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Guide to Java 8 Collectors: groupingByConcurrent()

Guide to Java 8 Collectors: groupingByConcurrent()

Introduction

A stream represents a sequence of elements and supports different kinds of operations that lead to the desired result. The source of a stream is usually a Collection or an Array, from which data is streamed from.

Streams differ from collections in several ways; most notably in that the streams are not a data structure that stores elements. They're functional in nature, and it's worth noting that operations on a stream produce a result and typically return another stream, but do not modify its source.

To "solidify" the changes, you collect the elements of a stream back into a Collection.

In this guide, we'll take a look at how to group Stream data in Java with Collectors.groupingBy()!

Collectors and Parallelism

Collectors represent implementations of the Collector interface, which implements various useful reduction operations, such as accumulating elements into collections, summarizing elements based on a specific parameter, etc.

All predefined implementations can be found within the Collectors class.

You can also very easily implement your own collector and use it instead of the predefined ones, though - you can get pretty far with the built-in collectors, as they cover the vast majority of cases in which you might want to use them.

To be able to use the class in our code we need to import it:

import static java.util.stream.Collectors.*;

Stream.collect() performs a mutable reduction operation on the elements of the stream.

A mutable reduction operation collects input elements into a mutable container, such as a Collection, as it processes the elements of the stream.

Parallel computing (parallelism) refers to the process of dividing a problem into two or more subproblems, solving those problems simultaneously, in parallel, with each subproblem being computed on a separate thread, and then combining all of the solutions to the subproblems in one uniform result.

One of the biggest challenges of implementing parallelism in programs that use collections is that the collections are non thread-safe, which means that multiple threads cannot manipulate a collection without introducing thread interference or memory consistency errors. What we also need to note is that parallelism is not necessarily faster performing than serial execution, although this heavily depends on the amount of data and the number of cores of the CPU.

Tying back into context, streams can be executed in serial or in parallel. When a stream executes in parallel, the Java runtime partitions the stream into multiple substreams. Operations are executed on independent substreams in parallel and then combined into a final result.

When creating a stream, it's always a serial stream unless stated otherwise, that is specifically parallel. To create a parallel stream we invoke the Collection.parallelStream(), which is a method found within the Collection interface.

Note: While using this method enables you to more easily implement parallelism, it is still your responsibility to determine if your application is suitable for parallelism at all, based on your knowledge of the hardware you're executing your code on.

Collectors.groupingByConcurrent()

Collectors.groupingByConcurrent() uses a multi-core architecture and is very similar to Collectors.groupingBy(), as it also behaves like the "GROUP BY" statement in SQL.

It groups objects by a given specific property and store the end result in a ConcurrentMap.

If you'd like to read more about groupingBy(), read our Guide to Java 8 Collectors: groupingBy()!

Let's define a simple class to use throughout the examples. It'll be a representation of a book, with a few fields:

public class Book {
private String title;
private String author;
private int releaseYear;

// Constructor, getters, setters, toString()
}

With the model in place, let's instantiate a list of a few books that we'll be working with:

List<Book> books = Arrays.asList(
new Book("The Lord of the Rings", "J.R.R. Tolkien", 1954),
new Book("The Hobbit", "J.R.R. Tolkien", 1937),
new Book("Animal Farm", "George Orwell", 1945),
new Book("Nineteen Eighty-Four", "George Orwell", 1949),
new Book("The Road to Wigan Pier", "George Orwell", 1937),
new Book("Lord of the Flies", "William Golding", 1954)
);

The groupingByConcurrent() has three overloads within the Collectors class. We'll be going through each one of them and explain the differences in implementation through examples along the way.

Let's start out with the simplest of them.

Collectors.groupingByConcurrent() with a Classifier Function

The first overload of this method takes only one argument - the classifier function:

public static <T,K> Collector<T,?,ConcurrentMap<K,List<T>>>
groupingByConcurrent(Function<? super T,? extends K> classifier)

This method returns a Collector that groups the input elements of type T according to the classification function. The classification function maps elements to a key of type K. The collector itself produces a ConcurrentMap<K, List<T>> whose keys represent the values we get by applying the classification function on the input, and whose corresponding values are Lists containing the input elements which map to the associated key.

This Collector is both concurrent and unordered. Being unordered, the collection operation doesn't preserve the order of the input by it's encounter. Being concurrent, the result container supports functions being called concurrently with the same result container from multiple threads.

This property isn't unique to this specific overload of the groupingByConcurrent() method, but applies to the other two overloads as well.

Let's go ahead and group the books by author:

ConcurrentMap<String, List<Book>> booksByAuthor = books.parallelStream()
.collect(Collectors.groupingByConcurrent(Book::getAuthor));

The collected elements will be unordered - but grouped. Running the same code will result in different sorts of the elements within the groups - but the sort of the groups themselves will be consistent:

{
J.R.R. Tolkien=[Book{author='J.R.R. Tolkien', title='The Hobbit', releaseYear=1937}, Book{author='J.R.R. Tolkien', title='The Lord of the Rings', releaseYear=1954}],
William Golding=[Book{author='William Golding', title='Lord of the Flies', releaseYear=1954}],
George Orwell=[Book{author='George Orwell', title='Animal Farm', releaseYear=1945}, Book{author='George Orwell', title='The Road to Wigan Pier', releaseYear=1937}, Book{author='George Orwell', title='Nineteen Eighty-Four', releaseYear=1949}]
}

Depending on how the threads in the CPU perform, and which finish their computation first - the Hobbit might appear after the Lord of the Rings, and vice versa.

Benchmarking Sequential and Concurrent Collectors.groupingBy()

While the difference between the regular groupingBy() and groupingByConcurrent() might not be obvious looking from afar - the underlying principle of grouping is significantly different.

When dealing with large amounts of books, with a decent processor, this approach may improve the performance significantly.

Let's generate a bunch of books and try grouping them sequentially and in parallel ...

List<Book> books = new ArrayList<>();
List<String> authorList = Arrays.asList(
"George Orwell",
"Nick Bostrom",
);

for (int i = 0; i < 100000; i++) {
String.valueOf(i),
authorList.get(new Random().ints(1, 1, authorList.size()).findFirst().getAsInt()),
1900));
}

long startTimeSequential = System.currentTimeMillis();
Map<String, List<Book>> booksByAuthorSequential = books.stream()
.collect(Collectors.groupingBy(Book::getAuthor));

long endTimeSequential = System.currentTimeMillis();
System.out.printf("Total time for sequential process: %sms\n",  (endTimeSequential-startTimeSequential));

long startTimeParallel = System.currentTimeMillis();
ConcurrentMap<String, List<Book>> booksByAuthorParallel = books.parallelStream()
.collect(Collectors.groupingByConcurrent(Book::getAuthor));
long endTimeParallel = System.currentTimeMillis();
System.out.printf("Total time for parallel process: %sms\n",  (endTimeParallel-startTimeParallel));

Depending on your system and CPU the sequential process might take longer or shorter than the parallel counterpart. This also heavily depends on the number of groups. If you have a few groups (fewer authors), the process of splitting them up and aggregating the results might offset the parallel approach enough to make it slower than the sequential approach.

Note: The fewer groups you're dealing with, the more likely it is for the sequential approach to outperform the parallel one, but this also highly depends on the CPU of the machine you're running the code on.

With just two authors, running this piece of code results in:

Total time for sequential process: 12ms
Total time for parallel process: 26ms

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While both processes took a really small amount of time to execute, considering the creation and grouping of 100k objects - the parallel process took significantly longer.

If we were to expand our list with a few more authors:

List <String> authorList = Arrays.asList(
"George Orwell",
"Nick Bostrom",
"Ray Kurzweil",
"J.R.R. Tolkien",
"Eliezer Yudkowsky",
"Stuart Russel",
"Max Tegmark",
"Anil Seth",
"Thomas Metzinger",
"Aurélien Geron",
"Max Lapan",
"Brian Greene",
"Frank Wilczek"
);

The results would be pretty similar:

Total time for sequential process: 13ms
Total time for parallel process: 19ms

However, if we expanded it significantly:

for (int i = 0; i < 10000; i++) {
}

Can you guess what happens now, with 10 thousand authors? Actually - the same thing:

Total time for sequential process: 19ms
Total time for parallel process: 33ms

But, if you run this code on another machine that can utilize threads more efficiently, you'll be greeted with:

Total time for sequential process: 97ms
Total time for parallel process: 52ms

Note: Concurrency isn't a silver bullet that always just works and makes code execute faster.

Collectors.groupingByConcurrent() with Classification Function and Downstream Collector

The second variation of the method takes two arguments - a classification function, and an additional, downstream collector:

public static <T,K,A,D> Collector<T,?,ConcurrentMap<K,D>>
groupingByConcurrent(Function<? super T,? extends K> classifier,
Collector<? super T,A,D> downstream)

This method returns a Collector that groups the input elements of type T according to the classification function, afterwards applying a reduction operation on the values associated with a given key using the specified downstream Collector.

The reduction operation "reduces" the data we've collected by applying an operation that's useful in a specific situation.

If you'd like to read more about reduction in Java in great detail - read our Java 8 Streams: Guide to reduce()!

