text:
# 4-VisualizeBench

This category of tools aims to enable machine learning practitioners and users to understand how machine-learning models may be attacked..
This includes:

# TextAttack- A Framework for Adversarial Attacks in Natural Language Processing

### Abstract

TextAttack is a library for generating natural language adversarial examples to fool natural language processing (NLP) models. TextAttack builds attacks from four components: a search method, goal function, transformation, and a set of constraints. Researchers can use these components to easily assemble new attacks. Individual components can be isolated and compared for easier ablation studies. TextAttack currently supports attacks on models trained for text classification and entailment across a variety of datasets. Additionally, TextAttack’s modular design makes it easily extensible to new NLP tasks, models, and attack strategies. TextAttack code and tutorials are available at this https URL.

It is a Python framework for adversarial attacks, data augmentation, and model training in NLP.

### Citations

@misc{morris2020textattack,
title={TextAttack: A Framework for Adversarial Attacks in Natural Language Processing},
author={John X. Morris and Eli Lifland and Jin Yong Yoo and Yanjun Qi},
year={2020},
eprint={2005.05909},
archivePrefix={arXiv},
primaryClass={cs.CL}
}

# Blackbox Generation of Adversarial Text Sequences

### Paper Arxiv

Published @ 2018 IEEE Security and Privacy Workshops (SPW), co-located with the 39th IEEE Symposium on Security and Privacy.

• Extended version @ PDF

### Abstract

Although various techniques have been proposed to generate adversarial samples for white-box attacks on text, little attention has been paid to a black-box attack, which is a more realistic scenario. In this paper, we present a novel algorithm, DeepWordBug, to effectively generate small text perturbations in a black-box setting that forces a deep-learning classifier to misclassify a text input. We develop novel scoring strategies to find the most important words to modify such that the deep classifier makes a wrong prediction. Simple character-level transformations are applied to the highest-ranked words in order to minimize the edit distance of the perturbation. We evaluated DeepWordBug on two real-world text datasets: Enron spam emails and IMDB movie reviews. Our experimental results indicate that DeepWordBug can reduce the classification accuracy from 99% to around 40% on Enron data and from 87% to about 26% on IMDB. Also, our experimental results strongly demonstrate that the generated adversarial sequences from a deep-learning model can similarly evade other deep models.

#### We build an interactive extension to visualize DeepWordbug:

• Interactive Live Demo @ ULR

### Citations

@INPROCEEDINGS{JiDeepWordBug18,
author={J. Gao and J. Lanchantin and M. L. Soffa and Y. Qi},
booktitle={2018 IEEE Security and Privacy Workshops (SPW)},
year={2018},
pages={50-56},
keywords={learning (artificial intelligence);pattern classification;program debugging;text analysis;deep learning classifiers;character-level transformations;IMDB movie reviews;Enron spam emails;real-world text datasets;scoring strategies;text input;text perturbations;DeepWordBug;black-box attack;adversarial text sequences;black-box generation;Perturbation methods;Machine learning;Task analysis;Recurrent neural networks;Prediction algorithms;Sentiment analysis;adversarial samples;black box attack;text classification;misclassification;word embedding;deep learning},
doi={10.1109/SPW.2018.00016},
month={May},}

# EvadeML-Zoo Our Benchmarking and Visualization AE Tool is released

### Tool Github URL

We have designed and implemented EvadeML-Zoo, a benchmarking and visualization tool for research on adversarial machine learning. The goal of EvadeML-Zoo is to ease the experimental setup and help researchers evaluate and verify their results.

EvadeML-Zoo has a modular architecture and is designed to make it easy to add new datasets, pre-trained target models, attack or defense algorithms. The code is open source under the MIT license.

We have integrated three popular datasets: MNIST, CIFAR-10 and ImageNet- ILSVRC with a simple and unified interface. We offer several representative pre-trained models with state-of-the-art accuracy for each dataset including two pre-trained models for ImageNet-ILSVRC: the heavy Inception-v3 and and the lightweight MobileNet. We use Keras to access the pre-trained models because it provides a simplified interface and it is compatible with TensorFlow, which is a flexible tool for implementing attack and defense techniques.

We have integrated several existing attack algorithms as baseline for the upcoming new methods, including FGSM, BIM, JSMA, Deepfool, Universal Adversarial Perturbations, and Carlini and Wagner’s algorithms.

We have integrated our “feature squeezing” based detection framework in this toolbox. Formulating detecting adversarial examples as a binary classification task, we first construct a balanced dataset with equal number of legitimate and adversarial examples, and then split it into training and test subsets. A detection method has full access to the training set but no access to the labels of the test set. We measure the TPR and FPR on the test set as the benchmark detection results. Our Feature Squeezing functions as the detection baseline. Users can easily add more detection methods using our framework.

Besides, the tool comes with an interactive web-based visualization module adapted from our previous ADVERSARIAL-PLAYGROUND package. This module enables better understanding of the impact of attack algorithms on the resulting adversarial sample; users may specify attack algorithm parameters for a variety of attack types and generate new samples on-demand. The interface displays the resulting adversarial example as compared to the original, classification likelihoods, and the influence of a target model throughout layers of the network.

### Citations

@inproceedings{Xu0Q18,
author    = {Weilin Xu and
David Evans and
Yanjun Qi},
title     = {Feature Squeezing: Detecting Adversarial Examples in Deep Neural Networks},
booktitle = {25th Annual Network and Distributed System Security Symposium, {NDSS}
2018, San Diego, California, USA, February 18-21, 2018},
year      = {2018},
crossref  = {DBLP:conf/ndss/2018},
timestamp = {Thu, 09 Aug 2018 10:57:16 +0200},
biburl    = {https://dblp.org/rec/bib/conf/ndss/Xu0Q18},
bibsource = {dblp computer science bibliography, https://dblp.org}
}

# Adversarial-Playground Paper Appear @ VizSec17

### Abstract

Recent studies have shown that attackers can force deep learning models to misclassify so-called “adversarial examples”: maliciously generated images formed by making imperceptible modifications to pixel values. With growing interest in deep learning for security applications, it is important for security experts and users of machine learning to recognize how learning systems may be attacked. Due to the complex nature of deep learning, it is challenging to understand how deep models can be fooled by adversarial examples. Thus, we present a web-based visualization tool, Adversarial-Playground, to demonstrate the efficacy of common adversarial methods against a convolutional neural network (CNN) system. Adversarial-Playground is educational, modular and interactive. (1) It enables non-experts to compare examples visually and to understand why an adversarial example can fool a CNN-based image classifier. (2) It can help security experts explore more vulnerability of deep learning as a software module. (3) Building an interactive visualization is challenging in this domain due to the large feature space of image classification (generating adversarial examples is slow in general and visualizing images are costly). Through multiple novel design choices, our tool can provide fast and accurate responses to user requests. Empirically, we find that our client-server division strategy reduced the response time by an average of 1.5 seconds per sample. Our other innovation, a faster variant of JSMA evasion algorithm, empirically performed twice as fast as JSMA and yet maintains a comparable evasion rate. Project source code and data from our experiments available at: GitHub

### Citations

@inproceedings{norton2017adversarial,
author={Norton, Andrew P and Qi, Yanjun},
booktitle={Visualization for Cyber Security (VizSec), 2017 IEEE Symposium on},
pages={1--4},
year={2017},
organization={IEEE}
}

### Abstract

With growing interest in adversarial machine learning, it is important for machine learning practitioners and users to understand how their models may be attacked. We propose a web-based visualization tool, \textit{Adversarial-Playground}, to demonstrate the efficacy of common adversarial methods against a deep neural network (DNN) model, built on top of the TensorFlow library. Adversarial-Playground provides users an efficient and effective experience in exploring techniques generating adversarial examples, which are inputs crafted by an adversary to fool a machine learning system. To enable Adversarial-Playground to generate quick and accurate responses for users, we use two primary tactics: (1) We propose a faster variant of the state-of-the-art Jacobian saliency map approach that maintains a comparable evasion rate. (2) Our visualization does not transmit the generated adversarial images to the client, but rather only the matrix describing the sample and the vector representing classification likelihoods.

