text:
# Who developed this equation, Pauli?

1. Jun 22, 2012

### PhilDSP

The earliest instance I've seen to this equation, involving the Pauli matrices, is from a 1967 paper by J. M. Levy-Leblond. But it seems general and useful enough that it must have been discovered earlier:

$(\sigma \cdot A)(\sigma \cdot B) = A \cdot B + i \sigma \cdot (A \times B)$

Last edited: Jun 22, 2012
2. Jun 22, 2012

### stevendaryl

Staff Emeritus
Actually, it turns out that this equation is mathematically the same as the equations of geometric algebra, developed by Clifford in the late 1800s, which in turn is a generalization of Hamilton's quaternions.

Pauli introduced spin matrices in 1925, but I'm not sure if he knew about geometric algebras, or whether it was a case of rediscovery.

3. Jun 22, 2012

### tiny-tim

Hi PhilDSP!
is 1846 early enough for you?

see http://www.emis.ams.org/classics/Hamilton/OnQuat.pdf for a reprint of william rowan hamilton's original papers

at page 19 of the .pdf (page 16 of the book) …

21. The fundamental rules of multiplication in this calculus give, in the recent notation, for the scalar and vector parts of the product of any two vectors, the expressions …
… the formula is given, albeit in different notation, essentially AB = S.AB + V.AB = -A.B + iA×B

(i think this comes originally from The London, Edinburgh and Dublin Philosophical Magazine and Journal of Science (3rd Series), vol. xxix (1846), pp. 26-31, but that volume doesn't seem to be on google e-books yet)

4. Jun 22, 2012

### PhilDSP

Thanks stevendaryl and tiny-tim!

Yes, I have *lots* of material on quaternions and quaternion algebra. But nothing from the original papers by Hamilton so the link provided will be interesting. Levy-Leblond used that equation (with Pauli matrices) to give a derivation of Pauli's linearization of the Schrödinger equation but I'm not sure if Pauli did also.

P.S. According to Wikipedia, Pauli's material is (1927) Zur Quantenmechanik des magnetischen Elektrons Zeitschrift für Physik (43) 601-623. Since I'd have to struggle to understand the paper in German, I'm hesitant to seek it out if it doesn't answer the question posed.

Last edited: Jun 22, 2012


text:
Computer Science

# Lab 6 - andor

## Activities

### roses_red.py

Some of the roses are green and blue! Bit has been tasked with painting the roses red.

### clear_green.py

Clear only the green squares next to a black square.

### scurry.py

Bit likes to scurry around.

He goes until the front, left, and right are all blocked.

If his front gets blocked, he turns in the direction that is open.

He paints empty squares green and green squares blue.

### remove_rocks.py

There are rocks (blue squares) in Bit’s strawberry patch. Remove them.

The green blocks mark the beginning and end of the strawberry patch. Only the rocks inside the patch should be removed. Do not remove the strawberries (red squares)!

### paint_and_exit.py

Bit needs to paint the room blue and then go to his green car.

roses_red.py6
clear_green.py6
scurry.py6
remove_rocks.py6
paint_and_exit.py6

format: plain


text:
# The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ? (0, 0) is ?

SHAIK AASIF AHAMED
8 years ago
Hello student,
The line passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0 is
ax + 2by + 3b+$\dpi{80} \lambda$(bx - 2ay - 3a)=0...............(1)
(a+b$\dpi{80} \lambda$)x+(2b-2a$\dpi{80} \lambda$)y+3b-3$\dpi{80} \lambda$a=0
As the line is parallel to x axis
a+b$\dpi{80} \lambda$=0 so $\dpi{80} \lambda$=(-b/a)
Putting $\dpi{80} \lambda$=(-b/a) in equation (1) we get
ax + 2by + 3b+(-a/b)(bx - 2ay - 3a)=0
y(2b+2a2/b)+3b+3a2/b=0
on simplifying we get y=-3/2
so it is 3/2 units below x-axis
Thanks and Regards
Shaik Aasif
Latika Leekha
8 years ago
The given lines are ax + 2by + 3b = 0 and bx – 2ay – 3a = 0.
Then the required equation is of the form
ax + 2by + 3b + l(bx – 2ay – 3a) = 0.
(a + bl)x + (2b – 2al)y + (3b-3a) = 0
Since, the lline is given to be parallel to x-axis, hence we must have the coefficient of x to be zero.
Hence, a+bl = 0
l = -a/b
So, the given equation reduces to (2b + 2a2/b)y + (3b + 3a2/b) = 0
This yields 2y + 3 = 0, which is a line parallel to x-axis and since, y = -3/2, so it lies below the x-axis at a distnce of 3/2 from the origin.
Thansk & Regards
Latika Leekha


text:
# XMLHttpRequest cannot load file. Cross origin requests are only supported for HTTP

I am getting the following error:

XMLHttpRequest cannot load file:///C:/Users/richa.agiwal/Desktop/get/rm_Library/templates/template_viewSettings.html. Cross origin requests are only supported for HTTP.

I realize that this question has been answered before, but I still have not found a solution to my problem. I tried running chrome.exe --allow-file-access-from-files from the command prompt, and moved the file to the local file system, but I still get the same error.

I appreciate any suggestions!

If you are doing something like writing HTML and Javascript in a code editor on your personal computer, and testing the output in your browser, you will probably get error messages about Cross Origin Requests. Your browser will render HTML and run Javascript, jQuery, angularJs in your browser without needing a server set up. But many web browsers are programed to watch for cross site attacks, and will block requests. You don't want just anyone being able to read your hard drive from your web browser. You can create a fully functioning web page using Notepad++ that will run Javascript, and frameworks like jQuery and angularJs; and test everything just by using the Notepad++ menu item, RUN, LAUNCH IN FIREFOX. That's a nice, easy way to start creating a web page, but when you start creating anything more than layout, css and simple page navigation, you need a local server set up on your machine.

## Here are some options that I use.

1. Test your web page locally on Firefox, then deploy to your host.
2. or: Run a local server

## Test on Firefox, Deploy to Host

1. Firefox currently allows Cross Origin Requests from files served from your hard drive
2. Your web hosting site will allow requests to files in folders as configured by the manifest file

## Run a Local Server

• Run a server on your computer, like Apache or Python
• Python isn't a server, but it will run a simple server

## Run a Local Server with Python

• On Windows: Open up the 'Command Prompt'. All Programs, Accessories, Command Prompt
• I always run the Command Prompt as Administrator. Right click the Command Prompt menu item and look for Run As Administrator
• Type the command: ipconfig and hit Enter.
• Look for: IPv4 Address . . . . . . . . 12.123.123.00
• There are websites that will also display your IP address

If you don't have Python, download and install it.

Using the 'Command Prompt' you must go to the folder where the files are that you want to serve as a webpage.

• If you need to get back to the C:\ Root directory - type cd/
• type cd Drive:\Folder\Folder\etc to get to the folder where your .Html file is (or php, etc)
• Check the path. type: path at the command prompt. You must see the path to the folder where python is located. For example, if python is in C:\Python27, then you must see that address in the paths that are listed.
• If the path to the Python directory is not in the path, you must set the path. type: help path and hit Enter. You will see help for path.
• Type something like: path c:\python27 %path%
• %path% keeps all your current paths. You don't want to wipe out all your current paths, just add a new path.
• Create the new path FROM the folder where you want to serve the files.
• Start the Python Server: Type: python -m SimpleHTTPServer port Where 'port' is the number of the port you want, for example python -m SimpleHTTPServer 1337
• If you leave the port empty, it defaults to port 8000
• If the Python server starts successfully, you will see a msg.

## Run You Web Application Locally

• Open a browser
• In the address line type: http://your IP address:port
• http://xxx.xxx.x.x:1337 or http://xx.xxx.xxx.xx:8000 for the default
• If the server is working, you will see a list of your files in the browser
• Click the file you want to serve, and it should display.

## More advanced solutions

• Install a code editor, web server, and other services that are integrated.

You can install Apache, PHP, Python, SQL, Debuggers etc. all separately on your machine, and then spend lots of time trying to figure out how to make them all work together, or look for a solution that combines all those things.

I like using XAMPP with NetBeans IDE. You can also install WAMP which provides a User Interface for managing and integrating Apache and other services.

• If you are not familiar with running commands from the MSDOS command prompt, you can get more info at Wikipedia: [link]en.wikipedia.org/wiki/List_of_DOS_commands – Sandy Good Dec 14 '13 at 1:36
• Is there a way of resolving this issue only with wamp? I am running wamp and still receiving this message. Apache, php and mysql are all up-to-date – user2883071 Dec 24 '14 at 18:52
• Unfortunately I haven't been using WAMP, and don't have an answer for you. Maybe ask a new question specific to your set up. – Sandy Good Dec 24 '14 at 18:59
• Is there a way to disable the cross origin check or add the file:// protocol to it? I'm working in a mobile webview that shows file://s that map to resources in the apk and I'm trying to load a babel js script which internally transforms <script type="text/babel" src="file://..."> to an XMLHttpRequest. – mtsvetkov Jan 5 '16 at 11:47
• Nevermind, fixed it by fidgeting with the app intents and allowed nav hrefs. On the desktop I added the cors headers to chrome for debugging. – mtsvetkov Jan 5 '16 at 12:37

Simple Solution

If you are working with pure html/js/css files.

Install this small server(link) app in chrome. Open the app and point the file location to your project directory.

