Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx

Busy week
• Lab 9 due Thursday at 5 pm:
L = { <M,s> : s ∈ L(M) and |L(M)| % 2 = 0}. For example,
suppose that L(M) = {aa}. Then
<M,aa> ∉ L because |L(M)| = 1, 1 % 2 = 1; If L(M) {a,aaa}
then <M,ℇ> ∉ L because ℇ ∉ L(M),
but <M,aaa> ∈ L. Prove that L ∉ D by reduction from H.
Your proof could implement R, the mapping reduction function,
as a Java or Python program in
the form demonstrated, which allows the user to configure
whether M halts on w.
We start the proof by creating R. We start are by thinking of
the input to ouput. What's the
input to the function R? Remember that R is a mapping
reduction from H to L. So the input to
R must be something that could be in H. <M,w> has the correct
syntactic form. Could R return
this: <M#,aaa>? think of terms of whether the output of R
could possible be in L.
R(<M,w>) =
1. Define M#:
1.a
1.x Run M on w
2. Return M#,aaa>

M halts on w, <M#,aaa> is in L. So if you make it through step
1.x, then M#,aaa satisfies the
characteristic function of L.
and if M does not halt on w, <M#,aaa> is not in L. If you
don't get through 1.x, then you lack
some component of the characteristic function.
• Online Pumping Theorem workshop Thursday at 9 pm: (10
extra credit points for full
participation)
• Lab 10 due Monday at noon:
L {<M,s> : s ∈ L(M), |L(M)| = 2}. Prove that L ∉ SD by a
reduction from ¬H}
• HW #3 due Friday May 8 at 2 pm.
• Midterm #3 Tuesday May 12
Rice Theorem and Semidecidability Page 1
Rice's Theorem and Semidecidability
Mapping Reducibility
Recall that a mapping reduction function ƒ transforms an L₁
membership question into an L₂
membership question:
∀ x∈ Σ* (x ∈ L₁ iff ƒ(x) ∈ L₂)
H = { <M,w> : M halts on w>
Hε = {<M> : M halts on ε}. That is the strings in the language
Hε are encodings of TMs

such tht the encoded TM halts when started on an empty input
tape.
if H ≤M Hε then if Hε ∈ D then H ∈ D; But we know that H is
not decidable.
And so by applying the modus tollens logic rule to the
implicaion,
Hε ∉ D
If there's a target langauge that we want to show is not
decidable, than we show that
by showing that a not decidable langauge, like H, is mapping
reducible to the target
language.
It seems more natural to go in the other direction, H ≤M Hε:
Reduce the problem of whether
you can decide if M will halt on some particular string to,
whether you can decide if M will halt
on any string. But we already know that H ∈ SD/D, and we are
using that knowlege to prove
something about Hε.
This proof is confusing at first because it uses two languages, H
and Hε, five different TMs, and
two string variables:
TM 1: Mₒ (the "Oracle"), the hypothetical machine that could
decide Hε. It may be helpful to
think of it as a source code analyzer, which claims to be able to
inspect any high-level TM
description and tell whether the machine would be an element of
the target language.
TM 2: R, which implements the mapping reduction function
from H to Hℇ: <M,w> → <M#>.
Note well: R maps <M,w>, a TM/string pair which has correct

syntax for an element of H, to
<M#>,which has correct syntax for an element Hε. Compare
this with ƒ in the definition of a
mapping reduction function. ∀ x∈ Σ* (x ∈ L₁ iff ƒ(x) ∈ L₂
R(<M,w>):
i. Define M#(x)
i.a Simulates M on w
i.b Enter a halting state
ii. Return <M#>
TM 3: M# The TM created by the mapping reduction function
R. Note that M# is analyzed
by Mₒ, but never run. This can be a very confusing point.l
TM 4: C is short for Composition, Mₒ + R. C = Mₒ(R<M,w>) =
Mₒ(<M#>). Mₒ, which
hypothetically can decide Hε, decides if the output of R is in
Hε. If Mₒ could do that, it could also
decide H.
TM 5: M The TM encoded as the first part of the <M,w> pair,
the input to R that has the correct
syntax for an element of H. It has no necessary relation to M#.
String 1: w The second part of the <M,w> pair. Could be any
string in ∑*.
String 2: x The contents of the M# input tape. Could be any
string in ∑*. x has no necessary
relationship to w.

