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# CSCI4390-6390 Assign5

## Assign5: Neural Nets (MLP)

Due Date: Nov 1, before midnight (11:59:59PM, Alofi Time; GMT-11)

You will use the Appliances energy prediction data set. You should ignore the first attribute, which is a date-time variable, and you should also remove the last attribute, which is a duplicate of the previous one. Use the first attribute (after removing the date-time variable), which denotes the Appliances Energy Use, as the response variable, with the remaining attributes as predictor variables. However, you have to discretize the response variable differently for the two sections as noted below, since CSCI4390 will implement binary classification, whereas CSCI6390 will implement multiclass classification via MLP.

### CSCI4390: Binary Classification

You will implement binary classification via the MLP training Algorithm 25.1 (Chapter 25, page 655). This algorithm assumes sigmoid activation for hidden layer, and squared error for the loss. However you have to use ReLU activation for the hidden layer, and use binary cross entropy for the loss. Consequently, you have to modify lines 10 and 11 appropriately so that line 10 uses cross-entropy loss, and line 11 uses ReLU derivative. See section 25.1.1 for derivative of ReLU, and section 25.2.3 for the derivative of binary cross-entropy loss function.

Note that the Appliances Energy Use attribute takes values in the range $[10,1080]$. However, for binary classification, we need only two values, so for the purpose of this assignment you should consider energy use less than or equal to 50 as the positive class (1), and energy use higher than 50 as negative class (0).

You should shuffle the data points before selecting 70% of the data training and 30% for testing, so that there is an equal mix of the classes in both.

### CSCI6390: Multiclass Classification

You will implement multiclass classification via the deep MLP algorithm in Algorithm 25.2 (Chapter 25, page 668). You should use ReLU activation for all hidden layers, and softmax for the output layer. See section 25.1.1 for the derivative of the activation functions, and end of section 25.4.3 for derivative of the multiclass cross-entropy function.

Note that the Appliances Energy Use attribute takes values in the range $[10,1080]$. However, for multiclass regression, we will convert these into four classes as follows: energy use less than or equal to 30 is class $c_1$, energy use greater than 30 but less than or equal to 50 is class $c_2$, energy use greater than 50 but less than or equal to 100 is class $c_3$, and finally energy use higher than 100 is class $c_4$. You need to do this conversion to create the categorical response variable, before you select the train (70%) and test (30%) subsets.

You should shuffle the data points before selecting 70% of the data training and 30% for testing, so that there is an equal mix of the classes in both.

### What to submit

• Write a scripy named as Assign5.py, which will be run as

Assign5.py FILENAME ETA MAXITER HIDDENSIZE for CSCI4390

Assign5.py FILENAME ETA MAXITER HIDDENSIZE NUMHIDDEN for CSCI6390

FILENAME is the datafile name, ETA is the step size $\eta$, MAXITER is the number of epochs to train the model, HIDDENSIZE is the size of the hidden layer, and NUMHIDDEN is the number of hidden layers (this is for CSCI6390). CSCI6390 students may assume that all hidden layers use the same size.

You should implement the algorithms using NumPy; you cannot use any of the deep learning libraries (like keras, pytorch or tensorflow, and so on).

Your script should print out the weight matrices and biases for each layer, and also the final accuracy value on the test data, where you use 70% of the data for training and 30% for testing.

Note that accuracy is defined as the fraction of correct class label predictions. So for each test point, you should predict its class as the one that has the highest probability, and then you should count how many test points are correctly predicted. Divide that number by the test data size to get the accuracy. Report the results for the combination of ETA and HIDDENSIZE (and NUMHIDDEN for CSCI6390) that yields the best accuracy.

• Submit a PDF file named Assign5.pdf that should include your answers to each of the questions (just cut and paste the output from python). Failure the submit the PDF will result in lost points.

• Submit the scripts and pdf file via submitty

### Policy on Academic Honesty

You are free to discuss how to tackle the assignment, but all coding must be your own. Please do not copy or modify code from anyone else, including code on the web. Any students caught violating the academic honesty principle will get an automatic F grade on the course and will be referred to the dean of students for disciplinary action.


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# Tag Info

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I'm using Windows 10 preview, and was able to install grub normally. Mount your ext4 partition to /mnt, bind /dev and /proc, cp /etc/resolv.conf, chroot into /mnt and run grub-install /dev/sda (or whatever HDD you need). Ask if you need more detailed instructions.

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The Linux password are stored in the /etc/password AND /etc/shadow which is a copy of the other one. The /etc/shadow file is readable and writable only by root. In these shadow files there is a field to hold the password expîration date but this field do not exist in the password file. Ref.: Linux tutorial: Password files EDIT: to clarify my answer. In ...

6

At least in my Ubuntu, the "too similar" messages cames out when: "...more than half of the characters are different ones...." (see below for details). thanks to the PAM support, as clearly explained in the @slhck answer. For other platform, where PAM is not used, the "too similar" messages comes out when: "...more than half of the characters are different ...

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This is a feature of the PAM system which is used in the background of the passwd tool. More specifically, pam_cracklib is a module for PAM which allows to reject passwords based on weaknesses. Since you supply both the old and the new password, they can be easily compared. The source code has various examples of what can be checked, e.g. whether a ...

4

Your symlink is broken. Delete it with rm netlogs. Recreate it using absolute paths, like this: ln -s /home/dir3/logs /home/dir1/netlogs The first argument is the existing real file or directory. The second argument is the link that will point to the first argument. Extra tip: lrwxrwxrwx 1 joasia None 9 Dec 27 14:06 netlogs -> dir1/logs The ...

1

You need to create your symbolic link with a path that will work wherever you are, either an absolute one like ln -s $HOME/dir3/logs$HOME/dir1/netlogs or better, a relative path with will allow the upper directory to be moved elsewhere without the link being broken. ln -s ../dir3/logs dir1/netlogs

1

You’re not using NFS but CIFS. CIFS hides the underlying file system structure. The appropriate command to mount the “software” share is as follows: mount -t cifs //x.x.x.x/software /mnt/software

0

The problem is not your Samba config here because the machines on the network can access your shares without problem. The error message clearly shows that a memory error was returned to smbclient from Windows. The only fix here is for your housemates to change the system memory settings on their Windows boxes. The memory usage for the share pool will be ...

0

I'm not an expert at all when it comes to SysVinit scripts. But from what I can tell, if you specify that a initscript should be started at only one runlevel (for example, runlevel 3), you will have S symlinks in /etc/rc3.d only, and K symlinks in all other /etc/rcX.d directories. Chances are, your script is getting called with stop on shutdown; the K ...

1

Some other answers have hinted that 4 GiB of memory is the minimum for a 64-bit o/s, but I have not found any support for that, and maybe some contraindication. I have a Lenovo R61 laptop which came with 1 GiB of RAM. I run Fedora spins of Linux, at first not realizing this has a 64-bit CPU, so I installed 32-bit Linux. Of course, Linux ran just fine, ...

0

"Connection reset by peer" means the TCP stream was abnormally closed from the other end. I think the most likely explanations are that the remote server process handling the connection has crashed, or else some network device (like a stateful firewall or load balancer) has decided to interfere with the connection. You need to debug this on the remote ...

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gmusicbrowser is the only application that could read my (600 GB) music collection without problems.

0

Your VirtualHost configuration indicates that apache accepts request from any of the interfaces in the host but it should have a Host: www.abctest.com header to select DocumentRoot as /websites/abctest.com/htdocs. Generally any other Host header will cause apache to use the default VirtualHost to be used. In case if you are trying to open the webpage with ...

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Tripwire is an open source tool for monitoring file changes.

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You can try this: find . -name 'flibble*' -ctime +90 -exec du -ch {} + | grep total

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Have you checked these fairly detailed documents http://as6edriver.sourceforge.net/Parallel-Port-Programming-HOWTO/parallel-port-programming-howto.html http://www.tldp.org/HOWTO/Serial-HOWTO.html

1

I found lpquota Perl script from http://perlwatch.net/perl-scripts/, this might be the one you're looking for. Also DK Tools have printqd daemon for enforcing print quotas on LPRng print systems. This is available in http://dktools.sourceforge.net/printqd.html, which doesn't appear to be related.

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You probably meant to use the -K flag instead of -k: -k checksum-mode valid values: all, noip, notcp, noudp, noicmp, and none -K logging-mode valid values: pcap, ascii, and none For more details see man snort.

0

Ok, so it was all the problem of the Netgear router (NETGEAR N450 WiFi DOCSIS 3.0 Cable Modem Router (N450-100NAS)) rather than SSH. When I switched back to Comcast's router (the ones you can rent) then the connection stopped dropping. Don't buy this router if you plan to expose a port from from your internal network to the world.

0

gzexe is an underappreciated awfulhack. There is a similar functionalty in one of the various zip incarnations - pkzip? If it can solve your problem, please let me know :). Being a presenile dummy, I would use a small perl stub to xor and unxor the script and pipe it to sh.

1

A POSIX-compliant shell script that requires openssl, curl and sed only; supporting AWS Signature Version 4, which is required for region eu-central-1 (Frankfurt) and recommended for the others: https://gist.github.com/vszakats/2917d28a951844ab80b1 #!/bin/sh # Copyright 2014 Viktor Szakats (vszakats.net) # Upload a file to Amazon AWS S3 using Signature ...

0

First of all, does the VM have an IP address? Many corporate networks will only give an IP address to registered MAC addresses, and the VM's network interface will have a MAC address not registered with the company. If you don't have an IP address, then you can try to use one that's not used by another computer. Most probably you would have to do all the ...