Goto the url shown in the app.

Edit: Smarter solution using Gulp

Step 1: To install Gulp. Run following command in your terminal.

npm install gulp-cli -g
npm install gulp -D

Step 2: Inside your project directory create a file named gulpfile.js. Copy the following content inside it.

var gulp        = require('gulp');
var bs          = require('browser-sync').create();

gulp.task('serve', [], () => {
bs.init({
server: {
baseDir: "./",
},
port: 5000,
});
});

Step 3: Install browser sync gulp plugin. Inside the same directory where gulpfile.js is present, run the following command

npm install browser-sync gulp --save-dev

Step 4: Start the server. Inside the same directory where gulpfile.js is present, run the following command

gulp serve

• Fantastic! Sandy's answer is comprehensive but your answer beats it hands down for 'make it work now' convenience. – Holf Jun 9 '15 at 19:41
• Brilliant! No need to remember fancy routines to make CORS work! – Stoyan Berov Nov 21 '15 at 10:02
• This is work with me. thank you :) – M.G Apr 12 '16 at 8:55
• Excellent little app. Thx :) – Mark May 29 '16 at 13:22
• brillilant , best and simpliest solution – firasKoubaa Aug 10 '16 at 8:01

This error is happening because you are just opening html documents directly from the browser. To fix this you will need to serve your code from a webserver and access it on localhost. If you have Apache setup, use it to serve your files. Some IDE's have built in web servers, like JetBrains IDE's, Eclipse...

If you have Node.Js setup then you can use http-server. Just run npm install http-server -g and you will be able to use it in terminal like http-server C:\location\to\app. Kirill Fuchs

I was facing this error while I deployed my Web API project locally and I was calling API project only with this URL given below:

localhost//myAPIProject

Since the error message says it is not http:// then I changed the URL and put a prefix http as given below and the error was gone.

http://localhost//myAPIProject

If you use the WebStorm Javascript IDE, you can just open your project from WebStorm in your browser. WebStorm will automatically start a server and you won't get any of these errors anymore, because you are now accessing the files with the allowed/supported protocols (HTTP).

Depends on your needs, but there is also a quick way to temporarily check your (dummy) JSON by saving your JSON on http://myjson.com. Copy the api link and paste that into your javascript code. Viola! When you want to deploy the codes, you must not forget to change that url in your codes!

Place your project folder into the htdocs of the Xampp directory Start your Apache server using xampp control panel then open a browser goto localhost/projectfolder then it start working

• This answer is not complete. – William Dunne Jan 26 '16 at 19:54
• hi William Dunne, what was missing in this let me known so that i can improve my answer.. – sunil KV Jan 27 '16 at 7:53
• You're assuming he already has XAMPP already installed, which doesn't seem to be stated in the original question, and if you're suggesting installing something new it is generally a good idea to explain why, and the cause behind the error - particularly for people in the future reading this. – William Dunne Jan 27 '16 at 8:42

## protected by Community♦Apr 13 '17 at 16:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?


text:
# Star Rock It

Geometry Level 2

Find the sum of the angles (in degrees) in a pentagram.

Do you know the general formula for an $n$-polygonal star?

×


text:
Cheat activated

Background

The game Grand Theft Auto: San Andreas went down in history also thanks to its wide selection of cheats. They're almost 90 and anyone who has ever touched this game, no doubt has tried them all!
One cheat is activated (on PC) typing in-game a secret keyword, and then boom, a jet pops out of thin air or perhaps all pedestrians look like Elvis Presley or some other rowdy effect...

They always come with this confirmation message:

Rockstar chose to store them hashed, so due to collisions, in addition to the intended ones there are many other strings that trigger each cheat.

Therefore I propose to solve this downside!

Write a full program that prints CHEAT ACTIVATED if and only if the last part of a string is a cheat code.

Cheat codes

THUGSARMOURY
PROFESSIONALSKIT
NUTTERSTOYS
INEEDSOMEHELP
TURNUPTHEHEAT
TURNDOWNTHEHEAT
PLEASANTLYWARM
TOODAMNHOT
DULLDULLDAY
STAYINANDWATCHTV
CANTSEEWHEREIMGOING
TIMEJUSTFLIESBY
SPEEDITUP
SLOWITDOWN
ROUGHNEIGHBOURHOOD
STOPPICKINGONME
SURROUNDEDBYNUTTERS
TIMETOKICKASS
OLDSPEEDDEMON
DOUGHNUTHANDICAP
JUSTTRYANDSTOPME
WHERESTHEFUNERAL
CELEBRITYSTATUS
TRUEGRIME
ALLCARSGOBOOM
STICKLIKEGLUE
GOODBYECRUELWORLD
DONTTRYANDSTOPME
ALLDRIVERSARECRIMINALS
PINKISTHENEWCOOL
SOLONGASITSBLACK
FLYINGFISH
WHOATEALLTHEPIES
BUFFMEUP
LEANANDMEAN
BLUESUEDESHOES
ATTACKOFTHEVILLAGEPEOPLE
LIFESABEACH
ONLYHOMIESALLOWED
BETTERSTAYINDOORS
NINJATOWN
LOVECONQUERSALL
EVERYONEISPOOR
EVERYONEISRICH
CHITTYCHITTYBANGBANG
CJPHONEHOME
JUMPJET
IWANTTOHOVER
TOUCHMYCARYOUDIE
SPEEDFREAK
BUBBLECARS
NIGHTPROWLER
DONTBRINGONTHENIGHT
SCOTTISHSUMMER
SANDINMYEARS
KANGAROO
NOONECANHURTME
MANFROMATLANTIS
LETSGOBASEJUMPING
ROCKETMAN
BRINGITON
STINGLIKEABEE
IAMNEVERHUNGRY
STATEOFEMERGENCY
CRAZYTOWN
TAKEACHILLPILL
FULLCLIP
GHOSTTOWN
HICKSVILLE
WANNABEINMYGANG
NOONECANSTOPUS
ROCKETMAYHEM
WORSHIPME
ICANGOALLNIGHT
PROFESSIONALKILLER
NATURALTALENT
OHDUDE
FOURWHEELFUN
ITSALLBULL
FLYINGTOSTUNT
MONSTERMASH

Input

• A string $$\s\$$ over the alphabet:
[A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z]

Output

• Print CHEAT ACTIVATED if there exists a cheat code $$\c\$$ such that $$\c\$$ is a suffix of $$\s\$$
• Nothing otherwise

This is , so the shortest code wins.

• Was there a reason you chose to override the default decision problem output choices? It doesn't appear to add much beyond being cute. The rules also seem to require that the input be all capital, and not allowed to be lower case - this is abnormal and you should probably draw more attention to it if you decide to keep it. May 26, 2020 at 20:01
• @FryAmTheEggman Yes you've nailed it, I just like the idea of recreating the original reply of the game. I'm aware it adds almost nothing, nevertheless it's suitable to some clever "recycling" of the part of the program that checks the input. For the second part I originally thought to allow all characters and make it case insensitive, then I saw that limit the input could only make the things easier and foster optimization. (I left out one cheat 18HOLES because was the only one with digits) Beside being abnormal, show me what problem could it lead and I'm willing to change it May 26, 2020 at 20:30
• @FryAmTheEggman Don't get me wrong, I've not justified a whim saying that it can be interesting to see how to handle that output. I really consider it can represent something different. Of course outside Code Golf this would only be a juxtaposition of a decision problem and the production of a specific string. And I totally agree that a good challenge is a dry one, one where everything is essential. But this doesn't want to be a default decision problem. And that output it's fundamental in it. May 26, 2020 at 21:24
• That is fine, I'd just like to say that I think it is unlikely your choices will improve the solutions that you receive. Generally, giving the golfer more freedom to choose I/O increases the quality of the responses you get, in addition to improving the chances you get of getting responses at all. May 26, 2020 at 22:05
• @FryAmTheEggman Ok thanks, I'll try to leave as it is, if it didn't get many responses I'll know why ahahah - Never mind, it's an experiment :) May 26, 2020 at 22:22

Python 3,  976  965 bytes

Saved 11 bytes thanks to @user202729

Nothing fancy here. This is a hand-crafted regular expression which was randomly shuffled to please zlib and save a few more bytes.

