Journal of Business Case Studies – Third Quarter 2006 Volume .docx

Journal of Business Case Studies – Third Quarter 2006 Volume
2, Number 3
65
Sexual Harassment
In The Workplace: Europe
John Lehman, (E-mail: [email protected]), University of Alaska
Fairbanks
ebecca McDonald returned to her office in Leverkusen Germany
still angry from her meeting with
Bertina Knies in Human Resources. Rebecca had presented an
open-and-shut sexual harassment
complaint, clearly supported both by company policies and the
Law, and Frau Knies had not only
refused to do anything about it, but had not very politely
insinuated that the whole thing was Rebecca’s fault. The
next step would probably be to appeal to Human Resources back
in Detroit, since the Germans clearly had no
understanding of the way things should work in a modern
society.
Rebecca had grown up in Flint, Michigan, where her father was
a supervisor for General Motors. As a first

generation college student, Rebecca had received a scholarship
from the University of Michigan, where she majored
in mechanical engineering and then went to work for one of the
major auto companies as a design engineer.
While women design engineers were a distinct minority in the
auto business, there were enough that Rebecca
did not feel out of place. While she was a student, she had
taken classes which covered the history and legal issues
involving women’s advancement in what had historically been
male professions, and she had been happy to find that
the company was sensitive to women’s needs, and that sexual
harassment was not a problem.
After five years in Detroit, she was given the opportunity to
work on a joint venture with Opa GmbH, a
German subsidiary of her company. The new position would be
a promotion, working with a team of German and
American engineers designing a new sports car. The
opportunity was especially attractive, since she was considering
a career shift into management, taking advantage of the
company policy to pay for an MBA for engineers whom it felt
had the potential to be effective technical managers. The
program was highly competitive, and the two years of
international experience would significantly improve her
chances, especially given the increasing number of

multinational mergers in the auto industry, and the need to work
with international partners. Besides, she had never
been outside of the United States except for visits to company
plants in Ontario.
Living in Germany had turned out to be a stressful experience.
While everyone at work spoke English, it was
difficult to get through the small problems of day to day life,
since store clerks, plumbers, and most of the other
people with whom she dealt spoke little if any English, and she
spoke no German. Neighbors and other people with
whom she dealt seemed often to be impolite, and life seemed to
be governed by rules of which she was not aware. For
example, one of her neighbors scolded her (in very poor
English) the first week for the way she put out her trash –
one was required to separate out glass, paper, and cans, and
dispose of them separately. Outside of CNN, there was
little English available on television, only one English
newspaper was available, and bookstores had a very limited
selection in English. Fortunately most road signs were in the
form of symbols, so at least driving was possible.
Another stressful aspect of living in Germany was the lack of
respect for modesty. People sunbathed nude in
parks; television and print advertisements featured nudity as
well. Worst, because it impacted her work environment,

was that unlike the United States, co-workers flaunted pictures
of nude women at work – it was like a return to the bad
old days of the 1950’s where women couldn’t enter male
workplaces without being regarded as sexual objects.
The problem was not that her co-workers put up nude pictures,
but rather that they brought objectionable
material to work and read it in front of her. The most popular
newspaper, Bild, had a picture of a nude woman on the
front page every day (and probably worse inside). Other
newspapers were no better; one or her colleagues even
commented that he preferred to read Express because the nude
pictures were of local girls. The final straw came when
she went into her supervisor’s office for a meeting and saw a
copy of the magazine Der Spiegel on his desk, the cover
R
2, Number 3
66
of which reproduced a French painting of a topless woman
waving a flag. At that point, she made an appointment
with Human Resources to complain formally about having to
work in an offensive work environment.

At her meeting with Bertina Knies in Human Resources,
Rebecca cited the American company policy on
sexual harassment (reproduced below). She pointed out that it
clearly prohibited “posters, cartoons, pictures, or
drawings” which had the “effect of interfering with an
individual's work performance or creating an intimidating,
hostile, or offensive work environment.” Being exposed to
pictures such as those in Bild or Der Spiegel in her
opinion clearly met the definition of sexual harassment.
To Rebecca’s dismay, Bertina was not at all supportive. She
pointed out that in Germany unwelcome sexual
advances or requests for favors were clearly unacceptable, but
that Rebecca should not expect Germans to cater to her
Puritan hang-ups. Bild, which Bertina personally deplored for
its right-wing Christian politics, was after all, the most
popular newspaper in Germany, and as Rebecca had observed,
its competitors also featured nudity. The painting on
the cover of Der Spiegel was one of the best known symbols of
liberty in the world; no one who knew anything about
art could consider it obscene. And basically, the Germans did
not consider nudity offensive, and so long as no one
asked Rebecca to do anything which she found unacceptable,
she should not try to impose her values on others. For

