In a paper, I saw a quiz question about auto-correlation of a WSS process than I can not understand. It says:
Let X(t) be WSS. Which of the following **can be** correct?
a) E[X(t1) X(t2)] = |t1 - t2|
b) E[X(t1) X(t2)] = max{t1 - t2,0}
c) E[X(t1) X(t2)] = max{1- |t1 - t2|,0}
The correct answer is c).
Now I don't really understand it. In all three cases, the auto-correlation function depends only on the difference t1 and t2 so they all satisfy this condition.
Thanks in advance
Related
I would like to solve problems combining boolean and integer logic in linear arithmetic with a SAT/SMT solver. At first glance, Z3 seems promising.
First of all, is it at all possible to solve the following problem? This answer makes it seem like it works.
int x,y,z
boolean a,b,c
( (3x + y - 2z >= 10) OR (A AND (NOT B OR C)) OR ((A == C) AND (x + y >= 5)) )
If so, how does Z3 solve this kind of problem in theory and is there any documentation about it?
I could think of two ways to solve this problem. One would be to convert the Boolean operations into a linear integer expression. Another solution I read about is to use the Nelson-Oppen Combination Method described in [Kro 08].
I found a corresponding documentation in chapter 3.2.2. Solving Arithmetical Fragments, Table 1 a listing of the implemented algorithms for a certain logic.
Yes, SMT solvers are quite good at solving problems of this sort. Your problem can be expressed using z3's Python interface like this:
from z3 import *
x, y, z = Ints('x y z')
A, B, C = Bools('A B C')
solve (Or(3*x + y - 2*z >= 10
, And(A, Or(Not(B), C))
, And(A == C, x + y >= 5)))
This prints:
[A = True, z = 3, y = 0, B = True, C = True, x = 5]
giving you a (not necessarily "the") model that satisfies your constraints.
SMT solvers can deal with integers, machine words (i.e., bit-vectors), reals, along with many other data types, and there are efficient procedures for combinations of linear-integer-arithmetic, booleans, uninterpreted-functions, bit-vectors amongst many others.
See http://smtlib.cs.uiowa.edu for many resources on SMT solving, including references to other work. Any given solver (i.e., z3, yices, cvc etc.) will be a collection of various algorithms, heuristics and tactics. It's hard to compare them directly as each shine in their own way for certain sublogics, but for the base set of linear-integer arithmetic, booleans, and bit-vectors, they should all perform fairly well. Looks like you already found some good references, so you can do further reading as necessary; though for most end users it's neither necessary nor that important to know how an SMT solver internally works.
I'm trying to solve project Euler problem 9 using prolog. I'm a 100% n00b with prolog and I'm just trying to understand the basics right now.
I'm trying to use findall to get all triplets that would add up to 1000, but I can't really figure out the syntax.
What I'm hoping for is something like:
pythag_trip(A, B, C, D) :- D is (A * A) + (B * B) + (C * C).
one_thou_pythag(A, B, C) :- pythag_trip(A, B, C, 1000).
product_trip(A, B, C, D) :- D is A * B * C.
findall([A, B, C], one_thou_pythag(A, B, C) , Bag)).
writeln(Bag).
I know that doesn't work because it's saying Bag is not instantiated. But there are still some basics that I don't understand about the language, too.
1: can I even do this? With multiple moving pieces at once? Can I find all triplets satisfying a condition? Do I need to go down a completely different like like using clpfd?
2: What is supposed to be going in that last argument where I put Bag?
3: Is it possible to create data types? I was thinking it might be good to create a triplet set type and an operation to get the pythagorean triplet sum of them if I have to find some way to generate all the possibilities on my own
Basically those questions and then, I could use some pointing in the right direction if anyone has tips
Sorry but I don't answer your questions. It seems to me that you're trying not a prolog-like approach.
You should try to solve it logically.
So do this problem from the top to bottom.
We want to have 3 numbers that sum to 1000.
between(1,1000,A), between(A,1000,B), between(B,1000,C), C is 1000-A-B.
In that case, we will have them sorted and we won't take permutations.
So let's go a step further. We want them to be pythagorem triplet.
between(1,1000,A), between(A,1000,B), between(B,1000,C), C is 1000-A-B, is_triplet(A,B,C).
But we don't have is_triplet/3 predicate, so let's create it
is_triplet(A,B,C) :- D is C*C - A*A -B*B, D=0.
And that's actually it.
So let's sum it up
is_triple(A, B, C) :- D is C*C - A*A - B*B, D = 0.
triplet(A,B,C) :- between(1,1000,A), between(A,1000,B), C is 1000-A-B, between(B,1000,C), C is 1000-A-B, is_triple(A,B,C).
When you call triplet(A,B,C) you should get an answer.
Notice one thing, that at the end I've swapped C is 1000-A-B with between(B,1000,C). It makes the program much faster, try to think why.
I'm following this paper to implement and Attentive Pooling Network to build a Question Answering system. In chapter 2.1, it speaks about the CNN layer:
where q_emb is a question where each token (word) has been embedded using word2vec. q_emb has shape (d, M). d is the dimension of the word embedding and M the length of the question. In a similar way, a_emb is the embedding of the answer with shape (d, L).
