Module 1: Course Overview
• Course: CSE 460
• Instructor: Dr. Eric Torng
• Grader/TA: To be determined
1
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
2
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?
3
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
4
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
5
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
6
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?
7
Module 2: Fundamental Concepts
• Problems
• Programs
– Programming languages
8
Problems
We view solving problems as the
main application for computer
programs
9
Definition
• A problem is a mapping or function between
a set of inputs and a set of outputs
• Example Problem: Sorting
(4,2,3,1)
(3,1,2,4)
(7,5,1)
(1,2,3)
Inputs
(1,2,3,4)
(1,5,7)
(1,2,3)
Outputs
10
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
11
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
12
Programs
Programs solve problems
13
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
14
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.
15
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
16
Maximum Element Problem
• Input
– integer n ≥ 1
– List of n integers
• Output
– The largest of the n integers
17
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);
}
18
Fundamental Theme
Exploring capabilities and limitations
of C++ programs
19
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
20
Question
• Is C++ general enough?
• Or is it possible that there exists some
problem P such that
– P can be solved by some program P in some
other reasonable programming language
– but P cannot be solved by any C++ program?
21
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
22
Summary *
• Problems
– When we talk about what programs can or
cannot “DO”, we mean what PROBLEMS can
or cannot be solved
23
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)
24
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
Subset 1
Set of Problems
Subset 2
Subset 3
25
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
26
Function Problems
• We will focus on problems where the
mapping from input to output is a function
Set of All Problems
Non-Function Problems Function Problems
27
General (Relation) Problem
• the mapping is a relation
– that is, more than one output is possible for a
given input
Inputs
Outputs
28
Criteria for Function Problems
• mapping is a function
– unique output for each input
Inputs
Outputs
29
Example Non-Function Problem
• Divisor Problem
– Input: Positive integer n
– Output: A positive integral divisor of n
9
Inputs
1
3
9
Outputs
30
Example Function Problems
• Sorting
• Multiplication Problem
– Input: 2 integers x and y
– Output: xy
2,5
Inputs
10
Outputs
31
Another Example *
• Maximum divisor problem
– Input: Positive integer n
– Output: size of maximum divisor of n smaller
than n
9
Inputs
1
3
9
Outputs
32
Decision Problems
• We will focus on function problems where
the correct answer is always yes or no
Set of Function Problems
Non-Decision Problems Decision Problems
33
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
Yes
No
Inputs
Outputs
34
Example
• Decision sorting
– Input: list of integers
– Yes/No question: Is the list in nondecreasing
order?
(1,3,2,4)
(1,2,3,4)
Inputs
Yes
No
Outputs
35
Another Example
• Decision multiplication
– Input: Three integers x, y, z
– Yes/No question: Is xy = z?
(3,5,14)
(3,5,15)
Inputs
Yes
No
Outputs
36
A Third Example *
• Decision Divisor Problem
– Input: Two integers x and y
– Yes/No question: Is y a divisor of x?
(14,5)
(14,7)
Inputs
Yes
No
Outputs
37
Focus on Decision Problems
Set of All Problems
Solvable Problems Unsolvable Problems
Other
Probs
Decision
Problems
• 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
38
Finite Domain Problems
• These problems have only a finite number of
inputs
Set of All Problems
Finite Domain Problems Infinite Domain Problems
39
Lack of Generality
Set of All Problems
Solvable Problems Unsolvable Problems
Infinite
Domain
Finite
Domain
Empty
• All finite domain problems can be solved
using “table lookup” idea
40
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”;
}
41
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
42
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
43
Analyzing Decision Problems
Can be defined by two sets
44
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
Set of All Legal Inputs
Yes Inputs
No Inputs
45
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
46
Typical Input Universes
• S*: The set of all finite length strings over finite
alphabet S
– 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
47
Language Recognition Problem
• Input Universe
– S* for some finite alphabet S
• Yes input instances
– Some set L subset of S*
• No input instances
– S* - L
• When S is understood, a language
recognition problem can be specified by just
stating what L is.
