text:
# URIʍ\i4j

L

L

ŏIXV
2006-04-14T01:01:00+09:00
̋LURIQ
http://www.7key.jp/rfc/2396/rfc2396_4.html#source
• RFC2396yUniform Resource Identifiers (URI): Generic Syntax -- August 1998z

## URIʍ\iaj

ŏIXV
2006-09-26T00:00:00+09:00
̋LURIQ
http://www.7key.jp/rfc/2396/rfc2396_4.html#translation

### 4. URI References

The term "URI-reference" is used here to denote the common usage of a
resource identifier.  A URI reference may be absolute or relative,
and may have additional information attached in the form of a
fragment identifier.  However, "the URI" that results from such a
reference includes only the absolute URI after the fragment
identifier (if any) is removed and after any relative URI is resolved
to its absolute form.  Although it is possible to limit the
discussion of URI syntax and semantics to that of the absolute
result, most usage of URI is within general URI references, and it is
impossible to obtain the URI from such a reference without also
parsing the fragment and resolving the relative form.

URI-reference = [ absoluteURI | relativeURI ] [ "#" fragment ]

ł́AURIQƂƂp\[Xʎq̋ʂgƒ`BURIQƂɂ͐ΎQƂƑΎQƂAǉ񂪃tOgʎq`ŕtĂꍇBAi΁jtOgʎq폜AXɑURIURI`ɉ߂AʂƂȂURI͐URIłBURI̍\ƈӖɂĂ̋c_URIɌ肷邱Ƃł邪AURȊ͈ʓIURIQƂ̒ŗp邽߁AtOgʎq̍\͂ƑΌ`̉ȂňʓIURIQƂURI𓾂邱Ƃ͕s\łB

The syntax for relative URI is a shortened form of that for absolute
URI, where some prefix of the URI is missing and certain path
components ("." and "..") have a special meaning when, and only when,
interpreting a relative path.  The relative URI syntax is defined in
Section 5.

URI̍\́AURIZkłB̍ېURȊOȗA߂ꂽpXR|[lgi"."".."j΃pX̉ߎɂ̂ݓʂȈӖBURI̍\ɂĂ5. URIQŒ`B

#### 4.1. Fragment Identifier

When a URI reference is used to perform a retrieval action on the
identified resource, the optional fragment identifier, separated from
the URI by a crosshatch ("#") character, consists of additional
reference information to be interpreted by the user agent after the
retrieval action has been successfully completed.  As such, it is not
part of a URI, but is often used in conjunction with a URI.

fragment      = *uric

URIQƂʂꂽ\[X̎擾sہAURI"#"ŋ؂邱ƂɂătOgʎqCӂŕt邱ƂłBtOgʎq́A\[X̎擾삪Ƀ[UG[WFg߂tIȎQƏłB̂߁AtOgʎqURIł͂ȂAURIƋɗp邱ƂȂȂB

The semantics of a fragment identifier is a property of the data
resulting from a retrieval action, regardless of the type of URI used
in the reference.  Therefore, the format and interpretation of
fragment identifiers is dependent on the media type [RFC2046] of the
retrieval result.  The character restrictions described in Section 2
for URI also apply to the fragment in a URI-reference.  Individual
media types may define additional restrictions or structure within
the fragment for specifying different types of "partial views" that
can be identified within that media type.

tOgʎq̈Ӗ́AQƂ̍ۂɗpꂽURǏ^Ɋւ炸A擾̌ʓꂽf[^̎łB̂߁AtOgʎq̌`Ɖ߂́Aꂽʂ̃fBA^Cv[RFC2046]ɈˑB2ɂĐGꂽURIɂ镶́AQƂURI̒̃tOgʎqɂKpBefBA^Cv́ÃfBA^CvŎʂłقȂ^"Q"ŵ߂ɁAtOgʎqɕtIȐ邢͍\`邱ƂłB

A fragment identifier is only meaningful when a URI reference is
intended for retrieval and the result of that retrieval is a document
for which the identified fragment is consistently defined.

URIQƂ擾ɍۂ̂ŁA擾̌ʓ镶łꂼ̕т`Ŏʂꍇ̂݁AtOgʎq͗LvȂ̂ƂȂB

#### 4.2. Same-document References

A URI reference that does not contain a URI is a reference to the
current document.  In other words, an empty URI reference within a
document is interpreted as a reference to the start of that document,
and a reference containing only a fragment identifier is a reference
to the identified fragment of that document.  Traversal of such a
reference should not result in an additional retrieval action.
However, if the URI reference occurs in a context that is always
intended to result in a new request, as in the case of HTML's FORM
element, then an empty URI reference represents the base URI of the
current document and should be replaced by that URI when transformed
into a request.

URI܂܂ȂURIQƂ́A݂̎̕QƂłB܂Ał̋URIQƂ邱Ƃ͂̐̕擪QƂ邱ƂƉ߂AtOgʎq݂̂܂ގQƂ́Aꕶ̒`ꂽQƂ邱ƂƉ߂B̂悤ȎQƂ̍ہAɎ擾Ă镶Ăю擾悤ȓsׂł͂ȂBAHTMLFORMvf̂悤ɁAURIQƂɐVȃNGXgs悤Î镶ŗpꂽꍇ́AURIQƂ͓ꕶ̊URI̎QƂ\ANGXg̍ۂɊURIւ̃NGXgƒu߂sׂłB

#### 4.3. Parsing a URI Reference

A URI reference is typically parsed according to the four main
components and fragment identifier in order to determine what
components are present and whether the reference is relative or
absolute.  The individual components are then parsed for their
subparts and, if not opaque, to verify their validity.

URIQƂ͒ʏAR|[lg邩ǂAQƂɊ֘A邩ǂAΎQƂǂ߂邽߂ɁA4̎ȃR|[lgƃtOgʎqɂĉ͂B̌AeR|[lg͕Ƃɉ߂AsłȂ΂̐mFB

Although the BNF defines what is allowed in each component, it is
ambiguous in terms of differentiating between an authority component
and a path component that begins with two slash characters.  The
greedy algorithm is used for disambiguation: the left-most matching
rule soaks up as much of the URI reference string as it is capable of
matching.  In other words, the authority component wins.

eR|[lgŔF߂`BNFŒ`Ă邪A@փR|[lg2̃XbVŎn܂pXR|[lgʂ_܂łB̂߁A}b`OKURIQƕ̍[̍Œ}b`ƂȂ悤ȁA茵ȃASYpB΁A@փR|[lgD悳ׂłB

Readers familiar with regular expressions should see Appendix B for a
concrete parsing example and test oracle.

K\ɖ邢̂ł΁A\̗͂̋ƎIȃASYɂċLqB. K\pURIQƂ̍\QƂ̂ƁB

L

Copyright (C) 2006  key@do.ai ŁF2006N0414 ŏIXVF2006N0926


text:
# American Institute of Mathematical Sciences

• Previous Article
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Synchronized and nonsymmetric phase-locked periodic solutions in a neteork of neurons with McCulloch-Pitts nonlinearity
2001, 2001(Special): 109-120. doi: 10.3934/proc.2001.2001.109

## Asymptotic behavior of positive solutions of the equation $\Delta u + K u^{\frac{n+2}{n-2}} = 0$ in $IR^n$ and positive scalar curvature

 1 Department of Mathematics, Rutgers University, Hill Center, 110 Frelinghuygen Road, Piscataway, NJ 08854, United States 2 Department of Mathematics, National University of Singapore, Singapore 117543, Singapore

Published  November 2013

Citation: Ka Luen Cheung, Man Chun Leung. Asymptotic behavior of positive solutions of the equation $\Delta u + K u^{\frac{n+2}{n-2}} = 0$ in $IR^n$ and positive scalar curvature. Conference Publications, 2001, 2001 (Special) : 109-120. doi: 10.3934/proc.2001.2001.109
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Impact Factor:

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# surfacegamma

Gamma value for different terrains

Since R2021a

## Syntax

``G = surfacegamma(TerrainType)``
``G = surfacegamma(TerrainType,FREQ)``
``surfacegamma``

## Description

````G = surfacegamma(TerrainType)` returns the γ value for the specified terrain. The γ value is for an operating frequency of 10 GHz.```

example

````G = surfacegamma(TerrainType,FREQ)` specifies the operating frequency of the system.```
````surfacegamma` displays several terrain types and their corresponding γ values. These γ values are for an operating frequency of 10 GHz.```

## Examples

collapse all

Determine the γ value for a wooded area, and then simulate the clutter return from that area. Assume the radar system uses a single cosine pattern antenna element and has an operating frequency of 300 MHz.

```fc = 300e6; g = surfacegamma('woods',fc); clutter = constantGammaClutter('Gamma',g, ... 'Sensor',phased.CosineAntennaElement, ... 'OperatingFrequency',fc); x = clutter(); r = (0:numel(x)-1)/(2*clutter.SampleRate) * ... clutter.PropagationSpeed; plot(r,abs(x)) xlabel('Range (m)') ylabel('Clutter Magnitude (V)') title('Clutter Return vs. Range')```

## Input Arguments

collapse all

Terrain type, specified as one of these:

• `"Sea State 3"`

• `"Sea State 5"`

• `"Woods"`

• `"Metropolitan"`

• `"Rugged Mountain"`

• `"Farmland"`

• `"Wooded Hill"`

• `"Flatland"`

Data Types: `char` | `string`

Operating frequency of radar system in hertz, specified as a scalar or a vector.

Data Types: `double`

## Output Arguments

collapse all

Gamma value, γ, in decibels, for constant-γ clutter model, returned as a scalar.

collapse all

### Gamma

A frequently used model for clutter simulation is the constant gamma model. This model uses a parameter, γ, to describe clutter characteristics of different types of terrain. Values of γ are derived from measurements.

## Algorithms

The γ values for the terrain types `"Sea State 3"`, `"Sea State 5"`, `"Woods"`, `"Metropolitan"`, and `"Rugged Mountain"` are from [2]. The γ values for the terrain types `"Farmland"`, `"Wooded Hill"`, and `"Flatland"` are from [3].

