text:
Chern classes of crystals

Research paper by Hélène Esnault, Atsushi Shiho

Indexed on: 21 Nov '15Published on: 21 Nov '15Published in: Mathematics - Algebraic Geometry

Abstract

The crystalline Chern classes of the value of a locally free crystal vanish on a smooth variety defined over a perfect field. Out of this we conclude new cases of de Jong's conjecture relating the geometric \'etale fundamental group of a smooth projective variety defined over a perfect field and the triviality of its category of isocrystals. We also discuss the case of the Gau{\ss}-Manin convergent isocrystal.

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# Talk:Bit numbering

WikiProject Computing
This article is within the scope of WikiProject Computing, a collaborative effort to improve the coverage of computers, computing, and information technology on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
$\sum_{i=0}^{w-1} bit(i) \cdot 2^i$
where $bit(i)$ denotes the value of the bit with number $i$ and $w$ denotes the number of bits in total.


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# Search by Topic

#### Resources tagged with Multiplication & division similar to What Do You Need?:

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### There are 160 results

Broad Topics > Calculations and Numerical Methods > Multiplication & division

### Square Subtraction

##### Stage: 2 Challenge Level:

Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?

### Repeaters

##### Stage: 3 Challenge Level:

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

### As Easy as 1,2,3

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When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .

### Divide it Out

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What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?

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What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.

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The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?

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56 406 is the product of two consecutive numbers. What are these two numbers?

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There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?

### Number Tracks

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Ben’s class were making cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?

### Which Is Quicker?

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Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?

### Multiply Multiples 3

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Have a go at balancing this equation. Can you find different ways of doing it?

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This challenge encourages you to explore dividing a three-digit number by a single-digit number.

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Can you work out some different ways to balance this equation?

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Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?

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This problem is designed to help children to learn, and to use, the two and three times tables.

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Can you find what the last two digits of the number $4^{1999}$ are?

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Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.

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I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?

### A First Product Sudoku

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Given the products of adjacent cells, can you complete this Sudoku?

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Here is a chance to play a version of the classic Countdown Game.

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Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?

### A Conversation Piece

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Take the number 6 469 693 230 and divide it by the first ten prime numbers and you'll find the most beautiful, most magic of all numbers. What is it?

### X Marks the Spot

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When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .

### A One in Seven Chance

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What is the remainder when 2^{164}is divided by 7?

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The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?

### The Pied Piper of Hamelin

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This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!

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48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?

### What's My Weight?

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There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?

### Zios and Zepts

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On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?

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Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?

### Rabbits in the Pen

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Using the statements, can you work out how many of each type of rabbit there are in these pens?

### What's in the Box?

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This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?

### Zargon Glasses

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Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?

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All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.

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Use your logical reasoning to work out how many cows and how many sheep there are in each field.

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### Mystery Matrix

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Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.

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This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!

### Four Goodness Sake

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Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?

### The Deca Tree

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Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.

### Arranging the Tables

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There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.

### Journeys in Numberland

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Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.

### How Old?

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Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?

### All the Digits

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This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?

### Ones Only

##### Stage: 3 Challenge Level:

Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.

### Doplication

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We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?

### ABC

##### Stage: 2 Challenge Level:

In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?

### Sam's Quick Sum

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What is the sum of all the three digit whole numbers?


text:
Indico has been upgraded to v2.3. Please see our blog for a list of improvements.

22-27 March 2015
Hotel do Bosque
Brazil/East timezone

## Gluon Schwinger-Dyson equation in the PT-BFM scheme

Not scheduled
3h 30m
Hotel do Bosque

#### Hotel do Bosque

Rodovia Mário Covas (Rio-Santos) BR - 101 Sul, Km 533, Angra dos Reis, RJ, Brazil
Poster Field theoretical approaches to QCD

### Speaker

Mr Antonio Maurício S. N. Ferreira (UNICAMP)

### Description

One of the most widely used approach for studying the IR sector of QCD are the Schwinger-Dyson equations. This infinite system of coupled non-linear integral equations for all Green’s functions of the theory is inherently non-perturbative and can accommodate phenomena such as chiral symmetry breaking and dynamical mass generation. In practice, however, their usefulness hinges crucially on one's ability to devise a self-consistent truncation scheme that would select a tractable subset of these equations, without compromising the physics one hopes to describe. Here in this work, we will review, the PT-BFM truncation scheme, which is based on the synthesis of the pinch technique (PT) and the background field method (BFM). We will show that this truncation scheme implements a drastic modification already at the level of the building blocks of the Schwinger-Dyson series, which after being rearranged, satisfy the Abelian Ward identities instead of the usual Slavnov-Taylor identity. As a result, gluonic and ghost contributions are separately transverse, within each order in the dressed-loop'' expansion. We will show explicitly, it is much easier to devise truncation schemes that manifestly preserve the transversality of the gluon self-energy.

### Presentation Materials

There are no materials yet.


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# Warning appeared when I ask a question, but I have a good history.

I just tried to ask a question and was presented with this warning:

It threatens to bar me from the site. If I was consistently ignoring the rules or being rude then fair play, but I've asked a few questions, and they have a reasonable score.

What's happening? I'm pretty sure I'm a good Internet citizen, so this seems very harsh.

• You also have one deleted question at a score of $-1$, but a warning with one closed question and one deleted with a score of $-1$ while having two positively scored questions, one at $+13$, seems exaggerated. However, the closed zero-score and the deleted negative-score questions are your newest questions, and newer questions get a greater weight in the algorithm, as far as I know. Perhaps the thresholds for the warning need a little tweaking. – Daniel Fischer Jun 5 '15 at 9:35
• I'd assumed that deleting a question was akin to putting your hand up and saying 'OK, my bad'. Unless there is a large number of deleted, destructive questions then it seems very stern indeed. Am I really in danger of being blocked from asking questions? – BanksySan Jun 5 '15 at 9:46
• This was deleted by the community bot, so it's not quite putting your hand up. But yes, one deleted question isn't much, even when 25% is a relatively high percentage. I don't know the algorithm, so I can't speak with certainty. But I've seen worse profiles that weren't blocked, so I would say you're not in immediate danger. However, if you ask a very badly received question, or a few badly received ones, that could put you below the threshold. So my advice would be to tread cautiously, but not be overly alarmed. – Daniel Fischer Jun 5 '15 at 9:54
• According to meta.SO: "The specific criteria for these aren't public largely for the same reasons the quality block criteria aren't public, but let's just say that folks who are nearing hitting a block will be getting a heads up and some instructions first." So, if I understand it correctly, we do not know exact criteria when a user is blocked from asking questions – Martin Sleziak Jun 5 '15 at 10:43
• Whatever the criteria may be, if the message appears with this track record, the criteria are wrong, or at the friendliest, "need tweaking". Simple as that. – Lord_Farin Jun 5 '15 at 14:12
• @Lord_Farin Half of the user's questions are either closed or deleted. A warning (not a block) seems reasonable. A warning does not mean that the user can't contribute good material; it's supposed to help them avoid contributing subpar material. – user147263 Jun 5 '15 at 15:41
• @HomegrownTomato I can see that. I just think that judgement ought to be postponed until like 4 "bad" questions/answers have appeared. – Lord_Farin Jun 5 '15 at 15:58
• A threat to be banned that soon, for very minor transgressions is rather draconian. Perhaps the algorithm used to spot baddies should be tweaked to include the quality of responses? The deleted question has a perfectly valid response, so someone did understand the question and the context and explained the answer as well. – BanksySan Jun 5 '15 at 19:10


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#### Archived

This topic is now archived and is closed to further replies.

# Yet Another Q3 Renderer Question

This topic is 5576 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi, i was wondering how i am meant to interpret the control points which share points with patches next to them a diagram or very good explenation would be help full, thanx (btw, i have the single patches working just not the 2+ side-by-side ones) special thanx to derilect@turbonet.com for the BiQuadratic Patch Class

##### Share on other sites
Are you sure you're in the correct forum?

[edited by - daerid on September 10, 2002 6:59:21 PM]

##### Share on other sites
hope so
its not OpenGL, or C++ ...
how do you move it?

##### Share on other sites
Ok. Say you have an M x N array of control points. It will be something like 5x3 or 9x7. I'll use a 5x3 example. What you want to do is use the first 3x3 section to create the first patch, and then use the last column from the first patch, along with the remaining two columns, for the second patch. Graphically, it would look something like this.

(1   2   [3)   4   5](6   7   [8)   9   10](11  12  [13)  14  15]

So your first patch is the 3x3 grid
1  2  36  7  811 12 13

And your second patch is the 3x3 grid
3  4  58  9  1013 14 15

Just render these individual patches as you normally do. With larger control point arrays (e.g. 9x7), you do essentially the same thing - break them up into 3x3 arrays.

Edit: Stupid formatting ...

[edited by - Martee on September 10, 2002 9:14:53 PM]


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# Skipping Formatting

By default formatting is always on but can be toggled with the following comments:

#! format: off
# Turns off formatting from this point onwards
...

#! format: on
# Turns formatting back on from this point onwards

These can be used throughout a file, or, for an entire file not be formatted add "format: off" at the top of the file:

#! format: off
#
# It doesn't actually matter if it's on
# the first line of the line but anything
# onwards will NOT be formatted.

module Foo
...
end

Note the formatter expects #! format: on and #! format: off to be on its own line and the whitespace to be an exact match.