Let's see an example of this variant of the method. As the downstream here we'll be using mapping(), which takes 2 parameters:

• A mapper - a function to be applied to the input elements and
• A downstream collector – a collector which will accept mapped values

Collectors.mapping() itself does a pretty straightforward job. It adapts a collector accepting elements of one type to accept a different type by applying a mapping function to each input element before accumulation. In our case, we'll map each Student to their name and return those names as a list.

Here we'll once again group our books by the author, but instead of using ConcurrentMap<String, List<Book> we'll use ConcurrentMap<String, List<String> and reduce our books into a simple string:

ConcurrentMap<String, List<String>> booksByAuthor = books.parallelStream()
.collect(Collectors.groupingByConcurrent(Book::getAuthor, Collectors.mapping(Book::getTitle, Collectors.toList())));

These are reductions of books, where we've reduced them to a title, though you could substitute this with any other reduction operation as well:

{
J.R.R. Tolkien=[The Lord of the Rings, The Hobbit],
William Golding=[Lord of the Flies],
George Orwell=[Nineteen Eighty-Four, The Road to Wigan Pier, Animal Farm]
}

Another very useful application of this overload is that our downstream function can be, well, another Collectors.groupingByConcurrent(). You can thus chain any number of groups, creating nested groups.

Let's group the books by their release year, but within those groups, we'll group the books by authors:

ConcurrentMap<Integer, ConcurrentMap<String, List<String>>> booksByAuthor = books.parallelStream()
.collect(Collectors.groupingByConcurrent(Book::getReleaseYear,
Collectors.groupingByConcurrent(Book::getAuthor, Collectors.mapping(Book::getTitle, Collectors.toList()))));

And get the following output:

{
1937={J.R.R. Tolkien=[The Hobbit], George Orwell=[The Road to Wigan Pier]},
1954={J.R.R. Tolkien=[The Lord of the Rings], William Golding=[Lord of the Flies]},
1945={George Orwell=[Animal Farm]},
1949={George Orwell=[Nineteen Eighty-Four]}
}

Collectors.groupingBy() with Classifier Function, Downstream Collector and Supplier

The third and last overload of this method takes three arguments. The first and the third are the same as in the previous overload, but the second argument is a supplier method.

The supplier method provides the specific ConcurrentMap implementation we want to use to contain our end result. We have two known classes that implement this interface - ConcurrentHashMap and ConcurrentSkipListMap:

public static <T,K,A,D,M extends ConcurrentMap<K,D>> Collector<T,?,M>
groupingByConcurrent(Function<? super T,? extends K> classifier,
Supplier<M> mapFactory,
Collector<? super T,A,D> downstream)

The return value of this method is the same as the previous overload as well. The only difference with this one is that the ConcurrentMap produced by the collector is created with the supplied factory function.

We'll be doing just one simple example for this overload, since everything is pretty much the same as the previous with the exception of the specified ConcurrentMap implementation:

ConcurrentMap<String, List<String>> booksByAuthor = books.parallelStream()
.collect(Collectors.groupingByConcurrent(Book::getAuthor,
ConcurrentHashMap::new,
Collectors.mapping(Book::getTitle, Collectors.toList())));

Conclusion

The Collectors class is a powerful one and allows us to collect streams into collections in various ways.

You can define your own collectors, but the built-in collectors can get you very far as they're generic and can be generalized to the vast majority of tasks you can think of.

In this guide, we've gone through a few examples of using the Collectors.groupingByConcurrent() method, that groups elements together given specific parameters and returns a ConcurrentMap.

By using this method instead of the non-concurrent Collectors.groupingBy() we can fully utilize the multi-core architecture, if the underlying hardware allows us to. However, while using this method enables you to more easily implement parallelism, it is still your responsibility to determine if your application is suitable for parallelism at all.

You've learned how to use the basic form, as well as forms with downstream collectors and suppliers to simplify code and run powerful yet simple functional operations on streams.

Last Updated: December 1st, 2021


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Shelf spac… As dynamic programming aims to reuse the code I know that it is necessary to use a recursive function, but when analyzing the problem I assumed that my answer field is in a matrix where the lines are referring to the number of refrigerators and the columns the stores. We study the profit maximization problem of a cognitive virtual network operator in a dynamic network environment. So there must be a faster way. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. In mathematical optimization, constrained optimization (in some contexts called constraint optimization) is the process of optimizing an objective function with respect to some variables in the presence of constraints on those variables. Both a general algebraic derivation of the problem and the optimality conditions and speciﬁc numerical examples are presented. From the remaining 720 we add (o 3, 300) for a marginal profit of 2.333%. A clever way to solve this problem is to break this problem into two subproblems. Then the solution is simply the sum of the solutions of the above two problems. THE FIRM’S PROFIT MAXIMIZATION PROBLEM These notes are intended to help you understand the ﬁrm’s problem of maximizing proﬁts given the available technology. Dynamic programming solves problems by combining the solutions to subproblems. CodeChef was created as a platform to help programmers make it big in the world of algorithms, computer programming, and programming contests.At CodeChef we work hard to revive the geek in you by hosting a programming contest at the start of the month and two smaller programming challenges at the middle and end of the month. are collecting terabytes of data on a daily basis, every decision in the brick and mortar stores is carefully thought through and analyzed. The researcher was constraint by time as time frame for the submission of this research was short for an expansive research. You need to output the maximum profit you can take, such that there are no two jobs in the subset with an overlapping time range. Let profit[t][i] represent maximum profit using at most t transactions up to day i (including day i). Market Value Maximization and Markov Dynamic Programming . Example. Dynamic Optimization Joshua Wilde, revised by Isabel ecu,T akTeshi Suzuki and María José Boccardi August 13, 2013 Up to this point, we have only considered constrained optimization problems at a single point in time. 13. Linear programming problemsare an important class of optimization problems, that helps to find the feasible region and optimize the solution in order to have the highest or lowest value of the function. But I am interested in this question, not 1-18. Linear Programming is a widely used mathematical modelling technique designed to help managers in planning and decisions making relative to resource allocation. Use of nous when moi is used in the subject. For dynamic programming problems in general, knowledge of the current state of the system conveys all the information about its previous behavior nec- essary for determining the optimal policy henceforth. Maximizing profit for given stock quotes. One important characteristic of this system is the state of the system evolves over time, producing a sequence of observations along the way. [8] [9] [10] In fact, Dijkstra's explanation of the logic behind the algorithm,[11] namely Problem 2. Reviews on Profit Maximization in the Bank Log in. Any expert developer will tell you that DP mastery involves lots of practice. Teunter R.H.: Determining Optimal Disassembly and Recovery Strategies. Who first called natural satellites "moons"? We first select to add (o 5, 1000) to our portofolio for a marginal profit of 2.4%. Why is a third body needed in the recombination of two hydrogen atoms? Analytics. CodeChef - A Platform for Aspiring Programmers. 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Scott Iowa State College Follow this and additional works at:https://lib.dr.iastate.edu/rtd Part of theAccounting Commons,Agricultural Economics Commons, and theEconomics Commons These problems, usually having a complex form, are disintegrated into smaller sub-problems whose optimal solutions lead to the optimal solution of the original problem. What is the application of rev in real life? Were there often intra-USSR wars? It provides a systematic procedure for determining the optimal com-bination of decisions. Stochastic Dynamic Programming for Wind Farm Power Maximization Yi Guo, Mario Rotea, Tyler Summers Abstract Wind plants can increase annual energy produc-tion with advanced control algorithms by coordinating the operating points of individual turbine controllers across the farm. "Proceedings of the IEEE International Conference on Systems, Man and Cybernetics" 2002, 5, pp. Space complexity is also O(n). A Profit-Maximization Dynamic Model for Supply Chain Planning. 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Each period the farmer has a stock of seeds. Then the relation is: profit[t][i] = max(profit[t][i-1], max(price[i] – price[j] + profit[t-1][j])) But the aim is to maximize the profit by buying a subset of these houses. In particular, assume that F(x) is concave, lies above the replacement line y = x if x E (0, K), F(0) = 0, F(K) = K, Su is the smallest positive x such that F'(x) = 1 and recall the equations For a total amount of 1720 this method works flawlessly. Active 3 years, 3 months ago. In the world of analytics, where retail giants like Walmart, Target etc.


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# What is a right way to compare AUCROC between logreg and logregs on segments?

I built a logistic regression on the entire sample (first model), as well as regressions on separate pieces (disjoint) of same sample (set of models). What is the best way to compare the effectiveness of the first model and the set of models (which are built on segments)? A common practice for evaluating the effectiveness of a set of models is some variant of the average AUCROC/GINI (does not matter) (as, for example, in the roc_auc_score for a multi-label case). Can it be permissible to calculate AUCROC for a union of predict_probas from the segments? And then compare that AUCROC vs AUCROC of first model?


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# CAT 2020 Quant Question [Slot 2] with Solution 01

Question

The sum of the perimeters of an equilateral triangle and a rectangle is 90 cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfy the relationship $R = {T^2}$. If the sides of the rectangle are in the ratio 1: 3, then the length, in cm, of the longer side of the rectangle, is

1. 24
2. 27
3. 21
4. 18
Option: 2
Solution:

Let the breadth of the rectangle be b.

Length of the rectangle $= 3b$

Let a be the side of the equilateral triangle.