### Citations

@inproceedings{norton2017adversarial,


text:
# Quantu physics Definition and 13 Discussions

No Wikipedia entry exists for this tag
1. ### Time Independence of the Momentum Uncertainty for a Free Particle Wave

Mine is a simple question, so I shall keep development at a minimum. If a particle is moving in the absence of a potential (##V(x) = 0##), then ##\frac{\langle\hat p \rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle=0## will require that the momentum expectation value remains...
2. ### I Is Planck's law of black-body radiation related to the energy to remove an electron from a surface?

First of all, Is ##\beta## given by the Planck's law of black-body is the amount of power contained in radiation emitted by a black body? I'm not sure to fully understand the law above. Does it means that if amount of power over all the frequencies is greater than the energy needed to remove an...
3. ### I Dimensions of quantum cell automata's state space

In the paper C. S. Lent and P. D. Tougaw, "A device architecture for computing with quantum dots," in Proceedings of the IEEE, vol. 85, no. 4, pp. 541-557, April 1997, doi: 10.1109/5.573 about quantum dots, it is stated that the basis vectors in the space of quantum states for a single cell...

7. ### A Definition of finite size, radiative and weak magnetism corrections

Hi all, I would like to understand the definition of finite size correction, radiative correction and weak magnetism correction, with their impacts on the beta spectrum. I'm not a physics student, thus I would like to seek for a help about the simple explanation that can be understand by...
8. ### A Applications of weak measurement of quantum mechanics in other domains

This is a surface level question and I don't want to go into detail. Imagine an algorithm which when used with a sensor output gives the statistical moments of a variable in nature (for example mean and standard deviation of a variable). The sensor measures this once in a while (like once in a...
9. ### B Cloud Chambers - I to know what particle this is

Hi, my son is fan of the Quantum Physics and we developed a cloud chamber. I'm attaching an image of particle sequence and I will like to find some help to know witch particle is. I will appreciate any help on it. Thanks
10. ### I Why the linear combination of eigenfunctions is not a solution of the TISE

The linear combination of the eigenfunctions gives solution to the Schrodinger equation. For a system with time independent Hamiltonian the Schrodinger Equation reduces to the Time independent Schrodinger equation(TISE), so this linear combination should be a solution of the TISE. It is not...
11. ### I Physical parameters for spin 1/2 particles

I am having trouble to understand what it means by "physically relevant real parameters" and how does it help us to specify a quantum system. Let say, we have a state of k half spin electrons? My guess is about the local phase of the spin, and this would make it 2^k parameters since each...
12. ### I On the Rydberg Constant and the Emission Lines

With regard to Rutherford's atomic model, and Rydberg's discovery in general for the hydrogen distribution lines, what does Rydberg's constant physically mean? Its unit is m ^ -1, as if it were a rate, but it was not clear to me its physical meaning. And why does it grow with atomic mass...
13. ### I Quantum mechanics supports solipsism?

There is a lot of information on the Internet that quantum physics supports solipsism and that physicists believe in solipsism. I only trust this forum and the people who are here, so I want to ask you: 1. Is it true that quantum physics says solipsism is true? If this is true, then only one...

format: plain


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## anonymous 3 years ago The screen at the movies is a square with a side length of 10ft. What is the measure of the diagonal of the screen to the nearest tenth of a foot?

• This Question is Open
1. anonymous

you mean as in a triangle?

2. anonymous

$10\sqrt{2}$ = 14.1 ft is the diagonal of the screen.


text:
1 answer

# How to recursive solve a fractional Knapsack problem in Java with log(n) time. Note!! And return...

###### Question:

How to recursive solve a fractional Knapsack problem in Java with log(n) time.

Note!! And return a boolean array of each being take(T) or not take(F)

this output shoud be: take=[0,1,1]

static double[] fractionalKnap(double[] values, int[] weights, int k) {

double[] take = new double[values.length];

return take;

}

0-1 Knapsack Problem Weight 10; Value 60; Weight 20; Value 100; Weight = 30; Value = 120; Weight = (20+10): Value = (100+60); Weight = (30+10): Value = (120+60); Weight = (30+20): Value = (120+100); Weight = (30+20+10) > 50 value1 = {60, 100, 120); weigh 3(10, 20, 30); W = 50; Solution: 220

## Answers

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text:
Simultaneous Resolvability in Families of Corona Product Graphs

Research paper by Yunior RamÃrez-;Cruz, Alejandro Estrada-;Moreno; Juan A. RodrÃguez-;VelÃ¡zquez

Indexed on: 26 Aug '16Published on: 19 Aug '16Published in: Bulletin of the Malaysian Mathematical Sciences Society

Abstract

Abstract Let $$\mathcal{G}$$ be a graph family defined on a common vertex set V and let d be a distance defined on every graph $$G\in \mathcal{G}$$ . A set $$S\subset V$$ is said to be a simultaneous metric generator for $$\mathcal{G}$$ if for every $$G\in \mathcal{G}$$ and every pair of different vertices $$u,v\in V$$ there exists $$s\in S$$ such that $$d(s,u)\ne d(s,v)$$ . The simultaneous metric dimension of $$\mathcal{G}$$ is the smallest integer k such that there is a simultaneous metric generator for $$\mathcal{G}$$ of cardinality k. We study the simultaneous metric dimension of families composed by corona product graphs. Specifically, we focus on the case of two particular distances defined on every $$G\in \mathcal{G}$$ , namely the geodesic distance $$d_G$$ and the distance $$d_{G,2}:V\times V\rightarrow \mathbb {N}\cup \{0\}$$ defined as $$d_{G,2}(x,y)=\min \{d_{G}(x,y),2\}$$ .AbstractLet $$\mathcal{G}$$ be a graph family defined on a common vertex set V and let d be a distance defined on every graph $$G\in \mathcal{G}$$ . A set $$S\subset V$$ is said to be a simultaneous metric generator for $$\mathcal{G}$$ if for every $$G\in \mathcal{G}$$ and every pair of different vertices $$u,v\in V$$ there exists $$s\in S$$ such that $$d(s,u)\ne d(s,v)$$ . The simultaneous metric dimension of $$\mathcal{G}$$ is the smallest integer k such that there is a simultaneous metric generator for $$\mathcal{G}$$ of cardinality k. We study the simultaneous metric dimension of families composed by corona product graphs. Specifically, we focus on the case of two particular distances defined on every $$G\in \mathcal{G}$$ , namely the geodesic distance $$d_G$$ and the distance $$d_{G,2}:V\times V\rightarrow \mathbb {N}\cup \{0\}$$ defined as $$d_{G,2}(x,y)=\min \{d_{G}(x,y),2\}$$ . $$\mathcal{G}$$ $$\mathcal{G}$$Vd $$G\in \mathcal{G}$$ $$G\in \mathcal{G}$$ $$S\subset V$$ $$S\subset V$$ $$\mathcal{G}$$ $$\mathcal{G}$$ $$G\in \mathcal{G}$$ $$G\in \mathcal{G}$$ $$u,v\in V$$ $$u,v\in V$$ $$s\in S$$ $$s\in S$$ $$d(s,u)\ne d(s,v)$$ $$d(s,u)\ne d(s,v)$$ $$\mathcal{G}$$ $$\mathcal{G}$$k $$\mathcal{G}$$ $$\mathcal{G}$$k $$G\in \mathcal{G}$$ $$G\in \mathcal{G}$$ $$d_G$$ $$d_G$$ $$d_{G,2}:V\times V\rightarrow \mathbb {N}\cup \{0\}$$ $$d_{G,2}:V\times V\rightarrow \mathbb {N}\cup \{0\}$$ $$d_{G,2}(x,y)=\min \{d_{G}(x,y),2\}$$ $$d_{G,2}(x,y)=\min \{d_{G}(x,y),2\}$$


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# zbMATH — the first resource for mathematics

$$Q$$-functions on quasimetric spaces and fixed points for multivalued maps. (English) Zbl 1215.49016
Summary: We discuss several properties of $$Q$$-functions in the sense of Al-Homidan et al. We prove that the partial metric induced by any $$T_0$$ weighted quasipseudometric space is a $$Q$$-function and show that both the Sorgenfrey line and the Kofner plane provide significant examples of quasimetric spaces for which the associated supremum metric is a $$Q$$-function. In this context we also obtain some fixed point results for multivalued maps by using Bianchini-Grandolfi gauge functions.