48
Language Recognition Problem *
• Traditional
Formulation
– Input
• A string x over some
finite alphabet S
– Task
• Is x in some language L
subset of S*?
• 3 set formulation
• Input Universe
– S* for a finite alphabet S
• Yes input instances
– Some set L subset of S*
• No input instances
– S* - L
• When S is understood, a
language recognition
problem can be specified
by just stating what L is.
49
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 S
– 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
50
Visualization *
Yes Inputs
Language L
Encoding
Scheme over
alphabet S
No Inputs
Original Decision
Problem
S* - L
Corresponding
Language Recognition
Problem
51
Analyzing Programs which Solve
Decision Problems
Four possible outcomes
52
Program Declaration
• Suppose a program P is designed to solve
some decision problem P. 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?
53
Program Declaration II
• Suppose a program P is designed to solve a
language recognition problem P. 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 S
– If S = {a,b}, then string will define variables consisting of only
a’s and b’s
– If S = {a, b, …, z, A, …, Z}, then string will define variables
consisting of any string of alphabet characters
54
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
55
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
56
Illustration
All Inputs
Y(P)
N(P)
C(P)
I(P)
57
Analyzing Programs and
Decision Problems
Distinguish the two carefully
58
Program solving a
decision problem
Y(P)
N(P)
C(P)
I(P)
• Formal Definition:
– A program P solves decision problem P if and only if
• The set of legal inputs for P is identical to the set of input
instances of P
• Y(P) is the same as the set of yes input instances for P
• N(P) is the same as the set of no input instances for P
– Otherwise, program P does not solve problem P
• Note C(P) and I(P) must be empty in order for P to solve
problem P
59
Solvable Problem
• A decision problem P is solvable if and only
if there exists some C++ program P which
solves P
– 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 P is unsolvable if and
only if all C++ programs P do not solve P
– Similar comment as above
60
Illustration of Solvability
Inputs of Program P
Y(P)
N(P)
Yes Inputs
No Inputs
C(P)
I(P)
Inputs of Problem P
61
Program halfsolving a problem
Y(P)
N(P)
C(P)
I(P)
• Formal Definition:
– A program P half-solves problem P if and only if
• The set of legal inputs for P is identical to the set of
input instances of P
• Y(P) is the same as the set of yes input instances for P
• N(P) union C(P) union I(P) is the same as the set of
no input instances for P
– Otherwise, program P does not half-solve
problem P
• Note C(P) and I(P) need not be empty
62
Half-solvable Problem
• A decision problem P is half-solvable if and
only if there exists some C++ program P
which half-solves P
– When the decision problem is a language
recognition problem for language L, we often
say that L is half-solvable
• A decision problem P is not half-solvable if
and only if all C++ programs P do not halfsolve P
63
Illustration of Half-Solvability *
Inputs of Program P
Y(P)
Yes Inputs
N(P)
C(P)
I(P)
No Inputs
Inputs of Problem P
64
Hierarchy of Decision Problems
All decision problems
Half-solvable
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.
65
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
66
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!
67
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
68
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
69
Preliminaries
Define set of problems
Observation about programs
70
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
71
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}
72
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
73
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 SP
• SP = {a, …, z, A, …, Z, 0, …, 9, white space,
punctuation}
74
Example *
int main(int A[], int n){
int i, max;
{26 characters including newline}
{13 characters including initial tab}
{1 character: newline}
if (n < 1)
return (“Illegal Input”);
max = A[0];
for (i = 1; i < n; i++)
if (A[i] > max)
max = A[i];
return (max);
}
{12 characters}
{28 characters including 2 tabs}
{13 characters}
{25 characters}
{18 characters}
{15 characters}
{15 characters}
{2 characters including newline}
75
Number of programs
• The set of legal C++ programs is clearly
infinite
• It is also no more than |SP*|
– SP = {a, …, z, A, …, Z, 0, …, 9, white space,
punctuation}
76
Goal
• Show that the number of languages L in
2{a}* is greater than the number of strings in
SP*
– SP = {a, …, z, A, …, Z, 0, …, 9, white space,
punctuation}
• Problem
– Both are infinite
77
How do we compare the relative
sizes of infinite sets?
