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Problem Detail: 

For every computable function $f$ does there exist a problem that can be solved at best in $\Theta(f(n))$ time or is there a computable function $f$ such that every problem that can be solved in $O(f(n))$ can also be solved in $o(f(n))$ time?

This question popped into my head yesterday. I've been thinking about it for a bit now, but can't figure it out. I don't really know how I'd google for this, so I'm asking here. Here's what I've come up with:

My first thought was that the answer is yes: For every computable function $f$ the problem "Output $f(n)$ dots" (or create a string with $f(n)$ dots or whatever) can obviously not be solved in $o(f(n))$ time. So we only need to show that it can be solved in $O(f(n))$ time. No problem, just take the following pseudo code:

x = f(n) for i from 1 to x:     output(".") 

Clearly that algorithm solves the stated problem. And it's runtime is obviously in $\Theta(f(n))$, so problem solved. That was easy, right? Except no, it isn't because you have to consider the cost of the first line. The above algorithm's runtime is only in $\Theta(f(n))$ if the time needed to calculate $f(n)$ is in $O(f(n))$. Clearly that's not true for all functions1.

So this approach didn't get me anywhere. I'd be grateful for anyone pointing me in the right direction to figure this out properly.


1 Consider for example the function $p(n) = \cases{1 & \text{if $n$ is prime} \\ 2 & \text{otherwise}}$. Clearly $O(p(n)) = O(1)$, but there is no algorithm that calculates $p$ in $O(1)$ time.

Asked By : sepp2k

Answered By : Alex ten Brink

By the Gap theorem (using the formulation from here, search for 'gap'), for any computable unbounded function $g : \mathbb{N} \rightarrow \mathbb{N}$, there exists some increasing (in fact, arbitrarily large) computable function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $DTIME(f(n)) = DTIME(g(f(n))$.

This answers your question in that there exists such an $f$ (infinitely many, in fact): for every computable function $g$ such that $g = o(n)$, there exists some increasing function $f$ such that all problems solvable in $O(f(n))$ time are also solvable in $O(g(f(n)) = o(f(n))$ time. Note that $f$ is not necessarily time-constructible - for the time-constructible case, see the answer by @RanG.

In the Wikipedia formulation (which requires that $g(x) \geq x$), then $g \circ f$ becomes your example, and $f$ needs to be $\omega(n)$ (so you go the other way around - 'problems solvable in $O(g(f(n))$ are also solvable in $O(g(n))$' is the interesting part).

The Wikipedia article does not note that $f$ is increasing and can in fact be arbitrarily large ($f(n) \geq g(n)$ for instance). The article that proves the gap theorem does mention and prove this (see here, for example).

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/1138

Problem Detail: 

I understand that the Cook-Levin theorem proved that any NP problem is reducible to an NP-complete problem, which signifies that if a polynomial-time algorithm for an NP-complete problem is found, it will mean that all problems in NP can be solved in polynomial time.

My question is: does this proof also hold the other way? I.e. if an exponential lower bound for an NP-complete problem is proved, would that mean that no problems in NP can be solved in polynomial time?

Asked By : haydnv

Answered By : D.W.

Not quite. If there's an exponential lower bound for a NP-complete problem, then it follows that no NP-complete problem can be solved in polynomial time. However, there will be some other NP problems (which aren't NP-complete) that can be solved in polynomial time.

If there's an exponential lower bound for any NP problem, then it follows that P != NP. In particular, an exponential lower bound for some problem means that the problem is not in P. So, if you have an exponential lower bound for some NP problem, that means the problem is in NP but not in P -- which means NP contains a problem that is not in P -- which means that NP != P.

It might help to recall that NP is a set of problems (the set of all search problems where YES answers have a polynomial-size witness that can be verified in polynomial time). Similarly, P is a set of problems (the set of all search problems that can be solved in polynomial time). So, if you've found one element of NP that is not an element of P, you can immediately conclude that those two sets are not the same sets.

Or, to put it another way: your second paragraph is the contrapositive of the statement in the first paragraph. If there is a NP problem with a non-polynomial lower bound, then not every problem in NP can be solved in polynomial time. If not every problem in NP can be solved in polynomial time, then it follows that no polynomial-time algorithm can be found for any NP-complete problem.

See What is the definition of P, NP, NP-complete and NP-hard? for more background.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/45417

Problem Detail: 

I'm looking for some clarification on some concepts/facts I came across while studying for a class.

I was reading the following wikipedia article. The below specific section and statement intrigued me when looking it over.

http://en.wikipedia.org/wiki/Computational_complexity_theory#Important_complexity_classes "It turns out that PSPACE = NPSPACE and EXPSPACE = NEXPSPACE by Savitch's theorem"

I also read that NTMs can be simulated by DTMs but that the shortest accepting computation of the DTM is exponential with respect to the shortest accepting computation of the target NTM.

My questions are:

1.) Are PSPACE and NPSPACE the set of all problems that require at least polynomial space to be solved on Deterministic and Non-deterministic Turing machines respectively?

2.) If so, is the actual size of the polynomial space required dependent on the size of the input?

3.) For P and NP, they are each the sets of problems that require at least polynomial time to be solved on DTMs and NTMs respectively correct?

4.) Is the reason that the shortest accepting computation of a DTM simulating a target NTM is exponential with respect to the shortest accepting computation of an NTM due to the exponential explosion of the number of configurations that an NTM supports as input grows for a given problem?

