text:
# Annuity problem viewed as life insurance in actuarial science

You are given $q_x=0.053$, $q_{x+1}=0.054$, $q_{x+2}=0.055$, $i_0=0.06$, $i_1=0.08$, $i_2=0.10$. If $\mathbf{b}=(1,2,3)$, find the vector $\mathbf{c=b*w_x}$

$(w_x)_1=v(0,1) q_x=(1.06)^{-1}(0.053)=0.05$

$(w_x)_2=v(1,2) q_{x+1}=(1.06)^{-1}(1.08)^{-1}(0.054)=0.0471$

$(w_x)_2=v(2,3) q_{x+2}=(1.06)^{-1}(1.08)^{-1}(1.10)^{-1}(0.055)=0.0436$

$\mathbf{c=b*w_x}=(1,2,3)*(0.05, 0.0471,0.0436)=(0.05,0.942,0..1308)$

But the answer is $(0.05, 0.1, 0.15)$

I think your mistake is in the discount factors $v(t,t+1)$. You should have \begin{align*} v(1,2) &= 1.08^{-1}\\ v(2,3) &= 1.1^{-1}. \end{align*} If you re-do the calculations using these definitions, you should get the right answer.
Note that in general, $v(s,t)$ denotes the discount factor from time $t$ back to time $s$. By writing $1.06^{-1}1.08^{-1}$, what you get is $v(0,2)$ (instead of $v(1,2)$, which is what you want).


text:
# Biber remove certain fields from citation

How can I remove certain fields from the citation in Biber?

My example document:

\documentclass[11pt]{article}
\usepackage[backend=biber,firstinits=true,citestyle=ieee]{biblatex}

\begin{document}

Dummy text \cite{Zaharia2012}.

\printbibliography
\end{document}

With an example library example_library.bib of:

@article{Zaharia2012,
abstract = {We present Resilient Distributed Datasets (RDDs), a distributed memory abstraction that lets programmers perform in-memory computations on large clusters in a fault-tolerant manner. RDDs are motivated by two types of applications that current computing frameworks handle inefficiently: iterative algorithms and interactive data mining tools. In both cases, keeping data in memory can improve performance by an order of magnitude. To achieve fault tolerance efficiently, RDDs provide a restricted form of shared memory, based on coarse-grained transformations rather than fine-grained updates to shared state. However, we show that RDDs are expressive enough to capture a wide class of computations, including recent specialized programming models for iterative jobs, such as Pregel, and new applications that these models do not capture. We have implemented RDDs in a system called Spark, which we evaluate through a variety of user applications and benchmarks.},
archivePrefix = {arXiv},
arxivId = {EECS-2011-82},
author = {Zaharia, Matei and Chowdhury, Mosharaf and Das, Tathagata and Dave, Ankur},
doi = {10.1111/j.1095-8649.2005.00662.x},
eprint = {EECS-2011-82},
isbn = {978-931971-92-8},
issn = {00221112},
journal = {NSDI'12 Proceedings of the 9th USENIX conference on Networked Systems Design and Implementation},
pmid = {2011},
title = {{Resilient distributed datasets: A fault-tolerant abstraction for in-memory cluster computing}},
year = {2012}
}

renders as:

How can I reduce the fields and only get an Output similarly to:

[1] M. Zaharia, M. Chowdhury, T. Das, et al. "Resilient distributed datasets: A fault-tolerant abstraction for in-memory cluster computing". In: NSDI'12 Proceedings of the 9th USENIX conference on Networked Systems Design and Implementation (2012).

\DeclareSourcemap{
\maps{
\map{
\step[fieldsource=entrykey, match=\regexp{\Actan\Z}, final]
\step[fieldset=url, null]
\step[fieldset=arxivId, null]
\step[fieldset=archivePrefix, null]
}
}
}
\AtEveryCitekey{\clearfield{archivePrefix}\clearlist{issn}}
\AtEveryBibitem{\clearlist{archivePrefix, arxivId, issn}}
\AtEveryBibitem{\clearlist{url}}
\AtEveryBibitem{\clearlist{issn}}

these additional fields are not cleared.

Thanks to the first \step the sourcemap shown in the question only applies to entries with the entry key ctan. Since the key of the example entry is Zaharia2012, the mapping does not apply.

The shown \clearlist and \clearfield statements wouldn't work for a variety of reasons. archivePrefix is a convenience/legacy alias for the proper field eprinttype, so only \clearfield{eprinttype} would work. issn and url are fields not lists, so they would need to be cleared with \clearfield. \clearlist{archivePrefix, arxivId, issn} doesn't work because \clearlist accepts only one list name and not a whole list of them.

Since you are already using sourcemaps I suggest you clear all fields with a sourcemap.

Try

\documentclass{article}
\usepackage[backend=biber, giveninits=true, citestyle=ieee]{biblatex}

\DeclareSourcemap{
\maps{
\map{
\step[fieldset=doi, null]
\step[fieldset=url, null]
\step[fieldset=eprint, null]
\step[fieldset=eprinttype, null]
\step[fieldset=arxivId, null]
\step[fieldset=archivePrefix, null]
\step[fieldset=issn, null]
\step[fieldset=isbn, null]
}
}
}

\begin{filecontents}{\jobname.bib}
@article{Zaharia2012,
archivePrefix = {arXiv},
arxivId       = {EECS-2011-82},
author        = {Zaharia, Matei and Chowdhury, Mosharaf
and Das, Tathagata and Dave, Ankur},
doi           = {10.1111/j.1095-8649.2005.00662.x},
eprint        = {EECS-2011-82},
isbn          = {978-931971-92-8},
issn          = {00221112},
journal       = {NSDI'12 Proceedings of the 9th USENIX conference
on Networked Systems Design and Implementation},
pmid          = {2011},
title         = {Resilient distributed datasets},
subtitle      = {A fault-tolerant abstraction for in-memory cluster computing},
year          = {2012}
}
\end{filecontents}

\begin{document}
Dummy text \cite{Zaharia2012}.

\printbibliography
\end{document}

I didn't change this in the example, but from what I can see it looks as if the entry should be an @inproceedings and not an @article.

In this particular case where you want to get rid of URLs, DOIs, eprints and ISSN/ISBN you could also use the style options url=false, doi=false, eprint=false, isbn=false,. So it would be enough to load biblatex as

\usepackage[backend=biber, citestyle=ieee, giveninits=true,
url=false, doi=false, eprint=false, isbn=false,]{biblatex}

But those options are only available for those few fields and not for more.