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Consistent estimator of percentile or Value-at-Risk

Given a sequence of i.i.d. random variables X_1,...,X_n I would like to estimate VaR_q(X) (see http://en.wikipedia.org/wiki/Value_at_risk#Mathematical_definition) for some 0

1) VaR_q(X)=X_(⌊qn⌋) and 2)VaR_q(X)= (1-(qn-⌊qn⌋))X_(⌊qn⌋)+(qn-⌊qn⌋)X_(⌊qn⌋+1)


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## Stochastic Gradient Descent: Strong convergence guarantees -- without parameter tuning

Seminar | August 30 | 4-5 p.m. | 60 Evans Hall

Rachel Ward, UT Austin

Stochastic Gradient Descent is the basic optimization algorithm behind powerful deep learning architectures which are becoming increasingly omnipresent in society. However, existing theoretical guarantees of convergence rely on knowing certain properties of the optimization problem such as maximal curvature and noise level which are not known a priori in practice. Thus, in practice, hyper parameters of the algorithm such as the stepsize are tuned by hand before training, taking days or weeks. In this talk, we discuss a modification of Stochastic Gradient Descent with an adaptive "on the fly" step size update known as AdaGrad which is used in practice but until now did not come with any theoretical guarantees. We provide the first such guarantees, showing that Stochastic Gradient Descent with AdaGrad converges to a near-stationary point of a smooth loss function, at a rate which nearly matches the "oracle" rate as if the curvature of the loss function and noise level on the stochastic gradients were known in advance. We also demonstrate its favorable empirical performance on deep learning problems compared to pre-tuned state-of-the-art algorithms.

CA, 5106422781


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RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PERSONAL OFFICE
 General information Latest issue Archive Search papers Search references RSS Latest issue Current issues Archive issues What is RSS

 Vestn. Yuzhno-Ural. Gos. Un-ta. Ser. Matem. Mekh. Fiz.: Year: Volume: Issue: Page: Find

 Vestn. Yuzhno-Ural. Gos. Un-ta. Ser. Matem. Mekh. Fiz., 2016, Volume 8, Issue 4, Pages 56–62 (Mi vyurm319)

Physics

Mathematical simulation of gas-phase synthesis of metal nanoparticles in a chamber with an arc discharge

A. E. Korenchenko, V. Jamal Jalal

South Ural State University, Chelyabinsk, Russian Federation

Abstract: Numerical simulation of the metallic nanoparticles synthesis in a chamber with an arc discharge were elaborated. The efficiency of the vapor phase condensation of metallic nanoclusters and nanopowders was determined by setting optimum process parameters, the possibilities of experimental estimation of which is limited. Mathematical and physical models were developed to perform computer analysis of the vapor phase condensation to describe the macroscopic characteristics of the process (temperature regime, gas mixture dynamics, diffusion and convective transport of clusters) with allowance for the properties of the components on a microscopic level. The classical Becker–Döring–Folmer–Weber thermodynamic nucleation theory was used for the description of the probability of atom-cluster aggregation. The distributions of macroscopic (temperature, pressure, velocity field) and microscopic (cluster size distribution) values in the chamber were obtained. It is found that the size distribution function of clusters deposited on the chamber walls has two peaks, the first — in the region of small clusters ($1$$50$ atoms) and the second — for the clusters containing more than $10^4$ atoms. Based on the results of numerical calculations the assumptions are made about the relationship between the type of size distribution function and characteristics of the process.

Keywords: metal nanoparticles, gas-phase synthesis, homogeneous nucleation.

 Funding Agency Grant Number Russian Foundation for Basic Research 15-03-04182_à

DOI: https://doi.org/10.14529/mmph160407

Full text: PDF file (369 kB)
References: PDF file   HTML file

Document Type: Article
UDC: 548.51

Citation: A. E. Korenchenko, V. Jamal Jalal, “Mathematical simulation of gas-phase synthesis of metal nanoparticles in a chamber with an arc discharge”, Vestn. Yuzhno-Ural. Gos. Un-ta. Ser. Matem. Mekh. Fiz., 8:4 (2016), 56–62

Citation in format AMSBIB
\Bibitem{KorJam16}
\by A.~E.~Korenchenko, V.~Jamal Jalal
\paper Mathematical simulation of gas-phase synthesis of metal nanoparticles in a chamber with an arc discharge
\jour Vestn. Yuzhno-Ural. Gos. Un-ta. Ser. Matem. Mekh. Fiz.
\yr 2016
\vol 8
\issue 4
\pages 56--62
\mathnet{http://mi.mathnet.ru/vyurm319}
\crossref{https://doi.org/10.14529/mmph160407}
\elib{http://elibrary.ru/item.asp?id=27157558}


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Known decays of pi_1(1600)-

(Only the first generation of decays, no cascades)

 Final state Branching


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# Convergence rate of Toda/Morse flow

Let $$A(t), A_0$$ be a $$n\times n$$ hermitian complex matrices and consider the following matrix flow

\begin{align} \frac{dA}{dt} &= \left [ C\circ A , A \right ] \\ A(0) &= A_0 \ . \end{align}

Here $$\circ$$ stands for Hadamard (pointwise) multiplication and the (antisymmetric) matrix $$C$$ has the form

$$C=\left(\begin{array}{ccccc} 0 & -1 & -1 & \cdots & -1\\ 1 & 0 & -1 & & \vdots\\ 1 & 1 & 0 & \ddots & -1\\ \vdots & & \ddots & \ddots & -1\\ 1 & 1 & \cdots & 1 & 0 \end{array}\right).$$

I am not 100% sure of the nomenclature Toda vs Morse flow, in any case the flow above or slight variation thereof, has also been considered in a few occasions on MO, for example here. It can be shown that the flow is isospectral and that as $$t\to\infty$$ , $$A(t)$$ converges to a diagonal matrix with the eigenvalues of $$A_0$$ ordered from lowest to largest.

My question is the following:

Is anything known about the convergence rate of $$\mathrm{diag}(A(t))$$ to the eigenvalues of $$A_0$$? Can we show that the convergence is bounded by some exponential (and in this case can we estimate the rates)?

If the question above is too complicated I would be equally interested in the convergence rate of the flow with the following modified $$C$$:

$$C_1=\left(\begin{array}{ccccc} 0 & -1 & -1 & \cdots & -1\\ 1 & 0 & 0 & 0 & \vdots\\ 1 & 0 & 0 & \ddots & 0\\ \vdots & & \ddots & \ddots & 0\\ 1 & 0 & \cdots & 0 & 0 \end{array}\right).$$

In case of the flow with $$C_1$$, it can be shown that the matrix element $$A_{1,1}(t)$$ converges to the lowest eigenvalue of $$A_0$$, $$\lambda_1$$.

For the $$C_1$$-flow, can it be shown that $$A_{1,1}(t)$$ converges exponentially fast to $$\lambda_1$$?

It is usually said that the Toda flow is a sort of continuous version of Lanczos diagonalization algorithm. This similarity is even more marked in case of the $$C_1$$ flow. In fact in the latter case one can show that the "diagonalization" amounts to a (continuous) series of rank-2 unitary rotation (reminiscent of Lanczos algorithm). Since the convergence of the Lanczos algorithm is exponential in the number of iteration, I formulate the conjecture that the convergence of the above flows is also exponential. Is this true? Can it be proven? I couldn't find anything in the literature linked to the MO answers.

• IN general, the strategy is to show that the correct diagonal matrix is the only stable fixed point, and hence the convergence is exponential. Are you looking to prove more than this ? – Piyush Grover Mar 24 '20 at 17:33
• @PiyushGrover that is exactly what I would like to prove or see a reference for – lcv Mar 24 '20 at 18:45
• It's easy to see that a diagonal $A$ is a fixed point, but how can you show from there that the convergence is exponential? (How do you prove the "hence" in your sentence?) – lcv Mar 24 '20 at 18:47
• There are several versions of Toda flow, for some there are results by Tomei et.al. that it gives continous version of QR algorithm. Hence the convergene is the same as for QR algorithm which is quite well-known. – Alexander Chervov Mar 24 '20 at 19:45
• google.com/… – Alexander Chervov Mar 24 '20 at 20:29


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Mathematics

# find the roots ? x² + 7x + 12 =0 x²-8x +12=0

We have to remember one simple pattern $\Delta =b^2-4ac$ 1. $x^2+7x+12=0$ in this case $b=7$ $a=1$ $c=12$ so: $\Delta =7^2-4*1*12$ $\Delta =49-48$ $\Delta =1$ $\Delta >0$ it means that there are 2 roots $\sqrt{\Delta} =1$ $x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$ $x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$ $x_{1}=\frac{-7+1}{2}$ $x_{1}=\frac{-6}{2}$ $x_{1}=-3$ $x_{2}=\frac{-7-1}{2}$ $x_{2}=\frac{-8}{2}$ $x_{2}=-4$ Roots are $x_1=-4$ and $x_2=-3$ 2. $x^2-8x +12=0$ in this case $a=1$ $b=-8$ $c=12$ $\Delta =(-8)^2-4*1*12$ $\Delta =64-48$ $\Delta =16$ $\Delta >0$ it means that there are 2 roots $\sqrt{\Delta}=4$ $x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$ $x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$ $x_{1}=\frac{8+4}{2}$ $x_{1}=\frac{12}{2}$ $x_{1}=6$ $x_{2}=\frac{8-4}{2}$ $x_{2}=\frac{4}{2}$ $x_{2}=2$ Roots:$x_1=6$ and $x_2=4$
$x^2 + 7x + 12 =0\\ x^2+4x+3x+12=0\\ x(x+4)+3(x+4)=0\\ (x+3)(x+4)=0\\ x=-3 \vee x=-4\\\\ x^2-8x+12=0\\ x^2-6x-2x+12=0\\ x(x-6)-2(x-6)=0\\ (x-2)(x-6)=0\\ x=2 \vee x=6$


text:
Search
Question: DelayedArray seeds: SolidRleArraySeed vs ChunkedRleArraySeed vs DataFrame of Rle's
0
23 days ago by
maltethodberg100
UCPH
maltethodberg100 wrote:

After reading the vignette for DelayedMatrixStats, I'm confused about the mention of two types of RleArray seeds: SolidRleArraySeed & ChunkedRleArraySeed

What's the difference between these two seeds and a simple DataFrame of Rle's as seed? Are there perfomance/memory differences between them? Are there any pro/cons associated with each of them?

modified 22 days ago by Peter Hickey440 • written 23 days ago by maltethodberg100
4
22 days ago by
Peter Hickey440
Johns Hopkins University, Baltimore, USA
Peter Hickey440 wrote:

I'm going to assume we're talking about representing 2-dimesionsal arrays.