Bijection (yes)
Proper subset (no)
78
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
79
Bijection Example
• Positive Integers Positive Even Integers
1
2
3
...
i
…
2
4
6
...
2i
...
80
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
81
Two sizes of infinity
Countable
Uncountable
82
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
83
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
84
Proof
85
(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 SP*
86
For any alphabet S, S* is
countably infinite
• Enumeration ordering
– All length 0 strings
• |S|0 = 1 string: l
– All length 1 strings
• |S| strings
– All length 2 strings
• |S|2 strings
–…
• Thus, SP* is countably infinite
87
Example with alphabet {a,b} *
• Length 0 strings
– 0 and l
• 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
88
(2) The set of languages in 2{a}*
is uncountably infinite
• Diagonalization proof technique
– “Algorithmic” proof
– Typically presented as a proof by contradiction
89
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[]
90
Visualizing D
List L[]
L[0]
L[1]
L[2]
Algorithm D
Language
D(L[]) not in
list L[]
L[3]
...
91
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
92
Visualizing One Possible L[ ] *
L[0]
L[1]
L[2]
L[3]
L[4]
l
a
aa aaa aaaa ...
IN
IN
IN
IN
IN
IN OUT IN OUT IN
OUT OUT OUT OUT OUT
OUT IN
IN OUT OUT
•#Rows is countably infinite
•Given
•#Cols is countably infinite
• {a}* is countably infinite
OUT IN OUT OUT IN
...
• Consider each string to be a feature
– A set contains or does not contain each string
93
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
D(L[]) l
a
L[0]
L[1]
L[2]
IN IN
OUT
L[3]
L[4]
OUT IN
aa aaa aaaa ...
IN
IN
IN
IN OUT
IN IN OUT IN
OUT OUT OUT
IN OUT OUT
IN OUT
IN OUT
OUT IN OUT OUT OUT
IN
...
94
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.
• …
95
Questions
L[0]
L[1]
L[2]
L[3]
L[4]
l
a
aa aaa aaaa ...
IN
IN
IN
IN
IN
IN OUT IN OUT IN
OUT OUT OUT OUT OUT
OUT IN
IN OUT OUT
OUT IN OUT OUT IN
...
• 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[])?
96
Visualization
All problems
Solvable Problems
The set of solvable problems is a proper subset of the set
of all problems.
97
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
98
Module 6
• Topics
– Program behavior problems
• Input of problem is a program/algorithm
• Definition of type program
– Program correctness
• Testing versus Proving
99
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?
100
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?
101
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
102
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 SP = {a, …, z,
A, …, Z, 0, …, 9, punctuation, white space}
• Note, many strings over SP are not legal programs
• We consider them to be programs that always crash
• Possible declaration of main procedure
– bool main(program P)
103
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?
104
Testing versus Analyzing
Test Inputs
x1
x2
x3
...
Program P
Program
P
Analyzer
Outputs
P(x1)
P(x2)
P(x3)
...
Analysis of
Program P
105
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
106
Module 7
• Halting Problem
– Fundamental program behavior problem
– A specific unsolvable problem
– Diagonalization technique revisited
• Proof more complex
107
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
108
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 x
4
109
Three key definitions
110
Definition of list L *
SP* is countably infinite where SP = {characters, digits, white
space, punctuation}
• Type program will be type string with SP as the alphabet
• Define L to be the strings in SP* listed in enumeration order
•
– length 0 strings first
– length 1 strings next
– …
• Every program is a string in SP
– For simplicity, consider only programs that have
• one input
• the type of this input is an unsigned
• Consider strings in SP* that are not legal programs to be
programs that always crash (and thus halt on all inputs)
111
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
112
Definition of program D
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 SP* 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 */
113
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 SP).