• It is still rendering the full bib resource including DOI and other deleted fields for me using PDF latex. – Georg Heiler Nov 16 '19 at 13:45
• Thanks for the hint. Indeed. That should be proceedings. – Georg Heiler Nov 16 '19 at 13:46
• @GeorgHeiler Oh yeah, I forgot the DOI. Sorry. Please try again. – moewe Nov 16 '19 at 13:46
• I still see imgur.com/a/3V9hKZk the full thing - not reduced by any of the fields which are nulled out. – Georg Heiler Nov 16 '19 at 13:50
• @GeorgHeiler Did you run LaTeX, Biber, LaTeX, LaTeX after you added the new code? Did you try the code in a new file in an empty folder? – moewe Nov 16 '19 at 13:51

format: plain


text:
# Julian Schwinger Net Worth

Julian Schwinger was born on February 12, 1918 in New York City, United States, is Physicist. Julian Seymour Schwinger was an American theoretical physicist who won the Nobel Prize in Physics in 1965 for his fundamental work in quantum electrodynamics. His fascination for physics grew while going through Encyclopedia Britannica at a very young age. Very soon he began to visit public libraries and read outside books on that subject. By the age of sixteen he had published his first paper on physics. However, his penchant for higher knowledge almost ruined his college education. It was Professor Isidor Isaac Rabi of Columbia University, who first recognized his talent and took him under his wings. Under his guidance, Schwinger not only earned his B.S. degree, but also his PhD. Very soon, he established himself as a brilliant theorist. However, he was fiercely independent and did not accept interference in his work. That is why he mostly worked at night. Although he dealt in diverse branches of physics, his works on quantum electrodynamics is specially noted. However, many believe that his students were his most enduring achievements. He left more than seventy students, through whom he continued to influence modern physics long after his death.
Julian Schwinger is a member of Scientists

## Age, Biography and Wiki

 Who is it? Physicist Birth Day February 12, 1918 Birth Place New York City, United States Age 102 YEARS OLD Died On July 16, 1994(1994-07-16) (aged 76)\nLos Angeles, California, U.S. Birth Sign Pisces Alma mater City College of New York Columbia University Known for Quantum electrodynamics Spin–statistics theorem Sigma model MacMahon Master theorem List of things named after Julian Schwinger Spouse(s) Clarice Carrol (m. 1947) Awards Albert Einstein Award (1951) National Medal of Science (1964) Nobel Prize in Physics (1965) Fields Physics Institutions University of California, Berkeley Purdue University Massachusetts Institute of Technology Harvard University University of California, Los Angeles Doctoral advisor Isidor Isaac Rabi Doctoral students Roy Glauber Ben R. Mottelson Sheldon Lee Glashow Walter Kohn Bryce DeWitt Daniel Kleitman Sam Edwards Gordon Baym Lowell S. Brown Stanley Deser Lawrence Paul Horwitz Margaret G. Kivelson

## Biography/Timeline

1929

Julian Seymour Schwinger was born in New York City, to Jewish parents originally from Poland, Belle (née Rosenfeld) and Benjamin Schwinger, a garment manufacturer, who had migrated to America. Both his father and his mother's parents were prosperous clothing manufacturers, although the family Business declined after the Wall Street Crash of 1929. The family followed the Orthodox Jewish tradition. He attended Townsend Harris High School and then the City College of New York as an undergraduate before transferring to Columbia University, where he received his B.A. in 1936 and his Ph.D. (overseen by Isidor Isaac Rabi) in 1939 at the age of 21. He worked at the University of California, Berkeley (under J. Robert Oppenheimer), and was later appointed to a position at Purdue University.

1941

After having worked with Oppenheimer, Schwinger's first regular academic appointment was at Purdue University in 1941. While on leave from Purdue, he worked at the Radiation Laboratory at MIT instead of at the Los Alamos National Laboratory during World War II. He provided theoretical support for the development of radar. After the war, Schwinger left Purdue for Harvard University, where he taught from 1945 to 1974. In 1966 he became the Eugene Higgins professor of physics at Harvard.

1960

In the 1960s, Schwinger formulated and analyzed what is now known as the Schwinger model, quantum electrodynamics in one space and one time dimension, the first Example of a confining theory. He was also the first to suggest an electroweak gauge theory, an SU(2) gauge group spontaneously broken to electromagnetic U(1) at long distances. This was extended by his student Sheldon Glashow into the accepted pattern of electroweak unification. He attempted to formulate a theory of quantum electrodynamics with point magnetic monopoles, a program which met with limited success because monopoles are strongly interacting when the quantum of charge is small.

1965

Schwinger was jointly awarded the Nobel Prize in Physics in 1965 for his work on quantum electrodynamics (QED), along with Richard Feynman and Shin'ichirō Tomonaga. Schwinger's awards and honors were numerous even before his Nobel win. They include the first Albert Einstein Award (1951), the U.S. National Medal of Science (1964), honorary D.Sc. degrees from Purdue University (1961) and Harvard University (1962), and the Nature of Light Award of the U.S. National Academy of Sciences (1949).

1972

Schwinger had a mixed relationship with his colleagues, because he always pursued independent research, different from mainstream fashion. In particular, Schwinger developed the source theory, a phenomenological theory for the physics of elementary particles, which is a predecessor of the modern effective field theory. It treats quantum fields as long-distance phenomena and uses auxiliary 'sources' that resemble currents in classical field theories. The source theory is a mathematically consistent field theory with clearly derived phenomenological results. The criticisms by his Harvard colleagues led Schwinger to leave the faculty in 1972 for UCLA. It is a story widely told that Steven Weinberg, who inherited Schwinger's paneled office in Lyman Laboratory, there found a pair of old shoes, with the implied message, "think you can fill these?". At UCLA, and for the rest of his career, Schwinger continued to develop the source theory and its various applications.

1989

After 1989 Schwinger took a keen interest in the non-mainstream research of cold fusion. He wrote eight theory papers about it. He resigned from the American Physical Society after their refusal to publish his papers. He felt that cold fusion research was being suppressed and academic freedom violated. He wrote: "The pressure for conformity is enormous. I have experienced it in editors’ rejection of submitted papers, based on venomous criticism of anonymous referees. The replacement of impartial reviewing by censorship will be the death of science."

1998

Schwinger died of pancreatic cancer. He is buried at Mount Auburn Cemetery; ${\displaystyle {\frac {\alpha }{2\pi }}}$ is engraved above his name on his tombstone. These symbols refer to his calculation of the correction ("anomalous") to the magnetic moment of the electron.


text:
It is currently 19 Sep 2019, 22:10

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# Two people were hired to mow a lawn for a total of $45. They  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Founder Joined: 18 Apr 2015 Posts: 8139 Followers: 157 Kudos [?]: 1710 [0], given: 7485 Two people were hired to mow a lawn for a total of$45. They [#permalink]  03 Sep 2019, 10:06
Expert's post
00:00

Question Stats:

66% (00:01) correct 33% (00:13) wrong based on 3 sessions
Two people were hired to mow a lawn for a total of $45. They completed the job with one person working for 1 hour and 20 minutes and the other working 40 minutes. If they split the$45 in proportion to the amount of time each spent working on the job, how much did the person who worked longer receive?

