text:
# Sort a list and write some English!

Your boss managed to read the secret hidden message. He didn't end up firing you, though, he just made you a secretary and forbade you from writing code.

But you're a programmer. You need to write code. You must code.

Therefore, your code needs to look as similar to English as possible, make sense, and look as little like code as possible. Your code should take a list of integers (either in a function, or STDIN), and return that list sorted (returning it, or STDOUT).

Any language can be used, but I'm looking for the most creative solution, (kudos if your code looks like a business letter).

This is a popularity contest!

• whitespace to the rescue? Sep 5 '14 at 19:26
• Or SPL. Sep 5 '14 at 20:06
• I thought of SPL too, but Chef may be fun as well :P Sep 7 '14 at 0:12
• Funny.. there are no ruby submissions yet! :P Sep 9 '14 at 8:22
• SPL might be possible, but it's severely limited on how big the list may be. I might attempt it at home tonight. Would be a really weird business letter to read though. Sep 10 '14 at 11:50

# GolfScript

Dear Boss Man.

It came to my attention that my keyboard needs replacement; the keys required to write the
symbols ~ $ . and } are not functioning properly. It's very difficult to work like this! Please instruct the IT department to exchange the faulty keyboard as soon as possible. Sincerely, Dennis Try it online! ### How it works • Undefined tokens (e.g., most English words) are noops in GolfScript. • . ; duplicates the input string and discards the copy. • ~$ evaluates the input string and sorts the resulting array.

•  inspects the array (for pretty printing).

• . and duplicates the output string and discards the copy.

• } is a "super comment" since it is unmatched; everything following it is ignored.

• I'm guessing you wrote this on your other keyboard? Sep 5 '14 at 20:23
• Perhaps he wrote it with the mouse keyboard hidden in the accessibility gaboodle? Sep 7 '14 at 19:34
• It's an inline comment... it's a block comment... it's a SUPER-COMMENT!
– Doorknob
Sep 7 '14 at 19:55
• @Geobits: Let's assume that "are not functioning properly" translates to "requires multiple attempts to eventually get the characters on the screen" Sep 8 '14 at 1:37
• Or he used Character Map. Sep 9 '14 at 2:42

# PHP

Defines a function called item that will sort an array that you pass it.

Dear Boss,
I have successfully discovered all brackets.
The ones marked with question marks are the ones which I am not sure about.
The ones marked with asterisks can be used both as an opening and closing delimiter.
(
)
{
}
[
]
>?
<?
/*
'*
"*

Thank you for reading my memo. In other news, the */ function item (#12 in the list of things that have an asterisk before them) was discovered recently, which I read on a local news site.
#12 is my favorite function item! Just thought you'd be interested. Sorry if this is too off-topic; here's some business stuff. Imagine that you had some (let's say you have a combination of $10 &$money) # of dollars (i.e. you have $10 +$money). Now, here's the important part. It's so important, I'll separate it from the rest of this message with the brackets I discovered:

{
#10 in my personal list of things to remember about money management is that you have to be careful. I still haven't been able to
sort #9 out yet (I bought the manual from someone else), but #9 also seems to be about being careful. You also have to guard the dollars
($money); #13 says that if you don't protect them by putting them in a bank or something, they might be stolen. } //-----------------------\\ // Signed, \\ // Your great employee \\ //-----------------------\\  It looks much better if you paste it into a text editor and resize the window so that you can see the entire thing on your screen at once. • "I have successfully discovered all brackets." Someone's going for employee of the month! Sep 6 '14 at 0:19 • -1: You missed « and ». (In seriousness, +1.) Sep 6 '14 at 21:19 • Wow man. :D Can't stop laughing – Ven Sep 8 '14 at 15:12 # Python 2 ''''''''''''''''''''''''''''''''''''''''''''' One of the phones in the office seems to be broken, so it needs to be fixed. '''''''''''''''''''' This memo was made while testing the functionality of its buttons since some of them didn't seem to work. '''''''' 1 (no alphabets) working with no problem '''''''' abc working with no problem ''''''''' def working (partially): please =( 'o" :\ at least I figured out that this was a problem! :') # list (phone number list) was missing as well, so I need to: #1 print (please, sorted (partially)) [1] # list, and [2] the memo #2 return #3 check (making-sure, ghi-jkl-etc. works) ''''''''''''''''''  • +1 love how you snuck in def to mean 'the letters on phone key 3'. Sep 6 '14 at 19:25 ## Python 2 ''' 5th of September 2014 Dear Boss, I am writing this Mail because i want to discuss the recent incident. I am Thankful that you did not fire me, and gave me a secretary position instead. I dont have any experience in being a sec retary and I am not very. good at writing mail, so please excuse any spelling, grammar or formating errors in this mail. I will certainly try to improve and do my best in my new job and will''' 'from now on strictly ';exec'''ute your orders. = Firstly i want to sincerly apologize my dumb behaviour,it was idiotic ( and childish ). I really ;apologize to you. it wont .happen again, sir. If i could only reverse the timeline (to undo it all). Then i would; I really promise to never write code in any form Can i talk about that in a meeting with you? Maybe today at'''[4::15]#in the afternoon?  Finding the right words was a real pain. Input: [1,7,4,3] Output: [1,3,4,7] • This looks like those emails I get from Nigeria lol Sep 8 '14 at 4:49 • @ChrisCirefice Well, these mails probably contain hidden code as well ;) The spacing and somewhat poor wording in my code is because every 15th character of the string after the exec forms a new string which is then executed. So the whole program is basically exec"y=input();y.sort();print y" Sep 8 '14 at 12:45 • The ''' multiline string at the top is a dead giveaway, sadly, but +1 for "from now on strictly ';exec'''ute your orders." and good effort :p Sep 12 '14 at 10:47 # NetLogo To Manager [IT] Let Sue sort it; then show Sue the end to the end  With indentation and capitalization changes, the code becomes clear(er). This defines a function called manager which takes a list as input and prints the list sorted. to manager [IT] let sue sort IT; then (semicolons introduce a comment) show sue the end to the end  • Definitely doesn't look like code. Sep 9 '14 at 17:40 • I like this one a lot. Sep 9 '14 at 18:11 # C Input as space-separated list through STDIN, output as space-separated list through STDOUT. Dear Boss, I have made for you a decision about my employment a t your company. At about noon a letter explaining this shall be presented to you, as I am a lazy person. As you are a fool, I will say no more. And I look forward to never seeing you again. Insincerely, Steward Pitt  This should be compiled with: gcc bossletter.c -o bossletter -Dam='+++' -Dwill='---' -Dthis='{' -Dhave=';' -Ddecision=',' -Dquit='*' -Dfor='(' -Dmy=')' -Dbrain='}' -Dlie=']' -Dnoon='-quit' -DD='' -Dto='D' -Dyou='D' -Dfool='you' -Das='to' -Dday='D' -Dno='fool' -Dcake='Pitt' -Dlook='you' -DPitt='the' -Dthe='as' -DBoss='a[9999 lie' -DAs='a have' -Dlazy='my' -Dperson='lazy have company' -Dyour='this' -DInsincerely='a' -Dcompany='b' -Demployment='int quit' -DDear='struct this employment a have brain b have' -Dbe='scanf for' -Dis='I[' -Dnever='for presented " " decision' -Dthat='4 decision' -Dagain="my have b" -Dt='decision quit about have' -Dshall='while for' -Dpresented='"%" to "d"' -Dletter='have brain' -DAt='a have return quit' -Dmore='have b' -DAnd='a have shall' -Dare='qsort for' -Dforward='--my printf' -Dexplaining='main for my' -Dsay='1 decision that made my day' -DSteward='Boss have no brain' -Dseeing='the cake is a lie'  The code expands to: struct{ int *a; } b; a[9999], I; made(a, about) int *a, *about; { b.a; return *about - *a; } main(){ while(scanf("%d", I++ + a)); b.a; qsort(a, I-- - 1, 4, made); b.a; while(I--) printf("%d ", I[a]); b.a, a[9999]; }  • +1 for preprocessor abuse. Of course with that sort of thing you could arguably make it read like any piece of code you want. Sep 6 '14 at 6:26 • You should have given meaning to keywords and variables and only used the preprocessor for special characters like braces. You didn't even make an effort. – bebe Sep 6 '14 at 15:24 • @bebe the more insolent the abuse, the better, IMHO :). – user16991 Sep 6 '14 at 19:52 • Next codegolf is to make this "source" compile to an implementation of Conway's Game of Life. -D golf! Sep 8 '14 at 8:12 • @fluffy I love preprocessor abuse in obfuscated C. You can make almost anything in it valid C code with enough #define/-Ds. Sep 8 '14 at 21:35 # Python 2 & 3 ### Unfortunately, the employees of today just can't stop using hashtags all over the place... #WritingALetter #Business Dear Boss, #SecondLine I found this scrap of paper on the floor. It said " def sortl(l): # define a function return sorted(l) # returns the list, sorted " #ScrapOfPaper Just thought I should let you know. #Honesty From Laurence  [Worth a try, I guess?] • Great excuse for using comments, but unfortunately it contains actual code – user16402 Sep 6 '14 at 15:15 • Add print "hello" at the end Sep 16 '15 at 20:50 # Dogescript It's basically English, right? shh oooot! my keyybr oad is brokn. i ne ed neew 1. such fixs much keys keys dose sort wow keys  translates to: // oooot! my keyybr oad is brokn. i ne ed neew 1. function fixs(keys) { keys.sort(); return keys; }  • I think your boss will fire you if you write like this... Jul 16 '16 at 10:36 # Brainfuck Hey boss! This is the public key that you asked! >,[[-[>>+<<-]>+>]<[<<]>,]+>[>+<-]>[>[>+<<->-]<[<<.>>-]<<[>>+<<-]>>+>>] Pretty cool huh?  Copyright note: Note: I copy pasted it from here. As far as I know, that codes from site are has copyright cc by-sa 3.0 with attribution. If this method is illegal, let me know and I will try to delete this. • "Your code needs to look as little as code as possible." Sep 6 '14 at 9:03 • It's a public key for encryption! Not code! Different! Sep 6 '14 at 9:03 • @Realdeo Don't forget the , on the second line will read more input after the program ran Sep 6 '14 at 9:19 • Good catch! Editing it Sep 6 '14 at 9:21 • With brainfuck it should be much easier to just incorporate the code in plain text. > for example can be used for quotes (that's how it works in emails) and so on. Sep 6 '14 at 10:42 # Detour Dear Boss Man I thought I had put your recovered files on drive z. In reality it was downloaded to the main drive, C://users/boss/recovery. Sorry for the confusion!  Interpreter Permalink How it works: The only important part is ://, and the line above it, z.. : means "put input here", the first / reflects it up to z, which is sort; it then goes up, wraps across the top, keeps going up from the bottom, then hits / from below, which bounces it left to the other /, which does the same thing, except this time wrapping around to ., or output. Simplified version: # Detour Dear Boss Man z. ://  Try it online! ## JavaScript I'm not sure if I can outsmart him, but I can throw him off my trail! /*Hello Bossman I feel the need to alert you to a pressing matter. Nothing Dan from Marketing's head seems to be */function/*ing anymore. I have a lot of evidence, but not enough time to */sort/* through it. Should I just send it to you directly */(/*although I'm sure you have about a thou*/s/*and other things to do*/)/*? Nah, here's it in a summary: Dan seems to be overly facinated by the */{/* key on his keyboard, and presses it many times a minute. He */return/*ed his new Galaxy */s/*4, after trying to eat it, multiple times.*/. /*When Dan was tasked to */sort/* through the Haberson account, he attempted to stab his computer multiple times with a ruler */(/*he was eventually deterred by Jenny) Everyone backed off for a few days, but we all remembered. Later, we approached him about the incident, he claimed hostile working conditions and that he was unable to */function(/*. I think he m*/a/*y need to see a psychatrist. None of us feel safe to approach him on the topic again*/, b/*rining up conflict is something most of us like to avoid. (except Jenny*/) {/*Dan was then moved to a different spot in the building, in an attempt to give him a new environment. Despite the move he */return/*ed to his old desk every day, and wrote */a - b/* on the walls of the new spot. We all just sort of accepted that and the layout planner gave up after a week or two. One week we got a new employee in the office. When she went to say Hi to Dan, he screamed a - b over and over. Not sure what his fascination is there. b-c is a much superior algorithm. We're not really sure what caused this, but one day he switched from the { key to the */} /*key. Dan switched back the next day. There is an ongoing betting pool on what will happen next (general categories mostly*/) /*Currently the betting for him stabbing an actual person is at almost$20,000.