H = { <M,w> : M halts on w>
Hε = {<M> : M halts on ε}. That is the strings in the language
Hε are encodings of TMs
such tht the encoded TM halts when started on an empty input
tape.
R is a mapping reduction from H to H_e.
R(<M,w>):
i. Define M#(x)
ii. Return <M#>
if there were an Oracle Mₒ that could decide H_e, then Mₒ(M#)
could decide H.
• <M,w> is in H: Mₒ accept/reject M#, and why? What would
M# do if M halts on w?
M# would proceed directly to i.b and halt. So, if M halts on w,
M# halts on everything.
Everything includes epsilon, so Mo accepts M# so because M#
halts on epsilon.
R(<M,w>):
i. Define M#(x)
ii. Return <M#>
• <M,w> is not in H:
i. Define M#(x)
i.a Simulates M on w -- loops forever, and never gets to 1.b

ii. Return <M#>
in this case what happens when x = epsilon. Will M# halt or
not?
However, no TM could decide H, so a TM such as Mₒ could not
possible exist.
C++
Sometimes C needs to do a little more work to use the verdict of
Mₒ to decide H.
Consider L₂ = { <M> : M accepts no even length string}, and
this proof using a reduction from
H to try to show that L₂ ∉ D.
R(<M, w>) =
1. Construct the description M#:
1.1. Erase the tape.
1.2. Write w on the tape.
1.3. Run M on w.
1.4. Accept.
2. Return <M#>
If Mₒ exists, then C = Mₒ(R(<M, w>)) = Mₒ(<M#>) decides H:
• <M, w> ∈ H: M halts on w.
L(M#) = ∑*; ∑* contains an infinite number of even length
strings, so Mₒ which can somehow
tell if L(M) contains any even length strings, would reject.

• <M, w> ∉ H: M does not halt on w.
L(M#) = Ø, and Ø contains no even length so Mₒ accepts.
The problem is that ¬C= ¬ Mₒ(<M#>) decides H: We need to
invert the verdict of Mₒ.
Rice's Theorem
In chapter 21, Rich considers a suite of languages which she
proves are not in D using a
reduction from H: Hε, HANY, HALL, A, Aε, AANY, AALL.
There's a pattern. According to Rice's theorem, no non-trivial
property of a language semi-
decided by a TM, such as whether ℇ ∈ L, is decidable.
A non-trivial property: not true or false for all SD languages.
Proof: Let P be any nontrivial property of the SD languages.
Ø is the language containing no strings. Either P(Ø) = True or
P(Ø) = False. Assume it is False
(a similar proof exists if it is True).
P(L) = { L contains an even length string}
P(Ø) = false
P'(L) = { |L| %2 = 0}
P'(Ø) = true
Since P is nontrivial, there is some SD language LT such that
P(LT) is True. Let MT be a Turing
machine that semidecides LT.

L = { <M> : P(L(M)) = True }
this is equivalent, in terms of our example property P, to saying
L = { <M> : L(M) contains an even length string }
LT = {aa}
R(<M, w>) = is a mapping reduction from H to L
1. Define M#:
1.1. Copy its input x to another track for later.
1.2. Erase the tape.
1.3. Write w on the tape.
1.4. Run M on w.
1.5 Put x back on the tape and simulate MT on x.
2. Return <M#>.
<M,w> ∈ H
L(M#) = …
Mₒ accepts/rejects <M#> because …
<M,w> ∉ H
L(M#) = …
Mₒ accepts/rejects <M#> because …
Appying Rice's Theorem
To use Rice’s Theorem to prove L ∉ D, all that's required is:
• Specify property P.
• The domain of P must be the set of languages that can be

semidecided by a TM (i.e, the SD
languages).
• Show that P is nontrivial (true for at least one SD language,
and false for at least one SD
language).
So, can Rice's Theorem be used to prove these languages are not
decidable?
1. {<M> : L(M) contains only even length strings}.
2. {<M> : L(M) contains an odd number of strings}.
3. {<M> : L(M) contains all strings that start with a}.
4. {<M> : L(M) is infinite}.
5. {<M> : L(M) is regular}.
6. {<M> : M contains an even number of states}.
7. {<M> : M has an odd number of symbols in its tape
alphabet}.
8. {<M> : M accepts ε within 100 steps}.
9. {<M>: M accepts ε}.
10. {<Ma, Mb> : L(Ma) = L(M1)}.
11. L¬R = { <M,s> : s ∈ L(M) and sR ∉ L(M) }. (The HW #3)
language.
Three Languages that look similar but are in different categories
L
1
= {a}.
L
2