0

Try putting a echo $PATH in the script, to ensure the path is what you think it is, and ensure the script itself can run (executable bit set, correct shebang #!/bin/sh or whatever in the first line). If nothing else works, you may try stracing the script invocation call. 0 ..i know its bandwidth and it will never congestion, but will lost packet sometimes. but there's congestion control in tcp, once packet loss, it will slow down, which is meaningless and loss performance. I don't think you understand the concept of congestion. Congestion can not only be happen in the network, but also in the network stack, that is ... 0 As I was still able to boot Windows using Hiren's Boot CD, I could reinstall the bootloader using a tool called EasyBCD. 0 You might have to still Mount it. Try to look into this, the Mount command. https://wiki.ubuntu.com/MountWindowsSharesPermanently Update: Better yet, try this command. sudo mount -t cifs //ip_of_host/share/path /local/mount/dir 0 Solved by installing grub without chrooting to system, then booting the system and installling grub again from there. 0 The probleme in your regex lies in the grouping of the second part: var regex_cron = "^((" + variants + ")(\\s+" + variants + "){4})$"; --- should be : var regex_cron = "^((" + variants + ")(\\s+(" + variants + ")){4})\$"; ! ! Because as you wrote it the ...

0

Run encfsw.exe with Administrator rights. (%Program Files℅\encfs4win\encfsw.exe)

1

This makes linux ignore the LVDS output altogether. That way it will not be used by all login managers, and won't appear in display configuration dialogs. Create a file /usr/share/X11/xorg.conf.d/50-disable-lvds.conf and write there: Section "Monitor" Identifier "lvds monitor" Option "ignore" "true" EndSection Section "Device" ...

2

The standard test for memory is Memtest86+. Since you're just testing basic functionality rather than long-term stability, you only need to let it run for the first few tests, rather than the full multi-hour test series. The standard test for the CPU is mprime in "torture test" mode. As a side benefit, it'll also test the CPU heatsink -- if the heatsink ...

1

You could always transfer files to the intermediate server and then copy it from there. However if the intermediate server has space limitations and cannot hold large files, then you need to create a tunnel through the intermediate server to the second server. Creating a tunnel through putty, 1) your ssh config should point the intermediate server like ...

3

Just thoughts - I've never actually tried anything like this - but could you not write your own script interpreter which contains a decoder engine? So the first line of your script is "#!/bin/cryptoshell", and everything else in the script file is encrypted by a separate script encoder engine. The question, though, is why you would want to do such a ...

1

Your problem is that you are moving, say, Downloads/2014-12-24-first to ~/docs/2014-12-24-first, and THEN trying to change the owner and permissions on Downloads/2014-12-24-first; which will not work, because you have already moved the file to ~/docs. Try re-ordering your -execs so the chmod/chown are first, and the mv last.

0

If I understand your scenario right, you are saying that you have to connect to one computer via SSH, and then connect to another computer from there? If so, why not log in to the first server, and then run scp mainserver:file.txt ./. Then log out, and run scp firstserver:file.txt ./ to get it from the first server. You can then log in to the first server ...

2

per the smbclient manpage, you need to use the mget command, with a mask and recursion and prompt set. Then cd to the directory you want to get recursively smbclient '\\server\share' mask "" recurse ON prompt OFF cd 'path\to\remote\dir' lcd '~/path/to/download/to/' mget * or, all on one line, smbclient '\\server\share' -N -c ...

0

Check out this somewhat old tomshardware article http://www.tomshardware.com/reviews/a-sturdy-companion,758-2.html Note that the circa 2004 drives were rated to take ~ 300g+ while not operating. Check all socketed components, cpu, mem and fire her up. I'd bet the laptop survived. G/L

9

Realistically by far the most likely damage to the system will be the hard drive, as its the only thing with moveable parts. If the HDD works without errors, its a fair bet the system will be fine. To test the HDD, start off with doing some kind of SMART test, then, if that passes, do some full testing with badblocks (but backup the data off the hard ...

11

My question is how I can make my script not readable/binary format? In general you can't easily do this in any way that isn't painful. Scripts are plain text files (not non-text binary) that have to be readable by a script interpreter. The script interpreter runs using the ID and permissions of the user running the script. Therefore the script has to ...

1

If /proc/mdstat exists you have the capability to set up software raid - although if it does not have anything meaningful it means you have not set it up. Software RAID under Linux does not require any particular hardware [ you could argue the ability to support 2 drives, but even thats debatable if you consider USB]

1

If You want to know capability to build hardware array. You can use lspci to view list of PCI controllers, determine type of disk controllers and for specs and manuals for it. UPD. Examples 1. # lspci |grep -iE "sata|sas|scsi|ide|disk" 00:1f.1 IDE interface: Intel Corporation 82801EB/ER (ICH5/ICH5R) IDE Controller (rev 02) 06:00.0 RAID bus controller: LSI ...

0

Your programs are all stored on your hard drive, but it's too slow to run them from there so the OS caches them (or the currently accessed parts of them) in ram when you use them. It takes time to copy this data to the ram. If you have the program installed to a ram-drive it can provide faster read speeds which means it will take less time to copy the ...

1

Programs and data are loaded form a drive into the system memory to execute. A RAM Drive is a virtual drive that exists entirely in system RAM. System RAM access, and data transfers to and from, are orders of magnitude faster than when dealing with a hard drive, so you aren't waiting nearly as long for the program (or data) to be loaded from the RAM ...

2

As it seems your bios may not support USB flash drives. I recommend you to use PLoP: Boot From a USB Drive Even if your BIOS Won’t Let You You’ve always got a trusty bootable USB flash drive with you to solve computer problems, but what if a PC’s BIOS won’t let you boot from USB? We’ll show you how to make a CD or floppy disk that will let you boot ...

0

You are using the diff utility correctly. It seems that the system calls are reporting that those files are directories instead. This is probably an issue somewhere else in the system, not in busybox diff. This all comes down to the following lines of code in editors/diff.c: if (S_ISDIR(stb[0].st_mode) && S_ISDIR(stb[1].st_mode)) ...

0

After some research, it looks like the answer is "no". What I'm wanting is basically to connect to IMAP in "disconnected mode" which fetchmail does not support (see the fetchmail FAQ). I will probably migrate over to offlineimap for this.

4

You're supposed to create the file yourself. It's in the tutorial, where it says: cat > /etc/ntp.conf << "EOF" # Asia server 0.asia.pool.ntp.org # Australia server 0.oceania.pool.ntp.org # Europe server 0.europe.pool.ntp.org # North America server 0.north-america.pool.ntp.org # South America server 2.south-america.pool.ntp.org driftfile ...

0

Check all allowed public keys by cat ~/.ssh/authorized_keys Find and remove your particular public keys to revoke access.

1

I can just help you with the dialog box. Create a .desktop file, that will run you executable script. You can find a quick introduction to .desktop files here or take a look at one of the files in /usr/share/applications. As for the rest of the questions please take a look at this question.

0

Shutdown actually shuts down most parts of the computer, where with reboot all the components are running. If you have hardware problems it could be, that the shutdown emptied some bad capacitors and after power on the system will work ok for a while. Look for some bad capacitors on your motherboard. You can find some descriptions how to find them here

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# Argmax word search

### [Equation Editor] Argmin and Argmax parameter

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12/14/2013 · hi there is one slight problem i have with equation editor and it that i cant write argmax and argmin parameter below the main function. for example i want to find i which minimize a function W=argmin_i f(i) the first i should be just below the argmin, but if i use _ it will be after thr argmin and below it. i also used the matrix section "2x1 emtry matrix" but it will reduce the size of the ...

### Arg max - Wikipedia

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ArgMax [f, cons, x ∈ reg] is effectively equivalent to ArgMax [f, cons ∧ x ∈ reg, x]. For x ∈ reg, the different coordinates can be referred to using Indexed [x, i]. ArgMax will return exact results if given exact input. If ArgMax is given an expression containing approximate numbers, it automatically calls NArgMax.

### ArgMax—Wolfram Language Documentation

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12/4/2013 · This video shows how to determine the minimum and maximum and also values associated to them (argmin and argmax as they are called). For example, the following program shows how users can enter ...

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1/31/2019 · Parameters: a: array_like. Input array. axis: int, optional. By default, the index is into the flattened array, otherwise along the specified axis. out: array ...

### numpy.argmax — NumPy v1.16 Manual - SciPy.org

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The axis in the argmax function argument, refers to the axis along which the array will be sliced. In another word, np.argmin(a,axis=0) is effectively the same as np.apply_along_axis(np.argmin, 0, a), that is to find out the minimum location for these sliced vectors along the axis=0.

### numpy: what is the logic of the argmin() and argmax ...

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hi there is one slight problem i have with equation editor and it that i cant write argmax and argmin parameter below the main function. ... Answered | 4 Replies ... I create a word document and create few formule with equation editor and save the file as .html file. When I open the same in any of the browsers, the formulae are ...

### Technet forums

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2 NLP Programming Tutorial 13 – Beam and A* Search Prediction Problems Given observable information X, find hidden Y Used in POS tagging, word segmentation, parsing Solving this argmax is “search” Until now, we mainly used the Viterbi algorithm argmax Y P(Y∣X)

### NLP Programming Tutorial 13 - Beam and A* Search

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Greedy Search Decoder. A simple approximation is to use a greedy search that selects the most likely word at each step in the output sequence. This approach has the benefit that it is very fast, but the quality of the final output sequences may be far from optimal.

### How to Implement a Beam Search Decoder for Natural ...

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4/26/2012 · Posts about ARGMAX written by TheDataSpecialist. A blogger was wondering, whether it was possible to retrieve a reference to the cell where the result of the functions MAX or LARGE was found.With the proper combination of INDEX and MATCH, this is indeed possible.

### ARGMAX | The Data Specialist

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Pre-trained models and datasets built by Google and the community

### tf.math.argmax | TensorFlow

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26 Responses to “how to write argmax in latex” ... If you want to know how to make extra \$\$\$, search for: Mrdalekjd methods for \$\$\$ on 25 Jul 2018 at 9:20 am KatlynJuicy. Hi. I see that you don’t update your blog too often. I know that writing posts is boring and time consuming. But did you know that there is a tool that allows you to ...

