Algorithms – one-to-many matching in two-part graphs?

Consider two sentences $ L $ (left) and $ R $ (right).
$ R $ Nodes have a capacity limit.
Every edge $ e $ has costs $ w (e) $,

I want to show everyone $ L $ Vertices to a node of $ R $ (One-to-many matching) with minimal total edge costs.

Every vertex in $ L $ must be mapped to a vertex in $ R $ (but every knot in $ R $ can be assigned to several $ L $-Node).

Examples: Consider the capacity of $ R $ Knot is $ 2 $,

1) This is NOT correct because a knot of $ L $ was not assigned to a node in $ R $,

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2) This is NOT correct because the capacity of a node is in $ R $ get hurt.

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3) This is correct. All $ L $ Nodes are assigned to a node in $ R $and the capacity of $ R $ Knot is not hurt.

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Any idea how I can solve that?

Reference request – What is known about the absence of strongly regular graphs srg (n, k, 0.2)?

Just two very regular graphs with parameters $ lambda = 0 $ (triangle free) and $ mu = 2 $ (any two non-adjacent vertices
have exactly two common neighbors) are known, see the Wikipedia page: the Clebsch diagram and the Sims-Gewirtz diagram.

I'm looking for information about the possible existence of other such graphics. For what values โ€‹โ€‹of $ n $ and $ k $ are you known? Not exist?

Integration – Is there a way to find "polynomrational functions" with turning points in their graphs?

For example, if the derivative of $ f (x) $ is$$ f & (39) (x) = frac {(x-1) ^ 2 (x-3)} {(x-2)} $$
then $ f (x) $ has a turning point at $ x = 1 $,

But $ f (x) $ is not a polynomial rational!

Is there a way to determine which antiderivatives are polynomial relations?

Shortest cycles in undirected graphs

Find the shortest cycle that contains a particular vertex on an undirected (and unweighted) chart $ s $It is usually said that a BFS is running $ s $ and the first time to find another visit, this is the smallest cycle length you want.

This does not seem to be true, quoted from https://www.cs.princeton.edu/courses/archive/spr08/cos226/exams/fin-f05-sol.pdf page 3:

Keep this in mind when you run BFS from $ s $ and stop once you visit a again
You may not get the vertex (using a previously unused edge)
shortest path contains $ s $,

Examples where this technique seems to be suggested: An efficient algorithm to find a shortest cycle including a specific vertex

Another technique, in which the shortest cycle in the entire diagram is determined by executing BFS from each vertex, seems to recognize only the shortest length +1 in a special case, as mentioned in this document: https: // link. springer.com/chapter /10.1007/978-3-540-78773-0_63 in the "unweighted case". on the other handHere (http://theory.stanford.edu/~virgi/cs267/lecture3.pdf) it is mentioned (first and second paragraph) that executing BFS results in every case from each vertex the shortest length (circumference). This is also mentioned in an archive (http://web.archive.org/web/20140513212046/https://webcourse.cs.technion.ac.il/234247/Winter2003-2004/ho/WCFiles/Girth.pdf) , ,

Which of all algorithms / methods applies to:

  1. Can you find the cycle with the shortest length in an undirected diagram?
  2. Find the shortest length cycle that passes through a known vertex $ s $ in an undirected graph?
  3. Where is the trap in each of my comparisons and contrasts above? (I can't believe some of the above are even wrong …)

Note: A randomized sub-cubic algorithm appears to exist in Raphael Yuster's publication "A shortest cycle for each vertex of a graph": http://research.haifa.ac.il/~raphy/papers/all-cycles .pdf

Get two Loees graphs in one

I have two records and I have to plot two on a chart. To create a loess, I need an independent variable (y) and another dependent variable (x), and I have two records, but the (y) is the same thing that changes for the (x) and I need the two regressions in the same graph, distinguishing the points corresponding to each pair and the curve of one color and another pair of a different color. I hope you understand me. This is the script I am using
Another problem is that the curve may not be tight and show strange behavior !!!! When I apply the summary, I get the statistics, but the square r is not shown. Is it possible to get the statistics?

x<-c(1:300)
y<-c(300-600)
lw1 = loess(y ~ x)
plot(y ~ x)
ggplot(lw1, aes(x, y)) + 
geom_point() +
geom_smooth(method = "loess", se = FALSE)
x<-c(20:100)
y<-c(40:120)
lw2 = loess(y ~ x)
plot(y ~ x)
ggplot(lw2, aes(x, y)) + 
geom_point() +
geom_smooth(method = "loess", se = FALSE)
summary(lw2)

dag – Is there a term for these "descendant" subgraphs of directed acyclic graphs?