= {<M> : a ∈ L(M)}.
L
3
= {<M> : L(M) = {a}
Proving that L ∉ SD
One way is to exploit the fact that if L ∈ SD/D then ¬L ∈ ¬SD,
where ¬L is the complement of
L. (If L and ¬D are both in SD, then both are also in D).
Another is to a reduction from ¬H. There is a wrinkle in that
we cannot use ¬C, as we did with
reductions from H to show that a language is not decidable.
That is because a semideciding Mₒ
could loop and there is no way to take the compliment of a loop.
L = {<M> : L(M) = {a} }. Complete the proof below that uses a
reduction from ¬H to show
that L ∉ SD.
R(<M, w>) is a reduction from ¬H to L. That is, <M, w> are
R's parameters, and R outputs
M#, which could be an element in L:
1. Define M#(x):
1.a If x = a, accept.
1.b Erase the tape.
1.c Write w.
1.d Run M on w.
1.e Accept x.
2. Return <M#>
If an Oracle Mₒ exists and semidecides L, then Mₒ(<M#>)
semidecides ¬H:

• <M,w> ∈ ¬H: Mₒ accepts <M#> because …
L(M#) ={a}, because M will not halt on w, so the only string
that M# will accept is a at step 1.
• <M,w> ∉ ¬H: Mₒ does not accept <M#> because …
L(M#) = ∑*, and ∑* contains other strings besides a. The
reason why L(M#) = ∑* is that M will
not halt on w, so M# will either accept at step 1 if x = a, or at
step 5 for all other strings in ∑*.
But no machine to semidecide ¬H can exist, so neither does Mₒ
L = {<M
a
, M
b
> : ε ∈ L(M
a
) – L(M
b
)}. That is, the strings in L are pairs of TM string encodings,
such that ε is in
the language accepted by the first encoded TM, but not the
second.
We can prove L ∈ ¬SD by a reduction from ¬H.
R( <M, w> ) is a reduction from ¬H to L:

1. Define M#1
1.a Accept
2. Define M#2
2.a Erase the tape.
2.b Write w on the tape.
2.c Simulate M on w.
2.d …
Return <M#1,M#2>.
Let Mₒ be an Oracle that semidecides L. Then Mₒ(
<M#1,M#2>) semidecides ¬H:
<M, w> ∈ ¬H: L(M#1) - L(M#2) = …, and Mₒ accepts
<M#1,M#2> because …
<M, w> ∉ ¬H: L(M#1) - L(M#2) = …, and Mₒ does not accept
<M#1,M#2> because …
L = {<M> : L(M) is context-free}.
Prove that L ∉ SD using a reduction from ¬H
R (<M,w>) =
1. Define M#
1.a Save x.
1.b Erase the input tape.
1.c Write w on the input tape.
1.d Simulate M on w.
1.e If x ∈ aⁿbⁿcⁿ, accept.
1.f Else loop.
2. Return <M#>.
What is the reasoning?

L = {<M> : L(M ) is infinite}.
¬H = {<M,w> : M does not halt on w}
R(<M,w>) is a reduction from ¬H to L:
1. Define M#(x)
1.a Save x.
1.b Erase the tape.
1.c Write w on the tape.
1.d Run M on w for a maximum of |x| steps.
1.e If M halted by step |x| then loop.
1.f Else accept x.
2. Return <M#>
If an Oracle Mₒ exists and semidecides L, then C =
Mₒ(R(<M,w>)) = Mₒ(<M#>) semidecides
¬H:
• <M,w> ∈ ¬H: M does not halt on w, so M# always makes it to
step 1.f and accepts ∑*. ∑* is
an infinte language, so Mₒ accepts <M#>.
• <M,w> ∉ ¬H: M halts on w. Suppose it does so in k steps.
Then M# loops on all strings of
length k or greater. It accepts strings of length less than k. But
that set is finite. So Oracle
does not accept. But no machine to semidecide ¬H can exist, so
neither does Oracle.
Note that M# could be very simple if we just inverted the
verdict of Mₒ:
1.a Save x.
1.b Erase the tape.
1.c Write w on the tape.
1.d Run M on w.
1.e Accept x.