Measurements provide values of γ for a system operating at 10 GHz. The γ value for a system operating at frequency f is:

`$\gamma ={\gamma }_{0}+5\mathrm{log}\left(\frac{f}{{f}_{0}}\right)$`

where γ0 is the value at frequency f0 = 10 GHz.

## References

[1] Barton, David. "Land Clutter Models for Radar Design and Analysis," Proceedings of the IEEE. Vol. 73, Number 2, February, 1985, pp. 198–204.

[2] Long, Maurice W. Radar Reflectivity of Land and Sea, 3rd Ed. Boston: Artech House, 2001.

[3] Nathanson, Fred E., J. Patrick Reilly, and Marvin N. Cohen. Radar Design Principles, 2nd Ed. Mendham, NJ: SciTech Publishing, 1999.

## Version History

Introduced in R2021a


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# For last 3 years, your investment returned: {+35%, +35%, -70%}. Which measure of central tendency would best describe your annual return?

Arithmetic Mean, Geometric Mean, Median or Standard Variance?

Say I started with $100. 1st Year =$135

2nd Year = $182.25 3rd Year = 54.675 100-54.675 = 45.325 = loss / 3 = roughly 15 = annual loss. Arthemtic Mean is 0 so thats no good. Median is 35 so thats no good. Standard Variance doesn't measure central tendency. Geometric Mean yields 44 so where am I going wrong? • I obtain -18% for the geometric mean return ($100(1 - (1.35\times1.35\times0.30)^{1/3})\$). How do you obtain 44? – whuber Oct 30 at 21:53
• Hey thanks for helping. Isn't the formula (35 * 35 * 70)^(1/3)? I thought I had to multiply all the numbers and n root it – Benny Oct 30 at 21:54
• You seem to be treating a loss of 70% as if it were a gain of 70%! (I wish I could do that with my investments ...) – whuber Oct 30 at 21:55
• Oh i c now..wow thanks for ur help. – Benny Oct 30 at 21:56
• Just to make sure, I can't measure central tendency with standard variance, correct? – Benny Oct 30 at 21:58


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Home > CC1 > Chapter 1 > Lesson 1.2.3 > Problem1-70

1-70.

How many pennies are represented by each expression below? Homework Help ✎

1. $3+4(5)$

Remember to multiply first, then add.

$23$ pennies

2. $4(3)+7$

This one is similar to the one above.

3. $2(3)+5+4(2)$

One student, Christy, did it this way:

$(2)(3) = 6$
$6 + 5 = 11$
$11 + 4 = 15$
$15 × 2 = 30$

What did Christy do wrong?

Make sure you do what is inside the parentheses first.

$19$ pennies


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# Rossby radius as horizontal scale

In the introduction of "Geographical and monthly variability of the first baroclinic Rossby radius of deformation in the South China Sea", the authors states:

In both theoretical and observational studies of large and meso-scale oceanic processes, the first baroclinic Rossby radius of deformation (R1) is an important quantity in determining horizontal scales which characterize eddy sizes, trapping scales for coastal jets and equatorial flows.

I am familiar with the fact that R1 can determine eddy sizes (e.g., this answer), but I am having an hard time to find references about trapping scales for coastal jets and equatorial flows.

Can someone, please, give me some explanations and/or references?

• For equatorial flows, please check for Matsuno 1966 Quasi-Geostrophic motions in the equatorial area. – Enver Apr 6 '18 at 17:29

Let the Rossby radius of deformation be denoted as $L_R$. According to Holton's Fifth Edition of "An introduction to Dynamic Meteorology," $$L_R=\frac{NH}{f_0}$$ where $N$ is the Brunt-Vaisala frequency, $H$ is the scale height, and $f_0$ is the Coriolis parameter.
For phenomena with length scale $L$, buoyancy can be neglected if $L<<L_R$.
In statically stable equatorial flows, $$\lim_{f_0\to 0} L_R=\infty$$
In certain coastal flows, $L \approx L_R$
So the author is saying that in equatorial flows, buoyancy can't be neglected, since the Rossby radius "blows up" and for coastal flows $L\approx L_R$


text:
f

How to write tcl generated xml file into a new file

Hi,

I have generated a xml file using xmlgen.
After running the tcl script the ouput is displayed on the terminal .
I want to take the terminal output into a xml file.

Any Suggestion ???/

Thanks
Ranjan

 0
11/25/2008 11:46:48 AM
comp.lang.tcl 23429 articles. 2 followers.

1 Replies
910 Views

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[PageSpeed] 30

On Nov 25, 12:46=A0pm, gyana behera <ranjan....@gmail.com> wrote:
> Hi,
>
> I have generated a xml file using xmlgen.
> After running the tcl script the ouput is displayed on the terminal .
> I want to take the terminal output into a xml file.
>
> Any Suggestion ???/
>
> Thanks
> Ranjan

Either redirect the output of the script to the xml-file. (xmlgen >
output.xml) or open the file in the script:

set f [open $filename w] then use puts$f instead of puts to output to the file and finally
close the file:

Komodo: Accessing widgets in the auto generated _ui.tcl from the associated .tcl file
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I'm just trying to write a short script that will search through web server log files and write out lines from the original files that contain a certain term to a new file. I did have the "puts" line of my code working, but after a few changes that doesn't work either! What am I doing wrong?? : There are a bunch of "ex2005##.log" files in the current directory. -------------------- #!c:\ruby\bin\ruby def find_r... How to write output of a tcl script to a file? Hi All I am a newbie in Tcl. And when it comes to file I/O, I get extremely confused. I am trying out below example from Tcl tutorial (Link ---> http://www.tcl.tk/man/tcl8.5/tutorial/Tcl25.html ) # # Report all the files and subdirectories in the current directory # For files: show the size # For directories: show that they _are_ directories # set dirs [glob -nocomplain -type d *] if {$dirs != {} } { puts "Directories: foreach d [lsort $dirs] { puts "$d" } } else { puts "(no subdirectories)" } set files [glob -nocomplain -type f *] if { $files != {} } { puts "Files: foreach f [lsort$files] { puts " [file size $f] -$f" } } else { puts "(no files)" } ################################ I want to write the output of this script directly to the new file. help me with that. And if anyone can provide a good tutorial with lots of examples about File I/O in Tcl, that'd be best. thanks in advance. On 13 nov, 08:23, kj <kinjaldesai200...@gmail.com> wrote: > Hi All > > I am a newbie in Tcl. And when it comes to file I/O, I get extremely > confused. > I am trying out below example from Tcl tutorial > (Link --->http://www.tcl.tk/man/tcl8.5/tutorial/Tcl25.html) > > # > # Report all the files and subdirectories in the current directory > # For files: show the size > # For directories: show that they _are_ directories > # > >...

UPnP xml file to WSDL xml file
Hello, could somebody help me to convert the below UPnP service file to WSDL web service file: <serviceList> <service> <serviceType> urn:schemas-upnp-org:service:PowerSwitch:1 </serviceType> <serviceId> urn:upnp-org:service:PowerSwitch:1 </serviceId> <controlURL>/upnp/control/power1</controlURL> <eventSubURL>/upnp/control/power1</eventSubURL> <SCPDURL>/blenderpowerSCPD.xml</SCPDURL> </service> thanks in advance bobby I tried to combine UPnP servive and blenderpowerSCPD.xml and come up with WSDL w...

how to make a file from the strings in a file and redirect the content to these new files
Here is scenari0 223 files copied to this this one big flat file. now how to make this file to 223 files. further explaination: below is excerpt from that one file under xxxxx you will see @TC_ID: :BIC4007 then another @TC_ID: :BIC40008 and has a 223 time with TC_ID: :BICxxx numbers How do I use such a script that makes a file named BIC40008, then another BIC40007 AND copying the content from line @TC_ID: :BIC40007 till one line before the next TC_ID: :BIC40008 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx @TC_ID: :BIC40007 @TC_TITLE: @ACT_IDS: @AUTHOR: @AUTO: @CLASS: @GROUP_ID: @RELEASES: @T...

Open folder, read file, close file, open new folder, read file close file etc....
Hello, My current directory has several folders within it. Each folder has the same type of file within it. How do i go from the current directory, into one of the folders, edit the file, then come out of the folder, then go into another folder, edit a file then come out of the folder then into a new one etc etc etc. If this does not make sense i apologise and will try to explain my problem a little better. Any help would be much appreciated. Cheers, Phil Phil: I'm not sure it makes sense. Personally I use the mouse and keyboard for doing this. Were you thinking of something else?...

Creating a TCL Channel from a C FILE* giving strange file handles
Hey Everybody! Having a problem with the Tcl C Library calls I cant seem to= figure out. I'm trying to create a channel from opened FILE *, and use it = in a TCL script that sources the module. So, what I am running so far is: =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D int openCFileCmd(ClientData clientData,=20 Tcl_Interp *interp, int objc, Tcl_Obj *CONST o= bjv[])=20 ...

