I'm wondering if Non-Deterministic Finite Automata can use an epsilon transition to reach an accepting state after computing all the given input? In a simple example where non-accepting state s1 has an arrow pointing at itself labelled 1 and an epsilon-labelled arrow pointing towards accepting state s2, given the input 111 would the automaton be able to process it like (s1,111)(s1,11)(s1,1)(s1,empty word)(s2,empty word) and thus accept the input?
Any help would be greatly appreciated.
Thanks
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I was solving the bloch sphere problem from IBM Qiskit textbook there I got a problem like this but it giving imaginary part with |0> . Is it even possible??
I am currently trying to find the longest cycle in a !directed! graph G = (V,E). For that, I first want to (have to) formulate an integer linear program (ILP). The objective function is in my opinion quite clear: min sum_{e in E} -w_e * x_e (I formulate it as a minimization problem for future work). Here, x_e is a binary variable which is 1 if I use this specific edge (/arc) and zero otherwise; w_e is the weight of edge e.
To assure that the solution is a simple cycle, I thought about having deg_{in}^{i} = deg_{out}^{i} = 1 for all nodes i in the solution set, i.e. there is exactly one arc entering node i and exactly one arc leaving node i, given that node i is part of the cycle.
Those constraints, however, are not enough to solve this problem, as it could still happen that the solution is a set of disjoint cycles, instead of one single cycle.
Does somebody have an idea how to solve the problem?
Thank you!!
You need to add TSP-like subtour-elimination constraints. See: https://yetanothermathprogrammingconsultant.blogspot.com/2020/02/longest-path-problem.html.
First of all a disclaimer: I'm posting this question here even if I realize it is quite maths heavy, because I have trouble figuring out on what other site it could belong.
I'm writing a 2d spaceships game where the player will have to select the ship's destination and a course will be automatically plotted.
Along with this, I'm offering various options to control the ship's acceleration while it gets there. All these options have to do with the target velocity at the destination.
One option is to select the desired destination and velocity vector, in which case the program will use cubic interpolation, since starting and target coordinates and velocity are available.
Another option is to just select the destination point, but let the game calculate the final velocity vector. This is done through quadratic interpolation (ie. acceleration is constant).
I would like to introduce another option: let the player select the destination and the maximum absolute value of the velocity vector, as in
sqrt( vf_x^2 + vf_y^2 ) <= Vf_max
I take I shall use a 3rd order polynomial to model the course in this case, but I'm having a pretty hard time figuring out the coefficients, since I miss one equation for each coordinate (the one given by velocity at destination). Furthermore, I'm confused as to how I should use the Vf_max constraint to help me figure out the missing coefficients.
I suspect it could be an optimization problem, but I'm quite ignorant about this topic.
Can anybody help me find a solution or point me in the right direction, please?
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I have a system of linear equations that make up an NxM matrix (i.e. Non-square) which I need to solve - or at least attempt to solve in order to show that there is no solution to the system. (more likely than not, there will be no solution)
As I understand it, if my matrix is not square (over or under-determined), then no exact solution can be found - am I correct in thinking this? Is there a way to transform my matrix into a square matrix in order to calculate the determinate, apply Gaussian Elimination, Cramer's rule, etc?
It may be worth mentioning that the coefficients of my unknowns may be zero, so in certain, rare cases it would be possible to have a zero-column or zero-row.
Whether or not your matrix is square is not what determines the solution space. It is the rank of the matrix compared to the number of columns that determines that (see the rank-nullity theorem). In general you can have zero, one or an infinite number of solutions to a linear system of equations, depending on its rank and nullity relationship.
To answer your question, however, you can use Gaussian elimination to find the rank of the matrix and, if this indicates that solutions exist, find a particular solution x0 and the nullspace Null(A) of the matrix. Then, you can describe all your solutions as x = x0 + xn, where xn represents any element of Null(A). For example, if a matrix is full rank its nullspace will be empty and the linear system will have at most one solution. If its rank is also equal to the number of rows, then you have one unique solution. If the nullspace is of dimension one, then your solution will be a line that passes through x0, any point on that line satisfying the linear equations.