Please do something about Dan. If anything start a reality TV show from the security tapes.

On a completely unrelated note, I love this new */}/* key, it's really neat looking. Did my keyboard always have this?
Sincerely,
Zeke*/

function sort(s) { return s.sort(function(a,b) { return a-b }); }

• Comment abuse kinda renders this more obvious. Sep 8 '14 at 21:39
• Although it's really obvious what's happening, I love the "Dan" story. I hope it is not based on actual events :p Oct 5 '20 at 15:52
• I kinda used your technique for my answer (uncommenting some parts to create the function)
– user100690
Mar 18 at 8:56

# LiveScript

Try figuring this one out...Been sending him emails like this the whole time to allow for easy, more versatile copy-paste.

# Mr. Boss,
#
#     So, I'm done with this job. It is driving
#     me crazy. We aren't allowed to have *any*
fun =#(. I am planning on leaving this
#     company while you all poorly attempt to
#     sort all this out. Also, I would like to
#     mention that a specific single bit in the
sort # for your in-house algorithm gets
#     inverted. Happy hunting while I have fun
#     programming for another company far more
#     grateful than yours.
#
# Best regards,

1. Its standard library, Prelude.ls, has a native sort function.
2. Note the two missing hashes (comments). The first is far more obvious than the second.

It parses as this: "fun = sort", and compiles to this: "var fun; fun = sort;"

• By the way, the syntax highlighter is a fail. One of the comments isn't parsed correctly. Sep 22 '14 at 22:01

# Vyxal

Dear Boss,

Quite worryingly, the printer has stopped working. Please bring a hammer, a battleaxe and a potentiometer.

Sincerely,
Joe

Try it Online!

By some miracle, Dear Bo leaves the original input on top. ss sorts it twice, , outputs, Q, halts the program, then the rest is unexecuted.

• I'm sorry, but who is Joe? Aug 18 at 11:15
• @exedraj JOE MAMA LMFAO
– user
Aug 19 at 13:44

# 05AB1E

Dear Boss‚

I hope I{qualify for this new function. I once again want to sincerely apologize for my earlier secret message.
Thank you for giving me the opportunity to work at this department instead.
I must admit it's hard to resist the urge to write code, but I'll do my best.

Thank you.

Kind regards from your new secretary,
Kevin Cruijssen

Try it online. (TIO uses the legacy version which is a bit faster. Works the same as the new 05AB1E version, though.)

Explanation:

05AB1E code ignores newlines and spaces outside of strings or compressed string/numbers (except when using the (if-)else-statement I just found out in an earlier attempt of making this program.. >.>), so let's do the same in this explanation.

DearBoss‚Ihope # These are all no-ops; they do execute, but won't affect the output:
D              #  Duplicates the (implicit) input-list
e             #  Calculates the number of permutations of each item in the list
#  (NOTE: this might time-out depending on the values in the input-list)
a            #  Check if these numbers are letters (becomes a list of 0s / falsey values)
r           #  Reverse the items on the stack
B          #  Base conversion
o         #  Raise 2 to the power of each number
s        #  Swap the top two items on the stack
s       #  Swap the top two items on the stack
‚      #  Pair the top two items
#  (NOTE: This is not a regular comma (,), since that would print the top of
#   the stack to STDOUT, which we of course don't want here.)
I     #  Push the input-list again
h    #  Convert each value to hexadecimal
o   #  Raise 2 to the power of each integer value
p  #  Check for each if it's a prime
e #  Calculate the number of permutations of each item again

I{q            # Then the actual program comes:
I              #  Push the input-list
{             #  Sort it
q            #  Stop the program
#   This makes everything that comes after it no-ops
#   And will output the top of the stack implicitly as result

Try it online with added debugger-mode to see all this in action.

• +1 for being natural.
– null
Sep 21 '19 at 5:28

# JavaScript (ES6)

      (a) => /*----------*\ <= (a)
|      Announcement      |
.   (a)    \*----------*/    (a)   .

sort (/* YOUR DOCUMENTS BEFORE 2/5 )*/

($,_) => /*--------*\ <= (_,$)
|         Warnings         |
//  (_-$) \*--------*/ ($-_)  //

/*( NO SMOKING IN OFFICE */)

Try it online!

Yes, I know the boss likes fancy decorations in his notices so I capitalized the texts and surrounded the texts with fancy symbols :)

PS. Why is the office still using TELNET?

Actually, removing comments and redundant blanks makes the stuff into (a)=>(a).sort(($,_)=>($-_)), and that's what the stuff intends ;P

# Groovy

'Dear Bossman,
I finally found the formula to calculate the salary for the new employees.
In the salary software just enter the following:'
print "${(args.toList()*.toInteger()).sort()}"' Sincerely, Employee'  • You're not supposed to be writing code. That's code. Sep 6 '14 at 6:56 • I quite like this one. It's just hiding the code in plain sight. – user16991 Sep 6 '14 at 19:55 • LOL you should put "I've found the Excel formula" then. Though I'm quite familiar with Excel, the reference never crossed my mind. – user16991 Sep 7 '14 at 9:30 • @kuroineko Excel formulas dont have such a programmy syntax, do they? Sep 7 '14 at 11:50 • 8 upvotes and 9 downvotes? Wow; this is controversial. Sep 9 '14 at 0:14 # Jelly, 91 bytes Gee Boss, Go to Starbuck. Whoops, forgot the (sorry for the dot, my keyboard's messed up) Ṣ  Try it online! # CJam Hi; Isaq \~O~/ (1) @O@ / \* *~50 Times! ;; WINNINGS ;; ;; ;; ;;]$10000O  ;;

p1 is "where Kings sit', according to Shakesphere or someone. All I know is that after 4 straight years
of earning second in the national clown contest, it sure feels great to be on top for once. My clown
ASCII art could use some work, but hopefully I did a solid representation of our troupe. I can't believe
this was the one year you were sick, but we went and earned the $100k in your absence. See you Monday, Ethan 'I remain just one thing, and one thing only, and that is a clown." -  Try it online Apparently my job is at a clown troupe that just won the grand prize, my boss's name is Isaq, and my signature didn't work and forgot to credit Charlie Chaplin. Notes: This was afwul. CJam throws an error on any character that isn't a command or whitespace, or when we try to take input but there isn't any, or you try to do a command but the stack's empty, etc. This means that knruvxylqo is unusable for the actual code (except the one q at the start to load in the array), and emth are also notably difficult. So of the famous rstlne from Wheel of Fortune, we have left... s. English is going to be hard. I'll be explaining what happens on each character, but not really why because there's so much to go through, especially with the clown. (also I don't know why a couple of the things work shh...) Explanation: I tried a few greetings to open us off (Greetings, Hello, etc.) and Hi was the only viable one. The others either threw errors or consumed input. Hi just pushes 17 and makes sure it's an int. The ; then throws that 17 away. Isaq gives us ["18"] by pushing 18, stringifying it, and putting it in an array. q then gives us our input array as a string. Row one of the clown: \ swaps our two elements. ~ takes "18" out of the array. O pushes an empty string. ~ makes it disappear. / throws away "18" but puts our input string into an array. Row two: ( takes our input out of the array, but leaves an empty string before it on the stack. 1 pushes 1. ) increments that 1. Row three I brute-forced after writing row 5, because I knew kinda what I needed it to do. @ rotates our array so the input is on bottom and empty string on top, O adds another empty string, and @ rotates it so 2 is on top. Row four: / gets rid of the 2, \ swaps the identical empty strings, and * gets rid of one of them. Out of the clown now. The next * gets rid of our other empty string, leaving just the input string. Finally, we parse it into a bunch of numbers with ~. 50 Times! ends up just pushing a 50 and a 0 when all is said and done. Now into the winnings block. ;; gets rid of our 50 and 0. WINNINGS takes advantage of the fact that CJam doesn't yell at you if you scream at it because capital letters are variables. Unfortunately, that means it also pushes a bunch of junk. The next 8 ;s take care of that. We're just back to our input numbers now. We'll use a not-so-sneaky ] to turn them into an array and $ to sort them. We just need to output with p now, but unfortunately for it to make sense we need some money first. We push 10000, plus an O because we need two things for the upcoming semicolons to destroy.