### how to write argmax in latex » from the desk of stinkpot

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What is an argmax? We'll define it for you. Meaning of the word Synonyms. Antonyms. Definitions. Rhymes. Sentences. Translations. Find Words. Word Forms. …

### What is an argmax - wordhippo.com

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Word Forms. Pronunciations. 8-letter words starting with ARGMAX. ATTENTION! Please see our ...

### 8-letter words starting with ARGMAX - wordhippo.com

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LANGUAGE MODELING FOR VOICE SEARCH: A MACHINE TRANSLATION APPROACH Xiao Li, Yun-Cheng Ju, Geoffrey Zweig, and Alex Acero Microsoft Research One …

### LANGUAGE MODELING FOR VOICE SEARCH: A MACHINE …

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Update: see Fast argmax in Python for the final word. Python doesn’t come with an argmax function built-in. That is, a function that tells you where a maximum is in an array. And I keep needing to write my own argmax function so I’m getting better and better at it. Here’s the best I could come ...

### Computing argmax fast in Python – Daniel Lemire's blog

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Recognized_Word = argmax. word . P(word)P(X| word) P(word) is a priori probability of word Obtained from our expectation of the relative frequency of occurrence of the word P(X|word) is the probability of X computed on the probability distribution function of word . 6.345 Automatic Speech Recognition Designing HMM-based speech recognition systems 4

$Computing argmax fast in Python – Daniel Lemire's blog$

### Part I: Designing HMM-based ASR systems

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Pandas DataFrame - Find row where values for column is maximal. Ask Question 169. 41. How can I find the row for which the value of a specific column is maximal? ... You just need the argmax() (now called idxmax) function. It's straightforward:

### python - Pandas DataFrame - Find row where values for ...

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The value assigned to each unique word key is simply an increasing integer count of the size of the dictionary. So, for instance, the most common word will receive the value 1, the second most common the value 2, the third most common word the value 3, and so …

### Word2Vec word embedding tutorial in Python and TensorFlow

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4/26/2012 · Search for: Recent Posts. ... ARGMAX and ARGMIN in Excel April 26, 2012 — TheDataSpecialist . A blogger was wondering, whether it was possible to retrieve a reference to the cell where the result of the functions MAX or LARGE was found. With the proper combination of INDEX and MATCH, this is indeed possible. ... Tags: ARGMAX, ARGMIN, Excel, ...

### ARGMAX and ARGMIN in Excel | The Data Specialist

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Speech Recognition as Bayesian Classification • Different words may occur with different frequency – Classification • Different words may occur with different frequency –

### Speech Recognition as Bayesian Classification Different ...

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Word Re-ordering and DP-based Search in Statistical Machine Translation ... The argmax operation denotes the search problem, ... troduce our novel concept to word re-ordering and a DP-based search ...

### Viterbi algorithm - Wikipedia

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argmax -2 points-1 points 0 points 11 months ago Okay, let me ask you, would you like to share your ssn, income, employment, dependents, address, assets, property you own information with me just because I can help you fill out a tax form?

### Word Re-ordering and DP-based Search in Statistical ...

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Multi-layer Recurrent Neural Networks (LSTM, RNN) for word-level language models in Python using TensorFlow. - hunkim/word-rnn-tensorflow

### TurboTax wants me to share my tax information with Intuit ...

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Lecture 9 LVCSR Search Michael Picheny, Bhuvana Ramabhadran, Stanley F. Chen, Markus Nussbaum-Thom Watson Group IBM T.J. Watson Research Center Yorktown Heights, New York, USA

### word-rnn-tensorflow/model.py at master · hunkim/word-rnn ...

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A Continuous Relaxation of Beam Search for End-to-end Training of Neural Sequence Models Kartik Goyal Carnegie Mellon University kartikgo@cs.cmu.edu Graham Neubig Carnegie Mellon University gneubig@cs.cmu.edu Chris Dyer Deepmind cdyer@google.com Taylor Berg-Kirkpatrick Carnegie Mellon University tberg@cs.cmu.edu Abstract Beam search is a ...

### Lecture 9 - LVCSR Search - Columbia University

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Sorting, searching, and counting ... argmax (a[, axis, out]) Returns the indices of the maximum values along an axis. nanargmax (a[, axis]) Return the indices of the maximum values in the specified axis ignoring NaNs. argmin (a[, axis, out])

### A Continuous Relaxation of Beam Search for End-to-end ...

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information not detected by linear offset may still be recoverable by a more sophisticated search method, and thus is actually encoded in the embedding. ... calculated as argmax d 2 V (sim (d;c a + b)). Here V is the vocabulary (excluding word vectors a ,b and c), and sim is a similarity measure, for which Mikolov and

### Sorting, searching, and counting — NumPy v1.13 Manual

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A Look at English-Inuktitut Word Alignment Fabrizio Gotti, Alexandre Patry, Guihong Cao, Philippe Langlais RALI Departement d’Informatique et de Recherche Op´ ´erationnelle Universit´e de ...

### Word Embeddings, Analogies, and Machine Learning: Beyond ...

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2 ˆ W argmax W L P O W P W LMSF Word Insertion Penalty But LM prob PW also from CS 224S at Stanford University. ... Fewer insertion errors Need wider search beam (since path scores larger) ... Word Insertion Penalty ...

### A Look at English-Inuktitut Word Alignment - researchgate.net

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As Pieter pointed out, the correct way to define argmin and argmax operators in LaTeX is: \usepackageamsmath \DeclareMathOperator*\argmaxarg\,max \DeclareMathOperator*\argminarg\,min Actually, using the amsopn package would be sufficient but it's loaded by amsmath internally, which is recommended for math typesetting anyway.

### 2 ˆ W argmax W L P O W P W LMSF Word Insertion Penalty But ...

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Using MS Equation. Some Math or Science teachers are in the habit of leaving empty spaces as they prepare work for their classes. After printing one copy, they will go back to the empty spaces and hand draw the mathematic or scientific symbols needed before printing a class set of the assignments or tests.

### math mode - Command for argmin or argmax? - TeX - LaTeX ...

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modeling [5, 6], to control the search space and cross-word context models [7], for improving noise robustness [8], to help drive dynamic pronunciation models [9, 10], and in language modeling [11, 12]. Use of prosody in speech understanding applications has been more extensive, and there are a growing number of applications that are being ...

### Using Microsoft Equation in Science and Math

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A Diversity-Promoting Objective Function for Neural Conversation Models ... a greedy approach or beam search can be adopted for word prediction. Greedy search selects the to-ken with the largest conditional probability, the em- ... Tˆ = argmax T logp(T|S) −logp(T)

### yEE Dept., University of Washington, Seattle, WA zAT&T ...

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A word search puzzle typically consists of a grid of letters in which words are hidden.. There are many varieties of word search puzzles. For the task at hand we will use a rectangular grid in which the words may be placed horizontally, vertically, or diagonally.

### A Diversity-Promoting Objective Function for Neural ...

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෢𝑾=argmax 𝑾 𝑾𝑨=argmax ... Guide the search algorithm (predict next word given history) Disambiguate between phrases which are acoustically similar Great wine vs Grey twine It assigns probability to a sequence of tokens to be finally recognized ... Deep Learning for Audio ...

### Word search - Rosetta Code

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The latest Tweets from argmax (@argmarx) Tweet with a location. You can add location information to your Tweets, such as your city or precise location, from the web and via third-party applications.

### Deep Learning for Audio - University Of Illinois

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we propose an alternative, word-based seg-mentor, which uses features based on com-pletewordsandwordsequences. Thegener-alized perceptron algorithm is used for dis-criminative training, and we use a beam-search decoder. Closed tests on the rst and second SIGHAN bakeoffs show that our sys-tem is competitive with the best in the litera-

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CONFIDENCE ESTIMATION, OOV DETECTION AND LANGUAGE ID USING PHONE-TO-WORD TRANSDUCTION AND PHONE-LEVEL ALIGNMENTS +Christopher White, ++Geoffrey Zweig, Lukas Burget, Petr Schwarz, Hynek Hermansky ... during search when …

### Chinese Segmentation with a Word-Based Perceptron Algorithm

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An overview of decoding techniques for large vocabulary continuous speech recognition Xavier L. Aubert† Philips Research Laboratories, Weisshausstrasse 2, 52066 Aachen, Germany Abstract A number of decoding strategies for large vocabulary continuous speech recognition (LVCSR) are examined from the viewpoint of their search space representation.

### CONFIDENCE ESTIMATION, OOV DETECTION AND LANGUAGE ID …

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Maximum Entropy Markov Models for Information Extraction and Segmentation Andrew McCallum, Dayne Freitag, and Fernando Pereira 17th International Conf. on Machine Learning, 2000 Presentation by Gyozo Gidofalvi Computer Science and Engineering Department University of California, San Diego gyozo@cs.ucsd.edu May 7, 2002

### An overview of decoding techniques for large vocabulary ...

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Overview I The Tagging Problem I Generative models, and the noisy-channel model, for supervised learning I Hidden Markov Model (HMM) taggers I Basic de nitions I Parameter estimation I …

### Maximum Entropy Markov Models for Information Extraction ...

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Need to determine if a beam search gives better results.

### Tagging Problems, and Hidden Markov Models

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Learning word models from continuous recordings Building word models from phoneme models Context-independent and context-dependent models Building decision trees Tied-state models Language Modelling Exercise: Training phoneme models Exercise: Training context-dependent models

Search-scanned-photos.html,Search-school-psychologist.html,Search-screen-ui.html,Search-sea-world.html,Search-sedimentary-rocks.html


text:
JB1625 one year ago Help

1. JB1625

2. JB1625

@hartnn

3. anonymous

put the value of z in the expression... $\frac{z-4}{-8}=\frac{-28-4}{-8}$ Now after this if you can't solve it, I can't really help you, it's pretty straightforward

4. JB1625

Ok thank you

5. anonymous

what do you think is the answer then?