It comes down to how the data are stored (i.e. run length encoded):

• A DataFrame with Rle columns is run length encoded column-wise with a separate Rle per column.
• A SolidRleArraySeed-backed DelayedArray is run length encoded column-wise but with one Rle for the entire dataset (think of it as wrapping around from the bottom of column 1 to the start of column 2, etc.).
• A ChunkedRleArraySeed-backed DelayedArray is run length encoded column-wise but in chunks of a fixed size.

I'll include some example code at in a follow-up comment that illustrates this (it's too long to include in one post).

The best choice depends on several factors, such as:

1. The run length patterns of your data
2. How you want to access the data (e.g. if you only require column-wise access, then data that is chunked per-column is theoretically more efficient)
3. Whether you want to use the broader DelayedArray framework (in which case, a DataFrame probably isn't the best choice).

Hope this helps,

Pete

1
suppressPackageStartupMessages(library(S4Vectors))
suppressPackageStartupMessages(library(DelayedArray))

nrow <- 200L
ncol <- 5L

# Generate some data that can be efficiently run length encoded.
x <- Rle(500L + sort(rpois(nrow * ncol, 10)))
x
#> integer-Rle of length 1000 with 21 runs
#>   Lengths:   1   2   8  27  32  60  98 121 ...  35  17   6   2   5   2   1
#>   Values : 501 502 503 504 505 506 507 508 ... 515 516 517 518 519 520 522

# Construct a DataFrame with the data in 5 columns
# (there's a bit of messing about to do this)
df <- DataFrame(as.list(split(x, rep(seq_len(ncol), each = nrow))))
# The data are stored as one Rle per column.
df@listData
#> $X1 #> integer-Rle of length 200 with 7 runs #> Lengths: 1 2 8 27 32 60 70 #> Values : 501 502 503 504 505 506 507 #> #>$X2
#> integer-Rle of length 200 with 3 runs
#>   Lengths:  28 121  51
#>   Values : 507 508 509
#>
#> $X3 #> integer-Rle of length 200 with 3 runs #> Lengths: 76 119 5 #> Values : 509 510 511 #> #>$X4
#> integer-Rle of length 200 with 2 runs
#>   Lengths: 105  95
#>   Values : 511 512
#>
#> $X5 #> integer-Rle of length 200 with 10 runs #> Lengths: 12 64 56 35 17 6 2 5 2 1 #> Values : 512 513 514 515 516 517 518 519 520 522 # Construct a SolidRleArraySeed-backed DelayedArray sras <- RleArray(x, dim = c(nrow, ncol)) # The data are in a single Rle. seed(sras) #> An object of class "SolidRleArraySeed" #> Slot "rle": #> integer-Rle of length 1000 with 21 runs #> Lengths: 1 2 8 27 32 60 98 121 ... 35 17 6 2 5 2 1 #> Values : 501 502 503 504 505 506 507 508 ... 515 516 517 518 519 520 522 #> #> Slot "DIM": #> [1] 200 5 #> #> Slot "DIMNAMES": #> [[1]] #> NULL #> #> [[2]] #> NULL # Construct a ChunkedRleArraySeed-backed DelayedArray with 1 chunk/column. cras <- RleArray(x, dim = c(nrow, ncol), chunksize = nrow) # The data are stored in an environment seed(cras) #> An object of class "ChunkedRleArraySeed" #> Slot "breakpoints": #> [1] 200 400 600 800 1000 #> #> Slot "type": #> [1] "integer" #> #> Slot "chunks": #> <environment: 0x565279e17b38=""> #> #> Slot "DIM": #> [1] 200 5 #> #> Slot "DIMNAMES": #> [[1]] #> NULL #> #> [[2]] #> NULL ls(seed(cras)@chunks) #> [1] "000001" "000002" "000003" "000004" "000005" # Let's take a look at the the values in the first chunk seed(cras)@chunks$000001
#> integer-Rle of length 200 with 7 runs
#>   Lengths:   1   2   8  27  32  60  70
#>   Values : 501 502 503 504 505 506 507
# The first chunk of cras is the same as the first column of df
# NOTE: This brushes over some tricksy stuff RleArray does to further compress
#       data where all values are in [0, 255]; see
#       https://github.com/Bioconductor/DelayedArray/blob/master/R/RleArray-class.R#L316
identical(seed(cras)@chunks$000001, df$X1)
#> [1] TRUE

# Construct a ChunkedRleArraySeed-backed DelayedArray with 10 chunks/column.
cras2 <- RleArray(x, dim = c(nrow, ncol), chunksize = nrow / 10)
# Now have 10-times as many chunks.
ls(seed(cras2)@chunks)
#>  [1] "000001" "000002" "000003" "000004" "000005" "000006" "000007"
#>  [8] "000008" "000009" "000010" "000011" "000012" "000013" "000014"
#> [15] "000015" "000016" "000017" "000018" "000019" "000020" "000021"
#> [22] "000022" "000023" "000024" "000025" "000026" "000027" "000028"
#> [29] "000029" "000030" "000031" "000032" "000033" "000034" "000035"
#> [36] "000036" "000037" "000038" "000039" "000040" "000041" "000042"
#> [43] "000043" "000044" "000045" "000046" "000047" "000048" "000049"
#> [50] "000050"

Thank you for the detailed explanation. A few follow up question:
- Why is it beneficial to store the data as a single Rle in SolidRleArraySeed? Doesn't that limit to the total size of the matrix since it spans all columns?

- If you chunk a ChunkedRleArray seed by column, how is that different from wrapping a DataFrame of Rle's in a DelayedArray?

1

Why is it beneficial to store the data as a single Rle in SolidRleArraySeed?

I think it's more the case that this is a simpler object to design and program with (Hervé Pagès will know best).

Doesn't that limit to the total size of the matrix since it spans all columns?

Yes, I think so.

If you chunk a ChunkedRleArray seed by column, how is that different from wrapping a DataFrame of Rle's in a DelayedArray?

Off the top of my head, there's probably no difference (but there may be weird corner cases). Oh, one difference is the tricksy thing I mentioned with a ChunkedRleArraySeed-backed DelayedArray - it may lead to a slightly smaller memory footprint for the ChunkedRleArraySeed-backed DelayedArray if all values are integers in [0, 255] (untested)

1

Hi,

The exact internal representation of an RleArray object is still a work-in-progress and subject to change. The 2 current *RleArraySeed classes reflect the 2 experimentations I did so far with this. ChunkedRleArraySeed came after SolidRleArraySeed and improves on it by replacing a single Rle by a collection of smaller Rle's (one per chunk) placed in an environment. This makes realization of an arbitrary DelayedArray object x as an RleArray object (with as(x, "RleArray")) more memory efficient. Also it can achieve slightly better compression when the array data is of type integer (e.g. count data) by storing the data of a given chunk in a raw-Rle instead of an integer-Rle if all the values in the chunk are >= 0 and <= 255. So ChunkedRleArraySeed is currently the default but SolidRleArraySeed has been kept around for backward compatibility with old serialized RleArray objects.

Compared to using a seed that is just a DataFrame with Rle columns, using a ChunkedRleArraySeed seed will probably not make much difference for read access. However using the former would be inefficient in the context of realization as it would trigger a copy of the entire data every time a block is written to it (realization is done block by block). ChunkedRleArraySeed was specifically designed for being an efficient realization sink.

Finally the plan for BioC 3.9 is to improve ChunkedRleArraySeed by allowing: (1) arbitrary chunk geometry, (2) have the Rle within each chunk run either along the columns or along the rows, and (3) generalize the use of raw-Rle to any chunk that contains 256 unique values or less (such chunk would also contain a small translation table to decode the data). This will provide opportunities to choose the chunk geometry and orientation (i.e. col-major or row-major) that is optimal for a given data set and for the typical access pattern to it.

H.

1

Doesn't that limit to the total size of the matrix since it spans all columns?

Indeed. I should add that one of the original motivations for coming up with ChunkedRleArraySeed was exactly this limitation. I wanted to be able to load the 1.3 Million Brain Cell Dataset (https://bioconductor.org/packages/release/data/experiment/html/TENxBrainData.html) in an RleArray object but using a SolidRleArraySeed seed wouldn't allow me to do this because it can only represent a data set of length <= 2^31 - 1.

Thanks for the detailed response! I'm considering updating the CAGEfightR-package from using dgCMatrix to using DelayedArray instead (giving the new addition of parallel processing and quick summaries from DelayedMatrixStats, particularly rowsum). Using RleMatrix seems like the obvious choice for replacement, but I'm hesitant to update if there are many planned changes coming up. Is it premature to start basing packages around RleMatrix?

I might change a little bit the interface of the RleArray() constructor function though (e.g. replacing the chunksize argument with chunkdim). Other than that, as long as you don't serialize your RleMatrix instances and your code doesn't try to access directly the internals of these objects, any change I'll make to their internals shouldn't affect you, at least in theory.


text:
# Is ω1 × βN normal?

Once upon a time I asked whether $\omega_1 \times \beta \mathbb{N}$ is normal. I got the answer no and a fairly convincing proof of this here

However I'm currently in a situation where I have three plausible proofs of plausible results at most two of which can be true, and of the three this is the one of which I'm currently the least sure (there are a bunch of details I haven't yet fully checked as I'm in the process of dusting off some of my recently rather unused knowledge about general topology), so was hoping someone could confirm.

