I'm trying to understand this algorithm the DFA minimization algorithm at http://www.cs.umd.edu/class/fall2009/cmsc330/lectures/discussion2.pdf where it says:
while until there is no change in the table contents:
For each pair of states (p,q) and each character a in the alphabet:
if Distinct(p,q) is empty and Distinct(δ(p,a), δ(q,a)) is not empty:
set distinct(p,q) to be x
The bit I don't understand is "Distinct(δ(p,a), δ(q,a))" I think I understand the transition function where δ(p,a) = whatever state is reached from p with input a. but with the following DFA:
http://i.stack.imgur.com/arZ8O.png
resulting in this table:
imgur.com/Vg38ZDN.png
shouldn't (c,b) also be marked as an x since distinct(δ(b,0), δ(c,0)) is not empty (d) ?
Distinct(δ(p,a), δ(q,a)) will only be non-empty if δ(p,a) and δ(q,a) are distinct. In your example, δ(b,0) and δ(c,0) are both d. Distinct(d, d) is empty since it doesn't make sense for d to be distinct with itself. Since Distinct(d, d) is empty, we don't mark Distinct(c, b).
In general, Distinct(p, p) where p is a state will always be empty. Better yet, we don't consider it because it doesn't make sense.
Related
Design a FA to accept L, where L = {Strings in which ‘a’ always appears tripled} over the input set Σ = {a, b}.
This is my solution (A is start state):
From what I understand the question doesn't say it should always contain "a",
it should also accept strings like {b,bb,bbb}. Is this correct?
Your solution appears right to me! It really depends on the interpretation of the question. I also interpreted that a's could also be separated, as long as they were always tripled.
abbabbbab should be accepted as a's are tripled always, come in threes.
You should make sure of that! I also did this FA in the case above, but my generic understanding of the question would be your interpretation which looks fine!
I am new to Z notation,
Lets say I have a function f defined as X |--> Y ,
where X is string and Y is number.
How can I get highest Y value in this function? Does 'loop' exist in formal method so I can solve it using loop?
I know there is recursion in Z notation, but based on the material provided, I only found it apply in multiset or bag, can it apply in function?
Any extra reference application of 'loop' or recursion application will be appreciated. Sorry for my English.
You can just use the predefined function max that takes a set of integers as input and returns the maximum number. The input values here are the range (the set of all values) of the function:
max(ran(f))
Please note that the maximum is not defined for empty sets.
Regarding your question about recursion or loops: You can actually define a function recursively but I think your question aims more at a way to compute something. This is not easily expressed in Z and this is IMO a good thing because it is used for specifications and it is not a programming language. Even if there wouldn't be a max or ran function, you could still specify the number m you are looking for by:
\exists s:String # (s,m):f /\
\forall s2:String, i2:Z # (s2,i2):f ==> i2 <= m
("m is a value of f, belonging to an s and all other values i2 of f are smaller or equal")
After getting used to the style it is usually far better to understand than any programming language (except your are trying to describe an algorithm itself and not its expected outcome).#
Just for reference: An example of a recursive definition (let's call it rmax) for the maximum would consist of a base case:
\forall e:Z # rmax({e}) = e
and a recursive case:
\forall e:Z; S:\pow(Z) #
S \noteq {} \land
rmax({e} \cup S) = \IF e > rmax(S) \THEN e \ELSE rmax(S)
But note that this is still not a "computation rule" of rmax because e in the second rule can be an arbitrary element of S. In more complex scenarios it might even be not obvious that the defined relation is a function at all because depending on the chosen elements different results could be computed.
Consider the DFA :
What will be δ(A,01) equal to ?
options:
A) {D}
B) {C,D}
C) {B,C,D}
D) {A,B,C,D}
The correct answer is option B) but I don't get how. Please some one explain me the steps to solve it and also in general how do we solve sit for any DFA and any transition?
Thanks.
B) Option is not Correct answer! for this transition graph.
In Transition Graph(TG) symbol ε means NULL-move (ε-move). There is two NULL-moves in TG.
One: (A) --- `ε` ---->(B)
Second: (A) --- `ε` ---->(C)
A ε-move means without consume any symbol you can change state. In your diagram from A to B, or A to C.
What will be δ(A,01) equal to ?
Question asked "what is path from state A if input is 01". (as I understand because there is only one final state)
01 can be processed in either of two ways.
(A) -- ε --->(B) -- 0 --> (B) -- 1 --> (D)
(A) -- ε --->(C) -- 0 --> (B) -- 1 --> (D)
Also, there is no other way to process string 01 even if you don't wants to reach final state.
[ANSWER]
So there is misprinting in question (or either you did).
You can learn how to remove NULL-move from transition graph. Also HOW TO WRITE REGULAR EXPRESSION FOR A DFA
If you remove null-moves from TG you will get three ways to accept 01.
EQUIVALENT TRANSITION GRAPH WITHOUT NULL-MOVE
Note there is three start-states in Graph.
Three ways:
{ABD}
{CBD}
{BBD}
In all option state-(B) has to be come.
Also, you have written Consider the DFA : is wrong. The TG is not deterministic because there is there is non-deterministic move δ(A,ε) and next states are either B or C.
Given:
I have no idea what the accepted language is.
From looking at it you can get several end results:
1.) bb
2.) ab(a,b)
3.) bbab(a, b)
4.) bbaaa
How to write regular expression for a DFA
In any automata, the purpose of state is like memory element. A state stores some information in automate like ON-OFF fan switch.
