement “L1,L2 in CARD-3, L1 union L2 in CARD-3”
– not (L1,L2 in CARD-3, L1 union L2 in CARD-3)
– L1,L2 in CARD-3, not (L1 union L2 in CARD-3)
– L1,L2 in CARD-3, L1 union L2 not in CARD-3
156
Proving/Disproving
• Which is easier and why?– Proving a closure property is true– Proving a closure property is false
157
Module 10
• Recursive and r.e. language classes– representing solvable and half-solvable problems
• Proofs of closure properties – for the set of recursive (solvable) languages– for the set of r.e. (half-solvable) languages
• Generic Element proof technique
158
RE and REC language classes
• REC– A solvable language is commonly referred to as
a recursive language for historical reasons– REC is defined to be the set of solvable or
recursive languages
• RE– A half-solvable language is commonly referred
to as a recursively enumerable or r.e. language– RE is defined to be the set of r.e. or half-
solvable languages
159
Closure Properties of REC *
• We now prove REC is closed under two set operations– Set Complement– Set Intersection
• In these proofs, we try to highlight intuition and common sense
160
REC and Set Complement
RECEVEN
All Languages
ODD• Even: set of even length strings
• Is Even solvable (recursive)?
• Give a program P that solves it.
• Complement of Even?– Odd: set of odd length strings
• Is Odd recursive (solvable)?
• Does this prove REC is closed under set complement?
• How is the program P’ that solves Odd related to the program P that solves Even?
161
P’ Illustration
PInput x Yes/No
P’
No/Yes
162
Code for P’
bool main(string y)
{
if (P (y)) return no; else return yes;
}
bool P (string y)
/* details deleted; key fact is P is guaranteed to halt on all inputs */
163
Set Complement Lemma
• If L is a solvable language, then L complement is a solvable language
• Proof– Let L be an arbitrary solvable language
• First line comes from For all L in REC
– Let P be the C++ program which solves L• P exists by definition of REC
164
– Modify P to form P’ as follows• Identical except at very end
• Complement answer – Yes → No– No → Yes
– Program P’ solves L complement• Halts on all inputs
• Answers correctly
– Thus L complement is solvable• Definition of solvable
proof continued
165
REC Closed Under Set Union
All Languages
L1
L2
L1 U L2
REC
• If L1 and L2 are solvable languages, then L1 U L2 is a solvable language
• Proof– Let L1 and L2 be arbitrary
solvable languages
– Let P1 and P2 be programs which solve L1 and L2, respectively
166
REC Closed Under Set Union
All Languages
L1
L2
L1 U L2
REC
– Construct program P3 from P1 and P2 as follows
– P3 solves L1 U L2 • Halts on all inputs• Answers correctly
– L1 U L2 is solvable
167
Yes/No
Yes/No
P3 Illustration
P1
P2
ORYes/No
P3
168
Code for P3
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;
}bool P1(string y) /* details deleted; key fact is P1 always halts. */
bool P2(string y) /* details deleted; key fact is P2 always halts. */
169
Other Closure Properties
• Unary Operations– Language Reversal– Kleene Star
• Binary Operations– Set Intersection– Set Difference– Symmetric Difference– Concatenation
170
Closure Properties of RE *
• We now try to prove RE is closed under the same two set operations– Set Union – Set Complement
• In these proofs– We define a more formal proof methodology– We gain more intuition about the differences
between solvable and half-solvable problems
171
RE Closed Under Set Union
• Expressing this closure property as an infinite set of facts– Let Li denote the ith r.e. language
• L1 intersect L1 is in RE
• L1 intersect L2 is in RE
• ...
• L2 intersect L1 is in RE
• ...