-
I fixed the link. –  Joel David Hamkins Apr 3 '10 at 12:01
Thanks. Sorry about that. My MathOverflow-fu is still rather white belt. –  David R. MacIver Apr 3 '10 at 12:16
What is $\omega_1$ and $\beta N$? –  Thomas Kragh Apr 3 '10 at 13:41
$\omega_1$ is the first uncountable ordinal number with the order topology. $\beta N$ should be $\beta \mathbb{N}$ (I'll fix that) and is the Stone-Cech compactification of the natural numbers with the discrete topology. –  David R. MacIver Apr 3 '10 at 14:01
By the way, I've proven one of the other two results false (or at least, the proof is wrong), so my belief in this one is reaffirmed. It would be good to have something to reference in a bibliography though if anyone can provide one. –  David R. MacIver Apr 3 '10 at 14:02

## 1 Answer

I have reread the proof and it's completely correct. The idea is that $\omega_1 \times \beta\mathbb{N}$ maps perfectly onto a non-normal space, and normality is preserved under perfect maps. Tamano's theorem says that $X$ is paracompact Hausdorff iff $X \times \beta X$ is normal, and we use that $\omega_1$ is not paracompact and $\beta \omega_1 = \omega_1 + 1$. But the direct proof as sketched is also correct (we can use the pushdown lemma to prove it).

All building blocks can be found in Engelking, e.g.

-
Thanks for double checking. As per comment, my reason for doubt had already evaporated. I'd also verified the steps myself from engelking. Now the only question is how to cite this. :-) –  David R. MacIver Apr 4 '10 at 9:39


text:
3 If not this, then the other...

It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$.

On the other hand, it is long known that other axioms, for example $GCH, CH, V=L, ...$ are not preserved by such extensions.

In a paper of Monro [1] the first paragraph speaks on a question by Dana Scott about why do some axioms get preserved by generic extensions and other not; it also says that there is an [almost obvious] equivalence between preservation in generic extensions, and preservation in Boolean-valued models.

The paper goes on to prove that some restricted versions of the axiom of choice are not preserved by generic extensions (and in a recent phone call it was explained to me how to break weak choice principles such $DC_\kappa$ quite easily in relatively simple generic extensions of models of $ZF$).

Lastly the paper says nothing about an answer to the question above, nor it cites any resource for a possible answer. However, the paper is quite old now, and in the past three decades some progress might have occurred.

Is there an answer for Scott's question: which axioms are preserved by all generic extensions?

Edit: If there is no "simple" and uniform answer to the above question, is there a possible answer to why is the Axiom of Choice preserved in generic extensions, while restricted versions are not?

Bibliography:

1. G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983), pp. 39-52
2 Corrected grammar

1

# Why does some axioms are preserved in generic extensions?

It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$.

On the other hand, it is long known that other axioms, for example $GCH, CH, V=L, ...$ are not preserved by such extensions.

In a paper of Monro [1] the first paragraph speaks on a question by Dana Scott about why do some axioms get preserved by generic extensions and other not; it also says that there is an [almost obvious] equivalence between preservation in generic extensions, and preservation in Boolean-valued models.

The paper goes on to prove that some restricted versions of the axiom of choice are not preserved by generic extensions (and in a recent phone call it was explained to me how to break weak choice principles easily in relatively simple generic extensions of models of $ZF$).

Lastly the paper says nothing about an answer to the question above, nor it cites any resource for a possible answer. However, the paper is quite old now, and in the past three decades some progress might have occurred.

Is there an answer for Scott's question: which axioms are preserved by all generic extensions?

Bibliography:

1. G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983), pp. 39-52


text:
## Chemistry: The Molecular Science (5th Edition)

The name of $KNO_2$ is : "Potassium nitrite"
1. Identify the positive ion, and name it. $K^+$ is the positive ion. So, the first name of the compound will be: "Potassium." 2. Identify the negative ion, and name it. $N{O_2}^-$ is the negative ion. As we can see in Table 2.2 (Page 58), this ion is called "Nitrite". 3. Putting those two names together: $K^+ + NO{_2}^- = KNO_2$ Potassium + nitrite = Potassium nitrite.


text:
# Math Help - which rule to use for this equation?

1. ## which rule to use for this equation?

y = x^4 - 4x^3 - 2 cos (x) + tan (x)

question asks to differentiate the following function

the answer i believe is 4x^3 - 16x^2 + 2 sin (x) + tan (x)

do i have the correct answer?

2. Originally Posted by JontyP
y = x^4 - 4x^3 - 2 cos (x) + tan (x)

question asks to differentiate the following function

the answer i believe is 4x^3 - 16x^2 + 2 sin (x) + tan (x)

do i have the correct answer?
no, the derivative of tan(x) is not tan(x)

why do you feel tan(x) should stay the same?

3. oops i missed out the last bit of the answer!

it should look like this
4x^3 - 12x^2 + 2 sin (x) + (1/cos^2(x))

i've changed the -16x^2 to a -12 too

4. Originally Posted by JontyP
oops i missed out the last bit of the answer!

it should look like this
4x^3 - 12x^2 + 2 sin (x) + (1/cos2(x))

i've changed the -16x^2 to a -12 too
that's correct now. by the way, we can write $\frac 1{\cos^2 x}$ as $\sec^2 x$


text:
# Homework Help: Fourier Transform of Distribution

1. Jan 17, 2014

### VVS

Hi,

I hope somebody can help me with this one.

1. The problem statement, all variables and given/known data
Compute the Fourier Transform of the distribution x-a

2. Relevant equations
The Fourier Transform of a distribution is just the distribution evaluated with the Fourier Transform of a test function.

3. The attempt at a solution
See this pdf View attachment Übung 27.pdf
I used integration by parts but now I am stuck, because I have to evaluate the integral of the Foureir Transform of the test function.

Thanks
VVS

2. Jan 17, 2014

### vanhees71

I'd rather regularize the distribution, e.g., defining
$$\tilde{f}_{\epsilon}(k)=\int_{\mathbb{R}} \mathrm{d} x \exp(-\mathrm{i} k x) (x-a) \exp(-\epsilon x^2).$$
Then you only need to know that
$$\delta_{\epsilon}(k)=\frac{1}{2 \sqrt{\pi \epsilon}} \exp \left (-\frac{k^2}{4 \epsilon} \right )$$
is a smoothed $\delta$ distribution, i.e.,
$$\lim_{\epsilon \rightarrow 0^+} \delta_{\epsilon}(k)=\delta(k).$$


text:
# HNN extension

In mathematics, the HNN extension is a basic construction of combinatorial group theory.

Introduced in a 1949 paper Embedding Theorems for Groups[1] by Graham Higman, B. H. Neumann and Hanna Neumann, it embeds a given group G into another group G' , in such a way that two given isomorphic subgroups of G are conjugate (through a given isomorphism) in G' .

## Construction

Let G be a group with presentation G = <S|R>, and let α : HK be an isomorphism between two subgroups of G. Let t be a new symbol not in S, and define

$G*_{\alpha} = \left \langle S,t \Big| R, tht^{-1}=\alpha(h), \forall h\in H \right \rangle.$

The group Gα is called the HNN extension of G relative to α. The original group G is called the base group for the construction, while the subgroups H and K are the associated subgroups. The new generator t is called the stable letter.

## Key properties

Since the presentation for Gα contains all the generators and relations from the presentation for G, there is a natural homomorphism, induced by the identification of generators, which takes G to Gα. Higman, Neumann and Neumann proved that this morphism is injective, that is, an embedding of G into Gα. A consequence is that two isomorphic subgroups of a given group are always conjugate in some overgroup; the desire to show this was the original motivation for the construction.

### Britton's Lemma

A key property of HNN-extensions is a normal form theorem known as Britton's Lemma.[2] Let Gα be as above and let w be the following product in Gα:

$w=g_0 t^{\varepsilon_1} g_1 t^{\varepsilon_2} \cdots g_{n-1} t^{\varepsilon_n}g_n, \qquad g_i \in G, \varepsilon_i = \pm 1.$

Then Britton's Lemma can be stated as follows:

Britton's Lemma. If w = 1 in Gα then

• either n = 0 and g0 = 1 in G
• or n > 0 and for some i ∈ {1, ..., n−1} one of the following holds:
1. εi = 1, εi+1 = −1, giH,
2. εi = −1, εi+1 = 1, giK.

In contrapositive terms, Britton's Lemma takes the following form:

Britton's Lemma (alternate form). If w is such that

• either n = 0 and g0 ≠ 1 ∈ G,
• or n > 0 and the product w does not contain substrings of the form tht−1, where hH and of the form t−1kt where kK,
then w ≠ 1 in Gα.

### Consequences of Britton's Lemma

Most basic properties of HNN-extensions follow from Britton's Lemma. These consequences include the following facts:

• The natural homomorphism from G to Gα is injective, so that we can think of Gα as containing G as a subgroup.
• Every element of finite order in Gα is conjugate to an element of G.
• Every finite subgroup of Gα is conjugate to a finite subgroup of G.
• If HG and KG then Gα contains a subgroup isomorphic to a free group of rank two.

## Applications

In terms of the fundamental group in algebraic topology, the HNN extension is the construction required to understand the fundamental group of a topological space X that has been 'glued back' on itself by a mapping f (see e.g. Surface bundle over the circle). That is, HNN extensions stand in relation of that aspect of the fundamental group, as free products with amalgamation do with respect to the Seifert-van Kampen theorem for gluing spaces X and Y along a connected common subspace. Between the two constructions essentially any geometric gluing can be described, from the point of view of the fundamental group.

HNN-extensions play a key role in Higman's proof of the Higman embedding theorem which states that every finitely generated recursively presented group can be homomorphically embedded in a finitely presented group. Most modern proofs of the Novikov-Boone theorem about the existence of a finitely presented group with algorithmically undecidable word problem also substantially use HNN-extensions.

