Consider there is a minimized DFA that accepts the language L. The problem is to find the minimum number of states in its complement.
Now if I take the complement of this DFA i;e if I make the non-final states as final and final states as non-final, do I also need to worry about minimizing this complemented DFA?
DFA - Deterministic Finite Automata
Lets start with that is accepted by a minimum DFA . Then we can retrieve the DFA of the complement (as you have mentioned). So let's derive a DFA which accepts which has the same amount of states as .
Now Lets assume that is not a minimum DFA of
we should be able to reduce the number of states in it further to get a DFA . But getting the complement of should give us a new DFA which accepts which is but has less states than making it not a minimum DFA of .
So our assumption that not being the miniumum of is wrong.
Related
I just started learning Theory of Computation this semester and a bit confused by the phrase "DFA for a language". If it is asked to construct a DFA for some collection of binary strings L, does it mean to find DFA M with L(M)=L or just $L(M)\supset L$?
Most compiler/theory courses tend to have different styles surrounding teaching definitions of deterministic finite automata and formal languages, but I'll try to make this description as agnostic as possible.
The phrase "DFA for a language" loosely means: a DFA which accepts every word in the language and rejects every word not in the language.
The way I was taught DFAs is to have final/accepting states and regular states which removes the necessity for an implicit error state.
This means that a DFA accepts a word if the state it is in at the end of input is accepting and it rejects the word if the state is not accepting.
Ex:
Let's define L as the language which contains an even number of 1s. These will be binary strings so the symbols are just 0 and 1.
00, 110, 111, 1111, etc are examples of words in this language. Notice that the empty string is in this language.
We can have two states in our DFA. The starting state, let's call it even 1s, is also an accepting state because 0 ones is even. The other state is odd 1s, this is not accepting.
As for transitions, when even 1s receives a 1, it transitions to odd 1s. And when odd 1s receives a 1, it transitions to even 1s.
Now, the number of 0s doesn't matter, so in either state, it transitions to itself.
Apologies for the double arrow, this website is great but I couldn't figure out how to separate the transitions between even 1s and odd 1s
Deterministic Finite Automaton (DFA)
In DFA, for each input symbol, one can determine the state to which the machine will move. Hence, it is called Deterministic Automaton. As it has a finite number of states, the machine is called a Deterministic Finite Machine or Deterministic Finite Automaton.
Formal Definition of a DFA
A DFA can be represented by a 5-tuple (Q, ∑, δ, q0, F) where −
-> Q is a finite set of states.
-> ∑ is a finite set of symbols called the alphabet.
-> δ is the transition function where δ: Q × ∑ → Q
-> q0 is the initial state from where any input is processed (q0 ∈ Q).
-> F is a set of final state/states of Q (F ⊆ Q).
Write your question in precise way. here DFA for a language means that you need to construct machine for particular language only not it's subset or superset. construct DFA maachine for which L(M)= L.
I am to construct a DFA from the intersection of two simpler DFAs. The first simpler DFA recognizes languages of all strings that have at least three 0s, and the second simpler language DFA recognizes languages of strings of at most two 1s. The alphabet is (0,1). I'm not sure how to construct a larger DFA combining the two. Thanks!
Here's a general idea:
The most straightforward way to do this is to have different paths for counting your 0s that are based on the number of 1s you've seen, such that they are "parallel" to each other. Move from one layer of the path to the next any time you see a 1, and then move from the last layer to a trap state if you see a third 1. Depending on the exact nature of the assignment you might be able to condense this, but once you have a basic layout you can determine that. Typically you can combine states from the first DFA with states in the second DFA to produce a smaller end result.
Here's a more mathematical explanation:
Constructing automata for the
intersection operation.
Assume we are
given two DFA M1 = (S1, q(1) 0 , T1,
F1) and M2 = (S2, q(2) 0 , T2, F2).
These two DFA recognize languages L1 =
L(M1) and L2 = L(M2). We want to
design a DFA M= (S, q0, T, F) that
recognizes the intersection L1 ∩L2. We
use the idea of constructing the DFA
for the union of languages. Given an
input w, we run M1 and M2 on w
simultaneously as we explained for the
union operation. Once we finish the
runs of M1 and of M2 on w, we look at
the resulting end states of these two
runs. If both end states are accepting
then we accept w, otherwise we reject
w.