My question is: how is the convolution done and how is it possible that W_1 and b_1 are the same for both the operations? In my opinion at least b_1 should have a different dimension in each case (and it should be a matrix, not a vector....).
At the moment I've implemented this operation in PyTorch:
### Input is a tensor of shape (batch_size, 1, M or L, d*k)
conv2 = nn.Conv2d(1, c, (d*k, 1))
I find that the authors of the paper are trusting the readers to assume/figure out a lot of things here. From what I read, here is what I could gather:
W1 should be a 1 X dk matrix because that is the only shape that would make sense in order to get Q as c X M matrix.
Assuming this, b1 need not be an matrix. From the above, you could get a c X 1 X M matrix which could be reshaped to c X M matrix easily and b1 could be a c X 1 vector which could be broadcasted and added to the rest of the matrix.
Since, c, d and k are hyper parameters, you could easily have the same W1 and b1 for both Q and A.
This is what I think so far, I will re read and edit in case anythings amiss.
if A and B are two events and P(A/B) = P(B/A) then I want to know how A and B are related to each other?. i.e. are they
1) exclusive events or
2) independent events or
3) exhaustive events
4) is it P(A) = P(B)
please let me know the answer with corresponding justification?
Rectify me if I am wrong anywhere
Bayes' Theorem states that:
P(A|B) = P(B|A) x P(A) / P(B)
So if P(A|B) = P(B|A) and is non-zero:
P(A) = P(B)
In DFA we can do the intersection of two automata by doing the cross product of the states of the two automata and accepting those states that are accepting in both the initial automata.
Union is performed similarly. How ever although i can do union in NFA easily using epsilon transition how do i do their intersection?
You can use the cross-product construction on NFAs just as you would DFAs. The only changes are how you'd handle ε-transitions. Specifically, for each state (qi, rj) in the cross-product automaton, you add an ε-transition from that state to each pair of states (qk, rj) where there's an ε-transition in the first machine from qi to qk and to each pair of states (qi, rk) where there's an ε-transition in the second machine from rj to rk.
Alternatively, you can always convert the NFAs into DFAs and then compute the cross product of those DFAs.
Hope this helps!
We can also use De Morgan's Laws: A intersection B = (A' U B')'
Taking the union of the compliments of the two NFA's is comparatively simpler, especially if you are used to the epsilon method of union.
There is a huge mistake in templatetypedef's answer.
The product automaton of L1 and L2 which are NFAs :
New states Q = product of the states of L1 and L2.
Now the transition function:
a is a symbol in the union of both automatons' alphabets
delta( (q1,q2) , a) = delta_L1(q1 , a) X delta_L2(q2 , a)
which means you should multiply the set that is the result of delta_L1(q1 , a) with the set that results from delta_L2(q1 , a).
The problem in the templatetypedef's answer is that the product result (qk ,rk) is not mentioned.
Probably a late answer, but since I had the similar problem today I felt like sharing it. Realise the meaning of intersection first. Here, it means that given the string e, e should be accepted by both automata.
Consider the folowing automata:
m1 accepting the language {w | w contains '11' as a substring}
m2 accepting the language {w | w contains '00' as a substring}
Intuitively, m = m1 ∩ m2 is the automaton accepting the strings containing both '11' and '00' as substrings. The idea is to simulate both automata simultaneously.
Let's now formally define the intersection.
m = (Q, Σ, Δ, q0, F)
Let's start by defining the states for m; this is, as mentioned above the Cartesian product of the states in m1 and m2. So, if we have a1, a2 as labels for the states in m1, and b1, b2 the states in m2, Q will consist of following states: a1b1, a2b1, a1b2, a2b2. The idea behind this product construction is to keep track of where we are in both m1 and m2.
Σ most likely remains the same, however in some cases they differ and we just take the union of alphabets in m1 and m2.
q0 is now the state in Q containing both the start state of m1 and the start state of m2. (a1b1, to give an example.)
F contains state s IF and only IF both states mentioned in s are accept states of m1, m2 respectively.
Last but not least, Δ; we define delta again in terms of the Cartesian product, as follows: Δ(a1b1, E) = Δ(m1)(a1, E) x Δ(m2)(b1, E), as also mentioned in one of the answers above (if I am not mistaken). The intuitive idea behind this construction for Δ is just to tear a1b1 apart and consider the states a1 and b1 in their original automaton. Now we 'iterate' each possible edge, let's pick E for example, and see where it brings us in the original automaton. After that, we glue these results together using the Cartesian product. If (a1, E) is present in m1 but not Δ(b1, E) in m2, then the edge will not exist in m; otherwise we'll have some kind of a union construction.
An alternative to constructing the product automaton is allowing more complicated acceptance criteria. Ordinarily, an NFA accepts an input string when it has reached any one of a set of accepting final states. That can be extended to boolean combinations of states. Specifically, you construct the automaton for the intersection like you do for the union, but consider the resulting automaton to accept an input string only when it is in (what corresponds to) accepting final states in both automata.