Ok, first off: a non-square system of equations can have an exact solution
[ 1 0 0 ][x] = [1]
[ 0 0 1 ][y] [1]
[z]
clearly has a solution (actually, it has an 1-dimensional family of solutions: x=z=1). Even if the system is overdetermined instead of underdetermined it may still have a solution:
[ 1 0 ][x] = [1]
[ 0 1 ][y] [1]
[ 1 1 ] [2]
(x=y=1). You may want to start by looking at least squares solution methods, which find the exact solution if one exists, and "the best" approximate solution (in some sense) if one does not.
Taking Ax = b, with A having m columns and n rows. We are not guaranteed to have one and only one solution, which in many cases is because we have more equations than unknowns (m bigger n). This could be because of repeated measurements, that we actually want because we are cautious about influence of noise.
If we observe that we can not find a solution that actually means, that there is no way to find b travelling the column space spanned by A. (As x is only taking a combination of the columns).
We can however ask for the point in the space spanned by A that is nearest to b. How can we find such a point? Walking on a plane the closest one can get to a point outside it, is to walk until you are right below. Geometrically speaking this is when our axis of sight is perpendicular to the plane.
Now that is something we can have a mathematical formulation of. A perpendicular vector reminds us of orthogonal projections. And that is what we are going to do. The simplest case tells us to do a.T b. But we can take the whole matrix A.T b.
For our equation let us apply the transformation to both sides: A.T Ax = A.T b.
Last step is to solve for x by taking the inverse of A.T A:
x = (A.T A)^-1 * A.T b
The least squares recommendation is a very good one.
I'll add that you can try a singular value decomposition (SVD) that will give you the best answer possible and provide information about the null space for free.
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In trying to solve a particular Project Euler question, I ran into difficulties with a particular mathematical formula. According to this web page (http://www.mathpages.com/home/kmath093.htm), the formula for determining the probability for rolling a sum, T, on a number of dice, n, each with number of sides, s, each numbered 1 to s, can be given as follows:
alt text http://www.freeimagehosting.net/uploads/8294d47194.gif
After I started getting nonsensical answers in my program, I started stepping through, and tried this for some specific values. In particular, I decided to try the formula for a sum T=20, for n=9 dice, each with s=4 sides. As the sum of 9 4-sided dice should give a bell-like curve of results, ranging from 4 to 36, a sum of 20 seems like it should be fairly (relatively speaking) likely. Dropping the values into the formula, I got:
alt text http://www.freeimagehosting.net/uploads/8e7b339e32.gif
Since j runs from 0 to 7, we must add over all j...but for most of these values, the result is 0, because at least one the choose formulae results are 0. The only values for j that seem to return non-0 results are 3 and 4. Dropping 3 and 4 into this formula, I got
alt text http://www.freeimagehosting.net/uploads/490f943fa5.gif
Which, when simplified, seemed to go to:
alt text http://www.freeimagehosting.net/uploads/603ca84541.gif
which eventually simplifies down to ~30.75. Now, as a probability, of course, 30.75 is way off...the probability must be between 0 and 1, so something has gone terribly wrong. But I'm not clear what it is.
Could I misunderstanding the formula? Very possible, though I'm not clear at all where the breakdown would be occuring. Could it be transcribed wrong on the web page? Also possible, but I've found it difficult to find another version of it online to check it against. Could I be just making a silly math error? Also possible...though my program comes up with a similar value, so I think it's more likely that I'm misunderstanding something.
Any hints?
(I would post this on MathOverflow.com, but I don't think it even comes close to being the kind of "postgraduate-level" mathematics that is required to survive there.)
Also: I definitely do not want the answer to the Project Euler question, and I suspect that other people that my stumble across this would feel the same way. I'm just trying to figure out where my math skills are breaking down.
According to mathworld (formula 9 is the relevant one), the formula from your source is wrong.
The correct formula is supposed to be n choose j, not n choose T. That'll really reduce the size of the values within the summation.
The mathworld formula uses k instead of j and p instead of T:
Take a look at article in wikipedia - Dice.
The formula here looks almost similar, but have one difference. I think it will solve your problem.
I'm going to have to show my ignorance here.... Isn't 9 choose 20 = 0? More generally, isn't n choose T going to always be 0 since T>=n? Perhaps I'm reading this formula incorrectly (I'm not a math expert), but looking at de Moive's work, I'm not sure how this formula was derived; it seems slightly off. You might try working up from Moive's original math, page 39, in the lemma.