Given,

$R = {T^2} \Rightarrow 3{b^2} = {\left( {\frac{{\sqrt 3 }}{4} \times {a^2}} \right)^2} \Rightarrow b = {a^2}/4$

Given,

$2(4b) + 3a = 90$

$\Rightarrow 8\left( {{a^2}/4} \right) + 3a - 90 = 0$

$\Rightarrow 2{a^2} + 3a - 90 = 0$

$\Rightarrow (a - 6)(2a + 15) = 0$

$\Rightarrow a = 6$

$\therefore b = 9$

$\Rightarrow 3b = 27$

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text:
Capturing Winning Ideas in Online Design Communities

Faez Ahmed and Mark Fuge, 20th ACM Conference on Computer-Supported Cooperative Work & Social Computing (2017).
Full text
DOI
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# Abstract

This paper describes how to find or filter high-quality ideas submitted by members collaborating together in online communities. Typical means of organizing community submissions, such as aggregating community or crowd votes, suffer from the cold-start problem, the rich-get-richer problem, and the sparsity problem. To circumvent those, our approach learns a ranking model that combines 1) community feedback, 2) idea uniqueness, and 3) text features—e.g., readability, coherence, semantics, etc. This model can then rank order submissions by expected quality, supporting community members in finding content that can inspire them and improve collaboration among members. As illustrative example, we demonstrate the model on OpenIDEO—a collaborative community where high-quality submissions are rewarded by winning design challenges. We find that the proposed ranking model finds winning ideas more effectively than existing ranking techniques (comment sorting), as measured using both Discounted Cumulative Gain and human perceptions of idea quality. We also identify the elements of winning ideas that were highly predictive of subsequent success: 1) engagement with community feedback, 2) submission length, and 3) a submission’s uniqueness. Ultimately, our approach enables community members and managers to more effectively manage creative stimuli created by large collaborative communities.

# BibTeX Citation

@inproceedings{ahmed:cscw_2016_winning,


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## 计算机代写|机器学习代写machine learning代考|COMP5318

2022年12月30日

couryes-lab™ 为您的留学生涯保驾护航 在代写机器学习 machine learning方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写机器学习 machine learning代写方面经验极为丰富，各种代写机器学习 machine learning相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
couryes™为您提供可以保分的包课服务

## 计算机代写|机器学习代写machine learning代考|Limiting Spectrum of the Null Model

As previously mentioned, it is convenient to start by investigating the null model innerproduct kernel matrix $\mathbf{K}=\mathbf{K}N$ with $$[\mathbf{K}]{i j}= \begin{cases}f\left(\mathbf{z}i^{\top} \mathbf{z}_j / \sqrt{p}\right) / \sqrt{p} & \text { for } i \neq j \ 0 & \text { for } i=j\end{cases}$$ for i.i.d. $\mathbf{z}_i \sim \mathcal{N}\left(\mathbf{0}, \mathbf{I}_p\right){ }^{10}$ We are, as usual, interested in the associated resolvent $$\mathbf{Q}(z) \equiv\left(\mathbf{K}-z \mathbf{I}_n\right)^{-1} \in \mathbb{R}^{n \times n} .$$ Following the Marčenko-Pastur and Bai-Silverstein approaches (in Theorems $2.4$ and 2.6, respectively), we first remove the $i$ th row and the $i$ th column of the symmetric matrix $\mathbf{K}$ to decompose it, up to permutation, as \begin{aligned} \mathbf{K} & =\left[\begin{array}{cc} \mathbf{K}{-i} & f\left(\mathbf{Z}{-i}^{\top} \mathbf{z}_i / \sqrt{p}\right) / \sqrt{p} \ f\left(\mathbf{z}_i^{\top} \mathbf{Z}{-i} / \sqrt{p}\right) / \sqrt{p} & 0 \end{array}\right] \ \text { with } \mathbf{K}{-i} & \equiv f\left(\mathbf{Z}{-i}^{\top} \mathbf{Z}{-i} / \sqrt{p}\right) / \sqrt{p}-\operatorname{diag}(\cdot) \in \mathbb{R}^{(n-1) \times(n-1)}, \end{aligned} (i.e., with zeros on the diagonal of $\left.\mathbf{K}{-i}\right)$, where $\mathbf{Z}{-i} \in \mathbb{R}^{p \times(n-1)}$ is the Gaussian matrix $\mathbf{Z}$ without the $i$ th column $\mathbf{z}_i$. As such, $\mathbf{K}{-i}$ is (i) independent of $\mathbf{z}i$, and is (ii) asymptotically close to $\mathbf{K}$ for $n$ large. We similarly define the resolvent of $\mathbf{K}{-i}$ as
$$\mathbf{Q}{-i} \equiv\left(\mathbf{K}{-i}-z \mathbf{I}{n-1}\right)^{-1} \in \mathbb{R}^{(n-1) \times(n-1)} .$$ With Lemma 2.6, the $(i, i)$ th diagonal entry of $\mathbf{Q}$ is given by $$[\mathbf{Q}]{i i}=\frac{1}{-z-\frac{1}{p} f\left(\mathbf{z}i^{\top} \mathbf{Z}{-i} / \sqrt{p}\right) \mathbf{Q}{-i} f\left(\mathbf{Z}{-1}^{\top} \mathbf{z}i / \sqrt{p}\right)} \equiv \frac{1}{-z-\delta_i},$$ where we recall that the diagonals of both $\mathbf{K}$ and $\mathbf{K}{-i}$ are zero. To evaluate the Stieltjes transform $m_n(z)=\frac{1}{n} \operatorname{tr} \mathbf{Q}(z)=\frac{1}{n} \sum_{i=1}^n \mathbf{Q}{i i}(z)$ of the spectral measure of $\mathbf{K}$, the key object is thus the (nonlinear) quadratic form $$\delta_i \equiv \frac{1}{p} f\left(\mathbf{z}_i^{\top} \mathbf{Z}{-i} / \sqrt{p}\right) \mathbf{Q}{-i} f\left(\mathbf{Z}{-i}^{\top} \mathbf{z}_i / \sqrt{p}\right)$$

## 计算机代写|机器学习代写machine learning代考|Properly Scaling Random Kernel Matrices

Having covered the analysis of the (pure-noise or null model) kernel matrix $\mathbf{K}N$, we present in this section the “information-plus-noise” random (asymptotic) equivalent for the kernel matrix $\mathbf{K}$, again under the nontrivial classification assumptions on the $k$-class mixture model defined in (4.20) (as for the $\alpha-\beta$ kernel studied in Section 4.2.4). The main idea for this “information-plus-noise” decomposition comes in two steps: (i) first, by an expansion of $\mathbf{x}_i^{\top} \mathbf{x}_j$ as a function of $\mathbf{z}_i, \mathbf{z}_j$ and the statistical mixture model parameters $\left{\boldsymbol{\mu}_a, \mathbf{E}_a\right}{a=1}^k$, the inner products $\mathbf{x}_i^{\top} \mathbf{x}_j$ are developed into successive orders of magnitudes with respect to $p$; this further allows for a Taylor expansion of $f\left(\mathbf{x}_i^{\top} \mathbf{x}_j / \sqrt{p}\right)$ for at least twice differentiable functions $f$ around its dominant term $f\left(\mathbf{z}_i^{\top} \mathbf{z}_j / \sqrt{p}\right)$. Then, (ii) relying on the orthogonal polynomial approach of the previous section, one may “linearize” the resulting matrix terms $\left{f\left(\mathbf{x}_i^{\top} \mathbf{x}_j / \sqrt{p}\right)\right},\left{f^{\prime}\left(\mathbf{x}_i^{\top} \mathbf{x}_j / \sqrt{p}\right)\right}$ and $\left{f^{\prime \prime}\left(\mathbf{x}_i^{\top} \mathbf{x}_j / \sqrt{p}\right)\right}$ (all terms corresponding to higher-order derivatives asymptotically vanish) and use Assumption 2 to extend the result to all square-summable $f$. The precise derivations may be found in Liao and Couillet [2019a].

The main conclusion is that the kernel matrix $\mathbf{K}$ asymptotically behaves like the sum $\tilde{\mathbf{K}}=\mathbf{K}_N+\tilde{\mathbf{K}}_I$ of the full-rank “noise” matrix $\mathbf{K}_N$ (characterized in Theorems $4.4$ and 4.5) and a low-rank “information” matrix $\tilde{\mathbf{K}}_I$. This is stated in the following theorem.