import zlib,base64,re
if re.search(zlib.decompress(base64.a85decode('''GaqK,9,=TU&D^"aoJ0#fKbc'=7=3*ZC@Sjrb9-nO,G4ij>guO6,*oNLi[tq[ZsOTo,XZG;oZsh/gQt)R:U?gtgs2QpS5&Cl,CM,c4B%<W5ZEuN=Sb$f!,mP%@TQ:&Dg(:q8*+N)Y?Q#F9kM'KZ&%\I2W#7oOUp1-0Ge3]q_o8g@lWflG*QMr*9[[i?a=U:auDF%8=66MDaNUUiPI8J#lO0\l@*E:%PS)VCHdD>/N=l1t*hVA<^ON+dhUtN/]3U\\tupZtkD$:DPFh-HW3PZ$AAiEFm(g;<YJ6I4"uMdQX16JBN/W.>QFrON6O,^B%Jl3Mn<d^IObA<=$[b;g-:O$Ga?$ch!SrhQNm;Led6I;8/VCeIra;1&X%b)&*]\XB4+g3FkTT#fc%HR8*)g95de$99CT;oVaps<Yn\im/BJu&A6-O8L3#Ha0p6]PKP2a7,,T6[K>'rQiC1R,U1_Ko/:-EAfrcS7ik6HfK>2:cMot]rkBY#!9d"ujd=<dn@]JT9h=mClIt.YL?*b"(ImA*!E/(\Gob*ghDs1dJ.;O"]CPGTo!=5g=9e2+caXT43.mu:*o8ai!KoaLJ5',M&K[V8d'?J=ISEG?GUZMbWXJ[iCde,Mu[(,9nY8%-73jGpPZFWBXSOTGg5:U<P2"/=V(g[PZJodG+cZuag^64TE.>i4i"BRD8=mu,Jga5Ip(0Z34oicZ"d'bi@2Bqe/D6Y*)Cc-#tb7!E4Q']mZ0F2M]&qrpj]qLqg0MLln_4Q7QA[9<@I^3OJAJ)p1?+NpZH)nYYp(.J''*9mJUg?9h.]o72ZKgN)90W>0sIe(IqbdBCsfj<Ma5N''')).decode(),input()):print('CHEAT ACTIVATED')  Try it online! • (I assume that you are more familiar with JavaScript, in that case (JavaScript doesn't have builtin for compression?) you can simply specify some language such as "JavaScript+<zlib library>". For demonstration purposes it isn't very hard to copy-paste the library source code into the header part.) May 27, 2020 at 6:22 • @DomenicoModica The whole point of a decision problem challenge is to output either true or false, no matter the nature of the task. Given the background story of this question, asking to output 'CHEAT ACTIVATED' instead (or nothing at all) makes some sense, so I don't really have a problem with that. But generally speaking, overriding the default rules (such as this one and these ones) is a bad idea. It's quite likely to decrease the number of answers to your question. May 28, 2020 at 0:26 • @DomenicoModica I'm not sure to understand your point about the compression. It all depends on the language used anyway. That's why you should not make any assumption like that and just let the people decide. May 28, 2020 at 9:03 • @DomenicoModica If static data or other helper functions are defined outside the main function, it must definitely be counted in the total byte score. Doing otherwise would be blatant cheating. May 29, 2020 at 23:36 • @DomenicoModica I'm afraid it might be a bit too late for that challenge, but keep that it mind for the next ones. ;-) May 30, 2020 at 19:03 05AB1E, 753 bytes .•ćÉÞû9‚2¬ú@iH°<]ĆA~Qò†ø‚”Kƒćž»”‚ŸǝD6øZ\§HÔØΔÚÿþjvĀ¤¼§….[Ωîužć“₅nÕbnιwU0›•#’ÿ¨¥€´y ‡ªsíÃ ÝæÿŠ… i‚Û€ä€Ô ‹‡€¾€€š¢ ‹‡„‹€€š¢ À®ly©Ã …«Â£…ß ÿ‚Ž ‹€€†€ƒ‰Ï…Û ÏÏ€È‚Ã›Î…ì €º‚‰f´‹€‹ ŠÍ€•€¾ ¥€€•„‹ À¸ÿ€´Â° ‹®â¢€‰€á á€€‹Ýæÿ €º€„»áí„ „ÏŠÍ«™n ÿÝæ‡´i¦§ €–€‡‚Æ¶¨ ‚‰‡Ü€ƒ‹®€á ‚Ãs€€¼§ ¨À‡‚ ‰¦ÿ €Ÿˆ™‚œÒÞ ¯š€É‚µ ©ß€èÿ ‚¿Þ¡ÿ‚ï ¬Í‡Ü€ƒ‹®€á €Ÿ¢ž€™žÜs ™Å€ˆ€€€¢îÁ ¸•ÿ®ísƒÏ °ä•Ç €Ðâì€Ÿ€€Èšs Ó´f€á€¾ ÞŸ€ƒ‹ˆ ‡—ÿŠë œÆ€‚€€™¨‚… ‚ìsaˆÁ €É€¨€ˆï¶ …ç‹€¹¶s ÿ‰à „Î¡¤ÿ€Ÿ ŽÇ€ˆ—‹ ŽÇ€ˆ›Í c•»tyc•»ty±»±» cjƒ¿€¨ ïÅ³• iƒ¾€„h‚Š ›ä€¯ƒ´€î—Ç ŠÍÿ Íˆˆ™ †æˆ×ÿ ¬ÍîŸ€‰€€†æ µ±Ž€ ³Ö€†€¯ÖÇ ÿ €¸€µ€©»å€á …Ž€šÿ ´Û‚œŠàëÌ ÕÕ…Ž i€·€œi‚µ îŸ€•€‰ st±µ€èaÍÝ i€Ü‚ÒàŠ ‚‹€‚™‰ ©„‰à ƒ¶acŽÏ¼Ã ‚è¡¹ iÅËˆ¬€‹ ·ä‰à ÿ´™e ÅË€ï€†€¯³£ €¸€µ€©‹®€ª ÕÕ€¿ÿ º°€á Ÿ™£Í i€©‚œ€Ÿ†æ ‡ªÃ½ ˆÃ¼Î ŠçÞ® †ì¦³ˆ¦ •»€€†â—É €ç€ŸÅÍ °ä€„ÿ ±¹mÌ¢’u#Å¿à‘Ç¥Ù‡‘×  Explanation: .•ćÉÞ...U0›• # Push compressed string "stunt hem hicksv atlantis kangaroo wler freak quer ninja suede nga cruel glue grime dough ters neighb dulldull ters thugs" # # Split it on spaces to a list  # Pop and push all strings separated to the stack ’ÿ¨¥...mÌ¢’ # Push dictionary string "ÿarmoury professionalskit nutÿtoys ineedsomehelp turnuptheheat turndowntheheat pleasantlywarm toodamnhot ÿday stayinandwatchtv cantseewhereimgoing timejustfliesby speeditup slowitdown roughÿourhood stoppickingonme surroundedbynutÿ timetokickass oldspeeddemon ÿnuthandicap notforpublicroads justtryandstopme wheresthefuneral celebritystatus trueÿ allcarsgoboom wheelsonlyplease sticklikeÿ goodbyeÿworld donttryandstopme alldriversarecriminals pinkisthenewcool soloÿsitsblack flyingfish whoateallthepies buffmeup leanandmean blueÿshoes attackofthevillagepeople lifesabeach onlyhomeisallowed betterstayindoors ÿtown loveconÿall everyoneispoor everyoneisrich chittychittybangbang cjphonehome jumpjet iwanttohover touchmycaryoudie speedÿ bubblecars nightproÿ dontbringonthenight scottishsummer sandinmyears ÿ noonecanhurtme manfromÿ letsgobasejumping rocketman idoasiplease bringiton stinglikeabee iamveryhungry stateofemergency crazytown takeachillpill fullclip iwannadriveby ghosttown ÿille wannabeinmygang noonecanstopus rocketmayÿ worshipme helloladies icangoallnight professionalkiller naturaltalent ohdude fourwheelfun hittheroadjack itsallbull flyingtoÿ monstermash", # where the ÿ are automatically filled with the strings on the stack u # Convert it to uppercase # # Split it on spaces Å¿ # Check for each word whether the (implicit) input-string ends with it # (1 if truthy; 0 if falsey) à # Check if any are truthy by taking the maximum ‘Ç¥Ù‡‘ # Push dictionary string "CHEAT ACTIVATED" × # Repeat the string the integer amount of times # (so once for truthy; and it becomes an empty string for falsey) # (after which the result is output implicitly)  See this 05AB1E tip of mine (sections How to use the dictionary? and How to compress strings not part of the dictionary?) to understand how compressed string .•ćÉÞ...U0›• and dictionary strings ’ÿ¨¥...mÌ¢’ and ‘Ç¥Ù‡‘ work. Retina 0.8.2, 1094 1082 bytes HE g Tg y ST a GO b TT c DO d IT e AS f LE h IN i AN j AR k LL l OU m RO n ES o ON p ER q ED r AT s HO t IC u jD w Al x ME z NIGHT = M(THUGSkMmRY|PnFoSIpAL(SKe|KILhR)|(SURRmNDrBYNUcq|NUcqaOY|TIzTOKuKf|NOTFORPUBLunAD|CEhBReYasU|xDRIVqSkECRIMiAL|BEcqaAYidOR|BUBBhCk|SwiMYEk|MjFnMsLjTI)S|(DUlDUlDA|TIzJUaFLIoB|IAMNEVqHUNGR|asEOFEMqGENC|IWjNADRIVEB)Y|(aOPPuKibN|TRUEGRI|CJPHpEt|WORSHIP)z|(JUaT|dNc)RYwaOPz|NOpECj(aOPUS|HURTz)|iErSOzgLP|T(URN(UP|dWN)ygs|OODAMNtT|mCHMYCkYmDIE|AKEACHIlPIl)|PhfjTLYWkM|aAYiwWsCHTV|CjTSEEWgREIMbiG|S(PEr(FREAK|eUP)|LOWedWN|OLpGfeSBLACK|COcISHSUMMq)|RmGHNEIGHBmRtOD|OLDSPErDEMp|dUGHNUTHwuAP|WgRoyFUNqAL|xCkSbBOOM|(WgELSpLYPhf|auKLIKEGLU|IdfIPhf|aiGLIKEABE|OHDUD)E|bODBYECRUELWORLD|PiKISyNEWCOOL|FLYiG(TOaUNT|FISH)|WtsExyPIo|BUFFzUP|hjwzj|BLUoUroto|AcACKOFyVIlAGEPEOPh|LIFoABEACH|pLYtMIoxOWr|LOVECpQUqSx|EVqYpEIS(POOR|RuH)|(CHIcY){2}BjGBjG|JUMPJET|IWjcOtVq|=PnWhR|dNTBRibNy=|KjGkOO|hTSbBfEJUMPiG|nCKETM(j|AYgM)|BRiGep|(Gta|CRAZY)TOWN|NiJsOWN|FUlCLIP|HuKSVILh|WjNABEiMYGjG|glOLADIo|ujbx=|NsURALTAhNT|FmRWgELFUN|HeynADJACK|eSxBUl|MpaqMfH)$
0

1
CHEAT ACTIVATED

Try it online!