How to capture the output from TCL prompt and save it to a file on Linux file system.
Hello All, I login to TCL prompt and run a simple query which lists few rows. I want to capture those rows and save it to a file on file system. I am working on a Sun Solaris ver. 5.9. Here is the sample query that i run from TCL prompt: >SELECT DISTINCT CATEGORY FROM DS_JOBS; And the Resultant set looks like this: Category............ ABC\ACEIQ\JOBS\C UST_SPCLTY ABC\ACEIQ\JOBS\C UST_SPC SSHIP\SEQ PLIC\CNTRL Any advise is greatly appreciated. Thanks Vini On Mar 22, 7:26=A0pm, Vini <vinay.renukun...@gmail.com> wrote: > Hello All, > > I login to TCL prompt and run a simple query which lists few rows. I > want to capture those rows and save it to a file on file system. I am > working on a Sun Solaris ver. 5.9. > > Here is the sample query that i run from TCL prompt: > > >SELECT DISTINCT CATEGORY FROM DS_JOBS; > > And the Resultant set looks like this: > > Category............ > > ABC\ACEIQ\JOBS\C > UST_SPCLTY > ABC\ACEIQ\JOBS\C > UST_SPC > SSHIP\SEQ > PLIC\CNTRL > > Any advise is greatly appreciated. Wrong group. You mean "SQL" rather than "TCL". -Alex On Mar 22, 2:30=A0pm, Alexandre Ferrieux <alexandre.ferri...@gmail.com> wrote: > On Mar 22, 7:26=A0pm, Vini <vinay.renukun...@gmail.com> wrote: > > > > > > > > > > > Hello All, > > > I login to TCL prompt and run a simple query which lists few rows. I > > want...

tcl-inotify: File & directory level notifications for Tcl. (extension)
This is an announcement for a relatively new Tcl project: tcl-inotify. Tcl-inotify is a Tcl interface to the linux kernel's inotify* service. It provides a Tcl package that enables Tcl scripts to monitor the file system changes at a file and directory level. A large number of file/directory actions can be monitored. A proc is executed asynchronously to notify for changes. * inotify is a Linux kernel subsystem that provides file system event notifications. It was written by John McCutchan with help from Robert Love and later Amy Griffis to replace dnotify. It was included in the mainline kernel starting from release 2.6.13, and could be compiled into 2.6.12 and possibly earlier releases by use of a patch. Its function is essentially an extension to filesystems to notice changes to the filesystem, and report those changes to applications. Home page: http://sourceforge.net/projects/tcl-inotify/ Man page: http://tcl-inotify.sourceforge.net/ Author: Alexandros Stergiakis alsterg ...

Anyone out there write a *.puz file using tcl?
Subject sez all. On Fri, 10 Apr 2009 07:40:55 -0700 (PDT), razajac wrote: >Subject sez all. Not to me, it doesn't. I used the extension .puz for problem files for my Tk Sudoku solver/assistant. Here's the contents of one such: ....9.6... 3...1...8 67..2..39 ...2...3.. 43..5..87 ...5...1.. 52..6..43 1...9...2 ....2.4... I'ts quite a hard puzzle; it defeated all my program's heuristics and forced me to add backtracking as a last resort. Is .puz used by some other file format or program? -- Jonathan Bromley, Consultant DOULOS - Developing Design Know-how VHDL * Verilog * SystemC * e * Perl * Tcl/Tk * Project Services Doulos Ltd., 22 Market Place, Ringwood, BH24 1AW, UK jonathan.bromley@MYCOMPANY.com http://www.MYCOMPANY.com The contents of this message may contain personal views which are not the views of Doulos Ltd., unless specifically stated. On Apr 11, 12:39=A0am, Jonathan Bromley <jonathan.brom...@MYCOMPANY.com> wrote: > On Fri, 10 Apr 2009 07:40:55 -0700 (PDT),razajacwrote: > >Subject sez all. > > Not to me, it doesn't. > > I used the extension .puz for problem files > for my Tk Sudoku solver/assistant. =A0Here's the > contents of one such: > > ...9.6... > 3...1...8 > 67..2..39 > ..2...3.. > 43..5..87 > ..5...1.. > 52..6..43 > 1...9...2 > ...2.4... > > I'ts quite a hard puzzle; it defeated all my program's > heuristics and forced me to add backtracking ...

Web resources about - How to write tcl generated xml file into a new file - comp.lang.tcl

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text:
# Parameters with trig functions

I'm having difficulty eliminating the parameter in the equations: $x = (tan^2\theta)$, $y = sec\theta$. The only strategy I know of for tackling trig parameters is to use the identity [$sin^2(x) + cos^2(x) = 1$] before setting that equal to some expression of $x + y$, but tangent gives me $x = \frac{sin^2\theta}{cos^2\theta}$, and I have no idea how to eliminate the denominator to get part of the identity. Am I just going about this completely wrong?

Thank you!

Squaring $y = \sec \theta$

$y^2 = \sec^2 \theta$

We know that,

$\sec^2 \theta - \tan^2 \theta = 1$

So we have,

$y^2 - x = 1$

Other method to drive. As you said,

$\sin^2 \theta + \cos^2 \theta = 1$

Divide above equation by $\cos^2 \theta$

$\tan^2 \theta + 1 = \sec^2 \theta$

$x + 1 = y^2$

• Mine pleasure.. – Kanwaljit Singh Jan 27 '17 at 3:07

Use,

$$\sin^2(\theta)+\cos^2(\theta)=1$$

Dividing both sides by $\cos^2 (\theta)$ gives:

$$\tan ^2 (\theta)+1=\sec^2 (\theta)$$

This should be enough to conclude.

$$x+1=y^2$$

• You are doing it completely wrong. – Kanwaljit Singh Jan 27 '17 at 3:03
• Why? @KanwaljitSingh – Ahmed S. Attaalla Jan 27 '17 at 3:04
• Uh...he did exactly the same thing you did, but with the $\tan^2$ on the other side of the equation. Saying "that's all wrong" without pointing out the location of the error is ... not very helpful, and against the spirit of the site. – John Hughes Jan 27 '17 at 3:05
• Thank you both for all your help! I missed the identity you both used, that makes a lot more sense! – JMartin Jan 27 '17 at 3:06
• @John Hughes actually his complete statement is wrong. That's why I wrote just that. – Kanwaljit Singh Jan 27 '17 at 3:22


text:
GITS 2015: CloudFS

This CTF was what I’d call an humbling experience; it was an absolutely great contest, don’t get me wrong, but damn! it was hard!, and since I’m not a CTF veteran, let me say that I learned an important thing: “There is no limit to the evilness people can put in their effort of creating challanging puzzles”.

Ok but let’s get back to this specific challenge. For 200 points we were asked to extract an hidden key from a pcap file.

The name of the challenge was an hint itself: cloudfs made me think about something that was used to transfer a file, and by opening a pcap all i could see were some HTTP traffic packets to wikipedia, and loots and loots of ICMP ping requests/responses.

Before diving into the pcap itself i ran through some usual checks you do to spot obvious things, and after having verified that nothing was appended at the bottom of the pcap file, and that it wasn’t a polyglot i started looking closer at the packets themselves.

Fresh from the Persistence competition I figured out pretty quickly that the icmp packets could have been used as a cover channel to hide something, and in fact if we look at the payload data from a couple of random icmp requests we could easly notice that the ping pattern was containing custom data.

The interesting things start at packet 1040 in the pcap file, where we can see the header for a bzip2 file

which was repeating onwards every 4 packets till the end of the pcap file.

I’ve then tried to extract the data payload of the four interesting packets and use binwalk to see if it could extract something useful.

# binwalk -e filtered.pcap

DECIMAL   	HEX       	DESCRIPTION
-------------------------------------------------------------------------------------------------------------------
20        	0x14      	bzip2 compressed data, block size = 900k

It was actually detecting and extracting a bzip2 file, the header was there, after all, but if i tried extracting its content it was informing me that the file was corrupted.

# bunzip2 14.bz2

bunzip2: Data integrity error when decompressing.
Input file = 14.bz2, output file = 14

It is possible that the compressed file(s) have become corrupted.
You can use the -tvv option to test integrity of such files.

You can use the bzip2recover' program to attempt to recover
data from undamaged sections of corrupted files.

bunzip2: Deleting output file 14, if it exists.

It took me a little while to notice that the pings were sent in an incorrect order, and that it was necessary to reorder them properly.

In fact, let’s look at the icmp packet identifier.

Here it’s possible to notice that the identifier are ordered as 13 14 16 15

I’ve tried to extract the packets payload data separately (after all they were only 4 packets) and reassemble them by hand.

# cat 13.raw 14.raw 15.raw 16.raw > reassembled.raw
# binwalk reassembled.raw -e

DECIMAL   	HEX       	DESCRIPTION
-------------------------------------------------------------------------------------------------------------------
20        	0x14      	bzip2 compressed data, block size = 900k

# ls -la
total 64
drwxr-xr-x 2 root root  4096 Jan 20 20:00 .
drwxr-xr-x 7 root root 36864 Jan 20 20:00 ..
-rw-r--r-- 1 root root 24556 Jan 20 20:00 14

# file 14
14: POSIX tar archive (GNU)

# mv 14 14.tar; tar xvf 14.tar
key
ping/ping.py

and appearently we got the tar archive containing both the key, as well as the python script that was used to create the cover channel

# cat key
key{WhyWouldYouEverUseThis}
`

And this one was one of the simpliest challenges in this contest… so well… congratulations to all the teams that solved a lot of them :)


text:
Timezone: »

Revisiting Bellman Errors for Offline Model Selection
Joshua Zitovsky · Rishabh Agarwal · Daniel de Marchi · Michael Kosorok

Applying offline reinforcement learning in real-world settings necessitates the ability to tune hyperparameters offline, a task known as $\textit{offline model selection}$. It is well-known that the empirical Bellman errors are poor predictors of value function estimation accuracy and policy performance. This has led researchers to abandon model selection procedures based on Bellman errors and instead focus on evaluating the expected return under policies of interest. The problem with this approach is that it can be very difficult to use an offline dataset generated by one policy to estimate the expected returns of a different policy. In contrast, we argue that Bellman errors can be useful for offline model selection, and that the discouraging results in past literature has been due to estimating and utilizing them incorrectly. We propose a new algorithm, $\textit{Supervised Bellman Validation}$, that estimates the expected squared Bellman error better than the empirical Bellman errors. We demonstrate the relative merits of our method over competing methods through both theoretical results and empirical results on datasets from the Atari benchmark. We hope that our results will challenge current attitudes and spur future research into Bellman errors and their utility in offline model selection.

#### Author Information

##### Rishabh Agarwal (Google Research, Brain Team)

My research work mainly revolves around deep reinforcement learning (RL), often with the goal of making RL methods suitable for real-world problems, and includes an outstanding paper award at NeurIPS.