– 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 l
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
…
114
Proof that H is not solvable
115
Argument Overview *
H is solvable
Definition of
Solvability
PH exists
H is NOT solvable
PH does NOT exist
D’s code
D exists
L is list of
all programs
D is on list L
D does NOT exist
D is NOT on list L
p → q is logically equivalent to (not q) → (not p)
116
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
117
Non-existent program behavior B
118
Visualizing List L *
P0
P1
P2
P3
P4
0
1
2
3
4
H
H
H
H
H
H NH
H
NH
H
NH NH NH NH NH
NH
H
H
H
H
H
NH NH
H
H
...
•#Rows is countably infinite
• Sp* is countably infinite
•#Cols is countably infinite
• Set of nonnegative 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
119
Diagonalization to specify B *
•We specify a non-existent program behavior B by using a unique feature
(number) to differentiate B from Pi
B
0
1
2
3
4
P0
NH
H
H
H
H
H
H NH
H
H
NH
H
P1
P2
P3
P4
...
NH NH NH
H NH NH
NH
H
H
H
H
H
NH
H NH
H
NH
H
...
120
Arguing D exhibits
program behavior B
121
Code for D
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 SP* 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 */
122
Visualization of D in action on
input y
• Program D with input y
– (type for y: unsigned)
0
1
2 ...
P0
P1
P2
H
H
H
H NH
H
NH NH
NH
y ...
...
Py
NH
H
...
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);
D
123
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.
• …
124
Alternate Proof
125
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
126
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
127
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
128
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
129
Placing the Halting Problem
All Problems
H
Solvable
130
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
131
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
132
Module 8
• Halting Problem revisited
– Universal Turing machine half-solves halting
problem
– A universal Turing machine is an operating
system/general purpose computer
133
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?
134
Example Input
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);
}
Nonnegative integer
4
135
Organization
Universal Turing machine’s Memory
Program P
Program P’s Memory
int i,Q;
4
Program Counter
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);
}
Line 1
136
Description of universal Turing
machine *
• Basic Loop
– Find current line of program P
– Execute current line of program P
Program P
• Update program P’s memory
– Update program counter
– Return to Top of Loop
Program P’s Memory
Program Counter
137
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
138
Debuggers
• How are debugger’s like gdb or ddd related
to universal Turing machines?
• How do debuggers simplify the debugging
process?
139
Placing the Halting Problem
All Problems
Half-solvable
H
Solvable
140
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
141
Module 9
• Closure Properties
– Definition
– Language class definition
• set of languages
– Closure properties and first-order logic
statements
• For all, there exists
142
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
143
Integers and Addition
2
7
5
Integers
144
Integers and Division
.4
2
5
Integers
145
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
146
Example Language Classes *
• In all these examples, we do not explicitly state what the
underlying alphabet S is
• Solvable languages (REC)
– Languages whose language recognition problem can be solved
• Half-solvable languages (RE)
– Languages whose language recognition problems can be halfsolved
• Finite languages
– Languages with a finite number of strings
• CARD-3
– Languages with at most 3 strings
147
Finite Sets and Set Union *
{a,b,aa}
{a,b,aa,bb}
{a,b,bb}
Finite Languages
All Languages
148
CARD-3 and Set Union
{a,b,aa,bb}
{a,b,aa}
{a,b,bb}
CARD-3
All Languages
149
Finite Languages and Set
Complement
{/\,aa,ba,bb,aaa,...}
{a,b,ab}
Finite Languages
All Languages
150
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 {l} is a finite language
{} union {0} is a finite language
...
{l} union {} is a finite language
...