After a simple p, we've printed our sorted array! It's not much of a letter though. 1 and is are harmless. We start a quote with ", then end it with ' to let us write in English unimpeded. We do the inverse with a quote at the end to close up, and the -throws what we have left on the stack away so it doesn't automatically output. This way, it looks like we intended to attribute the quote, but forgot.

• The only person here who knows the name of their boss. :)
– user92069
Jul 27 '20 at 8:09

# Zsh

I don't know whether the code or the creative writing was harder.

: RE: The state of the evaluation process; eval #problems'

: While I appreciate your enthusiasm, going to print
: this for the world to see isn't helpful in the long run.
:
: > People seem to just want that '$': It's a big problem : > which needs to be resolved. People need to buy into the : > company's team-based philosophy '{': ... ''}'' : : While I do not deny that people "just want that '$'",
: I absolutely disagree that it's a _bad_ thing. Employees
: should not be guilty for wanting to leave for greener
: pastures, so you should try and provide _some_ incentive
: to stay.
:
: > And yet everyone want's to just check "(y)" or '(n)'
: > without giving _any_ further feedback... You're too LAZY!
:
: I want to stop you here. Accusations like this against my
: team are _not_ to be tolerated. We do a _lot_ behind the
: scenes which you _clearly_ aren't aware of. Please refrain
: from '@-ing'' any of my team, and think with a bit more
: empathy next time, thank you.

-- Gamma :-}'

Try it online!

• Note: This will create files named "People" and "without" in the directory it is run from.

• From the first line, you can see the eval, followed by backticks.

• # comments continue to end-of-line, stopping quotes or backticks from being resolved, so we use : no-ops to indent.

• The standard pattern we make use of throughout this is  : lots of filler text.... : more filler text, then an important 'X' : back to the garbage

• At the end is one exception: Everything from @-ing to the last } is not no-op'd.

? JPY$14112 + (0 - Taxed) :This is the total price and what I will claim afterwards. ):0) (Attachment) || // Day 2 "Shinsaibashi Tower" [Conference, NOT FOR SHOPPING] I will be coming back on Day 2, so the trip will not cost too much hopefully. Best regards, The-former-programmer  Try it online! Okay this is an itinerary apparently. Another answer in different direction after one year. The really important part is Attachment => (Tennoji => Tennoji["sort"]((JPY$14112, Taxed) => JPY$14112 - Taxed))(Attachment) • Very clever answer, +1 – user100690 Mar 18 at 16:46 # Ly Dear BOss Ma&n. That jOb P&uts me in one of the best emotions that I've ever had! As for my recent incident, I promise that I will never touch code ever again. Thanks a lot for allowing me to be a secretary instead of firing me! Your faithful secretary, petStorm  Try it online! # JavaScript /* I always start my letters with /* because it means "slashstar", a great name for a weapon. So in case you forgot what an asterisk was, */asterisk/*is used as a sidenote in papers, and the function it serves */=/*is to get on your nerves when I write */papers/*. You made me a secretary */=>/* I really enjoy my job. Now, I need to sort the */papers/* in order because you like neat and tidy things. So I made a method */./* A guide to */sort/*ing follows: Start by finding the first paper */(/*the one on top), and then find the next one, */(/*wh*/a/*tever follows*/,/* nothing else. Then compar*/e/* them and place the longer one on top. Next, repeat (the process I told you before*/)/* */=>/* you should already see a clear pattern emerging. One of the best ways to compare, if you like automation, is */a-e/*, this returns the difference in characters. In the end, (having gone through all the papers*/)/*, you should have a sorted pile of papers. Regards, Secretary ;( */ A secretary who seems to enjoy his job on the outside gives his boss a method of sorting his messy pile of papers in an attempt to get his coding job back. • Hmmm... mystery: the snippet worked when I posted the answer but now it gives an error... – user100690 Mar 18 at 16:52 • Fixed now........ – user100690 Mar 18 at 16:53 # Pyth Dang those bounce back emails, I've been broke ever since the boss started charging us for every mistype. And he wonders why I wanted to quit in the first place... De$tination_unreachable:"Reason message:\
Company email servers charge for wasted electricity on bounce back emails:\
In this case, the charge was " $596.42; "\ Remember, this charge is employee specific, so make sure it doesn't happen again :3\ Please be more careful when typing your email addresses in the future.\ If you would like to refute this charge, your transaction ID is: " pSSQhaYk ";  # Javascript s=/* <k6cM L zE0 2]B^ cyF #hq% m(xg [ ; u e & & V 1 + T H h =$   T }cJF* H7Yj} *&P8        n^D>  N   ;  0O+   [G{
<   s >     w   < S  8        (   Q !   2     9     2    {
Ya8u  qf]?H Z   J 6   5       sRvz   **/  x=>x  //yE    c

.sort         (   (   a         ,          be)   =>a   -be   )/*(  DM@    r   E   v ]   9 g0$wh QBD 0 { 1 3x EK$ f        D   9 a   I 3   D P     @   !  > *  %6 hd I{ 9D {     *   %
P         )Di!D N x #       h%8>f       rM;5  }2d]  @   b g  =+ hTJ}  $;yC? H z s V S 1 U{zBV ;SRV i a s G 4 Q { 3 G T c H z l r 8 Z | a 0 * T C S 9 p +0oUy 1 I t 8 B ^ w h + O?P #%4 { ] l Z D M [ W LPTIx L * K r ] ffv ]3DUs p 0u1l Q-*:a -E{ ssD 0uLoS Aw@77 j ^QU V @ sF G 6 1 f A J n n & * + I m L n 5 y O % h C x N ! j k =)fy9 g]Y ;fvcN V R;0D (1wy+ d QrXc7 +@c} X { f+ o W z | 5 } - r P$  %  J       t   B   1 C  z    C
>   f  cVR    l         1   K       [5Vw  mvQ2g  j[K  +   # g2G-)   -   W   L f   (   J     H

WhA|F        6fi  }   F 0T4AI jmx#I       e   D  <tG  O   z
H         W   ( I   4   W     M         TZ  Z *   r z   e
]         g k W <   q   X     1         { Y r p   } t   Z
9         M  (  a   i   G     e         >  s{ 4   ^ 2 9 n
Y9S=y        xc \$  1P|  gOnp*   Q         0   b  s#P   [*/

Maybe he likes art? I mean, it's almost readable...

When comments and whitespace are removed you get s=x=>x.sort((a,be)=>a-be) and that's all that matters.

C#

// Dear Boss Man,

/*i have a */public/*shed a*/ List /*of*/ <string>/*s
to*/ Sort /*for you on our internal website
please take a look at the */(List /*of*/ <string>/*s and*/ check){
/*them for company restrictions. please also */check /*the */.Sort();/*ing of the     list*/
/*furthermore could you please */return /*an email to me in which you state your */check/*ing state*/;}

• Obvious comments :) The first line reads public List <String>`, without a lot of hassle :) Sep 9 '14 at 7:24


text:
# If A = <2 ,3 ,-4 >, B = <6 ,1 ,2 > and C=A-B, what is the angle between A and C?

Jan 3, 2016

${56.912}^{\circ}$

#### Explanation:

$C = A - B$
$= \left(2 , 3 , - 4\right) - \left(6 , 1 , 2\right)$
$= \left(- 4 , 2 , - 6\right)$.

Now to find the angle $\theta$ between any 2 vectors A and C, we may use either of the following 2 theorems involving algebraic products :

1. $A \times C = | | A | | \cdot | | C | | \cdot \sin \theta$
2. $A \cdot B = | | A | | \cdot | | B | | \cdot \cos \theta$

Since it is quicker to evaluate a Euclidean inner product as opposed to a vector cross product, I shall solve this problem using the latter theorem.

therefore cos theta = (A*C)/(|A||*||C||

$\therefore \theta = {\cos}^{- 1} \left(\frac{- 8 + 6 + 24}{\sqrt{{2}^{2} + {3}^{2} + {4}^{2}} \cdot \sqrt{{4}^{2} + {2}^{2} + {6}^{2}}}\right)$

$= {\cos}^{- 1} \left(\frac{22}{\sqrt{29} \cdot \sqrt{56}}\right)$

$= {56.912}^{\circ}$.

format: plain


text:
# DJI FPV Goggles Max Power option not available. Location: USA

I have a set of DJI FPV goggles but cannot change the Max Power option. It shows on the menu under the device section but I cannot select it.

• I am in the USA
• updated to latest firmware: V01.00.0500
• 2S 3000mAh battery
• .txt file on sd card narco_pwr.txt with pwr_2 attribute in txt file

• I have not powered or setup my Caddx Vista.

Max Power is set to 25mW and I cannot change it.

• Have you tried putting a text file called naco.txt in the SD root, containing only the number 1 (to set to US mode?) May 24 '20 at 22:30
• I see you state that you have not powered on or setup your Caddx Vista. Is it possible that you can't change these settings unless the goggles are bound to a powered up vista or air unit? I have not personally tested that yet, but that might be your issue. May 25 '20 at 4:19
• @5zero7rc, yes it is possible that the goggles need to bound with the Vista or air unit. But right now I do not have the supplies to test that, and am trying to get verification if this is so. May 25 '20 at 5:26
• @Kralc, I'm in the US, and it has 8ch, so I believe its already in US mode. But yes, I did add the naco.txt file with 1 within the txt file. May 25 '20 at 5:33


text:
# Constructor for skew polynomial rings¶

This module provides the function SkewPolynomialRing(), which constructs rings of univariate skew polynomials, and implements caching to prevent the same ring being created in memory multiple times (which is wasteful and breaks the general assumption in Sage that parents are unique).

AUTHOR:

• Xavier Caruso (2012-06-29): initial version
• Arpit Merchant (2016-08-04): improved docstrings, added doctests and refactored method
• Johan Rosenkilde (2016-08-03): changes to import format
sage.rings.polynomial.skew_polynomial_ring_constructor.SkewPolynomialRing(base_ring, base_ring_automorphism=None, names=None, sparse=False)

Return the globally unique skew polynomial ring with the given properties and variable names.

