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.
I've found a grammar online which I want to rewrite to BNF so I can use it in a grammatical evolution experiment. From what I've read online BNF is given by this form:
<symbol> := <expression> | <term>
...but I don't see where probabilities factor into it.
In a probabilistic context-free grammar (PCFG), every production also is assigned a probability. How you choose to write this probability is up to you; I don't know of a standard notation.
Generally, the probabilities are learned rather than assigned, so the representation issue doesn't come up; the system is given a normal CFG as well as a large corpus with corresponding parse trees, and it derives probabilities by analysing the parse trees.
Note that PCFGs are usually ambiguous. Probabilities are not used to decide whether a sentence is in the language but rather which parse is correct, so with an unambiguous grammar, the probabilities would be of little use.
I'm new to genetic algorithms and am writing code for the Traveling Salesman problem. I'm using cycle crossover to generate new offspring and I've found that this leads to some of the offspring retaining the same exact phenotype as one parent even when the two parents are different. Would translating the chromosomes avoid this?
By translate I mean a chromosome with phenotype ABCDE shifting over two to DEABC. They would be equivalent answers and have equal fitness, but might make more diverse offspring.
Is this worth it in the long run, or is it just wasting computing time?
Cycle crossover (CX) is based on the assumption that it's important to preserve the absolute position of cities (a city preferably inherits its position from either parent) and the preventive "translation" is against the spirit of CX.
Anyway multiple studies (e.g. 1) have shown that for TSP the key is to preserve the relative position of cities and the edges.
So it could work, but you have to experiment. Some form of mutation is another possibility.
Probably, if the characteristics of CX aren't satisfying, a different crossover operator is a better choice: staying with simple operators, one of the most successful is Order Crossover (e.g. 2).
L. Darrell Whitley, Timothy Starkweather, D'Ann Fuquay - Scheduling problems and traveling salesmen: The genetic edge recombination operator - 1989.
Pablo Moscato - On Genetic Crossover Operators for Relative Order Preservation.
does every algorithm have Big Omega?
Is it possible for algorithms to have both Big O and Big Omega (but not equal to each other- not Big Theta) ?
For instance Quicksort's Big O - O(n log n) But does it have Big Omega? If it does, how do i calculate it?
First, it is of paramount importance that one not confuse the bound with the case. A bound - like Big-Oh, Big-Omega, Big-Theta, etc. - says something about a rate of growth. A case says something about the kinds of input you're currently considering being processed by your algorithm.
Let's consider a very simple example to illustrate the distinction above. Consider the canonical "linear search" algorithm:
LinearSearch(list[1...n], target)
1. for i := 1 to n do
2. if list[i] = target then return i
3. return -1
There are three broad kinds of cases one might consider: best, worst, and average cases for inputs of size n. In the best case, what you're looking for is the first element in the list (really, within any fixed number of the start of the list). In such cases, it will take no more than some constant amount of time to find the element and return from the function. Therefore, the Big-Oh and Big-Omega happen to be the same for the best case: O(1) and Omega(1). When both O and Omega apply, we also say Theta, so this is Theta(1) as well.
In the worst case, the element is not in the list, and the algorithm must go through all n entries. Since f(n) = n happens to be a function that is bound from above and from below by the same class of functions (linear ones), this is Theta(n).
Average case analysis is usually a bit trickier. We need to define a probability space for viable inputs of length n. One might say that all valid inputs (where integers can be represented using 32 bits in unsigned mode, for instance) are equally probable. From that, one could work out the average performance of the algorithm as follows:
Find the probability that target is not represented in the list. Multiply by n.
Given that target is in the list at least once, find the probability that it appears at position k for each 1 <= k <= n. Multiply each P(k) by k.
Add up all of the above to get a function in terms of n.
Notice that in step 1 above, if the probability is non-zero, we will definitely get at least a linear function (exercise: we can never get more than a linear function). However, if the probability in step 1 is indeed zero, then the assignment of probabilities in step 2 makes all the difference in determining the complexity: you can have best-case behavior for some assignments, worst-case for others, and possibly end up with behavior that isn't the same as best (constant) or worst (linear).
Sometimes, we might speak loosely of a "general" or "universal" case, which considers all kinds of input (not just the best or the worst), but that doesn't give any particular weighting to inputs and doesn't take averages. In other words, you consider the performance of the algorithm in terms of an upper-bound on the worst-case, and a lower-bound on the best-case. This seems to be what you're doing.
Phew. Now, back to your question.
Are there functions which have different O and Omega bounds? Definitely. Consider the following function:
f(n) = 1 if n is odd, n if n is even.