Verify all matches

This solution is based on the observation that many letter pairs repeat several times in different cheat codes. We can replace those common pairs with lowercase letters (since the input will be in all uppercase) before executing the main regex.

Common letter pairs were generated using this program

• Can't that map be used in any way to write CHEAT ACTIVATED as Cgs ACTIVsr? May 27, 2020 at 23:47
• @DomenicoModica Unfortunately no. The lowercase letters are never converted back to the uppercase pairs that they represent May 28, 2020 at 0:15

ú↓╒E╢≤3░ü(⌡▬α▲Kw~Eæ║aù≡╕↓╠ækîª♥J╠⌂ÿb╞╝ø└╫£φ├╡▌ê╔ê;y9¿¿¢Ö⌡D◘MB╡ú:¢▌BÇ!ÜÅ?╫-◙¥·╘hM┌ûoƒPå¢¿δ╓x╓7╤½{α█hè▄ΔZ∙╙d≥♀'ô▼╘[πCò"┬╦•ù,╫▀╚╟TεÿRD°ß£{♠╕~>G╧ÜG⌠é~F~⌐?Iê→LA╟αa°♦V╔xhU∟δ¶╧▀Dïîc╫►τ8üîÄ╠⌂○╣*i☻▌ƒç╥¿ÇàM H─▬½÷bû>miWπ∞*ü\σ¶REuü╚∙╠K┌ΦîΣ╫←▒╜A+░Ñ@)nφ₧∞'i╤Su@↑⌂¥ynΓv█e-0y╧Å§∟8φ·k½▌↑ëK╡7╛ƒ └¥τ─~↕τ╫^sh<▀Ö¥4╟╔«▼»0┌▌ûÆ↨1ñFf┘ê▒}[d↔ätf,▀⌂qt‼≈Xⁿ←ìhà⌂ÇΦXDÿs═JéΔ♠ö╙═[═4○tºyBoz?e╔§▐Ñ│δ♣m♠°⌐ñ▐b╙|I₧◙Çq╧WNÿ:hYⁿ╝Ö┤ Ö╟╗φW@‼¡Δ¥♥£<Mÿt·╛~Ö◘U3∞P♂πΩ○U÷₧ï↨75-]├╢▒║3#│‼8^7.♦▼2↨8├Fⁿ♥>º√√─Ö¥┌Æ☼;∙Δ;+ï♪≥àΔñö╢íhåhò╤7e╣D┼┌τ·▐\½N⌂se0░╖1NmQQChs╕ë╒┌≈Å■╓{]&§╦π◄▲┘#?◙èáK?A▲≤û}╥5V#(¢⌠ü,∟δ$:åΓ↑╟ΘΘ±○«╠j┐≈o,v♦ì9╜æ-K│↑♀fwß☻δA┐┼à◄?║H√╖¿Æ↨á░Ö╗╣ΩÄú╢Mé=Ä▌↔→a°◘^j►╡┐t'◄wΔh←▬╜δt♫¡ïù«○├α7♣G▼K]H┬>♥á»φ9öù|╜Δ╥Éï¬¡.╚øKôªRø─Ç╞╗}₧4◙|▓íâM ╥╨uⁿYéΔß☻¶∟ëæ┤,á╔∞₧Hæ}♪↑ßvâúy[>lu7▌z╒sóÄO:L▬╦íDq§φiIΔ═ûδ▓^ff▒  Run and debug it It relies heavily on compressed string literals, and there's not too much interesting about it. JavaScript, 1200 bytes s=>/(ALL(CARSGOBOOM|DRIVERSARECRIMINALS)|ATTACKOFTHEVILLAGEPEOPLE|BETTERSTAYINDOORS|BLUESUEDESHOES|BRINGITON|BUFFMEUP|CELEBRITYSTATUS|DOUGHNUTHANDICAP|DULLDULLDAY|EVERYONEIS(POOR|RICH)|FLYING(FISH|TOSTUNT)|FOURWHEELFUN|FULLCLIP|GOODBYECRUELWORLD|HELLOLADIES|HICKSVILLE|IAMNEVERHUNGRY|INEEDSOMEHELP|ITSALLBULL|IWAN(NADRIVEBY|TTOHOVER)|JUMPJET|KANGAROO|LEANANDMEAN|LIFESABEACH|LOVECONQUERSALL|MANFROMATLANTIS|MONSTERMASH|NATURALTALENT|NIGHTPROWLER|NOONECAN(HURTME|STOPUS)|NOTFORPUBLICROADS|NUTTERSTOYS|OHDUDE|OLDSPEEDDEMON|ONLYHOMIESALLOWED|PINKISTHENEWCOOL|PLEASANTLYWARM|PROFESSIONAL(KILLER|SKIT)|ROCKETMA(N|YHEM)|ROUGHNEIGHBOURHOOD|SCOTTISHSUMMER|SLOWITDOWN|SPEED(FREAK|ITUP)|STA(TEOFEMERGENCY|YINANDWATCHTV)|STI(CKLIKEGLUE|NGLIKEABEE)|SURROUNDEDBYNUTTERS|TAKEACHILLPILL|THUGSARMOURY|TIME(JUSTFLIESBY|TOKICKASS)|TOODAMNHOT|TOUCHMYCARYOUDIE|TURN(DOWN|UP)THEHEAT|WHERESTHEFUNERAL|WHOATEALLTHEPIES|(HITTHEROADJ|SOLONGASITSBL)ACK|((CHITTY){2}BANGB|WANNABEINMYG)ANG|(BUBBLEC|SANDINMYE)ARS|(CANTSEEWHEREIMGO|LETSGOBASEJUMP)ING|(CJPHONEHO|(DON|JUS)TTRYANDSTOP|STOPPICKINGON|TRUEGRI|WORSHIP)ME|(DONTBRINGONTHE|ICANGOALL)NIGHT|(IDOASI|WHEELSONLY)PLEASE|(CRAZY|GHOST|NINJA)TOWN)$/.test(s)?'CHEAT ACTIVATED':''

Try it online!