Dillon’s Rule
Legal Framework
General Law – State prescribes numbers, functions of county
officials, revenue, spending, services
Home Rule Charters – greater flexibility
San Francisco – Single County and City (charter)
http://cao.lacounty.gov/forms/lacmap.
pdf

Look up -- Special Districts
Regional Districts in chapter 9, in
Governing California
Dillon’s Rule
Legal Framework
General Law – State prescribes numbers, functions of county
officials, revenue, spending, services
Home Rule Charters – greater flexibility
San Francisco – Single County and City (charter)
http://cao.lacounty.gov/forms/lacmap.
pdf

Look up -- Special Districts
Regional Districts in chapter 9, in
Governing California
GOVERNING CALIFORNIA, FIFTH EDITION
Copyright © 2014, W. W. Norton & Company
Art Slides
GOVERNING CALIFORNIA
FIFTH EDITION
by
ANAGNOSON • BONETTO
BUCK • DELEON • EMREY
KELLEHER • KOCH

Art Slides
GOVERNING CALIFORNIA
FIFTH EDITION
by
ANAGNOSON • BONETTO
BUCK • DELEON • EMREY
KELLEHER • KOCH
Governing California State and Local Government
What states require strict obedience to Dillon’s rule?
A. Alabama, Idaho, and Nebraska
B. California, Utah, and New York
C. Texas, California, and Arizona
D. Tennessee, Florida, and California

In order to deal with problems such as air pollution and waste
management that affect large
geographical areas in California,
A. regional governments have been formed.
B. special districts have been given authority to deal with these
types of problems.
C. a council-mayor system has been implemented.
D. secession movements in a number of cities have developed.
Which of the following local government has “City of Refuge
Laws”?
A. Roseville
B. San Francisco
C. Oxnard
D. Fountain Valley
The most common form of local government in California cities
is the
A. mayor-council system.
B. commission government.
C. city manager system.
D. councils of governments.
Which of the following is not a characteristic of reform
government at the local level?
A. At-large council elections
B. Nonpartisanship
C. City Manager plan

D. Concurrent elections
What does Dillon’s Rule establish?
A. The basic legal framework for relations between state and
local governments.
B. The basic legal framework for relations between the
executive and legislative branches.
C. The basic legal framework for relations between the judicial
and legislative branches.
D. The basic legal framework for relations between the
executive and judicial branches.
In California, school districts are governed by
A. the President of the United States.
B. locally elected boards of education.
C. Congress and the Department of Education.
D. the governor of California and the U.S. Supreme Court.
In California, what is the general purpose of local governments,
such as cities and counties?
A. Manage interstate highways, regulate foreign trade, and
maintain weapons and training for
security agencies such as Home Land Security.
B. Manage fires, clean streets, and protect the health, safety,
welfare, and overall quality of life of
all who live within their jurisdiction.
C. Regulate banks, alcohol and gun sales, print currency, and
exchange intelligence information

with various clandestine agencies.
D. Protect the border, regulate interstate commerce,negotiate
with foreign countries, andmonitor
the sale of prescription drugs, alcohol.
_________ are limited-purpose local governments filling the
need for services that counties and ci
A. Charter (or Home Rule) Counties
B. Unfunded mandates.
C. Special Districts
D. California Coastal Commission
The _________ has the authority to control all development
within the 1,000-yard-wide shoreline zone
along the entire California coast.
A. Charter (or Home Rule) Counties
B. Unfunded mandates.
C. Special Districts
D. California Coastal Commission
True / False -- In California, most cities are governed by the
mayor-council system
True / False -- All local elections in California are
nonpartisan

Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx

Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx

Recommended

Recommended

More Related Content

Similar to Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx

Similar to Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx (20)

More from clairbycraft

More from clairbycraft (20)

Recently uploaded

Recently uploaded (20)

Busy week • Lab 9 due Thursday at 5 pm L = { M,s s.docx