Theorem 4.6 (Random equivalent for properly scaling kernel, Liao and Couillet [2019a]). Let Assumption 2 hold and $\mathbf{K} \in \mathbb{R}^{n \times n}$ be the properly scaling kernel defined in (4.19) with $\mathbf{x}i=\mu_a+\left(\mathbf{I}_p+\mathbf{E}_a\right)^{\frac{1}{2}} \mathbf{z}_i$, for $\mathbf{z}_i$ having i.i.d. zero-mean, unit-variance and subexponential entries, $\mathbf{x}_i \in \mathcal{C}_a$ satisfying the following growth rate conditions $$\mathbf{M}=O{|\cdot|}(1), \mathbf{t}=\frac{1}{\sqrt{p}}\left{\operatorname{tr} \mathbf{E}a\right}{a=1}^k=O_{|\cdot|}(1), \mathbf{S}^{\circ}=\frac{1}{\sqrt{p}}\left{\operatorname{tr} \mathbf{E}a \mathbf{E}_b\right}{a, b=1}^k=O_{|\cdot|}(1) .$$
Then, as $n, p \rightarrow \infty$ with $p / n \rightarrow c \in(0, \infty)$,
$$|\mathbf{K}-\tilde{\mathbf{K}}| \stackrel{\text { a.s. }}{\longrightarrow} 0, \quad \tilde{\mathbf{K}}=\mathbf{K}_N+\mathbf{V A V}^{\top}$$
with $\mathbf{K}_N$ defined in (4.21) and
\begin{aligned} & \mathbf{A}=\left[\begin{array}{cc} a_1 \cdot \mathbf{M}^{\top} \mathbf{M}+\frac{a_2}{\sqrt{2}} \cdot\left(\mathbf{t 1}_k^{\top}+\mathbf{1}_k \mathbf{t}^{\top}+\mathbf{S}^{\circ}\right) & a_1 \mathbf{I}_k \ a_1 \mathbf{I}_k & \mathbf{0} \end{array}\right] \ & \end{aligned}
where we recall that $a_1$ and $a_2$ are the first two Hermite coefficients $a_1=\mathbb{E}[\xi f(\xi)]$ and $a_2=\mathbb{E}\left[\left(\xi^2-1\right) f(\xi)\right] / \sqrt{2}$ for $\xi \sim \mathcal{N}(0,1)$, as defined in (4.23).

# 机器学习代考

## 计算机代写|机器学习代写machine learning代考|Limiting Spectrum of the Null Model

$[\mathbf{K}] i j=\left{f\left(\mathbf{z} i^{\top} \mathbf{z}j / \sqrt{p}\right) / \sqrt{p} \quad\right.$ for $i \neq j 0 \quad$ for $i=j$ 对于 $\mathrm{iid} \mathbf{z}_i \sim \mathcal{N}\left(0, \mathbf{I}_p\right)^{10}$ 像往常一样，我们对相关的解决方案感兴 趣 $$\mathbf{Q}(z) \equiv\left(\mathbf{K}-z \mathbf{I}_n\right)^{-1} \in \mathbb{R}^{n \times n}$$ 遭循 Marčenko-Pastur 和 Bai-Silverstein 方法 (在定理中 $2.4$ 和 $2.6$ ，分别)，我们首先删除 $i$ 第行和 $i$ 对称矩阵的第列 $\mathbf{K}$ 分解它，直 到排列，如 (即，在对角线上有零 $\mathbf{K}-i$ )，在哪里 $\mathbf{Z}-i \in \mathbb{R}^{p \times(n-1)}$ 是高斯矩 阵 $\mathbf{Z}$ 没有 $i$ 第 列 $\mathbf{z}_i$. 像这样， $\mathbf{K}-i$ (i) 独立于 $\mathbf{z} i$, 并且 (ii) 渐近接近 $\mathbf{K}$ 为 了 $n$ 大。我们类似地定义的解决方案 $\mathbf{K}-i$ 作为 $$\mathbf{Q}-i \equiv(\mathbf{K}-i-z \mathbf{I} n-1)^{-1} \in \mathbb{R}^{(n-1) \times(n-1)} \text {. }$$ 使用引理 2.6，(i,i)的第 对角线条目 $\mathbf{Q}$ 是（谁） 给的 $$[\mathbf{Q}] i i=\frac{1}{-z-\frac{1}{p} f\left(\mathbf{z} i^{\top} \mathbf{Z}-i / \sqrt{p}\right) \mathbf{Q}-i f\left(\mathbf{Z}-1^{\top} \mathbf{z} i / \sqrt{p}\right)} \equiv$$ 我们记得两者的对角线 $\mathbf{K}$ 和 $\mathbf{K}-i$ 为雴。评估 Stieltjes 变换 $m_n(z)=\frac{1}{n} \operatorname{tr} \mathbf{Q}(z)=\frac{1}{n} \sum{i=1}^n \mathbf{Q} i i(z)$ 的光谱测量 $\mathbf{K}$ ，关键对象 因此是 (非线性) 二次形式
$$\delta_i \equiv \frac{1}{p} f\left(\mathbf{z}_i^{\top} \mathbf{Z}-i / \sqrt{p}\right) \mathbf{Q}-i f\left(\mathbf{Z}-i^{\top} \mathbf{z}_i / \sqrt{p}\right)$$

## 计算机代写|机器学习代写machine learning代考|Properly Scaling Random Kernel Matrices

$\backslash$ mathbf ${\mathrm{M}}=O{|\backslash \mathrm{Idot}|}(1)$, \mathbf ${\mathrm{t}}=\backslash \mathrm{frac}{1}{\backslash \mathrm{sqrt}{\mathrm{p}}} \backslash$ eft ${$ \operatorn

$$|\mathbf{K}-\tilde{\mathbf{K}}| \stackrel{\text { a.s. }}{\longrightarrow} 0, \quad \tilde{\mathbf{K}}=\mathbf{K}_N+\mathbf{V A} \mathbf{V}^{\top}$$

$$\mathbf{A}=\left[a_1 \cdot \mathbf{M}^{\top} \mathbf{M}+\frac{a_2}{\sqrt{2}} \cdot\left(\mathbf{t} \mathbf{1}_k^{\top}+\mathbf{1}_k \mathbf{t}^{\top}+\mathbf{S}^{\circ}\right) \quad a_1 \mathbf{I}_k a_1 \mathbf{I}_k\right.$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
The scale model of a skyscraper being built is 4.2 feet tall.
a. When it is finished, the skyscraper will be 525 meters tall. What
scale was used to make the model?

The scale that was used in the model can be calculated by first converting the height of the model from feet to meter in order to have consistency in the units.
height of model = (4.2 ft)(1 m/3.28 ft) = 1.28 m
Since it is said that this corresponds to 525 m actual height of the skyscraper, the scale used was:
1.28 m: 525 m
or simply,
32:13125

For this case we must take into account the following transformation of units:

1 meter = 3.28 feet

Applying the transformation of units we have:

$$(4.2feet)(\frac{1}{3.28}\frac{m}{feet}) = 1.28 m$$

Thus, the scale used for the model is given by:

1.28:525


text:
# Please state in 2-3 sentences the guidelines you should follow in order to work within your...

###### Question:

Please state in 2-3 sentences the guidelines you should follow in order to work within your scope of practice.

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The street department's storage building, which is used to store the rock, gravel, and salt used on the city's roadways, is built in the shape of a regular hexagonal pyramid. The altitude of the pyramid has the same length as any side of the base. If the volume of the interior is $11,972 \... 1 answer ##### The fixed setup overhead flexibleminus−budget variance is calculated as actual costs minus− flexibleminus−budget variance. True False The fixed setup overhead flexibleminus−budget variance is calculated as actual costs minus− flexibleminus−budget variance. True False... 5 answers ##### 11. The Department of Tourism reported that in May of 2017,approximately 63%, percent of the 500,000 tourists to the Philippines were from Asia; Suppose another organization had taken simple random sample of 600 of the tourists in that population, and assume that the reported 63% percent claim is accurate, what is the approximate probability that another organization'$ results were within 2 percentage points of the Department of Tourism'\$ results? 12. Suppose simple random sample of 15
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text:
# SPM Additional Mathematics 2017, Paper 2 (Question 13 & 15)

Question 13 (10 marks):
Table 2 shows the prices and the price indices of three types of ingredients P, Q and R, used in the production of a type of instant noodle.

(a) The price of ingredient Q increased by 20% from the year 2014 to the year 2016.
(i) State the value of x.
(ii) Find the value of y.

(b) Calculate the composite index for the cost of making the instant noodles for the year 2016 based on the year 2014.

(c) It is given that the composite index for the cost of making the instant noodles increased by 40% from the year 2012 to the year 2016.

(i)
Calculate the composite index for the cost of making the instant noodles in the year 2014 based on the year 2012.

(ii)
The cost of making a packet of instant noodle is 10 sen in the year 2012.
Find the maximum number of packet of instant noodles that can be produced using an allocation of RM80 in the year 2016.

Solution:
(a)(i)
x = 120

(a)(ii)
$\begin{array}{l}\frac{{P}_{2016}}{{P}_{2014}}×100=120\\ \frac{3.00}{y}×100=120\\ y=\frac{3.00}{120}×100\\ y=2.50\end{array}$

(b)

(c)(i)

(c)(ii)

Question 15 (10 marks):
Solution by scale drawing is not accepted.
Diagram 8 shows a quadrilateral PQRS on a horizontal plane.
VQSP is a pyramid such that PQ = 12 m and V is 5 m vertically above P.
Find
(a)QSR,
(b) the length, in m, of QS,
(c) the area, in m2, of inclined plane QVS.

Solution:
(a)
$\begin{array}{l}\frac{\mathrm{sin}\angle QSR}{20.5}=\frac{\mathrm{sin}{64}^{o}}{22}\\ \mathrm{sin}\angle QSR=\frac{\mathrm{sin}{64}^{o}}{22}×20.5\\ \mathrm{sin}\angle QSR=0.8375\\ \angle QSR={56}^{o}52‘\end{array}$

(b)

(c)


text:
Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memoryless binary symmetric channel with crossover probability p?