• It should be a complete program May 31, 2020 at 16:06

JavaScript (Node.js), 996 bytes

a=0n;process.argv[2].match(require(zlib).unzipSync(Buffer("\"T93)OC8m{J84WLB/+&GTui0kfVU#jJp;3(nOUQ))1b/-^*.t&O'V58#/0;D$- /te\"l_)RiO<j]tu=ck_:{b,bL^?Gm'^%I5M;I{g\\GA-\\kud?RvYIH8=nvGq|{S l-zt5?.(zt9thR%m.v=[|hJZW7^/1RYRTJ8+)_>V-k!)[c2y1Ra5];'BM+#+!n-2\\nl;libQ7NmmYjG*0$n=S;=<W)dm/(vF=Q'zuYjfx2p,OBrqGc%upq28Y65rxj@+]P6@s^O#A9G==H*bCz3G'rNADJOFs{8k~8#8.&Y|1=soo(JS>iVRy?Gj})\"?t NCjVmoUUDo*B98e-xYyA#X#El;vqDjJwNb8cE|~@N).v3TT{?Kk*M1zwy:}{$\\DPj\\GVv0lgP;5satf1\"K}4G:Jc1xt]OXCu^$#u&4?NA:Od)v+!+'1=TrUBx2vfh'~HFyXZ1BzK4(Dh_-_';/A}:SxM !3\"4qXIjrCu$]]r+$P,6_+PqH#IU9,><UJ4U#[wEb<,f%X*dr<#2V_U\"BjiPA]En0Ls\\yt61u #m@3rm6e]NMBR_\\La.7|9^)uh/$Ef.w^7{mn;~ug\\w3MDQPb$IR}_2A3qkc:V>rC9|xC/0g:_A'nuV<Q ugVj|1<(]*a~.jFlp$'Mw[gDF<E]5I$tbg\\10'xvwUBeWBk8g2( XjBhu*Y(KK}Yzu}(,YY6.5TcF)u9'=M[Lb4^ /g#<P3*>i}.Y4]m$2e1@c\\>dAW?ODdexO=jB$7!bJ'/|O:X,}]|qQ%zs;xVStjz[4~8?>\"m \"yR5 L+5)\\&(jZD6$<-<mb>^;j2<").map(n=>[a=a*95n+BigInt(n-32)]).map(n=>[(a>>=8n)%256n])))&&console.log('CHEAT ACTIVATED')  Try it online! a=0n; // Init process.argv[2] // Take Input .match( // Check require(zlib).unzipSync( // Uncompress Buffer("...") // For each char .map(n=>[a=a*95n+BigInt(n-32)]) // Add info to a .map(n=>[(a>>=8n)%256n]) // and take bytes out ) // [] convert BigInt to Number )&&console.log('CHEAT ACTIVATED') //Output  First time using Node for full program C (Linux GCC) Count: 1584 void f(char*s){char c[][30]={"THUGSARMOURY","PROFESSIONALSKIT","NUTTERSTOYS","INEEDSOMEHELP","TURNUPTHEHEAT","TURNDOWNTHEHEAT","PLEASANTLYWARM","TOODAMNHOT","DULLDULLDAY","STAYINANDWATCHTV","CANTSEEWHEREIMGOING","TIMEJUSTFLIESBY","SPEEDITUP","SLOWITDOWN","ROUGHNEIGHBOURHOOD","STOPPICKINGONME","SURROUNDEDBYNUTTERS","TIMETOKICKASS","OLDSPEEDDEMON","DOUGHNUTHANDICAP","NOTFORPUBLICROADS","JUSTTRYANDSTOPME","WHERESTHEFUNERAL","CELEBRITYSTATUS","TRUEGRIME","ALLCARSGOBOOM","WHEELSONLYPLEASE","STICKLIKEGLUE","GOODBYECRUELWORLD","DONTTRYANDSTOPME","ALLDRIVERSARECRIMINALS","PINKISTHENEWCOOL","SOLONGASITSBLACK","FLYINGFISH","WHOATEALLTHEPIES","BUFFMEUP","LEANANDMEAN","BLUESUEDESHOES","ATTACKOFTHEVILLAGEPEOPLE","LIFESABEACH","ONLYHOMIESALLOWED","BETTERSTAYINDOORS","NINJATOWN","LOVECONQUERSALL","EVERYONEISPOOR","EVERYONEISRICH","CHITTYCHITTYBANGBANG","CJPHONEHOME","JUMPJET","IWANTTOHOVER","TOUCHMYCARYOUDIE","SPEEDFREAK","BUBBLECARS","NIGHTPROWLER","DONTBRINGONTHENIGHT","SCOTTISHSUMMER","SANDINMYEARS","KANGAROO","NOONECANHURTME","MANFROMATLANTIS","LETSGOBASEJUMPING","ROCKETMAN","IDOASIPLEASE","BRINGITON","STINGLIKEABEE","IAMNEVERHUNGRY","STATEOFEMERGENCY","CRAZYTOWN","TAKEACHILLPILL","FULLCLIP","IWANNADRIVEBY","GHOSTTOWN","HICKSVILLE","WANNABEINMYGANG","NOONECANSTOPUS","ROCKETMAYHEM","WORSHIPME","HELLOLADIES","ICANGOALLNIGHT","PROFESSIONALKILLER","NATURALTALENT","OHDUDE","FOURWHEELFUN","HITTHEROADJACK","ITSALLBULL","FLYINGTOSTUNT","MONSTERMASH"};for(int i=0,j=0,k=0;i<87;i++){j=strlen(c[i]);k=strlen(s);if(k<j||strncmp(s+(k-j),c[i],j))continue;printf("CHEAT ACTIVATED\n");}}  Count w/out cheats themselves: 143 void f(char*s){for(int i=0,j=0,k=0;i<87;i++){j=strlen(c[i]);k=strlen(s);if(k<j||strncmp(s+(k-j),c[i],j))continue;printf("CHEAT ACTIVATED\n");}}  So if I wanted to improve mine, I'd need to find a better way to represent that data • 1544 bytes May 29, 2020 at 17:35 • Why don't you use a single string with a delimiter rather than a list of separate strings? For example, the first part could be "THUGSARMOURY/PROFESSIONALSKIT/NUTTERSTOYS..." May 30, 2020 at 18:04 Japt, 823 bytes Ö„t aå¡ê*pUè(kÂ žo|â‘(adjack|(g—¡|cŸzy|nˆja)Èu|fÒàg(‘Ð|t|f‰h)|Ü…hnÔAÌap|Ü‹(ç°GÊght|â2Øop´)|ÂlckÆøevÅNgepeop¤|¦fƒa¼a®|i(°n(nÃqveby|t‘—v€)|amšv€Ë¡ry|ÖÁoa¥night|tÑ¡ÞÑ|ã””lp)|æˆrQtÓƒ‰|n(å …ys|otfŽèFŠÞads|ÇÊ¯n(Ë†´|¡op«)|Âxà¤¤|ightpžw¤r)|ÊµsvÅH|àP‰(Îr|poŽ)|è wal(kÅHr|skŠ)|s(Ö-‰hsum´r|å€dbynÔjrs|áÊ‰kts½ack|pe‚(fœak|Ä¿)|t(…ee´rgcy|Ãà„d°t®tv|Åikeglue|Äikea¼e|áâ¶¨g´)|ÓqtÜµ|ÂnÛŒrs)|Î…hšighÞ€—od|goodbyecruelwŽld|b(ub½eÖÜ|Áªã §€ÇÔ|ãˆ E|luàÂoƒ|uff´up)|jÔÍã€…p´|fŒrØ‡Óô|c(jpÊÊÊà|Ê•yÖ’tyßAßA|e¤ßŠyÐa«|„t eØ€e‹goˆg)|è0b$ew¬ol|t(ruegÎX|‹e(j«tf¦ƒby|‘kick†s)|oo»mn—t|aê‚ZUpi¥|Ë¯‡Ú€y|Æ2myÖÝŒ¹e|¨n(Üµ|up)eÊ!)|ž×Fµ(n|y”m)|Ò „dÚ |§Êãà!»d|fªlc¦p|è "Ã§°rm|ÚŽ’rÚR|jumpjet|oldspe‚¸¶n|w(Æ¢ip´|„na¼ˆmyg„g|—ÂT¥epiƒ|â€ fÕal)|love¬nquÀ&¥|ohdu¸|Ê)oÓ©ƒ|(iº†i|Ø‡ls§)pÒ"e|a¥(ÖÜgoÞ¼|dãŽ€›e×Ëˆals)|ç9é»y|¤tsgoßPjuÛpg)\$u¹u

Try it online!

Explanation

...pUè...u¹u
...          u  // Compressed version of "CHEAT ACTIVATED"
p            // Repeat the string
Uè          // Number of matches of the regex on the input
...u    // Compressed version of regex from the python solution

The regex is from this python solution

JavaScript, no compression, 1356 1351 bytes

a=T=>~"THUGSARMOURY,PROFESSIONALSKIT,NUTTERSTOYS,INEEDSOMEHELP,TURNUPTHEHEAT,TURNDOWNTHEHEAT,PLEASANTLYWARM,TOODAMNHOT,DULLDULLDAY,STAYINANDWATCHTV,CANTSEEWHEREIMGOING,TIMEJUSTFLIESBY,SPEEDITUP,SLOWITDOWN,ROUGHNEIGHBOURHOOD,STOPPICKINGONME,SURROUNDEDBYNUTTERS,TIMETOKICKASS,OLDSPEEDDEMON,DOUGHNUTHANDICAP,NOTFORPUBLICROADS,JUSTTRYANDSTOPME,WHERESTHEFUNERAL,CELEBRITYSTATUS,TRUEGRIME,ALLCARSGOBOOM,WHEELSONLYPLEASE,STICKLIKEGLUE,GOODBYECRUELWORLD,DONTTRYANDSTOPME,ALLDRIVERSARECRIMINALS,PINKISTHENEWCOOL,SOLONGASITSBLACK,FLYINGFISH,WHOATEALLTHEPIES,BUFFMEUP,LEANANDMEAN,BLUESUEDESHOES,ATTACKOFTHEVILLAGEPEOPLE,LIFESABEACH,ONLYHOMIESALLOWED,BETTERSTAYINDOORS,NINJATOWN,LOVECONQUERSALL,EVERYONEISPOOR,EVERYONEISRICH,CHITTYCHITTYBANGBANG,CJPHONEHOME,JUMPJET,IWANTTOHOVER,TOUCHMYCARYOUDIE,SPEEDFREAK,BUBBLECARS,NIGHTPROWLER,DONTBRINGONTHENIGHT,SCOTTISHSUMMER,SANDINMYEARS,KANGAROO,NOONECANHURTME,MANFROMATLANTIS,LETSGOBASEJUMPING,ROCKETMAN,IDOASIPLEASE,BRINGITON,STINGLIKEABEE,IAMNEVERHUNGRY,STATEOFEMERGENCY,CRAZYTOWN,TAKEACHILLPILL,FULLCLIP,IWANNADRIVEBY,GHOSTTOWN,HICKSVILLE,WANNABEINMYGANG,NOONECANSTOPUS,ROCKETMAYHEM,WORSHIPME,HELLOLADIES,ICANGOALLNIGHT,PROFESSIONALKILLER,NATURALTALENT,OHDUDE,FOURWHEELFUN,HITTHEROADJACK,ITSALLBULL,FLYINGTOSTUNT,MONSTERMASH".split,.findIndex(E=>0<=T.indexOf(E)&T.indexOf(E)>=T.length-E.length)?"CHEAT ACTIVATED":""

Attempted using min<var<max in the array callback, but that resulted fale positives.

Pretty straightforward otherwise; I preferred not to use compression; can't be shorter that way.

5 bytes spared by @KevinCruijssen and my reading to JS golfing.