(A) $33.75 (B)$30.00

(C) $27.50 (D)$25.00

(E) $22.50 [Reveal] Spoiler: OA _________________ VP Joined: 20 Apr 2016 Posts: 1002 WE: Engineering (Energy and Utilities) Followers: 14 Kudos [?]: 796 [1] , given: 176 Re: Two people were hired to mow a lawn for a total of$45. They [#permalink]  07 Sep 2019, 19:28
1
KUDOS
Carcass wrote:
Two people were hired to mow a lawn for a total of $45. They completed the job with one person working for 1 hour and 20 minutes and the other working 40 minutes. If they split the$45 in proportion to the amount of time each spent working on the job, how much did the person who worked longer receive?

(A) $33.75 (B)$30.00

(C) $27.50 (D)$25.00

(E) $22.50 Let's convert hours to min; 1 hour 20 min = 80 mins Total time spent by the 2 people = 120 mins Total payment received =$45

Therefore, the person working for 80 mins receive = $$\frac{45}{120} *80$$= $30 and the person working for 40 mins receive = $$\frac{45}{120} *40$$=$15
_________________

If you found this post useful, please let me know by pressing the Kudos Button

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text:
# How to switch on error messages in mathematica 10?

I have mathematica 10 and 8 on different computers. Version 8 always shows me the error message about what goes wrong while I don't see any error message in version 10. I find it quite inconvenient as it take much time to figure out the problem. How to switch on error messages in version 10 ?

I remember that version 10 did show the warning messages, but somebody calculated something using my mathematica and all error messages were gone after that.

• Does executing On[] help? – Karsten 7. Mar 20 '15 at 7:36
• yes, it turns on some message if I know what kind of messages it is. Do I have to turn them on 1 by 1? – mastrok Mar 20 '15 at 8:23
• You can try going to the menu bar Edit ▶ Preferences... and click "Reset to Defaults" under the Messages tab. – Karsten 7. Mar 20 '15 at 9:03
• I solved it and thank you very much. – mastrok Mar 20 '15 at 9:11
• After your code add // Quiet. – David G. Stork Mar 20 '15 at 10:21


text:
### MaxMinMax problem and sparse equations over finite fields

Igor Semaev

##### Abstract

Asymptotical complexity of sparse equation systems over finite field $F_q$ is studied. Let the variable sets belong to a fixed family $\mathcal{X}=\{X_1,\ldots,X_m\}$ while the polynomials $f_i(X_i)$ are taken independently and uniformly at random from the set of all polynomials of degree $\leq q-1$ in each of the variables in $X_i$. In particular, for $|X_i|\le3$, $m=n$, we prove the average complexity of finding all solutions to $f_i(X_i)=0, i=1,\ldots,m$ by Gluing algorithm ( Semaev, Des. Codes Cryptogr., vol. 49 (2008), pp.47--60) is at most $q^{\frac{n}{5.7883}+O(\log n)}$ for arbitrary $\mathcal{X}$ and $q$. The proof results from a detailed analysis of 3-MaxMinMax problem, a novel problem for hyper-graphs.

Available format(s)
Publication info
Preprint. MINOR revision.
Keywords
complexity theoryequation solvingcryptanalysis
Contact author(s)
igor @ ii uib no
History
Short URL
https://ia.cr/2014/004

CC BY

BibTeX

@misc{cryptoeprint:2014/004,
author = {Igor Semaev},
title = {MaxMinMax problem  and  sparse equations over finite fields},
howpublished = {Cryptology ePrint Archive, Paper 2014/004},
year = {2014},
note = {\url{https://eprint.iacr.org/2014/004}},
url = {https://eprint.iacr.org/2014/004}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Faster way of reading csv to grid

I have following in Windows Forms .NET 3.5

It works fine for csv with records less than 10,000 but is slower for records above 30,000. Input csv file can can any records between 1 - 1,00,000 records

Code currently used :

/// <summary>
/// This will import file to the collection object
/// </summary>
private bool ImportFile()
{
try
{

String fName;
String textLine = string.Empty;
String[] splitLine;

// clear the grid view

accountsDataGridView.Rows.Clear();

fName = openFileDialog1.FileName;

if (System.IO.File.Exists(fName))
{

do
{
if (textLine != "")
{
splitLine = textLine.Split(',');
if (splitLine[0] != "" || splitLine[1] != "")
{
}
}
}
return true;
}
catch (Exception ex)
{
if (ex.Message.Contains("The process cannot access the file"))
{
MessageBox.Show("The file you are importing is open.", "Import Account", MessageBoxButtons.OK, MessageBoxIcon.Warning);
}
else
{
MessageBox.Show(ex.Message);
}

return false;
}

}

Sample Input file :
18906,Y
18908,Y
18909,Y
18910,Y
18912,N
18913,N

Need some advice on optimizing this code for fast reads & view in grid.

-
Do you have a multi core machine? Then you can parse in parallel. Does the CSV files contain redundant data? In this case it does make sense to intern strings to reduce memory load. The main issue why parsing does become slower is that you need more memory to hold the data. –  Alois Kraus Jan 4 '13 at 12:48
Never rely on Exception.Message to determine the type of an exception! Catch the right one instead. –  codesparkle Jan 4 '13 at 14:02
I haven't used it, but there's a CSV parser in the VB.NET runtime. It is Microsoft.VisualBasic.FileIO.TextFieldParser. It is probably not faster but will cater for quotes, etc. –  Mark Hurd Jan 4 '13 at 15:51

What's the use of presenting even 5,000 records to the user? Do you expect a user to scroll to row 4711, look at "18912,N", and say "Strange, why wasn't there a '12345,N' in row 1234?", or "Interesting, now let's see where the corresponding '54321,X" is to be found!"?

Your GUI should help the user to specify the comparatively small set(s) of records pertinent to his current problem/question; even if working with the data means "check every item in turn", it would be more human to present the records in small batches.

All this points to using your file as a database table bound to the controls of your GUI; then the problem of "reading a .csv fast" completely vanishes.

-
True but it is still a good idea to read the stuff at once at a decent speed. When reading a CSV file like this it you will get around 20MB/s which is far less than any hard disc can deliver. If you read it a second time you get it from the OS cache at ca. 500MB/s but you parse it still at 20MB/s. –  Alois Kraus Jan 4 '13 at 22:02

I am assuming it is OK to load everything into memory all at once, since you are doing that with the grid anyways. The fastest simple option I see is loading the data into a data object and binding it to the grid, rather than adding rows one at a time.

I tossed together a CSV file with 100,000 rows, all consisting of the following data:

"a", "a", "a", "a", 1234, 1234, 1234, 1234

(It was thrown together quick just for quick turn-around sake.)

  private void LoadDirect (string fName)
{
String textLine = string.Empty;
String[] splitLine;

// clear the grid view

CSVGrid.SuspendLayout ();

CSVGrid.Rows.Clear ();

if (System.IO.File.Exists (fName))
{

do
{
if (textLine != string.Empty)
{
splitLine = textLine.Split (',');
if (splitLine [0] != string.Empty || splitLine [1] != string.Empty)
{
}
}
} while (objReader.Peek () != -1);
}

CSVGrid.ResumeLayout ();
}

That consistently took around 21 seconds to populate the grid.