##### Michael Kosorok (University of North Carolina at Chapel Hill)

Michael R. Kosorok, PhD, is the W.R. Kenan, Jr. Distinguished Professor of Biostatistics and Professor of Statistics and Operations Research at the University of North Carolina at Chapel Hill. Research interests include reinforcement learning, precision medicine, and decision support.


text:
Flawless's blog

By Flawless, 8 years ago,

I was trying a Problem on Vertex Cover.I can't think of any good algo to do this. www.spoj.pl/problem/PT07X/ Can anyone give any hint to the method ? just a little hint ?

• 0

By Flawless, 8 years ago,

Hello Everyone. I was busy with my exams. before that i read on CROC entry that it will be open for Non- Russian to participate too ! I tried hard to find link to register for it..but i can't find the Registration link So can anyone provide link, i am looking forward to this..

Thanks a lot, Flawless (just another human )

• +4

By Flawless, 8 years ago,

What is the most efficient way to find Lowest Common Ancestor in Binary Tree ?

• +2

By Flawless, 8 years ago,

is it true that if you have added someone in your friend list on CF. you and him/her can never be in same room . ?

• +2

By Flawless, 8 years ago,

I know you can get an AC in problem with normally optimized prime sieve. but sometimes while doing problem on SPOJ. i noticed when i have execution time of 2 sec (for example), many good coders (Respected) have brought down their execution time to 0.5 sec around .

Here is what i have done for my optimized prime sieve.

Optimization i have done-
1. running main iteration loop to only square root of Limit.
2. avoid for even numbers.
3. not checking for multiple of 2 and 3
4. Prime numbers are of form 6k+1 or 6k+5.
5. in inner loop of j, insted of j+=i, i did j+=2*i. because i*i(as i will be odd for sure, i haven't run this loop for "2") will be odd for sure. so i*i+i will be odd for sure. so skip it by j+=2*i

Since even numbers and multiple of 3 are never marked. so they will never be checked while collecting all primes

 #define lim 10000009
vector<bool> isprime(lim,1);
void sieve()
{
int i,j,t,v,sq;
isprime[1]=isprime[0]=0;
sq=sqrt(lim);
for(i=5,t=2;i<=sq; )
{
if(isprime[i])
{
for(j=i*i;j<lim;j+=2*i)
isprime[j]=0; // 0 means composite- not prime , 1 means prime
}
i+=t;
t=6-t;
}
}

• +10

By Flawless, 8 years ago,

Respected Admin or Anyother coder i submitted this code for problem B. i am getting correct answer for Sample testcase but on codeforces i get "0" for this... i tested for many times and problem still prevails.. Many many advance thanks :)

 #include<iostream>
#include<cstdio>
#include<algorithm>
#define LL long long int
using namespace std;
int main()
{
LL n,t,i;
cin>>n>>t;
LL arr[n];
for(i=0;i<n;i++)
cin>>arr[i];
LL count=0,sum,start,j,ans,temp;
temp=0;
start=0;
count=0;
for(i=0;i<n;i++)
{
sum+=arr[i];
if(sum<=t)
temp++;
else
if(sum>t)
{
count=max(count,temp);
j=start;
temp++;
while(1)
{
sum-=arr[j];
j++;
temp--;
if(sum<=t)
break;
if(j>i)
break;

}
start=j;
}

}
count=max(count,temp);
cout<<count<<endl;
return 0;
}


text:
Loughborough University
Leicestershire, UK
LE11 3TU
+44 (0)1509 263171

# Loughborough University Institutional Repository

 Please use this identifier to cite or link to this item: https://dspace.lboro.ac.uk/2134/10954

 Title: Modelling thin film growth over realistic time scales Authors: Blackwell, Sabrina Issue Date: 2012 Publisher: © Sabrina Blackwell Abstract: Energy security and supply is a key problem for the UK in the coming years. Photovoltaics have an important role to play in this. In order for demand to be met, continued research into new materials and methods of production is necessary. By modelling deposition techniques using classical molecular dynamics (MD), an atomistic scale understanding can be obtained. Combining this with long time scale dynamics (LTSD) techniques allows us to also model diffusion and surface growth over realistic time scales. The LTSD technique applied throughout this project is an on-the-fly Kinetic Monte Carlo (otf-KMC) method, which determines diffusion pathways and barriers, in parallel, with no prior knowledge of the involved transitions. These simulation techniques allow parameters such as deposition energy, substrate bias and plasma pressure to be easily changed to gain understanding of their effects. During this project, growth via industrial scale deposition techniques has been simulated, including evaporation (thermal and electron beam), ion-beam assisted evaporation and reactive magnetron sputtering. Metal thin films, of interest due to their uses in reflectors in concentrator photovoltaics, electrical conductors in the monolithic interconnect processes and back contacts, were investigated using otf-KMC. Ag and Al film growth was simulated for around 0.3 seconds of real time. It was found that Ag has the ability to grow smooth surfaces, using several mechanisms including multiple-atom concerted motion, exchange mechanisms, and damage and repair mechanisms. Ag (111) and (100) surfaces grew dense, complete and crystalline films when sputtering was simulated, however, evaporation deposition produced incomplete layers. The inclusion of Ar in the ion-beam assisted evaporation of Ag (111) aided growth by transferring extra energy to the surface allowing increased diffusion and atomic mixing. Al (111) and (100), however, show different patterns. Growth by evaporation deposition and magnetron sputtering actually produced very similar results. The inclusion of the ion-beam assist on the (111) surface actually damaged the film, producing subsurface Ar clusters where Al atoms were displaced, creating voids throughout the film. Otf-KMC methods enabled the investigation of specific mechanisms allowing film growth and a very important transition enabling the smooth and complete Al film growth was found to be the Ehrlich-Schwoebel (ES) barrier. The ES barrier involves an atom dropping off a step edge of an island and this barrier was found to be much smaller for the Al surfaces, therefore allowing the more complete growth from both evaporation and sputtering. Metal oxides are also of great interest in the photovoltaic industry. The rutile TiO$_2$ (110) surface was investigated using single point depositions, high temperature MD and otf-KMC. Otf-KMC enabled the simulation for up to 9 seconds of real time, totally inaccessible using traditional simulation methods. Results concluded that the evaporation deposition process produced a void filled, incomplete structure, even with the use of a low energy ion-beam assist, this material is of interest for dye-sensitised solar cells where a dye is injected into the voids. Sputtering, however, produced dense and crystalline film, which is much more applicable to anti-reflective coatings where a crystalline structure is required. Mechanisms which enabled crystalline rutile to form were also investigated, highlighting Ti interstitial annealing in the presence of an O rich surface as an important rutile growth mechanism. ZnO, an inorganic compound with many uses including transparent conductive oxides, is investigated in the most stable wurtzite phase. The O-terminated (000$\bar{1}$) polar surface was used as the substrate for otf-KMC growth simulations, where around 1 second of real time was simulated. Evaporation deposition of a stoichiometric distribution of deposition species was found to produce the best quality film, however, a phase boundary was observed where an area of zinc blende forms within the wurtzite. Sputtering resulted in a denser, more complete and crystalline structure due to the higher deposition energy of arriving species, similar to the TiO$_2$ results. Post-annealing at 770K did not allow complete recrystallisation, resulting in films with stacking faults where monolayers formed in the zinc blende phase. Annealing at 920K, however, in some cases enabled the complete recrystallisation of films back into the wurtzite structure. Although, the higher annealing temperature did not always enable recrystallisation and in some cases both wurtzite and zinc blende phases existed in the same layer, resulting in a phase boundary. An important mechanism for the nucleation of ZnO growth was found to be the formation and vibration of Zn$_x$O$_y$ strings on the surface, which after hundreds of milliseconds formed the desired hexagonal structure. Combining MD and otf-KMC enabled the simulation of systems over very large time scales which were previously totally inaccessible. Key mechanisms occurring during the growth of metals and metal oxides were investigated, providing a much more precise understanding of how growth occurs. It is clear from the work that the deposition technique used plays a significant role on the resulting film quality and surface morphology and we are now able to provide an insight into the optimum conditions under which complete, crystalline layers can form. Description: A Doctoral Thesis. Submitted in partial fulfilment of the requirements for the award of Doctor of Philosophy of Loughborough University. URI: https://dspace.lboro.ac.uk/2134/10954 Appears in Collections: PhD Theses (Maths)

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# PSP Bibliography

 Notice: Clicking on the DOI link will open a new window with the original bibliographic entry from the publisher. Clicking on a single author will show all publications by the selected author. Clicking on a single keyword, will show all publications by the selected keyword.

Erratum: On the properties of Alfv\ enic switchbacks in the expanding solar wind: Three-dimensional numerical simulations [Phys. Plasmas 29, 072902 (2022)]

 Author Johnston, Zade; Squire, Jonathan; Mallet, Alfred; Meyrand, Romain; Keywords Parker Data Used; Astrophysics - Solar and Stellar Astrophysics; Physics - Plasma Physics Abstract Switchbacks — abrupt reversals of the magnetic field within the solar wind — have been ubiquitously observed by Parker Solar Probe (PSP). Their origin, whether from processes near the solar surface or within the solar wind itself, remains under debate, and likely has key implications for solar wind heating and acceleration. Here, using three-dimensional expanding box simulations, we examine the properties of switchbacks arising from the evolution of outwards-propagating Alfv\ en waves in the expanding solar wind in detail. Our goal is to provide testable predictions that can be used to differentiate between properties arising from solar surface processes and those from the in-situ evolution of Alfv\ en waves in switchback observations by PSP. We show how the inclusion of the Parker spiral causes magnetic field deflections within switchbacks to become asymmetric, preferentially deflecting in the plane of the Parker spiral and rotating in one direction towards the radial component of the mean field. The direction of the peak of the magnetic field distribution is also shown to be different from the mean field direction due to its highly skewed nature. Compressible properties of switchbacks are also explored, with magnetic-field-strength and density fluctuations being either correlated or anticorrelated depending on the value of $\beta$, agreeing with predictions from theory. We also measure dropouts in magnetic-field strength and density spikes at the boundaries of these synthetic switchbacks, both of which have been observed by PSP. The agreement of these properties with observations provide further support for the Alfv\ en wave model of switchbacks. Year of Publication 2022 Journal Physics of Plasmas Volume 29 Number of Pages 129902 Section Date Published dec ISBN URL DOI 10.1063/5.0133296


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# How do I find the Laurent series expansion?