Both HNN-extensions and amalgamated free products are basic building blocks in the Bass–Serre theory of groups acting on trees.[3]

The idea of HNN extension has been extended to other parts of abstract algebra, including Lie algebra theory.

## Generalizations

HNN extensions are elementary examples of fundamental groups of graphs of groups, and as such are of central importance in Bass–Serre theory.

## References

1. ^ Higman, Graham; B. H. Neumann; Hanna Neumann (1949). "Embedding Theorems for Groups" (PDF). Journal of the London Mathematical Society. s1-24 (4): 247–254. doi:10.1112/jlms/s1-24.4.247. Retrieved 2008-03-15.
2. ^ Roger C. Lyndon and Paul E. Schupp. Combinatorial Group Theory. Springer-Verlag, New York, 2001. "Classics in Mathematics" series, reprint of the 1977 edition. ISBN 978-3-540-41158-1; Ch. IV. Free Products and HNN Extensions.
3. ^ Jean-Pierre Serre. Trees. Translated from the French by John Stillwell. Springer-Verlag, Berlin-New York, 1980. ISBN 3-540-10103-9


text:
# Dynamic Programming#

Dynamic algorithms are a family of programs which (broadly speaking) share the feature of utilising solutions to subproblems in coming up with an optimal solution. We will discuss the conditions which problems need to satisfy to be solved dynamically but let us first have a look at some basic examples. The concept is quite challenging to digest, so there are many examples in this section.

## Simple Examples#

• Dynamic Fibonacci If you have read the section about recursion you probably know that the basic implementation for generating Fibonacci numbers is very inefficient ($$O(2^n)$$!). Now, as we generate the next numbers in the Fibonacci sequence, we will remember them for further use. The trade-off between memory and time is dramatic in this case. Compare the two versions of the function:

# For comparing the running times
import time

# Inefficient
def fib(n):
assert n >= 0
if n < 2:
return 1
else:
return fib(n-1) + fib(n-2)

# Dynamic version
def dynamicFib(n):
assert n >= 0
#prepare a table for memoizing
prevFib = 1
prevPrevFib = 1
temp = 0

#build up on your previous results
for i in range(2,n+1):
temp = prevFib + prevPrevFib
prevPrevFib = prevFib
prevFib = temp
return prevFib

start = time.time()
print(fib(32))
end = time.time()
print("Time for brute:" + str(end - start))

start = time.time()
print(dynamicFib(32))
end = time.time()
print("Time for dynamic:" + str(end - start))

3524578
Time for brute:0.8238296508789062
3524578
Time for dynamic:0.0

The time difference is enormous! As you can probably spot, dynamicFib is $$O(n)$$! With the use of three integer variables (prevFib, prevPrevFib and temp) we brought exponential time complexity down to linear. How did it happen? Let us now depict the work done by fib(5) on a graph:

Wow! This is quite a tree! However, the worst feature is that many of the nodes are repeated (e.g. node $$2$$ appears 3 times). These repeated results which are constantly recomputed bring us to the exponential complexity. Consider the dynamic solution graph:

Now, the number of nodes grows linearly with $$n$$. We have merged the subproblems to avoid redundancy.

• Shortest Path on a Grid Dynamic programming is often used for optimisation problems. Consider a square grid with numbers from 0 to 9 in each square. An example would be:

+-+-+-+-+
|1|0|8|4|
+-+-+-+-+
|3|5|1|0|
+-+-+-+-+
|6|8|9|3|
+-+-+-+-+
|1|2|4|5|
+-+-+-+-+

It is allowed to move only down or right. What is the value of the minimum path from the upper-left corner to the lower-right corner?

The initial approach to the problem might be to check all the possible paths and return the minimal one. This is exponential and too slow. To come up with a faster approach we need to find subproblems.

Let us imagine that we have reached the lower-right corner (hurray!). We could get there from the tile above it and left of it. This might at first seem like a Divide and Conquer problem, but let us keep on looking for an overlap:

In the simplest case of four tiles, we can already see that the upper-left tile is considered twice. We should then leverage this repetition. This overlapping generalises for larger grids. In our algorithm, we will remember the optimal path to the tiles already visited and build on this knowledge:

# grid is a square matrix
def shortestPathDownRight(grid):
# dictionary that will keep the optimal paths
# to the already visited tiles
table = {}
# n - length of the side of the square
n = len(grid)
assert n == 0 or len(grid[0]) == n
table[(0,0)] = grid[0][0]
# top and most left column have trival optimal paths
for i in range(1, n):
table[(0,i)] = table[(0,i-1)] + grid[0][i]
table[(i,0)] = table[(i-1,0)] + grid[i][0]
# the dynamic magic
for i in range(1,n):
for j in range(1,n):
table[(i,j)] = min(table[(i-1,j)],table[(i,j-1)]) + grid[i][j]

return table[(n-1,n-1)]

grid = [[1,0,8,4],[3,5,1,0],[6,8,9,3],[1,2,4,5]]
print(shortestPathDownRight(grid))

15

What is the time complexity of this algorithm? Based on the nested loop we can deduce that it is $$O(n^2)$$. Quite an improvement!

### Space Complexity#

We usually do not concern ourselves with the amount of memory a program utilises. Dynamic programs are an exception to this rule. The amount of memory they use can be a limiting factor for some machines, so we need to take it into account. In the case of dynamicFib this was $$O(1)$$ as we only needed to keep track of the last two Fibonacci numbers. In case of shortestPathDownRight we need to create a grid of size $$n \times n$$, so $$O(n^2)$$. The notion of space complexity is very similar to time complexity, so we will not discuss it in depth.

• Cutting Rods We are given a rod of integer length $$n$$ and a list of length prices $$n$$. The $$i$$th entry in the list corresponds to the profit we can get from selling the rod of length $$i$$. How should we cut the rod so we maximise our profit? The key to our dynamic algorithm will be the observation that:

We cut the rod at position $$k$$. Now we have a rod of lenght $$k$$ and $$n-k$$. Let us assume we know the maximum price for these two rods. Now we need to consider all the $$0 \leq k \leq \lfloor \frac{n}{2} \rfloor + 1$$ (the problem is symmetric so computing for $$\frac{n}{2} \leq k$$ would be redundant. For $$k=0$$ we just take prices[n]. The cutting introduces subproblems which are smaller than the initial problem and they are overlapping! Let us put this into code:

# For comparing the running times
import time
# For testing
from random import randint

def dynamicCutRod(n,prices):
# setting the initial values of variables
assert n >= 0 and n == len(prices)
# trival cases
if n == 0:
return 0
if n == 1:
return prices[0]

# setting up needed variables
table = {}
currLen = 2
table[0] = 0
table[1] = prices[0]
while currLen < n + 1:
# no cuts for a given len
table[currLen] = prices[currLen - 1]
# considering all possible cuts for a give len
for k in range(1,currLen//2 + 1):
# take the maximal one
if table[currLen] < table[k] + table[currLen - k]:
table[currLen] = table[k] + table[currLen - k]
currLen += 1
return table[n]

# for testing purposes
def bruteForceRecursive(n,prices):
assert n >=0
if n == 0:
return 0
if n == 1:
return prices[0]
currLen = n
res = prices[n-1]
for k in range(1,n//2 + 1):
res = max(res, bruteForceRecursive(k,prices) + bruteForceRecursive(n-k,prices))
return res

# testing
for i in range(1, 11):
prices = []
for j in range(i):
prices.append(randint(1,10))

assert bruteForceRecursive(len(prices),prices) == dynamicCutRod(len(prices),prices)

# comparing times
prices = []
for i in range(20):
prices.append(randint(1,10))

start = time.time()
print(bruteForceRecursive(len(prices),prices))
end = time.time()
print("Time for brute:" + str(end - start))

start = time.time()
print(dynamicCutRod(len(prices),prices))
end = time.time()
print("Time for dynamic:" + str(end - start))

66
Time for brute:1.457252025604248
66
Time for dynamic:0.0029768943786621094

Time complexity? For each $$0\leq i \leq n$$ we need to consider $$i$$ different cuts. This is the sum of arithmetic series so $$O(n^2)$$. We memoize only the optimal solutions to subproblems, therefore space complexity is $$O(n)$$.

## Characteristic features of dynamic problems#

All problems which can be tackled with the use of Dynamic Programming (DP) have two main features:

1. Optimal substructure: In broad terms, that solution to the problem can be formulated in terms of independent subproblems. Then you choose a solution/combination of optimal solutions to subproblems and show that this choice leads to an optimal global solution. For example, for the shortestPath problem we choose between the optimal solution to the tile above and tile to the left. For the rod problem, we choose between all possible $$n$$ cuts of a rod and assume we have solutions to the shorter rods.

2. Overlapping subproblems: The subproblems which we split the main question into should repeat. Eliminating the need to recompute the subproblems (by memoising them) should speed up our algorithm. For example, dynamicFib used two variables to keep the results of computing previous Fibonacci numbers.

## Common approaches to DP#

There are two main schools of thought when it comes to solving DP algorithms:

1. Memoization is using the memory-time trade-off. It remembers the results to the subproblems in a dictionary memo and recalls them when needed. If we have the result in memory, we take if from there, otherwise compute it.

2. Bottom-up approach starts with the trivial cases of the problem and builds on them. I relies on the the fact that subproblems can be sorted. So we first compute the optimal cutting for a rod of length 1 and then length 2 and so on…

Both usually lead to the same time complexities.