• Doesn't seem to work at all. Maybe something like this: Try it online! (the comment is not long enough to hold the whole string).
– Neil
May 27, 2020 at 19:06
• @Neil for the record, I tested with Deno and that one didn't let me create the function a without var that I left here out for brevity. But otherwise, console.log(a("AAAAAAAAAATOODAMNHOT")) works, for example. May 27, 2020 at 19:47
• Wait, sorry, there was an error indeed. Thanks May 27, 2020 at 20:01
• @KevinCruijssen thanks, single-mark operator was fine but 0<= is equivalent to -1<, not 1< - zero-inclusion is crucial because with indexOf returns, 0 is perfectly valid. May 28, 2020 at 13:32
• I know you don't want to use compression but you can save 70 bytes by getting rid of the commas (this is an edited version so that the link fits in a comment). May 28, 2020 at 23:58

Erlang (escript), 1322 bytes

a(X)->lists:any(fun(I)->lists:suffix(I,X)end,string:split("THUGSARMOURY
PROFESSIONALSKIT
NUTTERSTOYS
INEEDSOMEHELP
TURNUPTHEHEAT
TURNDOWNTHEHEAT
PLEASANTLYWARM
TOODAMNHOT
DULLDULLDAY
STAYINANDWATCHTV
CANTSEEWHEREIMGOING
TIMEJUSTFLIESBY
SPEEDITUP
SLOWITDOWN
ROUGHNEIGHBOURHOOD
STOPPICKINGONME
SURROUNDEDBYNUTTERS
TIMETOKICKASS
OLDSPEEDDEMON
DOUGHNUTHANDICAP
JUSTTRYANDSTOPME
WHERESTHEFUNERAL
CELEBRITYSTATUS
TRUEGRIME
ALLCARSGOBOOM
STICKLIKEGLUE
GOODBYECRUELWORLD
DONTTRYANDSTOPME
ALLDRIVERSARECRIMINALS
PINKISTHENEWCOOL
SOLONGASITSBLACK
FLYINGFISH
WHOATEALLTHEPIES
BUFFMEUP
LEANANDMEAN
BLUESUEDESHOES
ATTACKOFTHEVILLAGEPEOPLE
LIFESABEACH
ONLYHOMIESALLOWED
BETTERSTAYINDOORS
NINJATOWN
LOVECONQUERSALL
EVERYONEISPOOR
EVERYONEISRICH
CHITTYCHITTYBANGBANG
CJPHONEHOME
JUMPJET
IWANTTOHOVER
TOUCHMYCARYOUDIE
SPEEDFREAK
BUBBLECARS
NIGHTPROWLER
DONTBRINGONTHENIGHT
SCOTTISHSUMMER
SANDINMYEARS
KANGAROO
NOONECANHURTME
MANFROMATLANTIS
LETSGOBASEJUMPING
ROCKETMAN
BRINGITON
STINGLIKEABEE
IAMNEVERHUNGRY
STATEOFEMERGENCY
CRAZYTOWN
TAKEACHILLPILL
FULLCLIP
GHOSTTOWN
HICKSVILLE
WANNABEINMYGANG
NOONECANSTOPUS
ROCKETMAYHEM
WORSHIPME
ICANGOALLNIGHT
PROFESSIONALKILLER
NATURALTALENT
OHDUDE
FOURWHEELFUN
ITSALLBULL
FLYINGTOSTUNT
MONSTERMASH","
",all)).

Try it online!

• Fine, but it should be a complete program and the output: CHEAT ACTIVATED / <null> May 31, 2020 at 16:00


text:
# Requiring exactly $n_j$ slots for job $j$ (if scheduled)

Let $$x_{j}(t)=1$$ iff job $$j$$ is scheduled at time $$t$$. I want to say that if the job is scheduled at all, then it is scheduled at $$n_j$$ slots. I wrote this as:

$$x_{j}(t)\sum_{s=1}^{T}x_{j}(s)=n_jx_{j}(t).$$

Is it possible to write this as linear constraints?

If I am maximizing the number of jobs to be scheduled, can we do any better?

• I don't understand why you multiplied the left side by $x_j(t)$. I think it's not necessary. BTW, are $n_j$ constant values? If yes, you have a linear constraint, if no what is the domain for that values? – Oguz Toragay Apr 22 '20 at 0:45
• If I don't add the term in the left and $x_j(t)=0$, then $\sum\limits_{s=1}^{T}x_j(s)=0$ and the job will never be scheduled. – zdm Apr 22 '20 at 2:20
• You did not specify whether the slots need to be consecutive. The solutions from Rob and Siong do not require the slots to be consecutive, which hopefully is okay in the context of your problem. – prubin Apr 22 '20 at 15:30
• @prubin Requiring $n_j$ to be consecutive cannot be guaranteed with this formulation right? I mean I must redefine the variables $x_j(t)$? For example, $x_j(t)=1 \iff$ job $j$ is scheduled from $t$ to $t+n_j$? – zdm Apr 22 '20 at 17:45

If you need consecutive slots, one way to model it is to redefine $$x_j(t)$$ to indicate the start at time $$t$$ of $$n_h$$ consecutive slots occupied by $$j$$. The "$$n_j$$ or none" constraint then becomes $$\sum_{t=1}^{T-n_j+1} x_j(t)\le 1\quad\forall j$$or, if you need a binary indicator $$y_j$$ for whether $$j$$ gets scheduled, $$\sum_{t=1}^{T-n_j+1}x_j(t)=y_j\quad\forall j.$$

Introduce a binary variable $$y_j$$ to indicate whether $$x_j(t)>0$$ for some $$t$$, and impose linear constraints: $$\sum_{t=1}^T x_j(t) = n_j y_j$$

We need an indicator of whether a job is ever scheduled at all.

$$y_j=\begin{cases} 1, & \sum\limits_{t=1}^Tx_j(t)>0 \\ 0, &\sum\limits_{t=1}^Tx_j(t)=0\end{cases}$$

$$\sum_{t=1}^T x_j(t) = n_jy_j$$

Furthermore, you want to maximize the number of jobs.

\begin{align}\max&\quad\sum_j y_j\\\text{s.t.}&\quad\sum_{t=1}^T x_j(t) = n_jy_j\\&\quad x_j(t), y_j \in \{0,1\}.\end{align}


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# If n is a prime number greater than 2, is 1/x > 1?

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If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 21:03
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If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 Oct 2014, 03:29, edited 2 times in total.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 22:28
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

In such a case, 1/x>1 or x belongs to {0,1} -ve values again will not satisfy the equation.

Statement 2: x^(n-1) > x^2(n-1), for n being odd, n-1 will be even. Hence for the above equation the solution set will be {-1,1}
As mentioned above, the negative values of x will not hold the 1/x>1 equation. Hence the range is again {0,1}.

Am I doing sumthing wrong?

Regards,
Arpan
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 22:56
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a +ve proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X is a +ve proper fraction ?????

So, We have to find if X is a +ve Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a +ve proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a +ve proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

It can be written as ... .... $$x^{n+1} > x^{2n}$$

Therefore.. InSufficient ....

Hence, ...... A.......
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Last edited by manishuol on 09 May 2013, 00:42, edited 3 times in total.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 23:14
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan
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Last edited by arpanpatnaik on 08 May 2013, 23:35, edited 1 time in total.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 23:41
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The question is asking, is $$\frac{1}{x}$$>1 --> Is 0<x<1

From F.S 1, for n=3, we have$$x^3<x<x^{\frac{1}{3}}$$ . As $$x>x^3,$$ we have $$x(1-x^2)>0$$ --> x(x+1)(x-1)<0 .

Hence 0<x<1 OR x<-1. Again, from $$x<x^{\frac{1}{3}}$$, we can cube on both sides and get, $$x^3<x$$, which is the same as above. Thus, from F.S 1, we get :

0<x<1 OR x<-1. Insufficient.

From F.S 2, we know that for n=3, $$x^2 > x^4 --> x^2<1$$ --> -1<x<1. Clearly Insufficient.

Taking both together, we get 0<x<1 as the common range and a YES for the question stem. Sufficient.

C.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 23:45
arpanpatnaik wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As x>2 and is a prime number, x will always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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08 May 2013, 23:59
arpanpatnaik wrote:
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan

My answer is still A ...
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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09 May 2013, 00:06
vinaymimani wrote:
arpanpatnaik wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As n>2 and is a prime number, will n always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.

My Bad!! I missed that... Thanks for correcting me Vinay! I got the answer now! [C] it is
Can't believe I used the odd principle in Statement 2 and failed to use it in Statement 1. The range x < -1 will be valid and the solution set needs to be an intersection of Statement 1 and 2 i.e. x belongs to {0,1}.