Given a ring $$R$$ and a ring automorphism $$\sigma$$ of $$R$$, the ring of skew polynomials $$R[X, \sigma]$$ is the usual abelian group polynomial $$R[X]$$ equipped with the modification multiplication deduced from the rule $$X a = \sigma(a) X$$.

INPUT:

• base_ring – a commutative ring
• base_ring_automorphism – an automorphism of the base ring (also called twisting map)
• names – a string or a list of strings
• sparse – a boolean (default: False). Currently not supported.

Note

The current implementation of skew polynomial rings does not support derivations. Sparse skew polynomials and multivariate skew polynomials are also not implemented.

OUTPUT:

A univariate skew polynomial ring over base_ring twisted by base_ring_automorphism when names is a string with no commas (,) or a list of length 1. Otherwise we raise a NotImplementedError as multivariate skew polynomial rings are not yet implemented.

UNIQUENESS and IMMUTABILITY:

In Sage, there is exactly one skew polynomial ring for each triple (base ring, twisting map, name of the variable).

EXAMPLES of VARIABLE NAME CONTEXT:

sage: R.<t> = ZZ[]
sage: sigma = R.hom([t+1])
sage: S.<x> = SkewPolynomialRing(R, sigma); S
Skew Polynomial Ring in x over Univariate Polynomial Ring in t over Integer Ring
twisted by t |--> t + 1

The names of the variables defined above cannot be arbitrarily modified because each skew polynomial ring is unique in Sage and other objects in Sage could have pointers to that skew polynomial ring.

However, the variable can be changed within the scope of a with block using the localvars context:

sage: with localvars(S, ['y']):
....:     print(S)
Skew Polynomial Ring in y over Univariate Polynomial Ring in t over Integer Ring
twisted by t |--> t + 1

SQUARE BRACKETS NOTATION:

You can alternatively create a skew polynomial ring over $$R$$ twisted by base_ring_automorphism by writing R['varname', base_ring_automorphism].

EXAMPLES:

We first define the base ring:

sage: R.<t> = ZZ[]; R
Univariate Polynomial Ring in t over Integer Ring

and the twisting map:

sage: base_ring_automorphism = R.hom([t+1]); base_ring_automorphism
Ring endomorphism of Univariate Polynomial Ring in t over Integer Ring
Defn: t |--> t + 1

Now, we are ready to define the skew polynomial ring:

sage: S = SkewPolynomialRing(R, base_ring_automorphism, names='x'); S
Skew Polynomial Ring in x over Univariate Polynomial Ring in t over Integer Ring
twisted by t |--> t + 1

Use the diamond brackets notation to make the variable ready for use after you define the ring:

sage: S.<x> = SkewPolynomialRing(R, base_ring_automorphism)
sage: (x + t)^2
x^2 + (2*t + 1)*x + t^2

Here is an example with the square bracket notations:

sage: S.<x> = R['x', base_ring_automorphism]; S
Skew Polynomial Ring in x over Univariate Polynomial Ring in t over Integer Ring
twisted by t |--> t + 1

Rings with different variables names are different:

sage: R['x', base_ring_automorphism] == R['y', base_ring_automorphism]
False

Todo

• Sparse Skew Polynomial Ring
• Multivariate Skew Polynomial Ring


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Zbl 0968.47017
Chang, S.S.
Some results for asymptotically pseudo-contractive mappings and asymptotically nonexpansive mappings.
(English)
[J] Proc. Am. Math. Soc. 129, No.3, 845-853 (2001). ISSN 0002-9939; ISSN 1088-6826/e

Let $E$ be a real Banach space, $E^*$ the topological dual space of $E$, $\langle\cdot, \cdot\rangle$ the dual pair between $E$ and $E^*$, $D(T)$, $F(T)$ the domain of $T$ and the set of all fixed points of $T$, respectively, and $J: E\to 2^{E^*}$ the normalized duality mapping defined by $J(x)= \{f\in E^*:\langle x,f\rangle= \|x\|,\|f\|,\|f\|=\|x\|\}$, $x\in E$. A mapping $T: D(T)\subset E\to E$ is said to be\par (1) asymptotically nonexpansive if there exists a sequence $\{k_n\}$ in $(0,\infty)$ with $\lim_{n\to\infty} k_n= 1$ such that $\|T^n x- T^ny\|\le k_n\|x-y\|$ for all $x,y\in D(T)$ and $n= 1,2,\dots$\ ,\par (2) asymptotically pseudo-contractive if there exists a sequence $\{k_n\}$ in $(0,\infty)$ with $\lim_{n\to\infty} k=1$ and for any $x,y\in D(T)$ there exists $j(x- y)\in J(x- y)$ such that $\langle T^nx- T^ny, j(x- y)\rangle\le k_n\|x-y\|^2$ for all $n= 1,2,\dots$\ .\par In this paper some convergence theorems of modified Ishikawa and Mann iterative sequence with errors for asymptotically pseudo-contractive and asymptotically nonexpansive mappings in Banach spaces are obtained.\par Let $T: D(T)\subset E\to E$ be a mapping, let $D(T)$ be a nonempty convex subset of $E$, let $x_0\in D(T)$ be a given point, and let $\alpha_n$, $\beta_n$, $\gamma_n$ and $\delta_n$ be four sequences in $[0,1]$. Then the sequence $\{x_n\}$ defined by $x_{n+1}= (1-\alpha_n- \gamma_n) x_n+ \alpha_nT^n y_n+ \gamma_n u_n$, $y_n= (1-\beta_n- \delta_n) x_n+ \beta_n T^nx_n+ \delta_n v_n$ for all $n\ge 0$ is called the modifies Ishikawa iterative sequence with errors of $T$, where $u_n$ and $v_n$ are two bounded sequences in $D(T)$.\par If $\beta_n=0$ and $\delta_n= 0$, $n= 0,1,2,\dots$, then $y_n= x_n$. The sequence $x_{n+1}= (1- \alpha_n- \gamma_n) x_n+ \alpha_n T^n x_n+ \gamma_n u_n$, $n\ge 0$, is called the modified Mann iterative sequences with errors of $T$. Main result is the following:\par Theorem 1. Let $E$ be a real uniformly smooth Banach space, let $D$ be a non-empty bounded closed convex subset of $E$, let $T: D\to D$ be an asymptotically pseudo-contractive mapping with a sequence $\{k_n\}\in (0,\infty)$, $\lim_{n\to\infty}k_n= 1$, and let $F(T)\ne\emptyset$. Let $\alpha_n$, $\beta_n$, $\gamma_n$, and $\delta_n$ be four sequences in $[0,1]$ satisfying the following conditions:\par (i) $\alpha_n+\gamma_n\le 1$, $\beta_n+ \delta_n\le 1$;\par (ii) $\alpha_n\to 0$, $\beta_n\to 0$, $\delta_n\to 0$ $(n\to\infty)$;\par (iii) $\sum^\infty_{n=0} \alpha_n=\infty$, $\sum^\infty_{n=0} \gamma_n<\infty$.\par Let $x_0\in D$ be any given point and let $\{x_n\}$, $\{y_n\}$ be the modified Ishikawa iterative sequence with errors. Then:\par (1) If $\{x_n\}$ converges strongly to a fixed point $q$ of $T$ in $D$, there exists a nondecreasing function $\emptyset:[0,\infty)\to [0,\infty)$, $\emptyset(0)= 0$ such that $\langle T^n y_n- q, J(y_n- q)\rangle\le k_n\|y_n- q\|^2- \emptyset(\|y_n- q\|)$, for all $n\ge 0$.\par (2) Conversely, if there exists a strictly increasing function $\emptyset: [0,\infty)\to [0,\infty)$, $\emptyset(0)= 0$ satisfying preceding inequality, then $x_n\to q\in F(T)$.\par A similar result for asymptotically nonexpansive mappings is proved. If $\{x_n\}$ is the modified Mann iterative sequence with errors, a similar result for asymptotically pseudo-contractive mappings is obtained.\par This paper generalizes the results by [{\it K. Goebel} and {\it W. A. Kirk}, Proc. Am. Math. Soc. 35, No. 1, 171-174 (1972; Zbl 0256.47045), {\it W. A. Kirk}, Am. Math. Mont. 72, 1004-1006 (1965; Zbl 0141.32402), {\it Q. H. Liu}, Nonlinear Anal. Theory Methods Appl. 26, No. 11, 1835-1842 (1996; Zbl 0861.47047), {\it J. Schu}, J. Math. Anal. Appl. 15, No. 2, 407-413 (1991; Zbl 0734.47036)].
[V.Popa (Bacau)]
MSC 2000:
*47H05 Monotone operators (with respect to duality)
47J25 Methods for solving nonlinear operator equations (general)
47H10 Fixed point theorems for nonlinear operators on topol.linear spaces
47H09 Mappings defined by "shrinking" properties

Keywords: asymptotically nonexpansive mapping; asymptotically pseudo-contractive mapping; modified Isikawa iterative sequence with errors; modified Mann iterative sequence with error; fixed point; normalized duality mapping; asymptotically nonexpansive; asymptotically pseudo-contractive; Ishikawa and Mann iterative sequence with errors

Highlights
Master Server


text:
## Univariate Validation

Contributors:

Date created: | Last Updated:

: DOI | ARK

Category: Data

### Wiki

For univariate validation, 10,000 conditions were tested, consisting of 100 values for $\mu$ evenly distributed between 0 and 10, and 100 values for $\sigma$ evenly distributed between 0.25 and 10. For each condition, 1,000,000 data were drawn from a univariate normal distribution $X \sim \mathcal{N}(μ,σ^2)$. The mean and variance of the error between generated data and the origin were calculated...


text:
## College Algebra (6th Edition)

Solution set = $\displaystyle \{(2,\frac{1}{2})\}$
$D=$determinant of the coefficient matrix $=\left|\begin{array}{ll} 1 & 2\\ 3 & -4 \end{array}\right|=1(-4)-(2)(3)=-10$ $D_{x}=$ in D, replace the x column with the constants column $=\left|\begin{array}{ll} 3 & 2\\ 4 & -4 \end{array}\right|=-12-8=-20$ $D_{y}=$ in D, replace the x column with the constants column $=\left|\begin{array}{ll} 1 & 3\\ 3 & 4 \end{array}\right|=4-9=-5$ $x=\displaystyle \frac{D_{x}}{D}=\frac{-20}{-10}=2$ $y=\displaystyle \frac{D_{y}}{D}=\frac{-5}{-10}=\frac{1}{2}$ Solution set = $\displaystyle \{(2,\frac{1}{2})\}$


text:
# The distance between the sun and Neptune is approximately 2,800,000,000 miles, how do you write it in scientific notation?

