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Homework 6
Solution
CSCI 2670
October 11, 2005
3.7
The problem is with step 2.
The Turing machine has to evaluate the polynomial
at
every
possible combination of integer values for
x
1
, x
2
, …, x
k
.
Since there are an
infinite number of possibilities, the machine will never get past this step and accept if the
polynomial has some integer root.
3.15 b) Assume we have two decidable languages A and B.
We want to show that A
°
B is
also decidable.
Let M
1
and M
2
be two deciders such that L(M
1
) = A and L(M
2
) = B.
Consider the following Turing machine:
C = “On input
w
1
Let
l
=
length(w)
2
For each
i
= 0, 1, 2, …,
l
3
Let
w
l
= the leftmost
i
symbols in
w
4
Let
w
r
= the rightmost
li
symbols in
w
5
Run M
1
on input
w
l
6
If it accepts
7
Run M
2
on input
w
r
8
If it accepts
accept
9
Next
i
10 reject
”
Claim:
C decides A
°
B.
This is clear since if
w
is in A
°
B, there is some
i
for which
w
l
is in
A and
w
r
is in B.
When the loop reaches that
i
, C will accept.
If
w
is not in A
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This note was uploaded on 02/07/2011 for the course CS 501 taught by Professor Sm during the Spring '11 term at Indiana.
 Spring '11
 sm

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