example, Turkish employees sometimes objected to working
with women who did not wear head coverings or who
wore short sleeves; imposing their religious strictures on those
who did not share them would be a violation of
freedom of thought and religion. Rebecca’s request was no
different than requiring women to wear headscarves so as
not to offend Islamic immigrants. As an immigrant to Germany,
Rebecca could dress however she liked, but should
not expect Germans to adapt her old-fashioned values.
Clearly there was no point pursuing the issue with Human
Resources in Leverkusen. Equally clearly,
American courts long ago decided that depiction of nudity in
the workplace (or anyplace else outside the home) was
unacceptable. An appeal to Human Resources in Detroit would
be the next step. However, before doing that,
Rebecca had to consider both the chances of success, and the
possible impact on her future.
Excerpts from company policy on sexual harassment
Sexual harassment is a form of misconduct that undermines the
integrity of the employment relationship. All
employees have the right to work in an environment free from
all forms of discrimination and conduct which can
be considered harassing, coercive, or disruptive, including
sexual harassment. Anyone engaging in harassing

conduct will be subject to discipline, ranging from a warning to
termination.
Sexual harassment is defined as any unwanted physical, verbal
or visual sexual advances,requests for sexual
favors, and other sexually oriented conduct which is offensive
or objectionable to the recipient, including, but not
limited to: epithets, derogatory or suggestive comments, slurs or
gestures and offensive posters, cartoons, pictures,
or drawings.
When is conduct unwelcome or harassing? Unwelcome sexual
advances (either verbal or physical), requests for
favors and other verbal or physical conduct of a sexual nature
constitute sexual harassment when:
term or condition of employment (e.g.,
promotion, training, timekeeping or overtime assignments)
for making employment decisions (hiring,
promotion, termination)
individual's work performance or creating an
intimidating, hostile, or offensive work environment

2, Number 3
67
QUESTIONS
1. In a US work environment, would reading a newspaper or
magazine with nude pictures on the cover be
sexual harassment?
2. If so, would it be sexual harassment in Germany when
working for the same company? Why or why not?
3. Frau Knies argues that Rebecca’s request to prohibit Germans
from reading newspapers which offend her is
no different than prohibiting German women from dressing in
ways which offend Islamic immigrants
working in the plant. Is this a valid argument? Why or why
not?
4. What should Rebecca do?
5. What should Rebecca’s company have done to prevent this
problem?
Lab 9
Due Thursday, April 30 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
Reduction is Ubiquitous
● Calling Jen
Call Jen
Get hold of Jim
● Crisis detection via pizza orders
Show national crisis exists
Show spike in pizza orders at Pentagon
● Fixing dinner
Fix dinner
Fix entrée Fix salad Fix dessert

Suppose L₁ is reducible to L₂. Which of these statements are
true?
L₂ ∈ D → L₁ ∈ D
L₂ ∉ D → L₁ ∉ D
L₁ ∈ D → L₂ ∈ D
L₁ ∉ D → L₂ ∉ D
It may help to think about languages representing events from
everyday life, such as L₁ = Decide where the museum is L₂ =
Decide if the map app is working
A reduction R from L1 to L2 is one or more Turing
machines such that:
If there exists a Turing machine Oracle that decides (or
semidecides) L2, then the Turing machines in R can be
composed with Oracle to build a deciding (or a
semideciding) Turing machine for L1.
P £ P¢ means that P is reducible to P¢.
1. Using Reduction to prove L ∉ D
(R is a reduction from L1 to L2) Ù (L2 is in D) ® (L1 is in D)
If (L1 is in D) is false, then at least one of the two antecedents
of that implication must be false (invoking modus tollens). So:
If (R is a reduction from L1 to L2) is true,
then (L2 is in D) must be false.
1. Choose a language L1:

● that is already known not to be in D, and
● that can be reduced to L2.
2. Define the reduction R.
3. Describe the composition C of R with Oracle (the
hypothetical TM that can decide L2.)
4. Show that C does correctly decide L1 iff Oracle exists. We
do this by showing:
● R can be implemented by Turing machines,
● C is correct:
● If x Î L1, then C(x) accepts, and
● If x Ï L1, then C(x) rejects.
Showing that L2 is not in D:
H (known not to be in D) H in D But H not in D
R
L (a new language whose if L in D So L not in D
decidability we are
trying to determine)
modus tollens
Mapping Reductions
L1 is mapping reducible to L2 (L1 £M L2) iff there exists
some computable function f such that:

"xÎS* (x Î L1 « f(x) Î L2).
To decide whether x is in L1, we transform it, using f,
into a new object and ask whether that object is in L2.
Mapping reductions change a membership question about L1
into a membership question about L2.
A Block Diagram of C
The Oracle will accept L(M#) iff Hℇ ∈ D.
A clear declaration of the reduction “from” and “to” languages.
A clear description of R.
If R is doing anything nontrivial, argue that it can be
implemented as a TM.
Note that machine diagrams are not necessary or even sufficient
in these proofs. Use them as thought devices, where needed.
Run through the logic that demonstrates how the “from”
language is being decided by the composition of R and Oracle.
You must do both accepting and rejecting cases.
Declare that the reduction proves that your “to” language is not
in D.
Important Elements in a Reduction Proof
The right way to use reduction to show that L2 is not in D:
1. Given that L1 is not in D, L1
2. Reduce L1 to L2, i.e., show how to solve L1
(the known one) in terms of L2 (the unknown one) L2
Doing it wrong by reducing L2 (the unknown one to L1):

If there exists a machine M1 that solves H, then we could build
a
machine that solves L2 as follows:
1. Return (M1(<M, e>)).
This proves nothing. It’s an argument of the form:
If False then … everything is true.
The Most Common Mistake:
Doing the Reduction Backwards
Suppose that there are four languages W, X, Y, and Z. Each of
the languages may or may not be in SD. However, we know the
following about them:
• W ≤M X (There is a mapping reduction from W to X.)
• X ≤M Y (There is a mapping reduction from X to Y.)
Z ≤M Y There is a mapping reduction from Z to Y.
If W ∈ SD/D, is it possible that Y ∈ D?
W ≤M X; X ≤M Y; Z ≤M Y, and W ∈ SD/D.
Is it possible that Y ∈ D?
No, because the ≤M relationship is transitive, so W ≤M Y.
So if Y ∈ D, then W ∈ D.
Is it possible that W ∉ D and Z ∉ SD?
W ≤M X; X ≤M Y; Z ≤M Y.
Is it possible that W ∉ D and Z ∉ SD.

Yes. If Y ∉ SD, then it’s possible that none of the others are
either. On the other hand, if Y∈ D, then all the others must be
too.
Z ≤M Y There is a mapping reduction from Z to Y.
Is it true that if Y∈ D then ¬Z ∈ D?
W ≤M X; X ≤M Y; Z ≤M Y; Y ∈ D.
Is it always true that ¬Z ∈ D?
Yes. if Y ∈ D, then all the others are too, and D is closed
under compliment.
Theorem: He = {<M> : TM M halts on e} is not in D.
Proof: by reduction from H:
H = {<M, w> : TM M halts on input string w}
R
(?Oracle) He {<M> : TM M halts on e}
R is a mapping reduction from H to He:
R(<M, w>) =
1. Construct <M#>, where M#(x) operates as follows:
1.1. Erase the tape.
1.2. Write w on the tape.
1.3. Run M on w.
2. Return <M#>.
He = {<M> : TM M halts on e}

R(<M, w>) =
1.3. Run M on w.
2. Return <M#>.
If Oracle exists, C = Oracle(R(<M, w>)) decides H:
● C is correct: M# ignores its own input. It halts on
everything or
nothing. So:
● <M, w> Î H: M halts on w, so M# halts on everything.
In
particular, it halts on e. Oracle accepts M#.
● <M, w> Ï H: M does not halt on w, so M# halts on
nothing and
thus not on e. Oracle rejects M#.
Proof, Continued
R can be implemented as a Turing machine.
C is correct.
So, if Oracle exists:
C = Oracle(R(<M, w>)) decides H.
But no machine to decide H can exist.
So neither does Oracle.
Conclusion

If we could decide whether M halts on the specific string e, we
could solve the more general problem of deciding whether M
halts on an arbitrary input.
Clearly, the other way around is true: If we could solve H we
could decide whether M halts on any one particular string.
But doing a reduction in that direction would tell us nothing
about whether He was decidable.
The significant thing that we just saw in this proof is that there
also exists a reduction in the direction that does tell us that He
is
not decidable.
This Result is Somewhat Surprising
R
(Oracle) He {<M> : TM M halts on e}
R is a reduction from H to He:
R(<M, w>) =
1.1. Erase the tape. x is our name for the contents of M#'s
tape
1.3. Run M on w.
2. Return <M#>.
● Oracle (the hypothetical machine that could decide He).
● R (the machine that builds M#. Actually exists).