When constructing the new transition function, the easy way to think of it is by using pairs of states. For example, consider the following DFAs:
Now, we can start combining these by traversing both DFAs at the same time. For example, both start at state 1. Now what happens if we see an a as input? Well, DFA1 will go from 1->2, and DFA2 will go from 1->3. When combining, then, we can say that the intersection will go from state "1,1" (both DFAs are in state 1) to state "2,3". State 2 is an accept state in DFA1 and state 3 is an accept state in DFA2, so state "2,3" is an accept state in our new DFA3. We can repeat this for all states/transitions and end up with:
Does that make sense?
Reference: Images found in this assignment from Cornell University.
The simplest way would be using the 2DFA model: from the end state of the first DFA(the one testing for at least 3 zeros) jump to the start state of the second one, and reverse to the beginning of the input. Then let the second DFA test the string.
I'm thinking so, because the upper bound would be the 2^n, and given that these are both finite machines, the intersection for both the n-state NFA and the DFA with 2^n or less states will be valid.
Am I wrong here?
You're right. 2^n is an upper limit, so the generated DFA can't have more states than that limit. But it's the worst-case scenario. In most common scenarios there's less states than that in the resulting DFA. Sometimes it could be even less than in the original NFA.
But as far as I know, the algorithm to predict how many states the resulting DFA will actually have, doesn't exist yet. So if you'll find it, please let me know ;)
That is correct. As you probably already know, both DFAs and NFAs only accept regular languages. That means that they are equal in the languages they can accept. Also, the most primitive way of transforming a NFA to a DFA is with subset construction (also called powerset construction), where you simply create a state in the DFA for every combination of states in the NFA. This is called the powerset of states, which could at most be 2^n.
But, as mentioned by #SasQ that is the worst case scenario. Typically you will not end up with that many states if you use Hopcroft's algorithm or Brozowski's algorithm.
Suppose for various input strings an algorithm generates binary string with same number of 0's and 1's. The output for two different input strings may or may not be the same. Can we say anything about the space complexity of the algorithm?
The question isn't quite right.
Kolmogorov complexity K(x) doesn't apply to programs, it applies to a string x.
More specifically, the Kolmogorov complexity of a string x is the minimum program length needed to compute a particular string x.
It has been formally proven that one can't compute the Kolmogorov complexity of a string. In practice, you can approximate via an upper bound.
The following paper by Ferbus-Zanda and Griorieff gives you the theory http://arxiv.org/abs/1010.3201
An intuitive way of thinking about such an approximate upper bound is to consider the length of a compression program that can decompress to a particular string.
Applying this to your problem, the string you describe is a random binary one, doubled. The input string acts a seed for the random number generator.
Ignoring the kolmogorov complexity part of your question, and just looking at space complexity (ie. memory footprint) aspect as #templatetypedef did, the criteria you mention are so loose that all you can say is that the lower space bound for the algorithm is O(1) and the upper bound O(n), where n is the output.
No, I don't believe so. Consider the algorithm "print 01," which requires space Θ(1), and the algorithm "double the length of the input string, then print 01," which requires space Θ(n). Both algorithms meet the criteria you've provided, so just given those criteria you can't say anything about the space complexity of the algorithm.
Hope this helps!
I searched the site but did not find exactly what I was looking for... I wanted to generate a discrete random number from normal distribution.
For example, if I have a range from a minimum of 4 and a maximum of 10 and an average of 7. What code or function call ( Objective C preferred ) would I need to return a number in that range. Naturally, due to normal distribution more numbers returned would center round the average of 7.
As a second example, can the bell curve/distribution be skewed toward one end of the other? Lets say I need to generate a random number with a range of minimum of 4 and maximum of 10, and I want the majority of the numbers returned to center around the number 8 with a natural fall of based on a skewed bell curve.
Any help is greatly appreciated....
Anthony
What do you need this for? Can you do it the craps player's way?
Generate two random integers in the range of 2 to 5 (inclusive, of course) and add them together. Or flip a coin (0,1) six times and add 4 to the result.
Summing multiple dice produces a normal distribution (a "bell curve"), while eliminating high or low throws can be used to skew the distribution in various ways.