172
Generic Element or Template Proofs
• Since there are an infinite number of facts to prove, we cannot prove them all individually
• Instead, we create a single proof that proves each fact simultaneously
• I like to call these proofs generic element or template proofs
173
Basic Proof Ideas• Name your generic objects
– For example, L or L1 or L2
• Only use facts which apply to any relevant objects– For example, there must exist a program P1 that half-
solves L1
• Work from both ends of the proof– The first and last lines are often obvious, and we can
often work our way in
174
RE Closed Under Set UnionL1
L2
L1 U L2
• Let L1 and L2 be arbitrary r.e. languages
• L1 U L2 is an r.e. language
• There exist P1 and P2 s.t. Y(P1)=L1 and Y(P2)=L2
• There exists a program P that half-solves L1 U L2
• Construct program P3 from P1 and P2
• Prove Program P3 half-solves L1 U L2
175
• What code did we use for P3 when we worked with solvable languages?
Constructing Program P3
L1
L2
L1 U L2
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;}bool P1(string y) /* details deleted; key fact is P1 always halts. */
bool P2(string y) /* details deleted; key fact is P2 always halts. */
• Will this code work for half-solvable languages?
176
Proving P3 Is Correct
• 2 steps to showing P3 half-solves L1 U L2
– For all x in L1 U L2, must show P3
– For all x not in L1 U L2, must show P3
177
P3 works correctly?
– Let x be an arbitrary string in L1 U L2
• Note, this subproof is a generic element proof
– What are the two possibilities for x?• x is in L1
• x is in L2
– What does P3 do if x is in L1?• Does it matter if x is in or not in L2?
– What does P3 do if x is in L2?• Does it matter if x is in or not in L1?
– Does P3 work correctly on such x?• If not, what strings cause P3 a problem?
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;
}
178
P3 works correctly?
– Let x be an arbitrary string NOT in L1 U L2
• Note, this subproof is a generic element proof
– x is not in L1 AND x is not in L2
– What does P3 do on x in this case?
• What does P1 do on x?
• What does P2 do on x?
– Does P3 work correctly on such x?
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;
}
179
Code for Correct P3
bool main(string y){if ((P1(y) || P2(y)) return yes; /* P1 and P2 run in parallel (alternating execution) */
else return no;}bool P1(string y)
/* key fact is P1 only guaranteed to halt on yes input instances */
bool P2(string y) /* key fact is P2 only guaranteed to halt on yes input instances */
180
P3 works correctly?
– Let x be an arbitrary string in L1 U L2
• Note, this subproof is a generic element proof
– What are the two possibilities for x?• x is in L1
• x is in L2
– What does P3 do if x is in L1?• Does it matter if x is in or not in L2?
– What does P3 do if x is in L2?• Does it matter if x is in or not in L1?
– Does P3 work correctly on such x?• If not, what strings cause P3 a problem?
bool main(string y){ if ((P1(y) || P2(y)) return yes; else return no;}/* P1 and P2 run in parallel */
181
• First-order logic formulation for statement– RE is closed under set complement
• What this really means– Let Li denote the ith r.e. language
• L1 complement is in RE
• L2 complement is in RE
• ...
RE and Set complement L
LcRE
182
RE and Set complement
• Let L be an arbitrary r.e. language
• L complement is an r.e. language
• There exists P s.t. Y(P)=L
• There exists a program P’ which half-solves L complement
• Construct program P’ from P• Prove Program P’ half-solves L complement
LLc
RE
183
Constructing P’• What did we do in recursive case?
– Run P and then just complement answer at end• Accept → Reject• Reject → Accept
• Does this work in this case?– No. Why not?
• Does this prove that RE is not closed under set complement?
184
Other closure properties
• Unary Operations– Language reversal– Kleene Closure
• Binary operations– set intersection– concatenation
• Not closed– Set difference (on practice hw)
185
Pseudo Closure Property
• Lemma: If L and Lc are half-solvable, then L is solvable.
• Question: What about Lc?
186
High Level Proof
– Let L be an arbitrary language where L and Lc are both half-solvable
– Let P1 and P2 be the programs which half-solve L and Lc, respectively
– Construct program P3 from P1 and P2
– Argue P3 solves L
– L is solvable
187
Constructing P3
• Problem– Both P1 and P2 may loop on some input strings,
and we need P3 to halt on all input strings
• Key Observation– On all input strings, one of P1 and P2 is
guaranteed to halt. Why?