6. JB1625

4

7. anonymous

good job


text:
# Monte Carlo Methods in Practice

## Rendering the McBeth Chart using Monte Carlo Integration

In the lesson on Colors and Digital Images, we explained that colors could be represented as curves giving the amount of light at each wavelength (denoted with the greek letter lambda $$\lambda$$) in the visible spectrum (these curves are called spectral power distribution or SPD). Examples of such curves for three different materials are shown in Figure 1. As you can see, these curves can be interpreted as a function of wavelength: f($$\lambda$$). Generally, the visible spectrum is considered to be within the interval of 380 to 730 nanometers. To convert this curve back to tristimulus values X, Y, and Z, we need to calculate three one-dimensional integrals:

$$\begin{array}{l}X = \int_{\lambda1}^{\lambda2} spd(\lambda) \bar X(\lambda) d\lambda\\Y = \int_{\lambda1}^{\lambda2} spd(\lambda) \bar Y(\lambda) d\lambda\\Z = \int_{\lambda1}^{\lambda2} spd(\lambda) \bar Z(\lambda) d\lambda \end{array}.$$

Where the functions $$\bar X, \bar Y, \bar Z$$ are the CIE standard observer color matching functions. The resulting tristimulus values X, Y, and Z can then be converted to RGB. We explained this process in detail in the lesson on Color and Digital Images.

Hopefully, you have guessed that we were going to use Monte Carlo integration to solve these integrals. The SPDs we will be using is the spectral data for each of the 24 different colors used on a McBeth chart. Spectral data for these colors are readily available on the web. If you remember what we said about Monte Carlo integration in the previous chapter, this estimator is defined as:

$$\langle F^N \rangle = (b-a) \dfrac{1}{N} \sum_{i=0}^{N-1} f(X_i).$$

In our example, the interval [a,b] is  [380,730] the range of wavelength defining the visible spectrum. To approximate the XYX values for each color of this color checker, we need to choose a random wavelength within this interval, evaluate and multiply with each other the spd and the color matching functions at this random wavelength, repeat this process N times (where N is the number of samples), sum up all the results, divide the final number by N, and multiply it by (b-a). In pseudo-code we get:

int N = 32;
float lambdaMin = 380, lambdaMax = 730;
float X = 0, Y = 0, Z = 0;
for (int i = 0; i < N; ++i) { // for each sample
// select a random wavelength
float lambda = lambdaMin + drand48() * (lambdaMax - lambdaMin);
// evaluate the sod and the color-matching function at this wavelength,
// multiply the results, and add the contribution of these samples to the others
X += spd(lambda) * CIE_X(lambda);
Y += spd(lambda) * CIE_Y(lambda);
Z += spd(lambda) * CIE_Z(lambda);
}
// complete the integral, multiply by (b-a) and divide by N
X *= (lambdaMax - lambdaMin) / N;
Y *= (lambdaMax - lambdaMin) / N;
Z *= (lambdaMax - lambdaMin) / N;

Simple? Indeed. To illustrate what variance is, we will render each color of the MacBeth chart as a 64x64 square using MC integration for each pixel of that square (each pixel of a color bucket, is obtained by running a new Monte Carlo simulation and therefore each pixel is likely to be "slightly" different than the others. How different depends mostly on the number of samples N used in the simulation). Our final goal is to recreate something that looks like a real McBeth chart, but CG generated (and using Monte Carlo integration of course). In lesson 5, we provided the source code of a small program rendering a McBeth chart using a Riemann sum to compute the integrals. We will start from this program as it already contains the spectral data for the McBeth chart as well as the CIE color-matching functions. The heart of the code is the function in which the Monte Carlo integration is done.

unsigned int N = 32;

inline float linerp(const float *f, const short &i, const float &t, const int &max)
{ return f[i] * (1 - t) + f[std::min(max, i + 1)] * t; }

void monteCarloIntegration(const short &curveIndex, float &X, float &Y, float &Z)
{
float S = 0; // sum, used to normalize XYZ values
for (int i = 0; i < N; ++i) {
float lambda = drand48() * (lambdaMax - lambdaMin);
float b = lambda / 10;
short j = (short)b;
float t = b - j;
// interpolate
float fx = linerp(spd[curveIndex], j, t, nbins - 1);
b = lambda / 5;
j = (short)b;
t = b - j;
X += linerp(CIE_X, j, t, 2 * nbins - 1) * fx;
Y += linerp(CIE_Y, j, t, 2 * nbins - 1) * fx;
Z += linerp(CIE_Z, j, t, 2 * nbins - 1) * fx;
S += linerp(CIE_Y, j, t, 2 * nbins - 1);
}
// sum, normalization factor
S *= (lambdaMax - lambdaMin) / N;
// integral = (b-a) * 1/N * sum_{i=0}^{N-1} f(X_i) and normalize
X *= (lambdaMax - lambdaMin) / N / S;
Y *= (lambdaMax - lambdaMin) / N / S;
Z *= (lambdaMax - lambdaMin) / N / S;
}

As you can see with this code, we first sample a random value for the wavelength contained in the interval [380,730] (line 10). we need to convert this random wavelength to an index for performing a lookup into the CIE color matching functions (sampled every 5 nm) and McBeth spectral data (sampled every 10 nm) 1-D tables. Because the wavelength may fall between two samples in the table, it is best to perform a linear interpolation (interpolate the two nearest samples). Finally, X, Y, and Z are updated with the value of the CIE color-matching function and spectral data at this wavelength (lines 19-21). Because these values will need to be normalized (check lesson 5), we also need to compute the integral of the CIE color-matching function corresponding to the Y component. Here as well, we can use a Monte Carlo integration (line 22). Finally, the X, Y, and Z values are multiplied by (b-a) (in this example the minimum and maximum wavelength) and divided by N (lines 27-29). We also have to normalize them but this has nothing to do with an MC integration (it is just part of the process of converting spectral data to XYZ).

const double XYZ_to_RGB[][3] = {
{ 2.3706743, -0.9000405, -0.4706338},
{-0.5138850,  1.4253036,  0.0885814},
{ 0.0052982, -0.0146949,  1.0093968}
};

void XYZtoRGB(const float &X, const float &Y, const float &Z, float &r, float &g, float &b)
{
r = std::max(0., X * XYZ_to_RGB[0][0] + Y * XYZ_to_RGB[0][1] + Z * XYZ_to_RGB[0][2]);
g = std::max(0., X * XYZ_to_RGB[1][0] + Y * XYZ_to_RGB[1][1] + Z * XYZ_to_RGB[1][2]);
b = std::max(0., X * XYZ_to_RGB[2][0] + Y * XYZ_to_RGB[2][1] + Z * XYZ_to_RGB[2][2]);
}

int main(int argc, char **argv)
{
...
for (int y = 0, pixIndex = 0; y < height; ++y) {
short row = (short)(y / squareSize);
for (int x = 0; x < width; ++x, pixIndex++) {
short column = (short)(x / squareSize);
// compute each pixel of the square
float X = 0, Y = 0, Z = 0;
monteCarloIntegration(row * 6 + column, X, Y, Z);
float r, g, b;
XYZtoRGB(X, Y, Z, r, g, b);
buffer[pixIndex][0] += r;
buffer[pixIndex][1] += g;
buffer[pixIndex][2] += b;
}
}
...
}

The main function of the program loops over each pixel of the frame. For each pixel, it computes the row and column this pixel corresponds to in the McBeth chart (the squares of the chart are organized as a series of 4 rows and 6 columns. You can see an example of the McBeth chart in figure 2). The spectral data of that color will be used to compute a value for X, Y, and Z (lines 22 and 23). Then they are converted to RGB values, a process which we also explained in the lesson Introduction to Light, Color, and Color Space. Finally, these RGB values are stored in a buffer which is then save to disk once all the pixels are processed (your screen is likely to expect your image to be in sRGB so we also apply gamma to the pixel values before storing them to disk). The result for 32 samples can be seen in Figure 3. Now as you can see the result is far from being bad, but the image is quite noisy! This is the famous variance, we have been talking about in the lesson Mathematical Foundations of Monte Carlo Methods. Or what you call maybe more commonly noise when you look at your renders.

## Example of Progressive Rendering

To demonstrate another very nice property of Monte Carlo integration, we will modify our program so that it keeps refining the result by computing as many version of this image as we want (we will call these images passes) and averaging their results. In a way, it is the same as increasing the number of samples N. However if had increased N to 4096 for instance, the image might have taken a much longer time to render (in fact, technically 128 times longer than an image for which N = 32). Thus when N is small, we can see a result right away, and if we like what we see after this quick first pass, then we can let the program run and compute more passes until we are eventually satisfied with the image quality (that is when you think the noise is not distracting anymore). This method is known as progressive rendering. Imagine you need to deliver a sequence of images by a certain time. If X is the number of minutes you get when you divide the total number of rendering time you have by the total number of images you need to render, then you can let your program render each one of these images for no longer than X minutes. That way, you are certain that all your images will be rendered on time. They might have some noise, but you will have a complete sequence of images to look at, and, in some circumstances, this is often better than having an incomplete sequence of images even if they look much better. The fact that render time is predictable, is generally useful in a production environment. Also, if you like the sequence of images that you get after X hours of render time, you can re-launch the sequence and let it refine for another X number of hours. Rather than rendering each image from scratch again, the program can simply start from the existing renders and keep refining them by adding more passes. Therefore, no rendering resources are wasted! These are some of the great benefits of progressive rendering.

More formally, remember that a Monte Carlo integration is just like computing the sample mean of a random variable. In the lesson Mathematical Foundations of Monte Carlo Methods, we explained in detail that these samples' means (which are themselves random numbers) can also be averaged to produce more accurate results. In this example, a pass is nothing else than a sample mean. And the result of these passes can be averaged to produce a more accurate result (as with sample means). This is why Monte Carlo integration makes progressive rendering possible.