This question was previously asked in
GATE EC 2017 Official Paper: Shift 2
View all GATE EC Papers >

Option 3 :
Free
CT 1: Ratio and Proportion
2846
10 Questions 16 Marks 30 Mins

Detailed Solution

Concept:

Binary symmetric channel:

The channel capacity per manage symbol for which the unit is bits/symbol is defined as the maximum value of mutual information and can be written as;

Cm = max [I(x, y)]

When I(x, y) = Mutual information

Cm = 1 + plog2 P + (1 - p) log2 (1 - p) … (1)

Calculation:

From Equation-(1) the capacity is varying logarithmically.

Let, p = cross over probability ‘or’ probability of error.

For p = 0;

From Equation - (1),

Cm = 1;

For p $$= \frac{1}{2}$$

$${C_m} = 1 + \frac{1}{2}{\log _2}\left( {\frac{1}{2}} \right) + \left( {1 - \frac{1}{2}} \right){\log _2}\left( {1 - \frac{1}{2}} \right)$$

= 0

And for p = 1;

Cm = 1

Since the capacity varies logarithmically, the complete graph will be as shown:

Option (C) →


text:
# Uniform convergence (combinatorics)

For a class of predicates $H\,\!$ defined on a set $X\,\!$ and a set of samples $x=(x_{1},x_{2},\dots,x_{m})\,\!$, where $x_{i}\in X\,\!$, the empirical frequency of $h\in H\,\!$ on $x\,\!$ is $\widehat{Q_{x}}(h)=\frac{1}{m}|\{i:1\leq i\leq m,h(x_{i})=1\}|\,\!$. The Uniform Convergence Theorem states, roughly,that if $H\,\!$ is "simple" and we draw samples independently (with replacement) from $X\,\!$ according to a distribution $P\,\!$, then with high probability all the empirical frequency will be close to its expectation, where the expectation is given by $Q_{P}(h)=P\{y\in X:h(y)=1\}\,\!$. Here "simple" means that the Vapnik-Chernovenkis dimension of the class $H\,\!$ is small relative to the size of the sample.
In other words, a sufficiently simple collection of functions behaves roughly the same on a small random sample as it does on the distribution as a whole.

## Uniform convergence theorem statement[1]

If $H\,\!$ is a set of $\{0,1\}\,\!$-valued functions defined on a set $X\,\!$ and $P\,\!$ is a probability distribution on $X\,\!$ then for $\epsilon>0\,\!$ and $m\,\!$ a positive integer, we have,

$P^{m}\{|Q_{P}(h)-\widehat{Q_{x}}(h)|\geq\epsilon\,\!$ for some $h\in H\}\leq 4\Pi_{H}(2m)e^{-\frac{\epsilon^{2}m}{8}}.\,\!$

where, for any $x\in X^{m}\,\!$,

$Q_{P}(h)=P\{(y\in X:h(y)=1\}\,\!$,
$\widehat{Q_{x}}(h)=\frac{1}{m}|\{i:1\leq i\leq m,h(x_{i})=1\}|\,\!$ and $|x|=m\,\!$. $P^{m}\,\!$ indicates that the probability is taken over $x\,\!$ consisting of $m\,\!$ i.i.d. draws from the distribution $P\,\!$.

$\Pi_{H}\,\!$ is defined as: For any $\{0,1\}\,\!$-valued functions $H\,\!$ over $X\,\!$ and $D\subseteq X \,\!$,

$\Pi_{H}(D)=\{h\cap D:h\in H\}\,\!$.

And for any natural number $m\,\!$ the shattering number $\Pi_{H}(m)\,\!$ is defined as.

$\Pi_{H}(m)=max|\{h\cap D:|D|=m,h\in H\}|\,\!$.

From the point of Learning Theory one can consider $H\,\!$ to be the Concept/Hypothesis class defined over the instance set $X\,\!$. Before getting into the details of the proof of the theorem we will state Sauer's Lemma which we will need in our proof.

## Sauer–Shelah lemma

The Sauer–Shelah lemma[2] relates the shattering number $\Pi_{h}(m)\,\!$ to the VC Dimension.

Lemma: $\Pi_{H}(m)\leq\left( \frac{em}{d}\right)^{d}\,\!$, where $d\,\!$ is the VC Dimension of the concept class $H\,\!$.

Corollary: $\Pi_{H}(m)\leq m^{d}\,\!$.

## Proof of uniform convergence theorem [1]

Before we get into the details of the proof of the Uniform Convergence Theorem we will present a high level overview of the proof.

1. Symmetrization: We transform the problem of analyzing $|Q_{P}(h)-\widehat{Q}_{x}(h)|\geq\epsilon\,\!$ into the problem of analyzing $|\widehat{Q}_{r}(h)-\widehat{Q}_{s}(h)|\geq\epsilon/2\,\!$, where $r\,\!$ and $s\,\!$ are i.i.d samples of size $m\,\!$ drawn according to the distribution $P\,\!$. One can view $r\,\!$ as the original randomly drawn sample of length $m\,\!$, while $s\,\!$ may be thought as the testing sample which is used to estimate $Q_{P}(h)\,\!$.
2. Permutation: Since $r\,\!$ and $s\,\!$ are picked identically and independently, so swapping elements between them will not change the probability distribution on $r\,\!$ and $s\,\!$. So, we will try to bound the probability of $|\widehat{Q}_{r}(h)-\widehat{Q}_{s}(h)|\geq\epsilon/2\,\!$ for some $h \in H\,\!$ by considering the effect of a specific collection of permutations of the joint sample $x=r||s\,\!$. Specifically, we consider permutations $\sigma(x)\,\!$ which swap $x_i\,\!$ and $x_{m+i}\,\!$ in some subset of ${1,2,...,m}\,\!$. The symbol $r||s\,\!$ means the concatenation of $r\,\!$ and $s\,\!$.
3. Reduction to a finite class: We can now restrict the function class $H\,\!$ to a fixed joint sample and hence, if $H\,\!$ has finite VC Dimension, it reduces to the problem to one involving a finite function class.

We present the technical details of the proof.

### Symmetrization

Lemma: Let $V=\{x\in X^{m}:|Q_{P}(h)-\widehat{Q_{x}}(h)|\geq\epsilon\,\!$ for some $h\in H\}\,\!$ and

$R=\{(r,s)\in X^{m}\times X^{m}:|\widehat{Q_{r}}(h)-\widehat{Q_{s}}(h)|\geq\epsilon /2\,\!$ for some $h\in H\}\,\!$.

Then for $m\geq\frac{2}{\epsilon^{2}}\,\!$, $P^{m}(V)\leq 2P^{2m}(R)\,\!$.

Proof: By the triangle inequality,
if $|Q_{P}(h)-\widehat{Q_{r}}(h)|\geq\epsilon\,\!$ and $|Q_{P}(h)-\widehat{Q_{s}}(h)|\leq\epsilon /2\,\!$ then $|\widehat{Q_{r}}(h)-\widehat{Q_{s}}(h)|\geq\epsilon /2\,\!$.
Therefore,

$P^{2m}(R)\geq P^{2m}\{\exists h\in H,|Q_{P}(h)-\widehat{Q_{r}}(h)|\geq\epsilon\,\!$ and $|Q_{P}(h)-\widehat{Q_{s}}(h)|\leq\epsilon /2\}\,\!$
$=\int_{V}P^{m}\{s:\exists h\in H,|Q_{P}(h)-\widehat{Q_{r}}(h)|\geq\epsilon\,\!$ and $|Q_{P}(h)-\widehat{Q_{s}}(h)|\leq\epsilon /2\}dP^{m}(r)=A\,\!$ [since $r\,\!$ and $s\,\!$ are independent].

Now for $r\in V\,\!$ fix an $h\in H\,\!$ such that $|Q_{P}(h)-\widehat{Q_{r}}(h)|\geq\epsilon\,\!$. For this $h\,\!$, we shall show that

$P^{m}\{|Q_{P}(h)-\widehat{Q_{s}}(h)|\leq\frac{\epsilon}{2}\}\geq\frac{1}{2}\,\!$.

Thus for any $r\in V\,\!$, $A\geq\frac{P^{m}(V)}{2}\,\!$ and hence $P^{2m}(R)\geq\frac{P^{m}(V)}{2}\,\!$. And hence we perform the first step of our high level idea.
Notice, $m\cdot \widehat{Q_{s}}(h)\,\!$ is a binomial random variable with expectation $m\cdot Q_{P}(h)\,\!$ and variance $m\cdot Q_{P}(h)(1-Q_{P}(h))\,\!$. By Chebyshev's inequality we get,

$P^{m}\{|Q_{P}(h)-\widehat{Q_{s}(h)}|>\frac{\epsilon}{2}\}\leq\frac{m\cdot Q_{P}(h)(1-Q_{P}(h))}{(\epsilon m/2)^{2}}\leq\frac{1}{\epsilon^{2}m}\leq\frac{1}{2}\,\!$ for the mentioned bound on $m\,\!$. Here we use the fact that $x(1-x)\leq 1/4\,\!$ for $x\,\!$.

### Permutations

Let $\Gamma_{m}\,\!$ be the set of all permutations of $\{1,2,3,\dots,2m\}\,\!$ that swaps $i\,\!$ and $m+i\,\!$ $\forall i\,\!$ in some subset of $\{1,2,3,...,2m\}\,\!$.