I want to find the order of the pole of $$\sqrt{1+\frac{1}{z}}$$ at 0. But how can I expand this function at 0? Can I take the Taylor expansion of $$\sqrt{1+z}$$ and plug in $$\frac{1}{z}$$ in the place of z? And how can one find the Laurent series expansion of a function in general? I mean since the function diverges toward infinity at poles, how can the Laurent series be "defined"?

• The function $\sqrt{1+1/z}$ has branch points at $z=-1$ and $z=0$. So, you can develop the Laurent series outside the unit disk. Jan 17 at 16:28

You're quite right that to find the Laurent series expansion for $$f(z)=\sqrt{1+\frac1z}$$ about $$0,$$ we need only start with the Taylor series expansion of $$g(z)=\sqrt{1+z}$$ about $$0,$$ then note that $$f(z)=g(1/z).$$ In fact, this will show you that it isn't a pole, but rather an essential singularity.
As for how the Laurent series can be thought of as being defined, recall that a power series is defined wherever it converges. In this case, it converges at all values of $$z$$ such that $$|z|>1.$$
• The Taylor series of $g$ around $z = 0$ converges for $|z| < 1$, so the corresponding Laurent series of $f$ converges for $|z| > 1$, not for all non-zero $z$. $f$ does not have a Laurent series expansion on $0 < |z| < r$, $z = 0$ is not an essential singularity of $f$. Jan 17 at 20:38
• To be clear, the new center of expansion $z = \infty$ is not an essential singularity of $f$ either. Jan 17 at 22:52


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# Why does iterating in different ways produce different solutions?

Recently, I came across the following equation: $$2^x=4x$$ To solve it, I decided to iterate. Firstly I stated: $$x_{n+1}=\frac{2^{x_n}}{4},x_0=1$$ and found a solution of $x\approx 0.3099069324$.

Then I again rearranged it to: $$x_{n+1}=\log_{2}({4x_n}), x_0=1,$$ and achieved the solution $x=4$.

Despite trying many values, I was unable to get these iterations to find the other solution. What I would like to know is: what is the reasoning behind this?

I have seen a similar thing happen in the case of other iterations, where I iterate for a solution in different ways and get two different (correct) solutions - and it has perplexed me.

## 3 Answers

You are aiming at fixed points of two different functions.

Some fixed points are attractors and some are repellers.

If you graph your functions you see as you iterate, the points tend to the attractor and stay away from the repeller.

For $f(x) = \frac {2^x}{4}$ the point $x\approx 0.3099069324$ is an attractor and $x=4$ is a repeller.

For the other function it is the other way around.

• Attractors are cases where the absolute value of the derivative of the function being iterated is less than $1$. When you are close to a fixed point the distance to the fixed point is multiplied by about the derivative at each step. If that is less than $1$ in absolute value the iteration will converge. – Ross Millikan May 13 '18 at 21:00
• Thanks for a very informative comment. I should have mentioned this point in my answer. – Mohammad Riazi-Kermani May 13 '18 at 21:13
• I thought it would be helpful. You are welcome to incorporate it. – Ross Millikan May 13 '18 at 21:29

An "exact" solution may be obtained with use of the Lambert W-function which is defined by $W_{0}(x) e^{W_{0}(x)} = x$ for $x \geq -1$. This is seen by the following: \begin{align} 2^{x} &= 4x \\ 4 x &= e^{x \, \ln2} \\ (- x \, \ln2) \, e^{- x \, \ln2} &= - \frac{\ln2}{4} \\ - x \, \ln2 &= W_{0}\left( - \frac{\ln2}{4} \right) \\ x &= - \frac{1}{\ln2} \, W_{0}\left( - \frac{\ln2}{4} \right). \end{align} By calculation this gives a value of $x \approx .30990693238\cdots$.

• By $W_0$, do you mean the inverse Lambert W-function ? – onurcanbektas Jun 26 '18 at 3:46
• @onurcanbektas $W_{0}(x)$ is the Lambert W function valid for $x \geq -1$ and is the principle branch. – Leucippus Jun 26 '18 at 3:50
• What I meant is that the LHS is the image of $W(-x \ln(2))$, so I assumed you have applied the inverse function to both sides. – onurcanbektas Jun 26 '18 at 5:35

Sketch a graph for the two functions and you see that $y=\log_2 4x$ is greater than $y=x$ and the iteration tends to the greater solution while $y=2^x/4$ is smaller than $y=x$ and the iteration tends to the smaller solution.


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Обычная версия сайта

Totally dissipative dynamical processes

Communications on Pure and Applied Analysis. 2014. Vol. 13. No. 5. P. 1989-2004.
Chepyzhov V. V., Pata V., Conti M.

We discuss the existence of the global attractor for a family of processes $U_\sigma(t,\tau)$ acting on a metric space $X$ and depending on a symbol $\sigma$ belonging to some other metric space $\Sigma$. Such an attractor is uniform with respect to $\sigma\in\Sigma$, as well as with respect to the choice of the initial time $\tau\in\R$. The existence of the attractor is established for totally dissipative processes without any continuity assumption. When the process satisfies some additional (but rather mild) continuity-like hypotheses, a characterization of the attractor is given.


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# How do you simplify -(3a^6b^4)^3?

Mar 18, 2018

$- 27 {a}^{18} {b}^{12}$

#### Explanation:

$\textcolor{w h i t e}{=} - {\left(3 {a}^{6} {b}^{4}\right)}^{3}$

$= - \left(3 \times 3 \times 3\right) \cdot {a}^{18} \cdot {b}^{12}$

$= - 27 {a}^{18} \cdot {b}^{12}$

the rules of indices say that you will multiply.


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# How to figure out what numbers often appear together in a dataset?

I have a lottery style dataset we produce internally (example below). I am trying to figure out which numbers appear most frequently together. Example questions: What are the top 10 pair of numbers that appear most frequently together? What are the top 10 three numbers that appear most frequently together?

What methods/techniques would I need to use to answer these questions?

I was thinking clustering at first but a test run (ward) I performed, didn't return the results I was expecting.

• You might be able to mine a closely related thread for useful ideas. Provided you have few columns (only five are shown in the example), it would take little to adapt the solutions.
– whuber
Nov 12 '13 at 20:17
• On the question* "which pairs of numbers appear most often together", I'd start with cross-tabulations. $\quad\,\,\,$ $\,$* the 'question' is actually a host of somewhat different questions, depending on things like whether the pair (Num1=3,Num2=5) is considered the same as (Num1=5,Num2=3) (does order matter?), or whether it's considered the same as (Num1=3,Num5=5) (do the column labels matter?) Nov 12 '13 at 21:42
• @whuber...thanks for the link. That sounds like something that could be modified to answer my question. And I am using r. But not sure I have the necessary skill level yet to adapt that r code. I'll give it shot though. Nov 12 '13 at 22:25
• @Glen_b Great clarifying question. Order doesn't matter. The three examples you gave are all equivalent for the question I am asking. Also, the numbers are non-repeating for each row. Nov 12 '13 at 22:33
• Then it sounds like you could collapse all the pairwise cross-tabulations into a single table (add the counts from all pairwise tables), then add the counts from that single table to its transpose (since order doesn't matter), and take the lower triangular portion of that combined table (since its symmetric). Then find the biggest counts in that table ... and that gives the pairs that occur together the most. (If that's useful to you let me know and I'll write it up as an answer) Nov 12 '13 at 22:58

This question calls for a modification of the solution to a sequence counting problem: as noted in comments, it requests a cross-tabulation of co-occurrences of values.

I will illustrate a naive but effective modification with R code. First, let's introduce a small sample dataset to work with. It's in the usual matrix format, one case per row.

x <- matrix(c(3,5,7,10,13,
3,5,8,10,15,
2,5,10,11,18,
1,3,4,6,8,
2,4,6,12,14,
3,5,8,10,15),
ncol=5, byrow=TRUE)

This solution generates all possible combinations of $m$ items (per row) at a time and tabulates them:

m <- 3
x <- t(apply(x, 1, sort))
x0 <- apply(x, 1, combn, m=m)
y <- array(x0, c(m, length(x0)/(m*dim(x)[1]), dim(x)[1]))
ngrams <- apply(y, c(2,3), function(s) paste("(", paste(s, collapse=","), ")", sep=""))
z <- sort(table(as.factor(ngrams)), decreasing=TRUE)

The tabulation is in z, sorted by descending frequency. It is useful by itself or easily post-processed. Here are the first few entries of the example:

> head(z, 10)
(3,5,10) (3,10,15)  (3,5,15)   (3,5,8) ... (8,10,15)
3         2         2         2 ...         2

How efficient is this? For $p$ columns there are $\binom{p}{m}$ combinations to work out, which grows as $O(p^m)$ for fixed $m$: that's pretty bad, so we are limited to relatively small numbers of columns. To get a sense of the timing, repeat the preceding with a small random matrix and time it. Let's stick with values between $1$ and $20,$ say:

n.col <- 8       # Number of columns
n.cases <- 10^3  # Number of rows
x <- matrix(sample.int(20, size=n.col*n.cases, replace=TRUE), ncol=n.col)

The operation took two seconds to tabulate all $m=3$-combinations for $1000$ rows and $8$ columns. (It can go an order of magnitude faster by encoding the combinations numerically rather than as strings; this is limited to cases where $\binom{p}{m}$ is small enough to be represented exactly as an integer or float, limiting it to approximately $10^{16}$.) It scales linearly with the number of rows. (Increasing the number of possible values from $20$ to $20,000$ only slightly lengthened the time.) If that suggests overly long times to process a particular dataset, then a more sophisticated approach will be needed, perhaps utilizing results for very small $m$ to limit the higher-order combinations that are computed and counted.

