1.how can i get this automaton(non blocking and weighted)?
2.I want to see some simple examples.
Your original automaton is already a weighted one as stated in the problem. The new one adds to it loops for every letter in every state. Further it gives weight 0 to all the original transitions and a weight to the new ones that is not understandable from the information you give (what is \psi_{i,j}?).
You formally get this automaton by simply aplying the definitions you provide on the original automaton.
For example: original automaton over alphabet {a,b}, states {q (initial), p (final)}, transitions: (q, a, p) weight 1, (p, b, p) weight 2. In the new automaton you get (q, a, p) weight 0 and (p, b, p) with the weight that cannot be understood from your definitions. Further you get the new transitions (q, a, q), (q, b, q), and (p, a, p).
Related
Using Sagemath 9.2 on Windows 10
a1 = 9798722381116618056227637476565566484018606253194222755351973203508462742253522311154076194134700145275527578605535821781545038187843569198505993524407287520970070771172279404172004212310432500247465608472105231701909612623072343883942216806934904529600639676698348239426243771486521532222069409611514728756060897629936844695006373653175992634673678639333010508845045985607328371180356262460490393317997708599757357055386370808544031455931154122353547239678217006604692623467390849309525705453042722141078914816760002281629323554959490483823338710209710265138177331357093216148991708169324688688552846634664517554736
n = 27772857409875257529415990911214211975844307184430241451899407838750503024323367895540981606586709985980003435082116995888017731426634845808624796292507989171497629109450825818587383112280639037484593490692935998202437639626747133650990603333094513531505209954273004473567193235535061942991750932725808679249964667090723480397916715320876867803719301313440005075056481203859010490836599717523664197112053206745235908610484907715210436413015546671034478367679465233737115549451849810421017181842615880836253875862101545582922437858358265964489786463923280312860843031914516061327752183283528015684588796400861331354873
a2 = Mod(a1, n).sqrt()
I get the following
sage: print(a2)
sqrt9798722381116618056227637476565566484018606253194222755351973203508462742253522311154076194134700145275527578605535821781545038187843569198505993524407287520970070771172279404172004212310432500247465608472105231701909612623072343883942216806934904529600639676698348239426243771486521532222069409611514728756060897629936844695006373653175992634673678639333010508845045985607328371180356262460490393317997708599757357055386370808544031455931154122353547239678217006604692623467390849309525705453042722141078914816760002281629323554959490483823338710209710265138177331357093216148991708169324688688552846634664517554736
If you observe, a2 is prefixed with sqrt!
I don't see this with other roots, I calculated with Sage. What does this mean?
Is this a bug or does this have some other meaning?
It would appear n is prime (at least pseudo-prime):
sage: n.is_prime(proof=False)
True
Assuming it is, let us define the finite field in n elements:
sage: F = GF(n, proof=False)
and view a1 as an element A1 in F:
sage: A1 = F(a1)
Asking whether a1 is a square modulo n
amounts to asking whether A1 a square in F.
sage: A1.is_square()
False
It is not! So when we compute the square root of A1,
it has to be in a quadratic extension of F.
This is why when we ask Sage to compute this square root,
it gives it as a generator of that extension.
A natural name for this generator is "sqrt(n)",
which is what Sage uses.
Probably, when you computed other square roots,
they were square roots of numbers which were
squares modulo n, and therefore the square roots
could be computed in F, i.e. in ZZ / n ZZ,
without requiring a quadratic field extension.
I wonder if there is a fast algorithm, say (O(n^3)) for computing the cofactor matrix (or conjugate matrix) of a N*N square matrix. And yes one could first compute its determinant and inverse separately and then multiply them together. But how about this square matrix is non-invertible?
I am curious about the accepted answer here:Speed up python code for computing matrix cofactors
What would it mean by "This probably means that also for non-invertible matrixes, there is some clever way to calculate the cofactor (i.e., not use the mathematical formula that you use above, but some other equivalent definition)."?
Factorize M = L x D x U, whereL is lower triangular with ones on the main diagonal,U is upper triangular on the main diagonal, andD is diagonal.
You can use back-substitution as with Cholesky factorization, which is similar. Then,
M^{ -1 } = U^{ -1 } x D^{ -1 } x L^{ -1 }
and then transpose the cofactor matrix as :
Cof( M )^T = Det( U ) x Det( D ) x Det( L ) x M^{ -1 }.