Lemma: Let $R\,\!$ be any subset of $X^{2m}\,\!$ and $P\,\!$ any probability distribution on $X\,\!$. Then,

$P^{2m}(R)=E[Pr[\sigma(x)\in R]]\leq max_{x\in X^{2m}}(Pr[\sigma(x)\in R]),\,\!$

where the expectation is over $x\,\!$ chosen according to $P^{2m}\,\!$, and the probability is over $\sigma\,\!$ chosen uniformly from $\Gamma_{m}\,\!$.

Proof: For any $\sigma\in\Gamma_{m},\,\!$

$P^{2m}(R)=P^{2m}\{x:\sigma(x)\in R\}\,\!$ [since coordinate permutations preserve the product distribution $P^{2m}\,\!$].
$\therefore P^{2m}(R)=\int_{X^{2m}}1_{R}(x)dP^{2m}(x)\,\!$
$=\frac{1}{|\Gamma_{m}|}\sum_{\sigma\in\Gamma_{m}}\int_{X^{2m}}1_{R}(\sigma(x))dP^{2m}(x)\,\!$
$=\int_{X^{2m}}\frac{1}{|\Gamma_{m}|}\sum_{\sigma\in\Gamma_{m}}1_{R}(\sigma(x))dP^{2m}(x)\,\!$ [Because $|\Gamma_{m}|\,\!$ is finite]
$=\int_{X^{2m}}Pr[\sigma(x)\in R]dP^{2m}(x)$ [The expectation]
$\leq max_{x\in X^{2m}}(Pr[\sigma(x)\in R])\,\!$.

The maximum is guaranteed to exist since there is only a finite set of values that probability under a random permutation can take.

### Reduction to a finite class

Lemma: Basing on the previous lemma,

$max_{x\in X^{2m}}(Pr[\sigma(x)\in R])\leq 4\Pi_{H}(2m)e^{-\frac{\epsilon^{2}m}{8}}\,\!$.

Proof: Let us define $x=(x_{1},x_{2},...,x_{2m})\,\!$ and $t=|H|_{x}|\,\!$ which is atmost $\Pi_{H}(2m)\,\!$. This means there are functions $h_{1},h_{2},...,h_{t}\in H\,\!$ such that for any $h\in H,\exists i\,\!$ between $1\,\!$ and $t\,\!$ with $h_{i}(x_{k})=h(x_{k})\,\!$ for $1\leq k\leq 2m\,\!$.
We see that $\sigma(x)\in R\,\!$ iff for some $h\,\!$ in $H\,\!$ satisfies, $|\frac{1}{m}|\{1\leq i\leq m:h(x_{\sigma_{i}})=1\}|-\frac{1}{m}|\{m+1\leq i\leq 2m:h(x_{\sigma_{i}})=1\}||\geq\frac{\epsilon}{2}\,\!$. Hence if we define $w^{j}_{i}=1\,\!$ if $h_{j}(x_{i})=1\,\!$ and $w^{j}_{i}=0\,\!$ otherwise.
For $1\leq i\leq m\,\!$ and $1\leq j\leq t\,\!$, we have that $\sigma(x)\in R\,\!$ iff for some $j\,\!$ in ${1,...,t}\,\!$ satisfies $|\frac{1}{m}\left(\sum_{i}w^{j}_{\sigma(i)}-\sum_{i}w^{j}_{\sigma(m+i)}\right)|\geq\frac{\epsilon}{2}\,\!$. By union bound we get,

$Pr[\sigma(x)\in R]\leq t\cdot max\left(Pr[|\frac{1}{m}\left(\sum_{i}w^{j}_{\sigma_{i}}-\sum_{i}w^{j}_{\sigma_{m+i}}\right)|\geq\frac{\epsilon}{2}]\right)$
$\leq \Pi_{H}(2m)\cdot max\left(Pr[|\frac{1}{m}\left(\sum_{i}w^{j}_{\sigma_{i}}-\sum_{i}w^{j}_{\sigma_{m+i}}\right)|\geq\frac{\epsilon}{2}]\right)\,\!$.

Since, the distribution over the permutations $\sigma\,\!$ is uniform for each $i\,\!$, so $w^{j}_{\sigma_{i}}-w^{j}_{\sigma_{m+i}}\,\!$ equals $\pm |w^{j}_{i}-w^{j}_{m+i}|\,\!$, with equal probability.
Thus,

$Pr[|\frac{1}{m}\left(\sum_{i}\left(w^{j}_{\sigma_{i}}-w^{j}_{\sigma_{m+i}}\right)\right)|\geq\frac{\epsilon}{2}]=Pr[|\frac{1}{m}\left(\sum_{i}|w^{j}_{i}-w^{j}_{m+i}|\beta_{i}\right)|\geq\frac{\epsilon}{2}]\,\!$,

where the probability on the right is over $\beta_{i}\,\!$ and both the possibilities are equally likely. By Hoeffding's inequality, this is at most $2e^{-\frac{m\epsilon^{2}}{8}}\,\!$.

Finally, combining all the three parts of the proof we get the Uniform Convergence Theorem.


text:
# Complex exponential X delta function

1. Feb 22, 2012

### palex

1. Problem Statment:
Sketch the sequence x(n)=$\delta$(n) + exp(j$\theta$)$\delta$(n-1) + exp(j2$\theta$)$\delta$(n-2) + ...

3. Attempt at the Solution:
The angle theta is given in this case Can someone remind me of how to multiply a complex exponential by a delta function? This sequence represents impulse signals. Such multiplication yields a real and imaginary component. Would I ignore the imaginary component and essentially keep the cos(k$\theta$)? Thanks very much.

Last edited by a moderator: Feb 22, 2012
2. Feb 22, 2012

### cepheid

Staff Emeritus
Okay, so the terms in your sequence are complex valued. So the only way to really represent that is as two separate sequences. You can just use the Euler equation:

exp(jθ) = cos(θ) + jsin(θ)

So you could draw two separate sequences, each one representing the real part and the imaginary part x(n) respectively.

Alternatively, I suppose you could try to draw points in the complex plane.


text:
# visualization of 3D probability flow

I have a master equation for $P(N_A^+,N_B^+,N_C^+,t)$, with $N_A^+,N_B^+,N_C^+$ all discrete. The numerical integration is done by this Matlab program using Euler's method. Despite the crude Euler's method, the results from numerical integration are in agreement with the simulation, so the correctness of the program is not the focus.

I want to see how $P(N_A^+,N_B^+,N_C^+,t)$ evolves over time. Since $P(N_A^+,N_B^+,N_C^+,t)$ has 3 more coordinates in addition to time, it is really hard to visualize the changes. Ideally, I want to have something like

But I cannot find any function in Matlab that can does something similar. The closest thing I can find is Matlab's contourslice function.

By running the Matlab program mentioned above with

ABC_driver(0.5,100,20,200,10,true);

you will get something like the following:

You can guess what happens to $P(N_A^+,N_B^+,N_C^+,t)$, but it is nowhere as obvious as the atomic orbital graph.

My question is, how can I plot $P(N_A^+,N_B^+,N_C^+,t)$ in way similar to the atomic orbital density plot? Or can anyone suggest better way to visualize the time evolution of $P(N_A^+,N_B^+,N_C^+,t)$?

• Cool plot! Where did that come from? Mar 2 '14 at 19:29
• @meawoppl, the upper plot comes from here, the lower 6 plots are from contourslice in Matlab.
– wdg
Mar 3 '14 at 3:11
• dead link, fixed one here
– bhbr
Aug 14 '19 at 20:22

## 2 Answers

There is a nice volume rendering toolbox for MATLAB:

http://www.mathworks.com/matlabcentral/fileexchange/22940-vol3d-v2

I think you could tweak it for your purposes.

• looks very useful!
– wdg
Feb 28 '14 at 12:30
• I would encourage those in Python land with the same questions to checkout out mayavi2: docs.enthought.com/mayavi/mayavi Mar 2 '14 at 19:29

Programs like Visit and Paraview can do "volume rendering", which is what you show in your figure. You just need to export the data you have in a format that either of these programs can read.