188
Illustration
*
L
P1 halts
Lc
P2 halts
189
Construction and Proof
• P3’s Operation
– Run P1 and P2 in parallel on the input string x until one accepts x
• Guaranteed to occur given previous argument
• Also, only one program will accept any string x
– IF P1 is the accepting machine THEN yes ELSE no
190
P3 Illustration
P1
P2
Yes
Yes
P3
Input
Yes
No
191
Code for P3 *
bool main(string y)
{
parallel-execute(P1(y), P2(y)) until one returns yes;
if (P1(y)) return yes;
if (P2(Y)) return no;
}bool P1(string y) /* guaranteed to halt on strings in L*/
bool P2(string y) /* guaranteed to halt on strings in Lc */
192
RE and REC
REC
RE
All Languages
L Lc
193
RE and REC
REC
RE
All Languages
LLc
Lc
Lc
Are there any languages L in RE - REC?
194
RE and REC
REC
RE
All Languages
H
So where does Hc belong?
Hc
195
Closure Property Questions
• Which of the following imply L is solvable given REC is closed under set union?– L1 U L2 = L
– L1 U L = L2
– L U L2 = L1
• In all cases, L1 and L2 are known to be solvable
196
Closure Property Questions
• Which of the following imply L is NOT solvable given REC is closed under set union?– L1 U L2 = L
– L1 U L = L2
– L U L2 = L1
• In all cases, L1 is solvable and L2 is NOT solvable
197
Summary• Definition of REC and RE
• Proofs of some closure properties for both language classes– RE more complex
• Pseudo-closure Property
• RE is not closed under set complement
• Proving a language is or is not in a language class using closure properties
198
Module 11
• Proving more specific problems are not solvable
• Input transformation technique– Use subroutine theme to show that if one
problem is unsolvable, so is a second problem– Need to clearly differentiate between
• use of program as a subroutine and
• a program being an input to another program
199
Basic Idea/Technique
200
Proving a problem L is unsolvable
• Assume PL is a procedure that solves problem L– We have no idea how PL solves L
• Construct a program PH that solves H using PL as a subroutine– We use PL as a black box– (We could use any unsolvable problem in place of H)
• Argue PH solves H– But we know H is unsolvable
• Conclude that L is unsolvable– Otherwise PL would exist and then H would be solvable– L will be a problem about program behavior
201
Focusing on H
• In this module, we will typically use H, the Halting Problem, as our known unsolvable problem
• The technique generalizes to using any unsolvable problem L’ in place of H.– You would need to change the proofs to work with L’
instead of H, but in general it can be done
• The technique also can be applied to solvable problems to derive alternative consequences
• We focus on H to simplify the explanation
202
Constructing PH using PL
Answer-preserving input transformations and Program PT
203
PH has two subroutines
• There are many ways to construct PH using program PL that solves L
• We focus on one method in which PH consists of two subroutines– Procedure PL that solves L– Procedure PT which computes a function f that I
call an answer-preserving (or answer-reversing) input transformation
• More about this in a moment
204
Two Representations of PH *
PH
P,y Yes/NoPLY/NPT
PT(P,y)
bool PH (Program P, unsigned y) { return PL (PT (P,y));}
205
Answer-preserving input transformation PT
• Input– An input to H
• Output– An input to L such that
• yes inputs of H map to yes inputs of L• no inputs of H map to no inputs of L
• Note, PT must not loop when given any legal input to H
206
Why this works *
PH
PLPT
yes input to H yes input to L yes
no input to H no input to L no
We have assumed that PL solves L
bool PH (Program P, unsigned y) { return PL (PT (P,y));}
207
Answer-reversing input transformation PT
• Input– An input to H
• Output– An input to L such that
• yes inputs of H map to no inputs of L• no inputs of H map to yes inputs of L
• Note, PT must not loop when given any legal input to H
208
Why this works
PH
PLPT
yes input to H no input to L yes
no input to H yes input to L no
We have assumed that PL solves L
no
yes
bool PH (unsigned x) { return complement(PL (PT (P,y)));}
209
Yes→Yes and No→No
Domain of H
Yes inputsfor H
No inputsfor H
Yes inputsfor L
No inputsfor L
Domain of L
PLPT
PH
x PT(x) Yes/No
210
Notation and Terminology
• If there is such an answer-preserving (or answer-reversing) input transformation f (and the corresponding program PT), we say that H transforms to (many-one reduces to) L
• NotationH ≤ L
Domain of H
Yes inputs No inputs
Yes inputs No inputsDomain of L
211
Examples not involving the Halting Problem
212
Generalization
• While we focus on transforming H to other problems, the concept of transformation generalizes beyond H and beyond unsolvable program behavior problems
• We work with some solvable, language recognition problems to illustrate some aspects of the transformation process in the next few slides
213
Example 1
• L1 is the set of even length strings over {0,1}– What are the set of legal input instances and no inputs for
the L1 LRP?