● C (the composition of R with Oracle).
● M# (the machine we will pass as input to Oracle). Note that
we never run it. Think of the Oracle as like a source code
analyzer.
● M (the machine whose membership in H we are interested in
determining;
thus also an input to R along with w)
Note that x is the input to M#, and w is the input to M. Do not
confuse them!
To prove He ∉ D we consider 5 distinct TMs
R
(?Oracle) He {<M> : TM M halts on e}
H contains strings of the form:
(q00,a00,q01,a10,¬),(q00,a00,q01,a10,®),…,aaa
where aaa is one example of w ∈ ∑*.
He contains strings of the form:
(q00,a00,q01,a10,¬),(q00,a00,q01,a10,®),…
The language on which some M halts contains strings of some
arbitrary form, for example,
(letting S = {a, b}): aaa
How Many Languages Are We Dealing With?
Recall that a mapping reduction from L1 to L2 is a computable
function f where:

"xÎS* (x Î L1 « ƒ(x) Î L2)
The function ƒ transforms a membership question in L1 into
a membership question in L2.
When we use a mapping reduction, we return:
Oracle(f(x))
Note that Rich uses R as the name for ƒ in her reduction proofs.
Sometimes we need a more general ability to use Oracle
as a subroutine and then to do other computations after it
returns.
Sometimes Mapping Reducibility Isn’t Right
H = {< M, w> : TM M halts on input string w}
R
(?Oracle) L2 = {<M> : M accepts no even length strings}
R(<M, w>) =
1. Construct the description <M#>, where M#(x) operates as
follows:
1.3. Run M on w.
1.4. Accept.
2. Return <M#>.
If Oracle exists, then C = Oracle(R(<M, w>)) decides H:
● C is correct: M# ignores its own input. It accepts
everything or nothing,
depending on whether it makes it to step 1.4. So:
● <M, w> Î H: M halts on w. Oracle:

● <M, w> Ï H: M does not halt on w. Oracle:
Does C = Oracle(R(<M#>)) work in this proof that L₂ ∉ D?
H = {< M, w> : TM M halts on input string w}
R
(?Oracle) L2 = {<M> : M accepts no even length strings}
R(<M, w>) =
1. Construct the description <M#>, where M#(x) operates
as follows:
1.3. Run M on w.
1.4. Accept.
2. Return <M#>.
If Oracle exists, then C = ØOracle(R(<M, w>)) decides H:
● R and Ø can be implemented as Turing machines.
● C is correct:
● <M, w> Î H: M halts on w. M# accepts everything,
including some
even length strings. Oracle rejects so C accepts.
● <M, w> Ï H: M does not halt on w. M# gets stuck. So it
accepts
nothing, so no even length strings. Oracle accepts. So
C rejects.
But no machine to decide H can exist, so neither does Oracle.
It won't work without inverting the decision of the Oracle

We show that A is not in D by reduction from H.
R
(?Oracle) A = {<M, w > : w Î L(M) }
R(<M, w>) =
1. Construct the description <M#>:
1.3. Run M on w.
1.4. Accept
2. Return <M#, w>.
If an Oracle to decide A exists, then C = Oracle(R(<M, w>))
decides H:
● R can be implemented as a Turing machine.
● C is correct: M# accepts everything or nothing. So:
● <M, w> Î H: M halts on w, so M# accepts every possible
input x. In particular, it accepts x = w. So Oracle accepts <M#,
w>.
● <M, w > Ï H: M does not halt on w. M# gets stuck in
step 1.3 and so
accepts nothing. In particular, it does not accept x = w.
So Oracle rejects <M#, w>.
A = {<M, w> : w Î L(M)}
L = {<Ma, Mb> : e Î L(Ma) – L(Mb)}. That is, the strings in L
are pairs of TM string encodings, such that e 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( … ) 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>.
<M, w> Î ØH: L(M#1) - L(M#2) = …, and Oracle accepts
<M#1,M#2> because …
<M, w> Ï ØH: L(M#1) - L(M#2) = …, and Oracle rejects
<M#1,M#2> because …
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 Accept
Return <M#1, M#2>.
<M, w> Î ØH: L(M#1) - L(M#2) = ∑*, and Oracle accepts
<M#1,M#2> because ℇ ∈ ∑*.
<M, w> Ï ØH: L(M#1) - L(M#2) = Ø, and Oracle rejects
<M#1,M#2> because ℇ ∉ Ø.
The Problem ViewThe Language ViewStatusDoes TM M have
an even number of states?{<M> : M has an even number of