Jun 3, 2018

#### Answer:

$2.8 \cdot {10}^{9}$

#### Explanation:

When we write numbers in scientific notation, we want one non-zero digit before the decimal. This means we'll have the decimal between the $2$ and the $8$.

We need to loop the decimal $9$ times to the left. This will become our power of $10$.

Since we looped left, the exponent will be positive. We have

$2.8 \cdot {10}^{9}$

Hope this helps!


text:
Vectore area element?

1. May 6, 2009

-EquinoX-

1. The problem statement, all variables and given/known data

Let $$\vec{F} = xy\vec{i} + yz\vec{j} + xz\vec{k}$$ and C is the boundary of S, the surface z = 9 - x2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward. Use Stokes' Theorem to find [tex] \int\limits_c \vec{F} \cdot d\vec{r}.

2. Relevant equations

3. The attempt at a solution

well I've found the curlF at least.. I don't know what I should do now

2. May 6, 2009

gabbagabbahey

Re: stoke

Find an expression for the vector area element of S (it should have an x-component and a z-component)and integrate over the surface.

3. May 6, 2009

-EquinoX-

Re: stoke

vectore area element??

4. May 6, 2009

gabbagabbahey

Re: stoke

Yes, it's the product of the infinitesimal area element (usually denoted $dS$ or $da$) with the unit normal to the surface ($\vec{da}=\hat{n}da$). Have you not heard that term before?

For example, the outward vector area element for a spherical shell of radius $R$ is $\vec{da} =R^2 \sin \theta d \theta d \phi \hat{r}$, where $\theta$ is the polar angle, $\phi$ is the azimuthal angle, and $\hat{r}$ (sometimes written $\hat{e}_r$) is the radial unit vector.

Different authors use different notations.

Last edited: May 6, 2009
5. May 6, 2009

-EquinoX-

Re: stoke

yes, I've heard of it.. I now need to find the normal vector first... how can I do that...

6. May 6, 2009

gabbagabbahey

Re: stoke

Start by drawing a sketch of the surface, you should see that the outward normal to the surface is the same as the outward normal of the curve z=9-x^2. Parameterize that curve (I suggest using x=t) and find the tangent and normal in the usual ways.


text:
# All Questions

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text:
# Fit robot simulator to robot

I have odometry data $(x, y, angle)$ of a real two-wheeled robot, who received control commands $(forward speed, angular speed)$.

Now I want to code a motion model (in C++ (/ROS)) which should follow the same trajectory, given the same control commands.

Normally, the kinematics should look something like this:

\begin{align} v_{fwd} &= control_{fwd} \\ v_{ang} &= control_{ang} \\ x &= x_{old} + 0.5(control_{fwd} + v_{fwd,old}) * \cos(angle) * dt \\ y &= y_{old} + 0.5(control_{fwd} + v_{fwd,old}) * \sin(angle) * dt \\ angle &= angle_{old} + 0.5(control_{ang} + v_{ang,old}) * dt \end{align}

And I thought about just setting

\begin{align} v_{fwd} &= control_{fwd} + k_1 v_{fwd,old} + k_2 v_{fwd,old}^2 + k_3 v_{ang,old} + k_4 v_{ang,old}^2 \\ v_{ang} &= \text{ ...analog...} \\ x, y, angle &\text{ unchanged} \end{align}

and then just search the minimum of the squared distance of the computed trajetory to the real one - depending on the values of $k_i$. This would mean either a good optimization algorithm or just brute-forcing / randomly testing a lot of values.

Is this the way to go here? I tried the 2nd approach, but the results so far are not that good.

So, as you might guess now, I'm pretty new at this, so any help is appreciated.

• How great are the error values of wheel slip, and angle inaccuracy? The model could be perfect, but if your real-world errors are high, than the model will deviate strongly from reality, regardless of algorithm, unless this is taken into effect. – Mhz4.77 Apr 4 '14 at 1:41


text:
# 형광상관분광법을 이용한 광세기에 따른 유효 초점 부피 변화에 대한 연구

• Accepted : 2013.04.19
• Published : 2013.04.25

#### Abstract

Using fluorescence correlation spectroscopy, we analyzed the change of effective focal volume of a confocal system with light intensity. The fluorescence correlation spectroscopy system was home-built in accordance with the He-Ne laser with a wavelength of 632.8 nm, and two kinds of samples (AlexaFluor657 and Quantum dot655) suitable for the wavelength of the laser beam were used. For each sample, we analyzed and compared the correlation functions obtained while changing the intensity of the light source in a range of 1~50 ${\mu}W$. The result shows that the radius of the focal area increases linearly through the increase of particle number and diffusion time in response to an intensity change in weak light below 10 ${\mu}W$. In the higher intensity region (>10~15 ${\mu}W$), the increasing rate of particle number and diffusion time keep increasing but at a much slower rate. Through this result, it was also found that the radius increasing rate of the focal area was reduced however, the radius still increased slightly.

#### Acknowledgement

Supported by : 한국연구재단

#### References

1. P. Kask, P. Piksarv, M. Pooga, T. Mets, and E. Lippmaa, "Separation of the rotational contribution in fluorescence correlation experiments," Biophys. J. 55, 213-220 (1989). https://doi.org/10.1016/S0006-3495(89)82796-1
2. J. A. Hodgdon and F. H. Stillinger, "Stokes-Einstein iolation in glass-forming liquids," Phys, Rev. E 48, 207-213 (1993). https://doi.org/10.1103/PhysRevE.48.207
3. D. R. Lide, CRC Handbook of Chemistry and Physics (CRC Press, 1994), Chapter 6, p. 205.
4. K. Jacobson, E. D. Sheets, and R. Simson, "Revisiting the fluid mosaic model of membranes," Science 268, 1441-1442(1995). https://doi.org/10.1126/science.7770769
5. J. B. Pawley, Hand book of Biological Confocal Microscopy, 3rd ed. (Springer, New York, USA, 2006), p. 506.
6. K. Wang, X. Qiu, C. Dong, and J. Ren, "Single-molecule technology for rapid detection of DNA hybridization based on resonance light scattering of gold nanoparticles," Chem BioChem 8, 1126-1129 (2007).
7. J. Enderlein, I. Gregor, D. Patra, T. Dertinger, and U. B. Kaupp, "Performance of fluorescence correlation spectroscopy for measuring diffusion and concentration," ChemPhysChem 6, 2324-2336 (2005). https://doi.org/10.1002/cphc.200500414
8. J. Mertz, Introduction to Optical Microscopy (Robertand Company Publishers, Colorado, 2010), p. 269.
9. A. Cooper, Biophysical Chemistry (RSC, Cambridge, UK, 2005), Chapter 4.
10. S. H. Kim, T. Shin, and D. Kim, "Particles size measurement of silole nano-clusters by fluorescence correlation spectroscopy," J. Korean Phys. Soc. 56, 1264-1268 (2010). https://doi.org/10.3938/jkps.56.1264
11. P. Schwille, U. Haupts, S. Maiti, and W. W. Webb, "Molecular dynamics in living cells observed by fluorescence correlation spectroscopy with one- and two-photon excitation," Biophys. J. 77, 2251-2265 (1999). https://doi.org/10.1016/S0006-3495(99)77065-7
12. M. Ehrenberg and R. Rigler, "Rotational brownian motion and fluorescence intensify fluctuations," Chem. Phys. 4, 394(1974)
13. D. Magde, E. Elson, and W. W. Webb, "Thermodynamic fluctuation in a reacting system-measurement by fluorescence correlation spectroscopy," Phys. Rev. Lett. 29, 705-708 (1972). https://doi.org/10.1103/PhysRevLett.29.705
14. E. L. Elson and D. Magde, "Fluorescence correlation spectroscopy. I. conceptual basis and theory," Biopolymers 13, 1-27 (1974). https://doi.org/10.1002/bip.1974.360130102
15. J. R. Lakowicz, Principles of Fluorescence Spectroscopy, 3rd ed. (Springer Science+Business Media, New York, USA, 2006), p. 797.
16. O. Krichevsky and G. Bonnet, "Fluorescence correlation spectroscopy: the technique and its applications," Rep. Prog. Phys. 65, 251-297 (2002). https://doi.org/10.1088/0034-4885/65/2/203


text:
Can product of two singular matrices be invertible?

Suppose $A,B$ are square matrices of size $n\times n$. Can $AB$ be invertible, even though both $A$ and $B$ are singular (not invertible)?

And if not, does it follow that if $A_1 \times A_2 \times ... \times A_n$ is the invertible product of square matrices, then all factors $A_i$ are invertible?

• Are these square matrices? Nov 17, 2014 at 21:34
• @hardmath yes, sorry, I had to specify that. All matrices are of the same shape and are square matrices. Nov 18, 2014 at 7:49
• I'll edit that in. It's good to have the complete problem statement in the body of the Question, even if it repeats some of what the title/subject line says. Nov 18, 2014 at 11:17
• Note that the second part, while true, says something stronger than what one gets by induction on the first part. In the first part you assume neither of $A,B$ is invertible (both are singular), and ask if the product $AB$ can be invertible (it cannot). You need something stronger for the second part, namely that if either of $A,B$ is singular (not invertible), then the product is also singular. Nov 18, 2014 at 11:30
• @hardmath thanks. Nov 18, 2014 at 17:38

$$\det(A_1\cdot A_2\cdot\ldots\cdot A_n)\neq0\quad\Leftrightarrow\quad \det A_1\cdot\det A_2\cdot\ldots\cdot\det A_n\neq0\quad\Leftrightarrow\quad\det A_k\neq 0\quad \forall k$$

• Thank you, this is very educational! Nov 17, 2014 at 21:17
• This works only if the matrices are square. Nov 18, 2014 at 4:57
• @darijgrinberg Usually when one talks about invertible matrices, one refers to square matrices. Non-square matrix have no inverse: they can have a right-inverse or a left-inverse, but not both. Nov 18, 2014 at 6:50

Suppose $A$ and $B$ are such that $(AB)^{-1} = C$. Then $ABC = I$, so $A$ has inverse $BC$. Likewise, $CAB = I$, so $B$ has inverse $CA$.

