Camilla has made it clearly known that she does not intend to use the title Queen. Will you protest her decision ?

Why are you posting this question again?…

There are many more of the same posted.

Reposting the same questions over and over and over again is against rules of Answers.  The answer remains the same:

While I think the tradition of the wife’s assuming the titles and styles the proper thing to do, I realize that it is Camilla’s choice to make.At the time of their engagement, when Camilla made this announcement, royal scholars pointed out that no such title existed, that Camilla was trying to placate the rabid Diana-fans by choosing to be called a princess consort.

Is there a polynomial time algorithm for this decision problem?

Is there a factor in $M$ that is $>$ $1$, but $<$ $M$ that is NOT a factor of $N$?

False Result Example

$N$ = 8

$M$ = 16

1, 2, 4, 8, 16

There is no integer that is NOT a factor of $N$ that is $>$ $1$ but < $M$

True Result Example

$N$ = 2

$M$ = 26

1, 2, 13, 26

There is an integer $13$ which is NOT a factor of $N$ that is > 1 but < $M$


I have found a pseudo-polynomial solution, but is there a polynomial solution for this problem in the length of input?

Complexity theory – question about the decision of Turing Machine

I'm a little confused about using the lathe subset to prove the desirability of the lathe.

If I have a Turing machine M and we already know that M has a single stop state. If the machine is already using a string as input and reaches the q0 state, is it possible to prove that M is decidable by considering designing a new Turing machine that will stop at q0?

Thank you so much!

If a decision problem is in $P$, must finding the solution be possible in polynomial-time?

No, and the example you list is a classic example: as far as we know, factoring does not appear to be in $P$, but determining whether a number is prime is definitely in $P$.

Another example: Consider the game Hex. Consider the decision problem: given $n$, determine whether the first player has a winning strategy for Hex on a $n times n$ board. There is a corresponding function problem: given $n$, find such a winning strategy. Well, the decision problem is trivial (it is known that the answer is always “yes”), but the function problem is believed to be very hard (as far as we know).

number theory – Must a decision problem in $NP$ have a complement in $Co-NP$, if I can verify the solutions to in polynomial-time?

Goldbach’s Conjecture says every even integer $>$ $2$ can be expressed as the sum of two primes.

Let’s say $N$ is our input and its $10$. Which is an integer > 2 and is not odd.


1.Create list of numbers from $1,to~N$

2.Use prime-testing algorithm for creating a second list of prime numbers

3.Use my 2_sum solver that allows you to use primes twice that sum up to $N$

for j in range(list-of-primes)):
  if N-(list-of-primes(j)) in list-of-primes:

4.Verify solution efficently

if AKS-primality(N-(list-of-primes(j))):
    if AKS-primality(list-of-primes(j)):
        print('Solution is correct')


yes 7 + 3
Solution is correct


If the conjecture is true, then the answer will always be Yes. Does that mean it can’t be in $Co-NP$ because the answer is always Yes?

Decision Making – View items or units for in-game purchase conversion

Imagine a game in which you collect items in two sizes: A stands for 1 unit, B stands for 3 identical units.

When the player runs out of time, the field shows your meeting.

If the player can buy more time, it would be more convincing to convert the purchase:

1xB 3xA



Is it better to show the exact inventory or the larger number because it is the same amount?

np complete – Karp reduction of optimization problems to decision problems

If you take Cook reductions into account, the decision and optimization versions of the problems are polynomially reducible to each other.

If you focus on Cook reductions, there is a natural Karp reduction from the decision version of a problem to the optimization version. Is the opposite true?

If the optimization version of a problem is NP-Complete, can we conclude that the decision version is also NP-Complete? I am confused because the definition for NP-Complete uses Karp reductions.

Algorithms – Finding the lower limit using decision trees

One way to determine the lower limit of a comparison-based algorithm is to use the decision tree. You have two questions about this method:

1) We know that the height of the tree is a path connecting the root node to the most distant leaf node (longest path), which is the number of comparisons made from the root node to the leaf node. So when we draw the tree for a comparison-based algorithm, we simply have to find the worst-case time that corresponds to the largest path and therefore corresponds to the height of the tree. Now for each tree the height <= log2 (number of leaf nodes) is identical to Cworst (n) <= log2 (n), and now we have a lower limit for Cworst (n) and thus the lower limit of the problem = log2 (n) . Is my understanding correct?

2) What does it mean to have an inequality for Cworst (n) for a particular comparison problem? Does this mean that we can draw many trees for a particular comparison problem and each time the height of the path of the worst case scenario has a value that corresponds to equality? This means that we can draw many different trees for a particular problem?

Will Biden's decision to choose a running partner based on their gender, not just their qualifications, will improve or affect their chances of winning?

Overall, practicing positive action will do more harm than good to Biden. I think most of those interested in identity politics will vote for Biden regardless of who he chooses, so he gets few votes there, but I think many of the centrist swing voters are fed up with identity politics and want the best possible Person regardless of gender So he will lose many of these voters.

It's not just about the election of the Vice President. By taking positive action in his election as Vice President, Biden sends a strong message that he will generally support identity politics through equal opportunities if he wins. Again, I think most Americans are tired of identity politics in the middle of the street.