• 0-1 Knapsack The problem is about a backpack (knapsack) of an integer volume $$S$$, we are given lists spaces and vals which correspond to the size and value of the objects we might want to pack. The trick is to pack the objects of the highest accumulate value which fit into the backpack. We will consider prefixes of the sizes list in order of length and put the intermediate results in a table. Compare the bottom-up and memoized approaches:

# For comparing the running times
import time
# For testing
from random import randint

# bottom up approach
def knapSack(wt, val, W, n):
K = [[0 for x in range(W + 1)] for x in range(n + 1)]

# Build table K[][] in bottom up manner
for i in range(n + 1):
for w in range(W + 1):
if i == 0 or w == 0:
K[i][w] = 0
elif wt[i-1] <= w:
K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w])
else:
K[i][w] = K[i-1][w]

return K[n][W]

# memoized
memo = {}
def knapSackMemo(wt,val, W, n):
# base case
if n == 0 or W == 0:
memo[(n,W)] = 0
return 0
# check if memoized
if (n,W) in memo:
return memo[(n,W)]
# if not, calcuate
else:
if wt[n-1] <= W:
memo[(n,W)] = max(knapSackMemo(wt,val,W-wt[n-1],n-1)+val[n-1],
knapSackMemo(wt,val,W,n-1))
return memo[(n,W)]
else:
memo[(n,W)] = knapSackMemo(wt,val,W,n-1)
return memo[(n,W)]

# brute force for testing
def bruteForce(wt, val, W):
res = 0
# all combinations of the elements
for i in range(2**len(wt)):
sumSize = 0
sumVal = 0
for j in range(len(wt)):
if (i >> j) & 1:
sumSize += wt[j]
sumVal += val[j]
if sumSize > W:
sumVal = 0
break
res = max(sumVal, res)
return res

# testing
for _ in range(10):
sizes = []
vals = []
S = randint(0,200)
memo = {}
for _ in range(13):
sizes.append(randint(1,10))
vals.append(randint(1,20))

br = bruteForce(sizes,vals, S)
btup  = knapSack(sizes,vals,S,len(sizes))
mem = knapSackMemo(sizes,vals,S,len(sizes))

assert btup == br and mem == br

start = time.time()
print(bruteForce(sizes,vals, S))
end = time.time()
print("Time for brute:" + str(end - start))

start = time.time()
print(knapSack(sizes,vals,S,len(sizes)))
end = time.time()
print("Time for bottom-up:" + str(end - start))

memo = {}
start = time.time()
print(knapSackMemo(sizes,vals,S,len(sizes)))
end = time.time()
print("Time for memoized:" + str(end - start))

135
Time for brute:0.019960880279541016
135
Time for bottom-up:0.00098419189453125
135
Time for memoized:0.0009658336639404297

In this case, the memoized approach is the fastest as it does not usually require to consider all (n,W) pairs. However, both bottom-up and memoized approach have space (and time) complexity of $$O(nS)$$. The brute force has time complexity of $$O(2^n)$$, where $$n$$ is the length of the sizes list.

• Scheduling With all the chaos that COVID caused universities are going fully remote. The head of your department asked the lecturers to send the timeslots in which they can do lectures. This is a list timeslots which consist of pairs of integers - the start and end of the timeslot. If timeslots overlap, then we choose only one of them. Lectures can start at the same time the previous lecture finished. You aim to pick the subset of timeslots which assures that the sum of teaching hours is maximum. Return the maximum number of hours.

The question is similar to the previous one. We are given a list of items (timeslots) and need to come up with an optimisation. The value of each subset is the sum of hours of its elements. To speed things up in later stages, we will sort the timeslots by the ending time. This is done in $$O(nlog(n))$$. We will then consider the prefixes of the sorted array and memoize optimal results. memo[n] will store the maximum number of hours from the prefix of n length.

# For comparing the running times
import time
# For testing
from random import randint

# utility function to speed up the search
def binarySearch(timeslots,s,low,high):
if low >= high:
return low
mid  = (low + high) //2
if timeslots[mid][1] <= s:
return binarySearch(timeslots, s, mid+1, high)
else:
return binarySearch(timeslots, s, low, mid)

# init memo
memo = {}
# assumes that timeslots array is sorted by ending time
def schedule(timeslots, n):
# base case
if n == 0:
memo[0] = 0
return 0
# memoized case
elif n in memo:
return memo[n]
# else calculate
else:
s,e = timeslots[n-1]
# in log time
ind = min(binarySearch(timeslots,s,0,len(timeslots)),n-1)
memo[n] = max(schedule(timeslots,n-1), schedule(timeslots,ind) + (e - s))
return memo[n]

# brute force for testing
def bruteForce(timeslots):
res = 0
# all combinations of the elements
for i in range(2**len(timeslots)):
sumHours = 0
for j in range(len(timeslots)):
if (i >> j) & 1:
s, e = timeslots[j]
sumHours += e - s
# checking if a valid combination of timeslots
sumHours = 0
break
res = max(sumHours, res)
return res

# testing
for _ in range(10):
memo = {}
timeslots = []
for _ in range(12):
s = randint(0,100)
e = randint(s,100)
timeslots.append((s,e))
timeslots.sort(key = lambda slot: (slot[1],slot[0]))

br = bruteForce(timeslots)
mem = schedule(timeslots,len(timeslots))
assert br == mem

start = time.time()
print(bruteForce(timeslots))
end = time.time()
print("Time for brute:" + str(end - start))

memo = {}
start = time.time()
print(schedule(timeslots,len(timeslots)))
end = time.time()
print("Time for memo:" + str(end - start))

82
Time for brute:0.14516472816467285
82
Time for memo:0.0010349750518798828

Time complexity of this solution is $$O(nlog(n))$$. Why?

## Exercises#

• Tiling problem You are given a $$2 \times n$$ board and $$2 \times 1$$ tiles. Count the number of ways the tiles can be arranged (horizontally and vertically) on the board.

HINT: You have seen this algorithm before.

• Maximum sum subarray You are given an array with $$1$$ and $$-1$$ elements only. Find the array of the maximum sum in $$O(n)$$. Return this maximal sum. E.g. [-1,1,1,-1,1,1,1,-1] -> 4.

HINT: Consider the sums of prefixes of the array.

• Longest snake sequence You are given a grid of natural numbers. A snake is a sequence of numbers which goes down or to the right. The adjacent numbers in a snake differ by a maximum 1. Define a dynamic function which will return the length of the longest snake. E.g.

9, 6, 5, 2

8, 7, 6, 5

7, 3, 1, 6

1, 1, 1, 7

-> 7

## References#

• Cormen, Leiserson, Rivest, Stein, “Introduction to Algorithms”, Third Edition, MIT Press, 2009

• Prof. Erik Demaine, Prof. Srini Devadas, MIT OCW 6.006 “Introduction to Algorithms”, Fall 2011

• GeeksforGeeks, Find maximum length Snake sequence


text:
## Duke Mathematical Journal

### Definability of restricted theta functions and families of abelian varieties

#### Abstract

We consider some classical maps from the theory of abelian varieties and their moduli spaces, and we prove their definability on restricted domains in the o-minimal structure $\mathbb {R}_{\mathrm {an,\,exp}}$. In particular, we prove that the projective embedding of the moduli space of the principally polarized abelian variety $\operatorname{Sp}(2g,\mathbb {Z})\backslash \mathbb {H}_{g}$ is definable in $\mathbb {R}_{\mathrm {an,\,exp}}$ when restricted to Siegel’s fundamental set $\mathfrak {F}_{g}$. We also prove the definability on appropriate domains of embeddings of families of abelian varieties into projective spaces.

#### Article information

Source
Duke Math. J., Volume 162, Number 4 (2013), 731-765.

Dates
First available in Project Euclid: 15 March 2013

https://projecteuclid.org/euclid.dmj/1363355692

Digital Object Identifier
doi:10.1215/00127094-2080018

Mathematical Reviews number (MathSciNet)
MR3039679

Zentralblatt MATH identifier
1284.03215

#### Citation

Peterzil, Ya’acov; Starchenko, Sergei. Definability of restricted theta functions and families of abelian varieties. Duke Math. J. 162 (2013), no. 4, 731--765. doi:10.1215/00127094-2080018. https://projecteuclid.org/euclid.dmj/1363355692