(This relies on knowing that any left inverse is also a right inverse and vice versa.)

• I'm going to accept Adhvaitha's answer because it came first, but thank you too! Nov 17, 2014 at 21:16

If $A$ and $B$ are not invertible, then $AB$ is also not-invertible. More generally, $$\text{rank}(AB) \leq \max(\text{rank}(A), \text{rank}(B))$$ And the answer to your second question is yes as well by inducting the above argument.

• Oh, thanks! I couldn't find a way to prove invertibility of a matrix given by a large product, where I can only prove that factors are invertible, but not in that order. Now I can sleep soundly :) Nov 17, 2014 at 21:13
• In fact, rank(AB) is less than or equal to minimum of rank(A) and rank(B). Dec 10, 2021 at 6:47


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# An organic compound 'A' produces a gas with brisk effervescence when sodium bicarbonate is added to it. The compound 'A' contains the same number of carbons as in ethane. The compound 'A' reacts with ethanol to form a sweet smelling compound 'C' in presence of conc. H2SO4? acid. Then the name of compound A and the gas evolved is (1) Ethanoic acid ; CO2 (2) Methanoic acid ; CO2 (3) Ethanoic acid ; H2 (4) Methanoic acid ; H2

The compound A contains the same number of carbon as ethane.

Number of carbon atom in A compound = 2

When sodium bicarbonate is added A produces brisk effervescence. This is evidence that carbon dioxide is liberated.

CH3COOH + NaHCO3→ CH3COONa + CO2+ H2O

Sweet-smelling compounds are esters which are used in the perfume industry. Esters can be prepared by treating alcohol with an acid in the presence of conc.H2SO4. The process is called esterification.

Compound A (Ethanoic acid) + Alcohol ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⟩ Ester + Water

(Conc.H2SO4)

$CH_{3}COOH + C_{2}H_{5}OH \overset{Conc.H_{2}SO_{4}}{→} CH_{3}COOCH_{2}CH_{3} + H_{2}O$

Therefore Compound A is ethanoic acid and gas liberated is CO2.


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# Primitive roots

If Artin's conjecture on primitive roots is true, then 2 generates $(\mathbb{Z}/p\mathbb{Z})^\times$ for infinitely many primes $p$. Can one at least show that $(\mathbb{Z}/p\mathbb{Z})^\times$ is generated by 2 and 3 for infinitely many primes $p$?

-
Hej, I've changed $\mathbb{Z}_p^*$ to $(\mathbb{Z}/p\mathbb{Z})^\times$, because the generally accepted meaning of $\mathbb{Z}_p$ is the $p$-adic integers (though I don't think anyone was confused about what you meant), and because after changing that aspect, the asterisks gave me trouble with the markdown. –  Zev Chonoles Jan 28 '11 at 2:07
+1 by the way - neat question. –  Zev Chonoles Jan 28 '11 at 2:12
I am going to guess "probably not" - if you can find a prime $p$ and positive integers $j,k$ with $2^{j} \equiv 3^{k} \equiv 1 \, \mathrm{mod} \, p$ and $j*k < p-1$, then you've found a counterexample. –  David Hansen Jan 28 '11 at 2:31
Or indeed $\text{lcm}(j,k)<p-1$. But that would only be one counterexample - is there an intuitive reason why we should expect it to occur for all but finitely many primes? –  Zev Chonoles Jan 28 '11 at 2:35
@Hej: although your question is quite clear, I think it's natural that people are misreading it (as you mentioned in a comment below). Namely, a primitive root mod p is a generator of (Z/pZ)^*, so it is natural to (mis)read your question as: "Are there infinitely many primes p such that 2 or 3 is a primitive root mod p?" especially if you know about the result of Heath-Brown which is "slightly weaker" than this. You are weakening APR in a somewhat unexpected direction by allowing more than one generator. (Anyway, good question!) –  Pete L. Clark Jan 28 '11 at 4:05

I think the best approximation is due to Heath-Brown (Quart. J. Math. Oxford Ser. 37, 27-38.): for infinitely many primes p, one of 2,3,5 is a primitive root mod p.

Actually Heath-Brown's theorem works for any three primes in place of 2,3,5. You can find his paper online here (praise Google).

-
There is also a nice presentation by Murty, "Artin's conjecture for primitive roots", the Mathematical Intelligencer, vol 10 (4) (1988), 59-67. –  Andrés Caicedo Jan 28 '11 at 1:56
Thank you. I was aware of that result. Maybe my question was not clear. So what I am asking is whether the group generated by 2 and 3 in $\mathbb{Z}_p^*$ is all of $\mathbb{Z}_p^*$. Equivalently, is it true that for infinitely many primes $p$, every number $m$ is congruent to some $2^a3^b$ mod $p$, with $a,b$ natural. –  Hej Jan 28 '11 at 1:58
However, $n_1,\ldots,n_k\in\mathbb{Z}$ could still generate $\mathbb{Z}/p\mathbb{Z}^\times$ even if none of them is a primitive root mod $p$. It seems like Hej's question might not require such hard results. –  Zev Chonoles Jan 28 '11 at 2:00
Heath-Brown's proof goes by first showing that given three primes, they will generate $(\mathbb{Z}/p)^*$ for infinitely many $p$ and current technology does not allow us to reduce the number of primes to two. His method was inspired by an analogous result of Gupta and Murty on elliptic curves, where you ask if a given set of $m$ rational points (satisfying some obvious necessary condition) generates the group mod $p$ for infinitely many $p$ and this can be proved as long as $m$ is big enough (5 for CM and 8 for non-CM? I forget). The conjecture would be $m=1$, but current tech is not enough. –  Felipe Voloch Jan 28 '11 at 2:45
@Felipe: Amusing, I think we were composing our answers at the same time, with more-or-less the same material. –  Joe Silverman Jan 28 '11 at 3:03

This is an interesting question. More generally people have considered the following. Let $\Gamma$ be a subgroup of $\mathbf{Q}^*$ generated by $r$ primes. What can one say about $$N_\Gamma(X)=|\{p < X : \Gamma \bmod p \textrm{ generates } (\mathbf{Z}/p \mathbf{Z})^{\times}\}|.$$

There is also a natural elliptic analogue. Thus let $E/\mathbf{Q}$ be an elliptic curve and let $\Gamma\subset E(\mathbf{Q})$ be a subgroup of rank $r$. Then we can consider $$N_\Gamma(X)=|\{p < X : \Gamma \bmod p \textrm{ generates } E(\mathbf{Z}/p \mathbf{Z})\}|.$$ Gupta and Murty give a number of results, both conditional and unconditional, in their paper Primitive points on elliptic curves, Compositio Math. 58 (1986), 13–44. For example, if $r\ge6$ and $E$ has complex multiplication, then they prove unconditionally that $N_\Gamma(X)\gg X/(\log X)^2$. In the non-CM case, assuming the GRH, they prove that if $r\ge18$, then $N_\Gamma(X)\gg X/(\log X)$.

It would be interesting to investigate similar questions on higher dimensional algebraic groups, either abelian varieties of dimension $\ge2$, or even on $(\mathbf{Q})^{\times}\times(\mathbf{Q})^{\times}\times\cdots\times(\mathbf{Q})^{\times}$.

-

This is just a comment, but I'm unable to do it, so I post it as an answer:

$2^{11} \equiv 3^{11} \equiv 11 \, \mod 23$.

And a similar situation happens with other primes:

$47, 71, 73, 97, 167, 191, 193, 239, 241, 263, 307, 311, 313, ...$

although not always with $ord_{p}(2)=ord_{p}(3)=\frac{p-1}{2}$.

-
All well and good, but the question is asking for a proof that there are infinitely many primes for which this doesn't happen. –  Zev Chonoles Jan 28 '11 at 15:17
Also you mean $2^{11}\equiv 3^{11}\equiv1\bmod 23$ (not 11). –  Zev Chonoles Jan 28 '11 at 15:19


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# relativity

14. ### I Varying The Relativistic Action

In his book, Landau mentioned varying the relativistic lagrangian However, I do not understand how he got from varying the integral of ds to varying only the contravariant components. Would the general procedure not be varying $$\delta S = -mc\delta\int_a^b\frac{dx_idx^i}{\sqrt{ds}}$$ and...
15. ### I Acceleration Versus Gravity (General Relativity)

Hi, It is usually claimed that a person in an accelerating elevator with an acceleration equals to the gravity of the earth; this person cannot make any experiment that makes him know whether he is in the elevator or on the surface of earth. However, if this person project two light beams...
16. ### Physics Is gravitational wave research a smart direction to go in?

Is it a fair prediction to state that in the next several years or so, globally, there will be major investments into gravitational wave research, and many more ‘LIGOs’ being developed? Is it a good idea to venture into that area of physics?
17. ### B What would the universe be like with less mass and energy?

So I have sort of a conceptual question about the big bang and gravity. Imagine yourself in a universe, in which existed about the number of particles/energy in a 3X3 metre room at any given moment. This universe has the same laws of physics, constants and is identical in every way to our...
18. ### I Relativity and Electromagnetism

Hello, There is a common setup used when describing the intimate relationship between electricity and magnetism. I have a question about the setup. Setup: There is some long current-carrying wire. Outside of that wire, there is some test charge. In the first situation, the test charge is...
19. ### Time dilation in a magnetic field

Hi all. I was wondering if time is dilated whilst travelling in a stable magnetic field that is generated by the object travelling, and if so, does this vary if you reduce or intensify the magnetic field? Also, what happens if the object is generating two opposing magnetic fields, would...
20. ### I How to calculate the energy in gamma photons

how do I calculate the energy of to photons moving in the opposite direction. starting from the same origin.
21. ### Relative speed of two photons

1. Homework Statement What is the speed of a photon with respect to another photon if: the two photons are going in the same direction. they are going in opposite direction? 2. The attempt at a solution I think the answer to the first question should be zero and to the second one be 2xC...
22. ### A Electromagnetism for media in arbitrary motion

Hello To develop one interesting idea I need to be able to do calculations on (1) scattering of light from bodies in arbitrary motion, possibly at relativistic speeds; (2) Propagation of light in electromagnetic media that are in arbitrary motion (possibly relativistic). For example, I would...
23. ### I GPS relativity error - really?