• @ whuber. Thanks for your posted Answer. Why is the output of z showing a count of 5 for (3,5,10)? Based on the original matrix I was expecting to see a count of 3. This would come from rows x[c(1:2,6),]. If I enter m<-5 I would expect to see in the output of z (3,5,8,10,15) with a count of 2 and and 5 additional groups each with a count of 1. Nov 14 '13 at 18:31
• Sorry: although I checked that, I misread the output and did not catch the discrepancy. There was a typo in the code where "2" appeared but "m" should have (and R issued no warnings about a mismatch between the sizes of arrays x0 and y; it merely "recycled" the elements). I fixed that typo and pasted the new output. You might find it informative (and reassuring) to run the code and inspect the array y, which contains all the $m$-tuples for each row in x. As another check, sum(z) should equal $\binom{p}{m}$ times the number of rows in $x$: $\binom{5}{3}\times 6=60$ in the example.
– whuber
Nov 14 '13 at 18:41
• np. I was actually just inspecting y as you were writing your answer and thinking "I was not expecting to see 15 columns for m<-3..Should be choose(5,3) =10". But your edit fixed that. This is way cool! Thank you for very much! Nov 14 '13 at 18:55

You are not looking for clustering.

Instead, you are looking for frequent itemset mining. There are dozens of algorithms for this, the most widely known probably are APRIORI, FP-Growth and Eclat.

I use a simple Excel spreadsheet and load all the drawings and then do a data sort. When you do a data sort you can learn quickly which numbers will never and have never played together. I've also written Macros that can run pairs and 3 pairs.


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# A travelling field; Feynman lectures on Physics Volume 2

In Feynman's Lectures on Physics, volume 2, Chapter 18 (18-5), we look at the creation of a electromagnetic field, due to a moving infinite sheet of charge,

By looking at the Maxwell equations alone, it is not easy to see directly how to get the solution. So we will first show you what the answer is and then verify that it does indeed satisfy the equations. The answer is the following: The field B that we computed is, in fact, generated right next to the current sheet (for small x). It must be so, because if we make a tiny loop around the sheet, there is no room for any electric flux to go through it $$^*$$. But the field B out farther—for larger x—is, at first, zero. It stays zero for awhile, and then suddenly turns on. In short, we turn on the current and the magnetic field immediately next to it turns on to a constant value B ; then the turning on of B spreads out from the source region. After a certain time, there is a uniform magnetic field everywhere out to some value x, and then zero beyond. Because of the symmetry, it spreads in both the plus and minus x-directions.

Questions:

$$^*$$ What does he mean, by "there's no room for electric flux to go through" , why does it conclude, "magnetic field immediately next to it turns on to a constant value B"? How can this be, how can magnetic field suddenly attain a constant value, it has to increase from 0 shouldn't it ?


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# Slope compensation of MAX17498

I want to use MAX17498 in boost mode.

I have a problem setting the slope compensation. The slope (as I understand) is the gradient of the inductor's current ramp. The unit is [A/s] or [A/us]. However, the MAX17498's datasheet requires it as [mV/us] (S_E)

I guess that the MAX17498's current sensor characteristic missing to calculate this value.

How to calculate the slope resistor's value of MAX17498?


text:
## brandon125 is anybody good in personal finance? 10 months ago 10 months ago

1. goformit100

I am alwasys good in it

2. brandon125

ok here goes the question Dividends are paid out of profits, and: A. dividend payments must be approved by the firm's board of directors. B. dividends are guaranteed. C. dividends are paid before a firm's taxes are paid. D. dividends are usually paid twice a year.

3. brandon125

@goformit100

4. goformit100

D. dividends are usually paid twice a year.

5. brandon125

thnx man Which of the following statements is true about financial planning? A. Any kind of financial expert (such as a stockbroker, lawyer, or accountant) can help you develop a comprehensive financial plan. B. Once you have painstakingly developed a financial plan, it is not wise to change it. C. Once you have made a decision to buy an investment, there is no need for continued evaluation. D. You must always consider the tax consequences of selling your investments.

6. brandon125

@goformit100

7. goformit100

8. goformit100

Please post this question in eastion

9. goformit100

ok

10. Luis_Rivera

if i were good in personal finance, i wouldnt be here !

11. brandon125

eastion?? which is...?

12. brandon125

are you actually good?? @Luis_Rivera

13. amistre64

i dont think "D" is the most correct answer

14. Luis_Rivera

No, that's the point, I'm actually very bad !

15. brandon125

can you help @amistre64

16. amistre64

i could .... but the answer to your first question is not "D"

17. brandon125

then what is it?

18. brandon125

@amistre64

19. amistre64

you have 3 other options to choose from, and i perfer not to just hand out an answer. pick one and we will see if its a "best choice"

20. brandon125

a?

21. amistre64

.. good guess :) before a dividend is declared, the board of directors have to determine of they can offer one to begin with.

22. brandon125

so thats the right one?

23. amistre64

its the "best" one .... yes

24. brandon125

ok. can you help me on the second question?

25. amistre64

the second one you posted has 2 ridiculous options, and 2 options that sound legit, can you narrow it down to the 2 legit options for me?

26. brandon125

a and b are the bad ones?

27. amistre64

no, b is absurd ... but a reads like it could be legit

28. brandon125

so b and c no, right? and the right answer is D?

29. amistre64

i would lean towards D as the "best" answer. A sounds good, but is not really accurate. A financial planner needs to be adept all those fields.

30. brandon125

ok. i have on more.

31. brandon125

Which of the following is a true statement? A. The dividends-earnings ratio is a key factor that serious investors use to evaluate stock investments. B. The price-earnings ratio for a corporation must be studied for one period only. C. The price-earnings ratio is based on the company's dividends. D. The price-earnings ratio for one firm may be compared to the price-earnings ratio for all firms.

32. brandon125

@amistre64

33. amistre64

you are aware that these questions are designed to see how well you have studied the material right?

34. brandon125

yea but these 3 are the only ones i missed on a test. and im retaking it. and i dont find answer in textbook.

35. amistre64

do you recall ho w to determine a PE ratio?

36. brandon125

nope

37. amistre64

im sure thats in your textbook ....

38. amistre64

since 3 answers deal with PE ratio, id say A is out; which one did you pick to begin with, we can rule that one out then

39. brandon125

im going with C being the answer

40. amistre64

$P/E=\frac{market~price}{earnings~per~share}$ $P/E=\frac{market~price(\#shares)}{net~income}$ i dont think dividends play a big role in that

41. brandon125

so its D

42. amistre64

i like D the best.

43. brandon125

thnx bro!


text:
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## SBI PO Reasoning Ability test 36

1 . Directions (1-5): Study the following information carefully and answer the questions given below:

Seven persons – Q, R, S, T, U, V and W – are seated in a straight line facing north but not necessarily in the same order. R is fourth to the left of W. S is on the immediate right of W. V is second to the left of T. U and R are immediate neighbours

$Q.$ Who among the following is seated exactly in the middle?

 A.   T B.   Q C.   V D.   R
2 . How many persons are seated between W and R?

 A.   None B.   One C.   Two D.   Three
3 . Which of the following represents persons seated at the two extreme ends?

 A.   U and S B.   U and W C.   R and S D.   R and W
4 . What is Q’s position with respect to S?

 A.   Third to the right B.   Second to the left C.   third to the left D.   Second to the right
5 . Four of the following five are alike in a certain way based on their seating position in the above arrangement and so form a group. Which is the one that does not belong to that group?

 A.   RV B.   QT C.   WS D.   WT
6 . How many such pairs of letters are there in the word BANTER, each of which has as many letters between them in the word (in both forward and backward directions) as they have between them in the English alphabetical series?

 A.   None B.   One C.   Two D.   Three
7 . How many meaningful words can be formed with the letters LDFO, using each letter only once in each word?

 A.   None B.   One C.   Two D.   Three
8 . The positions of how many alphabets will remain unchanged if each of the alphabets in the word PROACTIVE is arranged in alphabetical order from left to right?

 A.   None B.   One C.   Two D.   Three
9 . Directions (9-10): Following questions are based on the five three-lettered words given below: SHE AND TWO WIT GUM

(Note: The words formed after performing the given operations may or may not be meaningful English words.)

$Q.$ If all the letters in each of the words are arranged in alphabetical order (within the word), how many words will remain unchanged ?

 A.   one B.   two C.   three D.   none
10 . According to the English alphabetical series, how many letters are there between the first letter of the second word and the first letter of the fifth word?

 A.   Two B.   one C.   None D.   More than three
Answers & Solutions

1 .
 Answer : Option B Explanation :
2 .
 Answer : Option D Explanation :
3 .
 Answer : Option A Explanation :
4 .
 Answer : Option C Explanation :
5 .
 Answer : Option D Explanation :
6 .
 Answer : Option C Explanation :
7 .
 Answer : Option B Explanation : Only one meaningful word can be formed with the letters LDFO, ie FOLD
8 .
 Answer : Option A Explanation : So, there is no letter which has not changed its position.
9 .
 Answer : Option D Explanation : $SHE\,\,\, AND \,\,\,TWO \,\,\,WIT \,\,\,GUM$ After rearrangement: $EHS\,\,\, ADN \,\,\,OTW \,\,\,ITW$ $GMU$. Thus, there is no such word whose position remains unchanged.
10 .
 Answer : Option D Explanation : First letter of the second word is ‘A’.First letter of the fifth word is ‘G’.A B C D E F GThere are five letters between A and G
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## SBI PO Puzzle Questions and Answers

1 . Directions: Study the following information carefully and answer the questions given below:
Eight family members Dhruv, Garima, Avinash, Varsha, Aakash, Deepti, Charu and Moksh are sitting around a square table in such a way that two persons sit on each of the four sides of the table facing the centre. Members sitting on opposite sides are exactly opposite each other. Aakash and Garima are exactly opposite each other. Deepti is immediately right to Garima. Dhruv and Moksh are sitting on the same side. Moksh is exactly opposite Avinash, who is to the immediate left of Varsha. Dhruv is towards right of Deepti.