To eliminate the possibility that the largest slowdown factor was loading from the file one line at a time, I produced a similar method which loads the entire file up front, then used a StringReader in place of the StreamReader inside the loop:

private void LoadBuffered (string fName)
{
String textLine = string.Empty;
String[] splitLine;

CSVGrid.SuspendLayout ();

// clear the grid view
CSVGrid.Rows.Clear ();

if (System.IO.File.Exists (fName))
{

do
{
if (textLine != string.Empty)
{
splitLine = textLine.Split (',');
if (splitLine [0] != string.Empty || splitLine [1] != string.Empty)
{
}
}
} while (strReader.Peek () != -1);
}
CSVGrid.ResumeLayout ();
}

Once again, this took around 21 seconds consistently.

Finally, I created a simplistic variant which loads all the data up front, places it into a System.Data.DataTable in the loop, then binds the table to the grid upon completion:

  private void LoadBound (string fName)
{
String textLine = string.Empty;
String[] splitLine;

// clear the grid view

var data = new DataTable ();
{
new DataColumn ("Column1"),
new DataColumn ("Column2"),
new DataColumn ("Column3"),
new DataColumn ("Column4"),
new DataColumn ("Column5",typeof (int)),
new DataColumn ("Column6",typeof (int)),
new DataColumn ("Column7",typeof (int)),
new DataColumn ("Column8",typeof (int)),
});

data.Rows.Clear ();

if (System.IO.File.Exists (fName))
{

do
{
if (textLine != string.Empty)
{
splitLine = textLine.Split (',');
if (splitLine [0] != string.Empty || splitLine [1] != string.Empty)
{
}
}
} while (strReader.Peek () != -1);
}

CSVGrid.DataSource = data;
}

This variant consistently takes around 0.8 seconds. Perhaps unsurprisingly, the real killer was adding items to the grid row-by-row.

It has the added benefit of isolating your file reading, parsing, and grid population operations into separate sections of code which could be split into methods/objects.

Of course, rather than using a System.Data.DataTable, I would prefer to use a well-structured object. The DataTable was merely used to build a quick example.

-
Nice analysis. I think you have nailed his perf issue. It should be forbidden to ask performance questions without showing a profiler screenshot. –  Alois Kraus Jan 5 '13 at 0:26

A few thoughts that could improve speed up what you're doing:

1. Use parallel processing to process multiple lines simultaneously. If you've got a multi-core machine, this would go some way to offloading the task of manipulating strings and adding them to a control. The BlockingCollection<T> would be well-suited to this task.

2. Buffer your read of the file using a FileStream. The IO itself is quite an overhead, so doing fewer-but-bigger reads could speed up the process considerably.

3. Don't actually read all the rows into the grid. Set up pagination so that you only load a portion of it until the user requests the next set of data. It's very unlikely a human will actually read 30,000 rows.

Here's an example of how to use the FileStream class and to handle the processing of the lines as an asynchronous process:

// Internally the collection uses a ConcurrentQueue<T> by default.
var queue = new BlockingCollection<string>();

// Set up an asynchronous operation to process the queue.
{
foreach(var line in queue.GetConsumingEnumerable())
{
if (textLine != "")
{
splitLine = textLine.Split(',');
if (splitLine[0] != "" || splitLine[1] != "")
{
accountsDataGridView.Invoke(
}
}
}
});

/* You should tweak the buffersize to get the balance you want between memory use
and speed of execution. */
var bufferSize = 8192; // 8k is the default size used by FileStream.

using(var fs =
new FileStream(
openFileDialog1.FileName,
FileMode.Open, // Open an existing file (don't create it)
bufferSize))
{
{
{
}

// Signal that there are no more items to be added.
}
}

As an aside, you probably want to perform the read of the CSV on a background thread, so the operation doesn't block the UI.

-

There isn't much to optimize in regards to speed, but following is much more readable. If it is too slow, it probably isn't the method reading the file, but your WinForm that needs to display >30k records.

    accountsDataGridView.Rows.Clear();
{
{
if (fields.Length == 2 && (fields[0] != "" || fields[1] != ""))
{
}
}
}

-

We use the GenericParser library. It's pretty easy to use, too.

Your code would look something like this:

GenericParserAdapter parser = new GenericParserAdapter(fName);

parser.ColumnDelimiter = ',';
parser.FirstRowSetsExpectedColumnCount = true;


text:
# Curvy arrows between two tables (with tikz?)

I try to connect two arbitrary tables with arrows. The code example below should demonstrate how: say we want to connect the "1" in A with the "5" in B, the "2" in A with the "7" in B and so on. My wish would be to have "curvy" arrows (as far as i know the tikz packages provides them) and not "straight" arrows.

Is it even possible to combine "ordinary" tabular elements with tikz nodes oder drawings? If yes, how? If not, is there a pure tikz way to solve that problem? Thanks for any advices or hints!

\documentclass[12pt,a4paper,twoside]{scrartcl}
\begin{document}
\begin{center}

\begin{tabular}{c c c c c c}
$A$ & 1 & 2 & 3 & 4 &
\end{tabular}

\vspace{15mm}

\begin{tabular}{c c c c c c c c c c }
$B$ & \fbox{5} & 6 & \fbox{7} & 8 & \fbox{9} & 10 & \fbox{11} & 12 &
\end{tabular}

\end{center}
\end{document}

-

Do you mean something like this:

## Code:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}

\newcommand*{\DrawArrow}[3][]{%
% #1 = draw options
% #2 = left point
% #3 = right point
\begin{tikzpicture}[overlay,remember picture]
\draw [very thick, -stealth, #1] ($(#2)+(0.25em,-0.3ex)$) to ($(#3)+(0.25em,2.5ex)$);
\end{tikzpicture}%
}%

\begin{document}
\begin{center}

\begin{tabular}{c c c c c c}
$A$ & \tikzmark{topA}1 & \tikzmark{topB}2 & \tikzmark{topC}3 & \tikzmark{topD}4 &
\end{tabular}

\vspace{15mm}

\begin{tabular}{c c c c c c c c c c }
$B$ & \fbox{\tikzmark{bottomA}5} & 6 & \fbox{\tikzmark{bottomB}7} & 8 & \fbox{\tikzmark{bottomC}9} & 10 & \fbox{\tikzmark{bottomD}11} & 12 &
\end{tabular}
\DrawArrow[red,   out=-90, in=90]{topA}{bottomA}
\DrawArrow[olive, out=-90, in=90]{topB}{bottomB}
\DrawArrow[blue,  out=-90, in=90]{topC}{bottomC}
\DrawArrow[brown, out=-90, in=90]{topD}{bottomD}
\end{center}
\end{document}

-
Wow! That's exactly my desired representation. Thank you very much for the quick response and the understandable code style! –  bobb_the_builder Mar 22 '13 at 22:05
Wait... I wasn't done... Solution now updated with a better macro interface. –  Peter Grill Mar 22 '13 at 22:11
The vertical line passing through the center of one row is not aligned with that of the other row. In other words, 2 and 8 should be in the same column. :D –  In PSTricks we trust Mar 22 '13 at 22:36
@Karl'sstudents: Not sure I understand. I am just connecting existing text, not adjusting it. The positions can be adjusted as needed. –  Peter Grill Mar 22 '13 at 22:40
I meant, 2 and 8 should be in the same column. In other words, 8 should be right below 2 such that 3 and 9 are connected with a straight line. By the way, it is not crucial issue. :-) –  In PSTricks we trust Mar 22 '13 at 22:44

With PSTricks.