If M is singular or nearly so, one element (or more) of D will be zero or nearly zero. Replace those elements with zero in the matrix product and 1 in the determinant, and use the above equation for the transpose cofactor matrix.
Given that fib(n)=fib(n-1)+fib(n-2) for n>1 and given that fib(0)=a, fib(1)=b (some a, b >0), which of the following is true?
fib(n) is
Select one or more:
a. O(n^2)
b. O(2^n)
c. O((1-sqrt 5)/2)^n)
d. O(n)
e. Answer depends on a and b.
f. O((1+sqrt 5)/2)^n)
Solving the Fibonacci sequence I got that:
fib(n)= 1/(sqrt 5) ((1+sqrt 5)/2)^n - 1/(sqrt 5) ((1-sqrt 5)/2)^n
But what would be the time complexity in this case? Would that mean the answers are c and f?
From your closed form of your formula, the term 1 / (sqrt 5) ((1 - sqrt 5) / 2)^n has limit 0 as n grows to infinity (|(1 - sqrt 5) / 2| < 1). Therefore we can ignore this term. Also since in time complexity theory we don't care about muliplication constants the following is true:
fib(n) = Θ(φ^n)
where φ = (1 + sqrt 5) / 2 a.k.a. the golden ratio constant.
So it's an exponential function and we can exclude a, d, e. We can exclude c since as was said it has limit 0. But answer b is also correct because φ < 2 and O expresses an upper bound.
Finally, the correct answers are:
b, f
Θ(φ^n) is correct when a=1 and b=1 or a=1 and b=2 . The value of φ depends on a and b.
For computing fib(n-1) and fib(n-2) if we compute them recursively complexity is exponential, but if we save two last values and use them, complexity is O(n) and not depends on a and b.
Hi recently i appeared in an aptitude,there was a problem that i realy cant understand please provide some idea, how to solve it.(and sorry to for poor English.)
(Question)-> Three candidates, Amar, Birendra and Chanchal stand for the local election. Opinion polls are
conducted and show that fraction a of the voters prefer Amar to Birendra, fraction b prefer Birendra to
Chanchal and fraction c prefer Chanchal to Amar. Which of the following is impossible?
(a) (a, b, c) = (0.51, 0.51, 0.51);
(b) (a, b, c) = (0.61, 0.71, 0.67);
(c) (a, b, c) = (0.68, 0.68, 0.68);
(d) (a, b, c) = (0.49, 0.49, 0.49);
(e) None of the above.
If you tried to list of possible preferences people can have
are either
ABC (means you prefer A to B, prefer B to C and therefore also prefer A to C)
ACB
BAC
BCA
CAB
CBA
in this case you'll find that each fraction of the population represents:
a=ABC+ACB+CAB
b=ABC+BAC+BCA
c=BCA+CAB+CBA
therefore a+b+c = 2(ABC+BCA+CAB)+ACB+BAC+CBA
as you notice not all groups within the population are repeated. we can therefore assume than (a+b+c) can never be more than twice the population since each member of the population is represented twice at the most.
out of the options C is the one where the sum is more than 2. and is therefore the impossible value.
Here's a few questions I had on a quiz in a class and just want to verify their correctness.
Grammar:
S -> ABC
A -> df | epsilon
B -> f | epsilon
C -> g | epsilon
1.) The Follow set of B contains g and epsilon (T/F)? Ans: F.
There is no epsilon in the Follow sets, correct? (Only $ aka end of input)
2.) The First set of S contains d, f, g, and epsilon (T/F)? Ans: T.
I said false for this because I thought First(S) = First(A), which g is not a part of. Who is correct?
You are correct. If epsilon is involved, it will be accounted for in the First set, not the Follow set. If it's possible for the production to end the string, then $ goes in the Follow set, not epsilon.
The quiz is correct. The production S can indeed start with any of d, f, and g, and it can also be started by the empty string. Consider the input string g. It matches S, right? A is satisfied by the empty string, B is satisfied by the empty string, and C is satisfied by g. Since A, B, and C are all satisfied, S is satisfied. The first character consumed by S is g, so g must be in First(S).