$Q.$ Which of the following statements is definitely true?

 A.   Charu is opposite varsha B.   Deepti is to the left of Garima C.   Avinash is towards the right of Aakash D.   Moksh is sitting opposite Dhruv
2 . Who is sitting opposite Dhruv?

 A.   Charu B.   Deepti C.   Varsha D.   Moksh
3 . Who is sitting opposite Deepti?

 A.   Moksh B.   Charu C.   Varsh D.   Charu or Varsha
4 . Who is next to Varsha in anti-clockwise direction?

 A.   Garima B.   Avinash C.   Dhruv D.   Deepti
5 . Which of the following pairs of persons has both the persons sitting on the same side with first person sitting to the left of second person?

 A.   Aakash - Charu B.   Moksh - Charu C.   Dhruv - Aakash D.   None of these
6 . Directions (6 - 10): Study the following information carefully and answer the questions given below:
Seven professionals A, B, C, D, E, F and G are practising their professions in different cities Chennai, Bangalore, Hyderabad, Mumbai, Ahmedabad, Jaipur and Bhubaneshwar, not necessarily in the same order. Each has a different profession –Doctor, Engineer, Pharmacist, Lawyer, Counsellor, Professor and Artist, not necessarily in the same order. A is a Pharmacist and practises in Bhubaneshwar. D practises in Bangalore but is not a Doctor or an Artist. The one who practises in Hyderabad is a Professor. G is a Counsellor and does not practise in Mumbai or Chennai. E is a Lawyer and practises in Ahmedabad. F practises in Chennai but is not an artist. C practises in Mumbai.

$Q.$ What is D’s profession?

 A.   Doctor B.   Professor C.   Engineer D.   Cannot be determined
7 . Who is the Professor?

 A.   B B.   C C.   D D.   E
8 . Which of the following combinations of profession and place is correct?

 A.   Pharmacist - Jaipur B.   Engineer - Chennai C.   Doctor - Bangalore D.   Artist - Mumbai
9 . Which of the following persons works in Jaipur?

 A.   B B.   G C.   C D.   B or G
10 . Who is the Doctor?

 A.   D B.   B C.   C D.   None of these
Answers & Solutions

1 .
 Answer : Option C Explanation :
2 .
 Answer : Option C Explanation :
3 .
 Answer : Option B Explanation :
4 .
 Answer : Option A Explanation :
5 .
 Answer : Option D Explanation :
6 .
 Answer : Option C Explanation :
7 .
 Answer : Option A Explanation :
8 .
 Answer : Option D Explanation :
9 .
 Answer : Option B Explanation :
10 .
 Answer : Option D Explanation :
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## SBI PO Reasoning Ability test 35

1 . Directions (1-5): Study the following information carefully to answer the given questions :

Eight persons – P, Q, R, S, T, U, V and W – are sitting around a circular table facing towards the centre but not necessarily in the same order. Q is sitting third to the left of W. Two persons are sitting between Q and P. R is sitting second to the left of S. S is not an immediate neighbour of W. T and U are immediate neighbours of each other. Only one person is sitting between U and V.

$Q.$ Who among the following is on the immediate left of W?

 A.   R B.   T C.   U D.   P
2 . How many persons are sitting between R and Q, if we count anticlockwise from R to Q?

 A.   None B.   One C.   Two D.   Three
3 . What is the position of V with respect to P?

 A.   second to the right B.   second to the left C.   third to the left D.   third to the right
4 . Who among the following is sitting second to the right of S?

 A.   V B.   R C.   Q D.   V
5 . If all the persons are asked to sit in clockwise direction in an alphabetical order starting from P, the position of how many persons will remain unchanged, excluding P?

 A.   Three B.   Two C.   None D.   Four
6 . Directions (6-10): In each question below are two or three statements followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance with commonly known facts and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts. Give answer

1) if only conclusion I follows.

2) if only conclusion II follows.

3) if either conclusion I or conclusion II follows.

4) if neither conclusion I nor conclusion II follows.

5) if both conclusions I and II follow.

$Q.$

$Statements:$

No colour is a paint.
No colour is varnish.

$Conclusions:$

I. No varnish is a paint.
II. At least some varnishes are paints.

 A.   if only Conclusion I is follows. B.   if only Conclusion II is follows. C.   if either I or II is follows. D.   if neither I nor II is follows.
7 . $Statements:$

All plugs are sockets.
All sockets are cards.

$Conclusions:$

I. All plugs are cards.
II. All sockets are plugs.

 A.   if only Conclusion I is follows. B.   if only Conclusion II is follows. C.   if either I or II is follows. D.   if neither I nor II is follows.
8 . $Statements:$

Some plants are mountains.
All plants are rivers.

$Conclusions:$

I. All mountains are plants.
II. At least some mountains are rivers

 A.   if only Conclusion I is follows. B.   if only Conclusion II is follows. C.   if either I or II is follows. D.   if neither I nor II is follows.
9 . $Statements:$

No month is a year.
Some years are weeks.

$Conclusions:$

I. No week is a year.
II. Some years are weeks.

 A.   if only Conclusion I is follows. B.   if only Conclusion II is follows. C.   if either I or II is follows. D.   if neither I nor II is follows.
10 . $Statements:$

All planets are stars.
All meteors are planets.
No orbit is a star.

$Conclusions:$

I. All meteors are stars.
II. No planet is an orbit.

 A.   if only Conclusion I is follows. B.   if only Conclusion II is follows. C.   if either I or II is follows. D.   if neither I nor II is follows.
Answers & Solutions

1 .
 Answer : Option B Explanation :
2 .
 Answer : Option D Explanation :
3 .
 Answer : Option A Explanation :
4 .
 Answer : Option C Explanation :
5 .
 Answer : Option C Explanation :
6 .
 Answer : Option C Explanation : No colour is a paint $\rightarrow$conversion $\rightarrow$ No paint is colour (E) + No colour is varnish (E) = E + E = No conclusion. But the conclusion make a complementary pair (IE). Hence, either conclusion I or II follows
7 .
 Answer : Option A Explanation : All plugs are sockets. (A) + All sockets are cards (A) = A + A = A = All plugs are cards. Hence, conclusion I follows. Again, All plugs are sockets $\rightarrow$ conversion Some sockets are plugs. Hence, conclusion II does not follow.
8 .
 Answer : Option B Explanation : Some plants are mountains $\rightarrow$ conversion $\rightarrow$ Some mountains are plants. Hence, conclusion I does not follow. Again, Some mountains are plants (I) + All plants are rivers (A) = I + A = I = Some mountains are rivers. Hence, conclusion II follows.
9 .
 Answer : Option D Explanation : Some years are weeks  conversion  Some weeks are years. Hence, conclusion I does not follow. The second conclusion is a restatement. Hence, II also does not follow.
10 .
 Answer : Option D Explanation : All meteors are planets (A) + All planets are stars (A) = A + A = A = All meteors are stars. Hence, conclusion I follows. Again, No orbit is a star $\rightarrow$ conversion $\rightarrow$ No star is an orbit. Now, All planets are stars (A) + No star is an orbit (E) = A + E = E = No planet is an orbit. Hence, conclusion II follows.
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## SBI PO Reasoning Ability test 34

1 . Directions (11-12): Study the following arrangement carefully and answer the questions given below:

$1\,\,\, 5 \,\,\,8 \,\,\,4 \,\,\,2 \,\,\,1 \,\,\,5 \,\,\,2 \,\,\,3 \,\,\,4 \,\,\,5 \,\,\,6 \,\,\,7 \,\,\,8 \,\,\,9 \,\,\,5 \,\,\,1 \,\,\,4 \,\,\,1 \,\,\,5 \,\,\,6 \,\,\,8 \,\,\,7 \,\,\,4$

$Q.$ How many such 4’s are there in the above arrangement, each of which is immediately preceded by a digit which has a numerical value more than four?

 A.   None B.   One C.   Two D.   Three
2 . If all the even digits are deleted from the above arrangement which of the following will be tenth from the right end of the arrangement?

 A.   5 B.   3 C.   1 D.   7
3 . S is mother of T. T is husband of J. N is son of J. Z is father-in-law of J. K is brother of N. How is K related to Z?

 A.   Son B.   Brother C.   Nephew D.   None of these
4 . According to Vijay’s mother, Vijay reached Mumbai after 13th January, 2013 but before 19th January, 2013. According to Vijay, he reached Mumbai after 16th January, 2013 but before 22nd January, 2013. If both of them are correct, then on which date did Vijay reach Mumbai?

 A.   16th January, 2013 B.   17th January, 2013 C.   18th January, 2013 D.   Cannot be determined
5 . In a certain code language ‘LIVES’ is written as ‘21&95’ and ‘LATER’ is written as ‘2%498'. How will ‘REAL’ be written in that code language?

 A.   49%1 B.   89&2 C.   89%2 D.   59%2
6 . Directions (6-10): Study the following information carefully to answer the questions given below :

There are six persons – J, K, L, M, N and O – each scored different marks. J got more marks than K and M. O got less marks than J. N got more marks than J but he did not get the highest marks. K got less marks than O but he did not get the lowest marks.

$Q.$ How many persons got more marks than O?