Regards,
Arpan
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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28 Oct 2014, 10:26
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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29 Oct 2014, 23:49
the question asks if 1/x > 1
or, 1/x - 1 > 0
or, (1-x) / x > 0
again for the above condition to be true;
(1-x) > 0 and x > 0 or (1-x) < 0 and x<0 (for x<0, 1-x can never be less than 0, hence ignore this)
basically the question asks if 0 < x < 1 <----------- this is what we need to find, does x lie between 0 and 1??

statement 1 is clearly insufficient

i've got a doubt here with statement 2.

x^(n-1) > x^(2n-2)
=> (x^(n-1)) - (x^(2n-2)) > 0
=> (x^n/x) - (x^2n/x^2) > 0
=> x^n/x (1-x^n/x) > 0 taking x^n/x common outside the bracket
=> for the above condition to be true
either or
x^n/x > 0 and (1- x^n/x) > 0 x^n/x < 0 and (1 - x^n/x) < 0
when x^n/x > 0, when x^n/x < 0, (1 - x^n/x) can never be less than 0
it means for (1 - x^n/x) > 0 hence, we can ignore this option
x^n/x has to be a fraction between 0 and 1
i.e. 0< x^n/x < 1
or, 0 < x^(n-1) < 1

when x^(n-1) is between 0 and 1, and n is a prime number greater than 2,
it implies that x is a fraction between 0 and 1 and x^(n-1) will always be positive and true for any value between 0 and 1.
in that case Statement 2 is satisfying what we are asked to find out. statement 2 should be sufficient.
correct me if i am wrong.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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30 Oct 2014, 07:52
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

Statement 1 -> will be true for negative integers and positive fractions. So X can either be a negative integer or positive fraction. If X is a negative integer 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 1 alone is insufficient.

Statement 2 -> Will be true for negative and positive fractions. If X is a negative fraction 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 2 alone is insufficient.

But for both statement 1 and 2 to be true, X can only be a positive fraction. Here 1/X will be greater than 1.

Hence both statements together are sufficient.

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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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01 Nov 2014, 07:12
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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01 Nov 2014, 07:38
thefibonacci wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.

Odd root from negative number is negative, not imaginary. For, example, $$\sqrt[3]{-8}=-2$$.

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.
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If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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01 Nov 2014, 07:47
Bunuel wrote:
thefibonacci wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.

Odd root from negative number is negative, not imaginary. For, example, $$\sqrt[3]{-8}=-2$$.

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.

Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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01 Nov 2014, 07:51
thefibonacci wrote:
Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29

No, that's generally true.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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20 Nov 2015, 05:29
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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11 May 2016, 12:45
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

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11 May 2016, 14:40
mattyahn wrote:
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

The OA is given under the spoiler and it's C, not B.

If x=-1/2 and n=3, $$x^{n-1} > x^{2n-2}$$ holds true: $$[(-\frac{1}{2})^{(3-1)} =\frac{1}{4}] > [\frac{1}{16}=(-\frac{1}{2})^{(2*3-2)}]$$
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

11 May 2016, 15:22
Bunuel, thanks - I overlooked the fact that the exponent will always be even given that n is prime greater than 2.
Man, this a tricky one.
Re: If n is a prime number greater than 2, is 1/x > 1?   [#permalink] 11 May 2016, 15:22
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# What's my mistake in finding $\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$?

I want to evaluate $$I=\int_0^\infty dx e^{-ax^2} \sin(b/x^2)$$ for $$a,b>0$$. A first simplification is to substitute $$y=x/\sqrt{a}$$ and define $$c=ab>0$$ to obtain $$I=\frac{1}{\sqrt{a}} \int_0^\infty e^{-x^2} \sin(c/x^2)$$ Now my idea was to use the Taylor series for the sine $$I=a^{-1/2} \int_0^\infty dx e^{-x^2} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\left(\frac{c}{x^2} \right)^{2k+1}$$ Now I interchange sum and integral although I have no justification $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \int_0^\infty dx e^{-x^2} \left(\frac{1}{x^2} \right)^{2k+1}$$ Substituting $$t=x^2$$ in the integral we obtain the gamma function $$I=a^{-1/2} \sum_{k=0}^\infty \frac{(-1)^kc^{2k+1}}{(2k+1)!} \frac{1}{2} \Gamma(-2k-1/2)$$ Using $$\Gamma \left({\frac{1}{2}}-n\right)={(-4)^{n}n! \over (2n)!}{\sqrt {\pi }}$$ (which can be shown using the reflection formula and the duplication formula for the Gamma function) with $$n=2k+1$$ I obtain $$I=a^{-1/2} \sqrt{\pi} \sum_{k=0}^\infty \frac{(-1)^k(-4c)^{2k+1}}{(4k+2)!} \frac{1}{2}$$ or $$I=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sum_{k=0}^\infty \frac{(-1)^k(4c)^{2k+1}}{(4k+2)!}=-\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \sinh(\sqrt{2c})$$ where I used wolfram alpha for the last series.

The problem: The above result is wrong. It should be (wolfram alpha and Gradshteyn) $$I=\frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c})\exp(-\sqrt{2c})$$

The question: Can someone spot my mistake? Was it interchanging the limits? I would also be interested in your solutions to the integral $$I$$ using other approaches.

• After the interchange, all the integrals diverge at $0$, aren't they? – user58697 Jul 29 at 19:58
• @user58697 Yes, you are right. Probably that is the mistake... Do you have an idea for an alternative approach to the original integral? – thomasfermi Jul 29 at 20:09

The Glasser's master theorem is a useful tool for the solution. First, use Euler's formula to decompose the sine term into the sum of exponentials. Then it boils down to computing the integral of the form

$$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - \frac{p}{x^2} \right) \, \mathrm{d}x.$$

Assume for a moment that $$a, p > 0$$. Then by completing the square, we get

$$J(p) = \int_{0}^{\infty} \exp\left( -a \left( x - \frac{\smash{\sqrt{p/a}}}{x} \right)^2 - 2\sqrt{ap} \right) \, \mathrm{d}x.$$

Then by the Glasser's master theorem and the gaussian integral, this evaluates to

$$J(p) = \int_{0}^{\infty} \exp\left( -a x^2 - 2\sqrt{ap} \right) \, \mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{ap}). \tag{*}$$

Although $$\text{(*)}$$ is originally proved for $$p > 0$$, both sides of $$\text{(*)}$$ define holomorphic functions for $$p$$ in the right-half plane $$\mathbb{H}_{\to} = \{z \in \mathbb{C} : \operatorname{Re}(z) > 0\}$$ and are continuous on the closed right-half plane $$\overline{\mathbb{H}_{\to}}$$. So by the identity theorem and continuity, $$\text{(*)}$$ extends to all of $$p \in \overline{\mathbb{H}_{\to}}$$. In particular, plugging $$p = \pm ib$$ for $$b > 0$$, we get

$$J(\pm ib) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-2\sqrt{\pm i ab}) = \frac{1}{2}\sqrt{\frac{\pi}{a}} \exp(-\sqrt{2c}(1\pm i)).$$

Therefore

$$I = \frac{J(-ib) - J(ib)}{2i} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \sin(\sqrt{2c}) \exp(-\sqrt{2c}).$$

• Thanks for your very nice solution. I had never heard of Glasser's theorem before! – thomasfermi Jul 30 at 22:03


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Creating surface meshes¶

This exercise uses PDAL to create surface meshes. PDAL is able to use a number of meshing filters: https://pdal.io/stages/filters.html#mesh. Three of these are ‘in the box’, without needing plugins compiled. These are 2D Delaunay triangulation, Greedy projection meshing and Poisson surface meshing.

In this exercise we’ll create a Poisson surface mesh - a watertight isosurface - from our input point cloud.

Exercise¶

We will create mesh models of a building and its surrounds using an entwine data input source.

After running each command, the output .ply file can be viewed in Meshlab or CloudCompare.

PDAL implements Mischa Kazhdan’s Poisson surface reconstruction algorithm. For details see [Kazhdan2006]_

Note

writers.ply will write out mesh vertices by default. In this exercise we set the attribute faces=”true”. Try using the ply writer without it. Also, if you’re using a machine with a lot of processing power, try increasing the depth parameter for a more detailed mesh.

Command¶

Invoke the following command, substituting accordingly, in your Conda Shell:

 1 2 3 4 5 pdal translate -i ept://http://act-2015-rgb.s3.amazonaws.com \ -o ./exercises/analysis/meshing/first-mesh.ply \ poisson --filters.poisson.depth=16 \ --readers.ept.bounds="([692738, 692967], [6092255,6092562])" \ --verbose 4
 1 2 3 4 5 pdal translate -i ept://http://act-2015-rgb.s3.amazonaws.com ^ -o ./exercises/analysis/meshing/first-mesh.ply ^ poisson --filters.poisson.depth=16 ^ --readers.ept.bounds="([692738, 692967], [6092255,6092562])" ^ --verbose 4

You can view the mesh in Cloud Compare, you should see something similar to

Filtering¶

If we want to just mesh a building, or just terrain, or both we can apply a range filter based on point classification. These data have ground labelled as class 2, and buildings as 6.

In this exercise we will create a poisson mesh surface of a building and the ground surrounding it, using the same data subset as above and adding a filters.range stage to limit the set of points used in mesh creation.