##### MSC:
 49J40 Variational inequalities 47H10 Fixed-point theorems
Full Text:
##### References:
 [1] Kada, O; Suzuki, T; Takahashi, W, Nonconvex minimization theorems and fixed point theorems in complete metric spaces, Mathematica Japonica, 44, 381-391, (1996) · Zbl 0897.54029 [2] Caristi, J; Kirk, WA, Geometric fixed point theory and inwardness conditions, No. 490, 74-83, (1975), Berlin, Germany · Zbl 0315.54052 [3] Ekeland, I, Nonconvex minimization problems, Bulletin of the American Mathematical Society, 1, 443-474, (1979) · Zbl 0441.49011 [4] Takahashi, W; Théra, MA (ed.); Baillon, JB (ed.), Existence theorems generalizing fixed point theorems for multivalued mappings, No. 252, 397-406, (1991), Harlow, UK · Zbl 0760.47029 [5] Al-Homidan, S; Ansari, QH; Yao, J-C, Some generalizations of Ekeland-type variational principle with applications to equilibrium problems and fixed point theory, Nonlinear Analysis: Theory, Methods & Applications, 69, 126-139, (2008) · Zbl 1142.49005 [6] Hussain, N; Shah, MH; Kutbi, MA, Coupled coincidence point theorems for nonlinear contractions in partially ordered quasi-metric spaces with a [inlineequation not available: see fulltext.]-function, No. 2011, 21, (2011) [7] Latif, A; Al-Mezel, SA, Fixed point results in quasimetric spaces, No. 2011, 8, (2011) · Zbl 1207.54061 [8] Alegre, C, Continuous operators on asymmetric normed spaces, Acta Mathematica Hungarica, 122, 357-372, (2009) · Zbl 1199.54165 [9] Ali-Akbari, M; Honari, B; Pourmahdian, M; Rezaii, MM, The space of formal balls and models of quasi-metric spaces, Mathematical Structures in Computer Science, 19, 337-355, (2009) · Zbl 1168.54012 [10] Cobzaş, S, Compact and precompact sets in asymmetric locally convex spaces, Topology and its Applications, 156, 1620-1629, (2009) · Zbl 1185.46001 [11] García-Raffi, LM; Romaguera, S; Sánchez-Pérez, EA, The goldstine theorem for asymmetric normed linear spaces, Topology and its Applications, 156, 2284-2291, (2009) · Zbl 1185.54028 [12] García-Raffi, LM; Romaguera, S; Schellekens, MP, Applications of the complexity space to the general probabilistic divide and conquer algorithms, Journal of Mathematical Analysis and Applications, 348, 346-355, (2008) · Zbl 1149.68080 [13] Heckmann, R, Approximation of metric spaces by partial metric spaces, Applied Categorical Structures, 7, 71-83, (1999) · Zbl 0993.54029 [14] Matthews SG: Partial metric topology. In Proceedings of the 14th Summer Conference on General Topology and Its Applications, 1994, Annals of the New York Academy of Sciences. Volume 728. The New York Academy of Sciences, New York, NY, USA; 183-197. · Zbl 0911.54025 [15] Romaguera, S, On computational models for the hyperspace, 277-294, (2009), New York, NY, USA [16] Romaguera, S; Schellekens, M, Partial metric monoids and semivaluation spaces, Topology and its Applications, 153, 948-962, (2005) · Zbl 1084.22002 [17] Romaguera, S; Tirado, P, The complexity probabilistic quasi-metric space, Journal of Mathematical Analysis and Applications, 376, 732-740, (2011) · Zbl 1227.54035 [18] Romaguera, S; Valero, O, A quantitative computational model for complete partial metric spaces via formal balls, Mathematical Structures in Computer Science, 19, 541-563, (2009) · Zbl 1172.06003 [19] Romaguera, S; Valero, O, Domain theoretic characterisations of quasi-metric completeness in terms of formal balls, Mathematical Structures in Computer Science, 20, 453-472, (2010) · Zbl 1193.54016 [20] Schellekens, M, The smyth completion: a common foundation for denotational semantics and complexity analysis, No. 1, 535-556, (1995), Amsterdam, The Netherlands [21] Schellekens, MP, A characterization of partial metrizability: domains are quantifiable, Theoretical Computer Science, 305, 409-432, (2003) · Zbl 1043.54011 [22] Waszkiewicz, P, Quantitative continuous domains, Applied Categorical Structures, 11, 41-67, (2003) · Zbl 1030.06005 [23] Waszkiewicz, P, Partial metrisability of continuous posets, Mathematical Structures in Computer Science, 16, 359-372, (2006) · Zbl 1103.06004 [24] Proinov, PD, A generalization of the Banach contraction principle with high order of convergence of successive approximations, Nonlinear Analysis: Theory, Methods & Applications, 67, 2361-2369, (2007) · Zbl 1130.54021 [25] Fletcher P, Lindgren WF: Quasi-Uniform Spaces, Lecture Notes in Pure and Applied Mathematics. Volume 77. Marcel Dekker, New York, NY, USA; 1982:viii+216. [26] Künzi, H-PA, Nonsymmetric distances and their associated topologies: about the origins of basic ideas in the area of asymmetric topology, No. 3, 853-968, (2001), Dordrecht, The Netherlands · Zbl 1002.54002 [27] Reilly, IL; Subrahmanyam, PV; Vamanamurthy, MK, Cauchy sequences in quasipseudometric spaces, Monatshefte für Mathematik, 93, 127-140, (1982) · Zbl 0472.54018 [28] Rakotch, E, A note on contractive mappings, Proceedings of the American Mathematical Society, 13, 459-465, (1962) · Zbl 0105.35202 [29] Bianchini, RM; Grandolfi, M, Trasformazioni di tipo contrattivo generalizzato in uno spazio metrico, Atti della Accademia Nazionale dei Lincei, 45, 212-216, (1968) · Zbl 0205.27202 [30] Proinov, PD, New general convergence theory for iterative processes and its applications to Newton-Kantorovich type theorems, Journal of Complexity, 26, 3-42, (2010) · Zbl 1185.65095 [31] Chang, TH, Common fixed point theorems for multivalued mappings, Mathematica Japonica, 41, 311-320, (1995) · Zbl 0840.47041
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Math Help - Langrange Multiplier Equations

1. ## Langrange Multiplier Equations

I was given the two equations f=x^2y^2z^2 and x^2+y^2+z^2=1
(I'll use 'u' to be equal to lambda)
and then set up my equations to be x^2y^2z^2-ux^2-uy^2-uz^2+u

taking the partial derivatives I got

fx=2x*y^2*z^2-2ux+1=0
fy=2x^2*y*z^2-2uy+1=0
fz=2x^2*y^2*z-2uz+1=0
fu=-x^2-y^2-z^2+1=0

I cant seem to find a way to solve this system of equations, can anyone? Or tell me what I did wrong?

Thanks!

2. Originally Posted by Reskal
I was given the two equations f=x^2y^2z^2 and x^2+y^2+z^2=1
(I'll use 'u' to be equal to lambda)
and then set up my equations to be x^2y^2z^2-ux^2-uy^2-uz^2+u

taking the partial derivatives I got

fx=2x*y^2*z^2-2ux+1=0
fy=2x^2*y*z^2-2uy+1=0
fz=2x^2*y^2*z-2uz+1=0
fu=-x^2-y^2-z^2+1=0

I cant seem to find a way to solve this system of equations, can anyone? Or tell me what I did wrong?

Thanks!
First off, I believe you have a mistake in your derivatives.

$
\frac{\partial}{\partial x} \left(x^2 y^2 z^2 - \lambda(x^2+y^2+z^2-1) \right) = 2x y^2 z^2 - 2 \lambda x = 2x\left(y^2 z^2 - \lambda\right)
$

Similarly for the other two. Notice that I factored. This gives rise to

$
2x\left(y^2 z^2 - \lambda\right) = 0
$

$2y\left(x^2 z^2 - \lambda\right) = 0$

$
2z\left(x^2 y^2 - \lambda\right) = 0
$

Now look at cases.

Let me ask, why use Lagrange multipliers when one of the variables can be easily eliminated?

3. Originally Posted by Danny
First off, I believe you have a mistake in your derivatives.

$
\frac{\partial}{\partial x} \left(x^2 y^2 z^2 - \lambda(x^2+y^2+z^2-1) \right) = 2x y^2 z^2 - 2 \lambda x = 2x\left(y^2 z^2 - \lambda\right)
$

Similarly for the other two. Notice that I factored. This gives rise to

$
2x\left(y^2 z^2 - \lambda\right) = 0
$

$2y\left(x^2 z^2 - \lambda\right) = 0$

$
2z\left(x^2 y^2 - \lambda\right) = 0
$

Now look at cases.