 A.   one B.   two C.   three D.   four
7 . Who among the following got the highest marks?

 A.   L B.   K C.   N D.   O
8 . If K got 65 marks then what may be the marks of M?

 A.   68 B.   72 C.   74 D.   62
9 . How many persons got less marks than N?

 A.   three B.   four C.   two D.   one
10 . Who among the following got the minimum marks?

 A.   L B.   O C.   J D.   M
Answers & Solutions

1 .
 Answer : Option C Explanation : 8 $4$, 7 $4$
2 .
 Answer : Option A Explanation : New arrangement$1\,\,\, 5 \,\,\,1 \,\,\,5 \,\,\,3 \,\,\,5 \,\,\,7 \,\,\,9 \,\,\,5 \,\,\,1 \,\,\,1 \,\,\,5 \,\,\,7$Hence, 10th from the right end is 5
3 .
 Answer : Option D Explanation :
4 .
 Answer : Option D Explanation : According to Vijay’s mother, possible dates are: 14, 15, 16, 17, 18 According to Vijay, possible dates are: 17, 18, 19, 20, 21 According to both, there are two possible dates: 17 and 18. Hence, we can’t determine the date on which Vijay reached Mumbai.
5 .
 Answer : Option C Explanation :
6 .
 Answer : Option C Explanation : L > N > J > O > K > M
7 .
 Answer : Option A Explanation : L > N > J > O > K > M
8 .
 Answer : Option D Explanation : L > N > J > O > K > M
9 .
 Answer : Option B Explanation : L > N > J > O > K > M
10 .
 Answer : Option D Explanation : L > N > J > O > K > M
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## SBI PO Reasoning Ability test 33

1 . Directions (1-5): Study the following information carefully and answer the questions given below:

In a certain code language ‘that can wait’ is written as ‘fa ge cl’, ‘come wait here’ is written as ‘ha lo ge’ and ‘can come later’ is written as ‘ne ha fa’

$Q.$ Which of the following is the code for ‘here’ ?

 A.   ge B.   ha C.   lo D.   ne
2 . The code ‘lo ge ha’ may represent

 A.   can come later B.   come later here C.   can wait here D.   come here wait
3 . Which of the following may be the code for ‘come here wait there’ ?

 A.   ra ha ge lo B.   fa cl ha ge C.   ge cl ha fa D.   lo ge ha ne
4 . Which of the following is the code for ‘later’ ?

 A.   cl B.   ge C.   fa D.   ne
5 . Which of the following is the code for ‘wait’ ?

 A.   fa B.   ge C.   cl D.   ha
6 . Directions (6-7): Study the following information carefully and answer the given questions:

The distance from point T to point S, which is in its south, is 9 metres. The distance from point U to point T, which is in its east, is 11 metres. The distance from point X to point W, which is in its north, is 4 metres. The distance from point V to point U, which is in its north, is 5 metres. The distance between point W and point V is 15 metres and point V is in its west. The distance between point Y and the point W, where W is in its south, is 5 metres.

$Q.$ Which of the following three points are in a straight line?

 A.   UTW B.   UTY C.   VWS D.   TWX
7 . What is the distance between points U and Y?

 A.   7 metres B.   9 metres C.   11 metres D.   15 metres
8 . Directions (8-10): Study the following arrangement carefully and answer the questions given below:

$1\,\,\, 5 \,\,\,8 \,\,\,4 \,\,\,2 \,\,\,1 \,\,\,5 \,\,\,2 \,\,\,3 \,\,\,4 \,\,\,5 \,\,\,6 \,\,\,7 \,\,\,8 \,\,\,9 \,\,\,5 \,\,\,1 \,\,\,4 \,\,\,1 \,\,\,5 \,\,\,6 \,\,\,8 \,\,\,7 \,\,\,4$

$Q.$ Which of the following is ninth to the left of twenty first from the left end of the above arrangement ?

 A.   7 B.   5 C.   6 D.   8
9 . How many such 5’s are there in the above arrangement, each of which is immediately preceded by an odd digit and immediately followed by an even digit?

 A.   None B.   one C.   two D.   three
10 . How many such 1’s are there in the above arrangement, each of which is immediately followed by a perfect square?

 A.   None B.   one C.   two D.   three
Answers & Solutions

1 .
 Answer : Option A Explanation : that can wait $\Rightarrow$fa ge cl ... (i)come wait here $\Rightarrow$ha lo ge ... (ii)can come later $\Rightarrow$ne ha fa ... (iii)From (i) and (ii), wait $\Rightarrow$ge ... (iv)From (i) and (iii), can $\Rightarrow$fa ... (v)From (i), (iv) and (v), that $\Rightarrow$cl ... (vi)From (ii) and (iii), come $\Rightarrow$ha ... (vii)From (ii), (iv) and (vii), here $\Rightarrow$lo ... (viii)From (iii), (v) and (vii), later $\Rightarrow$ ne ... (ix)
2 .
 Answer : Option D Explanation : that can wait $\Rightarrow$fa ge cl ... (i)come wait here $\Rightarrow$ha lo ge ... (ii)can come later $\Rightarrow$ne ha fa ... (iii)From (i) and (ii), wait $\Rightarrow$ge ... (iv)From (i) and (iii), can $\Rightarrow$fa ... (v)From (i), (iv) and (v), that $\Rightarrow$cl ... (vi)From (ii) and (iii), come $\Rightarrow$ha ... (vii)From (ii), (iv) and (vii), here $\Rightarrow$lo ... (viii)From (iii), (v) and (vii), later $\Rightarrow$ne ... (ix)
3 .
 Answer : Option A Explanation : that can wait $\Rightarrow$fa ge cl ... (i)come wait here $\Rightarrow$ha lo ge ... (ii)can come later $\Rightarrow$ne ha fa ... (iii)From (i) and (ii), wait $\Rightarrow$ge ... (iv)From (i) and (iii), can $\Rightarrow$fa ... (v)From (i), (iv) and (v), that $\Rightarrow$cl ... (vi)From (ii) and (iii), come $\Rightarrow$ha ... (vii)From (ii), (iv) and (vii), here $\Rightarrow$lo ... (viii)From (iii), (v) and (vii), later $\Rightarrow$ne ... (ix)
4 .
 Answer : Option D Explanation : that can wait $\Rightarrow$fa ge cl ... (i)come wait here $\Rightarrow$ha lo ge ... (ii)can come later $\Rightarrow$ne ha fa ... (iii)From (i) and (ii), wait $\Rightarrow$ge ... (iv)From (i) and (iii), can $\Rightarrow$fa ... (v)From (i), (iv) and (v), that $\Rightarrow$cl ... (vi)From (ii) and (iii), come $\Rightarrow$ha ... (vii)From (ii), (iv) and (vii), here $\Rightarrow$lo ... (viii)From (iii), (v) and (vii), later $\Rightarrow$ne ... (ix)
5 .
 Answer : Option B Explanation : that can wait $\Rightarrow$fa ge cl ... (i)come wait here $\Rightarrow$ha lo ge ... (ii)can come later $\Rightarrow$ne ha fa ... (iii)From (i) and (ii), wait $\Rightarrow$ge ... (iv)From (i) and (iii), can $\Rightarrow$fa ... (v)From (i), (iv) and (v), that $\Rightarrow$cl ... (vi)From (ii) and (iii), come $\Rightarrow$ha ... (vii)From (ii), (iv) and (vii), here $\Rightarrow$lo ... (viii)From (iii), (v) and (vii), later $\Rightarrow$ ne ... (ix)
6 .
 Answer : Option B Explanation :
7 .
 Answer : Option D Explanation : UY = VW = 15m
8 .
 Answer : Option C Explanation : Ninth to the left of twentyfirst from the left end = (21 – 9 ) = 12th from the left end = 6
9 .
 Answer : Option D Explanation : 158, 152, 156
10 .
 Answer : Option B Explanation : 14
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text:
# This “Peaks” My Interest6

Posted by Jiro Doke,

Jiro's pick this week is PeakFinder by Nate Yoder.

"What? Another peak finder?" you might say. Some of you may classify this as one of those utilities that has been created by many people over the years, like sudoku and waitbar. Well, peak finding happens to be something dear to my heart.

I have been using MATLAB for almost 10 years since my first year of graduate school. I initially learned by trying to decipher my advisor's code. One day, I was struggling to write some code for finding peaks in my data.

% Sample data
t = 0:0.01:10;
x = sin(2*t) - 3*cos(3.8*t);

That's when my advisor showed me his code:

dx = diff(x);          % get differences between consecutive points
pkIDX = (dx(1:end-1) >= 0) & (dx(2:end) < 0); % look for slope changes
pkIDX = [dx(1)<0, pkIDX, dx(end)>=0];         % deal with edges
plot(t, x, t(pkIDX), x(pkIDX), 'ro');

This was an eye-opener and was the moment I experienced the power of vector operation for the first time. The way I code in MATLAB had changed from that point on. ... So when I see "peak finding", it brings back memories.

There are quite a few File Exchange entries for finding peaks (and valleys), including two previous POTW selections: FPEAK and EXTREMA. But I really like peakfinder by Nate. Not only does his code deal with noisy data (my algorithm above will be useless if the signal is noisy), but also his coding practice is quite solid. He has a great help section, robust error-checking of input arguments, and variable input and output arguments for ease of use.

xNoise = x + 0.3*sin(40*t);   % add a few more bumps
peakfinder(xNoise);

I looked through a few peak finding entries, but I'm sure I may have missed some. Feel free to let me know of others you really like here.

Get the MATLAB code

Published with MATLAB® 7.9

Faris replied on : 1 of 6

hi Im using simulink and im having trouble getting peak time from my microphone input.. i tried using peak finder but to no avail. can you help me out?

Cody Anglin replied on : 3 of 6

Hello,
I have photovoltaic moduls. I need a current(I) and voltage (U) curve where you can see the Pmax (performance peak). The given values are the maximum values of the voltage and the maximum values of the current, the time and the modul temperature. Is it possible to get the performance curve out of these maximum values or at least the maximum power point?
Sincerely

UMP replied on : 4 of 6

Jiro Doke:How to find the minimum value using the same method for maxima value?I have try using dt=diff(t) but fail.Hope you will help me.Thank you.

UMP replied on : 5 of 6