## bitcoin core – Please check my server print to see if everything looks good or if the repeated mistakes mean that something is wrong

So I'm not sure if my Bitcoin server is syncing and everything is alright, or if the message "possibly obsolete tip detected, additional outgoing messages are being repeated" means I have to do something. The search for this string does not really give me an answer. I thought the full Bitcoin node installation would be a little more intuitive. Any help grateful

Solution Update:
My server did not connect to peers because my ROUTER had a FILTER set to Medium, which allowed only a few known ports. If I set to Low, peers will be displayed on my server instantly. Strange how the router has implemented this firewall in this way. The fact that I forwarded the port was not appreciated.

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## Proofing – How to prove the function of a recursive big theta without using repeated substitution, mastering the sentence, or having the closed form?

I have defined a function: $$V (j, k)$$ Where $$j, k in mathbb {N}$$ and $$t> 0 in mathbb {N}$$ and $$1 leq q leq j – 1$$, Note $$mathbb {N}$$ includes $$0$$,

$$V (j, k) = begin {cases} tj & k leq 2 tk & j leq 2 tjk + V (q, k / 2) + T (j – q, k / 2) & j , k> 2 end {cases}$$

I am not allowed to use repeated substitution and I want to prove this by induction. I can not use the main clause because the recursive part is not in that form. Any ideas how I can solve it with given limitations?

When I start induction: I fix $$j, q$$ and introduce $$k$$, Then the base case $$k = 0$$, Then $$V (j, 0) = tj$$, The question indicated that the function may be $$Theta (jk)$$ or maybe $$Theta (j ^ 2k ^ 2)$$ (but it does not necessarily have to be either).

I choose $$Theta (j, k)$$, In the base case, this would mean that I had to prove that $$tj = theta (j, k)$$ when $$j = 0$$, But if I start with the big-oh, I have to show it $$km leq mn = m cdot0 = 0$$ which I currently do not think possible.

I am not sure if I did the basic case wrong or if there is another approach.

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## Proving techniques – How to prove the function of a recursive Big-Oh without using repeated substitution, master phrase or closed form?

I have a function as defined $$V (j, k)$$ with two base cases with $$j, k$$ and the recursive part has an additional variable $$q$$ which it also uses. Also, $$1 leq q leq j – 1$$, The recursive part has the form: $$jk + V (q, k / 2) + V (j – q, k / 2)$$I am not allowed to use repeated substitution and I want to prove this by induction. I can not use the main clause because the recursive part is not in that form. Any ideas how I can solve it with given limitations?

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## Differential Equations – Question: How can multiple records with repeated experiments be fit to a model?

I have a question, for example from Oleksandr R. in How do I fit 3 records to a model of 4 differential equations?

Oleksandr R. made some fake data from a given model, and the values ​​of kinetic coefficients k1, k2 and k3 were given as follows:

``````sol = ParametricNDSolveValue({a'(t) == -k1 a(t) b(t) + k2 x(t),
a(0) == 1, b'(t) == -k1 a(t) b(t) + k2 x(t) - k3 b(t) x(t),
b(0) == 1, x'(t) == k1 a(t) b(t) - k2 x(t) - k3 b(t) x(t),
x(0) == 0}, {a, b, x}, {t, 0, 10}, {k1, k2, k3});

abscissae = Range(0., 10., 0.1);
ordinates =
With({k1 = 0.85, k2 = 0.15, k3 = 0.50},
Through(sol(k1, k2, k3)(abscissae), List));

data = ordinates +
RandomVariate(NormalDistribution(0, 0.1^2), Dimensions(ordinates));
ListLinePlot(data, DataRange -> {0, 10}, PlotRange -> All,
AxesOrigin -> {0, 0}, PlotStyle -> {Blue, Orange, Green})
`````` In the figure above, the curve of a is blue, orange is b, and green is x.

In the above article, Oleksandr R. has developed a way to adjust the three curves by adding an index of 1, 2 or 3 to the data set.

But what happens if you have multiple curves for the same substance, for example, if you perform several experiments with different initial concentrations? We can assume that the kinetic coefficients k1, k2, and k3 are constant and do not vary between experiments. This is a reasonable assumption, at least formally, especially when the concentrations are low. Is it then possible to adjust all data?