There are many popscience articles about relativity and how relativity is essential for GPS and in many is mentioned something like this: If GPS wasnt corrected for time dilation predicted by GRT, there would be an error of 12 kilometers in GPS positioning per day. But when I checked how GPS...
24. ### I Difference between an electromagnetic field and a photon?

So I'm kind of confused. The way I understand it, an electromagnetic field is just a regular electric field viewed from a relativistic point of view, meaning that since we see the charges moving relative to us, we feel like the particles and the fields created by them come closer together (I...
25. ### I Light's speed and relativity

If you were to travel alongside a train, as fast the train, to you the train would seem stationary. I read that if you were to travel along a photon of light, as fast as the speed of light, that photon would not seem stationary. Is this true? If so, why?
26. ### B High tower

The earth rotates around its own axis within 24 hours. Theoretically, if at the equator perform a tower with a height - H, ignoring its effect on the slowdown of the Earth's rotation (we assume that the material of low density), objects and other complications, at a certain height, the linear...
27. ### I Reconciling two observers in two frames

I am not a physicist—not even close—just a guy who, for some crazy reason, decided to try to understand some of the basics of relativity. I’d like to understand them well enough to be able to explain them (correctly) to another lay person. I’m trying to see how much I could explain without...
28. ### I Relativity Confirmed Again

A star and a black hole again confirm Relativity. https://www.space.com/41291-relativity-revealed-milky-way-core.html
29. ### B How can any physical body truly be at rest?

Can we truly have a rest frame or should it be a close to rest frame? Even if I'm stationary and sitting on my porch and the observer in the car passing is moving, I'm still not at 0 velocity. The earth is moving at 67,000 mph and the galaxy is moving at 250,000 mph. I'm never in a single...
30. ### I Circle tangential velocity approaching c

When a merry go rounds tangential velocity approach c, and horses look closer than when at rest, what is reduced, angular velocity or radius?


text:
# Question 1c0d4

Dec 12, 2014

The answer is $85.00 g$.

A solution's percent concentration is defined as the mass of solute divided by the total mass of the solution and multiplied by 100%.

c = (mass_(solute))/(mass_(solution)) * 100%, where

$m a s {s}_{s o l u t i o n} = m a s {s}_{s o l u t e} + m a s {s}_{w a t e r}$

Knowing that we have $100 g$ of solution, we can determine the mass of the solute by

m_(solute) = (c * m_(solution))/100 = (15.00 * 100.0)/(100%) = 15.00g

Therefore, the mass of water required for this solution is

${m}_{w a t e r} = {m}_{s o l u t i o n} - {m}_{s o l u t e} = 100.0 - 15.00 = 85.00 g$

This is a case where it is easy to eyeball the answer just because we are dealing with $100 g$ of solution...so a concentration of 20% would have required 80g of water and 20g of $H N {O}_{3}$, a concentration of 8%# would have required 8g of $H N {O}_{3}$ and $92 g$ of water, and so on...


text:
Linear algebra; conditions for spaces

1. Jan 6, 2008

Niles

[SOLVED] Linear algebra; conditions for spaces

1. The problem statement, all variables and given/known data
1) If I want to write a basis for R^3, what must the conditions for the three vectors? Must they be linearly independant or orthogonal or what?

2) If I want to write a basis for a supspace of R^3, what must the conditions for the vectors be? Must they only be linearly independant?

3) If I have a symmetric matrix A, I can find an orthogonal matrix S that satisfies S^(-1) *D*S, where D is a diagonal matrix consisting of eigenvalues.

If I have a diagonalizable matrix B, which is not symmetric, can I also find an orthogonal matrix S that satisfies what I wrote above?

3. The attempt at a solution
For the vectors spanning R^3, I believe they have to be orthogonal and linearly independant.

For the vectors spanning a subspace of R^3, I believe the vectors just have to be linearly independant. Correct?

Last edited: Jan 6, 2008
2. Jan 6, 2008

Dick

The conditions for the vectors in a basis is that they be linearly independent and span the required space. They don't have to be orthogonal unless the the problem says 'orthogonal basis'. Symmetric real matrices have the special property that eigenvectors corresponding to different eigenvalues are orthogonal. That's why you can construct an orthogonal matrix S. For a general diagonalizable matrix B, this is not true.

3. Jan 6, 2008

Niles

Cool, thanks :-)


text:
South Carolina General Assembly
119th Session, 2011-2012

Indicates Matter Stricken
Indicates New Matter

S. 971

STATUS INFORMATION

Senate Resolution
Document Path: l:\council\bills\rm\1278ahb11.docx

Introduced in the Senate on June 21, 2011
Adopted by the Senate on June 21, 2011

Summary: Roger Elmer Jermyn

HISTORY OF LEGISLATIVE ACTIONS

     Date      Body   Action Description with journal page number
-------------------------------------------------------------------------------
6/21/2011  Senate  Introduced and adopted (Senate Journal-page 8)

View the latest legislative information at the LPITS web site

VERSIONS OF THIS BILL

(Text matches printed bills. Document has been reformatted to meet World Wide Web specifications.)

### A SENATE RESOLUTION

TO EXPRESS THE PROFOUND SORROW OF THE MEMBERS OF THE SOUTH CAROLINA SENATE UPON THE DEATH OF ROGER ELMER JERMYN OF YORK COUNTY, AND TO EXTEND THE DEEPEST SYMPATHY TO HIS FAMILY AND MANY FRIENDS.

Whereas, the members of the South Carolina Senate were deeply saddened to learn of the death of Roger Elmer Jermyn of York County on January 1, 2011, at the venerable age of seventy-three; and

Whereas, a resident of Rock Hill, Mr. Jermyn was born a son of the late Robert Elmer Jermyn and the late Jessie Irving Jermyn in Elkton, Maryland; and

Whereas, this veteran of the Army National Guard had retired from Allied Signal Corporation, now Honeywell Corporation, with thirty-three years of service and also did chair caning for his son at A Lasting Touch; and

Whereas, giving back to the community was Roger Jermyn's mission in life, and he worked tirelessly to help those less fortunate, regardless of age, race, or creed. He strove to ensure that Rock Hill residents had knowledge of, and access to, mobility services, home- and community-based services, and consumer protections; and

Whereas, as a part of his service, Roger Jermyn was former Rock Hill chapter president of the AARP and was serving as treasurer. Under his leadership, the local AARP chapter performed an average six hundred fifty hours of community service per month. In 2006, he was the recipient of the prestigious AARP Andrus Award for his extensive community service. He was also a member of the Episcopal Church of Our Savior; and

Whereas, as all those who had the privilege of knowing and working with him will attest, Roger Jermyn was selfless in his service, an inspiring, courageous, and gracious soul who lived to help others; and

Whereas, Mr. Jermyn leaves to cherish his memory his beloved wife, Frances Hamilton Jermyn; his son, Keith Jermyn; his daughters, Lynne Dickert and Gayle Senn; three grandchildren; and a host of other relatives and friends. He will be missed. Now, therefore,

Be it resolved by the Senate:

That the members of the South Carolina Senate, by this resolution, express their profound sorrow upon the death of Roger Elmer Jermyn of York County and extend the deepest sympathy to his family and many friends.

Be it further resolved that a copy of this resolution be forwarded to Mrs. Frances Hamilton Jermyn for the family.

----XX----

This web page was last updated on Tuesday, December 10, 2013 at 10:06 A.M.


text:
Taiwanese Journal of Mathematics

A SHARP MAXIMAL INEQUALITY FOR A GEOMETRIC BROWNIAN MOTION

Cloud Makasu

Abstract

Let $X=(X_t)_{t\geq0}$ be a geometric Brownian motion with drift $\mu$ and volatility $\sigma\gt 0$, and let $Y=(Y_t)_{t\geq0}$ be the associated maximum process of $X$. Under certain conditions, we prove a sharp maximal inequality for the geometric Brownian motion. The method of proof is essentially based on explicit forms of the following optimal stopping problem: Find a stopping time $\tau^*$, if it exists, such that \begin{eqnarray*} \Phi(x,y):=\sup_{\tau}\mathbf{E}^{x,y}\left[Y_\tau-c\int_0^\tau X^\theta_sds\right],\quad c,\theta\gt 0 \end{eqnarray*} where the supremum is taken over all stopping times $\tau$ for the process $X$. The present result complements and extends a similar result proved by Graversen and Peskir [1].

Article information

Source
Taiwanese J. Math., Volume 19, Number 2 (2015), 585-591.

Dates
First available in Project Euclid: 4 July 2017

https://projecteuclid.org/euclid.twjm/1499133648

Digital Object Identifier
doi:10.11650/tjm.19.2015.2551

Mathematical Reviews number (MathSciNet)
MR3332315

Zentralblatt MATH identifier
1357.60044

Citation

Makasu, Cloud. A SHARP MAXIMAL INEQUALITY FOR A GEOMETRIC BROWNIAN MOTION. Taiwanese J. Math. 19 (2015), no. 2, 585--591. doi:10.11650/tjm.19.2015.2551. https://projecteuclid.org/euclid.twjm/1499133648

References

• S. E. Graversen and G. Peskir, Optimal stopping and maximal inequalities for geometric Brownian motion, J. Appl. Prob., 35 (1998), 856-872.
• G. Peskir, Optimal stopping of the maximum process: the maximality principle, Ann. Prob., 26 (1998), 1614-1640.
• L. A. Shepp and A. N. Shiryaev, The Russian option: reduced regret, Ann. Appl. Prob., 3 (1993), 631-640.
• L. A. Shepp and A. N. Shiryaev, A new look at the “Russian option”, Theory Prob. Appl., 39 (1994), 103-119.


text:
# Mbox Locking¶

The only standard way to lock an mbox is using a method called “dotlock”. This means that a file named <mailbox-name>.lock is created in the same directory as the mailbox being locked. This works pretty well when the mbox is locked for writing, but for reading it’s very inefficient. That’s why other locking methods have been used.