text:
## A note on finite dual frame pairs.(English)Zbl 1365.42023

Let $$\mathbb{K}^d$$ denote either one of the finite dimensional spaces $$\mathbb{R}^d$$ or $$\mathbb{C}^d$$. A sequence $$\{e_n\}_{n=1}^N\subset \mathbb{K}^d$$ is said to be a frame for $$\mathbb{K}^d$$, with frame bounds $$A,B>0$$, provided that $A\|x\|^2\leq \sum_{n=1}^N|\langle x,e_n\rangle|^2\leq B\|x\|^2,\qquad x\in \mathbb{K}^d.$ The frame $$\{e_n\}_{n=1}^N$$ is said to be tight if one can take $$A=B$$ in the above inequality.
For any given frame $$\{e_n\}_{n=1}^N$$, it is always possible to find at least one sequence $$\{f_n\}_{n=1}^N\subset \mathbb{K}^d$$ such that $x=\sum_{n=1}^N\langle x,e_n\rangle f_n=\sum_{n=1}^N\langle x,f_n\rangle e_n,\qquad x\in \mathbb{K}^d.$ This $$\{f_n\}_{n=1}^N$$ is also a frame and the sequences $$\{e_n\}_{n=1}^N$$ and $$\{f_n\}_{n=1}^N$$ are called dual frames. A tight frame $$\{e_n\}_{n=1}^N$$ has $$\{\frac{1}{A}e_n\}_{n=1}^N$$ for (its canonical) dual frame. The following important result characterizes the possible norms of elements of a tight frame:
Given a sequence $$\{a_n\}_{n=1}^N\subset [0,\infty)$$ with $$N\geq d$$, (1) there exists a tight frame $$\{e_n\}_{n=1}^N$$ for $$\mathbb{K}^d$$ such that $$a_n=\langle e_n,e_n\rangle$$, $$n=1,\ldots,N$$, if and only if (2) $$\max_{1\leq n\leq N}a_n\leq \frac{1}{d}\sum_{n=1}^Na_n$$.
Inequality (2) is known as the fundamental inequality of tight frames.
The paper extends this result to a pair of dual frames $$\{e_n\}_{n=1}^N$$ and $$\{f_n\}_{n=1}^N$$ by characterizing the possible values of $$\langle e_n,f_n\rangle$$. This is Theorem 3.1 of the paper, which states that given $$\{\alpha_n\}_{n=1}^N\subset\mathbb{K}$$ with $$N>d$$, the following statements are equivalent:
(1)
There exist dual frames $$\{e_n\}_{n=1}^N$$ and $$\{f_n\}_{n=1}^N$$ for $$\mathbb{K}^d$$ such that $$\alpha_n=\langle e_n,f_n\rangle$$, $$1\leq n\leq N$$.
(2)
There exists a tight frame $$\{g_n\}_{n=1}^N$$ and a corresponding dual frame $$\{h_n\}_{n=1}^N$$ for $$\mathbb{K}^d$$ such that $$\alpha_n=\langle g_n,h_n\rangle$$, $$1\leq n\leq N$$.
(3)
$$d=\sum_{n=1}^N\alpha_n$$.
The assumption $$N>d$$ is due to the fact that for $$\{e_n\}_{n=1}^N$$ to be a frame for $$\mathbb{K}^d$$ it is necessary that $$N\geq d$$, and if $$N=d$$, any frame $$\{e_n\}_{n=1}^N$$ is a basis and has a unique dual frame, namely, $$\{\|e_n\|^{-2}e_n\}_{n=1}^N$$, in which case $$\alpha_n=1$$ for all $$n$$.
The paper then studies to a good extent the frames $$\{e_n\}_{n=1}^N$$ for $$\mathbb{K}^d$$ having (what the authors name) the full range property, that is, those frames $$\{e_n\}_{n=1}^N$$ such that for every $$\{\alpha_n\}_{n=1}^N\subset\mathbb{K}$$ with $$\sum_{n=1}^N\alpha_n=d$$, one can find a dual frame $$\{f_n\}_{n=1}^N$$ satisfying that $$\alpha_n=\langle e_n,f_n\rangle$$ for $$1\leq n\leq N$$.

### MSC:

 42C15 General harmonic expansions, frames
Full Text:

### References:

 [1] John J. Benedetto, Alexander M. Powell, and Özgür Yılmaz, Sigma-Delta (\Sigma \Delta ) quantization and finite frames, IEEE Trans. Inform. Theory 52 (2006), no. 5, 1990 – 2005. · Zbl 1285.94014 [2] John J. Benedetto and Matthew Fickus, Finite normalized tight frames, Adv. Comput. Math. 18 (2003), no. 2-4, 357 – 385. Frames. · Zbl 1028.42022 [3] James Blum, Mark Lammers, Alexander M. Powell, and Özgür Yılmaz, Sobolev duals in frame theory and sigma-delta quantization, J. Fourier Anal. Appl. 16 (2010), no. 3, 365 – 381. · Zbl 1200.42019 [4] Peter G. Casazza, Matthew Fickus, Jelena Kovačević, Manuel T. Leon, and Janet C. Tremain, A physical interpretation of tight frames, Harmonic analysis and applications, Appl. Numer. Harmon. Anal., Birkhäuser Boston, Boston, MA, 2006, pp. 51 – 76. · Zbl 1129.42418 [5] Peter G. Casazza, Custom building finite frames, Wavelets, frames and operator theory, Contemp. Math., vol. 345, Amer. Math. Soc., Providence, RI, 2004, pp. 61 – 86. · Zbl 1082.42024 [6] P.G. Casazza and M. Leon, Frames with a given frame operator. Preprint. http://www.math.missouri.edu/$$\sim$$pete/pdf/PM2.pdf · Zbl 1225.42021 [7] Peter G. Casazza and Nicole Leonhard, Classes of finite equal norm Parseval frames, Frames and operator theory in analysis and signal processing, Contemp. Math., vol. 451, Amer. Math. Soc., Providence, RI, 2008, pp. 11 – 31. · Zbl 1210.42047 [8] Peter G. Casazza and Jelena Kovačević, Equal-norm tight frames with erasures, Adv. Comput. Math. 18 (2003), no. 2-4, 387 – 430. Frames. · Zbl 1035.42029 [9] Ole Christensen, An introduction to frames and Riesz bases, Applied and Numerical Harmonic Analysis, Birkhäuser Boston, Inc., Boston, MA, 2003. · Zbl 1017.42022 [10] Vivek K. Goyal, Jelena Kovačević, and Jonathan A. Kelner, Quantized frame expansions with erasures, Appl. Comput. Harmon. Anal. 10 (2001), no. 3, 203 – 233. · Zbl 0992.94009 [11] Mark Lammers and Anna Maeser, An uncertainty principle for finite frames, J. Math. Anal. Appl. 373 (2011), no. 1, 242 – 247. · Zbl 1203.42040 [12] Mark Lammers, Alexander M. Powell, and Özgür Yılmaz, Alternative dual frames for digital-to-analog conversion in sigma-delta quantization, Adv. Comput. Math. 32 (2010), no. 1, 73 – 102. · Zbl 1181.94050 [13] C.S. Güntürk, M. Lammers, A.M. Powell, R. Saab and Ö. Yılmaz, Sobolev duals for random frames and $$\Sigma \Delta$$ quantization of compressed sensing measurements. Preprint, 2010. · Zbl 1273.41020 [14] Shidong Li, On general frame decompositions, Numer. Funct. Anal. Optim. 16 (1995), no. 9-10, 1181 – 1191. · Zbl 0849.42023 [15] Amos Ron and Zuowei Shen, Weyl-Heisenberg frames and Riesz bases in \?$$_{2}$$(\?^{\?}), Duke Math. J. 89 (1997), no. 2, 237 – 282. · Zbl 0892.42017
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# Uniform convergence where speed of convergence varies with x?

I recently ran across a description of uniform convergence that claimed: "The difference between pointwise convergence and uniform convergence is that with pointwise convergence, the convergence of $f_n$ to $f$ can vary in speed at each point of $f$. With uniform convergence, the speed of convergence is roughly the same all across the domain."

Is this necessarily true? Taken strictly, all uniform convergence says is that, given an $\epsilon$, I can find an $N(\epsilon)$ such that $\forall n > N$, $|f_n(x) - f(x)|<\epsilon$ for all $x$ on the relevant domain. Meaning, if I go out far enough in the sequence, the entire function will be contained within an $\epsilon$-tube centered around $f$.

This doesn't seem to say anything about speed of convergence at different points: I can conceive of the sequence of functions converging very quickly at some points while "lagging behind" at others (but nonetheless eventually getting there), so that those laggers just require I take a larger $N$ for a given $\epsilon$. More precisely, while I may be able to find different $N$'s for different $x$, I can simply take the largest of these as my $N(\epsilon)$ (supposing it exists). I'm specifically thinking in contrast to situations like $f_n(x) = x^n, x \in [0,1]$, where we have a "stuck point" at $x = 1$ that screws everything up; or, similarly, in a Fourier series with the Gibbs phenomenon present, where the overshoot never in fact decays, it just moves outward toward the endpoints. (In both of these cases, there is no largest $N$ to take as my $N(\epsilon)$ independent of $x$, and the best we can do is pointwise convergence.)

So, I'm trying to construct examples where we have uniform convergence, but the speed of convergence still depends on x. My first thought was $f_n(x) = \frac{\sin^n(\pi x)}{n}$. Since $|\sin(x)| \leq 1$ , this clearly converges uniformly to $f(x) = 0$ (just let $N$ = $1/\epsilon$) -- but the closer $x$ is to an integer, the faster the convergence, with $x = \frac{k\pi}{2}, k$ odd, taking the longest to converge and in fact preventing any lower N from working.

Is this correct? Does this example work? Are there other, more interesting ones?

What's more: is there some extra condition I can add that will guarantee the same speed of convergence everywhere?