We have implemented a basic form of progressive rendering in our program. This time we won't display the result in an OpenGL window. The program runs entirely in a shell. It will keep rendering passes and accumulating the results in the buffer until the user terminates the program (using ctrl-c). The program will catch this signal and will call a function in which we will save the result of the buffer to disk before quitting. Note that in fact, after each passe, we will actually convert the content of the buffer to a proper image in which the content of the buffer is divided by the total number of passes (the buffer is like an accumulation buffer). It is the content of the image which is saved to disk, not the buffer. Here is the source code:

void handler(int s)
{
printf("Saving image after %d passes\n", npasses);
std::ofstream ofs;
ofs.open("./mcbeth.ppm");
ofs << "P6\n" << width << " " << height << "\n255\n";
for (uint32_t i = 0; i < width * height; ++i) {
unsigned char r, g, b;
r = (unsigned char)(std::min(1.f, pixels[i][0]) * 255);
g = (unsigned char)(std::min(1.f, pixels[i][1]) * 255);
b = (unsigned char)(std::min(1.f, pixels[i][2]) * 255);
ofs << r << g << b;
}
ofs.close();
exit(0);
}

int main(int argc, char **argv)
{
pixels = new color[width * height];
memset(pixels, 0x0, sizeof(color) * width * height);
buffer = new color [width * height];
memset(buffer, 0x0, sizeof(color) * width * height);
struct sigaction sigIntHandler;
sigIntHandler.sa_handler = handler;
sigIntHandler.sa_flags = 0;
sigaction(SIGINT, &sigIntHandler, NULL);
static const float gamma = 1 / 2.2;
while (1) {
for (int y = 0, pixIndex = 0; y < height; ++y) {
short row = (short)(y / squareSize);
for (int x = 0; x < width; ++x, pixIndex++) {
short column = (short)(x / squareSize);
// compute each pixel of the square
float X = 0, Y = 0, Z = 0;
monteCarloIntegration(row * 6 + column, X, Y, Z);
float r, g, b;
XYZtoRGB(X, Y, Z, r, g, b);
buffer[pixIndex][0] += r;
buffer[pixIndex][1] += g;
buffer[pixIndex][2] += b;
}
}
npasses++;
for (uint32_t i = 0; i < width * height; ++i) {
pixels[i][0] = powf(buffer[i][0] / npasses, gamma);
pixels[i][1] = powf(buffer[i][1] / npasses, gamma);
pixels[i][2] = powf(buffer[i][2] / npasses, gamma);
}
printf("npasses %3d (num samples %5d)\n", npasses, npasses * N);
}
delete [] pixels;
delete [] buffer;
return 0;
}

Let's look at some results. The following sequence of images shows the result after 1, 8, 16, and 32 passes (this corresponds to 32, 256, 512, and 1024 samples respectively).

Obviously, as expected the noise is reduced as the number of passes (i.e. the number of samples) increases. By zooming out on a section of these images, we can see the noise better. Keep in mind that in Monte Carlo, you need 4 times as many samples to reduce the noise (or variance) by 2.

As suggested already a couple of times throughout this lesson, the art of Monte Carlo rendering is mostly about finding ways of reducing this noise. We will talk about variance-reduction techniques later in this lesson.

## Conclusion

Hopefully, with this exercise, you better understand what Monte Carlo integration is, which as you can see is not only a simple idea but also really straightforward to implement. You also have a simple program that demonstrates the concept of progressive rendering (which is only possible because of the nature of the Monte Carlo algorithm). You hopefully now have a better understanding of where the noise (or variance if you use the correct technical term) comes from in your renders. And if you see this noise in our 3D renders, you can be sure that the renderer that produced this image, uses some sort of Monte Carlo integration.

## Exercice

Adapt the program to display the image each time a new pass is added in an OpenGL window.

Adapt the program to stop the render after a given time.


text:
The management of Sharrar Corporation would like to
investigate the possibility of basing its predetermined
overhead rate on activity at capacity rather than on the
estimated amount of activity for the year. The company’s
controller has provided an example to illustrate how this
new system would wo...


text:
# One of the parents of a cross has a mutation in its mitochondiria. In that cross, that parent is taken as a male. During segregation of F_(2) progen

11 views
in Biology
One of the parents of a cross has a mutation in its mitochondiria. In that cross, that parent is taken as a male. During segregation of F_(2) progenies that mutation is found in
A. fifty per cent of the progenies
B. one -third of the proggenies
C. none of the progenies
D. all the progenies

by (120k points)
Correct Answer - C


text:
• ### Study of fast neutron detector for COSINE-100 experiment(1805.03381)

May 9, 2018 hep-ex, physics.ins-det
A monitoring system for fast neutrons is planned in the COSINE experiment, a dark matter experiment with NaI crystals. We pursued several R&D approaches for a neutron detector using a liquid scintillator (LS). A pulse shape discrimination (PSD) technique is used for the identification of neutron events and the PSD properties of two different LS were compared. A good separation power between neutrons and $\gamma$ has been achieved for energies between 200 keVee to 1500 keVee. The combination of alumina adsorption, filtration, and water extraction is effective in purifying the LS, which leads to a reduction in the $\alpha$ contamination by $^{210}$Po of more than a factor of two. The measured activities of the internal $\alpha$ are 0.36$\pm$0.04~mBq/kg and 0.21$\pm$0.03~mBq/kg before and after purification, respectively.
• ### Background model for the NaI(Tl) crystals in COSINE-100(1804.05167)

April 14, 2018 physics.ins-det, astro-ph.IM
The COSINE-100 dark matter search experiment is an array of NaI(Tl) crystal detectors located in the Yangyang Underground Laboratory (Y2L). To understand measured backgrounds in the NaI(Tl) crystals we have performed Monte Carlo simulations using the Geant4 toolkit and developed background models for each crystal that consider contributions from both internal and external sources, including cosmogenic nuclides. The background models are based on comparisons of measurement data with Monte Carlo simulations that are guided by a campaign of material assays and are used to evaluate backgrounds and identify their sources. The average background level for the six crystals (70 kg total mass) that are studied is 3.5 counts/day/keV/kg in the (2-6) keV energy interval. The dominant contributors in this energy region are found to be $^{210}$Pb and $^3$H.
• ### Initial Performance of the COSINE-100 Experiment(1710.05299)

COSINE is a dark matter search experiment based on an array of low background NaI(Tl) crystals located at the Yangyang underground laboratory. The assembly of COSINE-100 was completed in the summer of 2016 and the detector is currently collecting physics quality data aimed at reproducing the DAMA/LIBRA experiment that reported an annual modulation signal. Stable operation has been achieved and will continue for at least two years. Here, we describe the design of COSINE-100, including the shielding arrangement, the configuration of the NaI(Tl) crystal detection elements, the veto systems, and the associated operational systems, and we show the current performance of the experiment.
• ### Muon detector for the COSINE-100 experiment(1712.02011)

The COSINE-100 dark matter search experiment has started taking physics data with the goal of performing an independent measurement of the annual modulation signal observed by DAMA/LIBRA. A muon detector was constructed by using plastic scintillator panels in the outermost layer of the shield surrounding the COSINE-100 detector. It is used to detect cosmic ray muons in order to understand the impact of the muon annual modulation on dark matter analysis. Assembly and initial performance test of each module have been performed at a ground laboratory. The installation of the detector in Yangyang Underground Laboratory (Y2L) was completed in the summer of 2016. Using three months of data, the muon underground flux was measured to be 328 $\pm$ 1(stat.)$\pm$ 10(syst.) muons/m$^2$/day. In this report, the assembly of the muon detector and the results from the analysis are presented.
• ### Understanding NaI(Tl) crystal background for dark matter searches(1703.01982)

June 21, 2017 physics.ins-det, astro-ph.IM
We have developed ultra-low-background NaI(Tl) crystals to reproduce the DAMA results with the ultimate goal of achieving purity levels that are comparable to or better than those of the DAMA/LIBRA crystals. Even though the achieved background level does not approach that of DAMA/LIBRA, it is crucial to have a quantitative understanding of the backgrounds. We have studied background simulations toward a deeper understanding of the backgrounds and developed background models for a 9.16-kg NaI(Tl) crystal used in the test arrangement. In this paper we describe the contributions of background sources quantitatively by performing Geant4 Monte Carlo simulations that are fitted to the measured data to quantify the unknown fractions of the background compositions. In the fitted results, the overall simulated background spectrum well describes the measured data with a 9.16-kg NaI(Tl) crystal and shows that the background sources are dominated by surface $^{210}$Pb and internal $^{40}$K in the 2 to 6-keV energy interval, which produce 2.4 counts/day/keV/kg (dru) and 0.5 dru, respectively.
• ### Performance of a prototype active veto system using liquid scintillator for a dark matter search experiment(1701.04514)

Jan. 17, 2017 physics.ins-det, astro-ph.IM
We report the performance of an active veto system using a liquid scintillator with NaI(Tl) crystals for use in a dark matter search experiment. When a NaI(Tl) crystal is immersed in the prototype detector, the detector tags 48% of the internal K-40 background in the 0-10 keV energy region. We also determined the tagging efficiency for events at 6-20 keV as 26.5 +/- 1.7% of the total events, which corresponds to 0.76 +/- 0.04 events/keV/kg/day. According to a simulation, approximately 60% of the background events from U, Th, and K radioisotopes in photomultiplier tubes are tagged at energies of 0-10 keV. Full shielding with a 40-cm-thick liquid scintillator can increase the tagging efficiency for both the internal K-40 and external background to approximately 80%.
• ### Understanding internal backgrounds of NaI(Tl) crystals toward a 200~kg array for the KIMS-NaI experiment(1510.04519)

The Korea Invisible Mass Search (KIMS) collaboration has developed low-background NaI(Tl) crystals that are suitable for the direct detection of WIMP dark matter. With experience built on the KIMS-CsI programs, the KIMS-NaI experiment will consist of a 200~kg NaI(Tl) crystal array surrounded by layers of shielding structures and will be operated at the Yangyang underground laboratory. The goal is to provide an unambiguous test of the DAMA/LIBRA's annual modulation signature. Measurements of six prototype crystals show progress in the reduction of internal contaminations of radioisotopes. Based on our understanding of these measurements, we expect to achieve a background level in the final detector configuration that is less than 1~count/day/keV/kg for recoil energies around 2~keV. The annual modulation sensitivity for the KIMS-NaI experiment shows that an unambiguous 7$\sigma$ test of the DAMA/LIBRA signature would be possible with a 600~kg$\cdot$year exposure with this system.
• ### Pulse-shape discrimination between electron and nuclear recoils in a NaI(Tl) crystal(1503.05253)

We report on the response of a high light-output NaI(Tl) crystal to nuclear recoils induced by neutrons from an Am-Be source and compare the results with the response to electron recoils produced by Compton scattered 662 keV $\gamma$-rays from a $^{137}$Cs source. The measured pulse-shape discrimination (PSD) power of the NaI(Tl) crystal is found to be significantly improved because of the high light output of the NaI(Tl) detector. We quantify the PSD power with a quality factor and estimate the sensitivity to the interaction rate for weakly interacting massive particles (WIMPs) with nucleons, and the result is compared with the annual modulation amplitude observed by the DAMA/LIBRA experiment. The sensitivity to spin-independent WIMP-nucleon interactions based on 100 kg$\cdot$year of data from NaI detectors is estimated with simulated experiments, using the standard halo model.


text:
# sklearn.preprocessing.RobustScaler¶

class sklearn.preprocessing.RobustScaler(*, with_centering=True, with_scaling=True, quantile_range=25.0, 75.0, copy=True, unit_variance=False)[source]

Scale features using statistics that are robust to outliers.