Let me ask, why use Lagrange multipliers when one of the variables can be easily eliminated?
The directions said to use Lagrange multipliers.

Useing the equations that you wrote does this imply that x,y,z are all equal to 0?

Thanks Danny for all your help.

4. Originally Posted by Reskal
The directions said to use Lagrange multipliers.

Useing the equations that you wrote does this imply that x,y,z are all equal to 0?

Thanks Danny for all your help.
The cases:

$x = 0$, any $y$ and $z$ satisfying $y^2+z^2=1$. Similarly for $y = 0$ and $z = 0$. The remaining case is when $x^2 y^2 = \lambda, \;\;x^2 z^2 = \lambda,\;\; y^2 z^2 = \lambda$ or

$
y = \pm x,\;\; z = \pm x,\;\; z = \pm y
$
all 8 cases. Sub these into $x^2+y^2+z^2 = 1$


text:
Physics Forums (http://www.physicsforums.com/index.php)
-   Quantum Physics (http://www.physicsforums.com/forumdisplay.php?f=62)
-   -   Bosons, Fermions and ? (http://www.physicsforums.com/showthread.php?t=152539)

 Kevin_spencer2 Jan22-07 09:08 AM

Bosons, Fermions and ??

I have heard in wikipedia (a joke?? ) that appart from Bosons and Fermions (types of particles) there were another kind of 'Probabilistic distribution' ???? i don't know how it was called but if we have the number of particles.

$$<n(T)>=\frac{1}{exp(\hbar \omega )-a}$$

a=1 for Bosons and -1 for Fermions, if other kind of particles exist, what would be their distribution?, thanx.

 StatMechGuy Jan22-07 10:53 AM

You might want to try looking into "anyons". Frank Wilczek coined the term for two dimensional field theories since the exclusion principle takes a different form in the reduced dimensions.

 Demystifier Jan23-07 05:20 AM

Try also with "generalized statistics".

 ZapperZ Jan23-07 05:42 AM

Quote:
 Quote by StatMechGuy (Post 1220343) You might want to try looking into "anyons". Frank Wilczek coined the term for two dimensional field theories since the exclusion principle takes a different form in the reduced dimensions.
In addition, note that as with the fractional charge phenomenon, anyons are also an "emergent" excitation that came out of the renormalized many-body effect. In Wilczek's paper, several references were made to Schrieffer's work (this is the "S" in BCS theory of superconductivity). Condensed matter systems are the ones that can clearly exhibit these fractional phenomena.

Zz.

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# ISO 148-1

## Metallic materials Charpy pendulum impact test Part 1: Test method

inactive
 Organization: ISO Publication Date: 15 February 2006 Status: inactive Page Count: 26 ICS Code (Mechanical testing of metals): 77.040.10

### Document History

October 15, 2016
Metallic materials - Charpy pendulum impact test - Part 1: Test method
This part of ISO 148 specifies the Charpy (V-notch and U-notch) pendulum impact test method for determining the energy absorbed in an impact test of metallic materials. This part of ISO 148 does not...
November 15, 2009
Metallic materials — Charpy pendulum impact test — Part 1: Test method
This part of ISO 148 specifies the Charpy pendulum impact (V-notch and U-notch) test method for determining the energy absorbed in an impact test of metallic materials. This part of ISO 148 does not...
ISO 148-1
February 15, 2006
Metallic materials Charpy pendulum impact test Part 1: Test method
A description is not available for this item.


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# Garland Words

A garland word is a word which can be strung together like a garland, because it ends with the same letters it starts with. These groups of letters can even overlap!

For example, underground is a garland word of order 3, because it starts and end with the same 3 characters, und. This means it could be strung together like undergroundergrounderground....

alfalfa is a garland word, too! It's of order 4. It starts and ends with alfa. It can be strung together like so: alfalfalfalfa.

A process which I call garlandifying is where once you determine the order n of a garland word, you take the original word and add the segment required to have it loop as a garland n times. So, since onion is an order 2 garland word, you would take onion, chop off the first 2 letters to get ion and add that to the end 2 times to get onionionion.

# Objective

Make a program or function which takes input from standard input or a function argument and prints out or returns the word, garlandified.

All words will be lowercase, and the highest possible order for a word is length(word) - 1.

# Example I/O

"onion"       --> "onionionion"
"jackhammer"  --> "jackhammer"
""            --> ""
"zvioz"       --> "zviozvioz"
"alfalfa"     --> "alfalfalfalfalfalfa"
"aaaa"        --> "aaaaaaa"

This is , so least number of bytes wins.

• Any N-letter word starts with the same N letters that it ends with. What is the maximum order that should be considered? – feersum Jul 15 '15 at 14:20
• @feersum The maximum order is the word's length - 1. Added that to the main post. – Kade Jul 15 '15 at 14:22
• Do I have to print out just the garland? or can I perhaps print it and an exception? – DeadChex Jul 15 '15 at 15:33
• @DeadChex There should be no exceptions. – Kade Jul 15 '15 at 15:34
• @LuisMendo It should work for arbitrarily long words. – Kade Jul 16 '15 at 11:34

# Pyth, 19 18 bytes

+z*>Kf!xz>zT1zl>zK

Try it online: Demonstration or Test harness

### Explanations:

+z*>Kf!xz>zT1zl>zK   implicit: z = input string
f      1        find the first number T >= 1, which satisfies:
>zT            all but the first T chars of z
xz               index of ^ in z
!                 == 0
K                store in K
the order is length(z) - K
>K        z       the last K chars
*                  repeated
l>zK   len(all but the last K chars) times
+z                   insert z at the beginning

# Python, 60 bytes

f=lambda s,i=1:s.find(s[i:])and f(s,i+1)or(len(s)-i)*s[:i]+s

Was hoping for better, but oh well. s.find works neatly here in place of not s.startswith.

# Retina, 58 bytes

.+
$0#$0
(.*)(.+)#.*\10#$1#$2-
+\w#(\w*)-
#$1-$1
#.*-
<empty line>

Each line should go to its own file but you can run the code as one file with the -s flag.

The four substitution pairs do the followings:

• Duplicate word so we can search for overlaps too.
• Append the word split up at order number of characters.
• Append the last part order times.
• Keep the original word and the lastly appended part and drop everything else.

The string states for the example onion:

onion
onion#onion
onion#onion#on#ion-
onion#onion##ion-ionion
onionionion

g s=[b>>a|(a,b)<-map(splitAts)[1..],and$zipWith(==)s b]!!0++s  Tests: λ: g "onion" == "onionionion" True λ: g "jackhammer" == "jackhammer" True λ: g "abracadabra" == "abracadabracadabracadabracadabracadabra" True λ: g "" == "" True λ: g "zvioz" == "zviozvioz" True λ: g "alfalfa" == "alfalfalfalfalfalfa" True λ: g "aaaa" == "aaaaaaa" True  # Java, 160 157 bytes static void g(String s){int i=s.length(),o;for(String p=s;i-->0;)if(s.endsWith(s.substring(0,i))){for(o=i;o-->0;)p+=s.substring(i);System.out.print(p);i=0;}}  Input/Output:  g("abracadabra"); --> "abracadabracadabracadabracadabracadabra"  Spaced and tabbed for readability: static void g(String s){ int i=s.length(),o; for(String p=s;i-->0;) if(s.endsWith(s.substring(0,i))){ for(o=i;o-->0;) p+=s.substring(i); System.out.print(p); i=0; } }  Suggestions welcome. • As a note to my self, the String ops can be moved into the for loop to save a byte or two on semicolons – DeadChex Jul 16 '15 at 5:16 • why not do i=0; ? – overactor Jul 16 '15 at 6:14 • @overactor where? The reason I use the length is because I want the full String and then I want to move towards none of it, with substring I don't think I can avoid using it in this method and will take the byte penalty for it – DeadChex Jul 16 '15 at 10:04 • I meant to break out of the outer for loop. – overactor Jul 16 '15 at 10:06 # Sed: 87 84 characters (83 characters code + 1 character command line option.) h s/(.*)./& \1/ T s/(.+) \1.*/ \1 \1/ t g q : s/^([^ ]+)(.*)[^ ]$/\1 \1\2/
t
s/ //g

Sample run:

bash-4.3$sed -r 'h;s/(.*)./& \1/;T;s/(.+) \1.*/ \1 \1/;t;g;q;:;s/^([^ ]+)(.*)[^ ]$/\1 \1\2/;t;s/ //g' <<< 'underground'
undergroundergroundergrounderground

• Automatic upvote of sed answer ;-). Follow this tip to drop 2 chars from your label definition and branch – Digital Trauma Jul 15 '15 at 17:50
• Tried, but I am afraid that advice is only for cases when you not have label-less jumps to the end of code. [A bit later…] Ok, thinking again, why I tried to process multiple input lines at once? – manatwork Jul 15 '15 at 17:57

# CJam, 24 23 bytes

q_:Q,{~)Q>Q\#!},W>~_Q>*

q_:Q                       e# Read the input, take a copy and store it in Q too
,{        },           e# Take the length of the input and filter [0 .. len - 1] array
~)                   e# Same as number * -1
Q>                 e# Take last number characters. Call this string S
Q\#!             e# See if Q starts with S. After the filter, we will only have
e# those numbers from [0 .. len - 1] array which are valid orders
W>~        e# Take the last order number, if exists.
_Q>*    e# Garlandify the input order times.