• L2 is the set of odd length strings over {0,1}– Same question as above
• Tasks– Give an answer-preserving input transformation f that
shows that L1 LRP ≤ L2 LRP
– Give a corresponding program PT that computes f
Domain of L1
Yes inputs No inputs
Yes inputs No inputsDomain of L2
214
Program PT
string main(string x)
{
return(x concatenate “0”);
}
215
Example 2
• L1 is the set of all strings over {0,1}
– What is the set of all inputs, yes inputs, no inputs for the L1 LRP?
• L2 is {0}– Same question as above
• Tasks– Give an answer-preserving input transformation f which
shows that the L1 LRP ≤ L2 LRP
– Give a corresponding program PT which computes f
Domain of L1
Yes inputs No inputs
Yes inputs No inputsDomain of L2
216
Program PT
string main(string x)
{
return( “0”);
}
217
Example 3
• L1 – Input: Java program P that takes as input an unsigned int– Yes/No Question: Does P halt on all legal inputs
• L2
– Input: C++ program P that takes as input an unsigned int– Yes/No Question: Does P halt on all legal inputs
• Tasks– Describe what an answer-preserving input transformation f
that shows that L1 ≤ L2 would be/do?
Domain of L1
Yes inputs No inputs
Yes inputs No inputsDomain of L2
218
Proving a program behavior problem L is unsolvable
219
Problem Definitions *
• Halting Problem H– Input
• Program QH that has one input of type unsigned int
• non-negative integer y that is input to program QH
– Yes/No Question• Does QH halt on y?
• Target Problem L– Input
• Program QL that has one input of type string
– Yes/No question• Does Y(QL) = the set of
even length strings?
• Assume program PL solves L
220
Construction review
PH
x Yes/No
•We are building a program PH to solve H
bool PH (unsigned x) { return}
PTPT(x)
•PH will use PT as a subroutine, and we must explicitly construct PT using specific properties of H and L
PT (P,y)
PLY/N
•PH will use PL as a subroutine
PL ( );
221
P’s and Q’s• Programs which are PART of program PH and
thus “executed” when PH executes– Program PT, an actual program we construct– Program PL, an assumed program which solves
problem L
• Programs which are INPUTS/OUTPUTS of programs PH, PL, and PT and which are not “executed” when PH executes– Programs QH, QL, and QYL
• code for QYL is available to PT
222
Two inputs for L *
• Target Problem L– Input
• Program Q that has one input of type string
– Yes/No question• Does Y(Q) = the set of
even length strings?
• Program PL
– Solves L– We don’t know how
• Consider the following program Q1
bool main(string z)
{while (1>0) ;}
– What does PL output when given Q1 as input?
• Consider the following program Q2
bool main(string z)
{ if ((z.length %2) = = 0) return (yes)
else return (no); }
– What does PL output when given Q2 as input?
223
Another input for L *
• Target Problem L– Input
• Program Q that has one input of type string
– Yes/No question• Does Y(Q) = the set of
even length strings?