This Scaler removes the median and scales the data according to the quantile range (defaults to IQR: Interquartile Range). The IQR is the range between the 1st quartile (25th quantile) and the 3rd quartile (75th quantile).

Centering and scaling happen independently on each feature by computing the relevant statistics on the samples in the training set. Median and interquartile range are then stored to be used on later data using the transform method.

Standardization of a dataset is a common requirement for many machine learning estimators. Typically this is done by removing the mean and scaling to unit variance. However, outliers can often influence the sample mean / variance in a negative way. In such cases, the median and the interquartile range often give better results.

New in version 0.17.

Read more in the User Guide.

Parameters
with_centeringbool, default=True

If True, center the data before scaling. This will cause transform to raise an exception when attempted on sparse matrices, because centering them entails building a dense matrix which in common use cases is likely to be too large to fit in memory.

with_scalingbool, default=True

If True, scale the data to interquartile range.

quantile_rangetuple (q_min, q_max), 0.0 < q_min < q_max < 100.0, default=(25.0, 75.0), == (1st quantile, 3rd quantile), == IQR

Quantile range used to calculate scale_.

New in version 0.18.

copybool, default=True

If False, try to avoid a copy and do inplace scaling instead. This is not guaranteed to always work inplace; e.g. if the data is not a NumPy array or scipy.sparse CSR matrix, a copy may still be returned.

unit_variancebool, default=False

If True, scale data so that normally distributed features have a variance of 1. In general, if the difference between the x-values of q_max and q_min for a standard normal distribution is greater than 1, the dataset will be scaled down. If less than 1, the dataset will be scaled up.

New in version 0.24.

Attributes
center_array of floats

The median value for each feature in the training set.

scale_array of floats

The (scaled) interquartile range for each feature in the training set.

New in version 0.17: scale_ attribute.

robust_scale

Equivalent function without the estimator API.

PCA

Further removes the linear correlation across features with ‘whiten=True’.

Notes

For a comparison of the different scalers, transformers, and normalizers, see examples/preprocessing/plot_all_scaling.py.

Examples

>>> from sklearn.preprocessing import RobustScaler
>>> X = [[ 1., -2.,  2.],
...      [ -2.,  1.,  3.],
...      [ 4.,  1., -2.]]
>>> transformer = RobustScaler().fit(X)
>>> transformer
RobustScaler()
>>> transformer.transform(X)
array([[ 0. , -2. ,  0. ],
[-1. ,  0. ,  0.4],
[ 1. ,  0. , -1.6]])

Methods

 fit(X[, y]) Compute the median and quantiles to be used for scaling. fit_transform(X[, y]) Fit to data, then transform it. get_params([deep]) Get parameters for this estimator. Scale back the data to the original representation set_params(**params) Set the parameters of this estimator. Center and scale the data.
fit(X, y=None)[source]

Compute the median and quantiles to be used for scaling.

Parameters
X{array-like, sparse matrix} of shape (n_samples, n_features)

The data used to compute the median and quantiles used for later scaling along the features axis.

yNone

Ignored.

Returns
selfobject

Fitted scaler.

fit_transform(X, y=None, **fit_params)[source]

Fit to data, then transform it.

Fits transformer to X and y with optional parameters fit_params and returns a transformed version of X.

Parameters
Xarray-like of shape (n_samples, n_features)

Input samples.

yarray-like of shape (n_samples,) or (n_samples, n_outputs), default=None

Target values (None for unsupervised transformations).

**fit_paramsdict

Returns
X_newndarray array of shape (n_samples, n_features_new)

Transformed array.

get_params(deep=True)[source]

Get parameters for this estimator.

Parameters
deepbool, default=True

If True, will return the parameters for this estimator and contained subobjects that are estimators.

Returns
paramsdict

Parameter names mapped to their values.

inverse_transform(X)[source]

Scale back the data to the original representation

Parameters
X{array-like, sparse matrix} of shape (n_samples, n_features)

The rescaled data to be transformed back.

Returns
X_tr{ndarray, sparse matrix} of shape (n_samples, n_features)

Transformed array.

set_params(**params)[source]

Set the parameters of this estimator.

The method works on simple estimators as well as on nested objects (such as Pipeline). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.

Parameters
**paramsdict

Estimator parameters.

Returns
selfestimator instance

Estimator instance.

transform(X)[source]

Center and scale the data.

Parameters
X{array-like, sparse matrix} of shape (n_samples, n_features)

The data used to scale along the specified axis.

Returns
X_tr{ndarray, sparse matrix} of shape (n_samples, n_features)

Transformed array.


text:
# Inversions of Insertion Sort and Bubble Sort

An array with bubblesort time $$\Theta(n)$$ is nothing but a sorted array like: A = 1 2 3 4 5
No swaps are done so only $$n - 1$$ comparisons.
An array with insertionsort run time $$\Theta(n^2)$$ is a reverse sorted array: A = 5 4 3 2 1
Each element is swapped in every iteration.
There cannot be any array that can have IS time as $$\Theta(n^2)$$ and BS time as $$\Theta(n)$$.
But how can i prove this in terms of their inversions?

Let's consider the swaps performed by insertion sort. In the $$i$$-th iteration, insertion sort repeatedly swaps the element $$x$$ originally in $$A[i]$$ with some element $$y$$ originally in $$A[1:i-1]$$ until the subarray $$A[1:i]$$ is sorted. Then, since we must have $$x for the swap to occur, we have that each swap induces an inversion in the original array. Notice that the $$i$$-th iteration takes time proportional to $$1+s_i$$ where $$s_i$$ is the number of swaps in the iteration.

If follows that insertion sort takes time $$O(n+\sum_{i=1}^n s_i) = O(n + s)$$, where $$s$$ is the number of inversions of $$A$$. Thus, when insertion sort takes time $$\Omega(n^2)$$ we must have $$s=\Omega(n^2)$$.

Finally, notice that at most one inversion can be removed by each swap performed by bubble sort, implying that bubble sort takes time $$\Omega(s)=\Omega(n^2)$$.


text:
More general invariance of the action functional

+ 7 like - 0 dislike
53 views

I will formulate my question in the classical case, where things are simplest.

Usually when one discusses a continuous symmetry of a theory, one means a one-parameter group of diffeomorphisms of the configuration space $M$ which fix the action functional $S:P\rightarrow \mathbb{R}$, where $P$ is the space of time evolutions, ie. differentiable paths in $M$. The idea is that, given some initial configuration $(x_0,v_0)\in TM$, there is a path in $P$ passing through $x_0$ with velocity $v_0$ and minimizing $S$ among all such paths. I will assume that this path is unique, which is almost always the case. Thus, if a diffeomorphism fixes $S$, it commutes with determining this path. One says that the physics is unchanged by taking the diffeomorphism.

Now here's the question: are there other diffeomorphisms which leave the physics unaltered? All one needs to do is ensure that the structure of the critical points of $S$ are unchanged by the diffeomorphism.

I'll be more particular. Write $P_{x_0,v_0}$ as the set of paths in $M$ passing through $x_0$ with velocity $v_0$. A diffeomorphism $\phi:M\rightarrow M$ is a symmetry of the theory $S:P\rightarrow \mathbb{R}$ iff for each $(x_0,v_0) \in TM$, $\gamma \in P_{x_0,v_0}$ is a critical point of $S|_{P_{x_0,v_0}}$ iff $\phi \circ \gamma$ is a critical point of $S|_{P_{\phi (x_0),\phi^* (v_0)}}$.

It is not obvious to me that this implies $S = S \circ \phi^{-1}$, where $\phi^{-1}$ is the induced map by postcomposition on $P$. If there are such symmetries, what can we say about Noether's theorem?

A perhaps analogous situation in the Hamiltonian formalism is in the correspondence between Hamiltonian flows and infinitesimal canonical transformations. Here, a vector field $X$ can be shown to be an infinitesimal canonical transformation iff its contraction with the Hamiltonian 2-form is closed. This contraction can be written as $df$ for some function $f$ (and hence $X$ as the Hamiltonian flow of $f$) in general iff $H^1(M)=0$. Is this analogous? What is the connection? It's been pointed out that this obstruction does not depend on the Hamiltonian, so is likely unrelated.

Thanks!

PS. If someone has more graffitichismo, tag away.

This post has been migrated from (A51.SE)

+ 3 like - 0 dislike

Let $\phi_s:P\rightarrow P$ be the induced diffeomorphism on the space of paths. You are assuming that the zero set $Zero(dS)$ coincides with the zero set $Zero(\phi_s^* dS)$. This does not even imply that $dS = \phi_s^* dS$, let alone $S = \phi_s^* S$.