Just to start it with something..

Try it online here

## Matlab: 9789 82 bytes

Function that uses a regular expression with lookbehind and a capture group:

function t=f(s)
n=sum(regexp(s,'(.*$)(?<=^\1.+)'))-1;t=[s(repmat(1:n,1,end-n)) s];  That sum is needed to handle the empty-string input (convert [] into 0). Examples: > f('onion'), f('jackhammer'), f('abracadabra'), f(''), f('zvioz'), f('alfalfa'), f('aaaa') ans = onionionion ans = jackhammer ans = abracadabracadabracadabracadabracadabra ans = Empty string: 1-by-0 ans = zviozvioz ans = alfalfalfalfalfalfa ans = aaaaaaa  # REGXY, 53 49 bytes Uses REGXY, a regex substitution based language //$'#/
/.(.+)#\1\K/#/
a/(#).(.*#)|#.*/$'$1$2/ //a  Overview: A number of regular expressions are applied. An example run would look like: onion (input) onion#onion (line 1 regex) onion#on#ion (line 2 regex - find the repeated section and separate with #) onionion#n#ion (line 3 regex - the length of the middle token is the garland order, remove a character and append the third token onto the original string on the left) onionionion##ion (line 4 regex is a pointer to line 3 - repeat the previous again) onionionion##ion (line 4 regex is a pointer to line 3 - strip everything after and including the #)  Detailed explanation The following is a line by line breakdown of the regexes: //$'#/

This is a regex substitution which matches the first empty string (i.e. the start of the string) and replaces it with everything to the right of the match ($') followed by a hash. For example, it will turn onion into onion#onion. /.(.+)#\1\K/#/  This line finds the section which overlaps by looking for a group of characters immediate preceding the # ((.+)) which are the same on the other side of the # (\1). The \K just means 'forget that I matched anything', meaning it will not actually be replaced in the substitution. This effectively, this means we just add a # to the position after the overlap has been found, turning onion#onion into onion#on#ion. a/(#).(.*#)|#.*/$'$1$2/

The initial 'a' is just a label for the regex. After this, we find the first # followed by a single character (.) and capture everything after this until the next # (.*#). We replace this with everything to the right of the match, i.e. the last token ($'), followed by a # ($1), followed by the second token less a character (we treat this as a counter, decreasing it each iteration). In the case of onion#on#ion, the two tokens we backreference on are shown in brackets, and the section the entire regex matches is between the pipes: onion|(#)o(n#)|ion. We then replace the bits we match (between the pipes) with $' (everything to the right of the match, i.e. 'ion'), then$1 (the #), then $2 (n#), meaning we end up with onion|(ion)(#)(n#)|ion (brackets show the three tokens in the replacement string). If the regex fails to match in the first alternation (everything before the pipe), we must have decreased our counter to zero, meaning there are no characters inside the second token. Instead, we look at the second part of the pattern, #.*. This simply replaces everything after the first # with $'$1$2. Since there are no backreferences created by this alternation, and there is nothing to the right of the match (.* matches until the end of the string), we terminate the match and return the result.

//a

This is just a pointer to the previous line, ensuring we continue to execute the regex substitution until it fails to match any more.

# jq 1.5: 91 characters

(87 characters code + 4 characters command line option.)

.+. as$t|[range(1;length)|select($t[:.]==$t[-.:])]|(max//0)as$i|[range($i)|$t[$i:]]|add  Sample run: bash-4.3$ jq -R -r -f judy.jq <<< 'underground'
undergroundergroundergrounderground

# rs, 51 48 bytes

(.+)/\1 \1
(.+)(.+) .+\1$/\1(\2)^^((^^\1_)) .*/  TAKE THAT, RETINA AND SED!!!!! ;) Cut off 3 bytes thanks to @randomra. Live demo and test cases. Note that the jackhammer test case is not there. There's a bug in the handling of spaces in the web interface that causes it to print incorrect output. The offline version of rs handles it correctly. 51-byte version: (.+)/\1 \1 ^(.+)(.+) (.+)\1$/\1(\2)^^((^^\1_))
.*/

Live demo and test cases for original.

• @randomra Updated. Thanks! – kirbyfan64sos Jul 16 '15 at 22:14

## JavaScript (ES6), 95 bytes

f=s=>{for(e=i=s.length;i&&e;)s+=s.slice(--i).repeat(!(e=!s.endsWith(s.slice(0,i)))*i);return s}

### Demo

Firefox only for now:

f = s => {
for (e = i = s.length; i && e;) s += s.slice(--i).repeat(!(e = !s.endsWith(s.slice(0, i))) * i);
return s
}

console.log = x => X.innerHTML += x + '\n';

console.log(f('onion'));
console.log(f('jackhammer'));
console.log(f(''));
console.log(f('zvioz'));
console.log(f('alfalfa'));
console.log(f('aaaa'));

## JavaScript (ES6), 82 bytes

g=(s,i=t=s.length)=>s.endsWith(c=s.slice(0,--i))?c+s.slice(i-t).repeat(i+1):g(s,i)

[Deleted my original answer, because I've now learned ES6 and was interested in finding a recursive solution to this challenge]

Example

g=(s,i=t=s.length)=>s.endsWith(c=s.slice(0,--i))?c+s.slice(i-t).repeat(i+1):g(s,i)

console.log(g('onion'));
console.log(g('jackhammer'));
console.log(g(''));
console.log(g('zvioz'));
console.log(g('alfalfa'));
console.log(g('aaaa'));

## CoffeeScript + ES6, 77 bytes

Same approach as my JavaScript answer.

f=(s,e=i=s.length)->s+=s[i..].repeat !(e=!s.endsWith s[...i])*i while--i&&e;s

## C

#include <stdio.h>
#include <string.h>

int main(int argc, char **argv) {
char *str   = NULL;
char *p     = NULL;
int len     = 0 ;
int i       = 0;
int j       = 0;
int k       = 0;
int loop    = 0;

if (argc == 1 )
return 0;

str = argv[1];
len = strlen(str);

if (len %2) {
loop = len/2 + 1;
}
else {
loop = len/2;
}

p = &str[len/2];
for (i = 0; i < loop ; i++) {
if (str[k] == *(p++)) {
k++;
}
else
k = 0;
}

printf("k = %d\n", k);
printf("%s", str);
p = &str[k];
for (j =0; j < k ; j++) {
printf("%s", p);
}
return 0;
}

# Golfed: 195 bytes - GCC

main(int c,char**a){
char *s=a[1],*p;int i=0,j=0,k=0,z,l=strlen(a[1]);
z=l%2?-~(l/2):l/2;p=&s[l/2];
for(;i<z;i++)k=s[k]==*(p++)?-~k:0;
printf("k=%d\n",k);puts(s);p= &s[k];
for(;j<k;j++)puts(p);}

• Welcome to Programming Puzzles and Code Golf! This question is code golf, so I suggest that you "golf" your code by removing unnecessary whitespace, etc., and then include the byte count of your code in the title of your post along with the language. – lirtosiast Jul 15 '15 at 23:08
• Got it. Thank you for the direction. I will keep it in mind next time. – Alam Jul 16 '15 at 3:51
• It's not too late to "golf" it. If you click on the "edit" button underneath your answer, you could still remove the unnecessary whitespace and add a byte count. – DJMcMayhem Jul 17 '15 at 19:47
• Isn't int implicit in (sufficiently old versions of) C? – Angew Jul 18 '15 at 11:56