states}DDoes TM M halt on w?H = {<M, w> : M
halts on w}SD/DDoes TM M halt on the empty tape?He =
{<M> : M halts on e}SD/DIs there any string on which TM M
halts?HANY = {<M> : there exists at least one string on which
TM M halts }SD/DDoes TM M halt on all strings?HALL =
{<M> : M halts on S*}ØSD Does TM M accept w?A = {<M, w>
: M accepts w}SD/DDoes TM M accept e?Ae = {<M> : M
accepts e}SD/DIs there any string that TM M accepts?AANY
{<M> : there exists at least one string that TM M accepts
}SD/D
Reduction workshop 4/30 (10 extra credit points)
HW #3 due 2pm Monday May 4.
Midterm #3 Tuesday May 5
Lab #9 due Friday midnight
L = { <M,s> : s ∈ L(M) and |L(M)| % 2 = 0}. For example,
suppose that L(M) = {aa}.
Then <M,aa> ∉ L bc |L(M)| = 1, 1 % 2 = 1; If L(M) {a,aaa}
then <M,ℇ> ∉ L bc ℇ ∉
L(M), but <M,aaa> ∈ L. Prove that L ∉ L 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
could R return <M#>? R, as a mapping reduction functions,
transforms the question
of whether <M,w> is in H to the question of whether R(<M,w>)
is an element
of L.

Reduction Overview
Reduction is ubiquitous — solving one problem in terms of
another.
Break down cleaning your house into components
• take out garbage, etc.
Reframe on problem in terms of another.
And remember our 520 perspective — any problem can be
defined in terms
of language recognition
L₁ = Decide where the museum is
L₂ = Decide if your map app is working correctly
Suppose L₁ is reducible to L₂. Which of these statements are
true?
√ L₂ ∈ D → L₁ ∈ D if you can tell the app is working then can
figure out
where the museum is.
This is the relationship we will use in 520, with a twist, to show
that some
new language Lᵩ ∉ D. We reason as follows:
1. Lᵩ ∈ D → H ∈ D #We demonstrate the Lᵩ is reducible to
H.
2. H ∉ D #We know this already
Reduction raw lecture notes Page 1

∴ Lᵩ ∉ D #Applying the Modus Tollens logic rule to
steps 1 and 2.
L₂ ∉ D → L₁ ∉ D (False)
— if the map app remains a mystery, there could be some
other way to locate the museum
√ L₁ ∉ D → L₂ ∉ D if you're not sure where the museum is then
you can't decide if the map app is working.
L₁ ∈ D → L₂ ∈ D (False)
— You could have located the museum usng a paper map
Proving if H ∉ D then L ∉ D
(R is a reduction from H to L) ∧ (L ∈ D) → (H ∈ D)
If H ∉ D is false, then at least one of the two antecedents of
that implication
must be false (invoking modus tollens). So:
If (R is a reduction from H to L) is true, then (L ∉ D) must be
false.
Don't do reduction backwards by reducing L to H: if H ∈ D then
L ∈ D
This will always be vacuously true.

Mapping Reduction—a transitive relationship
L₁ is mapping reducible to L₂ (L₁ ≤M L₂) iff there exists some
computable
function f such that:
∀ x∈ Σ* (x ∈ L₁ iff ƒ(x) ∈ L₂).
Mapping reductions transform a membership question about L₁
into a
membership question about L₂
Suppose that there are four languages W, X, Y, and Z. Each of
the
languages may or may not be in SD. However, we know the
following about
them:
• Z ≤M Y There is a mapping reduction from Z to Y.
If W ∈ SD/D, is it possible that Y ∈ D?
No, because the ≤M relationship is transitive, so W ≤M Y.
So if Y ∈ D, then W ∈ D
Is it possible that W ∉ D and Z ∉ SD.
Yes. If Y ∉ SD, then it’s possible that none of the others are
either. On the
other hand, if Y∈ D, then all the others must be too.
Y ∈ D. Is it always true that ¬Z ∈ D?
Yes. if Y ∈ D, then all the others are too, and D is closed
under compliment.