\documentclass[preview,border=12pt,12pt]{standalone}
\usepackage{pst-node}
\psset
{
mnode=r,
rowsep=2cm,
colsep=0.5cm,
}

\def\c{[mnode=circle] }

\begin{document}
\offinterlineskip
\begin{psmatrix}
& & A & 1 & 2 & 3 & 4\\
B & \c 5 & 6 & \c 7 & 8 & \c 9 & 10 & \c 11 & 12
\foreach \i/\j in {4/2,5/4,6/6,7/8}
{\nccurve[angleA=-90,angleB=90,nodesep=3pt]{->}{1,\i}{2,\j}}
\end{psmatrix}
\end{document}

## Warning:

• mnode for \psframebox is not available out of the box. We can circumvent it by wrapping each item with \psframebox manually, but it will take more keystrokes!
• mnode=circle will use \pscirclebox to wrap the item. Unfortunately, \pscirclebox has no option to adjust its radius. It results in the size of each circle will vary depending on the size of the contained item. Only framesep is available to adjust the border. Sad!
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@Karls students: looks very stylish too! Thanks for the pratical alternative! –  bobb_the_builder Mar 22 '13 at 23:15

Here's a very short solution using the tikz-cd package:

\documentclass{standalone}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}[out=-90, in=90, column sep=small, row sep=15mm]
& & A & 1 \arrow{dll} & 2 \arrow{dl} & 3 \arrow{d} & 4 \arrow{dr} \\
B & \fbox{5} & 6 & \fbox{7} & 8 & \fbox{9} & 10 & \fbox{11} & 12
\end{tikzcd}
\end{document}

The result is this:

Note that the table entries created by this method will automatically be in math mode. Any entries for which this behavior is undesired should have their text enclosed in a \text{} command.

-
@ Charles Staats: very clear and simple approach - nice! –  bobb_the_builder Mar 23 '13 at 0:43


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# Inequality and absolute value questions from my collection

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Inequality and absolute value questions from my collection [#permalink]  16 Nov 2009, 10:33
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Guys I didn't forget your request, just was collecting good questions to post.

So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#p1111747

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 07:39
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SOLUTIONS:

1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y – x = 3
(2) x^3< 0

First let's simplify given expression $$6*x*y = x^2*y + 9*y$$:

$$y*(x^2-6x+9)=0$$ --> $$y*(x-3)^2=0$$. Note here that we CAN NOT reduce this expression by $$y$$, as some of you did. Remember we are asked to determine the value of $$xy$$, and when reducing by $$y$$ you are assuming that $$y$$ doesn't equal to $$0$$. We don't know that.

Next: we can conclude that either $$x=3$$ or/and $$y=0$$. Which means that $$xy$$ equals to 0, when y=0 and x any value (including 3), OR $$xy=3*y$$ when y is not equal to zero, and x=3.

(1) $$y-x=3$$. If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case $$xy=18$$. But if y=0 then x=-3 and $$xy=0$$. Two possible scenarios. Not sufficient.

OR:

$$y-x=3$$ --> $$x=y-3$$ --> $$y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0$$ --> either $$y=0$$ or $$y=6$$ --> if $$y=0$$, then $$x=-3$$ and $$xy=0$$ $$or$$ if $$y=6$$, then $$x=3$$ and $$xy=18$$. Two different answers. Not sufficient.

(2) $$x^3<0$$. x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.

This one was quite tricky and was solved incorrectly by all of you.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 09:10
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5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 11:26
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7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 07:55
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3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 14:48
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11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.

5+3=8 and 5-3=2
OR -5-3=-8 and -5-(-3)=-2.

So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.

(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.

(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 07:47
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2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Note: as $$y=|x|+x$$ then $$y$$ is never negative. For $$x>{0}$$ then $$y=x+x=2x$$ and for $$x\leq{0}$$ then (when x is negative or zero) then $$y=-x+x=0$$.

(1) $$x<0$$ --> $$y=|x|+x=-x+x=0$$. Sufficient.

(2) $$y<1$$, as we concluded y is never negative, and we are given that $$y$$ is an integer, hence $$y=0$$. Sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 08:17
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4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
$$2x-2y=1$$ --> $$x=y+\frac{1}{2}$$
$$\frac{x}{y}>1$$ --> $$\frac{x-y}{y}>0$$ --> substitute x --> $$\frac{1}{y}>0$$ --> $$y$$ is positive, and as $$x=y+\frac{1}{2}$$, $$x$$ is positive too. Sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 16:09
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13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 11:34
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8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 15:42
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12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:
A. s is either positive or zero, as -s<=s;
B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.
But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well.
Consider: s=5, r=5 --> -5<=5<=5 |5|>=5
s=5, r=-5 --> -5<=-5<=5 |-5|>=5
Both statements are true with these values. Hence insufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 11:48
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10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 16:51
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lagomez wrote:
Just curious if my thinking is correct.

on the 2nd part I get y = -8 and y =14
Then I substituted the values into the first equation:
3|x^2-4|=-10
the answer will never give -10/3

do the same for 14
3|x^2-4|=12
x = 0

using my methodology I also got C, but is my thinking correct?

Well you can even not calculate for x. Statement 1 says that y must be greater than or equal to 2. Statement 2 gives 2 values of y -8 OR 14. Combining we get that y=14.

Remember we are asked to determine the value of y not x.
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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 09:25
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6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

(1) x+1>0 --> x>-1. As x is an integer x can take the following values 0,1,2,... But we know nothing about y. Not sufficient.

(2) xy>0. x and y have the same sign (both positive OR both negative) and neither x nor y is zero. Not sufficient.

(1)+(2) x is positive, as from (1) it's 0,1,2.. and from (2) x is not zero. Hence xy to be positive y also must be positive. Sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  16 Nov 2009, 11:42
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ahh..yes...fresh meat
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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 11:42
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9. Is n<0?
(1) -n=|-n|
(2) n^2=16

(1) -n=|-n|, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.

(2) n^2=16 --> n=4 or n=-4. Not sufficient.