The key is you are going for discrete numbers (and I hope you mean integers by that). Multiple dice throws famously generate a normal distribution. In fact, I think that's how we were first introduced to the Gaussian curve in school.
Of course the more throws, the more closely you approximate the bell curve. Rolling a single die gives a flat line. Rolling two dice just creates a ramp up and down that isn't terribly close to a bell. Six coin flips gets you closer.
So consider this...
If I understand your question correctly, you only have seven possible outcomes--the integers (4,5,6,7,8,9,10). You can set up an array of seven probabilities to approximate any distribution you like.
Many frameworks and libraries have this built-in.
Also, just like TokenMacGuy said a normal distribution isn't characterized by the interval it's defined on, but rather by two parameters: Mean μ and standard deviation σ. With both these parameters you can confine a certain quantile of the distribution to a certain interval, so that 95 % of all points fall in that interval. But resticting it completely to any interval other than (−∞, ∞) is impossible.
There are several methods to generate normal-distributed values from uniform random values (which is what most random or pseudorandom number generators are generating:
The Box-Muller transform is probably the easiest although not exactly fast to compute. Depending on the number of numbers you need, it should be sufficient, though and definitely very easy to write.
Another option is Marsaglia's Polar method which is usually faster1.
A third method is the Ziggurat algorithm which is considerably faster to compute but much more complex to program. In applications that really use a lot of random numbers it may be the best choice, though.
As a general advice, though: Don't write it yourself if you have access to a library that generates normal-distributed random numbers for you already.
For skewing your distribution I'd just use a regular normal distribution, choosing μ and σ appropriately for one side of your curve and then determine on which side of your wanted mean a point fell, stretching it appropriately to fit your desired distribution.
For generating only integers I'd suggest you just round towards the nearest integer when the random number happens to fall within your desired interval and reject it if it doesn't (drawing a new random number then). This way you won't artificially skew the distribution (such as you would if you were clamping the values at 4 or 10, respectively).
1 In testing with deliberately bad random number generators (yes, worse than RANDU) I've noticed that the polar method results in an endless loop, rejecting every sample. Won't happen with random numbers that fulfill the usual statistic expectations to them, though.
Yes, there are sophisticated mathematical solutions, but for "simple but practical" I'd go with Nosredna's comment. For a simple Java solution:
Random random=new Random();
public int bell7()
{
int n=4;
for (int x=0;x<6;++x)
n+=random.nextInt(2);
return n;
}
If you're not a Java person, Random.nextInt(n) returns a random integer between 0 and n-1. I think the rest should be similar to what you'd see in any programming language.
If the range was large, then instead of nextInt(2)'s I'd use a bigger number in there so there would be fewer iterations through the loop, depending on frequency of call and performance requirements.
Dan Dyer and Jay are exactly right. What you really want is a binomial distribution, not a normal distribution. The shape of a binomial distribution looks a lot like a normal distribution, but it is discrete and bounded whereas a normal distribution is continuous and unbounded.
Jay's code generates a binomial distribution with 6 trials and a 50% probability of success on each trial. If you want to "skew" your distribution, simply change the line that decides whether to add 1 to n so that the probability is something other than 50%.
The normal distribution is not described by its endpoints. Normally it's described by it's mean (which you have given to be 7) and its standard deviation. An important feature of this is that it is possible to get a value far outside the expected range from this distribution, although that will be vanishingly rare, the further you get from the mean.
The usual means for getting a value from a distribution is to generate a random value from a uniform distribution, which is quite easily done with, for example, rand(), and then use that as an argument to a cumulative distribution function, which maps probabilities to upper bounds. For the standard distribution, this function is
F(x) = 0.5 - 0.5*erf( (x-μ)/(σ * sqrt(2.0)))
where erf() is the error function which may be described by a taylor series:
erf(z) = 2.0/sqrt(2.0) * Σ∞n=0 ((-1)nz2n + 1)/(n!(2n + 1))
I'll leave it as an excercise to translate this into C.
If you prefer not to engage in the exercise, you might consider using the Gnu Scientific Library, which among many other features, has a technique to generate random numbers in one of many common distributions, of which the Gaussian Distribution (hint) is one.
Obviously, all of these functions return floating point values. You will have to use some rounding strategy to convert to a discrete value. A useful (but naive) approach is to simply downcast to integer.