• Program PL
– Solves L– We don’t know how
• Consider the following program QL with 2 procedures Q1 and QYL
bool main(string z) {
Q1(5); /* ignore return value */
return(QYL(z));}
bool Q1(unsigned x) {if (x > 3) return (no); else loop;
}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);
else return(no);}
• What does PL output when given QL as input?
224
Input and Output of PT *
• Input of PT (and H)
– Program QH
• one input of type unsigned int
– Non-negative integer y
• Program QL that is the output of PT
and input of Lbool main(string z) {
QH(y); /* QH and y come left-hand side */
/* ignore return value */
return(QYL(z));
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool QYL(string y) {
if ((y.length( ) % 2) = = 0) return (yes); else return(no);
}
QH,y PT QL
225
Example 1 *
Input to PT
Program QH
bool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}
Non-negative integer y5
Output of PT
Program QL
bool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 5; QH(y); return (QYL(z));}
QH,y PT QL
226
Example 2
Input to PT
Program QH
bool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}
Non-negative integer y3
Output of PT
Program QL
bool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 3; QH(y); return (QYL(z));}
QH,y PT QL
227
PT in more detail
228
Declaration of PT
• What is the return type of PT?– Type program1 (with one input of type string)
• What are the input parameters of PT?– The same as the input parameters to H; in this case,
• type program2 (with one input of type unsigned int)• unsigned int (input type to program2)
program1 main(program2 QH, unsigned y)
PLPT
PH
QH,y QL Yes/No
229
program1 main(program2 QH, unsigned y) {/* Will be viewing types program1 and program2 as STRINGS over the
program alphabet P */
program1 QL = replace-main-with-QH(QH);/* Insert line break */
QL += “\n”;
/* Insert QYL */
QL += “bool QYL(string z) {\n \t if ((z.length % 2) == 0) return (yes) else return (no);\n }”;
/* Add main routine of QL */
QL += “bool main(string z) {\n\t”; /* determined by L */
QL += “unsigned y =”
QL += convert-to-string(y);
QL += “;\n\t QH(y)\n\t return(QYL(z));\n}”;
return(QL);}
program1 replace-main-with-QH(program2 P) /* Details hidden */string convert-to-string(unsigned y) /* Details hidden */
Code for PT PLPT
PH
QH,y QL Yes/No
230
PT in action
PT
code for QYL
Program QHbool main(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}
Input y5
Program QLbool QH(unsigned y) { if (y ==5) return yes; else if (y ==4) return no; else while (1>0) {};}bool QYL(string z) { if ((z.length % 2) == 0) return (yes) else return (no);}bool main(string z) { unsigned y = 5; QH(y); return (QYL(z));}
PLPT
PH
QH,y QL Yes/No
231
Constructing QL (and thus PT)
How to choose QYL or QNL
232
Start with no input for H
• If QH, y is a no input to the Halting problem
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(Q?L(z)); /* yes or no? */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool Q?L(string y) {
}
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
233
Answer-preserving input transformation
• If QH, y is a no input to the Halting problem
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool QYL(string y) {
}
– Now choose a QYL (or QNL) that is a yes (or no) input instance to L
234
Make yes for H map to yes for L
• If QH, y is a no input to the Halting problem
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
– Now choose a QYL (or QNL) that is a yes (or no) input instance to L
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
235
Possible shortcutProgram QL bool main(string z) {
QH(y); /* ignore return value */if ((z.length( ) % 2) = = 0) return (yes);else return (no);
}
bool QH(unsigned x) {/* comes from left-hand side
}
Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
236
Another Example
237
Problem Definitions
• Halting Problem H– Input
• Program QH that has one input of type unsigned int
• non-negative integer y that is input to program QH
– Yes/No Question• Does QH halt on y?
• Target Problem L– Input
• Program QL that has one input of type string
– Yes/No question• Is Y(QL) finite?