# Groovy 7557 55 bytes

f={w->x=w;w.find{x-=it;!w.indexOf(x)};w+(w-x)*x.size()}

Amazing how coming back to something the day after can help

### Ungolfed:

f = {w ->

//Set x equal to w
x=w

//Loop through the characters of w until we return true
w.find {

//set x equal to x minus the first instance of the current character, i.e.     the word minus the first character
x-=it

//Returns the index of the first occurance of the string of chars x, when this is 0 (false) we want to return true, so negate it
!w.indexOf(x)
}

//When we've escaped the loop, if we've found a match return the word plus the word minus the match multiplied by the lengh of the match.
w+(w-x)*x.size()
}

In case somebody needs the code in JS to test it. Note: I traversed the string from end to increase efficiency:

"use strict";

var garlandify = function(inputString){
var stringLength = inputString.length;
var savedString = inputString;

for( var i=1; i<stringLength; i++ ){
var endIndex = Math.abs(i) * -1;
if( inputString.startsWith( inputString.substr(endIndex) ) ){
for( var j=1; j<=i; j++){
savedString += inputString.substr(i, stringLength );
}
console.log(savedString);
}
}
};

garlandify("onion");
`
• Welcome to the Programming Puzzles & Code Golf stack exchange! You don't need to worry about efficiency at all for code-golf, just the length of your program. So the slow, inefficient version may well be the best here (it can make a refreshing change from "real work"!). So remove the unnecessary whitespace, and use single-letter variable names - then read Tips for golfing in JavaScript. I think there's much you can do to golf this - but we do like to see the ungolfed, commented version if your algorithm is clever. Have fun! – Toby Speight Jul 16 '15 at 18:06


text:
## anonymous one year ago Open Solution Problem;Soft snow can be compressed by pressure about 3.0 ×10^3 Pa. What is the smallest area that a pair of snowshoes must have if they will enable a 80 kg person to walk over the snow without sinking in? Consider g = 9.8 m/s2.

1. anonymous

The first person who gets it earns a medal and fan

2. anonymous

First you have to consider the ?Newtonian force acting on the ground. This is caculated using Mg

3. anonymous

And use the equation Pressure=F/A(surface area in m^2) with inequality equation

4. DDCamp

First find the force the person will exert on the snow, using Newton's 2nd law: $F = ma \\ F = 88kg * 9.8 \frac{m}{s^2}$

5. anonymous

Correct! Newton's second law or simply mass*gravityt

6. anonymous

In fact you are calculating the normal force using that equation above

7. DDCamp

Use the equation for pressure:$P = F/A \\ 3000 Pa = 862.4 N / A \\ A = 862.4N / 3000 Pa \\ A = 0.287 m^2$

8. anonymous

Well done.


text:
# Counterexample for LTL - CTL equivalence [closed]

I have to find an example of a model where the LTL-formula $F G p \wedge F q$ is valid and the CTL-formula $EF AG p \wedge AF q$ is not valid. I found this example, but I'm not completely sure whether it's correct:

• Your example is incorrect, since the LTL formula doesn't hold: in the computation $s_0,s_1,s_2,s_3^\omega$ the formula $FGp$ doesn't hold. – Shaull Jan 27 '16 at 12:48
• Hmm yes, you're right. Do you know an example in which it does hold and the CTL-formula doesn't? Or would it be easier to find an example in which the CTL-formula does hold and the LTL-formula doesn't hold? – Pieter Verschaffelt Jan 27 '16 at 12:50
• Your question already includes a complete answer to the original problem but no question about this answer. Thus, only "yes/no" answers may remain, helping neither you nor future visitors. Please read related meta discussions here and here and adjust your question accordingly, for example by formulating a specific question about a single element of your answer you are uncertain about. Can you figure out what it is that you don't understand that's stopping you from determining whether your answer is correct? – David Richerby Jan 28 '16 at 1:06
• The actual problem is the fact that I don't really understand the difference between the A (all paths) in CTL and a normal LTL-formula which also holds for all paths? – Pieter Verschaffelt Jan 28 '16 at 7:27

## 1 Answer

Consider the following model: you have 3 states, $s_0,s_1,s_2$ with the transitions: $s_0\to s_0$, $s_0\to s_1$, $s_1\to s_2$ and $s_2\to s_2$ and the labels are $L(s_0)=\{p,q\}$, $L(s_1)=\emptyset$ and $L(s_2)=p$.

Then, every computation starts with $q$, so $Fq$ holds, and every infinite computation eventually gets stuck in $s_2$, or it is $s_0^\omega$, and both satisfy $FGp$, so the LTL formula holds.

However, it never holds that $AFq$, so the CTL formula does not hold.

• Thank you very much! :) I see now. Is there some trick that helps you find such differences between LTL and CTL or is it just plane reasoning? – Pieter Verschaffelt Jan 27 '16 at 12:58
• As with most problems, there is no "trick", just reasoning and some experience. Conceptually, it is possible to find such an example algorithmically: let $\phi$ be an LTL formula and $\psi$ be a CTL formula, then you can construct the CTL* formula $A\phi\wedge \neg \psi$, and then use CTL* satisfiabilityto find a model for the formula. But the complexity of this algorithm is terrible - 2EXPTIME, so it's not suited for solving exercises. – Shaull Jan 27 '16 at 13:33
• Thank you! I will practice more then :) Thank you for your help! – Pieter Verschaffelt Jan 27 '16 at 13:50
• I have one more little question: I don't quite understand why $AF q$ never holds? The computation starts with q, so at that point $AF q$ is satisfied? Or not? – Pieter Verschaffelt Jan 27 '16 at 14:08
• Hmm. I read your CTL formula as $EF(AGp\wedge AFq)$, in which case my example is fine. But if it's intended as $(EFAGp)\wedge AFq$, then my answer is incorrect, and I think that there is no (non-empty) model that will have the desired property - obviously $AFq$ has to hold, by the LTL formula, and if $EFAGp$ doesn't hold, then along every path, every state has a path from it in which $AGp$ doesn't hold, which means we can construct a path that does not satisfy $FGp$. – Shaull Jan 27 '16 at 16:10


text:
# What do you call a greedy algorithm that solves a combinatorial problem by optimizing the best k>1 choices altogether?

Suppose you have a problem which goal is to find the permutation of some set $$S$$ given in input that minimizes an objective function $$f$$ (for example the Traveling Salesman problem).

A trivial algorithm $$E(S)$$ that find the exact solution enumerates all the permutations and outputs the one that minimizes $$f$$. Its time complexity is $$O(n!)$$ where $$n$$ is the size of $$S$$.

A trivial greedy algorithm $$G(S)$$ that finds an approximation of the solution is:

out[0] = select a good starting item from S according to some heuristic h_1.
S = S - {out[0]}
for i=1 to n-1 do:
out[i] = select the next best element using some heuristic h_2
S = S - {out[i]}
return out

Where $$h_1$$ and $$h_2$$ are two heuristics. Its complexity in time is $$O(n^2)$$ assuming that $$h_2$$ runs in constant time.

Sometimes I mix the two techniques (enumeration and greedy) by selecting at each step the best $$k$$ items (instead of the best one) and enumerating all their permutations to find the one that locally minimizes $$f$$. Then I choose the best $$k$$ items among the remaining $$n-k$$ items and so on.

Here is the pseudocode (assuming $$n$$ is a multiple of $$k$$):

for i in 0 to n/k do:
X = select the best k items of S according to some heuristic h
S = S - X
out[i*k ... (i+1)*k-1] = E(X)
return out

Where $$E(X)$$ is algorithm that find the exact solution applied on a subset $$X \subset S$$ rather than on the whole $$S$$. This last algorithm finds an approximate solution and has a time complexity of $$O(\frac{n}{k}(n \log k + k! ))$$ assuming that $$h$$ can be computed in constant time. This complexity can be comparable to $$O(n^2)$$ if $$k$$ is small although according to my experience the performances can be way better than the greedy approach.

I don't think I invented this kind of optimization technique: do you know its name? Can you please include some theoretical references?