An example would be a free particle on $\mathbf{R}$. Let $S=\int \dot{x}(t)^2dt$ and consider the scaling transformation $x\rightarrow\exp(s) x$. Then the critical points are straight lines $x(t)=x_0+v_0t$ and the transformation clearly preserves them. On the other hand, the action gets multiplied by $\exp(2s)$.

To understand the differences between $Zero(dS)=Zero(\phi_s^* dS)$ and $dS=\phi_s^* dS$, consider the graph of $dS$ in $T^*P$. The first condition only fixes the intersection points with the zero section, while the second condition fixes the graph itself. Clearly, in the $C^\infty$ world you can adjust the behavior of $dS$ away from the intersection points as much as you like. In the holomorphic world it would be enough to remember the Taylor expansions around the critical points.

Finally, $dS=\phi_s^*dS$ does not imply that $S=\phi_s^*S$: you only know that $S=\phi_s^*S + c(s)$, where $c$ is a locally-constant function on $P$ which vanishes at $s=0$.

This post has been migrated from (A51.SE)
answered Dec 30, 2011 by (1,120 points)
Thanks for the simple counterexample.

This post has been migrated from (A51.SE)
+ 2 like - 0 dislike

Firstly, its enough for the variation of the action to be a total divergence (in the more general field theory case), i.e. in the case of mechanics - a time derivative. The classic example would be boost symmetry - transitions between frames of reference. Only problem it doesn't quite fit your framework since it depends explicitly on the time coordinate

Secondly, it's enough for this to hold on-shell, i.e. when the equations of motion are satisfied. In the field theory case the classic example for this is supersymmetry. Probably a mechanical (1-dimensional) analogue exists. However, this example lives in the slightly more general world of supermanifolds. Of course it's possible to construct artificial examples of this kind which fit your setting precisely - you can tweak the action functional almost any way you like away from critical points (just take care to avoid creating new critical points)

Thirdly, as the examples above show the statement "usually when one discusses a continuous symmetry of a theory, one means a one-parameter group of diffeomorphisms of the configuration space..." is not correct. Instead we can consider any time-local transformation on the history space. Btw, this entire discussion is equally relevant to discrete symmetries. Also one often considers multi-parameter groups but this is already semantics

I don't think your Hamiltonian analogy is correct since my examples above don't involve any topological obstructions. Btw, an example of a flow which is symplectic but not Hamiltonian is the time evolution of a particle on a circle under the influence of a constant force driving it e.g. clockwise everywhere, which is a system without Lagrangian formulation

This post has been migrated from (A51.SE)
answered Dec 30, 2011 by (1,725 points)
Ah, thanks. You've clarified quite a few of my thoughts. I wanted to focus on continuous symmetries because underlying all this I was thinking about its ramifications for Noether's theorem. Could you clarify what you mean by "time-local transformation on the history space"? I'm thinking some sort of automorphism of the space of possible on-shell worldlines, but I'm not sure what structure you'd want to preserve for each line.

This post has been migrated from (A51.SE)
Time-local means the result of the transformation at each point of time depends only on a small time-neighbourhood. This includes all transformations which can be expressed using time itself, configuration space coordinates and a finite number of derivatives of the configuration space coordinates, which is a popular definition of locality in physics texts. However, I believe the correct mathematical definition is the following. A "time local" transformation is a smooth automorphism of the _sheaf_ of (off-shell) histories. Some technicalities are involved in defining smoothness for infinite dim

This post has been migrated from (A51.SE)
Okay, that makes some sense. Now, returning to the (simplistic) formulation I had above, supposing the existence of a symmetry which preserves the critical structure of the action but not necessarily the action, what can we say about any conserved quantities arising therefrom?

This post has been migrated from (A51.SE)
If your transformation does not even preserve $dS$, there is no hope of getting any conservation laws.

This post has been migrated from (A51.SE)

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# Why is the variance of this random variable one?

I stumbled across the following problem in the paper "STOCHASTIC SPANNING TREE PROBLEM" of Ishii. There is a random variable defined by $$\frac{\sum_j (c_j x_j - \mu_jx_j)}{(\sum_j\sigma_j^2x_j^2)^{1/2}},$$ where each $$c_j$$ is distributed according to a normal distribution with mean $$\mu_i$$ and variance $$\sigma_j^2$$. It is said that this variable is distributed according to the standard normal distribution and I try to proof this for myself.

For the expected value I get zero, just by using its linearity and then the sums in the numerator cancel out. Now to the variance: I try to use the property that $$\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$$ where I already know that the second term is zero. Then I square the fraction and aim to get a nominator that cancels out to one with the denominator. Since the denominator is constant I drag it out of the expected value and get $$\frac{\mathbb{E} \left[ \left( \sum_j (c_j x_j) - \sum_j(\mu_jx_j) \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}$$ which I expand to $$\frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 -2 \left( \sum_j c_j x_j \right)\left( \sum_j \mu_j x_j \right) + \left(\sum_j\mu_jx_j \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}$$ where I again use the linearity of the expected value to get $$\frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 \right] - \left(\sum_j\mu_jx_j \right)^2}{\sum_j\sigma_j^2x_j^2} = \frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 - \left(\sum_j\mu_jx_j \right)^2\right]}{\sum_j\sigma_j^2x_j^2}$$ and now I am stuck and don't know what I miss.

I hope somebody can bring ligth into the dark.

T3 K14

• You need to use independence (or at least pairwise independence) of the $c_j$s. I assume they are independent? (If so, what you want is that variancenif sum = sum of variances when the covariance is 0) – Clement C. Jan 28 at 22:21

Independence of $$c_j$$ is needed.
$$(\sum_jc_jx_j)^2=\sum_jx_j^2c_j^2+\sum_j\sum_{i\ne j}x_ix_jc_ic_j$$
$$(\sum_j\mu_jx_j)^2=\sum_jx_j^2\mu_j^2+\sum_j\sum_{i\ne j}x_ix_j\mu_i\mu_j$$
For $$i\ne j$$, $$E(c_ic_j)=\mu_i\mu_j$$ (independence) so double sum over $$i,j$$ is zero.
$$E(x_j^2(c_j^2-\mu_j^2))=x_j^2(E(c_j^2)-\mu^2)=x_j^2\sigma_j^2$$, so sum over $$j$$ in the numerator is identical to the denominator.


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# Group Operation of points on a Montgomery elliptic Curve (project coordinates)

I was trying to implement the double and addition formulas for elliptic points on a Montgomery elliptic curve. I came across this weird thing which should definitely not be happening.

I took a point $P_1$, and doubled it twice. This should be $4P_1$. Let's call this point $P_4'$.

Next I doubled $P_1$ to get $P_2$. Then carried out $P_1 + P_2$ to get $P_3$, and then again added $P_3$ and $P_1$ to get $P_4$.

Now I was expecting both these points to be the same, since both are simply $4$ times $P_1$.

For example, let us take a curve $y^2 = x^3 + 6x^2 + x$ (Montgomery Curve), over $F_{2425967623052370772757633156976982469681}$

Let a point $P_1$ be $[2 :: 1]$ (projective coordinates, $y$ dropped)

Doubling this twice, gives the point $[339112225 :: 126909216]$ ($P_4'$).

Adding it stepwise to reach $4P_1$, gives the point $[106258781030400 :: 39766241378304]$

Question: Shouldn't $P_4$ be equal to $P_4'$ ? Since both are equal to $4P_1$?

I am using the following formulas for addition and doubling.

$$X_{m+n} = (z_{m-n}\cdot((x_m - z_m)\cdot(x_n + z_n) + (x_m + z_m)\cdot(x_n - z_n))^2 \mod{n})\\ Z_{m+n} = (x_{m-n}\cdot((x_m - z_m)\cdot(x_n + z_n) - (x_m + z_m)\cdot(x_n - z_n))^2 \mod{n})\\ X_{2n} = (((x_n + z_n)^2)\cdot((x_n - z_n)^2) \mod{n})\\ Z_{2n} = ((4\cdot x_n\cdot z_n)\cdot((x_n - z_n)^2 + {a+2\over4}\cdot 4\cdot x_n\cdot z_n) \mod{n})$$

These formulas were mentioned here http://en.wikipedia.org/wiki/Montgomery_curve

I have implemented a python version of the addition and double, and carried out the above arithmetic in this piece of code.

http://ideone.com/w68cKC

• Note that you can't have such point in the given field, as $2^3+6\cdot 2^2+2=34$ is a quadratic non-residue (WA) given modulo. If this matters, though. – Alexey Burdin Jun 5 '15 at 15:49
• Also note that $\dfrac{106258781030400}{39766241378304}$ $= \dfrac{339112225}{126909216}$ i.e. it's exactly the same point. – Alexey Burdin Jun 5 '15 at 15:55
• It was really silly of me not to notice that. I had been stuck on it for quite some time now, thanks a lot! – Kushagra Singh Jun 5 '15 at 15:58


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# Hydrogen Atom Radial Wave Function (limit problem)

1. Dec 4, 2009

### Master J

The Hydrogen Atom wave function.

With the substitution u(r) = r.R(r)

p=kr

We get a simplified version: d^2u/dp^2 = [1 - (p_0)/p + l(l + 1)/(p^2) ].u

Im sure some of you have seen that before.

Now, in the limit, p goes to infinity, I understand that we get u = A.exp[-p], but in the limit that p goes to 0, the answer is:

d^2u/dp^2 = [l(l + 1)/(p^2)].u

I dont get where that comes from. Surely the terms with p at the bottom explode and become infinite??

Last edited: Dec 4, 2009
2. Dec 4, 2009

### Bob_for_short

Yes, it is the biggest term in the square bracket. It should be multiplied by u too.

3. Dec 4, 2009

### Master J

But I dont get it. Why do the 2 terms with p at the bottom not become infinity???

4. Dec 4, 2009

### George Jones

Staff Emeritus
As p becomes small, the two terms become large at different rates, with the second term getting large faster. So, for small enough (but non-zero) p, the second term will be much larger than the first, and the first term can be ignored.