151
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
152
Example F-O logic statements *
• " L1,L2 in FINITE, L1 union L2 in FINITE
• " L1,L2 in CARD-3, L1 union L2 in CARD3
• " L in FINITE, Lc in FINITE
• " L in CARD-3, Lc in CARD-3
153
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
154
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
155
Proving/Disproving
• Which is easier and why?
– Proving a closure property is true
– Proving a closure property is false
156
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
157
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 halfsolvable languages
158
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
159
REC and Set Complement
REC
ODD
EVEN
All Languages
•
•
•
•
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?
160
P’ Illustration
P’
P
Input x
Yes/No
No/Yes
161
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 */
162
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
163
proof continued
– 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
164
REC Closed Under Set Union
L1
REC
L1 U L2
L2
All Languages
• 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
165
REC Closed Under Set Union
L1
REC
L1 U L2
– Construct program P3
from P1 and P2 as follows
L2
All Languages
– P3 solves L1 U L2
• Halts on all inputs
• Answers correctly
– L1 U L2 is solvable
166
P3 Illustration
P1
Yes/No
P3
P2
OR
Yes/No
Yes/No
167
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. */
168
Other Closure Properties
• Unary Operations
– Language Reversal
– Kleene Star
• Binary Operations
–
–
–
–
Set Intersection
Set Difference
Symmetric Difference
Concatenation
169
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
170
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
...
171
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
172
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
173
RE Closed Under Set Union
L1
L1 U L2
L2
• Let L1 and L2 be arbitrary r.e. languages
• There exist P1 and P2 s.t. Y(P1)=L1 and Y(P2)=L2
• Construct program P3 from P1 and P2
• Prove Program P3 half-solves L1 U L2
• There exists a program P that half-solves L1 U L2
• L1 U L2 is an r.e. language
174
Constructing Program P3
L1
L1 U L2
L2
• What code did we use for P3 when we
worked with solvable languages?
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?
175
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
176
P3 works correctly?
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;
}
– 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?
177
P3 works correctly?
bool main(string y) {
if (P1(y)) return yes;
else if (P2(y)) return yes;
else return no;
}
– 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?
178
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 */
179
P3 works correctly?
bool main(string y){
if ((P1(y) || P2(y)) return yes;
else return no;
}
/* P1 and P2 run in parallel */
– 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?
180
RE and Set
complement
Lc
RE
L
• 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
• ...
181
RE and Set
complement
Lc
RE
L
• Let L be an arbitrary r.e. language
• There exists P s.t. Y(P)=L
• Construct program P’ from P
• Prove Program P’ half-solves L complement
• There exists a program P’ which half-solves L
complement
• L complement is an r.e. language
182
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?
183
Other closure properties
• Unary Operations
– Language reversal
– Kleene Closure
• Binary operations
– set intersection
– concatenation
• Not closed
– Set difference (on practice hw)
184
Pseudo Closure Property
• Lemma: If L and Lc are half-solvable, then
L is solvable.
• Question: What about Lc?
185
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
186
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?
187
Illustration
S*
L
Lc
P1 halts
P2 halts
188
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
189
P3 Illustration
P1
Input
Yes
Yes
P3
P2
Yes
No
190
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 */
191
RE and REC
All Languages
RE
L
REC
Lc
192
RE and REC
All Languages
Lc
L
Lc
RE
Lc REC
Are there any languages L in RE - REC?
193
RE and REC
All Languages
RE
Hc
H
REC
So where does Hc belong?