For example, let's create two sets of fake data from different initial conditions. Follow the above code:

``````sol1 = ParametricNDSolveValue({a'(t) == -k1 a(t) b(t) + k2 x(t),
a(0) == 1, b'(t) == -k1 a(t) b(t) + k2 x(t) - k3 b(t) x(t),
b(0) == 1, x'(t) == k1 a(t) b(t) - k2 x(t) - k3 b(t) x(t),
x(0) == 0}, {a, b, x}, {t, 0, 10}, {k1, k2, k3});
sol2 = ParametricNDSolveValue({a'(t) == -k1 a(t) b(t) + k2 x(t),
a(0) == 0.75, b'(t) == -k1 a(t) b(t) + k2 x(t) - k3 b(t) x(t),
b(0) == 0.4, x'(t) == k1 a(t) b(t) - k2 x(t) - k3 b(t) x(t),
x(0) == 0}, {a, b, x}, {t, 0, 10}, {k1, k2, k3});
abscissae = Range(0., 10., 0.1);
ordinates1 =
With({k1 = 0.85, k2 = 0.15, k3 = 0.50},
Through(sol1(k1, k2, k3)(abscissae), List));
ordinates2 =
With({k1 = 0.85, k2 = 0.15, k3 = 0.50},
Through(sol2(k1, k2, k3)(abscissae), List));

data1 = ordinates1 +
RandomVariate(NormalDistribution(0, 0.1^2), Dimensions(ordinates1));
data2 = ordinates2 +
RandomVariate(NormalDistribution(0, 0.1^2), Dimensions(ordinates2));
ListLinePlot(Join(data1, data2), DataRange -> {0, 10},
PlotRange -> All, AxesOrigin -> {0, 0},
PlotStyle -> {Blue, Orange, Green})
`````` How do you adjust the six curves (three functions from two different experiments) simultaneously? How should Oleksandr's code be modified to achieve this? Thanks!!

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## sql – How to draw ORDER BY on repeated values ​​columns?

Suppose I do one `SELECT` something like that:

``````SELECT * FROM "users" ORDER BY "createdAt";
``````

In this context, if I have data that has the same value in the field `createdAt`How does SQL map these fields?

PS: I know if I have another field in the clause `ORDER BY` "I solve" this problem.

PS: I want to know how SQL reacts to such situations. Is the same order always displayed? Or is it random if the sort-clause values ​​have this type of "collision"?

## unity – Repeated nested game objects (UnityEngine.GameObject.gameObject)

I created one `UnityEngine.GameObject` and check its properties in debug mode.

``````GameObject obj = new GameObject();
``````

When I inspect the object, it has a similar property hierarchy as:

``````
active:true
activeInHierarchy:true
..
gameObject:New Game Object (UnityEngine.GameObject)
active:true
activeInHierarchy:true
..
gameObject:New Game Object (UnityEngine.GameObject)
active:true
activeInHierarchy:true
..
gameObject:New Game Object (UnityEngine.GameObject)
``````

When I'm doing something like a debugger console:

``````new UnityEngine.GameObject().gameObject.gameObject.gameObject;
``````

it would still return the same `New Game Object`,

Is the problem in my debugger or in Unity? I use Unity 2018.4.11f1.

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## Element repeated within the matrix – Pascal

For a long time I tried to create code that identifies the user and tells him when elements in a 5×5 random number matrix repeat in Pascal. I tried to search the matrix with one or two variables and also an entire matrix, but it did not work.
I have heard from some people that it is not necessary to have a code that is too big or too complex for the task. So my desperation rises.

I ask if anyone has a solution to the problem

PasteBin link for unsuccessful (and most commonly used) attempt – https://pastebin.com/uzPLACXe

## Using the CSS grid row: 1/3; in less is translated and repeated in .333

Here is the CSS

``````    .home-grid .grid-tall {
grid-row: 1/3 !important;
}
``````

When I put that into my _extend.less file and compile it, it calculates what it should do and translates it into .333333333 (repetition). Usually I like that, as it does the calc () math, which is nice. In this case, raster line is 1/3 but ONE of THREE columns.

Does anyone know how to get less to ignore the "/" or if not, how to format the raster line? Right now I'm putting the code into my less file (and let it work) and then put a quick style in my head that overrides "! Important" … I would rather not do that if I do not have one ,

## Repeated CSS class content. Structure CSS code

I'm a backend developer who works on a website and creates the CSS part. I like to write as little code as possible and use it across the site, many pages.

I want to be able to reuse some CSS code. For example, if I have this code:

``````.box1{
background: blue;
height: 200px;
align-items: center;
}
``````

and I have 5 other fields or more that I need to use the code: "align-items: center;" because it makes sense to have them that way

``````.box1, .box2, .box3, .box4, .box5{
align-items: center;
}
``````

and

``````.box1{
background: blue;
height: 200px;
}
``````

or declare a reusable class:

``````.box-align{
align-items: center;
}
``````

The problem I have is that I have to reuse so many codes in different parts of the webpage, and I want to write as little code as possible, which can have the advantage of reducing the size of the file and updating CSS easily and fast.

This also applies to colors:

``````.box1{
color: red;
}
``````

If I choose to declare a single class and reuse the class everywhere, my HTML may contain the following code:

``````

``````

but if I decide to make it so clean:

``````

``````

then I have to do:

``````.box1, .box2, .box3, .box4, .....{
align-items: center;
color: red;
text-decoration: underline;
.......
}
``````

The main problem is that these fields have different parents on different pages.
Does anyone have a clean code architecture for this?
Just as you can declare a function, variable or constant in OOP and reuse it throughout the project.

PS: I see so many sites that write CSS code and reuse it about 50 times, such as "text-align: center", and use or declare it in about 50 classes if it could be better optimized.

## Algorithm for the number of subsequences containing at most k numbers, where no element is repeated in each subsequence

For example, if the array is 2,2,3,3,5 and k = 3
There are a total of 18 subsequences
1 subsequence of length 0 (i.e., empty subsequence)
5 subsequences of length 1

8 subsequences with length 2

4 subsequences with length 3

how does it continue?

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