• Assume program PL solves L
238
Start with no input for H
• If QH, y is a no input to the Halting problem
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(Q?L(z)); /* yes or no? */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool Q?L(string y) {
}
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
239
Answer-reversing input transformation
• If QH, y is a no input to the Halting problem
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QNL(z)); /* no */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool QNL(string y) {
}
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
– Now choose a QYL (or QNL) that is a yes (or no) input instance to L
240
Make yes for H map to no for L
• If QH, y is a no input to the Halting problem
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QNL(z)); /* no */}
bool QH(unsigned x) {/* comes from left-hand side
}
bool QNL(string y) {if ((y.length( ) % 2) = = 0) return(yes);else return(no);
}
– Thus Y(QL) = {}
– QH loops on y
– Determine if this makes QL a no or yes input instance to L
– Now choose a QYL (or QNL) that is a yes (or no) input instance to L
241
Analyzing proposed transformations
4 possibilities
242
Problem Setup• Input of Transformation
• Program QH, unsigned x• Output of Transformation
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
• Problem L• Input: Program P• Yes/No Question: Is Y(P) =
{aa}?
• Question: Is the transformation on the left an answer-preserving or answer-reversing input transformation from H to problem L?
243
Key Step• Input of Transformation
• Program QH, unsigned x• Output of Transformation
• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
• Problem L• Input: Program P• Yes/No Question: Is Y(P) =
{aa}?
• The output of the transformation is the input to the problem.• Plug QL in for
program P above• Is Y(QL) = {aa}?
244
Is Y(QL) = {aa}?• Problem L
• Input: Program P• Yes/No Question: Is Y(P) =
{aa}?• Analysis
• If QH loops on x, Y(QL)={}• No input to H creates a QL that
is a no input for L• If QH halts on x, Y(QL) =
{even length strings}• Yes input to H creates a QL
that is a no input for L• Transformation does not work
• All inputs map to no inputs
• Input of Transformation• Program QH, unsigned x
• Output of Transformation• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
245
Three other problems• Problem L1
• Input: Program P• Yes/No Question: Is Y(P)
infinite?
• Problem L2• Input: Program P• Yes/No Question: Is Y(P)
finite?
• Problem L3• Input: Program P• Yes/No Question: Is Y(P) =
{} or is Y(P) infinite?
• Input of Transformation• Program QH, unsigned x
• Output of Transformation• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
246
Is Y(QL) infinite?• Problem L1
• Input: Program P• Yes/No Question: Is Y(P)
infinite?• Analysis
• If QH loops on x, Y(QL)={}• No input to H creates a QL
that is a no input for L• If QH halts on x, Y(QL) =
{even length strings}• Yes input to H creates a QL
that is a yes input for L• Transformation works
• Answer-preserving
• Input of Transformation• Program QH, unsigned x
• Output of Transformation• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
247
Is Y(QL) finite?• Problem L2
• Input: Program P• Yes/No Question: Is Y(P)
finite?• Analysis
• If QH loops on x, Y(QL)={}• No input to H creates a QL
that is a yes input for L• If QH halts on x, Y(QL) =
{even length strings}• Yes input to H creates a QL
that is a no input for L• Transformation works
• Answer-reversing
• Input of Transformation• Program QH, unsigned x
• Output of Transformation• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}
248
Is Y(QL) = {} or is Y(QL) infinite?• Problem L3
• Input: Program P• Yes/No Question: Is Y(P) =
{} or is Y(P) infinite?• Analysis
• If QH loops on x, Y(QL)={}• No input to H creates a QL
that is a yes input for L• If QH halts on x, Y(QL) =
{even length strings}• Yes input to H creates a QL
that is a yes input for L• Transformation does not work
• All inputs map to yes inputs
• Input of Transformation• Program QH, unsigned x
• Output of Transformation• Program QL bool main(string z) {
QH(y); /* ignore return value */
return(QYL(z)); /* yes or no */}
bool QH(unsigned x) {}
bool QYL(string y) {if ((y.length( ) % 2) = = 0) return (yes);else return (no);
}