#### References

• [1] W. L. Baily Jr., On the theory of $\theta$-functions, the moduli of abelian varieties, and the moduli of curves, Ann. of Math. (2) 75 (1962), 342–381.
• [2] W. L. Baily Jr. and A. Borel, Compactification of arithmetic quotients of bounded symmetric domains, Ann. of Math. (2) 84 (1966), 442–528.
• [3] C. Birkenhake and H. Lange, Complex Abelian Varieties, 2nd ed., Grundlehren Math. Wiss. 302, Springer, Berlin, 2004.
• [4] J.-B. Bost, “Introduction to compact Riemann surfaces, Jacobians, and abelian varieties” in From Number Theory to Physics (Les Houches, 1989), Springer, Berlin, 1992, 64–211.
• [5] A. M. Gabrièlov, Projections of semianalytic sets, Funkcional. Anal. i Priložen 2 (1968), 18–30.
• [6] J.-B. Hiriart-Urruty and C. Lemaréchal, Fundamentals of Convex Analysis, Grundlehren Text Ed., Springer, Berlin, 2001.
• [7] Jun-ichi Igusa, On the graded ring of theta-constants, II, Amer. J. Math. 88 (1966), 221–236.
• [8] J.-i. Igusa, Theta Functions, Grundlehren Math. Wiss. 194, Springer, New York, 1972.
• [9] H. Klingen, Introductory Lectures on Siegel Modular Forms, Cambridge Stud. Adv. Math. 20, Cambridge Univ. Press, Cambridge, 1990.
• [10] B. Klingler and A. Yafaev, The André–Oort conjecture, preprint, arXiv:1209.0936 [math.NT].
• [11] D. Mumford, Tata Lectures on Theta, I, Progr. Math. 28, Birkhäuser, Boston, 1983.
• [12] Y. Peterzil and S. Starchenko, Uniform definability of the Weierstrass $\wp$ functions and generalized tori of dimension one, Selecta Math. (N.S.) 10 (2004), 525–550.
• [13] Ya’acov Peterzil and Sergei Starchenko, Complex analytic geometry and analytic-geometric categories, J. Reine Angew. Math. 626 (2009), 39–74.
• [14] Y. Peterzil and S. Starchenko, Mild manifolds and a non-standard Riemann existence theorem, Selecta Math. (N.S.) 14 (2009), 275–298.
• [15] Ya’acov Peterzil and Sergei Starchenko, “Tame complex analysis and o-minimality” in Proceedings of the International Congress of Mathematicians, Vol. II, Hindustan Book Agency, 2010, 58–81.
• [16] J. Pila, Rational points of definable sets and results of André–Oort–Manin–Mumford type, Int. Math. Res. Not. IMRN 13 (2009), 2476–2507.
• [17] J. Pila, O-minimality and the André–Oort conjecture for $\mathbb{C}^{n}$, Ann. of Math. (2) 173 (2011), 1779–1840.
• [18] J. Pila and A. J. Wilkie, The rational points of a definable set, Duke Math. J. 133 (2006), 591–616.
• [19] C. L. Siegel, Lectures on Quadratic Forms, notes by K. G. Ramanathan, Tata Inst. Fund. Res. Lectures Math. 7, Tata Inst. Fund. Res., Bombay, 1967.
• [20] A. Terras, Harmonic Analysis on Symmetric Spaces and Applications, II, Springer, Berlin, 1988.
• [21] L. van den Dries, Tame Topology and O-Minimal Structures, London Math. Soc. Lecture Note Ser. 248, Cambridge Univ. Press, Cambridge, 1998.
• [22] L. van den Dries, A. Macintyre, and D. Marker, The elementary theory of restricted analytic fields with exponentiation, Ann. of Math. (2) 140 (1994), 183–205.
• [23] L. van den Dries and C. Miller, On the real exponential field with restricted analytic functions, Israel J. Math. 85 (1994), 19–56.
• [24] Lou van den Dries and Christopher Miller, Geometric categories and o-minimal structures, Duke Math. J. 84 (1996), 497–540.
• [25] G. van der Geer, “Siegel modular forms and their applications” in The 1-2-3 of Modular Forms (Nordfjordeid, Norway, 2004), Universitext, Springer, Berlin, 2008, 181–245.
• [26] G. M. Ziegler, Lectures on Polytopes, Grad. Texts in Math. 152, Springer, New York, 1995.


text:
# Statement of Cash Flows: Indirect Method

Wu Company's income statement for the year ended December 31, 2011, and its comparative balance sheets as of December 31, 2011 and 2010, are presented on the next page(attached documents). During 2011, Wu Company engaged in these transactions: a. Sold at a gain of $7,000 furniture and fixtures that cost$35,600, on which it had accumulated depreciation of $28,800. b. Purchased furniture and fixtures in the amount of$39,600. c. Paid a $20,000 note payable and borrowed$40,000 on a new note. d. Converted bonds payable in the amount of $100,000 into 4,000 shares of common stock. e. Declared and paid$6,000 in cash dividends. Required 1. Using the indirected method, prepare a statement of cash flows for Wu Company. Include a supporting schedule of noncash investing transactions and financing transactions. 2. What are the primary reasons for Wu Company's large increase in cash from 2010 to 2011, despite its low net income? 3. Compute and assess cash flow yield and free cash flow for 2011. Compare and contrast what these two performance measures tell you about Wu Company's cash-generating ability.

Manny v

Solution -

Cash Flow Statement

Operating Activities

Note Payable Paid                                                                                -\$20000

...


text:
## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE

#### View:

 Message: [ First | Previous | Next | Last ] By Topic: [ First | Previous | Next | Last ] By Author: [ First | Previous | Next | Last ] Font: Proportional Font

Subject:

Re: rsfs

From:

Date:

Thu, 27 Mar 1997 11:29:36 +0100

Content-Type:

text/plain

Parts/Attachments:

 text/plain (48 lines)
 Laurent Siebenmann wrote (commenting on Hans Aberg): > > The AMSFonts "Euler script" font is upright, and the TeX "calligraphic" > > font is slanted (but none are very "scripty"), so why not designing a new > > font? > > rsfs, Ralph Smith's Formal Script on CTAN is excellent, and so clearly > distinct from Euler script that the two can be used concurrently for > different things. Yes, and rsfs is enough scripty' I suppose, but it is still lacking the lowercase letters. This was one suggestion/topic in the article of Justin Ziegler (TeXnical working group on extended math font encoding), but that was about one and a half year ago and I haven't heard of that ever since. TeX's standard calligraphic' alphabet also lacks the lowercase (so I won't call that a proper font yet). Of course, uppercase is needed an used far more often than lowercase in these fonts. And may be lowercase letters in these three fonts (if they would exist) could be a bit confusing, when used alongside in the same text. May be there should be a fourth script-like font, or at least parts of an alphabet in such a style, to be used for some special one letter symbols, containing for example P for the power set (also mentioned in J. Ziegler's article) C to denote the set of continuous / continuous differentiable functions   (as in C^k(\mathbf{R}) or the like) O and o for the Landau symbols (denoting the order of magnitude of   a function) and so on, so that these would have special reserved symbols which do not use one of the other script styles (as these may be used to denote a certain concept). Here, again, a (imho) bad tradition' again interferes: the anglo-american usage of Fraktur R and I to denote real and imaginary part: if the Fraktur alphabet is used for some other concept, this could be quite confusing (German normally uses \mathrm{Re} and \mathrm{Im} for this). Johannes Kuester -- Johannes Kuester [log in to unmask] Mathematisches Institut der Technischen Universitaet Muenchen`