A Deterministic-Finite-Automata(DFA) called finite automata because finite amount of memory present in the form of states. For any Regular Language(RL) a DFA is always possible.
Let's see what information stored in the DFA (refer my colorful figure).
(note: In my explanation any number means zero or more times and Λ is null symbol)
State-1: is START state and information stored in it is even number of a has been come. And ZERO b.
Regular Expression(RE) for this state is = (aa)*.
State-4: Odd number of a has been come. And ZERO b.
Regular Expression for this state is = (aa)*a.
Figure: a BLUE states = EVEN number of a, and RED states = ODD number of a has been come.
NOTICE: Once first b has been come, move can't back to state-1 and state-4.
State-5: comes after Yellow b. Yellow b means b after odd numbers of a.
Once you gets b after odd numbers of a(at state-5) every thing is acceptable because there is self a loop for (b,a) at state-5.
You can write for state-5 : Yellow-b followed-by any string of a, b that is = Yellow-b (a + b)*
State-6: Just to differentiate whether odd a or even.
State-2: comes after even a then b then any number of b. = (aa)* bb*
State-3: comes after state-2 then first a then there is a loop via state-6.
We can write for state-3 comes = state-2 a (aa)* = (aa)*bb* a (aa)*
Because in our DFA, we have three final states so language accepted by DFA is union (+ in RE) of three RL (or three RE).
So the language accepted by the DFA is corresponding to three accepting states-2,3,5, And we can write like:
State-2 + state-3 + state-5
(aa)*bb* + (aa)*bb* a (aa)* + Yellow-b (a + b)*
I forgot to explain how Yellow-b comes?
ANSWER: Yellow-b is a b after state-4 or state-3. And we can write like:
Yellow-b = ( state-4 + state-3 ) b = ( (aa)*a + (aa)*bb* a (aa)* ) b
[ANSWER]
(aa)*bb* + (aa)*bb* a (aa)* + ( (aa)*a + (aa)*bb* a (aa)* ) b (a + b)*
English Description of Language: DFA accepts union of three languages
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's,
EVEN NUMBERs OF a's, FOLLOWED BY ONE OR MORE b's, FOLLOWED BY ODD NUMBERs OF a's.
A PREFIX STRING OF a AND b WITH ODD NUMBER OF a's, FOLLOWED BY b, FOLLOWED BY ANY STRING OF a AND b AND Λ.
English Description is complex but this the only way to describe the language. You can improve it by first convert given DFA into minimized DFA then write RE and description.
Also, there is a Derivative Method to find RE from a given Transition Graph using Arden's Theorem. I have explained here how to write a regular expression for a DFA using Arden's theorem. The transition graph must first be converted into a standard form without the null-move and single start state. But I prefer to learn Theory of computation by analysis instead of using the Mathematical derivation approach.
I guess this question isn't relevant anymore :) and it's probably better to guide you through it then just stating the answer, but I think I got a basic expression that covers it (it's probably minimizable), so i'll just write it down for future searchers
(aa)*b(b)* // for stoping at 2
U
(aa)*b(b)*a(aa)* // for stoping at 3
U
(aa)*b(b)*a(aa)*b((a)*(b)*)* // for stoping at 5 via 3
U
a(aa)*b((a)*(b)*)* // for stoping at 5 via 4
The examples (1 - 4) that you give there are not the language accepted by the DFA. They are merely strings that belong to the language that the DFA accepts. Therefore, they all fall in the same language.
If you want to figure out the regular expression that defines that DFA, you will need to do something called k-path induction, and you can read up on it here.
Is the language of all strings over the alphabet "a,b,c" with the same number of substrings "ab" & "ba" regular?
I believe the answer is NO, but it is hard to make a formal demonstration of it, even a NON formal demonstration.
Any ideas on how to approach this?
It's clearly not regular. How is an FA going to recognize (abc)^n c (cba)^n. Strings like this are in your language, right? The argument is a simple one based on the fact that there are infinitely many equivalence classes under the indistinguishability relation I_l.
The most common way to prove a language is NOT regular is using on of the Pumping Lemmas.
Using the lemma is a little tricky, since it has all those "exists" and so on. To prove a language L is not regular using the pumping lemma you have to prove that
for any integer p,
there is a word w in L of length n, with n>=p, such that
for all possible ways to decompose w as xyz, with len(xy) <= p and y non empty
there exists an i such that x(y^i)z (repeating the y bit i times) is NOT in L
whooo!
I'l l show how the proof looks for the "same number of as and bs" language. It should be straighfoward to convert to your case:
for any given p, we can make a word of length n = 2*p
a^p b^p (p a's followed by p b's)
any way you decompose this into xyz w/ |xy| <=p, y will only contain a's.
Thus, pumping the the y part will make the word have more as than bs,
thus NOT belonging to L.
If you need intuition on why this works, it follows from how you need to be able to count to arbritrarily large numbers to verify if a word belongs to one of these languages. However, Regular Languages are described by finite automata and no finite automata can represent the infinite ammount of states required to represent all the numbers. (The Wikipedia article should have a formal proof).
EDIT: It looks like you can't straight up use the pumping lemma in this particular case directly: if you always make y be one character long you can never make a word stop being accepted (aba becoming abbbba makes no difference and so on).
Just do the equivalence class approach suggested by Patrick87 - it will probably turn out to be cleaner than any of the dirty hacks you would need to do to make the pumping lemma applicable here.