(1)+(2) n is negative OR n equals to zero from (1), n is 4 or -4 from (2). --> n=-4, hence it's negative, sufficient.

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Re: Inequality and absolute value questions from my collection [#permalink]  20 Nov 2009, 13:45
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tihor wrote:
Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.

Thank you very much for this catch. +1. There was a typo. So you are right with the first one:
|x-1| < 1 means 0<x<2. Already edited the post.

As for the second one:
x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.

x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.

Hope it helps.
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Re: Inequality and absolute value questions from my collection [#permalink]  17 Nov 2009, 06:38
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3)
I) (x+y)^2=9a x^2+y^2=9a-2xy NS
II) (x-y)^2=a x^2+y^2=a+2xy NS
Together 2(x^2+y^2)=10a x^2+y^2=5a
If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E

4)
I don’t get the two clues; they seem to be mutually exclusive

5)
I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS
II) |3-y|=11 either y=-8 or y=14 NS
Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E

6)
I) x+1>0 so x={0, 1, 2, …} NS
II) xy>0 so x and y have the same sign and none of them is zero NS
Together, x={1, 2, 3, ..} and y has the same sign, hence C

7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1)
Reordering: x-y=0 or x+y=-4
I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S
II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D

8)a*b#0, hence a and b are both non-zero
I) |a*b|=a*b a and b have the same sign and the stem is always true S
II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A

9)
I) –n=|-n| n<=0 NS
II) n^2=16 n=+/-4 NS
Together n=-4 therefore C

10)n#0
I) n^2>16, so |n|>4 S
II) 1/|n|>n true for n<-1 NS, therefore A

11) Plugging in numbers I get B, but there’s no rime or reason to my solution

12)
I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero
II)|r|>=s obviously NS
Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C

13) x=(0:2) with 0 and 2 excluded
I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS
II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS
Together, for x=1.5 the stem is true, for x=2 it is false, hence E
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Re: Inequality and absolute value questions from my collection [#permalink]  18 Nov 2009, 16:26
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lagomez wrote:
would you think this is a 700+ question?

It's quite hard yes. Plus everyone solved it incorrectly, so I'd say it is 700+.
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Re: Inequality and absolute value questions from my collection [#permalink]  20 Nov 2009, 13:19
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Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is -2<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that -2<x<2 is true. Not sufficient.

Bunuel, two questions:
shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2?
secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2?
does this not translate into x<=0 or x<=2?

thank you very much for all the questions and solutions.
Re: Inequality and absolute value questions from my collection   [#permalink] 20 Nov 2009, 13:19

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# Natural numbers show "Stretching Similarity"!

Well we think that we have found a nice property of natural numbers which we are unable to explain in all details. So we published it without having any idea why it holds true. You can look at Figure 4 and related description given in the following paper:

http://www.nature.com/articles/srep14280

If you find any hint, let me know in the comments.

Cheers :)

Note by Snehal Shekatkar
1 year, 4 months ago

Sort by:

This is pretty awesome. It's going to take me a while to digest this paper. I don't normally study network theory, but I'm beginning to see how this sort of thing could be of value in some other branches of theoretical physics. Do I have any hint as to why this property of numbers seems to be true? You're kidding, right? I'll get back to you on that. · 1 year, 4 months ago

Sure.. contrary to my expectations, nobody thought that this was an interesting note! :D · 1 year, 4 months ago

Snehal, there's a difference between "not being interested", and "being glassy-eyed". It's a difficult read! It will take ME quite some time to grasp the subject. · 1 year, 4 months ago


text:
# Clustering as an approach to regression

Suppose $\varphi: X \rightarrow Y$ is a function between feature spaces that I want to model. I want to produce two easily computable functions $\phi_X: X \rightarrow \mathbb{R}^m$ and $\phi_Y: Y \rightarrow \mathbb{R}^n$ so the distance $|\phi_X(x) - \phi_X(x')|$ for $x, x'$ in $X$ is correlated as well as possible with $|\phi_Y(\varphi(x)) - \phi_Y(\varphi(x'))|$.

My question is simple: is there any research going on about questions like this? The problem seems awkwardly stuck between a clustering problem and a classification problem. I don't need an ironclad way of approaching the problem (there almost certainly isn't one), but I was wondering if the problem had any better classification than as a kinda-regression.

EDIT: As per D.W.'s comment, my condition is poorly stated as it stands. A condition that better expresses what I meant is to find a method that, given a point $x \in X$, a finite set of points $y_i \in Y$, and an assurance that $\varphi(x) = y_i$ for some $i$, to determine the most likely $y_i$ that fits this bill. But this is straight machine-ranking, solving the question.

• 1. I think you'll need to define your problem more carefully. For instance, $\phi_X(x)=0$ for all $x$ and $\phi_Y(y)=0$ for all $y$ provides an excellent correlation. I suspect you must have some additional requirements or constraints that you're not telling us about. 2. What kind of a space are $X,Y$? – D.W. Sep 22 '15 at 22:27
• Your first comment is a good point, and I reformulated the condition to be a bit more coherent. However, having done so, I realized that the answer to my question was simply machine ranking. Thanks for the push in the right direction! – Alexander Smith Sep 23 '15 at 2:26


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# Could we use both speed and heat to penetrate Europa's ice?

Could we fire a heavy, long (e.g., 50 meters), spear-shaped object into Europa's icy-surface from space and then, when momentum stops, gravity gradually moves weights down inside copper tubes the length of the object like electrical-generator components generating electricity to heat the tip for melting the rest of the way down?

• This seems a bit open ended. You can, given enough electrical energy, generate enough heat to melt your way through an arbitrary amount of ice. I think we will need you to provide bounds on the heating part. The spear-throwing approach is going to be limited by speed. once you get above the speed of sound, you're going to get in trouble rather quickly, but I can't figure out how to tell whether the whole system will work without bounds on how much electricity I can draw. – Cort Ammon Jan 22 at 23:14
• I will note that Europa is quite a lot smaller than the Earth, so the effects of generating heat via gravitational potentials (i.e. dropping things down tubes) is going to be limited. – Cort Ammon Jan 22 at 23:16
• @Dean A 10-30 km thick ice layer at -220 celsius isn't going to be penetrated by the bigest nuclear weapon we've ever made. Try calculating the energy needed to raise a 1 metre diameter collumn of ice that deep to melting temp. You'll be astonished. – Don Qualm Jan 22 at 23:30
• What is the speed of sound on Europa? Also isn't that the one we're supposed to stay away from? – Harper Jan 23 at 5:53
• A chemical energy source is would be more compact than a mechanical one. eg: thermite produces lots of heat without much change in volume. – Jasen Jan 23 at 9:44

The major limitation in this plan is going to be the energy of the objects that you let fall. All of the essential equations happen to have a cross sectional area term in them, so we can actually do some useful calculations: how far do you have to drop something to get the energy needed to melt a length of ice?