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Well, you wrote "roughly" the same speed, which i would take as: speed varies along the domain but is bounded. If you don't have "slowest" points, you don't have uniform convergence. –  wisefool Dec 24 '12 at 1:09
Point taken. But isn't it possible to have some points converge as slowly as I like, while others converge as quickly as I like, and still have uniform convergence? I guess what I'm getting at is: anything can happen inside the epsilon-tube, so long as the tube is shrinking at some rate. Right? But then uniform convergence doesn't seem as strong as it once did to me. –  AndrewG Dec 24 '12 at 1:23
Uniform convergence is related to the convergence in the supremum norm. So, it is related to the maximum and the minimum of the difference; as long as these two numbers go to zero, you have uniform convergence. If you set $v(x, m)=\min\{n : \forall k>n, |f_k(x)-f(x)|<1/m\}$, then you have uniform convergence if there exists $N(m)$ such that $v(x,m)<N(m)$ for every $x$ (i.e. $\|v(x,m)\|_\infty< N(m)$). In this sense, the speed is roughly constant with respect to $x$, as a bounded function is roughly constant. Obviously, the bounds can be as big as you wish, as long as they are finite. –  wisefool Dec 24 '12 at 13:45

The most extreme example I can construct is the sequence of functions $f_n:\mathbb{R}\to\mathbb{R}$ defined by $$f_n(x)=\begin{cases}\tfrac{1}{n} & \text{ if }x=0,\\\\ 0 & \text{ if }x\neq 0 \end{cases}$$ which converges uniformly to the zero function but where $x=0$ is the only laggard.


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# Moderate Geometry & Mensuration Solved QuestionAptitude Discussion

 Q. The trapezoid shown in the given figure represents a cross section of the rudder of a ship. If the distance from $A$ to $B$ is 13 feet, what is the area of the cross section of the rudder in square feet?
 ✖ A. 39 ✖ B. 40 ✔ C. 42 ✖ D. 45

Solution:
Option(C) is correct

The formula for calculating the area of a trapezoid is

$Area=\dfrac{1}{2}(\text{base1}+\text{base2})(height)$

The bases of the trapezoid are given as 2 feet and 5 feet, so only the height ($AQ$) needs to be found.
Since the dashed line $AB=13$ feet, and triangle $BQA$ is a right triangle, use the Pythagorean theorem to calculate $AQ$.Thus,

$AQ=\sqrt{{13^2-5^2}}$

$=\sqrt{144}$

$=12$

Substituting the values into the formula for calculatingthe area of a trapezoid:

$Area=\dfrac{1}{2}(2+5)(12)$

$= 42 \text{ square feet}$


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Impact of gas spin and Lyman–Werner flux on black hole seed formation in cosmological simulations: implications for direct collapse
ABSTRACT

Direct collapse black holes (BHs) are promising candidates for producing massive z ≳ 6 quasars, but their formation requires fine-tuned conditions. In this work, we use cosmological zoom simulations to study systematically the impact of requiring: (1) low gas angular momentum (spin), and (2) a minimum incident Lyman–Werner (LW) flux in order to form BH seeds. We probe the formation of seeds (with initial masses of $M_{\rm seed} \sim 10^4\!-\!10^6\, \mathrm{M}_{\odot }\, h^{-1})$ in haloes with a total mass >3000 × Mseed and a dense, metal-poor gas mass >5 × Mseed. Within this framework, we find that the seed-forming haloes have a prior history of star formation and metal enrichment, but they also contain pockets of dense, metal-poor gas. When seeding is further restricted to haloes with low gas spins, the number of seeds formed is suppressed by factors of ∼6 compared to the baseline model, regardless of the seed mass. Seed formation is much more strongly impacted if the dense, metal-poor gas is required to have a critical LW flux (Jcrit). Even for Jcrit values as low as 50J21, no $8\times 10^{5}~\mathrm{M}_{\odot }\, h^{-1}$ seeds are formed. While lower mass ($1.25\times 10^{4},1\times 10^{5}~\mathrm{M}_{\odot }\, h^{-1}$) seeds do form, they are strongly suppressed (by factors of ∼10–100) compared more »

Authors:
; ; ; ; ; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10382036
Journal Name:
Monthly Notices of the Royal Astronomical Society
Volume:
510
Issue:
1
Page Range or eLocation-ID:
p. 177-196
ISSN:
0035-8711
Publisher:
Oxford University Press
We explore implications of a range of black hole (BH) seeding prescriptions on the formation of the brightest $z$ ≳ 6 quasars in cosmological hydrodynamic simulations. The underlying galaxy formation model is the same as in the IllustrisTNG simulations. Using constrained initial conditions, we study the growth of BHs in rare overdense regions (forming $\gtrsim 10^{12}\, {\rm M}_{\odot }\,h^{-1}$ haloes by $z$ = 7) using a  (9 Mpc h−1)3 simulated volume. BH growth is maximal within haloes that are compact and have a low tidal field. For these haloes, we consider an array of gas-based seeding prescriptions wherein $M_{\mathrm{seed}}=10^4\!-\!10^6\, {\rm M}_{\odot }\,h^{-1}$ seeds are inserted in haloes above critical thresholds for halo mass and dense, metal-poor gas mass (defined as $\tilde{M}_{\mathrm{h}}$ and $\tilde{M}_{\mathrm{sf,mp}}$, respectively, in units of Mseed). We find that a seed model with $\tilde{M}_{\mathrm{sf,mp}}=5$ and $\tilde{M}_{\mathrm{h}}=3000$ successfully produces a $z$ ∼ 6 quasar with $\sim 10^9\, {\rm M}_{\odot }$ mass and ∼1047 erg s−1 luminosity. BH mergers play a crucial role at $z$ ≳ 9, causing an early boost in BH mass at a time when accretion-driven BH growth is negligible. With more stringent seeding conditions (e.g. $\tilde{M}_{\mathrm{sf,mp}}=1000$), the relative paucity of BH seeds results in a much lower merger rate. In this case, $z$more »
2. Abstract Deciphering the formation of supermassive black holes (SMBHs) is a key science goal for upcoming observational facilities. In many theoretical channels proposed so far, the seed formation depends crucially on local gas conditions. We systematically characterize the impact of a range of gas-based black hole seeding prescriptions on SMBH populations using cosmological simulations. Seeds of mass Mseed ∼ 103–106 M⊙ h−1 are placed in haloes that exceed critical thresholds for star-forming, metal-poor gas mass and halo mass (defined as $\tilde{M}_{\mathrm{sf,mp}}$ and $\tilde{M}_{\mathrm{h}}$, respectively, in units of Mseed). We quantify the impact of these parameters on the properties of z ≥ 7 SMBHs. Lower seed masses produce higher black hole merger rates (by factors of ∼10 and ∼1000 at z ∼ 7 and z ∼ 15, respectively). For fixed seed mass, we find that $\tilde{M}_{\mathrm{h}}$ has the strongest impact on the black hole population at high redshift (z ≳ 15, where a factor of 10 increase in $\tilde{M}_{\mathrm{h}}$ suppresses merger rates by ≳ 100). At lower redshift (z ≲ 15), we find that $\tilde{M}_{\mathrm{sf,mp}}$ has a larger impact on the black hole population. Increasing $\tilde{M}_{\mathrm{sf,mp}}$ from 5–150 suppresses the merger rates by factors of ∼8 at z ∼ 7–15. This suggests that themore »
Formation of supermassive black holes (BHs) remains a theoretical challenge. In many models, especially beginning from stellar relic ‘seeds,’ this requires sustained super-Eddington accretion. While studies have shown BHs can violate the Eddington limit on accretion disc scales given sufficient ‘fuelling’ from larger scales, what remains unclear is whether or not BHs can actually capture sufficient gas from their surrounding interstellar medium (ISM). We explore this in a suite of multiphysics high-resolution simulations of BH growth in magnetized, star-forming dense gas complexes including dynamical stellar feedback from radiation, stellar mass-loss, and supernovae, exploring populations of seeds with masses $\sim 1\!-\!10^{4}\, \mathrm{M}_{\odot }$. In this initial study, we neglect feedback from the BHs: so this sets a strong upper limit to the accretion rates seeds can sustain. We show that stellar feedback plays a key role. Complexes with gravitational pressure/surface density below $\sim 10^{3}\, \mathrm{M}_{\odot }\, {\rm pc^{-2}}$ are disrupted with low star formation efficiencies so provide poor environments for BH growth. But in denser cloud complexes, early stellar feedback does not rapidly destroy the clouds but does generate strong shocks and dense clumps, allowing $\sim 1{{\ \rm per\ cent}}$ of randomly initialized seeds to encounter a dense clump withmore »
4. ABSTRACT Possible formation scenarios of supermassive black holes (BHs) in the early universe include rapid growth from less massive seed BHs via super-Eddington accretion or runaway mergers, yet both of these scenarios would require seed BHs to efficiently sink to and be trapped in the Galactic Centre via dynamical friction. This may not be true for their complicated dynamics in clumpy high-z galaxies. In this work, we study this ‘sinking problem’ with state-of-the-art high-resolution cosmological simulations, combined with both direct N-body integration of seed BH trajectories and post-processing of randomly generated test particles with a newly developed dynamical friction estimator. We find that seed BHs less massive than $10^8\, \mathrm{M}_\odot$ (i.e. all but the already-supermassive seeds) cannot efficiently sink in typical high-z galaxies. We also discuss two possible solutions: dramatically increasing the number of seeds such that one seed can end up trapped in the Galactic Centre by chance, or seed BHs being embedded in dense structures (e.g. star clusters) with effective masses above the mass threshold. We discuss the limitations of both solutions.