194
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
195
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
196
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
197
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
198
Basic Idea/Technique
199
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
200
– L will be a problem about program behavior
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
201
Constructing PH using PL
Answer-preserving input
transformations and Program PT
202
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
203
Two Representations of PH *
P,y
PT
PT(P,y)
PL
Y/N
Yes/No
PH
bool PH (Program P, unsigned y) {
return PL (PT (P,y));
}
204
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
205
Why this works *
yes input to H
no input to H
PT
yes input to L
no input to L
PL
yes
no
PH
We have assumed that PL solves L
bool PH (Program P, unsigned y) {
return PL (PT (P,y));
}
206
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
207
Why this works
yes input to H
no input to H
PT
no input to L
yes input to L
PL
no
yes
yes
no
PH
We have assumed that PL solves L
bool PH (unsigned x) {
return complement(PL (PT (P,y)));
}
208
Yes→Yes and
No→No
x
PT PT(x) PL
PH
Yes/No
No inputs
for H
Yes inputs
for H
Domain of H
Yes inputs
for L
No inputs
for L
Domain of L
209
Notation and
Terminology
Yes inputs
No inputs
Domain of H
Yes inputs
No inputs
Domain of L
• 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
• Notation
H≤L
210
Examples not involving the
Halting Problem
211
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
212
Yes inputs
Example 1
No inputs
Domain of L1
Yes inputs
No inputs
Domain of L2
• 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
213
Program PT
string main(string x)
{
return(x concatenate “0”);
}
214
Yes inputs
Example 2
No inputs
Domain of L1
Yes inputs
No inputs
Domain of L2
• 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
215
Program PT
string main(string x)
{
return( “0”);
}
216
Yes inputs
Example 3
No inputs
Domain of L1
Yes inputs
No inputs
Domain of L2
• 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?
217
Proving a program behavior
problem L is unsolvable
218
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
219
Construction review
x
PT
PT(x)
PL
Y/N
Yes/No
PH
•We are building a program PH to solve H
•PH will use PL as a subroutine
•PH will use PT as a subroutine, and we must explicitly
construct PT using specific properties of H and L
bool PH (unsigned x) {
return PL (PT (P,y));
}
220
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
221
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?
222
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?
223
Input and Output of PT *
• Input of PT (and H)
• Program QL that is the output of PT
and input of L
– Program QH
• one input of type unsigned
int
– Non-negative integer y
QH,y
PT
QL
bool 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);
}
224
Example 1 *
QH,y
PT
QL
Input to PT
Output of PT
Program QH
bool main(unsigned y) {
if (y ==5) return yes;
else if (y ==4) return no;
else while (1>0) {};
}
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));
}
Non-negative integer y
5
225
Example 2
QH,y
PT
QL
Input to PT
Output of PT
Program QH
bool main(unsigned y) {
if (y ==5) return yes;
else if (y ==4) return no;
else while (1>0) {};
}
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));
}
Non-negative integer y
3
226
PT in more detail
227
QH,y
Declaration of PT
PT
QL
PL
Yes/No
PH
• 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)
228
Code for PT
QH,y
PT
QL
PL
Yes/No
PH
program1 main(program2 QH, unsigned y) {
/* Will be viewing types program1 and program2 as STRINGS over the
program alphabet SP */
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 */
229
QH,y
PT in action
PT
QL
PL
Yes/No
PH
Program QH
Program QL
bool main(unsigned y) {
if (y ==5) return yes;
else if (y ==4) return no;
else while (1>0) {};
}
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));
}
Input y
5
PT
code
for QYL
230
Constructing QL (and thus PT)
How to choose QYL or QNL
231
Start with no input for H
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(Q?L(z)); /* yes or no? */
– QH loops on y
– Thus Y(QL) = {}
– Determine if this
makes QL a no or yes
input instance to L
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool Q?L(string y) {
}
232
Answer-preserving input
transformation
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(QYL(z)); /* yes */
– QH loops on y
– Thus Y(QL) = {}
– 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
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool QYL(string y) {
}
233
Make yes for H map to yes for L
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(QYL(z)); /* yes */
– QH loops on y
– Thus Y(QL) = {}
– 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
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool QYL(string y) {
if ((y.length( ) % 2) = = 0) return (yes);
else return (no);
}
234
Possible shortcut
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);
Program 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
}
}
235
Another Example
236
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
237
Start with no input for H
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(Q?L(z)); /* yes or no? */
– QH loops on y
– Thus Y(QL) = {}
– Determine if this
makes QL a no or yes
input instance to L
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool Q?L(string y) {
}
238
Answer-reversing input transformation
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(QNL(z)); /* no */
– QH loops on y
– Thus Y(QL) = {}
– 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
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool QNL(string y) {
}
239
Make yes for H map to no for L
• Program QL
• If QH, y is a no input
bool main(string z) {
to the Halting problem
QH(y); /* ignore return value */
return(QNL(z)); /* no */
– QH loops on y
– Thus Y(QL) = {}
– 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
}
bool QH(unsigned x) {
/* comes from left-hand side
}
bool QNL(string y) {
if ((y.length( ) % 2) = = 0) return(yes);
else return(no);
}
240
Analyzing proposed
transformations
4 possibilities
241
Problem Setup
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L
•
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);
• Input: Program P
• Yes/No Question: Is Y(P) =
{aa}?