Let's start with density. The best object to let fall is going to be the densest thing you can find. Osmium, at $$22g/cc$$ ($$22000 kg/m^3$$) is the densest thing we can muster. From physics, we know that the potential energy due to gravity of an object is $$PE = mgh$$. Substituting in $$a$$ for cross sectional area, $$l$$ for the length of the Osmium slug, $$d$$ for the density of osmium, and $$L$$ for the length of the spear, we get $$PE=dalg(L-l)$$.

We also know that 1cc of ice takes a lot of heat to melt. Europa's ice is at -160C, and ice takes roughly 2J/cc-K to melt (actually it's 2J/g-K, which is a little different, but they'll be similar enough results for this back of the envelope exercise). That means it takes 320J to raise 1 cc of ice to 0C. We then need to impart 330J/cc or so to melt it (334J/g, technically... and again I'm handwaving the density because this is just back of the envelope). Put them together, and it takes $$660ax$$ joules of energy to melt a cross sectional area $$a$$ $$x$$ cm deep. To switch that over to meters of depth, it takes $$66000ax J$$ joules of energy if $$x$$ is written in meters.

These energies will be the same, so $$dalgL=66000ax$$. This makes it clear that the cross sectional areas cancel, yielding $$dlg(L-l)=66000x$$. If we're limited by total length, $$l+L$$, a bit of calculus will show that the best results we get are when the drop length and length of the Osmium slug are equal, $$l=(L-l)=\frac{L}{2}$$, so we get $$\frac{1}{4}dgL^2 = 66000x$$. The gravity on europa is 1.31m/s^2. Thus now we get $$7205L^2 = 66000x$$ or $$L=\sqrt{9.16x}$$. Since we handwaved a bunch of stuff, I'm going to round it slightly to $$L=3\sqrt x$$

What does this mean? To get through the 19,000m we need L=413m. Now we've been rounding a bunch, so let's do it once more. Let's call L=500m. This means you have a 250m long slug of Osmium, resting in a half-kilometer long object. That's what it takes to really melt the ice, assuming everything goes absolutely perfectly. Practically speaking, you'll run into issues melting ice fast enough to not waste energy into the surroundings. You're likely to be talking about a kilometer long object!

My personal recommendation is to land kilometer long spacecraft into ice sheets. Land them gently. Don't try to impact them. The structural effects caused by trying to plow through all that ice would be brutal. It is almost guaranteed that you will crumple those tubes you are relying on. Oh, and Osmium is very brittle, so you're likely to shatter your weights.

• Won't it be much higher than that? Not only do you need to melt the ice, you need to keep it melted long enough for the probe to pass. Making the object longer simply means more surface area to keep heated. Your formula isn't taking into consideration the heat loss. Not only do you need to melt the ice, you need to keep it melted in a -160C environment. – Thorne Jan 23 at 5:40
• you don't need to keep it melted so long as it does not stick to the surface of the spear. – Jasen Jan 23 at 9:42
• Excellent calculations and great way to tell this story... just make sure whoever is funding your project doesn't realize that melting ice does not mean it disappears from existence! – Tim Jan 23 at 17:10
• Thank you Cort Ammon for your level of technical knowledge backed by mathematics even though admittedly it's over my head. I'm very impressed by the imagination and creativity gone into the all the answers on this site. Given the degree of materials, the size and uncertainty of using a spear-like approach along with the density of the ice, I can see that it would not be practical to fire such an object in this way. I kind of got the idea from bunker-busting bombs and thought the generating system would be a minimally polluting one considering alternative Nuclear energy options. Thanks – Dean Jan 24 at 9:38

No

Heating the tip doesn't heat the rest of the spear and the ice will just close and hold the spear along the rest of the shaft.

I'm also doubtful a moving counterweight could generate enough energy to melt a path for the spear.

Why not just do it like NASA has already planned? Land a lander on the surface that runs on a nuclear battery and use the waste heat to melt the ice. The lander sinks into the ice and keeps melting while the water above refreezes.

Seems much easier than smashing a giant spear and hoping it will keep going based on a crude gravity based generator.

• I'm not so sure the gentle water-dwelling creatures under the ice will be impressed when the scaldingly hot, radioactive killer robot from Earth comes crashing through the SkyIce. – Oscar Bravo Jan 23 at 15:03
• After reconsideration, I agree that the 'spear' will likely get jammed etc. I like the idea of a lander, but I'm not keen on a nuclear battery sinking permanently into the pristine environment beneath the ice. Perhaps leave the battery on the surface and run a ultralight weight conducting cable of some kind down behind the sinking lander. Perhaps the sinking lander could be a shorter spear/missile shaped vessel. – Dean Jan 24 at 9:46

I don't think we need any special technique for getting through the ice. We have drilled about that deep through rock for oil right here on Earth.

I think there are two much bigger problems to deal with. First, think about how big the equipment for drilling through ~10km of ice will be. Now remember you have to get all that junk all the way out to Europa. And once it's out there, you have to land it; and Europa doesn't have an atmosphere so you can't use parachutes. Look at how much trouble SpaceX has had landing mostly-empty boosters. Now imagine trying to land an oil rig ~630 million kilometers away.

Second, what happens when your drill finally breaks through the bottom of the ice? The ocean down there must be highly pressurized; ~10km of ice weighs a lot after all, even in reduced gravity. So what's stopping that water from shooting up the bore hole at the speed of sound and destroying all your equipment?

• The Moon is likewise airless,and quite a number of vehicles - Surveyors, Apollo landers, Russian landers, the Chinese Chang'es - have landed there. Nor would you use an actual drill. Instead use hot water, as the recent project to drill into an Antarctic subglacial lake did. Except that you'd put a nuclear reactor in the probe, and just unreel a thin fiberoptic cable behind as it melts down. – jamesqf Jan 23 at 4:03
• You'd have to put together a team of the best dang drillers on Earth. You'd need a cowboy one, a wise-cracking skinny one, an over-eating one, an African-American one (who's into motorbikes), a clever, handsome one, and head up the team with a tough, heroic leader who's cooler than the ice they have to drill through. – Oscar Bravo Jan 23 at 15:08

Possibly. This is Worldbuilding.SE, not Space.SE. There could be a way to make your spear practical, we just need a realistic reason why it would (since the others have covered why it won't work outright).

Let's start off with the ice sheets. This is how NASA is envisioning the sheets and underlying ocean working. The red areas indicate thermal currents under the surface (we know they happen from the ice jets shot into space, but not much else)

And that ice sheet is estimated to be 19-25km thick by using the unique topology of its impact craters. But this tidbit is interesting

There is strong evidence that Europa’s icy shell is somewhat unstable and has been (or is) convecting. This means that blobs of deep crustal material rise upward toward the surface where they are sometimes exposed as domes several kilometers wide (think Lava Lamp, except that the blobs are soft solid material like Silly Putty). Any ocean material imbedded within the lower crust could then be exposed to the surface. This process could take thousands of years, and the exposure to Jupiter’s lethal radiation would be unfriendly to say the least! But at least we could investigate and sample what remains behind.