text:
Uncategorized

# characteristics of trees in points

They are smaller than Trees as they are less rigid than a single thick trunk. So, the value of all the vertices of the left sub-tree of an internal node $V$ are less than or equal to $V$ and the value of all the vertices of the right sub-tree of the internal node $V$ are greater than or equal to $V$. All trees have these key characteristics, although their stem shapes, wood composition, leaf shape, and … Light detection and ranging (LiDAR) has been applied to obtain the 3-D structural properties of … If a single vertex is left then it is the center of the tree and if two vertices joined by an edge is left then it is the bi-center of the tree. Leaves that are wilting, yellow or brown on newly planted trees. Research shows that within minutes of being surrounded by trees and green space, your blood pressure drops, your heart rate slows and your stress levels come down. Depending on the species of Dogwood you plant, you may have a short stout bush or a 25 foot tall tree. An ordered rooted tree is a rooted tree where the children of each internal vertex are ordered. Finally, we got two vertices ‘c’ and ‘d’ left, hence we stop the algorithm. Definition − A labeled tree is a tree the vertices of which are assigned unique numbers from 1 to n. We can count such trees for small values of n by hand so as to conjecture a general formula. The number of labeled trees of n number of vertices is $\frac {(2n)! There is a unique path between every pair of vertices in G. A tree with N number of vertices contains (N-1) number of edges. Pine trees are evergreen coniferous trees that are native to most countries in the Northern Hemisphere. 3. While humans can live longer than 80 years, trees … The number of labeled trees of n number of vertices is nn-2. America’s forests contain hundreds of different hardwood tree species. The maximum eccentricity is the tree diameter. A pine tree has distinct coverings, from its six-sided bark to its waxy needles. These evergreen trees are related to other coniferous trees… Difference between Definition and Declaration in Java. As there is single vertex, this tree has one center ‘c’ and the tree is a central tree. Smaller than Trees - Shrubs are usually less than 3–6 m (15–20 ft) tall. Tulsi Plant: Tulsi plant is considered as a holy and religious plant in India. Before deciding which tree species to plant, a grower should consider several points. Your tree likely has transplant shock! What is the new age definition of feminism? The Characteristics of Neem Trees Overview A native of India, the Neem tree is related to mahogany. The trees have broad leaves rather than needle-like leaves. A quadtree is a tree data structure in which each internal node has exactly four children. Binary Search tree is a binary tree which satisfies the following property −. If carefully treated, a mature Dogwood tree species, such as the Flowering Dogwood, may reach 40 feet in height. Definition − A Tree is a connected acyclic undirected graph. The most common tree … Tree - a woody perennial (a plant that lives many years), usually with a single trunk and growing more than 5 m in height. These are some of the most common characteristics to look for when identifying deciduous trees. Trees can live for many years. Definition − A Tree is a connected acyclic undirected graph. Thoughts about the three main similarities between humans and trees: American Hardwood trees are like humans in three distinct and profound ways: Both are mostly water Both have a peak life span of approximately 80 years Both are completely unique Let's talk about items #2 and #3. A rooted tree$G$is a connected acyclic graph with a special node that is called the root of the tree and every edge directly or indirectly originates from the root. The oil from birch bark, for example, has antiseptic properties. Trees boost soil fertility. Trees help clean the air we breathe, filter the water we drink, and provide habitat to over 80% of the world's terrestrial biodiversity. Tree is a discrete structure that represents hierarchical relationships between individual elements or nodes. There are two main types of trees: deciduous and evergreen. Tree is a discrete structure that represents hierarchical relationships between individual elements or nodes. Because the characteristics of Tree Points vary greatly between brands and disciplines, and they all interact to produce different fits, sometimes changing only the tree … In cold climates, this happens during the autumn so that the trees are bare throughout the winter. }$ (nth Catalan number). Height … A tree in which a parent has no more than two children is called a binary tree. Every tree is either central or bi-central. If every internal vertex of a rooted tree has exactly m children, it is called a full m-ary tree. A tree in which a parent has no more than two children is called a binary tree. Example − The following is an example of a tree −. Characteristics of Trees First, trees, shrubs, and woody vines are the only plants with woody growth, which makes them similar. There is a unique path between every pair of … The center of a tree is a vertex with minimal eccentricity. There are two main groups of trees—deciduous and evergreen. In the first post of the series, we listed the 5 characteristics of Tree Points (Width, Length, Angle, Flair, and Shape/Material), and discussed what Tree Width is and why it is super important to have the correct width for your horse. }{ (n+1)!n! Arborvitae (Thuja occidentalis ) An Arborvitae is a medium-sized forest tree that … The number of links from the root node to the deepest node is the height of the Binary Search Tree. Width (we will discuss this in depth below). Over 20 species of British trees and shrubs are known to have medicinal properties. Evergreen trees keep their leaves year-round. Example − The following is an example of a tree −. Angle. Characteristics of the different trees should be evaluated to determine which species will grow into high-quality trees on the available land and in the … In fact, about 40 percent of American trees … Binary Search tree is a binary tree which satisfies the following property −. The number of labeled trees of n number of vertices is $n^{n-2}$. Definition − An unlabeled tree is a tree the vertices of which are not assigned any numbers. In wider definitions, the taller palms, tree ferns, bananas, and bamboos are also trees. A tree in which a parent has no more than two children is called a binary tree. Tree and its Properties. Trees can be divided into two easily recognizable groups: conifers & broadleaf trees. Definition − A Tree is a connected acyclic undirected graph. If every internal vertex of a rooted tree has exactly m children, it is called a full m-ary tree. Tree is a discrete structure that represents hierarchical relationships between individual elements or nodes. Hardwood tree characteristics apply to many of the tree species in this country. They produce a fruit or nut, and often go dormant in the winter. Click on the map as close as possible to your intended planting site. If every internal vertex of a rooted tree has not more than m children, it is called an m-ary tree. Learn which trees are native to Ontario and the best species to plant where you live. Other varieties, such as oaks, have their leaves arrayed in alternating fashion along the twig. The eccentricity of a vertex $X$ in a tree $G$ is the maximum distance between the vertex $X$ and any other vertex of the tree. General features of the tree body As vascular plants, trees are organized into three major organs: the roots, the stems, and the leaves. As two vertices joined by an edge is left, this tree has bi-center ‘cd’ and the tree is bi-central. The tree displays medium growth, averaging between 13 and 24 inches annually. Most of the time, you can help your new tree recover … Finally we got a single vertex ‘c’ and we stop the algorithm. Sucker - A secondary shoot produced from the base or roots of a woody plant that gives rise to a new plant. On some deciduous trees, such as maples, the leaves are arranged opposite each other along the twig. A tree is a tall plant with a trunk and branches made of wood. Accurate extraction of the wood component of a tree is of great importance in visualizing trees. Shrubs typically branch near the ground and so have several narrow stems rather than a single Trunk. If a tree has only one center, it is called Central Tree and if a tree has only more than one centers, it is called Bi-central Tree. The sculptural form and bright green color of the winter foliage of coniferous trees are a striking sight in the winter landscape. Shape & Material. Step 1 − Remove all the vertices of degree 1 from the given tree and also remove their incident edges. In hot and dry climates, deciduous trees usually lose their leaves during the dry season. For example, if you have the right Tree Point Width, but any 1 of the other 4 Tree Point characteristics aren’t right for the horse, the tree isn’t going to be a good fit. Definition − A Tree is a connected acyclic undirected graph. Characteristics and Feature of Shrubs. The oldest tree ever discovered is approximately 5,000 years old and the oldest tree from the UK is about 1,000. Definition − An unlabeled tree is a tree the vertices of which are not assigned any numbers. The cell walls are composed of cellulose. Definition − A labeled tree is a tree the vertices of which are assigned unique numbers from 1 to n. We can count such trees for small values of n by hand so as to conjecture a general formula. What is definition of business finance and explain its types? Conifers come in a range of sizes, shapes and colors with more than 500 different trees included in the conifer classification. Deposited materials also make the heartwood of many tree species toxic to fungi and microorganisms, which is why lumber cut from tree heartwood is often more durable than the nutrient-rich sapwood. If every internal vertex of a rooted tree has not more than m children, it is called an m-ary tree. 2. Quadtrees are the two-dimensional analog of octrees and are most often used to partition a two-dimensional space by recursively subdividing it into four quadrants or regions. All of these characteristics affect the fit of the saddle … 4. If your tree has rainbow leaves, while other trees are green, your tree is stressed. Pines are resinous trees that are related to other types of conifers such as fir trees… many thin trees that result are sometimes called coppice-growth. See which of the top five problems your tree may have, then find the solution! The name conifer comes from the cones that grow on these trees as their form of ‘fruit.’ Of the 200 species of conifers in North America, pines, redwoods, firs, spruces, larches, cypress, and cedars are the most familiar ones. The maximum eccentricity is the tree diameter. }$(nth Catalan number). Binary Trees and Properties in Data Structures, Tournament Trees, Winner Trees and Loser Trees in Data Structure, Connectivity, Distance, and Spanning Trees. Use the following research questions as a guide. A tree with N number of vertices contains$(N-1)$number of edges. Many people think, for example, that so-called banana trees are trees, but in fact, they … The number of labeled trees of n number of vertices is$\frac {(2n)! $X$ in left sub-tree of vertex $V, Value(X) \le Value (V)$, $Y$ in right sub-tree of vertex $V, Value(Y) \ge Value (V)$. If m = 2, the rooted tree is called a binary tree. Length. The vertex which is of 0 … What is the definition of Hereditary and who Discovered it? So, the value of all the vertices of the left sub-tree of an internal node V are less than or equal to V and the value of all the vertices of the right sub-tree of the internal node V are greater than or equal to V. The number of links from the root node to the deepest node is the height of the Binary Search Tree. }{ (n+1)!n! There is a unique path between every pair of vertices in $G$. The way trees regulate water and hold the soil together … There is a unique path between every pair of vertices in G. A tree with N number of vertices contains (N-1) number of edges. X in left sub-tree of vertex V, Value(X) ≤ Value (V), Y in right sub-tree of vertex V, Value(Y) ≥ Value (V). Lack of Trunk - Most of the Shrubs lack Trunk. Pine tree needles serve many purposes, from protecting the tree from drought and fire to shielding its bark from the blazing sun. Flair. If you have a leaf like this, you will most likely have either a pinnately compound tree leaf or a leaf with multi- pinnate characteristics that form bi-pinnately compound tree leaves as illustrated and identified below. 1. Two labeled trees are isomorphic if their graphs are isomorphic and the corresponding points of the two trees have the same labels. Every tree is either central or bi-central. The vertex which is of 1 degree is called leaf node of the tree and the degree of an internal node is at least 2. Most deciduous trees, such as oaks and maples, are broad-leaved, and have broad, flat leaves. The eccentricity of a vertex X in a tree G is the maximum distance between the vertex X and any other vertex of the tree. The pine tree and its layers of thick needles are beneficial on many levels. The relatively small size between 1-2 cm, but some are up to 20 cm. A rooted tree G is a connected acyclic graph with a special node that is called the root of the tree and every edge directly or indirectly originates from the root. Step 2 − Repeat step 1 until either a single vertex or two vertices joined by an edge is left. The leaves are composed of a layer of cells in the bone except the … Why are trees important to the environment? An ordered rooted tree is a rooted tree where the children of each internal vertex are ordered. The deciduous trees include sugar maple, red maple, beech, red oak, and white ash.… forest Broad-leaved deciduous forests develop in middle-latitude climates, where there is an average temperature above 10 °C (50 °F) for at least six months every year and annual precipitation is above 400 mm (16 inches). Twig - a small limb; a branchlet of a tree. The four main parts of a tree are the roots, the trunk, the branches, and the leaves. Abstract: The wood structure is an important parameter that represents the geometrical and topological characteristics of trees. The leaves are the principal photosynthetic organs of most higher vascular plants. Or select a region for a larger-scale map. Trees benefit the … If a tree has only one center, it is called Central Tree and if a tree has only more than one centers, it is called Bi-central Tree. Apart from tall spruce trees that grow in coniferous forests, many dwarf spruces and spruce shrubs are suitable for garden landscapes. It is estimated that there are over 126 species of pine trees in the world. Deciduous trees lose their leaves in the fall. Two labeled trees are isomorphic if their graphs are isomorphic and the corresponding points of the two trees have the same labels. The center of a tree is a vertex with minimal eccentricity. These iconic characteristics of coniferous trees are adaptations to withstand the dry cold of northern climates and the accompanying challenges of water loss, water resupply and snow loads. There are 35 species of spruce trees belonging to the genus Picea. 5. Pine trees belong to the family Pinaceae and the genus Pinus. In botany, a tree is a perennial plant with an elongated stem, or trunk, supporting branches and leaves in most species.In some usages, the definition of a tree may be narrower, including only woody plants with secondary growth, plants that are usable as lumber or plants above a specified height. Find out the center/bi-center of the following tree −, At first, we will remove all vertices of degree 1 and also remove their incident edges and get the following tree −, Again, we will remove all vertices of degree 1 and also remove their incident edges and get the following tree −. Spruce trees are species of evergreen conifers that have sharp needle leaves. If $m = 2$, the rooted tree is called a binary tree. Deciduous trees lose all of their leaves for part of the year. Multicellular. This post is going to cover what Tree Point Length is, how it affects the Tree Width Forests provide jobs to over 1.6 billion people, absorb harmful carbon from the atmosphere, and are key ingredients in 25% of all medicines. The Tree Atlas . There are many trees and shrubs with pinnately compound leaves in North America. The vertex which is of 0 degree is called root of the tree. The data associated with a leaf cell varies by application, but … Christmas trees are an important part of North Carolina’s agricultural production. The vertex which is of 1 degree is called leaf node of the tree and the degree of an internal node is at least 2. Select one of these trees and, based on the characteristics of the tree, research to determine the type and species of the tree. The cells are eukaryotic. The vertex which is of 0 degree is called root of the tree. 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