Question: Is the
transformation on the
left an answerpreserving or answerreversing input
transformation from H
to problem L?
}
242
Key Step
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L
•
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);
• 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}?
}
243
Is Y(QL) = {aa}?
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L
• Input: Program P
• Yes/No Question: Is Y(P) =
{aa}?
• Analysis
• If QH loops on x, Y(QL)={}
bool main(string z) {
• No input to H creates a QL
QH(y); /* ignore return value */
that is a no input for L
return(QYL(z)); /* yes or no */
• If QH halts on x, Y(QL) =
}
{even length strings}
bool QH(unsigned x) {}
• Yes input to H creates a QL
bool QYL(string y) {
that is a no input for L
if ((y.length( ) % 2) = = 0) return (yes);
• Transformation does not work
else return (no);
• All inputs map to no inputs
}
244
Three other problems
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L1
• Input: Program P
• Yes/No Question: Is Y(P)
infinite?
• Problem L2
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);
• 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?
}
245
Is Y(QL) infinite?
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L1
• Input: Program P
• Yes/No Question: Is Y(P)
infinite?
• Analysis
• If QH loops on x, Y(QL)={}
bool main(string z) {
• No input to H creates a QL
QH(y); /* ignore return value */
that is a no input for L
return(QYL(z)); /* yes or no */
• If QH halts on x, Y(QL) =
}
{even length strings}
bool QH(unsigned x) {}
• Yes input to H creates a QL
bool QYL(string y) {
that is a yes input for L
if ((y.length( ) % 2) = = 0) return (yes); • Transformation works
else return (no);
• Answer-preserving
}
246
Is Y(QL) finite?
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L2
• Input: Program P
• Yes/No Question: Is Y(P)
finite?
• Analysis
• If QH loops on x, Y(QL)={}
bool main(string z) {
• No input to H creates a QL
QH(y); /* ignore return value */
that is a yes input for L
return(QYL(z)); /* yes or no */
• If QH halts on x, Y(QL) =
}
{even length strings}
bool QH(unsigned x) {}
• Yes input to H creates a QL
bool QYL(string y) {
that is a no input for L
if ((y.length( ) % 2) = = 0) return (yes); • Transformation works
else return (no);
• Answer-reversing
}
247
Is Y(QL) = {} or is Y(QL) infinite?
• Input of Transformation
• Program QH, unsigned x
• Output of Transformation
• Program QL
• Problem L3
• Input: Program P
• Yes/No Question: Is Y(P) =
{} or is Y(P) infinite?
• Analysis
• If QH loops on x, Y(QL)={}
bool main(string z) {
• No input to H creates a QL
QH(y); /* ignore return value */
that is a yes input for L
return(QYL(z)); /* yes or no */
• If QH halts on x, Y(QL) =
}
{even length strings}
bool QH(unsigned x) {}
• Yes input to H creates a QL
bool QYL(string y) {
that is a yes input for L
if ((y.length( ) % 2) = = 0) return (yes); • Transformation does not work
else return (no);
• All inputs map to yes inputs
}
248