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Homework answers / question archive / COMP 330 - Fall 2020 - Assignment 5 Due: 11:59pm Nov 21 st
COMP 330 - Fall 2020 - Assignment 5
Due: 11:59pm Nov 21 st.
General rules: In solving these questions you may consult books or other available notes, but you need to provide citations in that case. You can discuss high level ideas with each other, but each student must find and write her/his own solution. Copying solutions from any source, completely
or partially, allowing others to copy your work, will not be tolerated, and will be reported to the disciplinary office.
You should upload the pdf file (either typed, or a clear and readable scan) of your solution to mycourses.
A ≤T J and B ≤T J.
x is odd x is even
For every natural number x, if you start with x and iterate f, you obtain a sequence
x,f(x),f(f(x)),f(f(f(x))),....
Stop if you ever hit 1. For example, if x = 17, we get the sequence
It has been checked by computers that if we start with any number x < 1017, we will eventually hit 1 and terminate, but it is unknown whether this is true for every integer x. This is known
as Collatz’s conjecture[1]. Suppose that HALTTM were decidable, and R was a TM that was deciding HALTTM. Construct an Turing Machine M based on R such that it accepts ε if
Collatz’s conjecture is true (all numbers x hit 1 eventually), and rejects it if Collatz’s conjecture is false (there is some x that does not hit 1).
X = {hM,wi|M is a single-tape TM that never modifies the w part of the tape, and it accepts w }.
Is X decidable? Prove your answer.
where R is a decidable language. You need to prove that every language of this form is Turing recognizable, and that every Turing recognizable language can be described as above for some decidable language R.
8. (15 Points) Determine whether the following language is decidable or undecidable:
[1] It is interesting to view this in regard to Q4 in the previous assignment. It is fairly easy to construct a Turing Machine M with 100 states that given 1x as input, applies f recursively to this, and halts and accepts if we hit 1. Note that M is a decider if and only if Collatz’s conjecture is true, which currently is unknown.
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