CS52 - Spring 2017 - Class 23
which of these languages are regular?
- language consisting of repetitions of either 01 or 10
- Regular: (01|10)*
- language of strings with an equal number of 0s and 1s
- Not regular
- language over a's and b's of alternating a's starting with an a, i.e. every other character in the string is an a
- Regular: a((a|b)a)*|a((a|b)a)*(a|b)
- language of all words that are palindromes, i.e. read the same forward and reversed
- Not regular
- similar to DFAs/NFAs
- they have states with transitions defined over the alphabet
- they have a starting state and one or more final/accepting states
- key differences:
- Can *write* to the input tape as well as read
- We have two alphabets
- input alphabet: just like DFAs and NFAs
- tape alphabet: includes the input alphabet PLUS possibly additional characters
- Can move right *and* left on the input tape
- The input tape is infinite!
- it has "blank" characters everywhere beyond the input (in both directions)
- Has both accept and reject states
- if it ever enters either it *immediately* accepts/rejects
look at the 0n1n example (found in
- transitions are labeled with three things:
1) the input character read
2) the output character to write
3) which way to move the head (L, R or S)
- basic idea of this example:
- put Xs over the 0s. For each X you put on a 0, go find the next 1 and put a Y over it
- if you get to the point where everything is X and Y (and no 0s or 1s) you're done!
- slightly more detailed
- read a 0: put an X down and move right
- keep reading 0s and Ys to the right until you find a 1
- put a Y on that 1 and move left
- keep reading Ys and 0s until you find an X
- move right
- if you get to a point where the first character is a Y
- read all the Ys and move right
- make sure that you get a blank (and not a 1 or a 0). If so, we're done!
- JFLAP doesn't have specific reject states for Turing machines
- if it ever is in a state and there is no transition to take, it rejects
- to make it clear (and to be consistent with other formulations) we will make one or more reject states without any outgoing transitions
- if we want to reject, we will move to one of these states
look at 0n1n.v2 example
- what does it do? how?
- Instead of using X and Y, we just cover up the end 0's and 1's with blanks
- Cover up the first 0 with a blank
- Search for the 1s
- Go to the end of the 1s
- Cover the last 1 with a blank
look at eq1s0s (found in
look at palindrome (found in
- For this class, we'll just focus on accepting or rejecting strings
- However, since Turing machines can write, they can also provide output
- Look at simple_add (in
- accepts strings of the form: number+number
- where number is represented in *unary*
- unary numbers:
1 = 1
2 = 11
3 = 111
4 = 1111
5 = 11111
- in addition to accepting strings of this form, it also calculates the result!
- e.g. 11+111 should give us 11111
- it gives the result back by putting it on the tape
- in JFLAP we can run it in "transducer" mode which will also show the output
- JFLAP outputs whatever from the current position (when the tape accepts) to the blank on the right
- Look at fancy_add (in
- accepts strings of the form number(+number)* where number is represented in unary
- the output is the correct equation, e.g.
- e.g., 111+1111 would output 111+1111=1111111
- Could we write a Turing machine that simulates the running of another Turing machine?
- take as input two things:
1) some representation of the Turing machine
2) some input to run on the Turing machine input (i.e. the input Turing machine from 1)
- would then "run" the Turing machine it was simulating on the input
- this is called a "universal Turing machine"
- Consider a TM H that does the following:
- accept if M finishes (accepts or rejects) with input w
- reject if M never finishes with input w, i.e. runs forever
- this is called the halting problem
- this program comes up a lot in CS
- you run your program
- it doesn't finish
- will it or is it stuck in an infinite loop?
- for particular types of problems we can definitely answer this problem
- can we in general, though?
- Could we ever write this Turing machine?
To prove that we *cannot* we'll do a proof by contradiction
- We assume that we can, i.e. that we could construct H above
- And show that this results in a contradiction, meaning it couldn't be the case
- Let's construct the following Turing machine
Run H(M, [M]) where [M] is the string representation of M
- if it accepts (i.e. M(M) will finish), then loop infinitely
- if it rejects (i.e. M(M) will loop forever), then stop
- What happens when we run D(D)?
- Run H(D, [D])
- if it accepts, that implies that D(D) finishes
- however, if H(D, [D]) accepts, then we know by the definition of D that D(D) loops infinitely
- if it rejects, that implies that D(D) never finishes
- however, if H(D, [D]) rejects, then we know that D(D) halts
- We've reached a contradiction!
- If H says D finishes, then it loops infinitely
- If H says D loops, then it halts