The best case is "n is odd", in which case f is Theta(1); the worst case is "n is even", in which case f is Theta(n); and if we assume for the average case that we're talking about 32-bit unsigned integers, then f is Theta(n) in the average case, as well. However, if we talk about the "universal" case, then f is O(n) and Omega(1), and not Theta of anything. An algorithm whose runtime behaves according to f might be the following:
Strange(list[1...n], target)
1. if n is odd then return target
2. else return LinearSearch(list, target)
Now, a more interesting question might be whether there are algorithms for which some case (besides the "universal" case) cannot be assigned some valid Theta bound. This is interesting, but not overly so. The reason is that you, during your analysis, are allowed to choose the cases that constitutes best- and worst-case behavior. If your first choice for the case turns out not to have a Theta bound, you can simply exclude the inputs that are "abnormal" for your purposes. The case and the bound aren't completely independent, in that sense: you can often choose a case such that it has "good" bounds.
But can you always do it?
I don't know, but that's an interesting question.
Does every algorithm have a Big Omega?
Yes. Big Omega is a lower bound. Any algorithm can be said to take at least constant time, so any algorithm is Ω(1).
Does every algorithm have a Big O?
No. Big O is a upper bound. Algorithms that don't (reliably) terminate don't have a Big O.
An algorithm has an upper bound if we can say that, in the absolute worst case, the algorithm will not take longer than this. I'm pretty sure O(∞) is not valid notation.
When will the Big O and Big Omega of an algorithm be equal?
There is actually a special notation for when they can be equal: Big Theta (Θ).
They will be equal if the algorithm scales perfectly with the size of the input (meaning there aren't input sizes where the algorithm is suddenly a lot more efficient).
This is assuming we take Big O to be the smallest possible upper bound and Big Omega to be the largest possible lower bound. This is not actually required from the definition, but they're commonly informally treated as such. If you drop this assumption, you can find a Big O and Big Omega that aren't equal for any algorithm.
Brute force prime number checking (where we just loop through all smaller numbers and try to divide them into the target number) is perhaps a good example of when the smallest upper bound and largest lower bound are not equal.
Assume you have some number n. Let's also for the time being ignore the fact that bigger numbers take longer to divide (a similar argument holds when we take this into account, although the actual complexities would be different). And I'm also calculating the complexity based on the number itself instead of the size of the number (which can be the number of bits, and could change the analysis here quite a bit).
If n is divisible by 2 (or some other small prime), we can very quickly check whether it's prime with 1 division (or a constant number of divisions). So the largest lower bound would be Ω(1).
Now if n is prime, we'll need to try to divide n by each of the numbers up to sqrt(n) (I'll leave the reason we don't need to go higher than this as an exercise). This would take O(sqrt(n)), which would also then be our smallest upper bound.
So the algorithm would be Ω(1) and O(sqrt(n)).
Exact complexity also may be hard to calculate for some particularly complex algorithms. In such cases it may be much easier and acceptable to simply calculate some reasonably close lower and upper bounds and leave it at that. I don't however have an example on hand for this.
How does this relate to best case and worst case?
Do not confuse upper and lower bounds for best and worst case. This is a common mistake, and a bit confusing, but they're not the same. This is a whole other topic, but as a brief explanation:
The best and worst (and average) cases can be calculated for every single input size. The upper and lower bounds can then be used for each of those 3 cases (separately). You can think of each of those cases as a line on a graph with input size on the x-axis and time on the y-axis and then, for each of those lines, the upper and lower bounds are lines which need to be strictly above or below that line as the input size tends to infinity (this isn't 100% accurate, but it's a good basic idea).
Quick-sort has a worst-case of Θ(n2) (when we pick the worst possible pivot at every step) and a best-case of Θ(n log n) (when we pick good pivots). Note the use of Big Theta, meaning each of those are both lower and upper bounds.
Let's compare quick-sort with the above prime checking algorithm:
Say you have a given number n, and n is 53. Since it's prime, it will (always) take around sqrt(53) steps to determine whether it's prime. So the best and worst cases are all the same.
Say you want to sort some array of size n, and n is 53. Now those 53 elements can be arranged such that quick-sort ends up picking really bad pivots and run in around 532 steps (the worst case) or really good pivots and run in around 53 log 53 steps (the best case). So the best and worst cases are different.
Now take n as 54 for each of the above:
For prime checking, it will only take around 1 step to determine that 54 is prime. The best and worst cases are the same again, but they're different from what they were for 53.
For quick-sort, you'll again have a worst case of around 542 steps and a best case of around 54 log 54 steps.
So for quick-sort the worst case always takes around n2 steps and the best case always takes around n log n steps. So the lower and upper (or "tight") bound of the worst case is Θ(n2) and the tight bound of the best case is Θ(n log n).
For our prime checking, sometimes the worst case takes around sqrt(n) steps and sometimes it takes around 1 step. So the lower bound for the worse case would be Ω(1) and upper bound would be O(sqrt(n)). It would be the same for the best case.
Note that above I simply said "the algorithm would be Ω(1) and O(sqrt(n))". This is slightly ambiguous, as it's not clear whether the algorithm always takes the same amount of time for some input size, or the statement is referring to one of the best, average or worst case.
How do I calculate this?
It's hard to give general advice for this since proofs of bounds are greatly dependent on the algorithm. You'd need to analyse the algorithm similar to what I did above to figure out the worst and best cases.
Big O and Big Omega it can be calculated for every algorithm as you can see in Big-oh vs big-theta
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!