It’s important that all software that’s reading or writing to mboxes use the same locking settings. If they use different methods, they might read/write to an mbox while another process is modifying it, and see corrupted mails. If they use the same methods but in a different order, they can both end up in a deadlock.

If you want to know more details about locking, see Mbox Mailbox Format.

For Dovecot you can configure locking using the mbox_read_locks and mbox_write_locks settings. The defaults are:

mbox_read_locks = fcntl
mbox_write_locks = dotlock fcntl

Here’s a list of how to find out the locking settings for other software:

## Procmail¶

# procmail -v 2>&1|grep Locking
Locking strategies:     dotlocking, fcntl()

## Postfix¶

Postfix has two different ways to deliver to mboxes. One is the “mailbox” transport and another one is the “virtual” transport.

# postconf mailbox_delivery_lock
mailbox_delivery_lock = fcntl, dotlock
# postconf virtual_mailbox_lock
virtual_mailbox_lock = fcntl

In the above case, if you used the mailbox transport, you’d have to change Dovecot’s configuration to mbox_write_locks = fcntl dotlock or vice versa for Postfix.

If you used the virtual transport, it doesn’t really matter if the dotlock is missing, since the fcntl is common with Dovecot and Postfix.

## mutt¶

mutt -v|grep -i lock

## Debian¶

Debian’s policy specifies that all software should use “fcntl and then dotlock” locking, but this probably applies only to most commonly used software.


text:
## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.3 - Polynomials and Polynomial Functions - Exercise Set: 85

80,000 dollars

#### Work Step by Step

We are given the function $P(x)=45x-100000$, which models the relationship between the number of computer briefcases x that a company sells and the profit the company makes, $P(x)$. To find $P(4000)$, we can plug 4000 into the given function. $P(4000)=45(4000)-100000=180000-100000=80000$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# American Institute of Mathematical Sciences

August  2010, 4(3): 335-350. doi: 10.3934/ipi.2010.4.335

## On Levenberg-Marquardt-Kaczmarz iterative methods for solving systems of nonlinear ill-posed equations

 1 Fachbereich Mathematik, Johann Wolfgang Goethe Universität, Robert–Mayer–Str. 6–10, 60054 Frankfurt am Main, Germany 2 University of Graz, Institute for Mathematics and Scientific Computing, Heinrichstr. 36/III, A-8010 Graz, Austria 3 Department of Mathematics, Federal University of St. Catarina, P.O. Box 476, 88040-900 Florianópolis, Brazil

Received  July 2009 Revised  October 2009 Published  July 2010

In this article a modified Levenberg-Marquardt method coupled with a Kaczmarz strategy for obtaining stable solutions of nonlinear systems of ill-posed operator equations is investigated. We show that the proposed method is a convergent regularization method. Numerical tests are presented for a non-linear inverse doping problem based on a bipolar model.
Citation: Johann Baumeister, Barbara Kaltenbacher, Antonio Leitão. On Levenberg-Marquardt-Kaczmarz iterative methods for solving systems of nonlinear ill-posed equations. Inverse Problems & Imaging, 2010, 4 (3) : 335-350. doi: 10.3934/ipi.2010.4.335
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### vovuh's blog

By vovuh, history, 2 years ago, translation, ,

• +18

 » 2 years ago, # |   0
 » 2 years ago, # | ← Rev. 2 →   0 Can authors share the generator of testcase #36 in problem E?It seems heavy-light decomposition and some other solutions perform really bad on this testcase.
•  » » 2 years ago, # ^ |   +12 It is just full binary tree with n = 500000.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   +19 Thanks.I tested this case (pi = i / 2) during the contest. And my solution finished in ~1.2s. But I found the solution is 3 times slower if I shuffled the labels of vertices.It is so strange because I think the total operations in segment tree/Fenwick tree are same.UPD: If I visited the vertices in order, two consecutive operations will share many positions in Fenwick tree. These positions will be fetched in the cache. So it could be much faster than visiting the vertices randomly.
•  » » 2 years ago, # ^ |   0 Well, I've got an OK-submission with HLD which performs in 1.3 seconds on this case: 30478628.Or you can check out cdkrot's HLD submission 30469620 which does it in 1.152 s.
•  » » » 2 years ago, # ^ |   +14 Thanks. I will try to learn from your codes.I try to random shuffle array "by_h" in your code (like what my solution did during the contest). This submission got TLE. I think this result is surprising to me.So It's really important to make your solution cache-friendly :D
•  » » » » 2 years ago, # ^ |   0 Strange but true ¯\_(ツ)_/¯
 » 2 years ago, # |   0 Can someone please provide a more detailed explanation of Div2 D and their C++ code for reference? I know a Trie is to be used but I'm not able to implement it correctly.
•  » » 2 years ago, # ^ |   0 i didn't use trie just binary search works here :)http://codeforces.com/contest/861/submission/30487025i have stored all the distinct possible substrings formed from the each string...and then just sort this and finding which of the substring occured exactly once :)
•  » » 2 years ago, # ^ |   0 You can store each prefix in the trie first and then for each substring you remove the contribution of the present string and then see if the trie has any such substring. You just print the smallest substring which is not in the trie. My solution: http://codeforces.com/contest/861/submission/30489173Hope this helps :D
•  » » » 2 years ago, # ^ |   0 Can you explain me this part please "and then for each substring you remove the contribution of the present string and then see if the trie has any such substring" i will be highly thankful to you.
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   0 In my code, take a look at the last loop. The outermost loop traverses the strings one by one. For each string, I first remove all its substrings from the trie(If they were present then I will be taking them into consideration too, but I wanna check if any OTHER string other than my current string contains the substring or not). So, I remove all substrings of the current string, see if a substring is present in the trie and then again insert all the substrings of my current string. In case its still unclear, "current string" refers to the ith string I took, i.e, v[i].
 » 2 years ago, # |   0 I didn't get why in Div2 B problem the quantity of the flats on each floor is in [1,100], may someone explain please?
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 Because in worst case maximum flat number will not be greater than 100. Then maximum floor quantity will not exceed 100.
•  » » » 2 years ago, # ^ |   0 OK, but what about this sentence in the statement "Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors)"?
•  » » » » 2 years ago, # ^ |   +1 Oh, sorry, there was a mistake. I wanted to say "Then maximum flat quantity on each floor will not exceed 100."The condition about infinitely high building is no matter for you, because even if there is only one flat on each floor, maximum floor index we are interested in is not greater than 100 (and, of course, it is not getting greater with more flats on each floor).
•  » » » » » 2 years ago, # ^ | ← Rev. 4 →   0 I approximately got it. My mind closes some times :), so I'll back to this problem later.. Thank you very much.
 » 2 years ago, # |   +3 Here's a solution for div1e with just "brute force".The first step is to make the first observation in the tutorial. Let's consider a naive way to compute the value.We'll process all nodes with the same depth simultaneously.For each previous depth's nodes, let's suppose we have computed the number of children at the current depth that we are considering (I will show how to keep track of this later).For each node, we can naively iterate through its parents and add this stored value.To update the values, we can also do this naively. More specifically, for each node and ancestor pair, we will update the ancestor's value by adding number of children of node and subtract 1 (we subtract 1 since we no longer want to consider the current node, and we add number of children of this node to represent children at next depth). Of course, this solution is n^2. We can speed it up to pass with just two optimizations. If a node ever has 1 child, we delete this node, and connect the child directly to the parent. If a node has zero children, we'll delete this node (and recursively delete the parent if it now has 0 children). With these two simple optimizations, the running time becomes O(n sqrt n)! short justificationBasically, the idea is we are only keeping track of nodes whose sets of children at depth d are different, and for a particular node, you can show there are only at most O(sqrt n) parents. To prove this, we can think about how many nodes we need to get a new parent with a different set of children at the current depth.Here's a sample implementation with those two optimizations: http://codeforces.com/contest/860/submission/30449695 You can note that it's basically brute force but with these line added in various places: while (par[cur] != -1 && child[par[cur]] == 1) par[cur] = par[par[cur]];
 » 23 months ago, # |   0 could someone help me with problem D : my submission id is 35750991, it is showing runtime error, but dont know why.
•  » » 23 months ago, # ^ |   0 I think in the set called val in your code sometimes it could exceed it's initial size which is 70707 in this instruction val[mp[d]].insert(i); like if we have zeros in the the given numbers more than 70707


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Paper:

TR15-207 | 23rd December 2015 00:46

Finding Primitive Roots Pseudo-Deterministically

TR15-207
Authors: Ofer Grossman
Publication: 23rd December 2015 06:42
Keywords:

Abstract:

Pseudo-deterministic algorithms are randomized search algorithms which output unique solutions (i.e., with high probability they output the same solution on each execution). We present a pseudo-deterministic algorithm that, given a prime $p,$ finds a primitive root modulo $p$ in time $\exp(O(\sqrt{\log p \log \log p}))$. This improves upon the previous best known provable deterministic (and pseudo-deterministic) algorithm which runs in exponential time $p^{\frac{1}{4}+o(1)}.$ Our algorithm matches the problem's best known running time for Las Vegas algorithms which may output different primitive roots in different executions.

When the factorization of $p-1$ is known, as may be the case when generating primes with $p-1$ in factored form for use in certain applications, we present a pseudo-deterministic polynomial time algorithm for the case that each prime factor of $p-1$ is either of size at most $\log^c (p)$ or at least $p^{1/c}$ for some constant $c>0$. This is a significant improvement over a result of Gat and Goldwasser, which described a polynomial time pseudo-deterministic algorithm when the factorization of $p-1$ was of the form $kq$ for prime $q$ and $k= poly(\log p).$

We remark that the Generalized Riemann Hypothesis (GRH) implies that the smallest primitive root $g$ satisfies $g \le O(\log^6(p)).$ Therefore, assuming GRH, given the factorization of $p-1,$ the smallest primitive root can be found and verified deterministically by brute force in polynomial time.

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