As a simple example, consider

A = 1/x + 1/x^2

with x = 0.001. What does A equal? What does 1/x equal? What does 1/x^2 equal? Is 1/x^2 a good approximation to A?

5. Dec 6, 2009

### Master J

What happens to the one then?

Should the answer not be [1 +l(l + 1)/(p^2)].u

6. Dec 6, 2009

### George Jones

Staff Emeritus
Again, if p is small enough (but still nonzero), l(l + 1)/(p^2) will be much larger than one, so the one can be ignored.

7. Dec 7, 2009

### Master J

Thanks for the help, it is very much appreciated!!!

I figured out the last question anyway.

Now......sums!!!! Yay!

So its still the same ball game. 2 solutions have been found for the 2 limiting cases, as p goes to zero and infinity. Using these 2 regimes, we assume a solution that is the product of these to aforementioned solutions, and another function of p, which is to be determined, call it v(p) (while Im here, is this a general rule that one can assume that type of solution??)

Assuming a series solution:

v' = $$\Sigma$$ j.(c_j).(p)^(j-1)

the book equates this to $$\Sigma$$ (j+1).(c_j+1).(p)^j

where both sums are from zero to infinity, index of summation is j, and c is the coefficient. Im unsure about how the above equality is made, and why??

Then, we end up with a recursive formula for the coeffecients. But if this expansion is compatable with the previous asymptotic analsis, it must terminate after a finite j value, j_max. Why is this?

I understand there is alot there, any help is appreciated!

Last edited: Dec 7, 2009
8. Dec 7, 2009

### RedX

Is it a general rule to assume that type of solution? Well, it's mathematically legal. Usually what happens in these type of problems is that the asymptotic solution has two solutions, one of which blows up somewhere, e.g., $$e^{\pm r}$$ for the hydrogen atom. So you want to choose - and not + by writing $$R(r)=U(r)e^{-r}$$. When you solve U(r) in the new differential eqn by power series, you find that the solution of U(r) in general behaves like $$e^{2r}$$ at large r, which gives you back $$R(r)=e^{2r}e^{-r}=e^{r}$$, so the solution blows up for large r. This is to be expected in general, because you already established $$e^r$$ was a valid solution to the differential equation at large r. However, there are certain choices of the energy that shuts down the power series by making it a finite series, so that U(r) would be polynomial and not behave like an exponential at large r. This quantizes the energy.

The very same type of thing happens in the harmonic oscillator. You get two asymptotic solutions, one of which that blows up. You plug in the product of the one that doesn't blow up with an arbitrary function into the differential equation to get a new equation. In general, the solution to the new equation when solved in power series, multiplies with the well-behaved asymptotic solution to give the mis-behaved asymptotic solution. But certain values of energies cuts of the power series, so you avoid all this, and energy gets quantized.


text:
# Bulk Modulus calculation using Quantum ESPRESSO

### Tutorial

Quantum ESPRESSO has a package called ev.x that can be used to calculate the bulk modulus of a material. ev.x basically offers calculation of 4 equation of states, namely: These equation of states give the equilibrium volume (lattice constant in case of cubic) as well as the bulk modulus and it’s first derivative.

The procedure is pretty simple. You first need to calculate the energy of the system at different volumes of unit cell. So you would basically need to run several SCF calculations at different values of lattice parameters (thereby different volumes) and note down the corresponding energies in a text file. For cubic systems you can have the lattice parameter in the first column and the Energy in the second column, while for non-cubic systems, you’d note down the volumes in the first column.

A typical file would look like this:

Let’s say this text file is called: volvsE.txt

Now, we can use this to get the Murnaghan [1] equation of state which is very popular among the scientific community.
Murnaghan EOS is given by:

$\boxed{E(V)=E(V_0)+\frac{B_0V}{B'_0}\left[\frac{\left(V_0/V\right)^{B_0'}}{B'_0-1}+1\right]-\frac{B_0V_0}{B_0'-1} }$

Now you just need to run ev.x from the command-line (terminal) and supply inputs to the following prompts:
 [email protected]:~$/home/manas/qe-6.1/bin/ev.x Lattice parameter or Volume are in (au, Ang) > Angstrom Assuming Angstrom Enter type of bravais lattice (fcc, bcc, sc, noncubic) > sc Enter type of equation of state : 1=birch1, 2=birch2, 3=keane, 4=murnaghan > 4 Input file > volvsE.txt Output file > eos  Once, the calculation is done you should have file (eos.txt) that looks like this: This contains information of the equilibrium volume, bulk modulus and it’s derivative and the minimum energy. You can use the above generated data to create a neat plot of the equation of state (as you’d see in many publications), using gnuplot. The following shell script, plots the calculated data-points from file volvsE.txt as well as the fitted equation of state curve. [NOTE: This shell script is for cubic systems only. You’d need to modify the code to implement it for non-cubic systems] Create a file called script.sh and add the following code to it: #!/bin/bash a="$1"
k="$2" dk="$3"
emin="$4" V=$(echo "$a*$a*$a"|bc) echo " set terminal png size 1000,500 set output 'EOSplot.png' set xlabel 'Volume (Ang^3)' set ylabel 'Energy (Ry)' set title 'Equation of state' #set key box linestyle 1 f(x)= "$emin"+(x*10**(-30)*"$k"*10**9/"$dk"*(("$V"/x)**"$dk"/("$dk"-1)+1)-"$k"*10**9*"$V"*10**(-30)/("$dk"-1))/(1.6*10**(-19)*13.6)
plot f(x) w l title 'Murnaghan EOS fit', 'volvsE.txt' u 1:2 w points pointtype 7 pointsize 2 title 'Calculated data-points'
set terminal postscript enhanced color solid 22
set output 'EOSplot.eps'
set xlabel 'Volume (Ang^3)'
set ylabel 'Energy (Ry)'
set title 'Equaiton of state'
set autoscale
plot f(x) w l title 'Murnaghan EOS fit', 'volvsE.txt' u 1:2 w points pointtype 7 pointsize 2 title 'Calculated data-points'
set term x11" >EOSplotScript.p

Now save the file and open the terminal and run the command:
chmod u+x script.sh

Now simply run the script by providing the four arguments, in the following order: lattice parameter, bulk modulus, derivative of BM, and minimum energy.

Example: ./script.sh 5.5 81 3.9 -588.656

The above command will create a gnuplot script file called: EOSplotScript.p

This can then be run using gnuplot using the following command:
gnuplot ./EOSplotScript.p

When you run this command, it will generate a PNG image as well as an EPS file.

The output looks like this:

### References

[1] F.D. Murnaghan, Proceedings of the National Academy of Sciences 30, 244 (1944).


text:
# $\omega_1$ is $C^*$-embedded in $\omega_1+1$

consider the space $ω_1$ (the first uncountable ordinal) and $ω_1+1$ together with the order topology. is it true that

1. $ω_1$ is $G_\delta$-dense subspace of $ω_1+1$?
2. $\omega_1$ is $C^*$-embedded in $\omega_1+1$?

If both proposition are correct, then we conclude that $\omega_1$ is $C$-embedded in $\omega_1+1$.

thanks.

-

Every continuous $f:\omega_1\to\Bbb R$ is eventually constant, so $\omega_1$ is indeed $C$-embedded in $\omega_1+1$; you’ll find a proof of the first statement here in Dan Ma’s Topology Blog.
1. Yes, this is true, since any $G_\delta$ in $\omega_1+1$ containing the point $\omega_1$ contains an interval $(\alpha,\omega_1]$ for some $\alpha<\omega_1$.
2. This is also true; in fact, $\omega_1+1=\beta\omega_1$.
eventually constant means there exist a $a\in \omega_1$ such that $f$ is constant on $(a,\omega_1)$ ? – TXC Feb 12 '13 at 7:35
@TXC: Yes, exactly. And the extension that assigns $\omega_1$ that constant value is clearly continuous. – Brian M. Scott Feb 12 '13 at 7:36
with use this property "every continuous $f:ω_1→R$ is eventually constant" i cant show that $f$ can be countinuously extended to $\omega_1+1$. for example, $g(\omega_1)=1$ and $g(x)=c$ for $x\in \omega_1$ ($c$ is constant value), can be continuous extension of $f$? – TXC Feb 12 '13 at 13:24
@TXC: If $f: \omega_1 \rightarrow \mathbb{R}$ is continuous, there exists $\alpha < \omega_1$ and a constant $c \in \mathbb{R}$ such that $f(\beta) = c$ for all $\beta >= \alpha$. Now define $\bar{f}: \rightarrow \mathbb{R}$ by $\bar{f}(\alpha) = f(\alpha)$ for $\alpha < \omega_1$ and $\bar{f}(\omega_1) = c$. Then $\bar{f}$ extends $f$ and is continuous at $\omega_1$ too, as every neighbourhood of it contains one that is contained in $[\alpha,\rightarrow)$. So the extension is of course unique and uses the same $c$ as promised in the eventually constant fact. – Henno Brandsma Feb 12 '13 at 14:47
Is it true that $\omega_1$ is only proper dense subspace of $\omega_1+1$? – TXC Feb 13 '13 at 6:13


text:
# Ques. -8

Total number of functions $$f:\{1,2,3,4,5\} \to \{1,2,3,4,5 \}$$ that are onto and $$f(i) \neq i$$ is equal to :

×


text:
# How does Gradescope handle late assignment submissions?

Updated

When creating a student-uploaded assignment or programming assignment, you can check “allow late submissions” and set the due date as well as the “late due date.”

If you allow late submissions, students will see both the “due date” as well as the “late due date” for the assignment in their Gradescope dashboard. (Submitting an assignment)

If you choose not to allow late submissions, students will only see the due date. Instructors and TAs are able to upload submissions on behalf of students after the deadline from the Manage Submissions page by clicking “Upload submission” in the bottom action bar.

Time late will be displayed for late assignments on the grading interface (in days and hours) and on the Download Grades spreadsheet (in HH:MM:SS format). (Exporting grades)