It's kind of like that. But we'll use the usual computer science method to describe it in the language of binary relationships.

You are probably already familiar with binary relationships such as equality $ = $, less than or equal to $ le $subset $ subseteq $, and so on. Generally a binary relationship $ R $ about a sentence $ X $ is a subset $ R subseteq X times X $, If $ (x, y) in R $we call this as $ xRy $,

If $ forall x in X, xRx $, then $ R $ is reflexive, The relationships $ = $ and $ le $ are reflexive, however $ lt $ is not.

If $ for all x, y, z in X, xRy , Wedge , yRz Rightarrow xRz $, then $ R $ is transitive, Many relationships are transitive, including all of the above. If $ x le y $ and $ y le z $, then $ x le z $,

Given a relationship $ R $, the reflexive transitive closure of $ R $designated $ R ^ * $is the smallest relationship $ R ^ * $ so that $ R subseteq R ^ * $, and $ R ^ * $ is reflexive and transitive.

Interpret your graph as a binary relationship (since the edges don't seem to really matter to you, you're only interested in the amount of vertices). That's exactly what you want: $ xR ^ * y $ then and only if $ y $ is a "descendant" of your meaning $ x $,

If you look at the literature, you need to know another notation: the Transitive closure of $ R $designated $ R ^ + $is the smallest relationship $ R ^ + $ so that $ R subseteq R ^ + $, and $ R ^ + $ is transitive. Algorithms for calculating the transitive closure and the reflective transitive closure are related because they differ only in the "diagonal" entries: $ R ^ + cup left {(x, x) , | , x in X right } = R ^ * $,

There are several standard algorithms for calculating the RTC of a relationship. If the relationship is tight in the sense that it is feasible to represent it as a bit matrix, the Floyd-Warshall algorithm is one of the fastest practical algorithms; its term is $ Theta (| V | ^ 3) $ In theory, however, the inner loop on real hardware is quite fast because it involves a series of bit vector manipulations.

For sparse relationships, see Esko Nuutila's thesis, which contains a very good overview and some newer algorithms.

Graphs – Is there an algorithm to minimize the work set during a topological traverse?

I have a task dependency diagram that forms a DAG. I am looking for an algorithm to calculate the optimal topological traverse to minimize the "workload". Specifically, I define the current working set as the set of nodes that have been visited and that have at least one outgoing edge that is not traversed.

Even more useful would be an algorithm for calculating the optimal topological run that minimizes the number of cache flushes for a fixed-size cache for the workgroup.

Is there an algorithm to calculate this topological crossing? I can make it greedy from the topological order, but guess that's sub-optimal.

gt.geometric topology – the sanitary graphs of the sweetbread balls

To let $ p, q $ and $ r $ be positive integers. A sweetbread ball is a closed ball $ 3 $-Diversity defined by $$ Sigma (p, q, r) = {x ^ p + y ^ q + z ^ r = 0 } cap S ^ 5. $$

Its basic group is known for Milnor. It is always a rational area of โ€‹โ€‹homology. When $ p, q $ and $ r $ If coprimes continue to be selected in pairs, it is an integral homology sphere.

In this case the sanitary graph of a Brieskorn ball is easy to understand, see for example section 1 of Saveliev's book: Invariants of 3-sphere homology,

You have to find unique integers $ b, p & # 39 ;, q & # 39 ;, r & # 39; $ Solving the equation
begin {equation}
bpqr + p q qr + pq # r + pqr = = – 1
end {equation}

Where $ 1 leq p & # 39; leq p-1 $. $ 1 leq q & # 39; leq q-1 $ and $ 1 leq r & # 39; leq r-1 $, It is basically done by modifying these integers.

How about the rational case? Is it possible to find a clear representation for the sanitary diagram linked with Brieskorn balls?

How can I calculate sine signal frequencies from a sine + noise signal and its Fourier transform graphs?

I want to calculate the frequency of the two sinus peaks in the graphic below, but I really don't know how to do it.

left: sine signal + noise, right: sine frequencies

So far I only know that the signal is $ f (k) = sin (k) + n (k) $ The description shows that the length of the signal is 255 seconds with 255 discrete values. How can I calculate the peaks from the right side (i.e. the frequency of this sine wave should be a little under 50, measured on the left peak)?

Incidentally, the frequencies on the right are changed to (0.255) instead of (-128.127) to avoid confusion