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.
Related
I am just curious on how do I determine whether a simplified boolean expression is in a SOP form or POS form.
for example this question:
Question
the answer to this expression is : NOT B.D/ ⌝B.D
and this is in SOP form
Can anyone explain why?
I think this should be a 'philosophical' argument. ⌝B.D is the special case where the number of elements to be summed up becomes one.
You can think of ⌝B.D = ⌝B.D + ⌝B.B + ⌝D.D + 0.(anything) which makes it an SOP.
Terminology:
First the theory, you can further study on Wikipedia (DNF, CNF):
Sum of products = DNF (Disjunctive normal form) = disjunction (+) of conjunctions (·) ~ "the disjunction is not inside any bracket, but only
as the root operator(s)".
Product of sums = CNF (Conjunctive normal form) = conjuction of disjunctions ~ "the conjunction is not inside a bracket, but only
as the root operator(s)".
Full/Complete CNF/DNF = the terms (products/sums) contains all given variables in either direct or negated form; the terms are then maxterms/minterms.
Finding the right circles:
You can see, that the four circles in the Karnaugh map correspond to the four products in the original function in the same order (top to bottom, left to right).
Given function as a SOP:
The function is now in a form of sum of products, because you can literally see, that there are four products.
It is also in the form of sum of maxterms, because the four parts contain all variables in their direct or negated form.
f(a,b,c,d) = ¬a·¬b·¬c·d + ¬a·¬b·c·d + a·¬b·c·d + a·¬b·¬c·d
For example the first term: ¬a·¬b·¬c·d ~ if the variables a, b and c are logical zeros and only d is true, then the function's output is logical 1.
Minimized function as a SOP:
You can see, that the maxterms can be grouped and that creates the minimal sum of product: f(a,b,c,d) = ¬b·d, because all cells, where b is a logical 0 and d is a logical 1 are included.
The minimized function is indeed a SOP/DNF, because it certainly does contain only one product — the ¬b·d — and there is no + (disjunction) operator inside that product.
Minimized function as a POS:
The surprise might come when you realize, that circling and writing the function as a product of sums yields the same minimal form: f(a,b,c,d) = (¬b)·(d), because there are exactly two terms: ¬b (orange circle) and d (red circle).
Both are sums with only one operand. Because of that the minimized function is a product of sum.
Conclusion:
The minimized function f(a,b,c,d) = ¬b·d is both a SOP and a POS. You can check the correct solution by using wolframalpha.com.
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.
There's a certain over-verbosity that I have to engage in when writing certain Boolean expressions, at least with all the languages I've used, and I was wondering if there were any languages that let you write more concisely?
The way it goes is like this:
I want to find out if I have a Thing that can be either A, B, C, or D.
And I'd like to see if Thing is an A or a B.
The logical way for me to express this is
//1: true if Thing is an A or a B
Thing == (A || B)
Yet all the languages I know expect it to be written as
//2: true if Thing is an A or a B
Thing == A || Thing == B
Are there any languages that support 1? It doesn't seem problematic to me, unless Thing is a Boolean.
Yes. Icon does.
As a simple example, here is how to get the sum of all numbers less than 1000 that are divisble by three or five (the first problem of Project Euler).
procedure main ()
local result
local n
result := 0
every n := 1 to 999 do
if n % (3 | 5) == 0 then
result +:= n
write (result)
end
Note the n % (3 | 5) == 0 expression. I'm a bit fuzzy on the precise semantics, but in Icon, the concept of booleans is not like other languages. Every expression is a generator and it may pass (generating a value) or fail. When used in an if expression, a generator will continue to iterate until it passes or exhausts itself. In this case, n % (3 | 5) == 0 is a generator which uses another generator (3 | 5) to test if n is divisible by 3 or 5. (To be entirely technical, this isn't even syntactic sugar.)
Likewise, in Python (which was influenced by Icon) you can use the in statement to test for equality on multiple elements. It's a little weaker than Icon though (as in, you could not translate the modulo comparison above directly). In your case, you would write Thing in (A, B), which translates exactly to what you want.
There are other ways to express that condition without trying to add any magic to the conditional operators.
In Ruby, for example:
$> thing = "A"
=> "A"
$> ["A","B"].include? thing
=> true
I know you are looking for answers that have the functionality built into the language, but here are two other means that I find work better as they solve more problems and have been in use for many decades.
Have you considered using a preprocessor?
Also languages like Lisp have macros which is part of the language.
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.
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.