So the ice isn't static but shifting over time. Let's say we combined some elements here. The following has some plausibility to it (even if it is unlikely)

1. NASA detects an asteroid on a collision course with Europa with enough time to launch a probe
2. The asteroid strikes a thermal vent

If your spear followed close behind the asteroid, there could be just enough weakness in the ice at the impact crater to get your spear through.


text:
# What is non PV work? How is it related to Gibbs energy?

Recently, I was trying to understand more about Gibbs free energy and got stuck at this point. Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when $\Delta G$ = Non-$PV$ work. I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

At constant temperature and pressure, the maximum non expansion work is given by the change in the Gibbs Free Energy.

$$dW_{\text{max}}= dG$$

Proof:

At constant temperature and pressure,

• from the definition of enthalpy: $H= U+pV$
• first law of thermodynamics: $dU = dq+dw$

$\implies dH = dq+ dw + d(pV)$

Now,

$dG = dH - TdS -SdT \\\implies dG = dq + dw + d(pV)- TdS - SdT$

For isothermal process $dT =0$

$\implies dG = dq+dw+ d(pV)+ -TdS$

When the change is reversible,

$dq = dq_{\text{rev}}= TdS$ and $dw= dw_{\text{rev}}$

$\implies dG = dw_{\text{rev}}+d(pV)$

The work consists of expansion work, which for a reversible change is given by −pdV, and possibly some other kind of work (for instance, the electrical work of pushing electrons through a circuit or of raising a column of liquid); let's denote it by $dw_{\text{add},\text{rev}}$

$\implies dG= dw_{\text{add}, \text{rev}}+ Vdp$

For isobaric process, $dp =0$

$\implies dG = dw_{\text{add}, \text{rev}}$

You must be aware the the maximum work is obtained when the process is reversible, if not you may refer to this answer.

$\implies dG = dw_{\text{add}, \text{max}}$

Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when $\Delta G$ = Non-$PV$ work.

$\Delta S_{\text{total}}= \dfrac{-\Delta G_{\text{system}}}{T}$

Thus, for a reversible process $\Delta G_{\text{sys}}= 0$

And I have also proved to you above that $\Delta G_{\text{sys}}=$ maximum non expansion (non PV) work.

I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

PV work is only the work done during the change in volume. There can be different types of works that a system can do during a process, for instance: electrical work, work against gravity, work against friction, magnetic work, shaft work, etc. This is the concept of non expansion or non PV work.

Source:

• You proved $\Delta G_{sys=0}$ but also proved that $\Delta G_{sys}=$ maximum non expansion work , does that imply maximum non PV work=0? Apr 27 at 15:11


text:
?

Free Version
Moderate

Belief in Santa Claus

APSTAT-F3P1EL

A researcher wants to study the difference between $6$-year-old boys’ and girls’ belief in Santa Claus.

She asks $18$ randomly selected boys and $18$ randomly selected girls. Of those, $9$ boys and $13$ girls said they believe in Santa Claus.

The $95\%$ confidence interval for the difference in the response rates for the boys and girls was $-0.532$ to $0.088$ (boys to girls).

What can be inferred from this result?

A

It is highly likely that the same proportion of $6$-year-old boys and $6$-year-old girls believe in Santa Claus.

B

A significantly greater proportion of $6$-year-old girls believes in Santa Claus than do $6$-year-old boys.

C

No inference can be made because the sample sizes are much too small.

D

We cannot conclude that the proportion of $6$-year-old girls who believe in Santa Claus is different from the proportion of 6-year-old boys who believe in Santa Claus.

E

No inference can be made because this is a confidence interval, not a hypothesis test.


text:
# How to prove the integral formulae of the inverse path $\alpha^-$ and the product path $\alpha\beta$?

I need help with this problem:

Let $$f:S\subset\mathbb{R}^n\rightarrow\mathbb{R}$$ be continuous on $$S$$, and let $$\alpha:[a,b]\subset\mathbb{R}\rightarrow\mathbb{R}^n$$ and $$\beta:[c,d]\subset\mathbb{R}\rightarrow\mathbb{R}^n$$ be piecewise $$C^1$$ (continuously differentiable) paths in $$S$$, such that $$\alpha(b)=\beta(c)$$. Prove the following integral formulae concerning the inverse path $$\alpha^-$$ and the product path $$\alpha\beta$$: $$\int_{\alpha^-}f\ ds=\int_{\alpha}f\ ds$$ $$\int_{\alpha\beta} f \ ds=\int_{\alpha}f \ ds\ + \int_{\beta}f \ ds$$

I don't know if I'm correct, but I think that $$\alpha^-(t)=\alpha(-t)$$, so $$\alpha^-:[-b,-a]\rightarrow\mathbb{R}^n$$. Using this I tried to rpove the first one like this: $$\int_{\alpha^-}f\ ds=\int_{-b}^{-a} f(\alpha^-(t))\Vert \alpha{^-}'(t)\Vert dt=\int_{-b}^{-a} f(\alpha(-t))\vert(-1)\vert\Vert\alpha(-t)\Vert dt$$ let $$u=-t$$, thus $$du=-dt$$ $$-\int_a^b-f(\alpha(u))\Vert\alpha'(u)\Vert du=\int_a^bf(\alpha(u))\Vert\alpha'(u)\Vert dt=\int_\alpha f\ ds$$ Am I correct? For the second one, I don't know how to prove it. I think that $$\alpha\beta$$ woudl be the path that goess from $$a$$ to $$d$$, right?

The first one looks correct, for the second, you have to join the paths $$\alpha, \beta$$ into one parameterisation $$\gamma$$. One way to do this is by letting,

$$\gamma : [a, b + d - c] \to \mathbb{R}^n$$ with,

$$\gamma(t) = \begin{cases}\alpha(t), \ t \in [a,b] \\ \beta(t - b + c), \ t \in [b, b + d- c] \end{cases}$$

See here how I have combined the two paths into one interval from $$a$$ to $$b + d -c$$. I have not changed the paths at all here, so that $$\gamma = \alpha \beta$$. That means,

$$\int_\gamma f \ \mathrm{d}s = \int_a^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \int_a^b f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t + \int_b^{b+d-c} f(\gamma(t)) || \gamma'(t) || \ \mathrm{d}t = \dots$$

I'll let you finish the proof, at this point you have to use the definition of $$\gamma$$. You may also want to verify that $$\gamma$$ is in fact $$C^1$$ at every point except at $$t = b$$, which won't affect the integral since we only require $$\gamma$$ be piecewise $$C^1$$.

• I don't understand why the interval goes from $a$ to $b+d-c$. Also, why $gamma$ is not $C^1$ at $t=b$? – davidllerenav Apr 20 '19 at 3:22