I know for sure it is not beam search, because beam search never mixes the best $$k$$ solutions found at each step.

Thank you.

• I don't know of a name for this kind of heuristic. A slightly different type of heuristic for finding an optimal permutation ( = TSP tour), in which an initial permutation is formed somehow and then all subsets of $k$ elements are optimally permuted (keeping the remaining $n-k$ items fixed), repeating until no subset of $k$ elements yields an improvement, is simply called "$k$-opt". – j_random_hacker Mar 24 at 18:44


text:
# Adjusting Monte Carlo estimates to generate correct even moments - improving on antitthetic draws

If I want to generate a matrix 10,000 (row) samples of 3 uniform (uncorrelated) variables it is trivial to use antithetic draws to ensure the odd moments such as the mean equal their "true" value. Julia/Matlab code below:

w1=rand(5000,3)
w2 = 1 - w1
w = [w1;w2]

Each column of w will then have mean of 0.5. Can someone please tell me a method of ensuring the covariances equal their "true" values (covariance=0 and variance = 0.0833)? From looking at the MASS package in R it has a function "mvrnorm" that when used with argument "empirical=TRUE" returns multivariate normal samples whose mean and covariance match exactly their true values. I've failed on the web to find out how this is achieved. Any help much appreciated for either multivariate uniform or normal.

• When in doubt, read the source code. In R, it's freely available online (on CRAN), or by installing MASS and typing MASS::mvrnorm – shadowtalker Feb 25 '15 at 14:19
• You could read to source code for MASS::mvrnorm for more details, preferrably from the tarball as it will contain any comments in the source: cran.r-project.org/src/contrib/3.1.3/Recommended Plus look at the help for ?MASS::mvrnorm as it may contain a pointer to literature on the implementation. – Gavin Simpson Feb 25 '15 at 14:22
• Thanks for your help Gavin and ssdecontrol. Have looked at the code and can replicate (if not fully understand - will look into). Would tick your comments but not sure how to. Thanks again. Steve – Steve Kay Feb 25 '15 at 18:19
• Does this answer discuss what you need? – Glen_b Feb 26 '15 at 23:08

x = randn(100,3);


text:
ORCA
Online Research @ Cardiff

# Probing the formation of intermediate- to high-mass stars in protoclusters - II. Comparison between millimeter interferometric observations of NGC 2264-C and SPH simulations of a collapsing clump

 Peretto, Nicolas, Hennebelle, P. and Andre, P. 2007. Probing the formation of intermediate- to high-mass stars in protoclusters - II. Comparison between millimeter interferometric observations of NGC 2264-C and SPH simulations of a collapsing clump. Astronomy and Astrophysics 464 (3) , pp. 983-994. 10.1051/0004-6361:20065653

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## Abstract

Aims.The earliest phases of massive star formation in clusters are still poorly understood. Here, we test the hypothesis for high-mass star formation proposed in our earlier paper (Peretto et al. 2006) stating that a massive, ultra-dense core may be currently forming at the center of the collapsing NGC 2264-C protocluster via the gravitational coalescence of several intermediate-mass Class 0 objects. Methods.In order to confirm the physical validity of this hypothesis, we carried out IRAM Plateau de Bure interferometer observations of NGC 2264-C and performed SPH numerical simulations of the collapse of a Jeans-unstable, prolate dense clump. A detailed comparison between these hydrodynamic simulations and both our earlier IRAM 30 m observations and the new interferometer observations is presented. Results.Our Plateau de Bure observations provide evidence for disk emission in three of the six Class 0-like objects identified earlier with the 30 m in the NGC 2264-C clump. Furthermore, they reveal the presence of a new compact source (C-MM13) located only ~$10\,000$ AU away, but separated by ~1.1 km s-1 in (projected) velocity, from the most massive Class 0 object (C-MM3) lying at the very center of NGC 2264-C. Detailed comparison with our numerical SPH simulations supports the view that NGC 2264-C is an elongated cluster-forming clump in the process of collapsing and fragmenting along its long axis, leading to a strong dynamical interaction and possible protostar merger in the central region of the clump. The marked velocity difference observed between the two central objects C-MM3 and C-MM13, which can be reproduced in the simulations, is interpreted as an observational signature of this dynamical interaction. The present study also sets several quantitative constraints on the initial conditions of large-scale collapse in NGC 2264-C. Our hydrodynamic simulations indicate that the observed velocity pattern characterizes an early phase of protocluster collapse which survives for an only short period of time (i.e., $\le$ $1\times 10^5$ yr). To provide a good match to the observations the simulations require an initial ratio of turbulent to gravitational energy of only ~$5\%$, which strongly suggests that the NGC 2264-C clump is structured primarily by gravity rather than turbulence. The required "cold" initial conditions may result from rapid compression by an external trigger. Conclusions.We speculate that NGC 2264-C is not an isolated case but may point to key features of the initial phases of high-mass star formation in protoclusters.

Item Type: Article Publication Published Physics and Astronomy Q Science > QB Astronomy stars: formation -- stars: circumstellar matter -- stars: kinematics -- ISM: clouds -- ISM: kinematics and dynamics -- hydrodynamics Pdf uploaded in accordance with publisher's policy at http://www.sherpa.ac.uk/romeo/issn/0004-6361/ (accessed 17/04/2014) EDP Sciences 0004-6361 30 March 2016 04 Jun 2017 06:05 http://orca.cf.ac.uk/id/eprint/56284

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text:
# What is the purpose of this room on board the Engineer's ship?

Where is this room and what is it used for?

Technically it is not confirmed where this room is, whether it is on the ship or on the planet but in either case, I could not determine what the purpose of the room could have been?

• What is the giant head for?
• Did every ship carry one?
• Do they worship themselves?
• Why the mutable patterns on the ceiling?

The black mutagen is scattered around the room in containers that are OPEN. I am having a hard time understanding what purpose they could have sitting around the room. The spaces between the cannisters seems very small so I can't even seen the Engineers moving around this room very easily.

What would they use this room for? Speculation allowed but movie facts are desirable if possible.

• I'm pretty sure this room was not actually on the ship. The tunnels outside are Rough-hewn, and the floor is dirt. The ship is at the end of the LONG tunnel, far from the area first explored. – aslum Jun 15 '12 at 21:26
• If its not on the ship, then that makes it even stranger to me. How did it get there, why put it there and why surround it with Weapons of Genetic Mass Destruction? Arrrrggg! Too many questions! – Thaddeus Howze Jun 15 '12 at 23:24
• Um, it's a base/complex and the ship is docked. Remember there was more than one ship. If you pay close attention to the holographic map on Prometheus (not the star map David activates) you'll notice it is a huge dome shape (mimicing the huge structure we see from outside). There is a long tunnel, and later we see a hologram of the C shaped ship, which is docked at the end of the tunnel, halfway or so to where Prometheus landed. – aslum Jun 15 '12 at 23:49
• I think a better question is why they stored those things so poorly in that room. Just sitting there evenly spaced on the ground, or in some big pile. I'd think having them organized and secure would have been a better idea. – Gorchestopher H Jun 16 '12 at 3:01
• They show a stockroom of the canisters exactly like you describe a bit later in the film. Chilling. – Thaddeus Howze Jun 16 '12 at 3:40

I believe that there is a good chance the idol was merely meant to distract from the fact that the room is a death trap.

The room maybe similar to the ending room/construct of Revelation Space, where the aliens have lured the humans to a particular place with an origin mystery simply to assist the aliens in eradicating humanity.

Once the atmosphere hits the room the bio-agents are set in motion, the only thing that needs to happen is for the explorers to stick around long enough to get infected. But it does seem like an awful lot of canisters just to form a simple trap, though apparently they have quite a stockpile overall.

Perhaps the Engineers that built the bio-weapons facility also had rivals or enemies that they sought to guard against. Rigging a central room with a booby trap, might be useful as a security measure.

On the other-hand in the intro, the aliens appear to have had at one time a reverence for life and creating life. By that same token, maybe they attach a religious significance to causing death. Apparently, Scott considered having the Engineers be Jesus at some point but felt it was a little too blunt for mainstream audiences.

This explains the cave drawings 'inviting' humanity to track them down. Some fans have suggested that Jesus Christ was an Engineer and his death at the hands of man was the point they turned against humanity. In an interview with Movies.com Ridley Scott acknowledges that this back story was considered but neither confirms nor denies that this was indeed used in the final script:

Movies.com: "We have heard that it was scripted that the Engineers were targeting our planet for destruction because we had crucified one of their representatives, and that Jesus Christ might have been an alien. Was this ever considered?"