Anatomy of an H Reduction
L = { <M> : L(M) is a regular language}
For example, if L(M)= { w : w = aⁿbⁿ}, then <M> ∉ L, since
aⁿbⁿ is not a regular
language.
The textbook uses this mapping reduction from H to prove that
L ∉ D:
R(<M, w>) =
1. Define M#
1.1. If x ∈ aⁿbⁿ then accept, else:
1.2. Erase the input tape.
1.3. Write w on the input tape.
1.4. Simulate M on w.
1.5. Accept
2. Return <M#>.
To complete the proof, a good place to start is work out
equivalent TMs
for M# depending on whether or not <M,w> ∈ H (that is,
whether or not M halts
w),
and then to write characteristic functions for L(M#) for each
case.
• If <M,w> ∈ H, then M# is equivalent to M#H:
1.1. If x ∈ aⁿbⁿ then accept, else:
1.5. Accept
Thus M#H would accept aⁿbⁿ at step 1.1, and all other input at

step 1.5, so
L(M#H) = ∑*
• If <M,w> ∉ H, then M# is equivalent to M#¬H:
1.1 If x ∈ aⁿbⁿ then accept
Thus M#¬H would accept aⁿbⁿ at step 1.1 and nothing else, so
L(M#¬H) = aⁿbⁿ.
The next step is to consider whether a hypothetical TM
MORACLE that could decide L
would accept or reject <M#H> and <M#¬H>:
• MORACLE would accept <M#H> since L(M#H) = ∑* is a
regular language.
• MORACLE would reject <M#¬H> since L(M#¬H) = aⁿbⁿ, not
a regular language.
Thus if MORACLE could decide L(M#), in could also decide H.
But no TM could decide
H, so MORACLE could not possibly exist.
Dovetailing Exampe SD/D Proof # 1
L = {<M> : M rejects at least two even length strings}. Prove
that L ∈ SD/D.
Prove L is in SD. Verbal desctription:
Enumerate the strings for ∑* and then run M on the strings to
avoid the twin
problems of being infinte, and that M may not halt on

processing some string.
As soon as M rejects two even length strings, accept.
Pseudocode answer:
//Use tDovetailing to find two strings that tm rejects
accept/reject rejectsAtLeastTwotrings(TM tm) {
int acceptCount = 0;
SetofStrings SigmaStar = ∑*;
SetofCandidateStrings candidates = Ø;
while (true) {
String anotherStr = SigmaStar.pickdEvenLengthElement();
SigmaStar.remove(anotherStr);
candidates.add(anotherStr);
for (String str : candidates) {
// nextConfig remembers the current state and r/w head
// position, and returns the next state after executing
// one step in the computation on str.
//
State state = tm.nextConfig(str);
if (state == reject) {
if (++rejecttCount == 2)
return accept
} // if str rejected by tm
else if (state == accept) {
//remove str from further consideration
//
candidates.remove(str);
} //else if str accepted by tm
} //for all current candidate strings
} // while considering possible candidate strings
//but of course L ∉ D, so rejectsAtLeastTwotrings will

// never exectute this return statement
return false;
}
Complete the proof below that uses reduction to show that L ∉
D.
R(<M, w>) is a reduction from H to L, defining M# as follows:
1. Erase the tape.
2. Write w on the tape.
3. Run M on w.
4. Reject.
If Oracle exists and decides L, then Oracle(M#) decides H:
• <M,w> ∈ H: Oracle accepts (M#) because …
M# rejects ∑*, and |∑*| contains an infinite number of even
length strings
assuming ∑≠ Ø. M# rejects ∑* because it will always reach
step 4 after M halts
on w. Note that because M#'s input tape is erased in step 1, it
will reject
regardless of the input.
• <M,w> ∉ H: Oracle rejects (M#) because …
M# rejects Ø, and |Ø| = 0 < 2. M# rejects no even length strings
because if M
does not halt in w, M# will never reach step 4.

Journal of Business Case Studies – Third Quarter 2006 Volume .docx

Journal of Business Case Studies – Third Quarter 2006 Volume .docx

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Journal of Business Case Studies – Third Quarter 2006 Volume .docx