## Fit: How can I fit two sets of outdated data in an envelope?

My situation is similar to having a Cos and Sin function that would be combined with the envelope profile, but the envelope is offset horizontally by some dt.

It would be something like aCos (t + dt) + bSin (t + dt) = Over (t), I want to find the coefficients and the time change {a, b, dt}.

I have 2 datasets (lists) for the two input datasets and the dataset for the envelope. How can I fit this? The use of a NonLinearModelFit cannot be adjusted to so many functions at once.

## Time complexity: variant of several sets of known algorithms of subset summation problem

I have been working on the time analysis for a solver that I designed for the subset summation problem (multiple set variant), and determined that its time complexity depends on the count of repeated elements in the input.

The complexity of time is $$O (2 ^ {n / 2} cdot 0.75 ^ { frac {d / 2} {n}})$$ where $$d =$$ # of duplicates in the input instance (assuming both $$n$$ Y $$d$$ they are even)

For example when $$d = n / 2$$ so:

$$O (2 ^ n / 2} 0.75 ^ { frac {n / 4} {n}}) approx. Or (1.4142 0.93) approx. O (1,316 ^ n)$$

In addition to asking for comments, I am also looking for other known algorithms with similar behavior to compare approaches (I searched for them but did not find anything so far …)

## graphics: partition search with the maximum number of edges between sets

Given a graph (say in the form of an adjacency list), is there an algorithm to find a partition of vertices such that the number of edges between the two sets of the partition is the maximum possible?

For example, for the next set of edges of a chart with set of vertices $${1, 2, 3, 4, 5, 6 }$$:
$${(1, 2), (2, 3), (3, 1), (4, 5), (5, 6), (6, 4) }$$, a possible "maximum" partition is $${ {1, 3, 4, 6 }, {2, 5 } }$$ with $$4$$ borders between sets $${1, 3, 4, 6 }$$ Y $${2, 5 }$$.

## erd – Understand sets of ternary relationships from many to many

I am new to the ER model diagrams. I am a little confused when it comes to interpreting sets of ternary relationships like this:

Does this mean that each instance of Party relationship will have exactly one fighter, a wizard and a healer? If we were simply dealing with sets of binary relationships without key or total restrictions, each instance of the relationship would be linked to one entity from each set of entities.

But in the previous case, isn't it possible not to have an instance of a particular set of entities? E.g. if the party instance only had a fighter and a healer (and not a wizard)?

## The borel class of an accounting union of \$ G_ delta \$ -sets

Issue. Suppose a metrizable separable space $$X$$ is the accounting union $$X = bigcup_ {n in omega} X_n$$ of disjoint pairs $$G_ delta$$-sets $$X_n$$ in $$X$$ such that each $$X_n$$ it's an absolute $$F _ { sigma delta}$$-set. Is $$X$$ absolutely $$F _ { sigma delta}$$?

## nt.number theory – Eccentricity in the number of representations for sets too large to be Sidon sets

Leave $$A = {a_1 Be a set of integers. Leave $$r_A (n) = # {(a_i, a_j): a_i + a_j = n }$$ be the number of representations of $$n$$ as a sum of two elements of $$A$$. In the typical language, $$A$$ is a set of Sidon (or $$B_2$$ set) yes $$r_A (n) le 2$$ for all $$n$$. It is known that the maximum size of a Sidon set that is a subset of $${1,2, points, N }$$ is $$sqrt {N} (1 + or (1))$$.

My question, in general terms, is to ask if we can measure how often (and how much) $$r_A (n)$$ must exceed $$2$$Yeah $$A$$ contains at least $$(1+ epsilon) sqrt {N}$$ elements, for some $$epsilon> 0$$?

More specifically, let's leave $$E (A)$$ denote the "eccentricity" of $$A$$, given by
$$E (A) = sum_n max {r_A (n) -2.0 }.$$ Yes $$| A |> (1+ epsilon) sqrt {N}$$ for some $$epsilon> 0$$, then there must be some $$delta> 0$$ such that $$E (A)> delta N$$?

My impetus for asking this question comes from my attempts to understand the binary digits of $$sqrt {2}$$. It is currently known that the number of $$1$$is in the first $$N$$ binary digits of $$sqrt {2}$$ is $$ge sqrt {2N} (1 + or (1))$$, and for some infinite sequence of $$N$$This can be improved to $$ge sqrt {8N / pi} (1 + or (1))$$. However, this limit comes in part from assuming that the set of indexes of $$1$$It behaves like a set of Sidon, which is too big to really be. If you could show that $$E (A)> delta N$$, so I think a stronger lower limit could be tested.

## set theory: find minimal disjoint sets from a collection of overlapping sets

I have several sets that have overlapping elements.

``````e1, e2, e3, e4
e7,e9,e10
e1,e4
e2,e7
e3,e9
e10,e11,e12
e11,e12
``````

I want to divide the previous sets so that they don't overlap. So the output I hope is

``````e1, e4
e2
e3
e7
e9
e10
e11, e12
``````

What something / function can I write to do this?

## plotter – Contour 3D Plot of four data sets

I have four different functions that can generate four different data sets.

I want to be able to generate a single 3D contour of the result. I have tried many things and basically I don't get anything that looks remotely like what I want, a, b, c, d are constants in terms of data that x and z are axes.

Here is my code that is quite basic.

``````a = 2/Sqrt(3)*(Pi); b = 3/Sqrt(3)*(Pi); c = 2/Sqrt(2)*(Pi); d =
3/Sqrt(2)*(Pi); x1 = -(x*a); x2 = -(z*b); x3 = (x*
c); x4 = -(z*d); data1 =
Table({x1, x2}, {x, -10, 0, 1}, {z, -10, 0, 1}); data2 =
Table({x3, x4}, {x, 0, 10, 1}, {z, 0, 10, 1});
Show(ListPlot3D(data1), ListPlot3D(data2, PlotStyle -> Red))
``````

## plotting: PlotRange manually sets only one side of the plotrange

You can use `PlotRange -> {Automatic, {1, All}}`.

An example:

``````SeedRandom[1]
data = RandomReal[100, {100, 2}];
data = Append[data, {{500, 50}, {50, 500}}];

Row[{ListPlot[data,  ImageSize -> 300,
PlotLabel -> "PlotRange -> Automatic"],
ListPlot[data,  ImageSize -> 300,
PlotRange -> All, PlotLabel -> "PlotRange->All"]
ListPlot[data,  ImageSize -> 300,
PlotRange -> {Automatic, {70, All}},
PlotLabel -> "PlotRange -> {Automatic, {70, All}}"  ]},
Spacer[20]]
``````

This works in both directions. You can use `PlotRange -> {Automatic, {All, 50}}` so that the vertical plot range goes from the minimum data to 50:

``````ListPlot[data,  ImageSize -> 300,
PlotRange -> {Automatic, {All, 50}},
PlotLabel->"PlotRange -> {Automatic, {All, 50}}"  ]
``````

## You cannot create new documents within modern document sets in SP Online

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