Ridley Scott: "We definitely did, and then we thought it was a little too on the nose. But if you look at it as an 'our children misbehaving down there' scenario, there are moments where it looks like we've gone out of control, running around in armor and skirts, which of course would be the Roman Empire. And they were given a long run. A thousand years before their disintegration actually started to happen. And you can say, 'Let's send one more of our emissaries to see if he can stop it'. Guess what? They crucified him."

Either way the idol was probably built to inspire awe, but it's not clear whether the idol is intended for humans or Engineers, or whether it was built with a trap in mind.

What is the giant head for?

A trap and/or to inspire awe, religious or otherwise.

Did every ship carry one?

I don't think they were on the ship yet when they found the room.

Do they worship themselves?

As Orac would say this question is illogical since all religions possess some elements of self worship. But kidding aside, I would say yes in a sense, though as a god-form or a mortal I cannot say.

Why the mutable patterns on the ceiling?

Probably decorative art, whether murderous or religious I'm sure the Engineers felt they had to have a sense of style.

Another theory: The idol and room were for racial reunion, but that idea was abandoned in favor of a death trap:
What if the room was originally intended to be for a species reunion for the two species Engineer and man. But after some negative interaction with humanity the idea was abandoned and then the room was booby trapped when some of the Engineers started to have second thoughts about creating other intelligent species.

• Great answer, and (including "Another theory") it would confirm that the cave drawings on Earth were indeed inviting humans to come to the solar system (though the motive behind the invite is unclear). – Shane F. Jun 23 '12 at 18:40

--movie spoilers all over this text--

If you accept H. R. Giegers' original works and his commentary on it as canon, we know at least two things about the engineers:

1. Like most of H. R. Gieger's work they were or could be cyborgs or similar man-machine mixes. The pilot chair the engineer enter and gets enveloped in at the end is a replica of the fossilized pilot chair and pilot from the first alien movie. Giegers intention for it was to depict a pilot who was his pilot role to such a degree that he had fused with the machinery and ceased to be anything but the pilot.

2. They attached a religious significance to the alien monster. In Gieger's picture below, the overarching cyborg woman is a reference to the portrayal of the egyptian goddess Nut, which was the goddess of the sky and gave birth to several other gods. Gieger's picture then shows a goddess sending the alien eggs down from the skies.

It is then a small leap to imagine that the engineers, being already in a symbiotic relationship with their machines and revering the aliens, also were in some symbiosis with the aliens. In fact, have you noticed how there were only male engineers?

Gieger originally played a lot on allusions to sex. He was a designer for the first Alien film and his original design of the egg had the top split along a single slit in what became a comically obvious allusion to a vagina. (*) The face-hugger force the victim to deep-throat in an obvious rape simile and then the victim, often male, is forced to undergo a gestation and "give birth" to the alien. In short, the vagina-egg force the victim into taking the role of the receptive sex.

So imagine, here we have The Engineers, a species which has lost its females and now only reproduce via machines. (Plenty of cyborg fetuses in Gieger's art btw) But they revere as mystical the ancient "female", which is symbolically opposite both to males and machines. (**) In its contained revered form the engineer's "female" is abstracted away into a sticky oozing fluid with worms. It doesn't get more biologically icky than that after all and probably horrifies the engineer-machines like the depictions of the mutilated corpse on a cross are meant to do for christians. This goo doubles as paint for their mystical paintings.

A devout engineer may volunteer to "mate" with the goo. Only then do a true biological organism with its own lifecycle appear. Note that in Prometheus the protagonist only births a face-hugger, which must then infect another organism before an alien occurs. The aliens lay the eggs and have queens and drones. Recreating this two-sexes fully biological creature is seen as a holy act by some engineers.

We can guess at a schism early in the religious history of the engineers. They start out as in the opening scene of the movie, with the creation of biological life as the holy act. Then they split into two factions. One with a mechanical leaning which worshipped the machine-inspired alien monster and sent it from the stars to many destinations, preying on the life seeded by the original religion. (A semi-canon comic link an early extinction of life on earth with aliens) Then there's one faction with a sapient leaning which recreates life in their own image, including humans. Trope-wise they suffer from the "Gepetto Syndrome", which is sometimes used in ridicule about men who, unable to create life themselves, obsess with robotics or artificial life. Being a Gepetto they fall in love with their own creation, can find no fault in it or think they can just fix its faults like you fix a machine.

The reason the Engineers turned on us are then either that we were a failure in the eyes of the humanoid-makers and they turn on their own creation in anger (the movie characters' hypothesis I think) or that the engineers in Prometheus belong to the alien-faction and regard humans as sacrilege committed by the other faction. An interpretation is that the engineer awoken in the movie rips the head off David because his question implies that the engineer would be one of the heathen humanoid-makers. Prior to that the drowsy engineer hopes that the humans are there as devoted and willing vessels for the alien "incubation".

The Engineer ship is then one small part of a giant religious war machine where the alien-breeders seek to eradicate the works of the humanoid-breeders.

So what about the head? Others have pointed out that the head is not on the ship itself. The room is a church and they worship the original humanoid "mother species" which created the machines and their male-only descendants. The head is a "mother" (note machines are birthed in the head of the inventor) but is depicted as the male's faces with a few streaks of technology because the engineers have lost all recollection of how their females actually looked.

Ridley Scott: If you're reading this, I'll trade you the plot for a Gieger painting and a face-hugger plushy. :-)

(*) The vagina design was turned down by the producers, citing catholic audiences as a reason. (Really I think it was just too gross, but put it nicely to the artist :-) Gieger then changed the design by keeping the lips of the slit but making it into a cross instead. He writes that he was quite amused that he had hid the vagina by merging it with the sacred symbol of the prudes he re-designed it for.

(**) Female as base body and male as logical spirit is unfortunately a common underlying theme in much art, even today.

I believe the room originally had some religious significance to the Engineers but as more and more of the weapons were made they simply needed more storage space. They had the good common sense not to store the weapons on board the ships.

• But later in the film we see the weapons inside the C-shaped ship! – CamelBlues Jun 18 '12 at 18:52

I am not mistaken, the only thing we know about Chamber-Engineers interaction is how they ran there.

From the movie it seemed that they ran from something. They ran so badly that they dared to kill one of their own rather than let the doors open longer. So they thought that the sealed door will provide them with security (the chamber was not inspected completely but I doubt that other bodies were there so it actually saved them).

From this point of view room rather looks like laboratory or some kind of testing facility. This kind of room would provide them with security since it has to be properly isolated.

The purpose of the head is a mystery indeed; it is too big to be a device, which could be manipulated from the ground. What is strange for me is that the head has several stripes. The stripes look uncharacteristic for sculptures or any kind of art (why such imperfection). Also in this case I think it is some kind of a bigger device which can either move (open) or transmit. These stripes can be either power source or they can have mechanical origin.

And to fit my theory into the "mutable patterns on the ceiling": ). They could be some kind of shield. Or maybe evaporated black liquid (every liquid evaporates regardless the temperature) condensated on the ceiling.

Just speculating, but it's possible their quest to wipe out Earth had some sort of religious significance to them, in which case this room could function as some sort of shrine (it could also explain the artwork on the ceiling).

Or perhaps it was meant to symbolically show their superiority in a way that any earth-people entering the room of WMDs would know who it was that caused their destruction. I'm not sure / don't remember how they planned on distributing the mutagens, so again, this is just speculation.

Since all of the answers here are speculation, I'll add my own.

Perhaps the Engineers are some type of death-cult. We see this at the start of the film,

when the Engineer, on Earth, ingests the black oil. The oil kills him, but he falls into the water to populate the planet with the remnants of his DNA.

Possibly the room had religious significance in their death cult rituals, which would explain why the canisters are arranged around the head.

I think the big head is in the image of their creator. It appears to be bound. I think it is Fifield asked if they worshipped it. This might indicate that they have slain their creators. Now they're going after their creators' other creation, humans.

• Welcome to Science Fiction and Fantasy SE, take a look at our tour to get an idea of the site. Is there any evidence you could provide to aid your points of view? To make your answer a bit stronger? – Edlothiad Feb 12 '17 at 16:57