Command¶

Invoke the following command, substituting accordingly, in your Conda Shell:

 1 2 3 4 5 6 7 pdal translate -i ept://http://act-2015-rgb.s3.amazonaws.com \ -o ./exercises/analysis/meshing/building-exercise.ply \ range poisson \ --filters.range.limits="Classification[2:2],Classification[6:6]" \ --filters.poisson.depth=16 \ --readers.ept.bounds="([692738, 692967], [6092255,6092562])" \ --verbose 4
 1 2 3 4 5 6 7 pdal translate -i ept://http://act-2015-rgb.s3.amazonaws.com ^ -o ./exercises/analysis/meshing/building-exercise.ply ^ range poisson ^ --filters.range.limits="Classification[2:2],Classification[6:6]" ^ --filters.poisson.depth=16 ^ --readers.ept.bounds="([692738, 692967], [6092255,6092562])" ^ --verbose 4


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# Properties of Least Squares Regression Coefficients

## Presentation on theme: "Properties of Least Squares Regression Coefficients"— Presentation transcript:

Properties of Least Squares Regression Coefficients
In addition to the overall fit of the model, we now need to ask how accurate each individual estimated OLS coefficient is (R2 is a summary measure of the model as a whole – doesn’t distinguish between different variables) Measure statistical performance of estimated coefficients using 2 concepts Bias Efficiency

Unlike the estimated residual,
To do this need to make some assumptions about the behaviour of the (true) residual term that underlies our view of the world (Gauss-Markov assumptions) Unlike the estimated residual, the true residual u is never observed (remember the true state of the world Y=b0 +b1X + u is never observed, only an estimate of it So all we can ever do is make some assumptions about the behaviour of u

(NOT proved to be equal to zero unlike the OLS residual)
and the 1st assumption is that 1. E(ui) = 0 The expected (average or mean) value of the true residual is assumed to be zero (NOT proved to be equal to zero unlike the OLS residual) - sometimes positive, sometimes negative, but there is never any systematic behaviour in this random variable so that on average its value is zero

The 2nd assumption about the unknown true residuals is that
Var(ui / Xi) = σu2 = constant ie the spread of residual values is constant for all X values in the data set (homoskedasticity) - think of a value of the X variable and look at the different values of the residual at this value of X. The distribution of these residual values around this point should be no different than the distribution of residual values at any other value of X - this is a useful assumption since it implies that no particular value of X carries any more information about the behaviour of Y than any other

3. Cov(ui uj) = 0 for all i ≠ j (no autocorrelation) there is no systematic association between values of the residuals so knowledge of the value of one residual imparts no information about the value of any other residual.

4. Cov(X, ui) = 0 - there is zero covariance (association) between the residual and any value of X – ie X and the residual are independent so no value of X gives any information about the value of any residual - This means that we can distinguish the individual contributions of X and u in explaining Y (Note that this assumption is automatically satisfied if X is non-stochastic ie non-random so can be treated like a constant, measured with certainty) Given these 4 assumptions we can proceed to establish the properties of OLS estimates

The 1st desirable feature of any estimate of any coefficient is that it should, on average, be as accurate an estimate of the true coefficient as possible. Accuracy in this context is given by the “bias” This means that we would like the expected, or average, value of the estimator to equal the true (unknown) value for the population of interest ie if continually re-sampled and re-estimated the same model and plotted the distribution of estimates then would expect the mean value of these estimates to equal the true value (which would only be obtained if sampled everyone in the relevant population) In this case the estimator is said to be “unbiased”

Are OLS estimates biased?

Given a model Y=b0+b1X+u We now know the OLS estimate of the slope is calculated as Sub. In for Y=b0+b1X+u and using the rules of covariance (from problem set 0) Cov(X,Y) = Cov(X, b0+b1X+ u) = Cov(X, b0) + Cov(X, b1X) + Cov(X, u) Consider the 1st term Cov(X, b0) since b0 is a constant it has no variance no matter the value of X, so Cov(X, b0)=0

So Cov(X,Y) = 0 + Cov(X, b1X) + Cov(X, u)
Consider the 2nd term Cov(X, b1X) since b1 is a constant can take it outside the bracket so Cov(X, b1X) = b1Cov(X, X) (see problem set 0) = b1Var(X) (since Cov(X,X) is just Var(X) ) Hence Cov(X,Y) = 0+b1Var(X) + Cov(X,u) Sub. this into OLS slope formula Using rules for fractions

Given that now since Cov(X,u) is assumed =0, implies that Now need expected values to establish the extent of any bias. It follows that taking expectations And since term on right hand side is a constant so that, on average, the OLS estimate of the slope will be equal to the true (unknown) value ie OLS estimates are unbiased So don’t need to sample entire population since OLS on a sub-sample will give an unbiased estimate of the truth (Can show unbiased property also holds for ols estimate of constant – see problem set 2)

How to obtain estimates by OLS
Goodness of fit measure, R2 Bias & Efficiency of OLS Hypothesis testing

Precision of OLS Estimates
So this result means that if OLS done on a 100 different (random) samples would not expect to get same result every time – but the average of those estimates would equal the true value Given 2 (unbiased) estimates will prefer the one whose range of estimates are more concentrated around the true value. Measure efficiency of any estimate by its dispersion – based on the variance (or more usually its square root – the standard error) Can show (see Gujarati Chap. 3 for proof) that variance of the OLS estimates of the intercept and the slope are (where σ2u = Var(u) = variance of true (not estimated) residuals)

This formula makes intuitive sense since
1) the variance of the OLS estimate of the slope is proportional to the variance of the residuals, σ2u the more there is random unexplained behaviour in the population, the less precise the estimates 2) the larger the sample size, N, the lower (the more efficient) the variance of the OLS estimate more information means estimates likely to be more precise 3) the larger the variance in the X variable the more precise (efficient) the OLS estimates the more variation in X the more likely it is to capture any variation in the Y variable

Estimating σu2 In practice never know variation of true residuals used in the formula Can show, however, that an unbiased estimate of σ2u is given by which since means Or equivalently (learn this equation) Sub. this into (1) and (2) gives the formula for the precision of the OLS estimates.

So substitute into the equations below gives the working formula needed to calculate the precision of the OLS estimates. At same time usual to work with the square root to give standard errors of the estimates (standard deviation refers to the known variance, standard error refers to the estimated variance) (learn this)

Gauss-Markov Theorem Given Gauss-Markov assumptions 1-4 hold 1. E(ui) = 0 2. Var(ui / Xi) = σu2 = constant 3. 3. Cov(ui uj) = 0 for all i ≠ j 4. Cov(X, ui) = 0 then can prove a very important result: that OLS estimates are unbiased and will have the smallest variance of all (linear) unbiased estimators - there may be other ways of obtaining unbiased estimates, but OLS estimates will have the smallest standard errors regardless of the sample size


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30 years experience

# mg physical and chemical properties

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# Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Statement

Let ${\displaystyle f(x)}$ be a differentiable function satisfying the following properties:
          ${\displaystyle f(0)=1,f'(0)=-2,f''(0)=3;}$ and
${\displaystyle f'''(x)={\frac {6-x^{2}}{9-\cos(x)}}}$

Determine with proof and approximation A to ${\displaystyle f(2)}$ with the property that the error in the approximation is at most 1.


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## Differentially Private Bias-Term only Fine-tuning of Foundation Models

06 Oct 2022, 15:20 (modified: 21 Nov 2022, 04:57)TSRML2022Readers: Everyone
Keywords: deep learning, differential privacy, parameter efficient fine-tuning
TL;DR: We propose to only fine-tuning bias terms in large models in a differentially private manner, which is efficient and accurate.
Abstract: We study the problem of differentially private (DP) fine-tuning of large pre-trained models — a recent privacy-preserving approach suitable for solving downstream tasks with sensitive data. Existing work has demonstrated that high accuracy is possible under strong privacy constraint, yet requires significant computational overhead or modifications to the network architecture. We propose differentially private bias-term fine-tuning (DP-BiTFiT), which matches the state-of-the-art accuracy for DP algorithms and the efficiency of the standard BiTFiT. DP-BiTFiT is model agnostic (not modifying the network architecture), parameter efficient (only training about $0.1\%$ of the parameters), and computation efficient (almost removing the overhead caused by DP, in both the time and space complexity). On a wide range of tasks, DP-BiTFiT is $2\sim 30\times$ faster and uses $2\sim 8\times$ less memory than DP full fine-tuning, even faster than the standard full fine-tuning. This amazing efficiency enables us to conduct DP fine-tuning on language and vision tasks with long-sequence texts and high-resolution images, which were computationally difficult using existing methods.
4 Replies


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# In RSA. Why is $\phi(n)$ kept secret and $n$ is public?

I mean, $n$ can also be easily used to find the factors $p$ and $q$ right?

No, it is not easy! RSA is based on the difficulty of factoring the product $n=pq$ of two large prime numbers. But if you know $\varphi(n)$ for plain RSA you can compute the secret exponent $d=e^{-1}\bmod \varphi(n);\;$ and you can factor $n$ from the two equations $n=pq,\;\varphi(n)=(p-1)(q-1)$.


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# PACM Colloquium: The master equation and the convergence problem in mean-field games

Speaker:
François Delarue - Université Nice-Sophia Antipolis
Date:
Oct 26 2015 - 4:30pm
Event type:
PACM Colloquium
Room:
214 Fine Hall
Abstract:

We discuss the convergence, as $N$ tends to infinity, of a system of $N$ coupled Hamilton-Jacobi equations, called the Nash system. This system arises in differential game theory. We describe the limit problem in terms of the so-called master equation", a kind of second order partial differential equation stated on the space of probability measures. Our first main result is the well-posedness of the master equation. To do so, we first show the existence and uniqueness of a solution to the mean field game system with common noise", which consists in a coupled system made of a backward stochastic Hamilton-Jacobi equation and a forward stochastic Kolmogorov equation
and which plays the role of characteristics for the master equation. Our second main result is the convergence, in average, of the solution of the Nash system and a propagation of chaos property for the associated optimal trajectories".

Joint work with J.F. Chassagneux and D. Crisan; and P. Cardaliaguet, J.M. Lasry and P.L. Lions.