5.) My last and overarching question is: Are the differences in the set of problems that can be solved in polynomial time on DTMs versus NTMs related to time/space tradeoffs where DTMs can't run some polynomial NTM algorithms in polynomial time because they don't have the same "space" that an NTM has available to it?

I'd also appreciate any reading you can suggest to me on time/space tradeoffs and NTMs versus DTMs.

Asked By : Ethan Willis

Answered By : David Richerby

That's rather a lot of questions for one Stack Exchange post. One or two closely related questions per post is best. But here goes...

1.) Are PSPACE and NPSPACE the set of all problems that require at least polynomial space to be solved on Deterministic and Non-deterministic Turing machines respectively?

No. They're the set of all problems that require at most polynomial space to be solved on deterministic and nondeterministic Turing machines, respectively.

2.) If so, is the actual size of the polynomial space required dependent on the size of the input?

No. For any given machine, the polynomial is fixed. The input to the polynomial is the length of the input. (E.g., if the machine runs in space $3n^2+4$ and the input has length 3, the Turing machine will need at most 31 cells of storage on its work tape.)

3.) For P and NP, they are each the sets of problems that require at least polynomial time to be solved on DTMs and NTMs respectively correct?

Again, at most.

4.) Is the reason that the shortest accepting computation of a DTM simulating a target NTM is exponential with respect to the shortest accepting computation of an NTM due to the exponential explosion of the number of configurations that an NTM supports as input grows for a given problem?

We don't know for sure that there's any loss of efficiency at all when simulating a nondeterministic machine on a deterministic one. But the reason we don't know how to do it with better than exponential slowdown is, as you suggest, that we don't know any better technique than trying all the possible nondeterministic choices, one by one.

5.) My last and overarching question is: Are the differences in the set of problems that can be solved in polynomial time on DTMs versus NTMs related to time/space tradeoffs where DTMs can't run some polynomial NTM algorithms in polynomial time because they don't have the same "space" that an NTM has available to it?

Not really. It's because the nondeterministic machine in a sense computes lots of options in parallel and it accepts if any of those parallel options accepts. As I said above, the only way we know how to simulate all of these parallel computations is to do them sequentially, one after another. But there isn't any particular time/space trade-off: if a nondeterministic machine makes some computation in time $t(n)$ and space $s(n)$, a deterministic machine can do the same computation in time $2^O(t(n))$ and space $s(n)+O(t(n))$. The exponential blow-up in time is because a $q$-state nondeterministic machine might make up to $q$ nondeterministic choices (i.e., spawn up to $q$ "parallel threads") at each step of its execution; the smaller increase in space is because of the book-keeping required to keep track of which choices we tried at each step.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/22109

Problem Detail: 

I've just finished studying recursion at university. One thing that stood out for me however was that in both the lectures and in the practical we completed, all the tasks we were asked to do could be performed faster, and in less code, using iterative means.
This was something the lecturer confirmed.

Could somebody please give me some examples of situations when recursion is a better solution than iterative techniques? Additionally, are there any situations in which recursion is the only way to sole a problem?

Asked By : Andrew Martin

Answered By : jmite

There are no questions which can only be solved with recursion. This is because they can be solved with Turing machines, which don't use recursion.

The set of problems which can be solved with a TM are exactly the same as the ones which can be solved with recursion (or its formal model, the Lambda Calculus).

In particular, if you want to simulate recursion iteratively, the way to do this is to use a data structure called a stack (which simulates the call stack for functions).

As for algorithms that can be solved better using recursion, there are tons. I'm surprised that your recursive versions were longer, as recursion usually leads to less code. This is one of the reasons haskell is gaining popularity.

Consider the algorithm quicksort, for sorting lists. In rough pseudocode, it's as follows:

function quicksort(list)     if length(list) <= 1         return list            pivot = first element of list     lessList = []     equalList = []     greaterList = []     for each element in list:         if element < pivot, add to lessList         if element == pivot, add to equalList         if element > pivot, add to greater list    sortedLess = quicksort(lessList)    sortedGreater = quicksort(greaterList)    return sortedLess ++ equalList ++ sortedGreater 

where ++ means concatenation.

The code isn't purely functional, but by dividing the list into different parts, and sorting each sublist recursively, we get a very short $O(n\log n)$ sort.

Recursion is also very useful for recursive data structures. Often times you'll have a traversal on trees, of the following form:

function traverseTree(tree)     if (tree is a single node)         do something to that node     else, for each child of tree:          traverseTree(child) 
Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/9987

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  • Go to any link which are you required and your assignments will be display on your computer screen in a pdf format file or doc file.
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IGNOU Assignments New Syllabus Indira Gandhi national Open University (IGNOU) has published the notification for Assignments as IGNOU Assignments December 2017 for BA, BCA, BCOM, MBA, MCA, BED courses. Students who are pursuing these courses can download their Assignments from IGNOU Official website. For more current details about any information about IGNOU Assignments like steps to download its assignments 2017, Applicants may logon to our portal which is CLICK HERE TO DOWNLOAD DECEMBER 2017 ASSIGNMENTS.
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How to download IGNOU Assignments December 2017-18?

  • Candidates first of all log on to www.ignou.ac.in.
  • Go to student zone section of university home page �Student Zone� link.
  • A new page will be opened than go to �Assignments� link.
  • Again, a new page will be opened there are various courses assignments available.
  • Go to any link which are you required and your assignments will be display on your computer screen in a pdf format file or doc file.